Graduate Quantum Mechanics Class Notes

Leon Hostetler

May 4, 2019 Version 0.0 Contents

Preface v

1 The Origins of Quantum Mechanics1 1.1 Introduction...... 1 1.2 Bohr Quantization...... 2 1.3 Virial Theorem...... 5 1.4 Hierarchy of Lengths...... 10 1.5 The Correspondence Principle...... 10 1.6 Simple Variational Approximations...... 12 1.7 Summary: The Origins of Quantum Mechanics...... 16

2 Wave Mechanics 19 2.1 Inﬁnite Walls...... 19 2.2 Continuity Equation...... 22 2.3 Finite Walls...... 23 2.4 Propagating Waves...... 28 2.5 Resonance...... 29 2.6 Tunneling Waves...... 30 2.7 Level Density...... 32 2.8 Summary: Wave Mechanics...... 38

3 Wave Functions and Operators 41 3.1 Superposition Principle ...... 41 3.2 Dirac Delta Function...... 41 3.3 Representations of the Wave Function ...... 43 3.4 Operators...... 45 3.5 Green’s Functions ...... 49 3.6 Transformations ...... 50 3.7 Summary: Wave Functions and Operators...... 53

4 Formalism 56 4.1 Groups...... 56 4.2 Hilbert Space...... 57 4.3 Hermitian Operators...... 59 4.4 Commutators...... 62 4.5 Uncertainty...... 65 4.6 Summary: Formalism ...... 68

5 Quantum Dynamics 70 5.1 Unitary Transformations...... 70 5.2 Operator Dynamics and the Heisenberg Picture...... 72 5.3 Conservation Laws...... 77 5.4 Continuity Equation...... 78 Contents iii

5.5 Wave Packets...... 79 5.6 Decay of States...... 81 5.7 Survival Probability ...... 83 5.8 Time Reversal...... 84 5.9 Summary: Quantum Dynamics...... 86

6 General 1D Problems 88 6.1 General Properties...... 88 6.2 1D Harmonic Oscillator ...... 90 6.3 Multidimensional Harmonic Oscillators...... 94 6.4 Molecular Excitations ...... 95 6.5 Secondary Quantization...... 96 6.6 Heisenberg-Weyl Algebra ...... 97 6.7 Coherent States...... 100 6.8 Summary: General 1D Problems ...... 104

7 Magnetic Field 106 7.1 Minimal Coupling (Spinless Particles) ...... 106 7.2 Particle in a Magnetic Field...... 107 7.3 Gauge Invariance...... 110 7.4 Landau Levels ...... 111 7.5 Aharonov-Bohm Eﬀect...... 112 7.6 Summary: Magnetic Field...... 114

8 3D Motion 116 8.1 Angular Momentum Algebra ...... 116 8.2 Orbital Momentum and Spherical Functions ...... 120 8.3 Schrodinger Equation in Spherical Coordinates ...... 122 8.4 Hydrogen Atom...... 123 8.5 Symmetries and Conservation Laws...... 128 8.6 Summary: 3D Motion ...... 131

9 Spin and Angular Momentum Coupling 134 9.1 Spin-1/2...... 134 9.2 Angular Momentum Coupling...... 137 9.3 Atomic Splitting ...... 142 9.4 Wigner-Eckart Theorem...... 145 9.5 Electromagnetic Multipoles ...... 147 9.6 Summary: Spin and Angular Momentum Coupling ...... 150

10 Many-body Quantum Mechanics 153 10.1 Identical Particles ...... 153 10.2 Gibbs’ Paradox...... 154 10.3 Transposition Operator ...... 155 10.4 Bosons and Fermions...... 156 10.5 Second Quantization...... 158 10.6 Nuclear Conﬁgurations...... 161 10.7 Atomic Conﬁgurations...... 164 10.8 Summary: Many-body Quantum Mechanics ...... 169

11 Approximation Methods 173 11.1 Variational Method...... 173 11.2 The Two-State System...... 174 11.3 Diagonalization on Restricted Basis...... 175 iv Contents

11.4 Stationary Perturbation Theory...... 178 11.5 Time-dependent Perturbation Theory ...... 182 11.6 Periodic Perturbations...... 184 11.7 Summary: Approximation Methods...... 187

12 Atoms 189 12.1 Lamb Shift ...... 189 12.2 Polarizability ...... 192 12.3 Van der Waals Forces ...... 196 12.4 Summary: Atoms...... 198

13 Photons and the Electromagnetic Field 199 13.1 Hamiltonian of EM Field ...... 199 13.2 Quantization of the EM Field...... 201 13.3 Radiation...... 203 13.4 Electric Dipole Transitions...... 205 13.5 Multipole Transitions ...... 210 13.6 Photoabsorption ...... 212 13.7 Summary: Photons and the EM Field ...... 214

14 Relativistic Quantum Mechanics 216 14.1 Klein-Gordon Equation ...... 216 14.2 Four-vectors and Minkowski Space...... 220 14.3 Dirac Equation...... 222 14.4 Relativistic Corrections ...... 235 14.5 Summary: Relativistic Quantum Mechanics ...... 240

15 Scattering 243 15.1 Perturbation Theory...... 243 15.2 Potential Scattering ...... 245 15.3 Born (Series) Approximation ...... 248 15.4 The Method of Partial Waves...... 252 15.5 Scattering of Particles with Spin ...... 259 15.6 Scattering of Identical Particles...... 263 15.7 Summary: Scattering...... 264 Preface

About These Notes

My class notes can be found at www.leonhostetler.com/classnotes These are my class notes from a two-semester graduate course (PHY 851-852) on quantum mechanics taken at Michigan State University. The course was taught by Profes- sor Vladimir Zelevinsky. The primary textbooks used were “Quantum Physics” Volumes I and II both by Professor Zelevinsky. Please bear in mind that these notes will contain errors. Any errors are certainly my own. If you ﬁnd one, please email me at [email protected] with the name of the class notes, the page on which the error is found, and the nature of the error. This work is currently licensed under a Creative Commons Attribution-NonCommercial- NoDerivatives 4.0 International License. That means you are free to copy and distribute this document in whole for noncommercial use, but you are not allowed to distribute derivatives of this document or to copy and distribute it for commercial reasons.

Updates

Last Updated: May 4, 2019 Version 0.5 (May 4, 2019): Initial upload

Chapter 1

The Origins of Quantum Mechanics

In this text, we will primarily use Gaussian units. In these units, 4π0 → 1.

1.1 Introduction

Planck and the Blackbody Distribution

In classical mechanics, the action is the integral of the Lagrangian L dt along the trajectory taken by a particle. Planck’s constant h is the quantum of´ action, and has value

h = 6.626 × 10−34 Js = 6.626 × 10−27 ergs.

Planck introduced the quantum of action in 1900. He was interested in solving the problem of ﬁnding the distribution of the frequencies of radiation in a box due to temperature, if the box was in equilibrium. Classically, this resulted in several paradoxes. Instead of writing energy in terms of hν it is often more convenient to write it in terms of ~ω, where h = , ~ 2π is the reduced Planck constant.

Einstein and the Photoelectric Eﬀect

The photoelectric effect is the effect in which shining a light on a metal causes electrons to be ejected from the metal. The effect was explained by Einstein in 1905. There is a minimum energy, called a work function W or ionization energy that must be added to electrons in a metal to free the electron from the atom. The work function is specific to the material. Consider an electron in a metal. If E < 0, then the electron is bound to an atom. If E > 0, the electron is unbounded and undergoes free motion. At E = 0, we have the threshold energy. Consider the diagram shown below. To eject an electron from a metal, we need to add some energy W (i.e. the work function) to bring the bound electron’s energy up to the threshold. We need to add additional energy to tear the electron loose from the atom. This additional energy is the kinetic energy of the electron which is now a free particle. 2 The Origins of Quantum Mechanics

Paradoxically, increasing the intensity of the light does not increase the (kinetic) energy of the kicked oﬀ photons. Rather it simply increases the number of electrons that get ejected. As the plot below illustrates, the kinetic energy of the ejected photons depends on the frequency (i.e. energy) of the incident light, and not on the intensity of the light. Second, if the incident photon’s energy is less than the work function W of the metal, then no electrons are ejected.

If an electron in a metal absorbs a quantum of energy ~ω, then the maximum kinetic energy of the ejected photon is

Kmax = ~ω − W.

The photoelectric eﬀect is the ﬁrst example in which a quanta (in this case a photon) is considered a real particle.

Atomic Radiation and Spectroscopy At some point spectroscopic data was discovered to have a pattern called the combination principle, where the wavelengths of light emitted or absorbed from atoms was found to have the form 1 = T 0 − T . λ n n The T terms are called “spectral terms” and are labeled by integers n and n0. The spectral terms are diﬀerent for diﬀerent elements. For example, for the hydrogen atom,

1 1 1 = RH − . λ n02 n2

1.2 Bohr Quantization

The next major step in the origin of quantum mechanics was made by Niels Bohr in 1913. 1.2. Bohr Quantization 3

Bohr postulated that in an atom, there exist stable orbits in which electrons do not radiate energy. These stable orbits form a discrete i.e. quantized set. Bohr postulated that these stable orbits or stationary states satisfy

p dq = nh = 2πn , n = 0, 1, 2,... ˛ ~

On the left we have the action—the closed line integral over the period of the motion. So according to Bohr, the electron orbit is stable if the classical action over the period is equal to an integer times the quantum of action. If you plot a single period of the motion in the p vs. q plane, the trajectory forms a closed loop, and the integral p dq is simply the area enclosed by the loop. The trajectory must be periodic. If it is not,¸ for example, if the particle is simply scattering to inﬁnity rather than bouncing back and forth, then there is no quantization. There are combinations of quantities which do not change when you very slowly (i.e. adiabatically) change the parameters of the system. For example, if the system parameters p and q (which are the canonically conjugate momentum and coordinate) are changed very slowly, by this I mean τ >> T , where τ is the time over which the change takes place and T is the period of the motion, then the quantity p dq does not change. That is, it is adiabatically invariant. Suppose you have a hydrogen¸ atom in state n. Then you adiabatically change, and it stays in state n because there is no physical reason for it to jump to n ± 1. The phase space is (p, q). In 1D, our coordinate q is typically x. Then

p(x) dx = nh = 2πn , n = 0, 1, 2,... ˛ ~ This is an integral over the closed, classical trajectory. Often, the n = 0 case is forbidden. In 1D, this integral is simply the area enclosed by the trajectory.

In 1D, the energy, the sum of the kinetic and potential energy has the form p2 E = K(p) + U(x) = + U(x). 2m For a given energy, we can solve this for p(x), then

p2m [E − U(x)] dx = 2π n. ˛ ~

If the potential U(x) is given, this integral can typically be calculated—numerically if nothing else. In many cases, there is symmetry, and we only have to integrate over half of the trajectory and then multiply by 2. In higher dimensions, the problem is much more complicated unless the motion is separable in the coordinates. In multiple dimensions, the integral is a volume or hyper- volume rather than area. 4 The Origins of Quantum Mechanics

The de Broglie wave is motion related to p dx. The de Broglie wavelength of a particle is ¸ 2π λ = ~, p which implies 2π p(x) = ~ . λ(x) Then our quantization condition becomes

1 dx = n, ˛ λ(x)

where n is an integer for “stationary” waves. Example 1.2.1: 1D Inﬁnite Square Well

Consider the case of a particle in a 1D box of width a and with impenetrable walls. Since there is no potential in the box, the particle experiences no force (except an inﬁnite one when it hits a wall of the box). So within the box, velocity is constant, which implies the momentum p = mv is constant. So

p dx = p dx ˛ ˛ The particle simply ﬂies back and forth between the two walls of the box, so in a single period, its trajectory has a length of 2a. So dx = 2a and ¸ p dx = 2ap = 2amv = 2π n. ˛ ~ This implies that the allowed speeds of the particle are π n v = ~ . n am So in Bohr’s quantum mechanics, only the trajectories with these speeds survive. The allowed wavelengths are then

2π~ 2π~ 2π~ am 2a λn = = = = . pn mvn m π~n n The allowed energies are

1 π2 2 E = mv2 = ~ n2. n 2 n 2ma2 To get physical states, we have to exclude the case n = 0. Other than that, Bohr’s approach gives us the exact quantization. Notice that the energy spectrum becomes more and more rarefied, i.e., the energy levels are further and further apart, as n is increased. This is a property specific to the 1D infinite square well.

Exercise: Repeat the above example with a 2D inﬁnite square well of widths a and b, and for a 3D inﬁnite square well of widths a, b, and c. For these, the energy levels become closer and closer together as one goes up in energy. That is, the level density increases. 1.3. Virial Theorem 5

Example 1.2.2: Harmonic Oscillator p The harmonic oscillator potential is 1 1 U(x) = kx2 = mω2x2. 2 2 The total energy is p2 1 x E = + mω2x2. 2m 2 Dividing through by E gives us

p2 mω2x2 + = 1. 2mE 2E Figure 1.1: The elliptical “trajectory” of a 1D harmonic oscil- This is the equation of an ellipse as shown in Fig. (1.1). In general, an ellipse has lator. the equation x2 p2 + = 1, a2 b2 Tip and the area enclosed by the ellipse is πab. We can avoid the integration since we The virial theorem can only know the area of an ellipse. In our case, a2 = 2E/mω2 and b2 = 2mE, so be applied to ﬁnite motion. r For example, it cannot be 2E √ 2πE p dx = π 2mE = . used for a particle that scat- ˛ 2 mω ω ters away to inﬁnity. So the Bohr quantization condition p dq = 2πn~, implies that the stationary trajectories have energy ¸ En = n~ω. This is what we get from Bohr quantization. Notice that the energy levels are equidistant. In this case, Bohr quantization does not give us the correct energy levels, which for a harmonic oscillator are En = (n + 1/2)~ω. This is because Bohr quantization assumes classical turning points where E = U. The Bohr prescription gives us E0 = 0, but the correct ground state (or zero-point) energy is E0 = ~ω/2.

Exercise: Calculate the same for U(x) = Cx4, where C is some constant. The trajectory is no longer an ellipse, so we are forced to actually do the integral.

1.3 Virial Theorem

We will now take a look at the derivation of the classical virial theorem for ﬁnite motion. Suppose we have an arbitrary function f(t). If we wait for a long time T , then the average of f(t) is 1 T/2 f = lim f(t) dt. T →∞ T ˆ−T/2 Note that T here is not the period of some cycle—it is just a very long observation time. Suppose d f(t) = g(t). dt Plugging this into the equation above gives us

1 T T f = lim g − g − . T →∞ T 2 2 6 The Origins of Quantum Mechanics

Since coordinates (functions) and velocities (derivatives of those functions) stay ﬁnite for physical cases, we know that g(T/2) and g(−T/2) are both ﬁnite. Since T → ∞, this implies that the right side goes to zero. That is,

f = 0.

For the virial theorem, we apply this fact to the function f(t) = ~r · ~p. Then we know that in the long-term, the average of the derivative is zero

d (~r · ~p) = 0. dt Expanding the left-hand side gives us

d~r d~p ~p · ~p + ~r · = · ~p − ~r · ∇~ U, dt dt m d~p ~ ~ since dt = F = −∇U. This implies that

2K − ~r · ∇~ U = 0,

where K is the kinetic energy. This is the virial theorem

2K = ~r · ∇~ U.

Let’s look at an example. Suppose

U = Ars.

Then ∂U ∂U ∂U ∂U ∂r ∂r ∂r ∇~ U = , , = , , . ∂x ∂y ∂z ∂r ∂x ∂y ∂z But ∂r ∂ p x = x2 + y2 + z2 = , ∂x ∂x r and similarly for the other Cartesian derivatives. We ﬁnd that ∂U ∇~ U = rˆ. ∂r So for U = Ars, we get ∇~ U = sArs−1rˆ, and ~r · ∇~ U = sArs = sU. So in this case, the virial theorem implies that

2K = sU.

Example 1.3.1: Hydrogen atom

For the hydrogen atom, the problem becomes 3D, and we cannot make it 1D. However, it is spherically symmetric. The potential is

e2 U(r) = − . r 1.3. Virial Theorem 7

For more general “hydrogenic” atoms, e2 is replaced by Ze2 where Z is the atomic number.

For E < 0, the electron is bound in an elliptical orbit. In this regime, the energy levels are quantized, i.e. there are only certain allowed energies. For E > 0, the electron is unbounded. In this regime, the energy can be anything—it is no longer quantized. Remember, for potentials of the form U = Ars, the virial theorem tells us that 2K = sU. For the Coulomb ﬁeld, we have A = −e2 and s = −1, so we have total energy E = K + U = U/2 = −K. From 1 K = − U, 2 we have that mv2 e2 = , 2 2r or e2 v2 = . mr This is simply the virial theorem applied to the Coulomb potential. Note that we are assuming a circular orbit, in which case v and r are constant, and so K = K and U = U. Next, we look at the quantization condition

p dq = mv · 2πr = 2πn . ˛ ~ Here, we pulled p = mv outside the integral since mv is constant for a circular orbit. Then the integral dq = 2πr is simply the circumference of that orbit. This gives us ¸ n v = ~ . mr Comparing this result with our result v2 = e2/mr from the virial theorem, we ﬁnd the radii of allowed orbits 2n2 r = ~ . n me2 The case n = 0, is not allowed for the hydrogen atom. The ground state is n = 1, and the corresponding orbit is called the Bohr radius

2 r = ~ = a ' 0.529 A˚ = 0.529 × 10−10 m. 1 me2 8 The Origins of Quantum Mechanics

So we have that 2 rn = an , where a is the Bohr radius. If n >> 1 the radii are called Rydberg orbits, and they are of macroscopic size. The allowed energies are

1 1 e2 me4 1 13.6 eV En = Un = − = − 2 2 = − 2 . 2 2 rn 2~ n n Note, the quantity me4 1 Ry = ' 13.6 eV, 2~2 is a unit called the Rydberg. Note: We will measure temperature in units of energy with kB = 1

1 eV = 11 600 K ∼ 104 K.

Next, we look at the hydrogen spectrum. For energies E < 0, we have an inﬁnite (but countable) set of discrete, bound states. For energy E > 0 we have uncountably inﬁnite, continuous, unbound states (i.e. free electrons). Notice that as E → 0 from below, the discrete states come closer and closer to each other. In fact, at the threshold energy E = 0, we go from quantized states to continuous states smoothly.

For an electron (in hydrogen) to transition from state n to state n0, we have

4 4 E − E 0 1 me 1 1 me 1 1 ∆ω = n n = − − = − . ~ ~ 2~2 n2 n02 2~3 n02 n2 Thus, 2πc ∆λ = , ∆ω or 1 ∆ω me4 1 1 = = − . ∆λ 2πc 4πc~3 n02 n2 1.3. Virial Theorem 9

Experimentally, it is found that Tip 1 1 1 For atoms that are like hy- = RH 2 − 2 , ∆λ n0 n drogen, we can use the where same approach, but we re- 7 −1 2 2 RH = 1.096 78 × 10 m , place e with Ze everywhere. Z is the number of is the Rydberg constant for hydrogen. Does our theoretical value match? Al- protons in the nucleus. most. The theoretical value that we derived is 4 me 7 −1 R∞ = = 1.097 37 × 10 m . 4πc~3 The discrepancy is primarily due to the fact that in our theoretical derivation, we treated the proton as having inﬁnite mass, i.e. as sitting precisely at the center of the electron’s circular orbit. If instead of using m, we use the reduced mass memp/(me + mp), then we get the right value for RH . The Lyman series encompasses all electron transitions from any level down to the ground state n = 1. This set of transitions all have similar energy and are in the UV range. The Balmer series are all the transitions from higher energy levels down to the n = 2 energy levels. This set of transitions also have similar energy and are all in the visible range. All drops to n = 3 are part of the Paschen series, which are in the infrared range. Of course drops to n = 4 or higher states behave similarly, but these lowest three sets are the ones that we’ve named.

In our derivation, we excluded relativity. To ﬁgure out if the relativistic correction will be large or if it is negligible, we need to look at the speed of the electron. What is v/c for the electron? Its speed is

n~ n~me2 e2 1 vn = = 2 2 = . mrn m~ n ~ n So the ratio is v e2 1 1 n = = α , c ~c n n where 1 α ' , 137 is the ﬁne structure constant. This is a small number, but not an extremely small number, implying that relativistic eﬀects are not completely negligible for the electron in a hydrogen atom. 10 The Origins of Quantum Mechanics

1.4 Hierarchy of Lengths

In nature, we ﬁnd a hierarchy of lengths as we drill down deeper and deeper.

Bohr radius: 2 a = ~ ' 0.5 × 10−10 m. me2 Compton wavelength: This is the regime below which classical mechanics does not work and must be replaced with quantum mechanics.

λ = αa = ~ ' 3.8 × 10−13 m. C mc

Here, α = e2/~c is the ﬁne structure constant. Classical electron radius: This is considered the boundary of classical electrodynamics. e2 r = α2a = αλ = ' 2.8 × 10−15 m. cl C mc2 Planck length: This is a universal length scale that does not depend on particle mass or charge. It can only depend on ~, G, and c. To get the right unit of m, we can multiply λC and rcl and then take the square root. To ﬁgure out the correct combination of units, we compare the Coulomb and gravitational potentials, e2/r versus Gm2/r, then we see that e2/m2 → G. So the Planck length is r G ` = ~ ∼ 10−35 m. P c3

We will use the ﬁrst three of these length scales frequently.

Compton Eﬀect Suppose we have an electron that is initially at rest. It is struck by a photon of initial wavelength λi. The electron scatters oﬀ at some angle θ relative to its initial direction.

Some momentum is imparted to the electron, causing it to recoil. The energy imparted to the electron is equal to the energy lost by the photon. This implies that the photon’s wavelength is lengthened during this process by an amount

θ θ ∆λ = 4πλ sin2 = 4π ~ sin2 . C 2 mc 2

1.5 The Correspondence Principle

The correspondence principle states that classical mechanics should be a limit of quantum mechanics. Let’s consider the case of an electron’s orbital frequency. Recall the quantum-mechanical Rydberg energy of the hydrogen atom me4 1 En = − . 2~2 n2 1.5. The Correspondence Principle 11

The change in frequency when an electron transitions from the state n to the state n0 is

4 4 E − E 0 me 1 1 me 1 1 ∆ω = n n = − − = − . ~ 2~3 n2 n02 2~3 n02 n2 Consider the case when both n and n0 are large (i.e. n, n0 >> 1) and their diﬀerence is much smaller than their magnitude (i.e. ∆n = n − n0 << n and ∆n = n − n0 << n0). In this limit, me4 n2 − n02 me4 (n − n0)(n + n0) me4 2n ∆ω = = ' ∆n. 2~3 n2n02 2~3 n2n02 2~3 n4 So me4 ∆ω ' ∆n. ~3n3 On the other hand, the frequency Ω that corresponds to the classical motion of the electron is 2π 2π v Ω = = = , T 2πr/v r where T is the period of the electron’s orbit, v is the speed, and r is the radius. Bohr quantization tells us that

mvnrn = n~, where the left-hand side is the orbital angular momentum. This implies that

v n n me4 Ω = n = ~ = ~ = . n 2 2 2 2 3 3 rn mrn ~ n ~ n m me2 Comparing this result with ∆ω, tells us that

∆ω = Ωn∆n, which is what we expect from the correspondence principle. We can approach this more generally. Recall

(action) = p dq = 2πn , n ˛ n ~ where p pn = 2m(E − U), is the classical momentum on a trajectory corresponding to state n. Now, consider two neighboring trajectories corresponding to n and n + 1. 12 The Origins of Quantum Mechanics

If we diﬀerentiate both sides of the Bohr quantization condition with respect to n, we get d 2π = p dq. ~ dn ˛ n

This is the “distance” between the two neighboring states. If our turning points are an and bn, and assuming our position coordinate is x, then we can write

d d bn pn dq = 2 pn dx. dn ˛ dn ˆan

How do we diﬀerentiate this? We know that pn will generally depend on n, but addition- ally, the turning points an and bn, which are our integration limits, probably depend on n. This complicates the matter of diﬀerentiating the integral. However, we know that pn = 0 at the turning points, so it doesn’t matter—we can move the derivative inside the integral without worrying about the integration limits. At the turning points, the entire integral is zero. So

bn bn d d dpn pn dq = 2 pn dx = 2 dx. dn ˛ dn ˆan ˆan dn We can write the derivative as dp dp dE n = n n . dn dEn dn If U = 0, then E = p2/2m and

dE p mv = = = v, dp m m so dp 1 n = . dEn vn More generally, in Hamiltonian mechanics,

∂H =q. ˙ ∂p

Putting this back into our integral gives us

bn bn dpn dEn dEn 1 dEn 2π~ = 2 dx = 2 dx = Tn, ˆan dEn dn dn ˆan vn dn

where Tn is the total period of this orbit. Since T = 2π/Ω, where Ω is the classical frequency, we get dE n = Ω . dn ~ n

dEn But dn = ~ω, so in general, we get ω = Ω. That is, quantum frequency reduces to classical frequency.

1.6 Simple Variational Approximations

To estimate the ground state energy of a system 1. Write down the (usually simpliﬁed by applying relevant assumptions) Hamiltonian of the system containing a variational parameter which is to be optimized 1.6. Simple Variational Approximations 13

2. Minimize this energy by diﬀerentiating with respect to the unknown parameter and setting it equal to zero 3. Plug the minimal value of the parameter back into the energy to get the estimate of the ground state energy This is best illustrated with several examples. Memorize the numerical values of these quantities

−22 ~ ' 6.58 × 10 MeV · s 2 me ' 0.511 MeV/c 2 mp ' 938 MeV/c . Be sure to memorize these combinations of quantities along with their numerical values 2 ~ ˚ a0 = 2 ' 0.529 A (Bohr radius) mee e2 1 α = ' (Fine structure constant) ~c 137 1 α2m c2 = 1 Ry = 13.6 eV (Atomic energy unit) 2 e ~c ' 197 MeV · fm λ = ~ ' 3.86 × 10−9 m (Electron Compton wavelength) e mc e2 r = ' 2.82 × 10−11 m (Classical electron radius). e mc2

Example 1.6.1

Estimate the ground state energy of a hydrogenic atom containing Z protons in the nucleus and a single localized electron. We know the energy is a function of momentum and position H(p, r). The hydrogenic Hamiltonian has the form p2 Ze2 H = − . 2m r We don’t know what p or r are for the ground state, but we can express both using an unknown length scale parameter Λ. This corresponds to thinking of r ∼ Λ for the electron’s distance from the nucleus and λ ∼ Λ for the electron’s de Broglie wavelength. Then using the de Broglie relation, we can write

p ∼ ~ . Λ Making these two substitutions gives us 2 Ze2 H(p, r) ∼ E(Λ) = ~ − . 2mΛ2 Λ The ground state is the lowest allowed energy, so we minimize this with respect to our variational parameter Λ dE 2 Ze2 = − ~ + = 0. dΛ mΛ3 Λ2 This gives us 2 a Λ = ~ = 0 . Zme2 Z 14 The Origins of Quantum Mechanics

Plugging this back into the energy E(Λ) gives us our estimate for the ground state energy

2 4 2 2 mZ e 1 2 e 2 1 2 2 2 2 E0 ∼ − = − mZ c = − mα c · Z = −13.6 eV · Z . 2~2 2 ~c 2 This method happens to give us the exact ground state energy of hydrogen. It does not give us any excited state energies. With this approach we sought to minimize the energy, so we are only able approximate the minimum energy. Note that the kinetic energy of the electron goes as ∼ 1/Λ2, which has no minimum at finite Λ. Similarly, the potential energy goes as ∼ −1/Λ, which has no minimum at finite Λ. However, when we add the two together to get the total energy, we get a function with a definite minimum, hence why this approach works here. One assumption we made here, by not including angles in the Hamiltonian, is that the ground state is spherically symmetric.

Try this variational approach to ﬁnd the ground state energy of the harmonic oscillator and for a system with U(r) ∼ rs. Example 1.6.2

Estimate the ground state energy for an atom with Z protons in the nucleus and two electrons. The complication here is that our Hamiltonian must now account for the kinetic and potential energies of two electrons and also for the repulsive interaction between the two electrons p2 p2 Ze2 Ze2 Ae2 H(p, r) = 1 + 2 − − + , 2m 2m r1 r2 |r2 − r1| where A is some unknown that characterizes the strength of the repulsion between the two electrons. However, we want a much simpler estimate of the energy, so we have to make some assumptions. We will assume that both electrons are in a circular orbit at the same distance from the nucleus. Further, they will be on opposite sides of the nucleus. So we assume r1 = r2 = Λ, the distance between the two electrons is 2Λ, and both electrons have momentum p = ~/Λ. Then 2 2 Ze2 Ze2 e2 2 2Ze2 e2 E(Λ) = ~ + ~ − − + = ~ − + . 2mΛ2 2mΛ2 Λ Λ 2Λ mΛ2 Λ 2Λ Diﬀerentiating gives us dE 2 2 2e2 = − ~ + Z∗, dΛ mΛ3 Λ2 where 1 Z∗ = Z − . 4 Setting the derivative equal to zero gives us

2 Λ = ~ . me2Z∗ Plugging this back into the energy to get our estimate of the ground state energy gives us 2 me ∗ 2 ∗ 2 E0 ' − (Z ) = −2 · 13.6 eV · (Z ) . ~2 1.6. Simple Variational Approximations 15

This simple approach gives good results for small two-electron atoms like H− and He, and very good results for larger ones like Li+, Be2+,B3+ and so on. 16 The Origins of Quantum Mechanics

1.7 Summary: The Origins of Quantum Mechanics

Skills to Master • Understand the photoelectric eﬀect and calculate the maximum kinetic energy of an ejected photon • Calculate the allowed energies, speeds, wavelengths, radii, etc. for various systems using Bohr quantization • Understand the derivation of the virial theorem, and be able to apply it when solving problems • Know the hierarchy of lengths • Know the hydrogen atom and all calculations and how to adjust those calculations for hydrogenic atoms • For a hydrogenic atom, calculate the wavelength of the light emitted when the electron goes from n to n0 • Calculate the Compton wavelength shift for light scattering from an electron • Make simple variational approximations

Photoelectric Eﬀect function p(q) and identify its shape (is it a circle or an ellipse?) then you can use known formulas There is a minimum energy, called a work function W to calculate the area or ionization energy that must be added to electrons in • In 1D, the total energy is E = p2/2m + U(x). a metal to free the electron from the atom. The work For a given energy, you can solve this for p(x) function is speciﬁc to the material. and plug that into the integral to get If an electron in a metal absorbs a quantum of energy ~ω, then the maximum kinetic energy of the p2m [E − U(x)] dx = 2π n. ejected photon is ˛ ~

Kmax = ~ω − W. If U(x) is given, this integral can typically be calculated. In many cases, there is symmetry, and Bohr Quantization we only have to integrate over half of the trajectory and then multiply by 2. I.e. it becomes a Bohr postulated that in an atom, there exist stable or- 1D integral instead of a line integral bits in which electrons do not radiate energy. Bohr postulated that these stable orbits or stationary states The de Broglie wavelength of a particle is satisfy the quantization condition 2π~ h p dq = nh = 2πn , n = 0, 1, 2,... λ = = . ˛ ~ p p On the left we have the action—the closed line integral This equation is useful if we know the allowed momenta over the period of the motion. If you plot a single pe- and want to calculate the allowed wavelengths. It also riod of the motion in the p vs. q plane, the trajectory implies that our quantization condition can be written forms a closed loop, and the integral p dq is simply in the form 1 the area enclosed by the loop. The trajectory¸ must dx = n. be periodic. If it is not, for example, if the particle ˛ λ(x) is simply scattering to infinity rather than orbiting or bouncing back and forth, then there is no quantization. Virial Theorem There are several ways to calculate the value of the integral without actually doing a line integral. In any state of finite motion, e.g. periodic motion, the time average of the total time derivative of any dy- • If you know that p = mv is constant, then you namical variable (i.e. function of the coordinates or can take this outside the integral, then p dq = momenta) is zero. The virial theorem is a result of a p dq, where the remaining integral is the¸ length specific application of this general principle. For the of¸ the closed trajectory of the particle virial theorem, the dynamical variable is the dot prod- • The integral p dq is simply the area enclosed uct ~r · ~p, then according to the general principle, the within the plot¸ of p versus q. If you can find the time average of the time derivative of this quantity is 1.7. Summary: The Origins of Quantum Mechanics 17 zero. After some manipulation, this implies the virial in the UV range. Similarly, the Balmer series is all theorem the transitions from higher energy levels down to the 2K = ~r · ∇~ U. n = 2 energy levels and occurs in the visible range. All drops to n = 3 are part of the Paschen series, which A useful specific case is when U = Ars, then the are in the infrared range. virial theorem implies that For atoms that are like hydrogen, we can use the sU s 2 same approach, but we replace e2 with Ze2 everywhere. K = , E = + 1 U = + 1 K. 2 2 s Z is the number of protons in the nucleus.

Remember, the virial theorem can only be used if Variational Approximation the motion is finite as in periodic motion. It cannot be applied to a particle that scatters to infinity. To estimate the ground state energy of a system 1. Write down the (usually simplified by applying Hydrogen Atom relevant assumptions) Hamiltonian of the system containing a variational parameter which is to be For the hydrogen atom, the Coulomb potential is optimized 2. Minimize this energy by differentiating with re- e2 U(r) = − . spect to the unknown parameter and setting it r equal to zero Since the orbit of the electron is circular, the speed is 3. Plug the minimal value of the parameter back constant, so v = v, and we can apply the virial theorem into the energy to get the estimate of the ground to the potential get state energy

e2 v2 = . Miscellaneous mr Applying the Bohr quantization condition and using Remember the “hierarchy of lengths:” the fact that p is constant for a circular orbit, we get Bohr radius: n~ 2 v = . a = ~ ' 0.5 × 10−10 m. mr 0 me2 Comparing these two results implies that the radii of allowed orbits are Compton wavelength: This is the regime below which classical mechanics does not work and 2n2 r = ~ = a n2, must be replaced with quantum mechanics. n me2 0 where a is the Bohr radius, or the radius for n = 1. λ = αa = ~ ' 3.8 × 10−13 m. 0 e 0 mc The allowed energies are Here, α = e2/ c is the fine structure constant. 1 1 e2 me4 1 13.6 eV ~ En = Un = − = − 2 2 = − 2 . 2 2 rn 2~ n n Classical electron radius: This is considered the boundary of classical electrodynamics. If the electron transitions from state n to n0, then the change in angular frequency is 2 2 e −15 rcl = α a0 = αλC = ' 2.8 × 10 m. 4 mc2 E − E 0 me 1 1 ∆ω = n n = − . 3 02 2 ~ 2~ n n The Compton effect occurs when light scatters Then its change in wavelength is given by from an electron that is initially stationary. When the photon strikes the electron, it transfers some of its en- 1 ∆ω me4 1 1 ergy to the electron—causing it to move. Because it = = − . ∆λ 2πc 4πc~3 n02 n2 has lost energy, the photon scatters off at some angle θ with a change in wavelength ∆λ given by The Lyman series encompasses all electron transitions from any level down to the ground state n = 1. θ θ ∆λ = 4πλ sin2 = 4π ~ sin2 . This set of transitions all have similar energy and are e 2 mc 2 18 The Origins of Quantum Mechanics

Memorize the numerical values of these quantities

−22 ~ ' 6.58 × 10 MeV · s 2 me ' 0.511 MeV/c 2 mp ' 938 MeV/c .

Be sure to memorize these combinations of quantities along with their numerical values

2 ~ ˚ a0 = 2 ' 0.529 A (Bohr radius) mee e2 1 α = ' (Fine structure constant) ~c 137 1 α2m c2 = 1 Ry = 13.6 eV (Atomic energy unit) 2 e ~c ' 197 MeV · fm = 197 eV · nm λ = ~ ' 3.86 × 10−9 m (Electron Compton wavelength) e mc e2 r = ' 2.82 × 10−11 m (Classical electron radius) e mc2 2 Eion = 13.6 eV · Z , (Ionization energy of hydrogenic atom). Chapter 2

Wave Mechanics

For the classical free wave, i.e. no potential energy or force, the wave function in 1D has the form Ψ(x, t) = Aei(kx−ωt), where A is the amplitude, k is the wave number, and ω is the frequency. Now we consider the quantum-mechanical de Broglie waves which have wavelength

2π 2π λ = = ~. k p

Then ~ can be used in the mapping between classical quantities and quantum-mechanical quantities

p = ~k E = ~ω.

The intensity of the free wave is given by the square of the wave function

|Ψ|2 = |A|2 = const.

The probabilistic interpretation of the wave intensity comes from Max Born.

2.1 Inﬁnite Walls

Instead of a free wave, suppose we have a wave coming from the left (x < 0) and we have an impenetrable wall at x = 0. Impenetrable means the probability of finding the wave/particle at x > 0 is zero. When the wave hits the wall, it must get reflected, and this reflected wave will interfere with the source wave. 20 Wave Mechanics

We assume that ω is ﬁxed. Then the wave function in the region x < 0 is the sum of the right-moving source wave and the left-moving reﬂected wave

Ψ(x, t) = Aei(kx−ωt) + Bei(−kx−ωt) = e−iωt Aeikx + Be−ikx .

At the wall at x = 0, we have A + B = 0, which implies that B = −A. So our wave function is Ψ(x, t) = 2iA sin(kx)e−iωt. We often consider only the x part of the wave function, which in this case is

ψ(x) = Aeikx + Be−ikx,

where the first term is the incident wave and the second term is the reflected wave. The momentum of the incident wave is pinc = ~k. The momentum of the reflected wave is pref = −~k. Since the overall wave is a superposition of the incident and reflected waves, the overall wave could have momentum +~k or −~k, but we don’t know which. This is the embryo of the uncertainty principle. To satisfy the boundary condition ψ(0) = 0, we need B = −A, and then

ψ(x) = Aeikx − Ae−ikx = 2iA sin(kx).

Consider again the inﬁnite square well. Now we have our particle trapped between two inﬁnite potential walls.

We again have a free region bounded by an inﬁnite wall on the right, so we can use the same general form of the wave equation

Ψ(x, t) = A sin(kx)e−iωt = ψ(x)e−iωt,

where A is some unknown constant. However, now we also have a wall on the left side, so our boundary conditions are ψ(0) = ψ(a) = 0, which implies sin(ka) = 0 =⇒ ka = nπ, So nπ k = , n a is our quantization condition. This time our quantization condition came from the spatial boundary conditions. This quantization condition is the condition for having stationary waves. In general, n = −∞,..., ∞, but we can include the negative versions in the constant A, and just take n = 1, 2, 3,.... Inside the square well, the particle has kinetic energy but no potential energy, so

p2 2k2 2π2n2 E = = ~ = ~ . n 2m 2m 2ma2 2.1. Inﬁnite Walls 21

The set {kn} of allowed values of kn form a complete set, and we call it a “spectrum”. Our wave function in the inﬁnite square well is

−iEnt/ Ψ(x, t) = An sin(knx)e ~. Notice that a stationary state is a pure harmonic. The probability of finding the particle at a position x is proportional to the uncertainty dx 2 2 Prob = |An| sin (knx) dx. The total probability should be equal to one so a 1 = dx Prob. ˆ0 This is the process of normalization. Different stationary states are orthogonal in the sense that a ∗ dx ψn(x) ψn0 (x) = δnn0 . ˆ0 This assumes that the individual ψ are normalized. For an infinite square well in the usual region 0 < x < a, the potential is symmetric with respect to the center of the well at x = a/2. This implies that the stationary states are alternately symmetric and antisymmetric with respect to x = a/2. For the infinite square well, the ground state with n = 1 is symmetric with respect to x = a/2. The first excited state n = 2 is antisymmetric with respect to x = a/2.

A more convenient choice is to put the inﬁnite square well in the region −a/2 < x < a/2 so that it is centered at x = 0. Then, the n = 1 ground state is symmetric (i.e. with respect to the usual axis of symmetry x = 0), and the n = 2 state is antisymmetric, and so on. I.e., all odd n have symmetric wave functions, and all even n have antisymmetric wave functions. In general, if the potential is symmetric with respect to the middle of the region, then the wave functions alternate symmetric and antisymmetric. In 3D, if the potential depends only on r, then we have symmetry in terms of three diﬀerent coordinates. In general, integrating over two wave functions with opposite parity gives zero

dx ψ∗ ψ = 0. ˆ + − Recall that the normalization condition requires

dx |ψ (x)|2 = 1. ˆ n For the inﬁnite square well, this implies that r 2 A = . n a 22 Wave Mechanics

So for an inﬁnite square well of width a,

r 2 2 2 2 nπx i n π − Ent ~ Ψ(x, t) = sin e ~ ,E = , n = 1, 2, 3,.... a a n 2ma2

2.2 Continuity Equation

The probability to ﬁnd a particle does not change for an unperturbed system of particles. Total probability is 1, and it is conserved. A continuity equation is a consequence of a total quantity being conserved in a system. Consider a region of space with volume V , surface S, and surface normal element dS~. Then if ρ(~r, t) is the local probability density for a particle, then the probability of ﬁnding the particle in V is 3 ProbV = d r ρ(~r, t). ˆV Then ∂ 3 ProbV = − ~j · dS~ = − div ~j d r, ∂t ˛S ˆV where ~j is the probability current. This implies that ∂ ρ dV = − div ~j dV, ∂t ˆ ˆ which implies ∂ρ = −div ~j ∂t or ∂ρ + div ~j = 0. (2.1) ∂t This is the ‘local’ continuity equation when total ρ is conserved. For total volume, ∂ ρ dV = 0, ∂t ˆ which implies ρ dV = const. ˆ In 1D, Eq. (2.1) becomes ∂ρ ∂j + = 0. ∂t ∂x For a stationary state Ψ ∼ e−iEt/~,

ρ = |Ψ|2 = constant in time.

So in a stationary state, ∂ρ = 0. ∂t ∂j Plugging this into the 1D continuity equation gives us ∂x = 0, which implies j =const. In general, ~j = ρ~v, so for a plane wave, ~p ~k ~j = ρ~v = ρ = ρ~ . m m 2.3. Finite Walls 23

2.3 Finite Walls Tip

Even States For potential wells, E <

Now we consider a ﬁnite square well with a depth of V0 as shown below. 0 corresponds to bound states, and E > 0 corresponds to propagating particles.

The E is the binding energy. Notice that the potential is symmetric with respect to the center of the well, so the solutions are parity even/odd. We can just as well think of the well as being in the region −a/2 < x < a/2, which is more convenient. We will look for the even solutions ψ(x) = ψ(−x). For this speciﬁc potential—symmetric ﬁnite well—there is always at least one bound state. A typical wavelength is Λ ∼ a. By the de Broglie relation, the typical momentum must be p ∼ ~ ∼ ~. Λ a For a localized particle, i.e. a particle bound tightly to a deep well, the kinetic energy is

p2 2 K ∼ ∼ ~ . loc m ma2 For a shallow well, we have ka << 1, and the particle is very loosely bound. I.e. we cannot localize the particle because although it is bound, its wavelength stretches far into the classically forbidden region. Such a particle is called a quantum halo. The halo condition is Λ >> a. The shallow well condition is V ma2V 0 = 0 << 1. ~2/ma2 ~2 Let us look at the wave function inside the well, i.e. in the region −a/2 < x < a/2. Here the particle exists as a wave reﬂecting back and forth between the walls, so the general form is ψ(x) = Aeikx + Be−ikx. Since we are looking only for the even solutions for now, we require B = A, and so

ψ(x) = 2A cos(kx).

The general deﬁnition of the wave number is

p r2m k = = (E − V (x)). ~ ~2 To get this, simply solve p2 E = + V, 2m 24 Wave Mechanics

for p2, then make the substitution p = ~k. In our case, the wave number for the wave function inside the well is

r2m r2m k = (−E − (−V0)) = (V0 − E). ~2 ~2 Outside the well, i.e. in the classically forbidden region outside of −a/2 < x < a/2, the wave function is not zero as it would be if the potential walls were infinitely high. Since there is no reflection of the wave occuring in this region, the general form of the wave function is 0 ψ(x) = Ceik x, where the wave number is r r2m 2mE k0 = (−E − 0) = ±i . ~2 ~2 We now define k0 = iκ, then

ψ(x) = Cei(±iκ)x = Ce±κx.

Only the version with a negative sign is physical in the classically forbidden region at x > 0, so r 2mE ψ(x) = Ce−κx, κ = . ~2 We call the quantity 1/κ the penetration length. So our solution is ( 2A cos(kx) (inside) ψ(x) = . Ce−κx (outside) . Next, since the wave function must be continuous, we need to join the solutions at the boundary x = a/2. That means, we need to choose C so that we satisfy

ψleft = ψright ∂ψ ∂ψ = ∂x left ∂x right

at the boundary x = a/2. This gives us the system of equations

ka 2A cos = C−κa/2 2 ka −2Ak sin = −κC−κa/2. 2

If the determinant of this system is zero, then there is a solution. In our case, we can divide one equation by the other to get the condition

ka k tan = κ. (2.2) 2

If there is a solution to this equation, then a bound state exists. To ﬁnd such a solution, just plot both sides of this equation. Consider a shallow well and assume that E << V0. Then r 2mV0 k ≈ ≡ k0. ~2 2.3. Finite Walls 25

Since we are assuming a shallow well, V0 and thereby k0 are small. Then Eq. (2.2) implies that r k0a 2mE mV0a k0 ' = . 2 ~2 ~2 Then 2 m2V 2a2 mV 2a2 E = ~ 0 = 0 . 2m ~4 2~2 Notice that this implies that 2 E mV0a = 2 << 1, V0 2~ by the shallow well condition. The wavelength of the particle is r s 1 ~2 ~2 2~2 Λ ∼ = = 2 2 , κ 2mE 2m mV0 a which implies 2 Λ = ~ . mV0a By the shallow well condition, this implies

Λ ~2 = 2 >> 1. a mV0a So we have a halo particle, i.e. the particle lives mostly in the classically forbidden region. Can we get the same result using only physical arguments? Consider a generic potential well of width ∼ a and depth V0 as shown in the graphic below.

The classical energy is p2 E = + U(x), 2m where U(x) acts over a region of width ∼ a. We want to ﬁnd quantum states with

p ∼ ~ , Λ and we are expecting Λ >> a. Then our estimate of the energy is

2 a E ∼ ~ − V , 2mΛ2 Λ 0 where our estimate of the potential comes from the fact that the particle lives over a width Λ >> a, but U(x) only exists over a region ∼ a. I.e. −aV0/Λ is the average value 26 Wave Mechanics

of the potential over the region Λ if we assume U(x) = −V0 over a width a and U(x) = 0 elsewhere. To find the ground state energy, we minimize E by differentiating with respect to Λ and setting it equal to zero. We find that

~2 Λ ~2 Λ = , =⇒ = 2 >> 1. mV0a a mV0a So we get the same result using physical arguments and the variational approach. Plug- ging this value back into the expression for E and simplifying gives us ma2V 2 E = − 0 = −E. 2~2 So we get the same result for the energy also. Note: The variational approach is typically only accurate to an order of magnitude, but even that’s usually very useful. If we substitute V a → dx |V |, 0 ˆ then we get precise results and 2 m E = dx |U| . 2~2 ˆ If the potential is extremely narrow then it becomes a delta function

V (x) = −g δ(x),

with g > 0.

It turns out that this potential always has exactly one bound state with energy mg2 E = . 2~2 Its wave function has the form ψ(x) = Ae−k|x|.

Make sure you understand these two approaches for obtaining/estimating the wave function and energy for even states. 2.3. Finite Walls 27

Odd States Tip For the odd states, we have ψ(x) = −ψ(−x). Then ψ(0) = −ψ(0), which implies that there is a node in the middle, i.e. ψ(0) = 0. This implies that B = −A, and so the wave For non-symmetric wells, function inside the well has the form there is not always a bound state. ψ(x) = const · sin(kx).

Outside the well, i.e. outside of −a/2 < x < a/2, we have

r 2mE ψ(x) = Ce±κx, κ = . ~2 Next we need to smoothly join the two wave functions at the boundary x = a/2. The first time this can occur is when the energy is just barely negative, E = 0−. Then we need ka π = , 2 2 which implies ar2m π V0 = . 2 ~2 2 This is the condition for (the first) bound state. Note the odd wave functions for this particle in the finite well −a/2 < x < a/2 are zero at x = 0, the wave functions would be exactly the same if there were an infinite wall at x = 0. That is, the odd wave functions of the finite well in −a/2 < x < a/2 are the same as the wave functions of the half-finite square well drawn below.

The kind of well pictured here comes up in 3D whenever there is spherical symmetry. In 3D, the horizontal axis is r instead of x, and since r ≥ 0, this is equivalent to there being an inﬁnite wall at r = 0. For non-symmetric wells, like the one shown below, there is not always a bound state.

For the well shown above, ﬁnd the relation between U2 and U1 for the existence of a bound state. 28 Wave Mechanics

2.4 Propagating Waves

A propagating wave is an unbounded state.

In the region x < 0, we have an incident wave moving to the right and potentially a reflected wave moving to the left. So the form of the wave function is r 2mE ψ(x) = Aeikx + B−ikx, k = . ~2 In the region x > 0 we have only a transmitted wave r ik0x 0 2m ψ(x) = Ce , k = (E − V0). ~2 The coefficient A is arbitrary—it depends on the source of the particle. We have no condition to solve for A. But we do have conditions to solve for B and C. Next we join the wave functions of the two regions at the boundary x = 0. The condition that the wave function must be continuous there implies that A + B = C. The condition that the derivatives must be continuous there implies ik(A − B) = ik0C. This gives us k C = (A − B) = A + B, k0 which implies k − k0 B = A . k + k0 Notice that B = 0 if k = k0. I.e. if there’s no wall, there’s no reflection. Plugging this back in for C gives us 2k C = A . k + k0 Notice that C = A if k = k0. That is, if there’s no wall then the transmitted wave equals the reflected wave. In general, we assume that the particle source is on the left and that asymptotically, the potential goes to a constant value in both the left and the right directions. Only in the middle is there some localized disturbance. This implies that in the left region we have a incident and reflected waves and on the right we have a transmitted wave. There will be a reflected wave whenever there is any disturbance in the potential in the direction in which the incident wave is traveling. This is true whether the disturbance is of a barrier nature or of a well nature. The probability density current must be conserved. That is, we require

jinc + jref = jtrans. 2.5. Resonance 29

Remember, the general formula is Tip p ~k j = ρv = ρ = ρ , When working problems in- m m volving propagating waves, where ρ is the probability density of the wave and v is the wave speed. The probability always check that the prob- density is simply the complex magnitude squared of the coefficient of the wave, so in ability current is conserved. general, Alternatively, check that k the sum of the reflec- j = |wave coefficient|2 ~ . wave m tion and transmission coefficients is 1. In our case,

k j = |A|2 ~ inc m k k(k − k0)2 j = −|B|2 ~ = −|A|2 ~ ref m m (k + k0)2 k0 4k2k0 j = |C|2 ~ = |A|2 ~ . trans m m (k + k0)2

We ﬁnd that indeed the current on the left equals the current on the right

k (k − k0)2 j + j = |A|2 ~ 1 − = j . inc ref m (k + k0)2 trans

When working problems like this, we should always check that the current is conserved. Alternatively, we can look at the reflection coefficient and transmission coefficient |j | j R = ref ,T = trans . jinc jinc Probability conservation requires that

T + R = 1.

2.5 Resonance

Consider a generic potential well as shown below.

We have a diﬀerent wave function in each region. For x < 0, we have incident and reﬂected waves r 2mE ψ(x) = Aeikx + Be−ikx, k = . ~2 30 Wave Mechanics

In region 0 < x < a we also have incident and reflected waves since the potential changes Tip at the right end of this region, but now we have a different energy level Whenever there is a change r 0 0 2m(E − (−U)) in the potential, there will ik x −ik x 0 ψ(x) = Ce + De , k = 2 . be a reflected wave. ~ In region x > a, we only have a transmitted wave

ψ(x) = F eikx.

To solve, we match the wave functions and derivatives on the two boundaries. As this example illustrates, not only potential barriers, but also potential wells can cause scattering if the particle’s energy is too large for it to be in a bound state. For scattering from a potential well, the phenomenon of resonance can occur. This occurs when the wave inside the well interferes constructively with the wave coming from the left, and we end up with 100% transmission (i.e. T = 1). In this case, this will occur when k0a = nπ. The energy values satisfying this relation are called resonances.

2.6 Tunneling Waves

Next, we consider a potential barrier as shown below. The interesting case (i.e. tunneling) occurs when E < U.

We have a different wave function in each region. For x < 0, we have incident and reflected waves r 2mE ψ(x) = Aeikx + Be−ikx, k = . ~2 In region 0 < x < a we also have incident and reflected waves since the potential changes at the right end of this region, but now we have a different energy level r 0 0 2m(E − U) ψ(x) = Ceik x + De−ik x, k0 = . ~2 In region x > a, we only have a transmitted wave

ψ(x) = F eikx.

Note that k0 is now imaginary. We deﬁne

k0 = ±iκ.

Then in region 0 < x < a, our wave function becomes

ψ(x) = Ce−κx + Deκx. 2.6. Tunneling Waves 31

The reﬂection and transmission coeﬃcients are |B|2 |F |2 R = ,T = . |A|2 |A|2

We can think of Deκx as a (imaginary) wave being reﬂected at x = a and going downward to the left. Note that D ∼ e−2κa since the wave had to travel through the barrier before it even got reﬂected. So

T ∼ e−2κa.

Complicated Tunneling Potentials With more complicated potentials, we can use this result to estimate the tunneling probability. For example, we can slice a complicated potential into many separate barriers.

Then

a " a r # Y Y 2m T ∼ T ∼ e−2κiai ∼ exp −2 κ(x) dx = exp −2 dx (U(x) − E) . total i ˆ ˆ 2 i i 0 0 ~

Alpha Decay

With alpha decay, we are interested in the lifetime rather than the tunneling probability. Inside the well part of the potential, the electron bounces back and forth with frequency ω. Only very occasionally will the electron tunnel through the barrier. I.e. the electron will not tunnel through every time it reaches a. The tunneling probability is

2 a2 P (decay) = ωexp − p2m [U(x) − E]dx . κ ˆa1 This is not exact, but it gives a very good estimate of the tunneling probability. Then 1 lifetime = . P (decay) 32 Wave Mechanics

2.7 Level Density

In thermal equilibrium at temperature T , the distribution of particles is given by the Fermi-Dirac or Bose-Einstein distributions

1 f(ε) = , eα+βε ± 1

where the plus sign is taken for fermions, and the minus sign is taken for bosons.

Usually ∆ε, the diﬀerence between energy levels is very small, and we can approximate the ﬁlled states as continuous in E. In this continuum limit, an energy range E+∆E becomes E + dE and ∆N, which is the number of states in the range ∆E becomes dN. The probability for a particle to be in the energy range ∆E is

∆N f(ε).

The number of states in the interval E + dE is

dN = ρ(E) dE,

where ρ(E) is the level density. The total number of states less than energy E is

E N(E) = ρ(E) dE, ˆE0 so one way to calculate the level density is as

dN ρ(E) = . dE

In quantum mechanics, the probability of ﬁnding a particle at an exact point x is zero. Instead, we need to look at the probability of ﬁnding a particle in a region between x and x + dx. This probability is of course |ψ|2 dx. Here, |ψ|2 is the probability density, which is analogous to our level density. Another way to calculate the level density is as

X ρ(E) = δ(E − Ei). i

To see this, consider what happens if we integrate both sides over the energy

E E X ρ(E) dE = δ(E − E ) dE. ˆ ˆ i E0 E0 i 2.7. Level Density 33

The left side is now N(E). On the right, we can switch integration and summation to get

X E X δ(E − Ei) dE = 1. ˆE0 i i:Ei

The right side adds a 1 for every Ei that is less than E, so it gives us the total number of states N(E) less than E. Thus, our formula is justiﬁed. In general, the semi-classical density of states is dω ρ(E) = δ (E − ε(~r, ~p)) , ˆ (2π~)D where D is the number of dimensions, dω is the phase space element, and ε(~r, ~p) is the dispersion relation. For example, for 2D particles in a harmonic oscillator potential, the density of states would be ! dx dy dp dp p2 + p2 1 ρ(E) = x y δ E − x y − mω(x2 + y2) . ˆ (2π~)2 2m 2

Example 2.7.1

Consider an L×L×L box—a 3D inﬁnite square well containing particles. To get the level density of the energies in the box, we ﬁrst need the energy spectrum. The particles are just plane waves ψ(~r) = ei~p·~r/~, and the energy spectrum is given by the kinetic energies of the particles

p2 H = . 2m To satisfy the boundary conditions, we need the wave functions to be zero at the walls of the box. Then we will use periodic boundary conditions. We need

ψ(x + L, y, z) = ψ(x, y, z),

and similarly for y and z. So we need

eipx(L+x)/~ = eipxx/~.

This implies that eipxL/~ = 1, which means p L x = 2πn, n = 0, ±1, ±2,... ~ So our momenta can only take the quantized values 2π 2π 2π p = n ~, p = n ~, p = n ~. x 1 L y 2 L z 3 L In vector form, 2π ~p = ~ (n , n , n ) . L 1 2 3 We can write the possible momenta as lattice points. The number of states N(E) with energy less than E can be counted as X N(E) = 1.

~p:Ep≤E 34 Wave Mechanics

This is simply the sum of all momenta (i.e. lattice points) with energy less than E. Now we apply a trick. Since the n’s are integers, the diﬀerence ∆n between consecutive integers is always 1. So ∆n1 = ∆n2 = ∆n3 = 1. So we can write X N(E) = ∆n1∆n2∆n3.

~p:Ep≤E But we know that L ∆ni = ∆pi, 2π~ so this becomes 3 X L N(E) = ∆px∆py∆pz. 2π~ ~p:Ep≤E In the large L limit, ∆p ∼ dp, so we can write our sum as the integral L3 N(E) = d3~p . ˆ (2π~)3

To get the number of states with energy less than E, we use the fact that√ the energy at a given momentum is p2/2m. Thus, we want p2/2m ≤ E or p ≤ 2mE. Then V 3 N(E) = 3 d ~p. (2π~) ˆEp≤E This integral has an integrand of 1, so it simply gives the “volume”, and p = q 2 2 2 px + py + pz plays the role of the “radius”. So we are really just calculating the volume of a sphere V 4π 3 N(E) = (2mE) 2 . (2π~)3 3 An alternative approach to calculating the level density is to use the formula with the delta function X X p2 V p2 ρ(E) = δ (E − E ) = δ E − = d3~p δ E − . p 2m (2π )3 ˆ 2m ~p ~p ~ First, we integrate over the angles to get

∞ 2 V 2 p ρ(E) = 3 4π p δ E − dp. (2π~) ˆ0 2m Remember, in polar coordinates,

d3~r → 4πr2 dr d2~r → 2πr dr.

Next, we simplify the delta function. Since

δ(−x) = δx 1 δ(ax) = δx, |a| we can simplify it as p2 δ E − = 2m δ p2 − 2mE . 2m 2.7. Level Density 35

Using the rule, 1 δ(x2 − a2) = (δ(x − a) + δ(x + a)) , 2|a| we can simplify it further as

p2 m √ δ E − = 2m δ p2 − 2mE = √ δ(p − 2mE). 2m 2mE √ We neglected the term with δ(p + 2mE) here since it is outside the domain of integration and won’t contribute anyway. Then

V m ∞ √ √ 2 ρ(E) = 3 4π p δ(p − 2mE) dp. (2π~) 2mE ˆ0 V m = 4π √ 2mE. (2π~)3 2mE Then

E 2 E V 2m 1 V 4π 3 √ 2 2 N(E) = g(E) dE = 3 4π E dE = 3 (2mE) . ˆ0 (2π~) 2m ˆ0 (2π~) 3 For the case of a relativistic gas, we can calculate the level density using the same formula V 3 ρ(E) = d ~p δ (E − Ep) . (2π~)3 ˆ 2 But now, instead of using the energy Ep = p /2m, we would have to use the p 2 2 2 4 2 2 relativistic energy Ep = p c + m c . In the ultra-relativistic case, p c >> m2c4, then E ∼ pc.

In general, remember that

X L ∞ → dp, (1D) (2π ) ˆ ~p ~ −∞ X A → d2~p, (2D) (2π )2 ˆ ~p ~ X V → d3~p, (3D) (2π )3 ˆ ~p ~

where L is the length of the 1D region, A is the area of a 2D box, and V is the volume of a 3D box. For non-relativistic particles in a box,

r L 2m − 1 (1D): ρ(E) = ∼ E 2 , 2π~ E mA (2D): ρ(E) = ∼ E0, 2π~2 V 3 √ 1 (3D): ρ(E) = (2m) 2 E ∼ E 2 . 4π2~3 36 Wave Mechanics

For ultra-relativistic particles in a box, L (1D): ρ(E) = ∼ E0, π~c A (2D): ρ(E) = E ∼ E1, 2π~2c2 V (3D): ρ(E) = E2 ∼ E2. 2π2~3c3 What if instead of having free particles in a box, we want to calculate the level density for particles in some potential U(~r)? In this case, we assume the classical energy p2 ε(~p, ~r) = + U(~r), 2m is constant within a small element d3r d3p of phase space. Then the density of states is d3r d3p ρ(E) = δ (E − ε(~p, ~r)) . ˆ (2π~)3 Going to spherical coordinates, we can rearrange our integral as 4π p2 ρ(E) = d3r dp p2 δ E − − U(~r) . (2π~)3 ˆ ˆ 2m We can use the delta-function property,

X δ(x − xi) δ (g(x)) = , |g0(x )| i i

where the xi are the roots of g(x). In our case, the argument of the delta-function is a function of p, and so p2 g(p) = E − − U(~r), 2m which has the single root (for positive p) p p0 = 2m(E − U). The derivative is dg p = − . dp m So the delta-function property allows us to write p2 δ(p − p ) δ E − − U(~r) = 0 . 2m p0/m So our integral becomes

4π 3 2 δ(p − p0) ρ(E) = 3 d r dp p (2π~) ˆ ˆ p0/m 4π 3 = d r mp0. (2π~)3 ˆ

Plugging in p0 and simplifying gives us

m3/2 √ ρ(E) = √ d3r E − U. 2 3 2π ~ ˆE≥U Note that the integral is only over the classically allowed region—between the turning points deﬁned by E = U. Hence, this is called the semi-classical approach, since the true approach would include the (usually) small contributions from classically forbidden regions. To go further and integrate over r, we would need to know the actual potential U(~r). 2.7. Level Density 37

Example 2.7.2

Calculate the semi-classical level density for a particle of mass m trapped in a 3D harmonic oscillator potential 1 U = mω2r2. 2 The turning points E = U are r 2E r = ± ≡ ±R. mω2 Then the density of states is

m3/2 R √ ρ(E) = √ 4π dr r2 E − U 2 3 2π ~ ˆ0 m3/2 R r 1 = √ 4π dr r2 E − mω2r2 2 3 2π ~ ˆ0 2 E2 = . 2(~ω)3 38 Wave Mechanics

2.8 Summary: Wave Mechanics

Skills to Master

• Given a potential V (x) and a particle energy E, plot them and identify different characterists like bound states, propagating waves, reflection, and tunneling • Given a 1D potential, write down the wave functions and join them at the boundaries by applying continuity of the wave function and its derivative • Given a 1D potential well, calculate the bound state wave functions and energy levels • Given a 1D propagating wave, calculate the probability current densities and the reflection and transmission coefficients • Given a 1D potential barrier, calculate the wave functions, and the tunneling probability • Calculate the density of states for a quantum gas in a box

The wave function of a classical free wave moving The key difference between bound states and prop- toward positive x is agating waves is that bound states have discrete energy levels, and propagating waves have continuous energy i(kx−ωt) Ψ(x, t) = Ae , spectra. A particle that exists as a propagating wave is where A is the amplitude, k is the wave number, and completely delocalized. A particle that exists as a ω is the frequency. The intensity of the wave is given bound state in a deep potential well is localized. There by the square of the wave function can also be intermediate cases. A particle that is barely |Ψ|2 = |A|2. bound in a shallow potential well has a wavelength much larger than the width of the well, and thus, is not For quantum-mechanical waves, we use the same sym- localized. Such a particle whose wavelength stretches bolism but make the substitutions far into the classically forbidden region is called a quantum halo. p = ~k, E = ~ω. After plotting V (x) and E, separate the problem into different regions wherever V (x) and E cross each ikx The momentum of a wave Ae , is p = ~k. Remember other. In regions where E > V (x), you will have wave that a particle with momentum p corresponds to the functions consisting of a wave propagating to the right quantum-mechanical de Broglie wave with wavelength and possibly a reflected wave propagating to the left 2π 2π h (assuming your source is at x = −∞) λ = ~ = = . p k p ψ(x) = Aeikx + Be−ikx, The general approach for 1D problems is to plot where r the given potential V (x) and the constant energy E 2m k = [E − V (x)]. of the particle. In general, we can have the following ~2 possibilities: This is obtained from E = p2/2m + V (x) and making Bound States: If E < V (x) except for a region in the substitution p = ~k. In regions where E < V (x), the middle where E > V (x), then we have bound you will have exponentially decaying (imaginary) waves state(s) of the form ψ(x) = Aeκx + Be−κx, Propagating Waves: If E > V (x) for all x, then we have a purely propagating wave where r2m Reflection: If E > V (x) to the left or the right, and κ = 2 [V (x) − E]. E < V (x) on the other side, then we have prop- ~ agation combined with reflection The penetration length 1/κ quantifies how much the wave function extends into the classically forbidden re- Tunneling: If E > V (x) except for a region in the gion. Note, k and κ are the same provided that you middle where E < V (x), then we have propaga- order E and V (x) such that the radical doesn’t become tion combined with tunneling imaginary. 2.8. Summary: Wave Mechanics 39

Once you have written down a wave function for The probability density current must be conserved. each region, tie them all together by enforcing that the That is, we require wave functions and their derivatives are continuous at each boundary. This will eliminate all of the coeffi- jinc + jref = jtrans. cients but one. The final coefficient is eliminated by normalizing the overall wave function. Alternatively, we can look at the reflection and Note: If a region is bounded by one or more in- transmission coefficients finite walls, use a sine wave function as opposed to a |jref | jtrans complex exponential. For example, if your wall is at R = ,T = , j j x = a, you can use sin(k[x − a]) which is zero at x = a, inc inc which is what you want. which give the probability of reflection and transmission. Probability conservation requires that Bound States T + R = 1. A symmetric potential well always has at least one bound state. An asymmetric well may or may not have For scattering from a potential well (i.e. inverted bound state(s). barrier), the phenomenon of resonance can occur. This In general, if the potential is symmetric with re- occurs when the wave inside the well interferes con- spect to the middle of a well, then the wave functions structively with the wave coming from the left, and we alternate symmetric and antisymmetric. In that case, end up with 100% transmission (i.e. T = 1). For a we may search for only the even (or odd) solutions by flat-bottomed well, this occurs when imposing ψ(x) = ψ(−x) (or ψ(x) = −ψ(−x)) on the wave function from the beginning. ka = nπ. For an infinite square well of width a, remember the wave function The energy values satisfying this relation are called resonances. r 2 nπx i − Ent Ψ(x, t) = sin e ~ , a a Tunneling and the energy levels Tunneling occurs when E > V (x) except for a small region in the middle where E < V (x). For this type of 2 2 2 n π ~ problem, we are generally interested in the probability En = , n = 1, 2, 3,... 2ma2 of tunneling, i.e. the transmission probability. For a potential barrier of height associated with κ Propagating Waves and width a, transmission is exponentially suppressed Propagating waves occur when E > V (x) for all x. If as −2κa there is a small region in which this is not true, then T ∼ e . we would have tunneling. For propagating waves, we have a continuous energy spectrum as opposed to the Quantum Statistics discrete spectrum associated with bound states. Thus, for propagating waves, we are not interested in the en- In thermal equilibrium at temperature T , the distribu- ergy levels. We are typically interested in reflection tion of particles is given by the Fermi-Dirac or Bose- and transmission probabilities. Einstein distributions The probability density current of a wave is 1 f(ε) = α+βε , p k e ± 1 j = ρv = ρ = ρ~ , m m where the plus sign is taken for fermions, and the minus sign is taken for bosons. where ρ is the probability density of the wave and v is For a gas of free particles in a D-dimensional box the wave speed. The probability density is simply the of length L in each dimension, the level density or den- complex magnitude squared of the coefficient of the sity of states is wave, so in general, D k L D j = |wave coefficient|2 ~ . ρ(E) = d p δ (E − E(p)) , wave m 2π~ ˆ 40 Wave Mechanics where E(p) is the energy of the individual particles. If the particles in the box are not free, but rather, For non-relativistic free particles, E(p) = p2/2m, then are subject to a potential U(~r), then this becomes dDr dDp L D p2 D ρ(E) = D δ (E − E(~p, ~r)) , ρNR(E) = d p δ E − . ˆ (2π~) 2π~ ˆ 2m For 1, 2, and 3 dimensions, this simplifies to where, in the non-relativistic case,

r 2 L 2m − 1 p (1D): ρ(E) = ∼ E 2 , E(~p, ~r) = + U(~r). 2π~ E 2m mA (2D): ρ(E) = ∼ E0, In 3D, this simplifies to 2π~2 V 3 √ 1 3/2 (3D): ρ(E) = (2m) 2 E ∼ E 2 . m √ 2 3 ρ(E) = √ d3r E − U. 4π ~ 2 3 2π Ê≥U For ultra-relativistic particles, E(p) = pc, then the den- ~ sity of states is The number of particles in the box with energy D less than E is L D ρUR(E) = d p δ (E − pc) , 2π~ ˆ E N(E) = dE0 ρ(E0). For 1, 2, and 3 dimensions, this simplifies to Ê0 L (1D): ρ(E) = ∼ E0, π~c A (2D): ρ(E) = E ∼ E1, 2π~2c2 V (3D): ρ(E) = E2 ∼ E2. 2π2~3c3 Chapter 3

Wave Functions and Operators

3.1 Superposition Principle

By the superposition principle, a general wave function can be written as a Fourier expansion X Ψ(~r, t) = cnΨn(~r, t). n

We can choose the Ψn to be stationary states with energy En, then

−iEnt/ Ψn(~r, t) = ψn(~r)e ~. The initial state is then X Ψ(~r, 0) = cnψn(~r). n

This expansion is possible because the set {ψn(~r)} is complete. The whole dynamics of the system is predetermined by the initial state. Essentially, quantum dynamics is the dynamics of phases only. Completeness is related to orthogonality. States with diﬀerent energies are orthogonal to each other, which means

( 0 ∗ 0 if n 6= n 0 (volume) ψnψn = 0 . ˆ (norm)n if n = n . Normalization requires that

(volume) |ψ |2 = 1. ˆ n

For a free particle, ψ ∼ ei~k·~r, and then |ei~k·~r|2 = 1 and the integral over all of space yields inﬁnity, so it is not normalizable. Thus, such a state is said to not belong in our space.

3.2 Dirac Delta Function

The Dirac delta function is not a true function. Rather, it is a “distribution” or a “generalized function”. It is deﬁned as an integral operator

b ( 0 0 0 f(x ) if x ∈ (a, b) dx f(x)δ(x − x ) = 0 . ˆa 0 if x ∈/ (a, b). If f(x) = 1, we have the special case

b ( 0 0 1 if x ∈ (a, b) dx δ(x − x ) = 0 . ˆa 0 if x ∈/ (a, b). 42 Wave Functions and Operators

There are a number of definitions of the delta function in terms of limits. The simplest definition is where you start with a piecewise function that is equal to 1/a in the region [−a/2, a/2] and zero everywhere else. This way the total area (i.e. the integral over it) is 1. Then you halve the region and double the height, so that the total area of one is maintained. Then the delta function is the limit as this process is repeated indefinitely.

Gaussian Limit Another deﬁnition of the delta function is as the limit

lim G(x, x0; σ) = δ(x − x0), σ→0

of the Gaussian distribution

1 2 2 −(x−x0) /2σ G(x, x0; σ) = √ e . 2πσ2

The area under the Gaussian is always 1 since it is normalized.

Plane Wave Limit Another approach is as the limit of plane waves. ( K ∞ if x = 0 lim dk eikx = = 2πδ(x). K→∞ ˆ−K 0 if x 6= 0.

This is a superposition of all possible plane waves.

Breit-Wigner Limit Another deﬁnition is as ( 1 η ∞ if x = x0 lim 2 2 = = δ(x − x0). η→0 π (x − x0) + η 0 if x 6= x0 3.3. Representations of the Wave Function 43

Properties An important property of the delta function is that

δ(x) δ(ax) = . |a|

To prove this, we start by noting that ( ∞ if x = 0 δ(ax) = . 0 if x 6= 0

This implies that δ(ax) ∝ δ(x). To find the proportionality factor, we have to normalize ε aε dξ ξ dx f(x)δ(ax) = f δ(ξ). ˆ−ε ˆ−aε a a Here we made the substitution ξ = ax. But now the right side can be simplified as aε dξ ξ 1 aε 1 f(0) f δ(ξ) = f(0) dξ δ(ξ) = f(0) sign(a) = . ˆ−aε a a a ˆ−aε a |a| If the delta function occurs at the boundary of our region, then the integral receives a factor of one-half b 1 dx f(x)δ(x − a) = f(a). â 2

What if we have δ(g(x))? Suppose xi are the zeros of g(x), i.e. g(xi) = 0. Then near x = xi, we can perform the expansion

0 0 g(x) ≈ g(xi) + (x − xi)g (xi) = (x − xi)g (xi). Then X dx δ(g(x))f(x) = dx δ ((x − x )g0(x )) f(x). ˆ ˆ i i i Each zero of g(x) inside the region we are integrating over contributes to the delta function, so we get

X f(xi) dx δ(g(x))f(x) = . ˆ |g0(x )| i i This will be useful, for example, in scattering problems. Another useful rule is 1 δ(x2 − a2) = (δ(x − a) + δ(x + a)) . 2|a|

3.3 Representations of the Wave Function

So far, we’ve looked at the coordinate (i.e. x) representation of the wave function. There are inﬁnitely many representations we can use, and they are all equivalent. For example, with a Fourier expansion, we can write ψ(x) as a sum of stationary states ψn(x), X ψ(x) = cnψn(x). n 44 Wave Functions and Operators

Then for t > 0, we have

X −iEnt/ ψ(x, t) = cnψn(x)e ~. n

Now all of the information is contained in the set of coeﬃcients cn

ψ(x) → {ψn(x)} → {cn}. We say that ψ(x) is the coordinate representation of our wave function and the set of coeﬃcients {cn} form the energy representation of our wave function. For {cn} to completely characterize our wave function, the set {ψn(x)} of stationary states must be orthogonal and complete. If the states are normalized such that

dx |ψ (x)|2 = 1, ˆ n then the orthogonality condition is

∗ dx ψ (x)ψ 0 (x) = δ 0 , ˆ n n n,n where the integrals are over whatever region your system lives in. We can use this orthogonality condition to ﬁnd the coeﬃcients {cn}, by starting with the initial state of our system X ψ(x) = cnψn(x), n ∗ then multiplying both sides by ψm(x) and integrating X dx ψ∗ (x)ψ(x) = dx ψ∗ (x) c ψ (x) ˆ m ˆ m n n n X = c dx ψ∗ (x)ψ (x) n ˆ m n n X = cnδnm n

= cm. This is just Fourier analysis. So the coeﬃcients are

c = dx ψ∗ (x)ψ(x), n ˆ n

∗ where ψ(x) is the initial state of the wave function. We can think of ψn(x) as the projection of ψ(x) onto the nth normal mode. Plugging our deﬁnition of the coeﬃcients back into the initial state gives us X ψ(x) = cnψn(x) n X = ψ (x) dx0 ψ∗ (x0)ψ(x0) n ˆ n n X = dx0 ψ(x0) ψ (x)ψ∗ (x0) ˆ n n n = dx0 ψ(x0) δ(x − x0) ˆ = ψ(x). 3.4. Operators 45

Completeness is the requirement that

X ∗ 0 0 ψn(x)ψn(x ) = δ(x − x ). n We can also expand the wave function ψ(x) as plane waves

ψ(x) = A dp eipx/~φ(p) ˆ

φ(p) = B dx e−ipx/~ψ(x). ˆ Taking one and plugging it into the other gives us

0 ψ(x) = A dp eipx/~B dx0 e−ipx /~ψ(x0) ˆ ˆ

0 = AB dx0 dp eip(x−x )/~ ψ(x0) ˆ ˆ δ(x − x0) = AB dx0 2π ψ(x0) ˆ |i/~| = 2π AB dx0 δ(x − x0)ψ(x0) ~ ˆ = 2π~ABψ(x). This implies that 1 AB = . 2π~ There’s√ a choice about how to split the 2π~ between A and B. Some authors set A = B = 1/ 2π~, but we will use the convention 1 A = ,B = 1. 2π~ Then 1 ψ(x) = dp eipx/~φ(p) 2π~ ˆ φ(p) = dx e−ipx/~ψ(x). ˆ

3.4 Operators

An operator is just the symbolic representation of a measurement process. For now, we only have coordinatex ˆ and momentump ˆ operators. The probability of ﬁnding the particle between x and x + dx is

P (x, x + dx) = w(x)dx = |ψ(x)|2dx.

In general, the expectation value of a quantity O is

hOi = dx O w(x), ˆ where w(x) is the probability distribution. In quantum mechanics, this becomes

hOiˆ = dx ψ∗(x) Oˆ ψ(x). ˆ 46 Wave Functions and Operators

Two important expectation values are the mean position and the mean-square position

hxi = dx ψ∗(x)x ˆ ψ(x) ˆ hx2i = dx ψ∗(x)x ˆ2 ψ(x). ˆ

For momentum, hpi and hp2i have analogous forms. The deviation is D E D E (x − hxi)2 = x2 − 2x hxi + hxi2

= hx2i − 2 hxi hxi + hxi2 = hx2i − hxi2 .

The uncertainty in x is calculated as

q ∆x = hx2i − hxi2.

In the momentum representation, the wave function ψ(x) is written as

dp ψ(x) = eipx/~φ(p), ˆ 2π~ where φ(p) = dx e−ipx/~ψ(x). ˆ The probability of ﬁnding the particle to have momentum in some interval is

P (p, p + dp) = |φ(p)|2dp.

In the momentum representation, the momentum operator isp ˆ = p. The expectation value of the momentum is dp dp hpi = p|φ(p)|2 = φ∗(p)p ˆ φ(p). ˆ 2π~ ˆ 2π~ Can we obtain hpi from the coordinate representation? If we start with

dp hpi = φ∗(p)p ˆ φ(p), ˆ 2π~ and plug in the integral deﬁnitions of φ(p), we get dp 0 hpi = dx eipx/~ψ∗(x) pˆ dx0 e−ipx /~ψ(x0) . ˆ 2π~ ˆ ˆ This can be rearranged as dp 0 hpi = dx ψ∗(x) dx0 peip(x−x )/~ ψ(x0). ˆ ˆ ˆ 2π~ We can write this as ∗ 0 ∂ dp ip(x−x0)/ 0 hpi = dx ψ (x) dx −i~ e ~ ψ(x ). ˆ ˆ ∂x ˆ 2π~ 3.4. Operators 47

But now the innermost integral is just the delta function, and so ∂ hpi = dx ψ∗(x) dx0 −i δ(x − x0) ψ(x0) ˆ ˆ ~∂x ∂ = dx ψ∗(x) −i dx0 δ(x − x0)ψ(x0) ˆ ~∂x ˆ ∂ = dx ψ∗(x) −i ψ(x). ˆ ~∂x But now we have an expression of the form

h(operator)i = dx ψ∗(x)(operator)ψ(x), ˆ and this implies that the momentum operator is ∂ pˆ = −i , ~∂x in the coordinate representation. Consider a state ψ(x) that is localized in position. That is, if you take a position measurement by applying the position operatorx ˆ, you get a deﬁnite value x0 for the particle’s position xψˆ (x) = x0ψ(x). The function ψ(x) does not change under the measurement of position. The function of x that satisﬁes all this is ψ(x) = δ(x − x0). This is the position space eigenstate of the position operator since

xδˆ (x − x0) = x0δ(x − x0). Suppose φ(p) is the momentum representation of the same function ψ(x). Then

xφˆ (p) = x0φ(p).

∂ But in the momentum representation,x ˆ = i~ ∂p , so this equation is now a simple partial diﬀerential equation, and we ﬁnd that

φ(p) = e−ix0p/~.

This is the eigenfunction ofx ˆ in the position representation. Now let us consider a plane wave, which is a state that is localized in momentum space as opposed to a state that is localized in position space. Then

pφˆ (p) = p0φ(p), which implies that

φ(p) = δ(p − p0). This is a plane wave in momentum space. That is, it is the eigenfunction ofp ˆ in the momentum representation. What is it in position space? If ψ(x) describes the same plane wave, then pψˆ (x) = p0ψ(x). ∂ But in position space,p ˆ = −i~ ∂x , so this equation is a simple PDE ∂ −i ψ(x) = p ψ(x). ~∂x 0 48 Wave Functions and Operators

We ﬁnd that ψ(x) = eip0x/~.

This is the eigenfunction ofp ˆ in the position representation. Consider again the state φ(p) = e−ix0p/~,

which is localized in position but not in momentum. Note that x0 is constant—only p is allowed to vary. Notice that |φ(p)|2 = const no matter what value p has. So the momentum distribution gives us a ﬂat line. In other words, this state is completely delocalized in momentum. Similarly, if we consider the state

ψ(x) = eip0x/~,

which is localized in momentum, we have that |ψ(x)|2 = const, so it is completely delocalized in position. In general, physical states are intermediate between these two extremes of position localization and momentum localization. To express energy in terms of operators, we just place hats on quantities. Kinetic energy K = p2/2m becomes pˆ2 Kˆ = . 2m Potential energy, we can write as Uˆ(~r). Then total energy is the Hamiltonian

pˆ2 Hˆ = Kˆ + Uˆ = + Uˆ(~r). 2m

For stationary states ψ(x), i.e. states with deﬁnite energy, we know that

Hˆ ψ(x) = E ψ(x).

This gives us the time-independent Schrodinger equation

2 − ~ ∇2 + U(~r) ψ(~r) = E ψ(~r). 2m

This is an eigenvalue equation since it has the form

Qˆ ψ = q ψ,

where Qˆ is an operator, q is an eigenvalue of the operator, and ψ is an eigenfunction of the operator. How do we get the time-dependent Schrodinger equation? Recall that

X −iEnt/ Ψ(~r, t) = cnψn(~r) e ~. n

Applying the Hamilton operator gives us

X −iEnt/ Hˆ Ψ(~r, t) = cn Hˆ ψn(~r) e ~. n

But we know that Hˆ ψn(~r) = En ψn(~r), so this simpliﬁes as

X −iEnt/ Hˆ Ψ(~r, t) = cn En ψn(~r) e ~. n 3.5. Green’s Functions 49

But we can also write this as ! ∂ X ∂ Hˆ Ψ(~r, t) = i c ψ (~r) e−iEnt/~ = i Ψ(~r, t). ~∂t n n ~∂t n So the time-dependent Schrodinger equation is

∂ Hˆ Ψ(~r, t) = i Ψ(~r, t). ~∂t

To summarize, in 1D our operators in the position and momentum representations are as shown in the table below:

Coord. Repr. Mom. Repr.

∂ xˆ x i~ ∂p

∂ pˆ −i~ ∂x p

2 2 2 ˆ ~ ∂ p ∂ H − 2m ∂x2 + U(x) 2m + U i~ ∂p

In 3D, these operators become

Coord. Repr. Mom. Repr.

~r ~r i~∇~p

~p −i~∇ ~p

2 ˆ ~ 2 ~p·~p H − 2m ∇ + U(~r) 2m + U (i~∇~p)

3.5 Green’s Functions

Given an arbitrary linear differential operator L, Green’s function G(x, y) is the function that satisfies the equation L G(x, y) = δ(x − y). Then the solution u(x) to a differential equation of the form

Lu(x) = f(x), can be written in the integral form

u(x) = G(x, y) f(y) dy. ˆ This formalism is useful to rewrite certain diﬀerential equations in integral form. Consider the time-independent Schrodinger equation in one dimension

2 d2ψ − ~ + U(x)ψ(x) = Eψ(x). 2m dx2 Suppose we know that E < 0, then we can rewrite this as

d2 2m|E| 2m − ψ(x) = U(x)ψ(x). dx2 ~2 ~2 50 Wave Functions and Operators

This is now in the form Lu(x) = f(x), with

u(x) = ψ(x) d2 2m|E| L = − dx2 ~2 2m f(x) = U(x)ψ(x). ~2 So in the integral form,

2m ψ(x) = G(x, y) U(y)ψ(y) dy, ~2 ˆ where G(x, y) is the solution to

d2 p2m|E| − κ2 G(x, y) = δ(x − y), κ = . dx2 ~ You can verify that the solution to this ODE is 1 G(x, y) = − e−κ|x−y|, 2κ by plugging this in and calculating the derivatives.1 This is the Green’s function, and plugging it back into the integral gives us

m ψ(x) = − e−κ|x−y| U(y)ψ(y) dy. κ~2 ˆ This is the integral form of the Schrodinger equation.

3.6 Transformations

We will now look at operators as simple geometric transformations. Consider a state ψ(x). Now consider a simple displacement by a distance a to create a new state ψ0(x) = ψ(x − a).

In operator notation, we would write this as

Dˆ(a) ψ(x) = ψ(x − a),

1Keep in mind how delta functions behave and can be simplified. For example, δ(x − y)h(x − y) = δ(x − y)h(0) for any function h. 3.6. Transformations 51 where Dˆ(a) is the displacement operator. Consider an infinitesimal displacement δa. Then Dˆ(δa) ψ(x) = ψ(x − δa). Since δa is small, we can expand in a Taylor series ∂ψ Dˆ(δa) ψ(x) = ψ(x) − δa + ··· ∂x a Since ∂ = i pˆ, we can write this as ∂x ~ i δa i Dˆ(δa) ψ(x) ' ψ(x) − pψˆ = 1 − δa pˆ ψ(x). ~ ~ In 3D, we would write δa p as δ~a · ~p. What if the displacement is not infinitesimal? Then we would have to include all the terms in the Taylor expansion

2 2 ∞ n n ∂ψ a ∂ ψ X (−a) ∂ ψ − i a·pˆ ψ(x − a) = ψ(x) − a + + ··· = = e ~ ψ(x). ∂x 2 ∂x2 n! ∂xn n=0 So in general, we can write the displacement operator in terms of the momentum operator as − i ~a·~p Dˆ(~a) = e ~ . In this case, the momentum operator ~p is called a generator of displacements. Suppose we have two consecutive shifts, then

− i ~b·~p − i ~a·~p − i (~a+~b)·~p Dˆ(~b)Dˆ(~a) = e ~ e ~ = e ~ = Dˆ(~b + ~a). In this case, order does not matter. The displacement operation is commutative Dˆ(~b)Dˆ(~a) = Dˆ(~a)Dˆ(~b). Notice that the eigenfunctions of Dˆ are plane waves

i ~p ·~r ψ(~r) = e ~ 0 , since i ~p ·(~r−~a) − i ~p ·~a Dˆ(~a) ψ(~r) = e ~ 0 = e ~ 0 ψ(~r). Now consider rotation in a plane by an angle α

Rˆz(α) ψ(φ) = ψ(φ − α), where Rˆz(α) is the rotation operator which rotates a state at angle φ to an angle φ + α. Here, ψ(φ) could be a function of other variables as well—any variables which do not change under this rotation. 52 Wave Functions and Operators

Following the same process, we find that ∂ Rˆ (δα) = 1 − δα , z ∂φ for infinitesimal rotations, and −α ∂ Rˆz(α) = e ∂φ , for finite rotations. The z-component of angular momentum is

[~r × ~p]z =x ˆpˆy − yˆpˆx. In polar coordinates, x = r cos φ, y = r sin φ, so we can write ∂ ∂x ∂ ∂y ∂ ∂ ∂ ∂ ∂ i i i = + = −r sin φ +r cos φ = x −y = xpy− ypx = [~r × ~p]z . ∂φ ∂φ ∂x ∂φ ∂y ∂x ∂y ∂y ∂x ~ ~ ~ So we can write angular momentum as 1 L~ = [~r × ~p] , ~ where the factor of 1/~ was included to make it dimensionless. Then ∂ i = Lˆz, ∂φ ~ or ∂ Lˆ = −i . z ~∂φ

The eigenfunctions of Lˆz must be purely exponential in φ.

Lˆzψ = mψ,

implies that the eigenfunctions of Lˆz are ψ(φ) = eimφ. Thus, we can write the rotation operator in terms of the angular momentum operator as ∂ i ˆ ˆ −α ∂φ − αLz Rz(α) = e = e ~ . imφ Since the eigenfunctions of Lz are e , and Rz(α) can be written as a sum of the powers imφ of Lz, we know that the eigenfunctions of Rz(α) are also e . One feature of rotation, is that rotating by 2π leaves the angle invariant. Thus imφ im(φ−2π) −i2πm imφ imφ Rˆz(2π)e = e = e e = e , and therefore, e−i2πm = 1. This implies that m = 0, ±1, ±2,... is an integer. Hence, orbital momentum is quantized purely as a geometrical result of rotation in space. Recall that translation is a commutative operation. What about rotation?

Rˆz(α)Rˆz(β) = Rˆz(α + β) = Rˆz(β + α) = Rˆz(β)Rˆz(α). So rotation is commutative provided that both rotations are about the same axis (z in this example). The time evolution operator is

− i tHˆ Uˆ(t) = e ~ . 3.7. Summary: Wave Functions and Operators 53

3.7 Summary: Wave Functions and Operators

Skills to Master • Given the initial state of a system, write it in the energy representation and add the time dependence • Be able to switch between the coordinate and momentum representations of a wave function • Know the basic operatorsx ˆ,p ˆ, and Hˆ in both the position and momentum representations • Know the eigenfunctions of the basic operatorsx ˆ,p ˆ, and Hˆ in both the position and momentum representations • Calculate the expectation value of operators in the position and momentum representations • Calculate the uncertainty of position and momentum • Be familiar with the displacement and rotation operators and the association of generators with transformations • Be familiar with the diﬀerent deﬁnitions of the Dirac delta function • Be able to simplify problems involving the Dirac delta function

Wave Function Representations Operators By the superposition principle, a general wave function The three basic operators in both the position and mo- can be written as a Fourier expansion mentum representations are tabulated below. X Ψ(x, t) = cnΨn(x, t). n Coord. Repr. Mom. Repr.

We can choose the Ψn to be stationary states with en- xˆ x i ∂ ergy En, then ~ ∂p

Ψ (x, t) = ψ (x)e−iEnt/~. ∂ n n pˆ −i~ ∂x p

2 2 2 The initial state is then ˆ ~ ∂ p ∂ H − 2 + U(x) + U i~ X 2m ∂x 2m ∂p Ψ(x, 0) = cnψn(x). n To find the eigenfunction of an operator in some Fourier analysis gives us the coefficients representation, you solve the eigenvalue problem in that representation. For example, to find the eigen- ˆ cn = dx Ψ(x, 0) ψn(x). functions of some operator O in the momentum repre- ˆ sentation, we would solve the eigenvalue problem Here, ψ(x) is the coordinate representation of our wave Oˆ φ(p) = O φ(p). function, and the {cn} form the energy representation. We can also write our wave function in terms of For the energy operator Hˆ , the eigenvalue problem plane waves and switch between the coordinate and to solve, i.e., Hˆ ψ(x) = E ψ(x), is the time-independent momentum representations by using Schrodinger equation (in 1D) dp ipx/~ ψ(x) = e φ(p) 2 ∂2 ˆ 2π~ − ~ + U(x) ψ(x) = E ψ(x). 2m ∂x2 φ(p) = dx e−ipx/~ψ(x). ˆ The time-dependent Schrodinger equation Alternatively, we can write them in terms of k instead ∂ of p as Hˆ Ψ(x, t) = i Ψ(x, t), ~∂t dk ψ(x) = eikxφ(k) ˆ 2π can be obtained by using the full time-dependent wave function in the eigenvalue problem instead of just the φ(p) = dx e−ikxψ(x). ˆ time-independent stationary state ψ(x). 54 Wave Functions and Operators

We can write the angular momentum as Transformations 1 L~ = [~r × ~p] . To denote the displacement of a wave function from a ~ position x to a position x − a, we write Expanding the cross product allows us to write the an- ˆ gular momentum in terms of position and momentum D(a) ψ(x) = ψ(x − a), operators. The z-component of the angular momentum is where ∂ ˆ − i ~a·~p Lˆ = −i . D(a) = e ~ . z ~∂φ Its eigenfunctions are is the displacement operator. Similarly, to denote rotation of the state about the ψ(φ) = eimφ. z-axis by an angle α, we write It is often useful to write Rˆz(α) ψ(φ) = ψ(φ − α),

Lˆi = εijk xˆjpˆk. where ∂ i ˆ ˆ −α ∂φ − αLz Expectation Values Rz(α) = e = e ~ . In the position representation, the probability of ﬁnd- is the rotation operator. The eigenfunctions of the ro- ing the particle between x and x + dx is tation operator are

P (x, x + dx) = w(x)dx = |ψ(x)|2dx. ψ(φ) = eimφ, In the momentum representation, the probability of just as for the Lˆ operator. ﬁnding the particle to have momentum in some inter- z The fact that rotation by 2π leaves an angle un- val is changed implies that m = 0, ±1, ±2,... I.e. orbital P (p, p + dp) = |φ(p)|2dp. angular momentum is quantized purely as a geometri- In the position representation, the expectation cal result of rotation invariance. value of an operator Oˆ is The time evolution operator is

ˆ ∗ ˆ − i tHˆ hOi = dx ψ (x) O ψ(x). Uˆ(t) = e ~ . ˆ Two important expectation values are the mean position and the mean-square position Dirac Delta Function The Dirac delta function is hxi = dx ψ∗(x)x ˆ ψ(x) ˆ ( ∞ if x = x0 hx2i = dx ψ∗(x)x ˆ2 ψ(x). δ(x − x0) = . ˆ 0 if x 6= x0.

For momentum, hpi and hp2i have analogous forms. More specifically, it is defined by the integral operator The uncertainty in x is calculated as b ( q 2 f(x0) if x0 ∈ (a, b) ∆x = hx2i − hxi . dx f(x)δ(x − x0) = . â 0 if x0 ∈/ (a, b). We can calculate expectation values in either the position or momentum representations. For example, Other definitions include in the position representation, 1 2 2 −(x−x0) /2σ δ(x − x0) = lim √ e ∗ ∂ σ→0 2πσ2 hpi = dx ψ (x) −i~ ψ(x), ˆ ∂x K dk ik(x−x0) δ(x − x0) = lim e and in the momentum representation K→∞ ˆ−K 2π 1 a dp ∗ δ(x − x0) = lim 2 2 . hpi = φ (p) p φ(p). a→0 π (x − x0) + a ˆ 2π~ 3.7. Summary: Wave Functions and Operators 55

Useful properties include Miscellaneous You should know the following integrals

δ(x) ∞ r δ(ax) = 2 π |a| e−ax dx = ˆ a 1 −∞ 2 2 ∞ r δ(x − a ) = (δ(x − a) + δ(x + a)) 2 π 2|a| e−a(x−x0) dx = ˆ a b 1 −∞ dx f(x)δ(x − a) = f(a) ∞ −ax2 ˆa 2 xe dx = 0 ˆ−∞ X f(xi) dx f(x) δ g(x) = . ∞ r ˆ |g0(x )| 2 −ax2 1 π i i x e dx = ˆ−∞ 2a a ∞ r −ax2+bx π b2/4a In the last one, the x are the zeros of g(x). e dx = e . i ˆ−∞ a Chapter 4

Formalism

4.1 Groups

A group is a set G of elements {g1, g2, . . . , gN } associated with a binary group operation called a “product”, such that the product of any two elements of the group is also an element of the group gi · gk ∈ G. The operation must be associative

gi · gk · gl = gi · (gk · gl) = (gi · gk) · gl.

There must be a “unit” element 1 ∈ G such that

1 · gn = gn · 1 = gn.

−1 Finally, for each element gn ∈ G, there must be an inverse element g ∈ G, such that

g · g−1 = g−1 · g = 1.

The simplest single element group is the number 1. This is trivial. The simplest 2- element group is {1,P } where 1 is the unit element, and P is inversion. The multiplication table for this group is unique and shown below.

For 3-element groups, the multiplication table is also unique, but already for 4-element groups, there are two possible multiplication tables. For example, translations Dˆ(~a) form a continuous group. The group “product” is two translations Dˆ(~a)Dˆ(~b) = Dˆ(~a +~b) = Dˆ(~b)Dˆ(~a). Notice that the product Dˆ(~a)Dˆ(~b) of two group elements is also an element of the group: Dˆ(~a + ~b). Also, notice that the group product is commutative. Such groups are called Abelian. In this group, the unit element is 1 = Dˆ(0) since

Dˆ(0)Dˆ(~a) = Dˆ(~a). 4.2. Hilbert Space 57

The inverse operation is h i−1 Dˆ(~a) = Dˆ(−~a), since Dˆ(−~a)Dˆ(~a) = Dˆ(0) = 1. Any plane wave transformed by Dˆ is mapped to itself with a diﬀerent phase. I.e. after any translation, the resulting object is still a plane wave. So we say the plane waves form a representation of the group of translations. In fact, we only need a single plane wave since any plane wave can be transformed into any other plane wave by translation. So we say the plane wave is an irreducible representation of the group of translations.

4.2 Hilbert Space

We are interested in the Hilbert space formed by the set of all quantum states associated with a given Hamiltonian and set of boundary conditions. A state in this Hilbert space is written in the form |Ai and can be thought of as a vector. Consider vectors in 3D. These form a Hilbert space. Given any vector ~v, we have a basis {~en}, which is a minimal set of independent vectors. It is useful but not necessary that these basis vectors are orthogonal. Then we can write the vector in terms of the basis as X ~v = ~envn. n

For an orthonormal basis {~en}, the dot product of the basis vectors is

~en · ~en0 = δnn0 .

Projection is given by X ~en0 · ~v = (~en0 · ~en) vn = vn0 . n So we can write X X ~v = ~envn = ~en (~en · ~v) , n n and deﬁne the unit operator as X 1 = ~en(~en · . n This is the condition of completeness. Now we apply this approach to our quantum states. We denote the inner (scalar) product of two states |Ai and |Bi as hA|Bi. We can expand any state as a linear combination of basis states |Bi = a1 |C1i + a2 |C2i + ··· .

In general, the coeﬃcients an are complex. The inner product is linear with respect to the right side. I.e.

hA|a1C1 + a2C2 + · · ·i = a1 hA|C1i + a2 hA|C2i + ···

If we reverse the order of the states, we get complex conjugation

hA|Bi = hB|Ai∗ .

The scalar square is always non-negative. I.e. for any state |Ai,

hA|Ai ≥ 0, 58 Formalism

Right now, the result ψB(~r) is in the B basis, but we have not speciﬁed what basis that is. Suppose it is the coordinate basis (|Bi → |~r 0i), then we might write it as X ψ (~r) = h~r|~r 0i h~r 0|Ai = d3~r 0δ(~r − ~r 0)ψ (~r 0). A ˆ A ~r 0 What does each part mean? The sum itself is a sum over position basis states X → d3~r 0. ˆ ~r 0 Next, we have the position eigenstates

h~r|~r 0i → δ(~r − ~r 0).

Finally, we have the state |Ai in the position basis

0 0 h~r |Ai → ψA(~r ). We could also use the momentum basis: |Bi → |~pi, then

X d3p h~r|Ai = h~r|~pi h~p|Ai = ei~p·~r/~φ (~p). ˆ (2π )3 A ~p ~ Now the sum is over the momentum basis states X d3p → , ˆ (2π )3 ~p ~ then we have the momentum eigenstates

h~r|~pi → ei~p·~r/~.

Finally, we have the state A~ in the momentum basis

If |Bi is the coordinate basis, then

4.3 Hermitian Operators

In quantum mechanics, Hermitian operators correspond to measurements. Assume Qˆ is a linear operator. Then given a state

|Ai = a1 |A1i + a2 |A2i , applying the operator to it gives us

Qˆ |Ai = a1Qˆ |A1i + a2Qˆ |A2i .

The matrix elements of Qˆ are

QBA = hB|Qˆ|Ai .

In matrix form, it looks like   Q11 Q12 ···   ˆ  ..  Q =  Q21 . .   . . QNN

For an operator Qˆ, we can deﬁne its Hermitian conjugate Qˆ†. The Hermitian conjugate is the transpose and complex conjugate. I.e.,

Qˆ† = QˆT ∗.

This is true in any basis. The elements are related as

∗ hA|Qˆ†|Bi = hB|Qˆ|Ai .

In matrix form, it would look like

 ∗ ∗  Q11 Q21 ···  .  Qˆ† =  ∗ .. .  Q12   .  . ∗ . QNN

An operator Qˆ is Hermitian if

Qˆ† = Qˆ ← Hermitian.

Consider the position operator

x = dx ψ∗(x)x ˆ ψ (x), 12 ˆ 1 2 60 Formalism

and its Hermitian conjugate

x † = dx ψ∗(x)x ˆ ψ (x). 12 ˆ 2 1 Sox ˆ is a Hermitian operator. What aboutp ˆ?

d dψ pˆ = dx ψ∗(x)p ˆ ψ (x) = dx ψ∗(x) −i ψ (x) = −i dx ψ∗ 2 . 12 ˆ 1 2 ˆ 1 ~dx 2 ~ ˆ 1 dx

On the other hand, dψ pˆ † = −i dx ψ∗ 1 . 12 ~ ˆ 2 dx Can we get the two to look the same? If we integrate the ﬁrst one by parts, we can write it as dψ upper dψ ∗ 2 ∗ ∗ 1 pˆ12 = −i~ dx ψ1 = −i~ ψ1 ψ2 − i~ dx ψ2 . ˆ dx lower ˆ dx

So we see thatp ˆ12 is not Hermitian unless the boundary term is zero upper ∗ −i~ ψ1 ψ2 → 0. lower In practical cases, the wave function does go to zero in the distance, and we can treatp ˆ as a Hermitian operator. Why are Hermitian operators special? Suppose the eigenvalues of a Hermitian operator Qˆ are q, and the associated eigenstates are |Ai. Then we can write

Qˆ |Ai = q |Ai .

There are two important theorems about Hermitian operators: 1. Eigenvalues q of Hermitian Qˆ are real 2. Eigenstates with diﬀerent eigenvalues (i.e. q0 6= q) are orthogonal To prove the ﬁrst one, we left multiply the equation above by hA|, then

hA|Qˆ|Ai = q hA|Ai ≥ 0.

Since Qˆ is Hermitian, we also have

hA|Qˆ†|Ai = q∗ hA|Ai .

This implies that q = q∗, which implies that q must be real. To prove the second one, we start with

Qˆ |ψi = q |ψi .

Multiplying from the left by a (diﬀerent) eigenstate,

hψ0|Qˆ|ψi = q hψ0|ψi .

Taking the Hermitian conjugate of both sides gives us

hψ|Qˆ†|ψ0i = q∗ hψ|ψ0i .

But since we already know that Qˆ is Hermitian and hence q is real, this simpliﬁes to

hψ|Qˆ|ψ0i = q hψ|ψ0i . 4.3. Hermitian Operators 61

On the other hand, if we start with

Qˆ |ψ0i = q0 |ψ0i , then hψ|Qˆ|ψ0i = q0 hψ|ψ0i . Comparing this with the previous result, we see that

(q − q0) hψ0|ψi = 0.

So either, q = q0 or hψ0|ψi = 0. I.e. if the states have diﬀerent eigenvalues (i.e. q 6= q0), then the states must be orthogonal (i.e. hψ0|ψi = 0). If diﬀerent states have the same q, they are said to be degenerate. Suppose we have a general state |ψi which is an eigenstate of Qˆ with eigenvalue q

Qˆ |ψi = q |ψi .

The set {q} is called the spectrum of the operator Qˆ. We can multiply from the left by another basis state |n0i

0 X 0 X hn |Qˆ|ni = q cn hn |ni = q cnδnn0 . n

We can write the matrix elements of Qˆ as

0 Qnn0 = hn |Qˆ|ni . Then X X 0 X cnQnn0 = q cn hn|n i = q cnδnn0 . n n This implies that X 0 cn (Qnn0 − qδnn0 ) = 0, n = 1,...,N. n

This is a system of linear equations. We are not interested in the trivial solution cn = 0. Then non-trivial solutions are given by

Det (Qnn0 − qδnn0 ) = 0.

It has N roots qi, all of which are real provided that Qˆ is Hermitian. After a solution is found, we can plug it back into the system of equations and repeat to get all of the solutions in terms of one. Finally, normalization can be applied. This process is called diagonalization. If we then write Qˆ in its eigenbasis, it is diagonal   q1 ˆ  .  Qψψ0 =  .. .   qN 62 Formalism

Tip Consider an operator Qˆ in its original basis When expanding a commu-   Q Q ... tator, always apply it to an 11 12   arbitrary function. ˆ  ..  Q =  Q21 . .   . . QNN

The Hermitian conjugate is obtained by transposing the matrix and taking the complex conjugate of all the elements

 ∗ ∗  Q11 Q21 ...  .  Qˆ† =  ∗ .. .  Q12   .  . ∗ . QNN

If Qˆ† = Qˆ, we say the operator is Hermitian. Notice that for a Hermitian operator, the diagonal elements must be real. An arbitrary complex N × N matrix has 2N 2 parameters—N 2 real parts and N 2 imaginary parts. A Hermitian N × N matrix has N(N + 1)/2 parameters. This is because the diagonal elements are real and the oﬀ- diagonal elements in the upper triangle are the complex conjugates of the elements in the lower triangle.

4.4 Commutators

Given operators Q,ˆ R,ˆ S,...ˆ , the commutator is

h i Q,ˆ Rˆ = QˆRˆ − RˆQ.ˆ

The anticommutator is the same but with a plus sign

h i Q,ˆ Rˆ = QˆRˆ + RˆQ.ˆ +

A useful relation between the two is h i h i Q,ˆ Rˆ + Q,ˆ Rˆ = 2QˆR.ˆ +

Algebraic properties of the commutator include

h i h i Q,ˆ Rˆ = − R,ˆ Qˆ h i h i h i Q,ˆ Rˆ + Sˆ = Q,ˆ Rˆ + Q,ˆ Sˆ h i h i h i Q,ˆ RˆSˆ = Q,ˆ Rˆ Sˆ + Rˆ Q,ˆ Sˆ h h ii h h ii h h ii 0 = Q,ˆ R,ˆ Sˆ + R,ˆ S,ˆ Qˆ + S,ˆ Q,ˆ Rˆ .

The last one is called the Jacobi identity. When expanding a commutator, always apply it to an arbitrary function in the same (e.g. coordinate) representation. This helps you remember to use the product rule as necessary. 4.4. Commutators 63

Example 4.4.1

Determine the commutator ofx ˆ andp ˆ.

d [ˆx, pˆ] ψ(x) = x,ˆ −i ψ(x) ~dx d d = x −i ψ − −i xψ ~dx ~dx = i~ψ(x). Notice that we used the product rule in going from the second to the third line. So the commutator is [ˆx, pˆ] = i~. We could also have expanded it in the momentum represetation.

d [ˆx, pˆ] φ(p) = i , pˆ φ(p) ~dp d d = i (pφ) − p i φ ~dp ~dp = i~φ(p). So we again get [ˆx, pˆ] = i~.

In general, the form of the commutator depends on the representation. However, in the example above, [ˆx, pˆ] = i~, in both the coordinate and momentum representations. More generally, for higher dimensions,

[ˆxi, pˆj] = i~δij.

Having the following commutators memorized is useful

2 [ˆx, pˆ ] = 2i~pˆ 2 [ˆp, xˆ ] = −2i~x.ˆ

More, generally,

dg [ˆx, g(ˆp)] = i ~dp df [ˆp, f(ˆx)] = −i . ~dx

Recall that ˆ 1 Lx = (~r × ~p)x , ~ and similarly for the y and z components. In general, we can write components of cross products in terms of the Levi-Civita tensor as (~a ×~b)i = εijkajbk. 64 Formalism

So we can write

Lˆi = εijk xˆjpˆk. Then h i 1 Lˆx, Lˆy = [ypz − zpy, zpx − xpz] ~2 1 = (y[pz, z]px + [z, pz]pyx) ~2 1 = (−i~ypx + i~pyx) ~2 i = (xpy − ypx) ~ = iLˆz.

Similarly for the other components. All together, h i Lˆx, Lˆy = iLˆz h i Lˆy, Lˆz = iLˆx h i Lˆz, Lˆx = iLˆy.

Using the completely antisymmetric rank-3 tensor εxyz, which has values

εxyz = εyzx = εzxy = +1

εyxz = εzyx = εxzy = −1,

and zero for other permutations of x, y, z, we can write the following

h i Lˆi, Lˆj = iεijkLˆk.

We can also derive h i Lˆi, xˆj = iεijkxˆk h i Lˆi, pˆj = iεijkpˆk.

Recall that 1 L~ = ~r × ~p, ~ is a generator of rotation. The magnitude of the angular momentum operator squared is

2 ~ ~ 2 2 2 L = L · L = Lx + Ly + Lz. Then hˆ ˆ2i hˆ ˆ2 ˆ2 ˆ2i Lx, L = Lx, Lx + Ly + Lz .

Consider only a single component of Lˆ2:

hˆ ˆ2i hˆ ˆ i ˆ ˆ hˆ ˆ i ˆ ˆ ˆ ˆ Lx, Ly = Lx, Ly Ly + Ly Lx, Ly = i LzLy + LyLz .

Now, doing this for all components of Lˆ2, we ﬁnd that

hˆ ˆ2i hˆ ˆ2 ˆ2 ˆ2i ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Lx, L = Lx, Lx + Ly + Lz = i LzLy + LyLz − i LzLy + LyLz = 0. 4.5. Uncertainty 65

So Lˆ and Lˆ2 commute. So x Tip h 2i Lˆi,L = 0. Two observables Aˆ and Bˆ We can also combine all components and write it as can be measured simultaneously if and only if the op- h i L~ ,L2 = 0. erators commute.

In general, h i Lˆi, f = 0, for any scalar function f. Speciﬁc examples include

h 2i h 2i h 2i Lî, ~r = Lî, pˆ = Lî, L~ = 0.

On the other hand, h i Lî, Vˆj = iεijk Vk, if V~ is a vector. Another useful fact is that h i Lî, Hˆ = 0, if the potential in the Hamiltonian Hˆ is a central potential. More generally, if you have generators gi (e.g. Lˆx, Lˆy, Lˆz) of some group, then the operator Cˆ that commutes with all generators is called the Casimir operator h i C,ˆ gî = 0.

In our case, Cˆ = Lˆ2. In a d-dimensional space, there are d(d−1)/2 generators of rotation. This is the same as the number of independent planes in the space. In 3D, there are 3(3 − 1)/2 = 3 generators {Lˆx, Lˆy, Lˆz}. It just so happens in 3D, that the number of generators is also the same as the number of dimensions. This is not generally true.

4.5 Uncertainty

Assume two observables, given by the operators Aˆ and Bˆ, can be measured simultaneously for the state |ψi. I.e. One measurement does not perturb ψ enough with respect to the other measurement that the ﬁrst measurement destroys the possibility of the second. Then

Aˆ |ψi = a |ψi Bˆ |ψi = b |ψi .

The commutator of the two operators is

[A,ˆ Bˆ]ψ = (AˆBˆ − BˆAˆ)ψ = Abˆ |ψi − Baˆ |ψi = ba |ψi − ab |ψi = 0.

So a state |ψi which is an eigenstate of two diﬀerent operators Aˆ and Bˆ is also an eigenstate of the operator [A,ˆ Bˆ] with eigenvalue 0. I.e. Aˆ and Bˆ can be simultaneously measured if [A,ˆ Bˆ] = 0. To have a complete set of common eigenstates of two operators Aˆ and Bˆ, it is necessary and suﬃcient that [A,ˆ Bˆ] = 0. 66 Formalism

Suppose Aˆ |ψi = a |ψi, such that the a are non-degenerate, and further suppose that [A,ˆ Bˆ] = 0. Then by commutativity, we know that

AˆBˆ |ψi = BˆAˆ |ψi .

I.e. not only are the |ψi eigenvectors of Aˆ with eigenvalues a, but also, the Bˆ |ψi are eigenvectors of Aˆ with the same eigenvalues. Since the eigenvalues a are non-degenerate, this can only mean the sets of vectors |ψi and Bˆ |ψi are linearly dependent. If Bˆ |ψi = b |ψi, then the |ψi are also eigenvectors of Bˆ. Now, instead of the eigenvalues a being non-degenerate, suppose they are d-fold degenerate. Then in the d-dimensional space, we can always diagonalize Bˆ. The diagonalization will give us the speciﬁc linear combinations which are eigenvectors of Bˆ whereas any linear combination in this subspace is automatically an eigenvector of Aˆ. The most interesting case occurs when two Hermitian operators Aˆ and Bˆ do not commute. I.e. suppose

Aˆ† = A,ˆ Bˆ† = B,ˆ [A,ˆ Bˆ] 6= 0.

Then the Hermitian conjugate of the commutator is

† † [A,ˆ Bˆ] = AˆBˆ − BˆAˆ = Bˆ†Aˆ† − Aˆ†Bˆ† = BˆAˆ − AˆBˆ = [B,ˆ Aˆ] = −[A,ˆ Bˆ].

Notice the negative sign at the end. In general, we can write an anti-Hermitian operator as the imaginary unit times a Hermitian operator. In our case, we can write

[A,ˆ Bˆ] = iC,ˆ

where α is an arbitrary real number. Then

0 ≤ hψ0|ψ0i D E ˆ0 ˆ0 ˆ0 ˆ0 = A + iαB ψ A + iαB ψ D E ˆ0 ˆ0 ˆ0 ˆ0 = ψ A − iαB A + iαB ψ D E 0 2 ˆ0 ˆ0 ˆ0 ˆ0 2 ˆ0 2 = ψ A + iα A B − B A + α B ψ D E 0 2 ˆ 2 ˆ0 2 = ψ A + iα iC + α B ψ D E 0 2 ˆ 2 ˆ0 2 = ψ A − αC + α B ψ 2 2 2 = (∆A)ψ − α hCiψ + α (∆B)ψ . In the last line, we used, for example,

ˆ0 2 2 2 hψ|A |ψi = hψ|(A − a) |ψi = (∆A)ψ. 4.5. Uncertainty 67

This is uncertainty. So we have the quadratic equation in α

2 2 2 (∆A)ψ − α hCiψ + α (∆B)ψ ≥ 0. Since it is positive, the only roots of this quadratic are complex. The condition for such a quadratic to have complex roots for α is

4(∆A)2(∆B)2 ≥ hCi2 .

This implies 1 1 |∆A||∆B| ≥ | hCi | = h[A,ˆ Bˆ]i . 2 2 This is often just written as

1 D E ∆A ∆B ≥ [A,ˆ Bˆ] . 2

For example, for position and momentum, we have 1 1 ∆x∆p ≥ |[ˆx, pˆ| = |i | = ~. 2 2 ~ 2

2 2 Recall the state which gives the minimum uncertainty is the Gaussian ψ ∼ e−(x−x0) /4σ . For the components of angular momentum, we have 1 ∆L ∆L ≥ hL i . x y 2 z 68 Formalism

4.6 Summary: Formalism

Skills to Master • Determine if an operator is Hermitian • Calculate the commutator or anticommutator of a pair of operators • Determine if a pair of operators have simultaneous eigenvalues • Determine the uncertainty relation for a pair of operators

Hermitian Operators Algebraic properties of the commutator include In quantum mechanics, observables correspond to Her- mitian operators. h i h i Q,ˆ Rˆ = − R,ˆ Qˆ Hermitian conjugation, in terms of linear algebra h i h i h i is the combined operation of transposition and com- Q,ˆ Rˆ + Sˆ = Q,ˆ Rˆ + Q,ˆ Sˆ plex conjugation. For an operator Qˆ, the Hermitian h i h i h i conjugate is Q,ˆ RˆSˆ = Q,ˆ Rˆ Sˆ + Rˆ Q,ˆ Sˆ T Qˆ† = Qˆ∗ . h h ii h h ii h h ii 0 = Q,ˆ R,ˆ Sˆ + R,ˆ S,ˆ Qˆ + S,ˆ Q,ˆ Rˆ . A discrete operator Qˆ can be written in matrix form. Then if the eigenvectors of Qˆ are |Ai, The last one is called the Jacobi identity.   Q11 Q12 ··· When expanding a commutator, always apply it   to an arbitrary function in the same (e.g. coordinate) ˆ ˆ  ..  QBA = hB|Q|Ai , Q =  Q21 . . representation. This helps you remember to use the  .  product rule as necessary. . Q NN The following commutators are useful to have Similarly, the matrix elements and matrix repre- memorized: sentation of its Hermitian conjugate are

 ∗ ∗  [ˆxi, pˆj] = i δij Q11 Q21 ··· ~ ∗  .  dg hA|Qˆ†|Bi = hB|Qˆ|Ai , Qˆ† =  Q∗ .. . [ˆx, g(ˆp)] = i  12  ~dp  .  . Q∗ df NN [ˆp, f(ˆx)] = −i ~dx An operator is Hermitian if it equals its Hermitian h i Lˆ , xˆ = iε xˆ conjugate i j ijk k Qˆ† = Qˆ ← Hermitian. h i Lî, pˆj = iεijk pˆk Two important facts about Hermitian operators hˆ ˆ i ˆ are: Li, Lj = iεijk Lk 1. Eigenvalues q of Hermitian Qˆ are real hˆ 2i 2. Eigenstates with different eigenvalues (i.e. q0 6= Li, ~r = 0 q) are orthogonal h 2i Lî, pˆ = 0

h 2i Commutators Lˆi, L~ = 0 Given operators Qˆ and Rˆ, their commutator is h i Q,ˆ Rˆ = QˆRˆ − RˆQ.ˆ For a central potential, Their anticommutator is the same but with a plus sign

h ˆ ˆi ˆ ˆ ˆ ˆ h i Q, R = QR + RQ. Lˆi, Hˆ = 0. + 4.6. Summary: Formalism 69

Uncertainty an uncertainty relation

Two observables Aˆ and Bˆ can be measured simultane- 1 D E ∆A ∆B ≥ [A,ˆ Bˆ] . ously if and only if the operators commute 2

h i For example, A,ˆ Bˆ = 0. 1 ∆x ∆p ≥ h|i |i = ~ 2 ~ 2 If they do not commute, then the two quantities are 1 ∆L ∆L ≥ hL i . not simultaneously observable, and we can write down x y 2 z Chapter 5

Quantum Dynamics

5.1 Unitary Transformations

A unitary operator Uˆ is one whose Hermitian conjugate equals its inverse

Uˆ † = Uˆ −1 =⇒ Uˆ †Uˆ = UˆUˆ †.

Recall the displacement, rotation, and time evolution operators

− i ~a·~p i Dˆ(~a) = e ~ ' 1 − δ~a · ~p ~ −iαLˆz Rˆz(α) = e − i tHˆ Uˆ(t) = e ~ .

The generators ~p, Lˆz, and Hˆ are Hermitian, but Dˆ, Rˆz, and Uˆ are unitary. Given a scalar product hψ|ψ0i, performing a unitary transformation on both gives us

hUψˆ |Uψˆ 0i = hψ|Uˆ †Uψˆ 0i = hψ|ψ0i .

So a unitary operator preserves scalar products. If we have an operator Qˆ and a state |ψi, then since 1 = Uˆ −1Uˆ, we can write Uˆ Qˆ |ψi = UˆQˆUˆ −1Uˆ |ψi = UˆQˆUˆ −1 Uˆ |ψi .

If Uˆ is a unitary operator, then Uˆ −1Uˆ = Uˆ †Uˆ = 1 and the transformation

|ψi → Uˆ |ψi Qˆ → UˆQÛˆ †, is a unitary transformation. The so-called S-matrix, Sˆfi, is an operator that is used often in scattering problems. 2 It takes an initial state i at −∞ to a final state f at +∞. Then |Sfi| is the probability that i → f. If we take the initial state and sum over all possible final states we should get 1, so X 2 X ∗ X † † 1 = |Sfi| = SfiSfi = Sif Sfi = S S ii . f f f

This implies that Sˆfi must be a unitary operator. Let’s consider the eigenstates of unitary operators. The eigenvalue equation has the form Uˆ |ψi = u |ψi . 5.1. Unitary Transformations 71

For example, for the displacement operator, the eigenstates are plane waves

i ~p ·~r − i ~a·~p i ~p ·~r Dˆ(~a)e ~ 0 = e ~ 0 e ~ 0 .

− i ~a·~p In this case, the eigenvalues are u = e ~ 0 . For rotation, we have

imφ im(φ−α) Rˆz(α)e = e , and the eigenvalues are u = e−imα. For the time evolution operator,

ˆ − i Et U(t)ψE(0) = e ~ ψE(0),

− i Et which has eigenvalue u = e ~ . Notice that for all three of these unitary operators, the eigenvalues are just phases. This is a general property. Note that

hUψˆ |Uψˆ i = u∗u hψ|ψi .

On the other hand hUψˆ |Uψˆ i = hψ|Uˆ †Uˆ|ψi = hψ|ψi . Therefore, u∗u = 1. That is, all eigenvalues of unitary operators have an absolute value of 1, i.e. |u|2 = 1. So in general, we can write the eigenvalues u of a unitary operator in the form u = eig, where g ∈ R. We now deﬁne a Hermitian operator Gˆ which has eigenvalues g

Gˆ |ψi = g |ψi .

Then any unitary operator can be written in the form

Uˆ = eiGˆ , where Gˆ is called the generator. Example 5.1.1

Consider the transformation xˆ → xˆ(a) = Dˆ(a)x ˆ Dˆ −1(a), where − i apˆ Dˆ(a) = e ~ , is the translation operator. What does the transformed operatorx ˆ(a) look like? We have − i apˆ i apˆ xˆ(a) = e ~ xˆ e ~ . The standard way of working with something like this is in terms of derivatives with respect to a parameter. If there is no parameter, insert one, and then set it equal to 1 in the end. In this case, we can work with the parameter a. Diﬀerentiating both sides with respect to a gives us dxˆ − i apˆ i i i apˆ = e ~ − pˆxˆ +x ˆ pˆ e ~ da ~ ~ i − i apˆ i apˆ = e ~ [ˆx, pˆ] e ~ ~ i − i apˆ i apˆ = e ~ (i~) e ~ ~ = −1. 72 Quantum Dynamics

Solving this diﬀerential equation gives us

xˆ(a) = −a +x ˆ(0) =x ˆ − a,

illustrating thatx ˆ(a) is a translation ofx ˆ by a.

5.2 Operator Dynamics and the Heisenberg Picture

In the Schrodinger picture, the wave functions evolve in time according to

− i tHˆ Ψ(t) = Uˆ(t)ψ = e ~ ψ.

That is, time dependence is added by modifying the wave function—the operators stay ﬁxed. In the Heisenberg picture, the wave function stays ﬁxed, and it’s the operators that evolve in time. Consider the quantity hΨ(t)|Qˆ|Ψ(t)i . This is in the Schrodinger picture since it shows the wave functions Ψ as being time dependent whereas the operator Qˆ is not. We can show the Schrodinger time-dependence by replacing the wave function Ψ by the initial wave function and the time-evolution operator Uˆ(t) hΨ(t)|Qˆ|Ψ(t)i = hUˆ(t)Ψ(0)|Qˆ|Uˆ(t)Ψ(0)i . We can rearrange the right side as D E ˆ ˆ † ˆ ˆ hΨ(t)|Q|Ψ(t)i = Ψ(0) U (t)QU(t) Ψ(0) .

But now we have a time-independent wave function Ψ(0) and a time-dependent operator Uˆ †(t)QÛˆ(t) D E D E ˆ † ˆ ˆ ˆ Ψ(0) U (t)QU(t) Ψ(0) = Ψ QH (t) Ψ , where † QˆH (t) = Uˆ (t) Qˆ Uˆ(t), is the Heisenberg version of Qˆ. In general, a Heisenberg operator QˆH (t) is obtained from a Schrodinger operator Qˆ as ˆ ˆ † ˆ ˆ i tHˆ ˆ − i tHˆ QH (t) = U (t) Q U(t) = e ~ Q e ~ . The standard way of simplifying a transformation like this is to differentiate with respect to the parameter in the exponentials. In this case, our parameter is t, so we differentiate both sides with respect to t ˆ ˆ dQH i i tHˆ − i tHˆ i tHˆ ∂Q − i tHˆ i i tHˆ − i tHˆ = Heˆ ~ Qeˆ ~ + e ~ e ~ − e ~ QˆHeˆ ~ dt ~ ∂t ~ ˆ i tHˆ ∂Q − i tHˆ i i tHˆ − i tHˆ i tHˆ − i tHˆ = e ~ e ~ + e ~ Hˆ Qeˆ ~ − e ~ QˆHeˆ ~ ∂t ~ ! ∂Qˆ i i = + Hˆ Qˆ − QˆHˆ ∂t H H H ~ ~ ! ∂Qˆ i = + Hˆ Qˆ − QˆHˆ . ∂t H H ~ 5.2. Operator Dynamics and the Heisenberg Picture 73

This gives us the Heisenberg equation, which gives the time evolution of a Heisenberg operator ! dQˆ ∂Qˆ 1 h i H = + Q,ˆ Hˆ . dt ∂t i H H ~

To get the Heisenberg equation of motion for an operator Qˆ, you would ∂Qˆ 1. Calculate the partial time derivative ∂t . This is typically zero 2. Calculate the commutator [Q,ˆ Hˆ ]. Note, Qˆ and Hˆ are just the standard Schrodinger versions of those operators 3. Put them into the equations above. If there remain any operators Oˆ on the right side, write them as Oˆ(t) or as OˆH to indicate that these are time-dependent Heisen- berg operators 4. Solve the remaining diﬀerential equation The Heisenberg equation in full expression would be

ˆ ˆ ! dQH i tHˆ ∂Q − i tHˆ 1 i tHˆ h i − i tHˆ = e ~ e ~ + e ~ Q,ˆ Hˆ e ~ . dt ∂t i~

Example 5.2.1

Consider a particle in one dimension where the Hamiltonian is

pˆ2 Hˆ = + U(ˆx). 2m The operatorsp ˆ andx ˆ don’t have time-dependence, so the ﬁrst term in the Heisenberg equation is zero, and for the Heisenberg equations for these operators, we get

dxˆH 1 h i 1 = x,ˆ Hˆ = pˆH dt i~ H m dpˆH 1 h ˆ i = p,ˆ H = − (∇U)H = FH . dt i~ H So the equations look the same as they do in classical mechanics, e.g. the time derivative of the momentum is the force.

In general,

dx d 1 H = hx i = hp i dt dt H m H dp d H = hp i = hF i . dt dt H H

In 3D, ~p 2 Hˆ = + U(~r), 2m and then d 1 hψ|~r |ψi = h~r˙ i = h~p i dt H H m H d hψ|~p |ψi = h~p˙ i = hF~ i . dt H H H 74 Quantum Dynamics

More importantly,

˙ ~pH ~rH = m (5.1) ˙ ~ ~pH = F H = −∇U(~rH ).

That is, the equations have the exact same form as the equations of motion of classical mechanics. Keep in mind that, in general,

hF i= 6 F (hxi) .

That is, the average force is not generally the force at the wave function’s average position (i.e. its center). In 3D, hF~i= 6 F~ (h~ri) . It is only for the three cases: free motion, uniform ﬁeld, and harmonic oscillator, that these equations are the same as the classical equations for the center of the wave packet. For other cases, there will be corrections. If we expand the force F~ near the center h~ri in a Taylor expansion, then ! D E 1 D E ∂2F~ hF~(~r)i = F~ (h~ri) + xi − hxii xj − hxji + ··· . 2 ∂xi∂xj h~ri

The ﬁrst derivative term in the Taylor expansion is zero since h~r − h~rii = 0. We can approximate the average force as

hF~i hF~(~r)i ∼ , R2 where R is the typical size of U(~r). The quantum correction is of the order

λ2 ∼ . R2 The semi-classical region is where λ2 << 1. R2

Example 5.2.2

Calculate the Heisenberg operatorsx ˆ(t) andp ˆ(t) for a particle of mass m and charge e moving along a uniform electrostatic ﬁeld E which lies in the x-direction. The general formula for the Heisenberg operator is ! dQˆ ∂Qˆ 1 h i H = + Q,ˆ Hˆ . dt ∂t i H H ~

We have our Schrodinger operatorsx ˆ andp ˆ which take the place of Qˆ, and we want to calculatex ˆ(t) andp ˆ(t) which take the place of QˆH . The Schrodinger operatorsx ˆ andp ˆ have no explicit time dependence, so in both cases, the ﬁrst term on the right is zero. So all we have to calculate are the commutators. We also need to know the Hamiltonian. For a uniform electric ﬁeld in the x 5.2. Operator Dynamics and the Heisenberg Picture 75

direction, we know that U(x) = −eEx, so the Hamiltonian is

pˆ2 Hˆ = − eEx.ˆ 2m Forp ˆ(t), we get

dpˆ(t) ∂pˆ 1 h i = + p,ˆ Hˆ dt ∂t H i~ H 1 pˆ2 = p,ˆ − eExˆ i~ 2m H 1 = (i~eE)H i~ = eE.

Integrating gives us pˆ(t) =p ˆ(0) + eEt. Note thatp ˆ(0) ≡ pˆ is the original time-independent Schrodinger operator. Forx ˆ(t), we get

dxˆ(t) ∂xˆ 1 h i = + x,ˆ Hˆ dt ∂t H i~ H 1 pˆ2 = x,ˆ − eExˆ i~ 2m H 1 pˆ = i~ i~ m H 1 = pˆ(t). m Plugging inp ˆ(t) from above and integrating gives us

pˆ(0) eE xˆ(t) =x ˆ(0) + t + t2. m 2m A faster approach is to apply our knowledge of Eq. (5.1). The force on our particle with charge e due to the uniform electric ﬁeld E is just FH = eE. Therefore, the particle is accelerated as in Newtonian physics, and

pˆ(t) =p ˆ(0) + eEt pˆ(0) eE xˆ(t) =x ˆ(0) + t + t2. m 2m

When simplifying Heisenberg equations, it is very useful to have the following commutators memorized

2 [ˆx, pˆ ] = 2i~pˆ 2 [ˆp, xˆ ] = −2i~x.ˆ

Example 5.2.3

Calculate the Heisenberg operators for the harmonic oscillator, which has 76 Quantum Dynamics

Hamiltonian pˆ2 1 Hˆ = + mω2xˆ2. 2m 2 We have two operators—ˆx andp ˆ in the Hamiltonian, so these are the Heisen- berg operators we want to ﬁnd. Applying the Heisenberg equation for each of them gives us

1 h i 1 1 h i pˆ(t) x˙ = x,ˆ Hˆ = x,ˆ pˆ2 = i~ H i~ 2m H m 1 h i 1 1 h i p˙ = p,ˆ Hˆ = mω2 p,ˆ xˆ2 = −mω2 xˆ(t). i~ H i~ 2 H Now we have a pair of coupled ODEs. To solve this system, we can diﬀerentiate x˙ and plug inp ˙ to get p˙ x¨ = = −ω2 xˆ(t). m Doing the same thing for the other equation gives us

p¨ = −ω2 pˆ(t).

The general solution to this pair of second-order ODEs is

xˆ(t) = Aˆ cos(ωt) + Bˆ sin(ωt) pˆ(t) = Cˆ cos(ωt) + Dˆ sin(ωt).

To find the specific solution, we look at initial conditions. Clearly,x ˆ(0) = Aˆ. We definex ˆ(0) ≡ xˆ. Similarly,p ˆ(0) =p ˆ = Cˆ. Differentiating our general solution forx ˆ(t), we find that xˆ˙(0) = Bωˆ . But from our system of ODEs, we also know that xˆ˙(0) =p ˆ(0)/m. Thus, Bˆ =p/mω ˆ . Similarly, we find that Dˆ = −mωxˆ. So our final solution is pˆ xˆ(t) =x ˆ cos(ωt) + sin(ωt) mω pˆ(t) =p ˆcos(ωt) − mωxˆ sin(ωt).

Now we look at the stationary states of a discrete spectrum in the Heisenberg picture. Assuming no explicit dependence on t in QˆH , then the expectation value of the Heisenberg equation is * + dQˆ Dh iE D E i H = Qˆ , Hˆ = ψ(0) Qˆ Hˆ − Hˆ Qˆ ψ(0) . ~ dt H H H

In the last term, we can operate the ﬁrst Hˆ toward the left and the second Hˆ toward the right, and we get

D E D E ˆ ˆ ˆ ˆ ˆ ˆ ψ(0) QH H − HQH ψ(0) = ψ(0) QH E − EQH ψ(0) = 0.

Thus, * + dQˆ H = 0, dt

in any stationary state (i.e. has deﬁnite energy) of a discrete spectrum. 5.3. Conservation Laws 77

Recall the virial theorem d h~r · ~pi = 0 dt ~p · ~p + h~r · (−∇U)i = 0 m hKi = 2 h~r · ∇Ui , for any stationary state of the discrete spectrum. The classical Hamiltonian equations of motion are ∂H ∂H q˙i = , p˙i = − . ∂pi ∂qi For a function f(q, p, t), the total time derivative is

df ∂f X ∂f ∂H ∂f ∂H = + − . dt ∂t ∂q ∂p ∂p ∂q i i i i i For any pair A and B, the Poisson bracket is deﬁned as

X ∂A ∂B ∂A ∂B {A, B} = − . ∂q ∂p ∂p ∂q i i i i i Thus, df ∂f = + {f, H} . dt ∂t This is analogous to the Heisenberg equation of quantum mechanics, which has a commutator instead of a Poisson bracket. Note that the Poisson bracket has the same algebraic properties as the commutator. For example,

{A, B} = − {B,A} {A, BC} = {A, B} C + B {A, C} 0 = {A, {B,C}} + {B, {C,A}} + {C, {A, B}} .

The correspondence between the commutator and the Poisson bracket is h i A,ˆ Bˆ → i~ {A, B} .

5.3 Conservation Laws

Recall the Heisenberg equation for an operator Qˆ

dQˆH ∂QˆH h i i~ = i~ + H,ˆ Qˆ . dt ∂t H

∂QˆH Assuming QˆH has no explicit time dependence, then = 0. If we also assume that h i ∂t H,ˆ Qˆ = 0, then H dQˆ H = 0 =⇒ Qˆ is conserved. dt H So if an operator does not explicitly depend on time and it commutes with Hˆ , then the associated observable is conserved. Consider for stationary states |ni and |mi

hn|[Q,ˆ Hˆ ]|ni = (En − Em) Qmn. 78 Quantum Dynamics

If Qˆ is conserved, then this quantity is zero

hn|[Q,ˆ Hˆ ]|ni = (En − Em) Qmn = 0.

So, assuming Q is conserved, the operator matrix element Qmn is nonzero only inside the degenerate space where En = Em. Conversely, if Qˆ is conserved, then ∂Qˆ h i = 0, and Q,ˆ Hˆ = 0, ∂t

and we get a multiplet with E = const since (En − Em)Qmn = 0. The time-dependent transition amplitude, whether Qˆ is conserved or not is

i i ˆ X ∗ Emt − Ent X ∗ iωmnt hψ2(t)|Q|ψ1(t)i = cme ~ cne ~ hm|Q|ni = cmcne Qmn. m,n m,n

If Qˆ is conserved, the only matrix elements Qmn that survive are those with the same energy, and there is no time dependence—whether or not they are stationary states and

X ∗ cmcnQmn = const in time. m,n When the symmetry of a state is lower than the symmetry of the Hamiltonian, we have what is called spontaneous symmetry breaking. The Goldstone-Nambu theorem tells us that if you have spontaneous breaking of continuum symmetry, then you get an excited state. A relatively unknown constant of motion is the Laplace-Runge-Lenz vector ~r M~ = ~ ~p × L~ − L~ × ~p − α . 2m r This is for any potential U = α/r. Keep in mind that all the vectors here are operators. It can be shown that h i M~ , Hˆ = 0. This constant of motion is why bound orbits are closed, i.e. why the orbiting body returns to the exact same point once every orbit. Bertrand’s theorem tells us that there are only two potentials which give closed bound state orbits—the Coulomb/Kepler potential, and the 3D harmonic oscillator potential.

5.4 Continuity Equation

Recall the continuity equation ∂ρ = −div ~j. ∂t In our problems, ρ = |ψ|2. What is the current? From the Schrodinger equation, ∂ψ ∂ψ∗ i = Hψ,ˆ −i = Hˆ ∗ψ∗. ~ ∂t ~ ∂t Then ∂ρ ∂ψ ∂ψ∗ = ψ∗ + ψ ∂t ∂t ∂t 1 1 = ψ∗ Hψˆ + ψ −Hˆ ∗ψ∗ i~ i~ 1 = ψ∗Hψˆ − ψHˆ ∗ψ∗ . i~ 5.5. Wave Packets 79

We cannot relate this to current ~j in the general case, however, we can do it in speciﬁc cases. Consider if Hˆ is real, then Hˆ = Hˆ ∗ and we get

∂ρ 1 − 2∇2 − 2∇2 = ψ∗ ~ ψ − ψ ~ ψ∗ = ~ ψ∗∇2ψ − ψ∇2ψ∗ . ∂t i~ 2m 2m 2mi The right-hand side is indeed the divergence of a vector, ∂ρ = ∇~j, ∂t where ~j = ~ (ψ∗∇ψ − ψ∇ψ∗) . 2mi √ √ Suppose ψ = ρeiφ, then ∇ψ = i ρ∇φ · eiφ. In this case, then

√ √ √ √ ~j = ~ ρe−iφi ρ∇φ · eiφ − ρeiφ −i ρ∇φ · e−iφ = ~ ρ∇φ. 2mi m

√ i ~p·~r For example, if ψ = ρe ~ , then ∇φ is momentum, and the current is

~p ~j = ρ = ρ~v. m

5.5 Wave Packets

The plane wave eigenfunction has the form

i(kx−ωt) ψk(x, t) = Ae .

However, for the free particle, this is not normalizable. That is, a free particle cannot be described as a plane wave with deﬁnite momentum. What we have to do is think of the free particle as a whole set of plane waves (each with diﬀerent wave numbers k) chosen such that there is constructive interference in a small region and destructive interference everywhere else. That is, we want the superposition to give us a “bump” in one small region and be zero everywhere else. This clump of plane waves is called a wave packet, and its wave function (in 1D) has the form ∞ dk Ψ(x, t) = A(k) ei(kx−ω(k)t), ˆ−∞ 2π where p E(k) k = , ω(k) = . ~ ~ The A(k) gives the amplitude of the wave packet. You can think of A(k) as ∆k concen- trated around some k0. In classical physics, a free particle is localized in space. In quantum mechanics, particles are described by wave functions which are localized in momentum but are not localized in space. The wave packet gives us a wave function that is partially localized in both space and in momentum, so it gives us a compromise. When describing particles as wave packets, we can speak simultaneously of the particle’s position and its momentum, albeit with a little uncertainty in both. The initial state of the wave packet is

∞ dk ikx Ψ0 = Ψ(x, 0) = A(k) e . ˆ−∞ 2π 80 Quantum Dynamics

To ﬁnd A(k), we perform a Fourier transform ∞ −ikx A(k) = dx Ψ0e . ˆ−∞ The group velocity of the wave packet

dω(k) v = , g dk is the velocity of the clump of plane waves as this clump propagates along. The phase velocity ω(k) v = , ph k is the velocity of the single plane wave with wave number k that forms part of the wave packet. For a free particle, p2 2k2 k2 E = = ~ =⇒ ω(k) = ~ . 2m 2m 2m For narrow ∆k, (if ∆k is not narrow, then it doesn’t really make sense to speak of a wave packet) we can expand ω(k) about k0 as dω ω(k) ≈ ω(k ) + (k − k ) , 0 0 dk k0 and we can write eikx = eik0x+i(k−k0)x, then " !# dω Ψ(x, t) ≈ eik0x−iω(k0)t dk A(k) exp i(k − k ) x − t ˆ 0 dk k0 dω = eik0x−iω(k0)tf x − t . dk k0 This shows how the wave packet has an envelope.

The group velocity of the signal is given by dω dE v = = . g dk dp Note that ∆p ∆k ∼ , ~ and recall that ∆x∆p ≥ ~/2. If there is uncertainty ∆p, then there will be a distribution ∆v of wave velocities composing the wave packet ∆p ∆v ∼ . m 5.6. Decay of States 81

After some time t, ∆p ∆v · t ∼ (∆x) ∼ t ∼ ~ t. new m m∆x That is, the size changes with time since the wave packet is composed of waves with diﬀerent speeds. The spreading of the wave packet is given by t (∆x)2 ∼ ~ . spread m The total width is the width at t = 0 plus the time-dependent spreading 2 2 2 (∆x) = (∆x)t=0 + (∆x)spread .

Stationary Phase Approximation 5.6 Decay of States

Real stationary excited states don’t exist in nature. All of them decay at some point. Only ground states of isolated systems are truly stationary in the sense that they have inﬁnite lifetime. The probability amplitude is

i γ − E0t− t Ψ(t) ∼ e ~ 2 , then dN = −γN, dt where γ is the decay rate, and the number of decays per unit time is N(t) = N(0)e−γt. This process is typically radioactive decay. For atomic radiation, we can estimate via classical electrodynamics that e2ω2 e2 ω r γ ∼ ∼ ω ∼ ω 0 , mc3 mc2 c λ −15 −7 7 where r0 ∼ 10 m, and r0/λ ∼ 10 for visible light. So an electron orbits ∼ 10 times before radiating and dropping to the next lower energy state. Mean lifetime is given by 1 τ = . γ Half-life is given by ln 2 t = . 1/2 γ Since ω = E/~, we can write the wave function as ∞ −iωt −iω t− γ t dω e Ψ(t) ∼ e 0 2 = i , ˆ 2π iγ −∞ ω − ω0 + 2 where t > 0. This can be calculated using the method of residues by integrating over a semi-circle of radius R in the complex plane and then letting R → ∞. Note that e−iωt = e−i(<(ω)+i=(ω))t, where =(ω) < 0 and t > 0. Then we have a pole at E iγ ω = 0 − . ~ 2 This pole is in the lower part of the complex plane, so we put our semi-circle contour in the lower half. 82 Quantum Dynamics

Then we apply the residue method, to get

∞ dω e−iωt i E iγ i = 2πi × (−1)exp − 0 − t . E0 iγ ˆ−∞ 2π ω − + 2π ~ 2 ~ 2 Now we can interpret this as 1 Prob(ω) ∝ 2 . ω − E0 + iγ ~ 2 Then the normalized probability distribution is

Γ 1 w(E) = 2 2 , 2π (E − E0) + Γ /4

where Γ = ~γ, and E = ~ω. The function w(E) is a Lorentzian function and often called the Breit-Wigner formula. This function is normalized so that

∞ dE w(E) = 1. ˆ−∞

Since an excited state is unstable, its energy cannot have a deﬁnite value. Rather, its energy follows the Breit-Wigner distribution w(E). Note that Γ = γ = ~, ~ τ is the full width at half maximum (i.e. FWHM) of w(E) and is often called the level width. Parts of the distribution w(E) are unphysical since we cannot actually have energies going down to −∞. We can only have energies going down to some ground state. However, w(E) is good near the center. Notice that lim w(E) = δ(E − E0). Γ→0 5.7. Survival Probability 83

The Doppler width is defined as v Γ ∼ ω. D c ~ This is larger than the natural width Γ. This results in an experimental problem called Doppler broadening. Another problem is collisional broadening. The Mossbauer effect occurs when the atom under study is in a solid such as a crystal lattice. This effect occurs because of recoil-less radiation absorption.

5.7 Survival Probability

Given an initial state |ψ(0)i, we observe what happens at a later time |ψ(t)i. What is the probability for the system to not decay? I.e. what is the survival probability that |ψ(0)i → |ψ(t)i? survival probability = P (t) = |hψ(0)|ψ(t)i|2 . We can write − i Htˆ |ψ(t)i = e ~ |ψ(0)i , then D i E 2 − Htˆ P (t) = ψ(0) e ~ ψ(0) . Now we do a second-order expansion of the exponential operator in the middle to get 2 i 1 2 2 P (t) ≈ ψ(0) 1 − tHˆ − t Hˆ ψ(0) ~ 2~2 i 1 i 1 = ψ(0) 1 + tHˆ − t2Hˆ 2 ψ(0) ψ(0) 1 − tHˆ − t2Hˆ 2 ψ(0) ~ 2~2 ~ 2~2 t2 2 1 1 = 1 + hHˆ i − t2 hHˆ 2i − t2 hHˆ 2i ~2 2~2 2~2 t2 2 = 1 − hHˆ 2i − hHˆ i . ~2 Note that we dropped the O(t3) terms. Thus we can write the survival probability in terms of the uncertainty of the energy at t = 0, squared

2 2 t 2 P (t) = |hψ(0)|ψ(t)i| = 1 − (∆E)t=0 . ~2 Note that ∆E ∼ Γ ∼ γ ∼ ~, ~ τ where τ is the lifetime. The ∆E here is often called the uncertainty for energy

∆E τ ∼ ~. However, keep in mind that this is not the same as the uncertainty ∆x and ∆p. The “uncertainty” ∆E is like an uncertainty, but its source is the ﬁnite lifetime of the state. For a wave packet, ∂E ∆E ∼ ∆p ∼ v∆p, ∂p where v is the average or group velocity of the packet. We can also write

∆x ∼ ~ , ∆p 84 Quantum Dynamics

then v v ∆E ∼ v∆p ∼ ~ ∼ ~ ∼ ~ . ∆x v∆t ∆t This gives us another form of energy “uncertainty”

∆E ∆t ∼ ~.

For the uncertainty relation ∆E τ ∼ ~, the time τ is the lifetime. For the uncertainty ∆E ∆t ∼ ~, the time ∆t is the “passing” time. I.e. how long it takes the wave packet to pass by.

5.8 Time Reversal

We have an initial state at t = 0 at point A with coordinate and momentum qi, pi. The trajectory evolves to some ﬁnal time t at point B with coordinate and momentum qf , pf . 0 The system is said to be time reversible if you can start at point B with qi = qf and 0 pi = pf and end at point A along the exact same trajectory and with the same trip time—the only diﬀerence being the opposite sign of the velocity.

The classical equations of motion are ∂H ∂H q˙i = , p˙i = − . ∂pi ∂qi

For time reversal,q ˙i and pi both change sign and neitherp ˙i nor qi change sign. Thus for a time reversible system, H must be invariant. For a charge in an EM field, e m~v˙ = eE~ + ~v × B~ . c The left side is invariant under time reversal. The first term on the right is invariant under time reversal. However, the second term is not invariant under time reversal unless we include the source of the magnetic field as part of the systen. Now we look at the quantum case. We have some Hˆ and ψ such that

∂ψ(t) i = Hψˆ (t). ~ ∂t

We are only considering the case in which Hˆ itself is independent of time. Applying time reversal gives us

∂ψ(−t) i = Hψˆ (−t) ~ ∂(−t) ∂ψ(−t) −i = Hψˆ (−t). ~ ∂t 5.8. Time Reversal 85

To bring it back into the original form, we can take the complex conjugate to get

∂ψ∗(−t) i = Hˆ ∗ψ∗(−t). ~ ∂t

So the time-reversed function ψ˜(t) = ψ∗(−t) satisﬁes the Schrodinger equation with the new Hamiltonian Hˆ ∗. For example, a plane wave under time reversal,

i(~k·~r− E t) i(−~k·~r− −E (−t)) −i(~k·~r+ E t) e ~ → e ~ = e ~ .

What happens with a transition amplitude?

hψ (t)|Hˆ |ψ (t)i = dτ ψ∗(t) Hˆ ψ (t). 2 1 ˆ 2 1 Here, the dτ denotes the integration measure over the variables, whatever they are. Under time-reversal, we get

∗ dτ ψ (−t) Hˆ ∗ ψ∗(−t) = dτ Hψˆ (−t) ψ (−t) = dτ ψ∗(−t) Hˆ † ψ (−t). ˆ 2 1 ˆ 1 2 ˆ 1 2

If Hˆ is Hermitian, then we can replace Hˆ † with Hˆ . We can deﬁne the time-reversal operator

Tˆ = UˆKˆ Oˆt, where Uˆ operates on any spin variables, Kˆ is the complex conjugation operator, and Oˆt reverses the sign of any explicit t. For example, this operator makes the following transformations:

~r → ~r ~p = −i~∇ → i~∇ = −~p L~ → −L~ S~ → S~. 86 Quantum Dynamics

5.9 Summary: Quantum Dynamics

Skills to Master • Determine whether or not an operator is unitary • Simplify unitary transformations • Calculate the equations of motion for Heisenberg operators • Determine if a Heisenberg operator represents a conserved quantity • Understand wave packets, and calculate the spreading of a wave packet as a function of time • Solve radioactive decay problems • Calculate survival probabilities

Unitary Transformations time. In the Heisenberg picture, the wave functions are static, and it’s the operators that depend on time. A unitary operator Uˆ is one whose Hermitian conjugate In general, a Heisenberg operator Qˆ (t) is ob- equals its inverse H tained from a Schrodinger operator Qˆ as ˆ † ˆ −1 ˆ † ˆ ˆ ˆ † U = U =⇒ U U = UU . ˆ ˆ † ˆ ˆ i tHˆ ˆ − i tHˆ QH (t) = U (t) Q U(t) = e ~ Q e ~ . Any unitary operator can be written in the form This implies the Heisenberg equation, which gives the Uˆ = eiGˆ , time evolution of a Heisenberg operator ! ˆ dQˆ ∂Qˆ 1 h i where G is an operator called a generator. If u is an H = + Q,ˆ Hˆ . eigenvalue of Uˆ, then dt ∂t i H H ~ |u|2 = u∗u = 1. To get the Heisenberg equation of motion for an operator Qˆ, you would Thus, the eigenvalues u can always be written in the ˆ 1. Calculate the partial time derivative ∂Q . This is form ∂t typically zero u = eig, 2. Calculate the commutator [Q,ˆ Hˆ ]. Note, Qˆ and where g is the eigenvalue of the generator Gˆ associated Hˆ are just the standard Schrodinger versions of with the unitary operator Uˆ. those operators We will often encounter unitary transformations 3. Put them into the equations above. If there re- of the form main any operators Oˆ on the right side, write them as Oˆ(t) or as Oˆ to indicate that these are ˆ ˆ ˆ ˆ † iGˆ ˆ −iGˆ H QH = U Q U = e Qe . time-dependent Heisenberg operators The standard way to simplify something like the right- 4. Solve the remaining differential equation ˆ 2 hand side is to: For a standard Hamiltonian H = ~p /2m + U(~r), 1. Differentiate both sides with respect to some pa- ~p ˆ ~r˙ = H rameter in the exponents that is not also in Q. If H m there is no parameter, insert one and then set it ~p˙ = F~ = −∇U(~r ). equal to 1 in the end. H H H 2. Rearrange the result so the right hand side looks That is, the equations have the exact same form as in ˆ the same, but now with Q replaced by a commu- classical mechanics. ˆ tator involving Q For any stationary state (i.e. has definite energy) 3. Simplify the commutator, and then the exponen- of a discrete spectrum, tials cancel each other * + 4. Solve the remaining differential equation dQˆ H = 0. dt The Heisenberg Picture This implies the virial theorem In the Schrodinger picture of quantum mechanics, the operators are static and the wave functions depend on hKi = 2 h~r · ∇Ui . 5.9. Summary: Quantum Dynamics 87

Given a Heisenberg operator QˆH , if and only if A common type of transition is radioactive decay, ˆ where the rate of change of the number of particles is ∂QH h ˆ î = 0, and H, Q = 0, dN ∂t = −γN, then dt ˆ dQH where γ is the decay rate, and the number of decays = 0 =⇒ QˆH is conserved. dt per unit time is So if an operator does not explicitly depend on time −γt and it commutes with Hˆ , then the associated observ- N(t) = N(0)e . able is conserved. Mean lifetime is given by 1 Wave Packets τ = . γ In 1D, the wave function of a wave packet is ∞ Half-life is given by dk i(kx−ω(k)t) Ψ(x, t) = A(k) e , ln 2 ˆ−∞ 2π t = . 1/2 γ where p E(k) k = , ω(k) = . Since an excited state is unstable, its energy can- ~ ~ not have a definite value. Rather, its energy follows For a free wave packet, E = p2/2m. the normalized Breit-Wigner distribution The initial state of the wave packet is Γ 1 ∞ dk w(E) = 2 2 , Ψ = Ψ(x, 0) = A(k) eikx. 2π (E − E0) + Γ /4 0 ˆ 2π −∞ where To find A(k), we perform a Fourier transform ~ Γ = ~γ = , ∞ τ −ikx A(k) = dx Ψ0e . is the full width at half maximum (i.e. FWHM) of ˆ −∞ w(E) and is often called the level width. The group velocity of the wave packet dω(k) Survival Probability vg = , dk Given an initial state |ψ(0)i and a final state |ψ(t)i, is the velocity of the clump of plane waves as this clump the survival probability that |ψ(0)i → |ψ(t)i is propagates along. The phase velocity 2 ω(k) P (t) = |hψ(0)|ψ(t)i| . vph = , k − i Htˆ By writing |ψ(t)i = e ~ |ψ(0)i, and then ex- is the velocity of the single plane wave with wave num- panding |hψ(0)|ψ(t)i|2 in a series of Hˆ , we can write ber k that forms part of the wave packet. the survival probability in terms of the uncertainty of Since a wave packet is composed of waves of dif- the energy at t = 0, squared ferent speeds, it tends to spread out over time. This 2 2 spreading is given by t ˆ 2 ˆ 2 t 2 P (t) = 1 − 2 hH i − hHi = 1 − 2 (∆E)t=0 . t ~ ~ (∆x)2 ∼ ~ . spread m This gives us an energy uncertainty relation in The total width of the wave packet at time t is the terms of the lifetime of an excited state width at t = 0 plus the time-dependent spreading ∆E τ ∼ ~. (∆x)2 = (∆x)2 + (∆x)2 . t=0 spread For a wave packet, we find a similar uncertainty relation for the energy, Decaying of Excited States ∆E ∆t ∼ ~. Only ground states of isolated systems are truly stationary in the sense that they have infinite lifetime. All but now the time ∆t is the time it takes the wave packet excited states will undergo transitions at some point. to pass by. Chapter 6

General 1D Problems

6.1 General Properties

For now we consider stationary wave functions satisfying the eigenvalue problem

Hψˆ = Eψ, with Hamiltonians pˆ2 Hˆ = + U(ˆx). 2m In the coordinate representation, this gives us the second order ODE

2 d2ψ − ~ + U(x) ψ = Eψ. 2m dx2 The standard approach is to introduce 2m k2(x) = [E − U(x)] , ~2 then ψ00 + k2ψ = 0. Given two functions f(x) and g(x), the Wronskian is the 2 × 2 determinant

f g W = = fg0 − f 0g. 0 0 f g

Then the derivative of the Wronskian is dW = fg00 − f 00g. dx

Suppose you have two solutions ψ1 and ψ2 (perhaps with diﬀerent energies) so that

00 2 ψ1 + k ψ1 = 0 00 2 ψ2 + k ψ2 = 0.

Multiply the ﬁrst equation by ψ2 and the second by ψ1 and then add/subtract the two to get

00 00 2 2 ψ1ψ2 − ψ2ψ1 + k2 − k1 ψ1ψ2 = 0

00 00 2m ψ1ψ2 − ψ2ψ1 + (E2 − E1) ψ1ψ2 = 0. ~2 6.1. General Properties 89

Notice that U(x) dropped out. Let the Wronskian be

0 0 W12 = ψ1ψ2 − ψ1ψ2, then its derivative is dW12 2m = (E1 − E2) ψ1ψ2. dx ~2

If there is degeneracy, E1 = E2, then W12 is constant (since its derivative is zero). Can this happen? Consider an arbitrary potential with E > U.

On the far left (i.e. x → −∞), we have the source wave going to the right and reﬂected wave going to the left ikx −ikx ψ1 = Ae + Be . The second solution corresponds to the source being on the right and the wave traveling to the left. There’s no reﬂected wave on the left side if the source is on the right side,

−ikx ψ2 = De .

Then the Wronskian is

0 0 W12 = ψ1ψ2 − ψ1ψ2 = Aeikx + Be−ikx −ikDe−ikx − ik Aeikx + Be−ikx De−ikx = −2ikAD = constant.

So the Wronskian can indeed be constant. Consider the two cases shown below. The dotted lines represent two diﬀerent possible energies. At the higher energy, the wave function oscillates until it reaches the potential wall where it decays to zero. At the lower energy, there is an oscillating bound state.

dW12 Suppose E1 = E2, which implies dx = 0, which implies that W12 = constant. In both cases shown above (reflection and bound state), the wave function goes to zero when x → ∞. Then W12 = 0 = constant. If W = 0, then the solutions can only differ by a constant. I.e. they are not linearly independent. So if you have W = 0, you only need to find a single solution. 90 General 1D Problems

Now consider two solutions with diﬀerent energies—E1 and E2 > E1. Any wave function (except for the ground state) has nodes (i.e. zeros). Suppose a and b are two consecutive zeros of ψ1 so that

ψ1(a) = ψ1(b) = 0.

Since the zeros are consecutive, we know that ψ1 does not change sign in the interval (a, b). What happens to the Wronskian? At x = a, the Wronskian is

0 0 0 W12(a) = ψ1(a)ψ2(a) − ψ1(a)ψ2(a) = −ψ1(a)ψ2(a). 0 Assume ψ2 6= 0 in the interval (a, b), and ψ1(a) > 0. Then W12(a) is negative. On the other hand, at x = b, the Wronskian is

0 0 0 W12(b) = ψ1(b)ψ2(b) − ψ1(b)ψ2(b) = −ψ1(b)ψ2(b). 0 Since ψ1(a) > 0 (i.e. ψ1 is increasing at a), then since b is the next zero, we know that 0 ψ (b) < 0. So W12(b) is positive. This means W12 changes sign in the interval [a, b]. But this gives us a contradiction. Our assumption that ψ2 6= 0 in (a, b) must be wrong. Therefore, ψ2 = 0 in between any roots of ψ1. This is the oscillation theorem. Between any pair of roots of ψ1, there is a root of ψ2. If you have one solution for dW/dx = 0, you can ﬁnd another easily by solving the ﬁrst order ODE 0 0 ψ1ψ2 − ψ1ψ2 = const.

6.2 1D Harmonic Oscillator

We will illustrate a standard approach for solving 1D quantum problems by solving the harmonic oscillator problem. p2 1 H = + mω2x2. 2m 2 Step 1: Draw the potential. Below we draw the potential. We know that all states are bound states, and the energy levels are equidistant 1 E = ω n + . n ~ 2 Drawing the potential allows us to make qualitative conclusions. 6.2. 1D Harmonic Oscillator 91

Step 2: Write down the Schrodinger equation and simplify

2 d2ψ 1 − ~ + mω2x2 ψ = Eψ, 2m dx2 2 or ψ00 + k2ψ = 0, where 2m k2(x) = [E − U(x)] . ~2 Try to get ride of any coeﬃcients by rescaling as necessary. Let r rmω x = ~ ξ ↔ ξ = x, mω ~ then ξ is dimensionless. Then

d d rmω d2 mω d2 = , = . dx dξ ~ dx2 ~ dξ2 Then the Schrodinger equation becomes

d2ψ + εψ − ξ2ψ = 0, dξ2 where 2E ε = , ~ω is dimensionless. Step 3: Find singular points of the second order ODE. In our case, the singular points occur at ξ → ±∞. We know the wave function should go to zero at ±∞. At ξ → ±∞, our ODE becomes

ψ00 − ξ2ψ = 0, since ε is ﬁnite. This has solution 2 ψ = e±ξ /2. Thus, this is the asymptotic solution of our original ODE. The solution with the positive sign blows up at ξ → ∞, so we discard it. Our asymptotic solution is then

2 ψ = e−ξ /2.

Step 4: The general solution should be the asymptotic solution times another function with no singularities. In practice, the other function is always a polynomial. So we write our solution as

2 ψ(ξ) = v(ξ)e−ξ /2, where v(ξ) is some polynomial that we want to ﬁnd. The derivatives are v0 ψ0 = − ξ ψ v v00 v0 ψ00 = − 1 − 2ξ + ξ2 ψ. v v 92 General 1D Problems

Plugging this along with ψ back into the original ODE and simplifying gives us

v00 − 2ξv0 + (ε − 1)v = 0.

This is all we can do in the asymptotics. Step 5: Guess a solution. Try v = const =⇒ v0 = v00 = 0. Then (ε − 1)v = 0 =⇒ ε = 1.

Since ε = 2E/~ω, this gives us the ground state energy ω E = ~ . 2 This is in fact a solution. v = const is an eigenfunction of the eigenvalue equation. Then the ground state wave function is

−ξ2/2 −ξ2/2 − mω x2 ψgs = ve = e = e 2~ .

The next level polynomial is v = Cξ. We try this as a solution. Then v0 = C, v00 = 0, and we get −2Cξ + (ε − 1)Cξ = 0, which is a solution if ε = 3. 2 Next we can try a second order polynomial v = C0 +C2ξ . We know there’s no linear term because ψ should be an even function for the third solution. Then we get 1 −4C + (ε − 1)C = 0,C = − C . 2 2 2 2 0 We can keep going like this—trying higher and higher order polynomials to generate simple solutions. Step 6: Solve the ODE and get the general solution in terms of recurrence relations. Recall that the general ODE is

v00 − 2ξv0 + (ε − 1)v = 0.

The most general polynomial is X n v(ξ) = cnξ . n Taking derivatives and plugging back into the ODE gives us

X n−2 n−1 n cn n(n − 1)ξ − 2nξξ + (ε − 1)ξ = 0. n After re-indexing so each term is multiplied by ξn, we get

cn+2(n + 2)(n + 1) − cn(2n + 1 − ε) = 0,

which gives us the recurrence relation 2n + 1 − ε c = c . n+2 (n + 2)(n + 1) n

There are two possibilities: 6.2. 1D Harmonic Oscillator 93

1. Either the series eventually terminates with cn 6= 0 but cn+2 = 0 for some n. Then

1 ε = 2n + 1 =⇒ E = ω n + . n ~ 2

2. Or the series is inﬁnite. Then we get a solution that is non-physical—it blows up at ±∞. As n → ∞, 2n 2 c ≈ c = c . n+2 n n2 n n Consider the series ∞ 2k 2 X ξ eξ = . (6.1) k! k

The cn, which occurs when 2k = n is

ξ2 c = , n (n/2)! and ξn+2 1 1 2 1 cn+2 = n+2 ∝ n = n ≈ . 2 ! 2 + 1 ! 2 + 1 (n/2)! n (n/2)! Then c 2 n+2 = . cn n Thus, the series given by Eq. (6.1) has the same asymptotic behavior as the solution to our ODE. But we already know that Eq. (6.1) is nonphysical. The only way out is to conclude that the solution to our ODE cannot be infinite—the series must terminate. That is, only finite polynomial solutions have physical meaning. Our final solution to the 1D harmonic oscillator is r −ξ2/2 mω ψn(ξ) = vn(ξ) e , ξ = x, ~ where the vn are called Hermite polynomials. The energy levels are given by

1 E = ω n + . n ~ 2

The Hermite polynomials are deﬁned as

n 2 d 2 H (ξ) = (−1)neξ e−ξ . n dξn

The ﬁrst few are

H0 = 1

H1 = 2ξ 2 H2 = 4ξ − 2 . . . .

The Hermite polynomials can also be found using the generating functions

2 2 F (ξ, t) = eξ t−t , 94 General 1D Problems

then ∂nF (ξ, t) Hn(ξ) = n . ∂t t=0 2 The Hermite polynomials are orthogonal with weight e−ξ , meaning

∞ −ξ2 √ n e Hn(ξ) Hm(ξ) dξ = δnm π 2 n!. ˆ−∞ So the orthogonal wave functions are

1 −ξ2/2 ψn(ξ) = √ Hn(ξ) e . π1/4 2nn! Converting to x gives us

1/4 r mω 1 mω − mω x2 ψn(x) = √ Hn x e 2~ . π~ 2nn! ~

6.3 Multidimensional Harmonic Oscillators

The quantized EM ﬁeld can be treated as a ﬁeld of multi-dimensional oscillators. Crystal lattices of solid state physics is another place where multidimensional harmonic oscillators come up. The Hamiltonian of a multi-dimensional harmonic oscillator is

X pˆ2 1 Hˆ = a + m ω2xˆ2 . 2m 2 a a a a a

The Hamiltonian contains parameters a and m. So why is the energy En independent of m? We can make a change of variables

0 pâ 0 √ pâ → pâ = √ , xâ → xâ = maxâ. ma

With this transformation, which does not violate [p0, x0] = −i~, the Hamiltonian becomes

X (ˆp0)2 1 Hˆ = + ω2(ˆx0)2 . 2 2 a This change of variables is an example of a canonical transformation. The wave function of the entire system is Y ψ ({xa}) = ψna (xa). a Notice the separation of variables, which allows us to write the overall wave function as a product of single oscillator wave functions. Then overall energy is

X 1 E = ω n + . ~ a a 2 a When a few frequencies coincide, we can have resonance. The simplest example is a 2D harmonic oscillator with ωx = ωy. We can also have higher order resonances nxωx = nyωy with period T = 2π/ω,

nx ny ny = → Ty = Tx. Tx Ty nx 6.4. Molecular Excitations 95

Consider the simplest case—two degrees of freedom and equal frequencies (i.e. ωx = ωy = ω). Then 1 1 E = ω n + + ω n + = ω (n + n + 1) . ~ x x 2 ~ y y 2 ~ x y

Now E depends on the sum N = nx + ny

E = ~ω (N + 1) .

In this sense, resonance signals degeneracy. If N = 0, then (nx, ny) = (0, 0), so no degeneracy. If N = 1, then (nx, ny) = (1, 0) or (0, 1), so there’s a degeneracy of 2. If N = 2, we can have (2, 0), (0, 2), (1, 1), so there’ a degeneracy of 3. In general, the degeneracy is N + 1. Note, for a given N, there is rotational symmetry, and within this degenerate subspace, Lˆz is conserved.

6.4 Molecular Excitations

For a diatomic molecule, there are three kinds of excitations: • electronic • vibrational • rotational Suppose the mass of electrons is m and of ions is M. Then the Born-Oppenheimer parameter is the very small parameter m << 1. M If the equilibrium size of the molecule is ∼ a, then the typical energy of electron excitations is 2 E ∼ ~ . e ma2 The vibrational energy is

2 E ∼ ~ ∼ kL2, vib ML2 where typically, L << a. If L ∼ a, then Evib ∼ Ee, then

2 2 ka2 ∼ ~ =⇒ k ∼ ~ . ma2 ma4 This gives us an estimate of the elastic constant k. The frequency of vibration is r r k 2 r m ω ∼ ∼ ~ ∼ ~ . vib M mMa4 ma2 M Then the energy is 2 r m r m E ∼ ω ∼ ~ ∼ E . vib ~ vib ma2 M e M The rotational energy is ( `)2 E ∼ ~ , rot I where I ∼ Ma2 implies 2`2 m E ∼ ~ ∼ E , rot Ma2 e M 96 General 1D Problems

since ` ∼ 1. So we have the hierarchy

r m m E : E : E = 1 : : . e vib rot M M What is the number of vibrational modes for an n-atom molecule? For an n-atom molecule, we know that there are a total of 3n degrees of freedom. For motion as a whole (i.e. translation), there are three degrees of freedom. For rotational motion there are also 3 degrees of freedom. This leaves 3n − 6 degrees of vibrational freedom. For example, for the water molecule, which consists of n = 3 atoms, there are a total of 3n = 9 degrees of freedom

3 (translational) + 3 (rotational) + 3 (vibrational) = 9.

For a linear molecule, however, there are only two rotational degrees of freedom. For example, for the carbon dioxide molecules, there are

3 (translational) + 2 (rotational) + 4 (vibrational) = 9.

From statistical mechanics, we know that each quadratic term in the Hamiltonian 1 contributes 2 to the heat capacity. For translation, there are three quadratic terms in the 2 2 2 Hamiltonian—(px + py + pz)/(2m). So both water and carbon dioxide get a contribution 3 of 2 to the heat capacity from their translational degrees of freedom. For rotation, there is also a quadratic term for each degree of freedom. For vibration, both the kinetic and the potential parts contain quadratic terms for each degree of freedom, so for water, then the heat capacity is 3 3 + + 3 = 6. 2 2 For carbon dioxide, the heat capacity (a high temperatures) is 3 13 + 1 + 4 = . 2 2

6.5 Secondary Quantization

Consider a pair of operators Aˆ and Bˆ. Assume that h i A,ˆ Bˆ = λA,ˆ

where λ is some complex number. Then the number λ, if it exists, constructs the ladder of allowed states. Suppose

Bˆ |bi = b |bi ,

So Aˆ takes us from a state with eigenvalue b to a state with eigenvalue b − λ. We can reapply Aˆ to get a state with b − 2λ, and so on. 6.6. Heisenberg-Weyl Algebra 97

Now consider the Hermitian conjugate

h i† h i A,ˆ Bˆ = λ∗Aˆ† = − Aˆ†, Bˆ† .

Thus h i Aˆ†, Bˆ† = −λ∗Aˆ†.

This is again a ladder operator, but now with step value −λ∗ instead of λ. Note that Bˆ† has eigenvalues b∗.

If Bˆ is Hermitian, then b and λ are real, and we get a single ladder going in both directions instead of two separate and oppositely-directed ladders. If Bˆ† = Bˆ, we can also write

hb0|AˆBˆ − BˆAˆ|bi = λ hb0|Aˆ|bi , which implies (b0 − b − λ) hb0|Aˆ|bi = 0. I.e. the matrix elements are nonzero only for

b0 = b + λ.

This is just another way of showing the ladder construction.

6.6 Heisenberg-Weyl Algebra ˆ This algebra consists only of the variablesx ˆ,p ˆ, and ` and the condition [ˆx, pˆ] = i~.√ We start by introducing an arbitrary real parameter ν > 0 with dimension mω. Then we deﬁne the ladder operators

1 i aˆ = √ νxˆ + pˆ 2~ ν 1 i aˆ† = √ νxˆ − pˆ . 2~ ν 98 General 1D Problems

The inverse relations are r 1 xˆ = ~ aˆ +a ˆ† 2 ν r pˆ = −i ~ν aˆ − aˆ† . 2 The commutator is [ˆa, aˆ†] = 1. We deﬁne the number operator as

Nˆ =a ˆ†aˆ = Nˆ †.

Then [â, Nˆ] = [â, aˆ†aˆ] = [â, aˆ†]â =a, ˆ and [â†, Nˆ] = −aˆ†. So these satisfy the requirements for ladder operators. Recall that ladder operators have the form h i A,ˆ Bˆ = λA.ˆ

The eigenvalue equation is Nˆ |ni = n |ni . Since Nˆ is Hermitian, we know the eigenvalues n are real.

What are the eigenvalues n? All we know about them at this point is that they are real numbers. For an arbitrary state |ψi,

since the norm squared of any state includinga ˆ |ψi is non-negative. This implies that the ladder cannot go downwards indeﬁnitely. It must end at n = 0. I.e. there must be some lowest state such that Nˆ |lowesti = 0. This implies that n ∈ Z and n ≥ 0. If it were not an integer, it could drop down the ladder past zero, without hitting zero. However, it must hit zero in order to terminate. We assume the |ni are normalized, then hn|ni = 1. 6.6. Heisenberg-Weyl Algebra 99

The ladder-down operator gives us

aˆ |ni = µn |n − 1i , where µn = hn − 1|aˆ|ni . The ladder-up operator gives us

† aˆ |ni =µ ˜n |n + 1i , where † ∗ ∗ µ˜n = hn + 1|aˆ |ni = hn|aˆ|n + 1i = µn+1. By inserting a set of states, we can write

ˆ † † ∗ 2 n = hn|N|ni = hn|aˆ aˆ|ni = hn|aˆ |n − 1i hn − 1|aˆ|ni =µ ˜n−1µn = µnµn = |µn| .

Thus, 2 n = |µn| . Since the phase is arbitrary and nonphysical, we can take it to be 1, then √ µn = n. and √ µ˜n = n + 1. So the ladder operators act on stationary states in the following way: √ aˆ |ni = n |n − 1i √ aˆ† |ni = n + 1 |n + 1i .

We can write 1 i i 1 1 1 Nˆ =a ˆ†aˆ = νxˆ − pˆ νxˆ + pˆ = ν2xˆ2 + pˆ2 − . 2~ ν ν 2~ ν2 2 √ For the harmonic oscillator, ν = mω, then

1 1 1 Nˆ = mωxˆ2 + pˆ2 − . 2~ mω 2 So we can write the Hamiltonian of the harmonic oscillator as pˆ2 1 1 1 Hˆ = + mω2xˆ2 = ωNˆ + ω = ω Nˆ + . 2m 2 ~ 2~ ~ 2 √ Sincea ˆ |ni = n |n − 1i anda ˆ |0i = 0, we have the ﬁrst order ODE

1 √ i √ mω xˆ − √ pˆ |0i = 0, 2~ mω or √ d mω x + √~ ψ (x) = 0, mω dx 0 This has the solution − mω x2 ψ0(x) = Ae 2~ . 100 General 1D Problems

We can normalize this then all higher states, which are obtained by applying the ladder-up operator, will be automatically normalized. Recall that the nth eigenstate is

1 n |ni = √ aˆ† |0i . n! For any state, 1 ν2xˆ2 + pˆ2 ≥ , ν2 ~

for any ν ∈ R.√ This is a stronger version of the uncertainty principle. For the harmonic oscillator, ν = mω, then hp2i mω hx2i + ≥ . mω ~ For the 1D harmonic oscillator, the important results to remember are:

1 √ i aˆ = √ mωxˆ + √ pˆ 2~ mω 1 √ i aˆ† = √ mωxˆ − √ pˆ 2~ mω Nˆ =a ˆ†aˆ Nˆ |ni = n |ni a,ˆ aˆ† = 1 1 Hˆ = ω Nˆ + ~ 2 aˆ |0i = 0 1 n |ni = √ aˆ† |0i n! √ aˆ |ni = n |n − 1i √ aˆ† |ni = n + 1 |n + 1i .

6.7 Coherent States

Coherent states |αi are states most like the states of the classical harmonic oscillator. Coherent states are eigenstates of the operatora ˆ,

aˆ |αi = α |αi .

The operatora ˆ† has no such eigenstates. The number of these states is the same as the number of points in the complex plane. This is many more than the number of stationary states which are indexed by integers. The stationary states are a complete set, so we can expand coherent states as X |αi = cn(α) |ni . n Then X X √ X aˆ |αi =a ˆ cn(α) |ni = cn n |n − 1i = α cn(α) |ni . n n n Consider X √ X cn n |n − 1i = αcn |ni . n n 6.7. Coherent States 101

If we expand both sides and equate coeﬃcients, we get √ c1 1 = αc0 √ c2 2 = αc1 √ c3 3 = αc2 . . . .

Thus α c1 = √ c0 1 α2 c2 = √ c0 1 · 2 α3 c3 = √ c0 1 · 2 · 3 . . . . αn cn = √ c0. n! So ∞ X αn |αi = c0 √ |ni . n=0 n!

To get c0, we normalize.

∗ n0 n 2 X (α ) 0 X α 1 = hα|αi = |c0| √ hn | √ |ni 0 n0 n ! n n! 2n X |α| 2 = |c |2 = |c |2e|α| . 0 n! 0 n Thus −|α|2/2 c0 = e . So ∞ n 2 X α |αi = e−|α| /2 √ |ni . n=0 n! We can also write n n ∞ n aˆ† ∞ αaˆ† 2 X α 2 X |αi = e−|α| /2 √ √ |0i = e−|α| /2 |0i , n! n=0 n! n! n=0 which simpliﬁes to 2 ˆ |αi = e−|α| /2+αa† |0i . When α = 0, |αi = |n = 0i. This is the only state for which |αi equals a stationary state |ni. The mean value of Nˆ is

hα|Nˆ|αi = hα|aˆ†aˆ|αi = |α|2 = n.

So |α|2 is just the mean value of the number of quanta. 102 General 1D Problems

The probability is

2n n 2 |α| n Prob(n) = |hn|αi|2 = e−|α| = e−n . n! n! So the quanta are distributed according to a Poisson distribution. The maximum of the distribution occurs at n. What is the width ∆n? In general,

(∆n)2 = n2 − n2.

To calculate n2 = hα|Nˆ 2|αi = hα|aˆ†aˆ aˆ†aˆ|αi , we start by bringing it into normal form, which is when all creation operatorsa ˆ† are on the left and the annihilation operatorsa ˆ are on the right:a ˆ†aˆ† ··· aâˆ. To do this, we use the commutator [â, aˆ†] =a âˆ† − aˆ†aˆ = 1, which allows us to write

n2 = hα|aˆ† aˆ†aˆ + 1 aˆ|αi = (α∗)2 α2 + α∗α = |α|4 + |α|2 = n2 + n.

Thus (∆n)2 = n, which implies that q 2 (∆n) 1 = √ . n n The states |αi are overcomplete, meaning there are too many of them. In other words, states |αi with diﬀerent values of α are not orthogonal to each other. In general,

∗ n0 n 2 2 X (β ) α hβ|αi = e−|α| /2−|β| /2 √ √ hn0|ni 0 n,n0 n ! n! ∗ n0 n −|α|2/2−|β|2/2 X (β ) α = e √ √ δn,n0 0 n,n0 n ! n! 1 X (β∗α)n = exp − (|α|2 + |β|2) 2 n! n 1 = exp − (|α|2 + |β|2) + β∗α . 2

The absolute value of the overlap is

2 |hβ|αi|2 = exp −(|α|2 + |β|2) + β∗α + βα∗ = e−|β−α| .

If |β − α| is large, then |βi and |αi are practically orthogonal. Now we add the time dependence.

∞ n 2 α i −|α| /2 X − Ent |α; t > 0i = e √ e ~ |ni , n=0 n!

where 1 E = ω n + . n ~ 2 6.7. Coherent States 103

Plugging this in gives us

|α; t = 0i → |αe−iωti .

r xˆ = ~ aˆ + aˆ† 2mω r mω pˆ = −i ~ aˆ − aˆ† . 2 We can write these in terms α as r r 2 xˆ = ~ (α + α∗) = ~ Re(α) 2mω mω r mω √ pˆ = −i ~ (α − α∗) = 2 mω Im(α). 2 ~ Then the area element is r 2 √ dx dp = ~ d(Re(α)) 2 mω d(Im(α)) = 2 d2α. mω ~ ~ So d2α dx dp |αi hα| = |αi hα| . ˆ π ˆ 2π~ Compare this with the Bohr quantization recipe from the beginning of this book. 104 General 1D Problems

6.8 Summary: General 1D Problems

Skills to Master • Solve the 1D Schrodinger equation for a given potential in terms of a recurrence relation between the coeﬃcients of a power series • Be comfortable using the ladder operators for various calculations involving a 1D harmonic oscillator • Estimate the heat capacity of various molecules via their degrees of freedom

General 1D Problems For the 1D harmonic oscillator, the solution is The oscillation theorem tells us that given two solutions (with different energies) ψ1 and ψ2 to the 1/4 r mω 1 mω − mω x2 Schrodinger equation, then between any pair of roots ψn(x) = √ Hn x e 2~ , π 2nn! of ψ1, there is a root of ψ2. ~ ~ The general approach for solving 1D problems is: 1. Draw the potential V (x). This allows us to make where the H are Hermite polynomials. The first sev- qualitative conclusions n eral are 2. Write down the Schrodinger equation and simplify ~2 d2ψ − + V (x) ψ = Eψ, H = 1 2m dx2 0 Try to get rid of any coefficients by rescaling as H1 = 2ξ 2 necessary. H2 = 4ξ − 2. 3. Find the singular points of the second order ODE, i.e. get the asymptotic solutions. Typi- cally, these occur at x = ±∞ or at r = 0 and Remember the ground state wave function for a Har- r = ∞. monic oscillator, which is the Gaussian 4. The general solution should be the asymptotic solution (or the product of the asymptotic so- 1/4 lutions) times an unknown function f with no mω − mω x2 ψ0(x) = e 2~ . singularities. In practice, the unknown function π~ f is always a polynomial. Write the wavefunc- tion ψ in this new form, and plug it back into the Schrodinger equation to get a simpler ODE. At this point, if you’re only seeking the ground state wave function, you can guess a solution (e.g. Ladder Operators f = const) 5. Otherwise, write the most general polynomial The general ladder operator relation is X f(x) = c xn, n h i n A,ˆ Bˆ = λA,ˆ take its derivatives, and plug it into the ODE. 6. Write the solution as a recurrence relation between, e.g., cn+2 and cn where λ is some complex number. Then the operator 7. Examine the asymptotic behavior of the power Aˆ takes us down the ladder with steps λ. The oper- series given above. What happens when n → ∞? ator Aˆ† takes us up the ladder with steps λ∗. If Bˆ is Is the series infinite or does it eventually termi- Hermitian, then λ is real, and we get a single ladder nate? Typically, the series will terminate with going both up and down rather than two separate and cn+2 = 0 but cn 6= 0 at some nmax. Then the oppositely-directed ladders.

recurrence relation implies cnmax = 0, which can For the 1D harmonic oscillator, the ladder-down be solved for the energy levels En (aka annihilation) operator isa ˆ, and the ladder-up (aka 6.8. Summary: General 1D Problems 105

† creation) operator isa ˆ . This is harmonic approximation. Typically, Hˆpert is some function ofx ˆ andp ˆ, which you can write in terms 1 √ i aˆ = √ mωxˆ + √ pˆ ofa ˆ anda ˆ† and solve. 2~ mω 1 √ i aˆ† = √ mωxˆ − √ pˆ Coherent States 2~ mω r Coherent states are eigenstates of the operatora ˆ, xˆ = ~ aˆ +a ˆ† 2mω aˆ |αi = α |αi . r mω pˆ = −i ~ aˆ − aˆ† Coherent states |αi are important because they are the 2 states most like the states of the classical harmonic os- † Nˆ =a ˆ aˆ cillator. The number of these states is the same as Nˆ |ni = n |ni the number of points in the complex plane. Note, the † † operatora ˆ has no such eigenstates. a,ˆ aˆ = 1 We can expand the coherent states |αi as a linear 1 combination of stationary states of the 1D harmonic Hˆ = ω Nˆ + ~ 2 oscillator |ni X aˆ |0i = 0 |αi = cn(α) |ni . n 1 n |ni = √ aˆ† |0i n! If we expand both sides and equate coeﬃcients, and √ normalize to get c , we ﬁnd that aˆ |ni = n |n − 1i 0 √ † ∞ n aˆ |ni = n + 1 |n + 1i . 2 X α |αi = e−|α| /2 √ |ni . n! To calculate quantities like hf(ˆx, pˆ)i for a har- n=0 monic oscillator, write f(ˆx, pˆ) in terms of the ladder By writing the |ni in terms of the ground state |0i of † operatorsa ˆ anda ˆ . the harmonic oscillator, we can write the above in the The virial theorem for harmonic oscillators states exponential form that 1 1 1 2 ˆ† hKˆ i = hVˆ i = E = ω n + , |αi = e−|α| /2+αa |0i . 2 n 2~ 2 where Kˆ is the kinetic energy operator, and Vˆ is the Molecular Excitations potential energy. Thus, For a diatomic molecule, there are three kinds of exci- pˆ2 1 1 tations: electronic, vibrational, and rotational. Their n n = n mω2xˆ2 n = E . 2m 2 2 n contribution to the total energy forms the hierarchy

This makes it much easier to calculate hpˆ2i and hxˆ2i as r m m E : E : E = 1 : : , opposed to using the ladder operators. ele vib rot M M Given a Hamiltonian that is the Hamiltonian of a harmonic oscillator plus a small perturbation where m is the mass of the electron, and M is the mass of the ion. ˆ ˆ ˆ H = HHO + λHpert, For an n-atom molecule, we know that there are a total of 3n degrees of freedom. For motion as a whole where λ is a small parameter, you can estimate the (i.e. translation), there are three degrees of freedom. energy levels by assuming the eigenstates of Hˆ are the For rotational motion there are also 3 degrees of free- eigenstates |ni of the harmonic oscillator. Note, in fact, dom. This leaves 3n−6 degrees of vibrational freedom. the eigenstates of Hˆ are |ni, but the eigenstates of HO Each quadratic term in the Hamiltonian con- Hˆ deviate from this due to the perturbation. However, tributes 1 to the heat capacity. For translation, in ﬁrst-order perturbation theory, we assume that |ni 2 there are three quadratic terms in the Hamiltonian— are the eigenstates of the whole Hamiltonian Hˆ . Then (p2 + p2 + p2)/(2m). For rotation, there is also a we can estimate the energy levels as x y z quadratic term for each degree of freedom. For vibra- 1 tion, both the kinetic and the potential parts contain E = hHi ≈ hn|H|ni = ω n + + λ hn|H |ni . n ~ 2 pert quadratic terms for each degree of freedom. Chapter 7

Magnetic Field

7.1 Minimal Coupling (Spinless Particles)

Consider the Euler-Lagrange equation d ∂L ∂L = . dt ∂q˙i ∂qi To get the correct Lorentz force for a charged particle in a magnetic ﬁeld, the Lagrangian is 1 e L(~r, ~v) = m~v 2 − eφ + ~v · A~ , 2 c where φ is the electric potential, and A~(~r, t) is the vector potential. Recall that the magnetic ﬁeld is the curl of the vector potential

B~ = ∇ × A~ = curl A~. Then the Euler-Lagrange equation gives us d ∂L ∂L = dt ∂~v ∂~r d e ∂ e m~v + A~ = −eφ + ~v · A~ . dt c ∂~r c Going to components of the vectors (j is summed over), we have

e e ∂Ai ∂φ e ∂Aj mv˙i + A˙ i + x˙ j = −e + vj c c ∂xj ∂xi c ∂xi ∂φ e e ∂Aj ∂Ai mv˙i = −e − A˙ i + vj − ∂xi c c ∂xi ∂xj ∂φ e e h i mv˙i = −e − A˙ i + vj εijk curl A~ . ∂xi c c k

Since h i h i vjεijk curl A~ → ~v × curl A~ = ~v × B~ , k i i putting everything back into vector form gives us e ˙ e h i m~v˙ = −e ∇φ − A~ + ~v × B~ . c c The electric ﬁeld is deﬁned as

1 dA~ E~ = −∇φ − , c dt 7.2. Particle in a Magnetic Field 107 so we can write the equation of motion as

e h i F~ = m~v˙ = eE~ + ~v × B~ . c

This is the familiar form of the Lorentz force. The Hamiltonian is related to the Lagrangian via

X ∂L H = q˙ − L ∂q˙ i i i ∂L = · ~v − L ∂~v e 1 e = m~v 2 + ~v · A~ − m~v 2 − eφ + ~v · A~ c 2 c 1 = m~v 2 + eφ. 2 We deﬁne the canonical momentum as

∂L e ~p = = m~v + A~. ∂~v c

Then 1 e ~v = ~p − A~ . m c Plugging this in gives us the minimal coupling Hamiltonian

1 e 2 H = ~p − A~ + eφ. 2m c

Notice that the Hamiltonian is now a function only of the momentum and coordinates. Note that

m~v ← kinetic momentum e ~p = m~v + A~ ← canonical momentum c

7.2 Particle in a Magnetic Field

For a particle in a magnetic ﬁeld, 1 e 2 H = ~p − A~ + eφ. 2m c

Because of the square, H contains both linear and quadratic terms in B~ . We will assume that B~ is a small perturbation. We can expand this as

~p 2 H = + H(1) + H(2) + eφ, 2m where e Hˆ (1) = − ~p · A~ + A~ · ~p 2mc e2 h i2 H(2) = A~(~r) . 2mc2 108 Magnetic Field

The linear terms are e Hˆ (1) = − ~p · A~ + A~ · ~p . 2mc Remember that ~p and A~ are operators, so we can’t assume that the two terms above are the same. If we act the ﬁrst term on a wave function, using the product rule as necessary, we get ~p · A~ ψ = −i~ ∇ · A~ ψ + A~ · (~pψ) = A~ · (~pψ) ,

since the divergence of A~ (i.e. ∇ · A~) is zero. So e Hˆ (1) = − A~ · ~p . mc Due to gauge invariance, there are many equivalent A~. A convenient choice here is ˆ 1 A~ = B~ × ~rˆ , 2 where B~ is a time-independent constant vector. Then e e e ˆ Hˆ (1) = − B~ × ~rˆ · ~pˆ = − B~ · ~rˆ × ~pˆ = − ~` · B~ = −~µ · B~ , 2mc 2mc 2mc~ ` where e ~µ = g `ˆ= ~ `,ˆ ` ` 2mc

is the orbital magnetic moment, and g` is the gyromagnetic ratio. For the electron, the gyromagnetic moment is the Bohr magneton

e~ g` = = µB. 2mec Because of the c in the denominator, the eﬀect on the energy of this term is very small. The only number that needs to be remembered for numerical calculations is B Energy = µ B = 0.6 × 10−8 eV . B 1 Gauss In addition to the orbital magnetic moment, there’s also the spin magnetic moment when orbital motion is zero ~µs = gs~s,

where gs is the spin gyromagnetic ratio. For the electron, gs = 2g`. There are corrections, however: gs = 2g`(1 + α/2π), where α is the ﬁne structure constant. Overall, the linear term is

ˆ (1) ~ ~ H = −~µ` · B − ~µs · B.

If the spin magnetic moments in the system cancel each other out, as they do for the two electrons in helium, then there is no ~µs. For the proton and neutron,

e~ µp = 2.79 2mpc e~ µn = −1.91 . 2mpc 7.2. Particle in a Magnetic Field 109

The quantity e~/(2mpc) is called the nuclear magneton. If you have many electrons, sum over all of them. If a lot of electron spins are aligned, you can have a large overall magnetic moment. This is the phenomenon of paramagnetism. Even if the magnetic moments cancel out, there will be second-order corrections. Plugging in our convenient choice for A~ and assuming that B~ lies along the z-axis,

e2 h i2 e2 h i2 H(2) = A~(~r) = B~ × ~r . 2mc2 8mc2 This simpliﬁes to e2 H(2) = B2 xˆ2 +y ˆ2 . 8mc2 This is approximately

e2 2 e2B2 H(2) ≈ B2 hr2i = hr2i , 8mc2 3 12mc2 which is approximately equal to the area perpendicular to B. This is always positive. If the first-order effects cancel, then the effect of B~ is always an increase to the energy. This is the phenomenon of diamagnetism. Notice that the second-order term is independent of ~p. Thus, we can think of it as a correction to the potential. Consider a particle in free motion (i.e. zero potential) with Hamiltonian 1 Hˆ = mvˆ2, 2 where, recall that e m~v = ~p − A~. c Now we consider the first and second-order B terms together since there’s no external potential to compare with. We will assume that the magnetic field is in the z direction. First we consider the commutator 1 h e e i [ˆv , vˆ ] = pˆ − Aˆ , pˆ − Aˆ x y m2 x c x y c y e = − [p ,A ] + [A , p ] m2c x y x y e = − [p ,A ] − [p ,A ] m2c x y y x e ∂A ∂A = − −i y + i x m2c ~ ∂x ~ ∂y i e ∂A ∂A = ~ y − x m2c ∂x ∂y i e = ~ curl A~ m2c z i e = ~ B . m2c z

The component vz along the magnetic field does not feel the effect of the magnetic field. Now, consider new variables mc Pˆ = mvˆ , Qˆ = vˆ . y eB x Then m2c [Q, P ] = [ˆv , vˆ ] = i . eB x y ~ 110 Magnetic Field

Now consider the Hamiltonian in the new coordinates 1 Hˆ = m v2 + v2 + v2 2 x y z ! 1 1 Pˆ2 (eB)2 = mvˆ2 + m + Qˆ2 2 z 2 m2 (mc)2 Pˆ2 1 eB 2 = + m Qˆ2. 2m 2 mc This is a harmonic oscillator with frequency in the plane perpendicular to B of |e|B ω = . c mc

This is the cyclotron frequency. The vz term is ignored since it corresponds to free motion in the z direction. It is equivalent to p2/2m in normal momentum. The total energy is then p2 1 E(p , n) = z + ω n + , z 2m ~ c 2 where the second term gives the oscillator levels in the plane perpendicular to the magnetic ﬁeld. These levels are called Landau levels. If B = 0, we have three quantum numbers— px, py, and pz. If B 6= 0, we have only two quantum numbers—pz and n. What happened with the third quantum number? This is related to gauge invariance, and the answer lies in the fact that all levels are degenerate.

7.3 Gauge Invariance

Recall that the electric and magnetic fields are defined in terms of the scalar and vector potentials as 1 ∂A~ E~ = −∇φ − , B~ = ∇ × A~. c ∂t There are many, in fact infinitely many, potentials that give the same fields. Consider the gauge transformation

1 ∂f φ → φ0 = φ − c ∂t A~ → A~ 0 = A~ + ∇f,

where f(~r, t) is a “gauge function”. Under this transformation,

B~ → B~ 0 = B~ E~ → E~ 0 = E~. In general, we select f so that your equations become simple. For a uniform ﬁeld B~ = (0, 0,B), there are many ways to choose 1 A~ = B~ × ~r . 2 Three useful possibilities are

1 A~ = (−By, Bx, 0) ← symmetric gauge 2 A~ = (0, Bx, 0) ← x gauge A~ = (−By, 0, 0) ← y gauge 7.4. Landau Levels 111

This is all for the classical case. Now we try to ﬁnd the analogues in the quantum case. The Schrodinger equation is

∂Ψ 1 e 2 i = Hˆ Ψ = ~p − A~ + eφ Ψ. ~ ∂t 2m c

Performing the gauge transformation,

∂Ψ 1 e e 2 e ∂f i = Hˆ Ψ = ~p − A~ − ∇f + eφ − Ψ. ~ ∂t 2m c c c ∂t

This implies that the wave function transforms as

Ψ → Ψ0 = Ψeiγ(~r,t), where e γ(~r, t) = f(~r, t). ~c So we can make a gauge transformation which leaves the Schrodinger equation invariant, but it comes at the cost of changing the wave function by a phase that depends on position and time. Quantum mechanics is invariant under global phase change, i.e. a phase change that is independent of position and time. What we are adding is a local phase change since it depends on position and time. Electric and magnetic ﬁelds are introduced as a tool to maintain gauge invariance for local phase changes. If gauge invariance is true in nature, this implies that the quanta of E~ and B~ (i.e. photons) are massless. This all implies that potentials are more fundamental than ﬁelds.

7.4 Landau Levels

For a static and uniform magnetic ﬁeld and no electric ﬁeld, the Hamiltonian is

1 e 2 1 e 2 1 e 2 1 e 2 H = ~p − A~ = pˆ − Aˆ + pˆ − Aˆ + pˆ − Aˆ . 2m c 2m x c x 2m y c y 2m z c z

Assume that B~ = Bzˆ and all motion occurs in the xy-plane. Using the y-gauge, the Hamiltonian simpliﬁes to

1 eB 2 pˆ2 pˆ H = pˆ + yˆ + y + z . 2m x c 2m 2m

Let cp y ≡ − x , 0 eB then the Hamiltonian for the perpendicular motion is

pˆ2 1 eB 2 H = y + (y − y )2. ⊥ 2m 2m c 0

This is the Hamiltonian of a harmonic oscillator centered at y0 which depends on the quantum number px. We can separate the wave function as

i (pz z+pxx) ψ(pz, px, n) = e ~ ψn(y − y0), 112 Magnetic Field

where ψn(y − y0) is the wave function of a 1D harmonic oscillator centered at y = y0. The energy of the system is then

1 E = E + ω n + , z ~ c 2

where e |e|B ~ω = − B~ , ω = . mc c mc 2 The motion in the z-direction is free motion, so Ez = pz/2m. Here we have two quantum numbers—n and pz. However, in the absence of a magnetic ﬁeld, i.e. if free motion were allowed in all directions, then we would have three quantum numbers—px, py, and pz. Apparently, by adding the magnetic ﬁeld, we force the states of free motion in the x and y directions into discrete levels labeled by n. These levels are called Landau levels. These energy levels are degenerate. The size of the ground state wave function is

r s ~ ~c −4 1 |y − y0| ∼ ∼ ∼ 3 × 10 cm√ , mωc |e|B B

where B is in units of Gauss. Suppose an experiment takes place in a box with dimensions Lx × Ly × Lz. Then cp 0 < x < L , eB y We can estimate the degeneracy of the Landau levels, i.e. estimate the number of states allowed in each level as

∆x∆px Lx |e|B LxLy|e|B |e|Φ Φ → ∆y0 = = = , 2π~ 2π~ c 2π~c 2π~c Φ0

where LxLy is the area perpendicular to the magnetic ﬁeld, and Φ = LxLyB is the magnetic ﬂux. The quantity Φ N⊥ = , Φ0 is the maximum number of electrons we can have per Landau level. The constant 2π c Φ = ~ ≈ 4.14 × 10−7 Gs · cm2 = 4.14 × 10−15 Wb, 0 |e|

is the “quantum of magnetic flux”. The number of electrons that can be added to a level depends on how many units of magnetic flux there are. So if our magnetic field is strong enough, we can put all available electrons into the lowest Landau levels.

7.5 Aharonov-Bohm Eﬀect

In classical mechanics, the magnetic vector potential is just a mathematical tool. However, in quantum mechanics, at least in some cases, it comes an observable. Consider a long solenoid containing a magnetic field. There is no magnetic field outside the solenoid. There is an electric field everywhere. Suppose we have a particle source such that we shoot particles past the solenoid containing the magnetic field. We restrict the possible paths of the particles to two paths—one on each side of the solenoid. The paths are labeled (1) and (2). 7.5. Aharonov-Bohm Effect 113

The ﬂux is A~ · d~` = B~ · dS~ = Φ. ˆ ˆ Outside the solenoid, B~ = 0, which implies curl A~ = 0 outside the solenoid, and we can write the vector potential as the gradient of a scalar function χ

A~ = −∇χ.

We make the transformation

A~ → A~ 0 + ∇f, f = χ e ψ → ψ0 = ψeiγ , γ = f. ~c This introduces a phase change

b e e e ∇γ · d~` = ∇f · d~` = ∇χ · d~` = − A~ · d~`. ˆa ~c ˆ ~c ˆ ~c ˆ Now consider the two possible trajectories (1) and (2) that can be taken by the particles. What is the diﬀerence in the phase? " # e e eΦ − A~ · d~` − A~ · d~` = A~ · d~` = . ~c ˆ(1) ˆ(2) ~c ˛ ~c

Surprisingly, this is not zero. The phase change depends on A~ even though the particles never feel the magnetic field—the magnetic field exists only inside the solenoid and the particles never travel through that region. We can think of the solenoid as a “hole” or a boundary condition which affects the allowed wave functions. This effect has been measured experimentally. 114 Magnetic Field

7.6 Summary: Magnetic Field

Skills to Master • Write down the Hamiltonian for a particle in an electric or magnetic ﬁeld • Use gauge invariance to select a gauge that makes your Hamiltonian as simple as possible • Complete the square or perform a canonical transformation to write a Hamiltonian as the Hamil- tonian of a harmonic oscillator • Calculate the energy levels for a particle in an electric or magnetic ﬁeld

Hamiltonian in a Magnetic Field is the orbital magnetic moment, and g` is the gyromagnetic ratio and Given an electric potential φ and a vector potential A~, ~µ = g ~s, the magnetic and electric fields are s s is the spin magnetic moment, and g is the spin gy- ~ ~ s B = ∇ × A romagnetic ratio. For the electron, the gyromagnetic 1 dA~ moment is the Bohr magneton g` = µB, and gs = 2g`. E~ = −∇φ − . c dt The effect on the energy of the linear term is very small. The only number that needs to be remembered for nu- The Lorentz force on a particle with charge e is merical calculations is e h i F~ = m~v˙ = eE~ + ~v × B~ . B c Energy = µ B = 0.6 × 10−8 eV . B 1 Gauss The minimal coupling Hamiltonian for a charged particle in an electromagnetic field is Assuming that B~ lies along the z direction, the second- order term in A~ is 1 e 2 H = ~p − A~ + eφ, 2 2 2m c (2) e B 2 2 H = 2 xˆ +y ˆ . where we use the canonical momentum 8mc e This is always positive. If the first-order effects cancel, ~p = m~v + A~, c then the effect of B~ is always an increase to the energy. This is the phenomenon of diamagnetism. instead of the kinetic momentum m~v. From the canon- To show that operators are conserved, show that ical momentum, we have that they commute with the Hamiltonian. 1 e To find the spectrum of eigenvalues of some opera- ~v = ~p − A~ . m c tor, we typically transform it into a harmonic oscillator of some frequency Ω. ~ If B is a time-independent and constant vector We can make a canonical transformation from the then a convenient representation is variables q, p to new variables Q, P , provided that 1 A~ = B~ × ~r.ˆ h ˆ î 2 Q, P = i~. Then we can expand the above Hamiltonian, keeping in For example, given a Hamiltonian H(q, p), we often mind that ~p and A~ are operators and don’t necessarily want to find Qˆ and Pˆ both in terms ofq ˆ andp ˆ such commute, to get that ˆ2 ~p 2 P 1 2 ˆ2 H = + H(1) + H(2) + eφ, H(Q, P ) = + Ω Q , 2m 2 2 where the parameter Ω is found by enforcing the con- where the first order terms in A~ simplify to dition [Q,ˆ Pˆ] = i~. Once we have a Hamiltonian in this form, we can read off the energy spectrum as that of a Hˆ (1) = −~µ · B~ − ~µ · B~ , ` s harmonic oscillator where e~ 1 ~µ = g `ˆ= `,ˆ En = ~Ω n + . ` ` 2mc 2 7.6. Summary: Magnetic Field 115

Gauge Invariance and for the y gauge A~ = (−By, 0, 0), it is The fields are defined in terms of scalar and vector po- " 2 # tentials. However, the potentials are not unique. That 1 eB 2 2 H = pˆx + yˆ +p ˆ +p ˆ . is, there are many potentials which result in the same 2m c y z physical fields. This gives us the freedom to choose the most convenient potentials for a given problem. A gauge transformation is a transformation of the Landau Levels form If a particle is in free motion (in a magnetic field ori- 0 1 ∂f ˆ 1 2 1 2 φ → φ = φ − ented in the z-direction), then H = 2 mvˆ = 2 m(ˆvx + c ∂t 2 2 vˆy +v ˆz ). Using the above definition of velocity in a A~ → A~ 0 = A~ + ∇f, magnetic field, and then making the transformation where f(~r, t) is a “gauge function” chosen so that the 1 equations become simple. If we perform a gauge trans- Pˆ = mvˆy, Qˆ = vˆx, formation on the Schrodinger equation, we find that ωc the wave function transforms as where 0 iγ(~r,t) Ψ → Ψ = Ψe , |e|B ω = where c e mc γ(~r, t) = f(~r, t). ~c is the cyclotron frequency, then the Hamiltonian be- Three useful gauges are comes ~ 1 Pˆ2 1 A = (−By, Bx, 0) ← symmetric gauge Hˆ = + mω2Qˆ2. 2 2m 2 c A~ = (0, Bx, 0) ← x gauge This is a harmonic oscillator with total energy A~ = (−By, 0, 0) ← y gauge 2 When choosing a gauge A~, use the one that most sim- pz 1 E(pz, n) = + ~ωc n + . plifies the problem. I.e., choose the one that leaves 2m 2 your Hamiltonian with the fewest variables. The Hamiltonian for a particle in a static magnetic I.e. a free particle in a magnetic field oriented in the field is, for the x-gauge A~ = (0, Bx, 0) z-direction, undergoes free motion in the z-direction, and harmonic oscillation in the xy plane. In the sec- " 2 # 1 2 eB 2 ond term, the energy levels of the oscillation, are called H = pˆ + pˆy − xˆ +p ˆ , 2m x c z Landau levels. Chapter 8

3D Motion

8.1 Angular Momentum Algebra

Recall that i(`ˆ·nˆ)α R~n(α) = e . is a small rotation by an angle α about the axis ~n. Now, this becomes

i(J~·~n)α R~n(α) = e , where J~ = L~ + S~, is the total angular momentum. If α is inﬁnitesimal, then ei(J~·~n)α ∼ 1 − i J~ · ~n δα.

If S is a scalar operator, then h i Jˆi, Sˆ = 0.

I.e. scalars are invariant under rotation. For example, the dot product of two vectors ~u and ~v is a scalar, so it should commute with components of the angular momentum. Let’s check. h ˆ i h ˆ i Ji, (~u · ~v)j = Ji, uˆjvˆj h i h i =u ˆj Jˆi, vˆj + Jˆi, uˆj vˆj

=u ˆj iεijkvˆk + iεijkuˆk vˆj

= iεijkuˆkvˆj + iεikjuˆk vˆj

= iεijkuˆkvˆj − iεijkuˆk vˆj = 0.

If V~ is a vector with components Vi, then h i Jˆi, Vˆj = iεijkVˆk.

For example, [Jˆx, Vˆy] = iVˆz. By deﬁnition, objects that transform like this are called vectors. The total angular momentum J~ is itself a vector since h i Jˆi, Jˆj = iεijkJˆk. 8.1. Angular Momentum Algebra 117

We say that angular momentum is an algebra with three generators. It is an SU(2) algebra. This is not quantum mechanics. This is simply the geometry of rotations. Next, consider a second-rank tensor T consisting of 9 components aibj which are constructed from the Cartesian components of two vectors ~a and ~b. We can write the components of such a tensor in terms of a symmetric part, and an antisymmetric part

Tij = Sij + Aij, where 1 S = (a b + a b ), ij 2 i j j i are the elements of a symmetric matrix, and

1 A = (a b − a b ), ij 2 i j j i are the elements of an anti-symmetric matrix. The symmetric part has six components, and the anti-symmetric part has 3 components. The anti-symmetric part is equivalent to ~a ×~b. The symmetric part can be further reduced since it contains a (invariant) scalar. This scalar is its trace, which corresponds to the dot product ~a ·~b Tr S = ~a ·~b. By subtracting oﬀ this part, we can write the tensor T in the irreducible representation

1 2 1 1 T = T + T − δ Tr T + δ Tr T + (T − T ) . ij 2 ij ji 3 ij 3 ij 2 ij ji

This is an explicit identity for any rank-2 tensor. The right-hand side breaks the tensor into three independent parts which transform only among themselves. This is an example of a reduction to an irreducible representation. Since Jˆ2 is a scalar, we know that

h 2i Jˆi, Jˆ = 0.

This implies that Jˆ2 is a Casimir operator. This is the only Casimir operator in this group. Other functions of Jˆ2 also commute with J~, but they are trivial since they give us nothing new. Now we consider the ladder operators

Jˆ± = Jˆx ± iJˆy.

We choose z to be the quantization axis. I.e. Jˆz is certain and Jˆx and Jˆy are not—they precess about the z-axis. 118 3D Motion

Then h i h i h i Jˆ±, Jˆz = Jˆx, Jˆz ± i Jˆy, Jˆz

2 = −iJˆy ± i Jˆx = ∓Jˆx − iJˆy = ∓ Jˆx ± iJˆy

= ∓Jˆ±.

Recall that a commutation relation like

[A,ˆ Bˆ] = λA,ˆ

implies that Aˆ is a ladder operator, and λ is the step-size. In our case, Jˆ± must be a ladder operator with steps λ = ±1. In fact, Jˆ− is the annihilating operator for Jˆz, and Jˆ+ is the creation operator for Jˆz. We will assume that M are the eigenvalues of Jˆz.

In our case, the Casimir operator is

ˆ ˆ2 ˆ2 ˆ2 ˆ2 ˆ2 C = J = Jx + Jy + Jz ≥ Jz .

2 This implies that M ≤ C, which implies that there will always be an Mmin and an Mmax, i.e. a bottom and a top rung of our ladder. Consider the operators

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 ˆ2 J+J− + J−J+ = Jx + iJy Jx − iJy + Jx − iJy Jx + iJy = 2 Jx + Jy .

This allows us to write 1 Cˆ = Jˆ2 = Jˆ Jˆ + Jˆ Jˆ + Jˆ2. 2 + − − + z Note that h i h i h i h i Jˆ−, Jˆ+ = Jˆx − iJˆy, Jˆx + iJˆy = i Jˆx, Jˆy − i Jˆy, Jˆx = −2Jˆz,

which implies that Jˆ−Jˆ+ = Jˆ+Jˆ− − 2Jˆz. Thus, ˆ ˆ ˆ ˆ ˆ2 ˆ ˆ ˆ ˆ2 C = J+J− − Jz + Jz = J−J+ + Jz + Jz . This implies that 2 2 C = Mmax + Mmax = −Mmin + Mmin, which implies Mmin = −Mmax,

since Mmin < Mmax is the only valid solution. 8.1. Angular Momentum Algebra 119

We now deﬁne Mmax ≡ J, where J is without the hat or the vector symbol. Then the eigenvalues of Jˆ2 are

C = J + J 2 = J(J + 1).

The number of steps in our ladder is

Mmax − Mmin = J − (−J) = 2J, where J can be an integer or a half-integer. For pure orbital momentum, the half-integers are not allowed. For pure spin, we can have both integers and half-integer steps. Remember that M are the eigenvalues of the z component of angular momentum, Jˆz, and the largest of these is Mmax = J. The eigenvalues of total angular momentum squared, Jˆ2, are J(J + 1), and

2 2 J = Mmax < J(J + 1). The values of J can in general be integers or half-integers 1 3 J = 0, , 1, ,.... 2 2 The values of M are M = −J, −J + 1,...,J − 1, J. For a given J, there are 2J + 1 possible values of M. SU(2) is the simplest nontrivial representation for a group with two elementary “objects”, e.g., spin = ±1/2. Higher dimensional angular momentum can be constructed from this. To include flavor, we would have to go to SU(3). Consider a state |JMi with definite values of the total angular momentum and z component. We know 0 0 hJM|J M i = δJJ 0 δMM 0 . We want to find the matrix elements ˆ (+) hJ, M + 1|J+|JMi ≡ µM ˆ (−) hJ, M − 1|J−|JMi ≡ µM . Then ∗ (−) ˆ ∗ (+) µM = hJM|J+|J, M − 1i = µM−1 . We can write ˆ ˆ ˆ ˆ2 ˆ hJM|C|JMi = J(J + 1) = hJM|J−J+ + Jz + Jz|JMi ˆ ˆ ˆ2 ˆ = hJM|J−J+|JMi + hJM|Jz |JMi + hJM|Jz|JMi (+) (−) 2 = µM µM+1 + M + M ∗ (+) (+) 2 = µM µM + M + M 2 (+) 2 = µM + M + M. So 2 (+) µM = J(J + 1) − M(M + 1), which implies

(+) p p µM = J(J + 1) − M(M + 1) = (J − M)(J + M + 1), 120 3D Motion

which is zero at the highest rung of the ladder. Similarly,

(−) p µM = (J + M)(J − M + 1),

is zero at the lowest rung of the ladder.

8.2 Orbital Momentum and Spherical Functions

Assume a spherically symmetric potential U(r), then the angular dependence of the Schrodinger equation occurs only in the kinetic term

~p 2 2 = − ~ ∇2. 2m 2m We will use spherical coordinates r, θ, φ where

1 ∂ ∂ ~` 2 ∇2 = r2 − . r2 ∂r ∂r r2 Recall that ∂ `ˆ = −i , z ∂φ and its eigenfunctions are eimφ. For angular momentum, remember that its quantum numbers m and ` are integers, and ` = 0 denotes s-states, ` = 1 denotes p-states, and ` = 2 denotes d-states. ˆ2 We want to derive the spherical functions Y`m(θ, φ) which are eigenfunctions of ` ˆ and `z. At this point, we know that

imφ Y`m ∝ e × (some function of θ).

We also know that

ˆ2 ` Y`m = `(` + 1)Y`m ˆ `zY`m = mY`m,

with ` = 0, 1, 2,... and m = −`, . . . , `. For a point charge at the origin, the potential is 1/r, and the charge density is δ(~r). The Laplace equation tells us that

1 ∇2 = −4πδ(~r). r

If r 6= 0, then 1 ∇2 = 0. r 8.2. Orbital Momentum and Spherical Functions 121

In Cartesian coordinates, ∂ 1 1 ∂r 1 ∂ p 1 x sin θ cos φ = − = − x2 + y2 + z2 = − = − ∂x r r2 ∂x r2 ∂x r2 r r2 ∂ 1 1 ∂r 1 ∂ p 1 y sin θ sin φ = − = − x2 + y2 + z2 = − = − ∂y r r2 ∂y r2 ∂y r2 r r2 ∂ 1 1 ∂r 1 ∂ p 1 z cos θ = − = − x2 + y2 + z2 = − = − . ∂z r r2 ∂z r2 ∂z r2 r r2 Notice that ∂ ∂ 1 1 1 1 ± i ≡ ∇ = − (x ± iy) = − sin θ e±iφ. ∂x ∂y r ± r r3 r2 If we diﬀerentiate repeatedly, we increase the number of iφ in the exponent, and we increase the power of r in the denominator. For any set of constant vectors {~ek}, 1 F (θ, φ) (~e · ∇) ··· (~e · ∇) = ` . 1 ` r r`+1 This function is again a solution of the Laplace operator, and

F 1 ∂ ∂ 1 `ˆ2 ∇2 ` = r2 F − F r`+1 r2 ∂r ∂r r`+1 ` r`+3 ` 1 ∂ 1 `ˆ2 = r2(−1)(` + 1) F − F r2 ∂r r`+2 ` r`+3 ` 1 ∂ 1 `ˆ2 = −(` + 1) F − F r2 ∂r r` ` r`+3 ` 1 `ˆ2 = `(` + 1) F − F = 0, r`+3 ` r`+3 ` which implies 2 ~` F` = `(` + 1)F`, verifying that the F` are the spherical functions Y`m that we were looking for. We can deﬁne the spherical functions by the relation 1 Y (θ, φ) = const · r`+1 (∇ )m ∇`−1 . `m ± z r The spherical harmonics form a complete set with orthogonality condition

∗ sin θ dθ dφ Y Y 0 0 = δ 0 δ 0 , ˆ `m ` m `` mm and completeness relation

X ∗ 0 0 Y`m(~n)Y`m(~n ) = δ(~n − ~n ), `,m where x y z ~n = (sin θ cos φ, sin θ sin φ, cos θ) = , , . r r r Remember that ` is an integer, and m = −`, . . . , `. For example, 1 Y00 = √ , 4π and  1 √ (nx − iny) if m = −1 r 3  2 Y = n if m = 0 1m 4π z − √1 (n + in ) if m = 1 2 x y 122 3D Motion

For a vector V~ , we can also introduce spherical components. Instead of Cartesian components V~ = (Vx,Vy,Vz), we write V~ = (V0,V±1). In this notation, we would write 1 ±iφ ~n = nm = cos θ, ∓√ sin θ e . 2 Then r 3 Y = n . 1m 4π m

An important property of the spherical functions Y`m is that they have alternating parity (−1)`. We also have the relation

m ∗ Y`m = (−1) Y`−m. Note that for θ = 0, Y`m(θ, φ) = Y`0(θ)δm0.

That is, if θ = 0, then Y`m has no φ dependence. The Legendre polynomials are related to the spherical functions via r 2` + 1 P (cos θ) = Y (θ). ` 4π `0

` Because of parity, P`(θ = 0) = 1 and P`(θ = π) = (−1) .

8.3 Schrodinger Equation in Spherical Coordinates

Suppose we have a spherically-symmetric potential U(r). When it is spherically-symmetric, U(r) is a scalar function and therefore commutes with angular momentum. This implies that ` and m are conserved. Then we can separate variables as

ψ(~r) = ψ(r, θ, φ) = Y`m(θ, φ)R`(r).

The Schrodinger equation

2 − ~ ∇2 + U(r) ψ = Eψ, 2m can be written in the form ∇2 + k2 ψ = 0, where 2m k2 = k2(r) = [E − U(r)] . ~2 Expanding the Laplacian ∇2 in spherical coordinates, we have 1 ∂ ∂ 1 2m r2 − `(` + 1) + [E − U(r)] R = 0. r2 ∂r ∂r r2 ~2

Notice that the other factor in ψ—namely, Y`m—has canceled from the equation. Rear- ranging, we have

1 ∂ ∂ 2m 2`(` + 1) r2 + E − U − ~ R = 0. r2 ∂r ∂r ~2 2mr2 It is useful to deﬁne u(r) R(r) = . r 8.4. Hydrogen Atom 123

Don’t mistake this u(r) with the potential U(r). Plugging this in and simplifying gives us `(` + 1) 2m u00 − u + [E − U(r)]u = 0. r2 ~2 Thus, we already see that the energy will not depend on the quantum number m, rather, only on `. I.e. E = E(`). This implies that there will always be degeneracy. In this case, there will be 2` + 1 degenerate states. In the continuum limit r → ∞, the second term in the ODE disappears. The potential also disappears. So in this asymptotic limit, 2mE u00 + u = 0. ~2 The solution to this ODE is a spherical wave

e±ikr u ∼ e±ikr =⇒ R ∼ . r What happens in the limit r → 0? Assuming that U(r) doesn’t have a singularity stronger than 1/r2, then when r → 0, the ODE reduces to

`(` + 1) u00 − = 0. r2 This is the eﬀective ODE in the vicinity of the origin, and it has the solution

u ∼ r`+1 =⇒ R ∼ r`.

8.4 Hydrogen Atom

In spherical coordinates, the wave function becomes

ψ(~r) = R(r) Y`m(~n), where u(r) R(r) = , r and r > 0. Then the Schrodinger equation becomes

d2u 2mE 2mU(r) u `(` + 1) + u − − u = 0. dr2 ~2 ~2 r2 To generalize this beyond hydrogen to ions of other atoms, we let the potential be

Ze2 U(r) = − , r where Z is the charge of the nucleus. Then we deﬁne

2mε 2mZe2 ε ≡ −E, κ2 ≡ , ξ ≡ . ~2 ~2κ When r → ∞, u ∼ e−κr, so κ is the inverse of the penetration length. Then the Schrodinger equation becomes

u00 u `(` + 1) − u + ξ − u = 0. κ2 κr κ2r2 124 3D Motion

Next, we rescale the coordinates with ρ = κr to get

d2u u `(` + 1) − u + ξ − u = 0. dρ2 ρ ρ2 Or we can write it in terms of operators acting on u as

d2 ξ `(` + 1) − 1 + − u = 0. dρ2 ρ ρ2

To solve this ODE, we need to consider the singular points. In this case, we have two singular points—when r → 0 and when r → ∞. In the limit r → 0,

R ∼ r` =⇒ u ∼ r`+1.

In the limit r → ∞, the ODE reduces to u00 − κ2u = 0, and so

e−κr R ∼ =⇒ u ∼ e−κr. r So to remove these limiting behaviors from the ODE, we write

u(ρ) = ρ`+1e−ρv(ρ),

where v(ρ) is some polynomial function of ρ. The ﬁrst and second derivatives of this are

du ` + 1 v0 = − 1 + u dρ ρ v d2u ` + 1 v00 (v0)2 ` + 1 v0 2 = − + − u + − 1 + u. dρ2 ρ2 v v2 ρ v

Plugging u and these two derivatives back into the ODE and simplifying gives us

ρv00 + 2(` + 1)v0 − 2ρv0 − 2(` + 1)v + ξv = 0.

We can already see a simple solution. If v is constant, then its derivatives are zero, and we have ξ = 2(` + 1). Now we look for series solutions in powers of ρ

X k v(ρ) = ckρ . k=0

Then the ODE becomes X ρk(k − 1)ρk−2 + 2(` + 1)kρk−1 − 2ρkρk−1 − 2(` + 1)ρk + ξρk = 0 k X k(k − 1)ρk−1 + 2(` + 1)kρk−1 − 2kρk − 2(` + 1)ρk + ξρk = 0. k This implies that

ck+1 [(k + 1)k + 2(` + 1)(k + 1)] − ck [2k + 2(` + 1) − ξ] = 0,

or 2(k + ` + 1) − ξ c = c . k+1 (k + 1)(k + 2` + 2) k 8.4. Hydrogen Atom 125

Either the series is inﬁnite or it terminates at some k. If the series is inﬁnite, then analyzing its asymptotic behaviour tells us that the solution is nonphysical

u ∼ eκr.

Thus, the series must terminate at some ﬁnal k with ck 6= 0 and ck+1 = 0. Then

ξ = 2(k + ` + 1) = 2n, where n = k + ` + 1. This implies quantization. Let N be the k-value of the last nonzero term, then n = N + ` + 1. We know that √ 2mZe2 2mZe2 2mZe2 2mZ2e4 ξ = = = √ =⇒ ε = . 2 q 2 2 ~ κ 2 2mε ~ ε ~ ξ 2 ~ ~ Plugging in ξ = 2n, we get the hydrogenic energy levels

2mZ2e4 mZ2e4 Z2 εn = = = · 1 Ry, ~24n2 2~2n2 n2 or in terms of E, 2mZ2e4 mZ2e4 Z2 En = − = − = − · 1 Ry. ~24n2 2~2n2 n2 Since the energy depends on only a single quantum number n, we know there is a lot of degeneracy. For the ground state energy level, we have

n = 1,N = 0, ` = 0, s-state, degen. = 2.

This is the s-state and it has a degeneracy of 2 if we include the spin degeneracy which allows 2 electrons in each state. For the ﬁrst excited state,

n = 2,N = 1, ` = 0, 2s-state, degen. = 2 n = 2,N = 0, ` = 1, 2p-state, degen. = 6.

For the second excited state,

n = 3,N = 2, ` = 0, 3s-state, degen. = 2 n = 3,N = 1, ` = 1, 3p-state, degen. = 6 n = 3,N = 0, ` = 2, 3d-state, degen. = 10.

Notice that the total degeneracy of the nth level is 2n2. We know that r 2 2mεn me Z Z κn = 2 = 2 = , ~ ~ n a0n where a0 is the Bohr radius. Then the penetration length for the nth state is 1 a n = 0 . κn Z The ground state wave function is

−ρ −κr −Zr/a0 ψg.s ∼ e = e = e . 126 3D Motion

Normalizing this according to |ψ|2r2 dr dΩ = 1, ˆ gives us s Z3 −Zr/a0 −Zr/a0 ψg.s = ψn`m = ψ100 = Y00e = 3 e . πa0 We can derive excited state wave functions in the same manner. In summary, for a hydrogenic atom with z-axis quantization, we have a wave function of the form ψn`m(r, θ, φ) = Y`m(θ, φ)Rn`(r), where the radial part of the wave function has the form

` −κr Rn`(r) = const · r e vn`(r),

` −κr where r gives the behavior at the origin, e gives the behavior at inﬁnity, and vn`(r) is a polynomial that interpolates between the two boundaries. The energy levels are mZ2e4 1 En` = − , 2~2 n2 where n = N + ` + 1 and m is the reduced mass of the particles. The degeneracy of the energy levels is 2n2, where the factor of 2 is due to the two spin states at each n-level. The ground state wave function is s 3 Z −rZ/a ψg.s. = ψ100(r) = 3 e . a0π For the ground state of hydrogen, with Z = 1,

2 2 2 2 hx i = hy i = hz i = a0, and 2 2 ~ h~p i = 2 . a0 The uncertainty in position and momentum is

a0~ ~ 1 (∆x ∆px)g.s = √ = √ > ~. 3a0 3 2 The Runge-Lenz vector is a constant ~r M~ = − ~ pˆ × `ˆ− `ˆ× pˆ . r 2mZe2 Remember that ~` does not commute with any vector. Since M~ is conserved, [M~ , Hˆ ] = 0. However, since [M~ , `ˆ] 6= 0, we cannot simultaneously measure both ~` and M~ . Corrections to the hydrogenic solution have to be made due to the following: 1. Spin of the electron 2. Relativistic corrections 3. Finite size of the nucleus 4. Spin of the nucleus 5. Magnetic moment µ of the nucleus 6. Deformation (from sphericity) of the nucleus Spin splits the states pairwise. Now, j = ` ± 1/2. This small relativistic splitting is called the “ﬁne structure”. 8.4. Hydrogen Atom 127

Energy Structure It is important to know the general structure of the hydrogen energy levels.

Basic Structure The basic structure is given by the main quantum number n as shown here:

Fine Structure The ﬁne structure consists of the relativistic corrections to the basic structure. Each of the above levels are split by total angular momentum j. For example, the n = 2 level is now split into a pair of closely-spaced levels:

The ﬁne structure splitting is due primarily to the spin-orbit interaction between the electron’s orbital momentum and spin. The energy of the splitting is proportional to (Zα)2 2 En(Zα) 1 3 ∆E = 1 − . n j + 2 4n

For example, the energy diﬀerence between the 2p3/2 and the 2s1/2 and 2p1/2 levels is ∆E ≈ 4.5 × 10−5 eV.

Hyperfine Structure The hyperfine structure is additional splitting caused by interaction between the spin of the nucleus and the magnetic field of the electron. The energy levels are now split by

F = J + I = L + S + I, where J is the total angular momentum of the electron, and I is the total spin of the nucleus. Together, the basic, fine, and hyperfine structures given us the “natural” spectrum of the hydrogen atom at finer and finer resolution. Additional splitting can be forced by, e.g., applying a magnetic field (Zeeman effect) or electric field (Stark effect) to the atom. 128 3D Motion

8.5 Symmetries and Conservation Laws

In general, for an operator Qˆ, if it is independent of time and it commutes with the Hamiltonian, then the associated observable is conserved

∂Qˆ h i = 0 and Q,ˆ Hˆ = 0 =⇒ Qˆ conserved. ∂t The hard way of determining whether a quantity is conserved is to calculate the commutator with the Hamiltonian. The easy way is to use deduce the conservation laws from the symmetry of the problem. This can be done for simple quantities like momentum, angular momentum, and energy. In general, for quantum operators with classical counterparts, the conservation laws are the same, and for the same reason. For example, in quantum mechanics, translational invariance implies conservation of momentum, just like it does in classical mechanics.

Energy In any closed system, the Hamiltonian does not have explicit time dependence, and this implies conservation of energy. Note, if there is an external potential or ﬁeld that is changing in time, then energy is not conserved, since whatever is causing the potential to change is outside of the system.

Momentum Translational symmetry implies that linear momentum is conserved. If the problem has translational invariance in direction i, then pi is conserved. If the problem is translationally invariant in all directions, then ~p is conserved.

Angular Momentum ˆ ˆ2 All three components `i of angular momentum along with ` are conserved in a rotationally invariant system. So if the system looks the same in all directions, then ~` and `ˆ2 are conserved. ˆ˙ ˆ More generally, if ` = 0, then `i is conserved. We know that

ˆ˙ 1 hˆ i ` = ì, Hˆ = ~r × F~ . i~ i ˆ So ì is conserved if the torque component (~r × F~)i is zero. Recall that ~` commutes with any scalar function of coordinates and momenta. In general, `ˆ2 is conserved if and only if the potential is rotationally invariant, i.e. the ˆ potential does not depend on the angles. The components ì, however, may be conserved even if the potential is not rotationally invariant.

Parity

If h i Pˆ, Hˆ = 0,

where Pˆ is the parity operator, then parity is said to be conserved. In general, if the potential is invariant under inversion

U(~r) = U(−~r),

then parity is conserved. In other words, reﬂection (through the origin) symmetry implies parity conservation. 8.5. Symmetries and Conservation Laws 129

In particle interactions, parity is believed to be conserved except in weak interactions. Parity is multiplicative when it comes to calculating the parity of a system from the intrinsic parities of the subsystems. For a system of two particles, the overall parity is ` P = P1P2(−1) , where P1 and P2 are the intrinsic parities of the particles, and ` is the relative or mu- tual orbital angular momentum of the two particles. The relative angular momentum is so called because a single particle doesn’t have orbital angular momentum (only spin), whereas a system of two particles could be orbiting each other and thus have orbital angular momentum relative to some point. Example 8.5.1

Consider the decay of a spin-1/2 particle X into a pion (spin-0 and parity −1) and a nucleon (spin-1/2 and parity +1)

X −→ π + N.

What is the angular distribution of the pion and nucleon products if: 1. X has parity −1 and parity is conserved in the interaction 2. X has parity +1 and parity is conserved in the interaction 3. parity is not conserved in the interaction In all three cases, we require angular momentum to be conserved, and in the first two cases, we require parity to be conserved. Case 1: The parity of the left side is −1, and so the overall parity of the right side ` ` P = PπPN (−1) = (−1)(+1)(−1) , must also be −1. This implies that ` is even. The lowest possibility is ` = 0, which satisfies angular momentum conservation. The left side has total angular momentum Ji = +1/2 (assuming it is spin-up), and the right side has total angular momentum Jf = ` + s = 0 + 1/2 = 1/2 provided that the spin of the nucleon is in the same direction as the spin of X. In this case, we are assuming both are spin-up. Since ` = 0, the pion and nucleon have zero relative angular momentum. Thus, they simply fly away directly opposite each other from X’s reference frame. That is, their distribution is isotropic. Formally, the angular wave function of the product system is just the spinor corresponding to X

ψΩ = χ. The angular distribution is

2 ∗ ∗ |ψΩ| = ψΩψΩ = χ χ = const. Case 2: The parity of the left side is +1, and so the overall parity of the right side ` ` P = PπPN (−1) = (−1)(+1)(−1) , must also be +1. This implies that ` is odd. The lowest possibility is ` = 1, which satisﬁes angular momentum conservation provided that the spin of the nucleon is opposite the spin of X. If X is spin-up, then Ji = +1/2. If the nucleon is spin-down, then Jf = ` + s = 1 − 1/2 = +1/2. The general angular wave function corresponding to ` = 1 is ψΩ = (~σ · ~n) χ, where χ is a spinor corresponding to X. Then the angular distribution is

2 ∗ 2 ∗ ∗ |ψΩ| = ψΩψΩ = (~σ · ~n) χ χ = χ χ = const. 130 3D Motion

So again the distribution is isotropic. Case 3: There are two possible values for parity—+1 and −1. If parity is not conserved, then the right side could have either of these possible parities. In other words, the angular wave function is now a superposition of the two cases P = +1 and P = −1. So, most generally,

ψΩ = aχ + b (~σ · ~n) χ,

where a and b are constants. Now the angular distribution will depend on the angles (via ~n), so it is not isotropic. 8.6. Summary: 3D Motion 131

8.6 Summary: 3D Motion

Skills to Master • Know the commutator of angular momentum with scalar or vector operators 2 • Know the eigenvalue relations for the angular momentum operators Jˆ , Jˆz, and Jˆ± • Know the ﬁrst several spherical harmonics • Know the properties of the spherical harmonics ˆ2 ˆ • Know the eigenvalues relations for the spherical harmonics (with ` and `z) • Understand the derivation of the hydrogen atom solution • Use separation of variables and power series to solve a PDE in spherical coordinates

Angular Momentum Algebra However, if J~ = L~ , i.e. for purely orbital momentum, then the half-integer values are not allowed. If J~ = S~, Remember that the rotation operator for rotation i.e., for pure spin, both integers and half-integers are about an axis ~n by an angle α is allowed. The allowed values of M are i(J~·~n)α R~n(α) = e , M = −J, −J + 1,...,J − 1, J, where the total angular momentum is such that J~ = L~ + S~. J ≡ Mmax. If S is a scalar operator, then For a given J, there are 2J + 1 possible values of M. h ˆ ˆi Ji, S = 0. The ladder operators for angular momentum are

For example, since the total angular momentum Jˆ = Jˆ ± iJˆ . squared operator Jˆ2 is a scalar, we know that ± x y h i ˆ ˆ2 ~ Ji, J = 0. A scalar operator, like J · ~n does not Note, it also works to replace J by L for the orbital mo- change the quantum numbers of the state it is acting mentum or S for the spin. We can use these definitions on. to define two new operators If V~ is a vector with components Vi, then 1 h ˆ ˆ i ˆ Jˆ = Jˆ + Jˆ Ji, Vj = iεijkVk. x 2 + − 1 Jˆ = Jˆ − Jˆ . For example, the total angular momentum J~ is itself a y 2i + − h i vector since Jî, Jˆj = iεijkJˆk. Typically, when working with angular momentum, Given a state |JMi with definite z-component M, ap- we take the z-axis to be the quantization axis. Then plying these operators lets us go up or down the ladder 2 of states to |J, M ± 1i. The eigenvalue relations are Jˆz and Jˆ are certain, but Jˆx and Jˆy are uncertain and precess about the z-axis. For a state with definite ˆ p total momentum and definite z-component, we denote J± |JMi = J(J + 1) − M(M ± 1) |JM ± 1i the total momentum projections by J and the z-axis = p(J ∓ M)(J ± M + 1) |JM ± 1i . projections by M. That is, for a state |JMi, we have the eigenvalue relations The unit vector in spherical coordinates is Jˆ |JMi = M |JMi z x y z ˆ2 ~n = (sin θ cos φ, sin θ sin φ, cos θ) = , , . J |JMi = J(J + 1) |JMi . r r r

In general, the values of J can be integers or half- For example, integers 1 3 J = 0, , 1, ,.... ~ ˆ ˆ ˆ 2 2 J · ~n = Jx sin θ cos φ + Jy sin θ sin φ + Jz cos θ. 132 3D Motion

Spherical Functions A spherically symmetric potential U(r) is a scalar function and therefore commutes with angular momen- In spherical coordinates, the eigenfunctions of `ˆ2 and tum. This implies that ` and m are conserved. Then ` , i.e. the states |`mi with definite total orbital mo- z we can separate variables as mentum and definite z-component of orbital momentum are the spherical functions Y`m(θ, φ). That is, we ψn`m(~r) = ψn`m(r, θ, φ) = Y`m(θ, φ)R`(r). have the eigenvalue relations Then the Schrodinger equation reduces to ˆ2 ` Y`m = `(` + 1)Y`m `(` + 1) 2m u00 − u + [E − U(r)]u = 0, ˆ 2 2 `zY`m = mY`m, r ~ where where ` = 0, 1, 2,... and m = −`, . . . , `. u(r) R(r) = . In general, r The m refers to the mass of the orbiting electron. If imφ Y`m ∝ e × (some function of θ). the particle is not an electron, replace this mass with the reduced mass of the particle and nucleus. The first several spherical functions are To simplify things, we define 1 2 2 2mε 2mZe Y00 = √ ε ≡ −E, κ ≡ , ξ ≡ , 4π ~2 ~2κ  1 √ (nx − iny) if m = −1 and rescale the coordinates as ρ = κr, then our ODE r 3  2 Y = n if m = 0 becomes 1m 4π z − √1 (n + in ) if m = 1 d2 ξ `(` + 1) 2 x y − 1 + − u = 0. dρ2 ρ ρ2 The spherical harmonics form a complete set with In the limit, r → ∞, the ODE has the solution the orthogonality condition u ∼ e−κr. ∗ sin θ dθ dφ Y Y 0 0 = δ 0 δ 0 . ˆ `m ` m `` mm In the limit, r → 0, it has the solution `+1 The spherical functions have alternating parity (−1)`. u ∼ r . We also have the relation So we can write the solution to the ODE as

m ∗ `+1 −ρ Y`m = (−1) Y`−m, u(ρ) = ρ e v(ρ), which tells us that the spherical functions for m and where v(ρ) is a power series in ρ X −m are related by complex conjugation and a negative v(ρ) = c ρk. sign if m is odd. k k=0 The angular distribution of a wave function ψ(r, θ, φ) = R(r) Ω(θ, φ) is given by |Ω|2. Then we find the recurrence relation To find ` of a given wave function ψ, write ψ in 2(k + ` + 1) − ξ ck+1 = ck. terms of the spherical functions Y`m, then read off `. (k + 1)(k + 2` + 2) Then j = ` + s where s = ±1/2 for electrons. We can show that the series must terminate for all physical solutions. This implies quantized energy lev- Hydrogenic Atom els mZ2e4 Z2 A “hydrogenic” or “hydrogen-like” atom is an atom E = − = − · 1 Ry, n 2 2n2 n2 with a nucleus of atomic number Z (Z = 1 for hydro- ~ where n = 1, 2, 3,.... Replace m with the reduced mass gen) and a single orbiting, negatively-charged particle of the system. (electron for hydrogen). Since the energy depends on a single quantum For a hydrogenic atom, the potential is number n, there is a lot of degeneracy. If we include Ze2 spin degeneracy, then the degeneracy of the nth state U(r) = − . 2 r is 2n . Remember that ` = 0, 1, . . . n − 1. 8.6. Summary: 3D Motion 133

The penetration length for the nth state is where again, the m should be the reduced mass of the 1 a n system. = 0 , The normalized ground state wave function is κn Z where s 2 Z3 ~ −Zr/a0 ψ100 = e . a0 = 2 . 3 me πa0 is the Bohr radius. Note, the m in the Bohr radius formula refers speciﬁcally to the electron mass. If you For the ground state of hydrogen (i.e. Z = 1), have a hydrogenic atom where the orbiting particle is something other than an electron, you have to replace 2 2 2 2 this m with the reduced mass of the system whenever hx i = hy i = hz i = a0 2 you use a0 in a formula. 2 ~ h~p i = 2 The radius is a0 2 2 a0 1 n ~ 1 2 ∆x ∆p = √ ~ = √~ > . rn = = n a0, x ~ mZe2 Z 3a0 3 2 Chapter 9

Spin and Angular Momentum Coupling

9.1 Spin-1/2

The lowest nontrivial system with rotational properties is spin-1/2. It has two states: Sz = ±1/2. Hence the “2” in SU(2). For a given z-axis, the z-component of the spin is Sz = ±1/2, and we denote the two eigenstates as up and down

↑= χ+ (Sz = +1/2), ↓= χ− (Sz = −1/2).

The general spin function, or spinor, is a superposition of these two states " # a χ = aχ+ + bχ− ≡ , b which is normalized such that |a|2 + |b|2 = 1. Then the probability of ﬁnding the particle with spin up or down is

Prob(↑) = a2, Prob(↓) = b2.

Recall the Pauli matrices " # " # " # 0 1 0 −i 1 0 σx = , σy = , σx = . 1 0 i 0 0 −1

The square of any of them is equal to the identity matrix,

" # 2 2 2 1 0 σx = σy = σz = 1 = . 0 1

More generally, we have the useful identity

σiσj = δij1 + iεijkσk, where the ﬁrst term, δij, is symmetric and real, and the second term, iεijkσk, is antisymmetric and imaginary. Their commutation relation is

[σi, σj] = 2iεijkσk. 9.1. Spin-1/2 135

We deﬁne the Pauli vector as the 3D vector whose components are the Pauli matrices   σx   ~σ =  σy .   σz Its eigenvalues are ±1. Then a useful relation is (~a · ~σ) ~b · ~σ = aiσibjσj

= aibj (δij + iεijkσk) = ~a ·~b + i ~a ×~b · ~σ.

If ~n is a unit vector, then (~n · ~σ)2 = 1. The spin operator for spin-1/2 particles is 1 S~ = ~σ. 2

Since S~ is a vector, we know that

[Sˆi, Sˆj] = iεijkSˆk. The operator that rotates spin wave functions about the axis ~n by an angle α is

∞ α k −iα(S~·~n) −i α (~σ·~n) X −i 2 (~σ · ~n) Rˆ (α) = e = e 2 = . ~n k! k=0 We can write this as α α Rˆ (α) = cos − i(~σ · ~n) sin . ~n 2 2 The first term comes from the even terms of the series since (~n · ~σ)n = 1 for even n. The second term comes from the odd terms of the series. We want to find the spinor χ for which S~ · ~n = +1/2. That is, we want to solve 1 (S~ · ~n)χ = χ, 2 for χ. This is equivalent to finding the spinor for which ~σ · ~n = 1. What we’re doing is selecting particles with a given polarization. With z-axis quantization, " # a χ = , b where a is the amplitude for Sz = +1/2, and b is the amplitude for Sz = −1/2. Then " # " # a a (~σ · ~n)χ = (~σ · ~n) = λ b b " # " # a a (σxnx + σyny + σznz) = λ b b " #" # " # n n − in a a z x y = λ , nx + iny −nz b b 136 Spin and Angular Momentum Coupling

where we are looking for λ = 1. This gives us the system of equations

nza + (nx − iny)b = λa

(nx + iny)a − nzb = λb.

2 2 2 2 This implies nx + ny + nz = λ , and since ~n is a unit vector, λ = ±1. In spherical coordinates, we can write nx = sin θ cos φ, ny = sin θ sin φ, and nz = cos θ. Then we ﬁnd that b n + in sin θ cos φ + i sin θ sin φ sin θeiφ sin θ eiφ = x y = = = 2 . a 1 + n 1 + cos θ 1 + cos θ θ z cos 2 For normalization, we require |a|2 + |b|2 = 1, so a convenient choice for our spinor is " # " # a cos θ e−iφ/2 χ = = 2 = χ . θ iφ/2 ~n(θ,φ) b sin 2 e

1 This is the spinor for a particle with spin + 2 in the direction ~n(θ, φ). Note that 1 θ Prob S~ · ~n = + = |a|2 = cos2 . 2 2 What this means is that if the spin is oriented along ~n, then a detector oriented along 2 θ the z-axis will detect a fraction cos 2 of the full intensity. For spin-1/2 particles, we can always ﬁnd a direction in which it is completely polarized, i.e., has spin equal to +1/2. Remember that when we discuss spinors " # a χ = , b

we typically assume the z-axis is selected for the basis of quantization. What this means is that the eigenvectors of the spin operator S~ are " # " # 1 0 |↑i = χ+ = , |↓i = χ− = . 0 1

Then " # " # a cos θ e−iφ/2 χ = = 2 = χ , θ iφ/2 ~n(θ,φ) b sin 2 e 1 is the spinor of a particle polarized such that it has spin Sz = + 2 along the ~n direction. But, recall that due to quantization, no matter which direction along which we measure, 1 1 we will ﬁnd Sz = + 2 or Sz = − 2 . Because of our choice of basis, if we measure along the 1 z-axis we will ﬁnd Sz = + 2 with probability θ Prob(↑) = |a|2 = cos2 , 2

1 and we will ﬁnd Sz = − 2 with probability θ Prob(↓) = |b|2 = sin2 . 2 More generally, we can write " # cos θ e−iφ/2 χ = eiγ 2 , θ iφ/2 sin 2 e 9.2. Angular Momentum Coupling 137 where γ is an arbitrary phase. A common choice is γ = φ/2, but we will use the choice eiγ = 1. For a spinor oriented in the direction ~n, the transmission coeﬃcient in the direction ~n 0 is 1 + ~n · ~n 0 T 0 = . ~n~n 2 The spin projection operator projecting in the direction given by the unit vector ~n is " −iφ # ˆ ˆ 1 1 cos θ e sin θ S~n = S · ~n = ~σ · ~n = . 2 2 eiφ sin θ − cos θ

9.2 Angular Momentum Coupling

Angular momentum coupling or angular momentum addition is how multiple angular momenta add together. Due to the quantization of angular momentum, this is not a simple matter of adding two momenta arithmetically. Consider a system consisting of two subsystems with angular momenta j1 and j2. The systems can be rotated together or separately. If the systems are interacting with each other, then their angular momenta are coupled, and we cannot rotate the systems independently. There are two ways to think of coupled angular momenta. The ﬁrst is to think of the systems as being decoupled. I.e. if two systems are decoupled, how do we add their momenta to get the momentum of the combined system?

~ ~ In this scheme, the angular momentum vectors j1 and j2 of the subsystems precess independently about the z-axis. The total momentum

~ ~ ~ J = j1 + j2, is not constant in this scheme. In this scheme, we denote the state of the system by

|j1, m1; j2, m2i .

The second scheme is to treat the subsystems as coupled. Now, the whole system ~ ~ is precessing about the z-axis. More precisely, the angular momenta j1 and j2 of the subsystems are precessing about the total angular momentum

~ ~ ~ J = j1 + j2, whose magnitude J is now constant. The total angular momentum vector J~ is now precessing about the z-axis (while being precessed about by the angular momenta of the 138 Spin and Angular Momentum Coupling

subsystems).

Where before, the projections m1 and m2 of the subsystem angular momenta on the z-axis were constant, in this scheme, they are not constant, but their sum

M = m1 + m2, is constant. In this scheme, we denote the state of the system by

|j1j2; JMi . ~ ~ ~ ~ In this scheme, J is now the effective quantization axis. Squaring both sides of j2 = J−j1, rearrangin, and applying them to a state |J, M; j1, j2i, we find that the state |J, M; j1, j2i has definite projections 1 ~j · J~ = [J(J + 1) + j (j + 1) − j (j + 1)] , 1 2 1 1 2 2 ~ ~ and similarly for projections j2 · J. Both the coupled and decoupled schemes span the entire space—they are simply two different representations of the same system. Suppose we have two systems with angular ~ ~ momenta (spin, orbital, or both) j1 and j2. What are the possible total angular momenta ~ ~ ~ J = j1 + j2? We know that there are

j1+j2 X (2J + 1) = (2j1 + 1) × (2j2 + 1),

J=|j1−j2| projections (i.e. substates with total angular momentum J). This is the number of states, i.e., the dimension of the system. For example, suppose you have a system composed of two spin-1/2 particles both with no orbital momentum so that j1 = j2 = 1/2. Then the largest possible momentum of the combined system is J = j1 + j2 = 1. Counting down by positive integers, we can also have J = 0, which occurs when the two spins cancel each other. So the total states are X (2J + 1) = 3 + 1 = 4. J=0,1 This tells us that we’ll have four states consisting of a triplet and a singlet. As it must, this matches

(2j1 + 1) × (2j2 + 1) = (2[1/2] + 1) × (2[1/2] + 1) = 4.

If we have two systems |j1, m1i and |j2, m2i and you want to write down the states in the coupled representation |j1, j2, J, Mi. One way to approach this is to draw a table 9.2. Angular Momentum Coupling 139

with 2j1 + 1 columns numbered by m1 which ranges −j1 ≤ m1 ≤ j1, and 2j2 + 1 rows numbered by m2 which ranges −j2 ≤ m2 ≤ j2. We will assume here that j1 ≥ j2.

The z-components of angular momentum add directly, so

M = m1 + m2.

So each square of our table can be labeled with a deﬁnite value of M. A given state in the representation |j1, j2, J, Mi has a deﬁnite value of M. So does each diagonal of our table. So each state |j1, j2, J, Mi is a superposition of the states in a diagonal of the table. We start with the upper right corner of the table, which has M = j1 + j2, which is the maximum possible total projection. There is only one state, which can have M = j1 + j2, and that’s the state with

J = Mmax = j1 + j2.

This corner piece is already a diagonal of the table, so we have our ﬁrst state in the coupled representation |j1, j2; JMi = |j1 + j2, j1 + j2i . If we were adding two spin-1/2 particles, we might write this as

|1/2, 1/2; 11i = |↑↑i .

Then we move down to the diagonal with

M = Jmax − 1 = j1 + j2 − 1.

To do this, we apply the lowering operator to the previously found state. 140 Spin and Angular Momentum Coupling

The decoupled and coupled schemes are related via X |j m , j m i = CJM |j j ; JMi, 1 1 2 2 j1m1,j2m2 1 2 J,M

where CJM = hj j ; JM|j m , j m i, j1m1,j2m2 1 2 1 1 2 2 are the Clebsch-Gordon coefficients. These coefficients can have arbitrary phases attached. We will use the Condon-Shortley convention which uses no phases. Since the coefficients have no complex parts, we can also write

CJM = hj m , j m |j j ; JMi. j1m1,j2m2 1 1 2 2 1 2 The completeness relations for the Clebsch-Gordon coeﬃcients are

0 0 X JM J 0M 0 hJM|J M i = C C = δ 0 δ 0 , j1m1,j2m2 j1m1,j2m2 JJ MM m1,m2

and 0 0 X JM JM hJM|J M i = C C 0 0 = δm m0 δm m0 . j1m1,j2m2 j1m1,j2m2 1 1 2 2 J,M Going back to our table, consider the top-right corner:

From the box labelled A, we get

|j m , j m i = |j j , j j i = CJM |j j ; JMi = |j j ; j + j j + j i, 1 1 2 2 1 1 2 2 j1j1j2j2 1 2 1 2 1 2 1 2

since m = j , m = j , J = M = m + m = j + j and Cj1+j2 j1+j2 = 1. On the left, 1 1 2 2 1 2 1 2 j1j1j2j2 we have the state in the decoupled scheme, and on the right, we have the state in the coupled scheme. Suppose we want the coupled state with M − 1 z-component. To get that, we apply a lowering operator to the state above. First we apply the overall lowering operator Jˆ− to the coupled state. p Jˆ−|j1j2; j1 + j2 j1 + j2i = (J + M)(J − M + 1)|j1j2; j1 + j2 j1 + j2 − 1i p = 2(j1 + j2)|j1j2; j1 + j2 j1 + j2 − 1i.

We should get the same answer if we apply the individual lowering operators to the decoupled state since Jˆ− = ˆj1− + ˆj2−. With these, we get p ˆj1− + ˆj2− |j1j1, j2j2i = (j1 + j1)(j1 − j1 + 1)|j1j1 − 1, j2j2i p + (j2 + j2)(j2 − j2 + 1)|j1j1, j2j2 − 1i p p = 2j1|j1j1 − 1, j2j2i + 2j2|j1j1, j2j2 − 1i. 9.2. Angular Momentum Coupling 141

Equating the results from the diﬀerent lowering operators gives us the state we were looking. s s j1 j2 |j1j2; j1 + j2 j1 + j2 − 1i = |j1j1 − 1, j2j2i + |j1j1, j2j2 − 1i. j1 + j2 j1 + j2

The left-hand side gives the state in the coupled scheme, and the right-hand side gives the state in the decoupled scheme. In terms of our table, the right-hand side is the sum of the states in cells B and C. This equation also implies the two Clebsch-Gordon coeﬃcients s s j j Cj1+j2 j1+j2−1 = 1 ,Cj1+j2 j1+j2−1 = 2 . j1j1−1 j2j2 j1j1 j2j2−1 j1 + j2 j1 + j2

The state we obtained via the lowering operators is a symmetric state. There is one more state in the multiplet—an antisymmetric state. To obtain it, we just write down the state that is orthogonal: s s j2 j1 |j1j2; j1 + j2 − 1 j1 + j2 − 1i = |j1j1, j2j2 − 1i − |j1j1 − 1, j2j2i. j1 + j2 j1 + j2

We could also have the minus sign on the other term. It doesn’t change the physics. Example 9.2.1

Suppose we want to add two spin-1/2 particles, i.e., two particles with s1 = s2 = 1/2. The total number of states is

(2s1 + 1) × (2s2 + 1) = 4, so we want a 2 × 2 table.

1 1 m1 = − 2 m1 = + 2

1 ↓↑ ↑↑ m2 = + 2

1 ↓↓ ↑↓ m2 = − 2

In the decoupled basis, the top-right corner of the table gives us the state, in the decoupled basis 1 1 1 1 |s1m1, s2m2i = , ≡ |↑↑i . 2 2 2 2 So for our ﬁrst state, we have 1 1 ; 11 = |↑↑i , 2 2

where the left-hand side is written in the coupled representation |s1s2; SMi, and the right-hand side is written in the decoupled representation. This is a symmetric state, meaning that exchanging the two particles doesn’t change anything. We can obtain the other two symmetric states by applying the lowering operator to get 1 1 |↓↑i + |↑↓i ; 10 = √ 2 2 2 1 1 ; 1 − 1 = |↓↓i . 2 2 142 Spin and Angular Momentum Coupling

The remaining state is the antisymmetric state 1 1 |↓↑i − |↑↓i ; 00 = √ . 2 2 2

We can prove that this state has S = 0 by applying Sˆ+, Sˆ−, and Sˆz and ﬁnding that all three are zero.

An alternative to the Clebsch-Gordon coeﬃcients, is the Wigner 3j-symbol. It is related to the Clebsch-Gordon coeﬃcient via   j1 j2 j3 (−1)j1−j2−m3   = √ Cj3−m3 .   j1m1 j2m2   2j3 + 1 m1 m2 m3

Alternatively, we can write this as  

j1 j2 J j1−j2+M   (−1) JM   = √ C .   2J + 1 j1m1 j2m2 m1 m2 −M

Properties include m1 + m2 − M = 0, the triangular inequality |j1 − j2| ≤ J ≤ j1 + j2, and the symmetries       j j J j J j J j j  1 2   2 1   1 2    =   =         m1 m2 −M m2 −M m1 −M m1 m2   j1 j2 J j +j +j   = (−1) 1 2 3     −m1 −m2 M

9.3 Atomic Splitting

For external magnetic ﬁeld B~ , recall that the interaction energy of magnetic moment ~ with the ﬁeld is −~µB · B. The orbital magnetic moment of an electron is e ~µ = g `,ˆ g = ~ . ` ` ` 2mc The spin magnetic moment of an electron is

e α ~µ = g S,ˆ g = ~ 1 + + ··· . s s s mc 2π Assuming minimal coupling, then e ~p → ~p − A~(~r). c 9.3. Atomic Splitting 143

Then for spin-1/2 particles, recalling that (~a · ~σ)2 = ~a2, we can write the kinetic energy for spin as

~p 2 (~σ · ~p)2 Kˆ = → 2m 2m 2 e ~ ~σ · ~p − c A = 2m σ σ e e = i j pˆ − Aˆ pˆ − Aˆ 2m i c i j c j 1 e e = (δ + iε σ ) pˆ − Aˆ pˆ − Aˆ 2m ij ijk k i c i j c j i e e = ε σ pˆ − Aˆ pˆ − Aˆ . 2m ijk k i c i j c j

The εijk is antisymmetric, so only the cross terms survive in the product. ie Kˆ = − ε σ pˆ Aˆ + Aˆ pˆ 2mc ijk k i j i j e~ ∂Aj = − εijkσk . 2mc ∂xi But ∂Aj εijkσk = Bk, ∂xi and σk/2 → S~, so e Kˆ = − ~ S~ · B~ = −g S~ · B~ . mc s For an electron in a magnetic ﬁeld B~ aligned in the z-direction, recall that the Landau levels are 1 |e|B E = ω n + , ω = . n ~ c 2 c mc Now we need to add the spin contribution to the energy. The total energy of a Landau level is now 1 e 1 e E = ω n + − ~ σ B = ω n + ∓ ~ B. n ~ c 2 mc z ~ c 2 2mc We write this as 1 E = ω n + ∓ ω B. n ~ c 2 larmor where the Larmor frequency is |e| 1 ω = ~ = ω . larmor 2mc 2 c We see that every Landau level is split in half by the spin contribution. 144 Spin and Angular Momentum Coupling

Since the split levels overlap, every Landau level (except the ground state) is now degenerate with two levels with Sz = ±1/2. ˆ If we have a system of particles with individual orbital angular momenta ì and spin sî, then the total orbital angular momentum of the system is X ˆ Lˆ = ì, i the total spin of the system is X Sˆ = sî, i and the total angular momentum of the coupled system is

J~ = L~ + S~.

This term is split into fine structure. In the states |J, M; j1, j2i each J has 2J + 1 projections. I.e. for a given system with a specified J, there are a total of 2J + 1 possible substates it can be in. If the system is in a magnetic field then there is further splitting into subsubstates. Generally, if J~1 + J~2 = J~3, then there are 2J˜ + 1 number of allowed J3, where J˜ = min(J1,J2). For example, for L~ + S~ = J~, there are 2L + 1 components if L < S. Otherwise, if S < L, then there are 2S + 1 possible values.

Given J, L, S for an atomic state, the standard spectroscopic notation is to denote the state by (2S+1) LJ , where L is replaced by the corresponding s, p, d, . . .. For every electron, the magnetic moment is speciﬁed by g` and gs X ~µ = g`~`i = g`L~ . i More generally, ~µ = g`L~ + gsS~. For a coupled system, we can write ~µ = gJ~, 9.4. Wigner-Eckart Theorem 145

where the Lande factor g is some function of g` and gs. We ﬁnd that

~µ · J~ = gJ~ · J~ = gJ(J + 1) = g`L~ · J~ + gsS~ · J~.

We know that L~ 2 = J~ 2 + S~ 2 − 2 J~ · S~ , which implies L(L + 1) = J(J + 1) + S(S + 1) − 2 J~ · S~ .

Similarly, we can square S~ = J~ − L~ and solve for L~ · J~. Then g g ~µ · J~ = s [J(J + 1) + S(S + 1) − L(L + 1)] + ` [J(J + 1) + L(L + 1) − S(S + 1)] . 2 2 Therefore, g + g g − g L(L + 1) − S(S + 1) g = ` s + ` s . 2 2 J(J + 1)

For a magnetic ﬁeld B~ = Bzˆ, we have −~µ · B~ = −µzB, then ~µ = gJ~ becomes

µz = gM.

This is for a weak magnetic ﬁeld. It gets more complicated for a strong magnetic ﬁeld. The value listed in tables is conventionally g times J = Mmax. I.e, the value listed in a table for µ will be µtable = gJ. For example, the magnetic moment of an electron is listed in tables as g e µ = g s = s = g = ~ . ele s 2 ` 2mc

Nuclear spin, I~, can also contribute. In that case, the total angular momentum of the entire system consisting of the electrons and the nucleus is

F~ = J~ + I~ = L~ + S~ + I~.

Including nuclear spin gives us the hyperﬁne structure of hydrogen. For example, for hydrogen in its ground state, the electron has orbital angular momentum ~` = 0, and spin se = 1/2, for total value J = 1/2. The proton has spin sp = 1/2 for total nuclear spin of I = 1/2. This implies that the total angular momentum of the atom can be F = 1 or F = 0. Thus the ground state splits as shown below:

This splitting gives us the famous 21cm. line.

9.4 Wigner-Eckart Theorem

In general, for a function |ψJMai, where J and M are the angular momentum quantum numbers, and a encompasses any other quantum numbers, we can write

D E 0 0 0 0 0 0 ψJMa ψJ M a = δJJ δMM f(a, a ), 146 Spin and Angular Momentum Coupling

where f is some function of a and a0. We can write this more compactly as

0 0 0 0 hJMa|J M a i = δJJ 0 δMM 0 f(a, a ). We want to look at the matrix elements

0 0 0 hJMa|Tˆλµ|J M a i,

of a generic tensor Tˆλµ, where λ is the angular momentum and µ is the z-component. Note that,

Scalar T : λ = 0, µ = 0 Vector T : λ = 1, µ = −1, 0, 1 Tensor T : λ = 2, µ = −2, −1, 0, 1, 2.

If we couple two states, one with JM and the other with 00, then our table is a single line:

Then JM CJM 00 = 1. A state |00i is a scalar function. A scalar operator acting on any system does not change its rotational properties. For a scalar operator Sˆ,

X 0 Sˆ |JMai = Xa0 |JMa i , a0

where the Xa0 are coeﬃcients. Then

0 0 0 0 hJ M a | Sˆ |JMai = δJJ 0 δMM 0 hJMa |Sˆ|JMai . If you rotate the system, M changes, but Sˆ doesn’t know anything about M. Therefore, the braket on the right cannot depend on M. An example of a vector operator is

r 3 Y (θ, φ) = n , 1m 4π m where m = −1, 0, 1, with ( ∓ √1 (n ± in ) if m = ±1 2 x y nm = . nz if m = 0

In general, we can write a vector V~ in spherical components ( ∓ √1 (V ± iV ) if m = ±1 2 x y Vˆm = Vz if m = 0

∝ Y1m.

For a generic tensor operator Tˆ, we can write its 9 components in the form 1 1 2 1 Tˆ = δ Tr(Tˆ) + Tˆ + Tˆ − δ Tr(Tˆ) + Tˆ − Tˆ . ij 3 ij 2 ij ji 3 ij 2 ij ji 9.5. Electromagnetic Multipoles 147

A scalar operator corresponds to J = 0, a vector operator corresponds to J = 1, and a rank-2 tensor operator corresponds to J = 2. The Wigner-Eckart theorem tells us that

ˆ 0 0 0 JM JM hJMa|Tλµ|J M a i = CJ 0M 0 λµhJMa|ψJM i = CJ 0M 0 λµhJa|ψJ i, where the braket on the right is called the reduced matrix element. In other words, the matrix elements of spherical tensor operators in angular momentum eigenstates can be written as a Clebsch-Gordon coeﬃcient times something that does not depend on the angular momentum orientation M. This is often written in the 3j notation as   J λ J 0 ˆ 0 0 0 J−M   0 0 hJMa|Tλµ|J M a i = (−1)   hJa||Tλ||J a i .   −M µ M 0 where M = µ + M 0. For a vector   J 1 J 0 ˆ J−M 0   ˆ hJM a|V1µ|JMai = (−1)   hJa||V ||Jai .   −M 0 µ M

For example,   J 1 J 0 ˆ J−M 0   ˆ hJM a|Jµ|JMai = (−1)   hJa||J||Jai .   −M 0 µ M

We can divide one by the other to get a proportionality ratio

V~ = vJ~.

Using V~ · J~ = vJˆ2 = vJ(J + 1), we ﬁnd that

V~ · J~ v = . J(J + 1)

So V~ · J~ V~ = J~. J(J + 1) An example of such a relation is ~µ = gJ~, where now, the proportionality constant v is the Lande g factor.

9.5 Electromagnetic Multipoles

We can write the electric potential of an electrostatic system as a sum of multipoles of various orders. I.e., we can write an electrostatic potential in terms of a multipole expansion X 4π 1 φ(~r) = Y ∗ (~n)Mˆ , 2` + 1 r`+1 `m `m `m 148 Spin and Angular Momentum Coupling

where the multipole moments are ˆ X ` M`m = eirî Y`m(~ni). i For example, the monopole moment is the scalar quantity 1 X total charge of the system M00 = √ ei = √ . 4π i 4π For the dipole moment, ` = 1, and it has three components corresponding to m = −1, 0, 1. I.e., it is a vector. r r X 3 3 X M = e r (n ) = (e ~r ) . 1m i i 4π i m 4π i i m i i On the right, the quantity in parentheses corresponds to the dipole operator d~. There is a similar multipole expansion for the magnetic potential. In the following table, we list the first several allowed electric and magnetic multipoles for quantized systems with different values of angular momentum J. The ones with a “+” are allowed, and the ones with a “−” are not allowed by rotational symmetry. For example, M2 denotes the magnetic multipole of order 2, otherwise called the magnetic quadrupole.

J E0 M0 E1 M1 E2 M2 E3 M3

0 + + − − − − − −

1 2 + + + + − − − −

1 + + + + + + − −

3 2 + + + + + + + +

2 + + + + + + + +

If there is parity conservation, then odd electric multipoles are forbidden and even magnetic multipoles are forbidden. Then we get the following table:

J E0 M0 E1 M1 E2 M2 E3 M3

0 + − − − − − − −

1 2 + − − + − − − −

1 + − − + + − − −

3 2 + − − + + − − +

2 + − − + + − − +

0 Now, we apply the Wigner-Eckart theorem for the matrix elements hJMa|Mλµ|JM ai. We will consider time reversal D ~ ~E D E D E d · J JMa|d~|JM 0a = JMa|J~|JM 0a . J(J + 1) 9.5. Electromagnetic Multipoles 149

Then under time reversal, D E D E D E JMa|d~· J~|JMa → J − Ma|d~· (−J~)|J − Ma − JMa|d~· J~|JMa = 0.

So for rotational symmetry and time-reversal symmetry (but not enforcing parity conservation), our table looks like:

J E0 M0 E1 M1 E2 M2 E3 M3

0 + + − − − − − −

1 2 + + − + − − − −

1 + + − + + − − −

3 2 + + − + + − − +

2 + + − + + − − + 150 Spin and Angular Momentum Coupling

9.6 Summary: Spin and Angular Momentum Coupling

Skills to Master • Know the Pauli matrices and their properties • Know the Pauli vector and its properties • Given a spinor, calculate the probability of measuring Sz = 1/2 along the z-direction • Given an arbitrary direction, write down a spinor polarized along that direction • Add the angular momenta of two particles by writing the coupled states in terms of the decoupled states • Calculate the energy level splitting due to spin-orbit coupling for a particle in a magnetic ﬁeld • Calculate the Lande g-factor and the energy level splitting given an atomic term (in spectroscopic notation)

Spin-1/2 The spin operator for spin-1/2 particles is Remember the Pauli matrices ~ 1 " # " # " # S = ~σ. 0 1 0 −i 1 0 2 σx = , σy = , σx = . 1 0 i 0 0 −1 The operator that rotates spin wave functions about the axis ~n by an angle α is The Pauli vector is defined as ~σ = (σx, σy, σz). Useful relations include (~n is a unit vector) ˆ −iα(S~·~n) α α R~n(α) = e = cos − i(~σ · ~n) sin . 2 2 2 2 2 σx = σy = σz = 1 σ σ = δ 1 + iε σ The spin projection operator projecting in the direc- i j ij ijk k tion given by the unit vector ~n is [σi, σj] = 2iεijkσk " # ~ ~ ~ −iφ (~a · ~σ) b · ~σ = ~a · b + i ~a × b · ~σ ˆ ˆ 1 1 cos θ e sin θ S~n = S · ~n = ~σ · ~n = . 2 2 eiφ sin θ − cos θ (~n · ~σ)2 = 1 ~σ × ~σ = 2i~σ For spin-1/2 particles, we can always find a direc- ~σ · (~σ × ~σ) = 6i1. tion in which the particle definitely has spin equal to +1/2. The spinor of a particle polarized such that it When simplifying things involving the Levi-Civita 1 has spin Sz = + along the ~n(θ, φ) direction is symbol and Kronecker delta, it may be helpful to know 2 the identity " # " θ −iφ/2 # a cos 2 e εjkiεjk` = 2δi`. χ~n(θ,φ) = = . b sin θ eiφ/2 For a spin-1/2 particle, the z-component of the 2 spin has two possible values—S = ±1/2, and we de- z This spinor has a definite value of S = + 1 if we mea- note the two eigenstates as up and down z 2 sure along ~n(θ, φ). If, however, we measure along the 1 2 ↑= χ+ (Sz = +1/2), ↓= χ− (Sz = −1/2). z-axis, then we get Sz = + 2 with probability |a| and 1 2 Sz = − with probability |b| . The general spin function, or spinor, is a superposition 2 of these two states " # Angular Momentum Coupling a χ = aχ+ + bχ− ≡ , Given two particles one with angular momentum j b 1 (orbital angular momentum, spin, or the sum of both) which is normalized such that |a|2 + |b|2 = 1. Then and the other with angular momentum j2, then if the the probability of finding the particle with spin up or spins are decoupled, we write the state of the system of down is both particles as

2 2 Prob(↑) = a , Prob(↓) = b . |j1m1, j2m2i . 9.6. Summary: Spin and Angular Momentum Coupling 151

For spin-1/2 particles, we use a shorthand with arrows. For adding two spin-1/2 particles, the relations 1 1 1 1 For example, for |s1m1, s2m2i = | 2 − 2 , 2 2 i, we would between the coupled and decoupled representation are: use the shorthand |↓↑i. 1 1 If the spins are interacting, then we use the coupled ; 11 = |↑↑i 2 2 representation 1 1 |↓↑i + |↑↓i |j1j2; JMi , ; 10 = √ 2 2 2 where J is the total momentum of the two particles, 1 1 ; 1 − 1 = |↓↓i and M is the z-component of the total momentum. 2 2 The possible values of the total momentum of the cou- 1 1 |↓↑i − |↑↓i pled system are given by |j1 − j2| ≤ J ≤ j1 + j2. For ; 00 = √ . 2 2 2 example, for two spin-1/2 particles, j1 = j2 = 1/2, so J = 0, 1. In the coupled representation, for a state Atomic Splitting with a given J, M = m1 + m2 has possible values M = −J, . . . J. For example, for two spin-1/2 parti- An electron in a magnetic field will undergo har- cles, there are three states with J = 1. These three monic oscillation—producing the Landau levels. Be- states are differentiated by M = −1, 0, 1. yond that, the magnetic moment of an electron will A typical task is, given two particles with speci- also interact with the magnetic field—producing fur- fied j1 and j2, to write down the possible states in the ther splitting of the energy levels. coupled representation in terms of the states in the de- In general, the energy of the interaction between a coupled representation. The number of possible states, magnetic moment produced by a charged particle with i.e., the dimension of the system is angular momentum and magnetic field is ~ j1+j2 −~µ · B. X (2J + 1) = (2j1 + 1) × (2j2 + 1). The orbital magnetic moment for an electron is J=|j1−j2| ˆ e~ ~µ = g``, g` = . This is the number of states in both the coupled and ` 2mc decoupled representations, and the task is to relate the The spin magnetic moment of an electron is states in the different representations. ˆ 1. Determine the dimension of the system—create ~µs = gsS, gs ' 2g`. a table. For example, for an electron in a magnetic field, the 2. List the possible states in the decoupled repre- spin contribution to the energy is −gsS~ · B~ . Then the sentation corresponding to each cell of the table total energy of the electron due to the magnetic field 3. Start with the top-right corner of the table. This (Landau + spin contribution) is is the state with maximum z-projection for both 1 e~ 1 1 particles, so m1 = j1 and m2 = j2. Then E = ω n + ∓ B = ω n + ∓ ω , n ~ c 2 2mc ~ c 2 2~ c M = m1 + m2 = j1 + j2. Then the only state in the coupled representation for which this is pos- where the cyclotron frequency is sible is the state with J = M = j + j , so 1 2 |e|B ω = . c mc |j1j2, j1 + j2 j1 + j2i = |j1j1, j2j2i . So the spin contribution splits every Landau level in For example, if our particles are a pair of spin-1/2 half with a degeneracy of two. 1 1 1 1 1 1 For a coupled system of particles with angular mo- particles, this is the state | 2 2 ; 11i = | 2 2 , 2 2 i ≡ |↑↑i. mentum (e.g. two electrons in a helium atom), or even 4. Obtain the rest of the states in the same mul- for a single particle with both spin and orbital momen- tiplet by applying the lowering operator to the tum, to calculate the energy contribution −~µ · B~ we state found above (and then also to each new would first calculate the total magnetic moment state). This will yield the rest of the states with ~µ = gJ~ = g`L~ + gsS~, total momentum J = j1 + j2. 5. To get the remaining states, you have to con- where struct other linear combinations of the decoupled g + g g − g L(L + 1) − S(S + 1) g = ` s + ` s , states. 2 2 J(J + 1) 152 Spin and Angular Momentum Coupling is the Lande g-factor. where L is replaced by the corresponding s, p, d, . . .. To (2S+1) Given J, L, S for an atomic state, the standard find the splitting of a given atomic term LJ de- spectroscopic notation is to denote the state by duce S, L, and J and calculate the Lande g-factor, and (2S+1) then E = −~µ · B~ = gJB. LJ , Chapter 10

Many-body Quantum Mechanics

10.1 Identical Particles

In classical mechanics, the energy of a system of identical particles is just the sum of the individual kinetic energies and potential energies of their interactions

2 X ~pa X + Uab. 2ma a a6=b

Dealing with identical particles in classical mechanics is straightforward and makes sense. However, in quantum mechanics, systems of identical particles are more tricky to deal with due to interference and the quantum statistics of identical particles. For example, in classical mechanics, the scattering of two particles a and b from each other, shown in the image below, makes perfect sense.

However, in quantum mechanics, this scattering is indistinguishable from the scattering shown below:

In fact, in quantum mechanics, all we know is that two particles a and b went into a scattering region, and two particles came out, but we don’t know which is which. 154 Many-body Quantum Mechanics

So when calculating the scattering probability in quantum mechanics, we have to consider both cases. Suppose the amplitude for the ﬁrst kind of scattering is A1 and the amplitude for the second kind is A2. Then the total scattering amplitude is the sum of the two amplitudes A1 + A2. Remember, in quantum mechanics, it’s the probability amplitudes that add rather than the probabilities themselves.

10.2 Gibbs’ Paradox

Consider two identical systems (same particle number N, and same temperature, pressure, and volume) adjacent to each other and separated by an impermeable wall. Now remove the wall between the two systems. Since the systems were identical with the wall in place, removing the wall does not change the temperature or the pressure of the system. Now we have a single combined system with temperature and pressure T and P , but now with volume 2V and 2N particles.

What happens to the entropy when we do this? If the particles are identical, there is no change in entropy. However, if the particles are different, then the change in entropy per particle is ∆S = ln 2. N Now suppose we make the particles less and less different. Then as long as there remains a difference, no matter how small, the entropy change per particle due to the mixing is ln 2. As soon as the difference between the particles becomes zero, the entropy change becomes zero. So if we continuously vary the difference between the particles, the entropy change discountinuously changes from ln 2 to 0 as the difference between the particles becomes zero. It was suggested that quantum mechanics resolves this paradox since there is no continuous parameter that distinguishes particles—only discrete quantum numbers. How- ever, this is wrong. As we saw earlier, there do exist such continuously variable parameters. One example is the spin polarization of the particles. It turns out that varying a continuous parameter like spin polarization makes the paradox disappear. I.e., the entropy per particle is no longer discontinuous. If the particles of the two systems are or- thogonally polarized, then ∆S/N = ln 2 when they are mixed. If the particles of the two systems have the same polarization, then ∆S/N = 0 when they are mixed. Now, there is 10.3. Transposition Operator 155 a continuous transition from ln 2 to 0 as the spin polarization is varied from orthogonal to parallel.

10.3 Transposition Operator

For a system with many particles, the Hamiltonian is now a function of many variables Hˆ = Hˆ (a, b, c, . . .). If the particles are identical, we can swap two particles, and the Hamiltonian remains the same. For example, Hˆ (a, b, c, . . .) = Hˆ (b, a, c, . . .).

We deﬁne the transposition operator Pˆab which swaps particles a and b

Pˆabψ(a, b) = ψ(b, a).

If the particles are identical, then Pˆabψ = ψ, so h i Pˆab, Hˆ = 0.

There are many possible transposition operators Pˆ12, Pˆ23, Pˆ13,... Interestingly, they don’t commute with each other. Suppose we have a wave function of three particles ψ(a, b, c). Then Pˆ12Pˆ23ψ(a, b, c) = Pˆ12ψ(a, c, b) = ψ(c, a, b). However, if we switch the order of the transposition operators, we get

Pˆ23Pˆ12ψ(a, b, c) = Pˆ23ψ(b, a, c) = ψ(b, c, a). The ﬁnal results are not the same. Since the order of the Pˆ matters, we know they don’t commute. Notice that Pˆ12Pˆ23 = Pˆ13Pˆ23. This is the “algebra” of these permutations. Suppose you have a wave function ψ(a, b, c) which is an eigenfunction of all Pˆab. I.e., all of the possible transposition operators for this system have simultaneous eigenvalues. This is the case for quantum particles.

Pˆ12ψ = λ12ψ

Pˆ23ψ = λ23ψ

Pˆ13ψ = λ13ψ. Then by the relation we found earlier,

λ12λ23 = λ13λ23.

For any transposition operator Pâb, ˆ2 Pab = 1. Thus, in our case, it must be that all the eigenvalues λ are +1 or −1. Then the relation we found implies that λ12 = λ13 = λ23. That is, the eigenvalues are all the same—either all +1 or all −1. There are only two classes of allowed wave functions of identical particles. The wave functions are either completely symmetric or completely antisymmetric. Quantum field theory shows that all particles with integer spin have wave functions that are completely symmetric. These are called bosons. All particles with half-integer spin have wave functions that are completely antisymmetric. These are called fermions. This fact, called the spin-statistics theorem is a result of the quantization of the field. 156 Many-body Quantum Mechanics

10.4 Bosons and Fermions

Consider a gas of particles in a box. Recall the de Broglie wavelength λ ∼ ~/p. If the energy of a particle is ε ∼ p2/m, then √ √ p ∼ mε ∼ mT , and λ ∼ √~ . mT As the temperature T of the gas goes down, the de Broglie wavelength increases. Eventu- ally, the wave packets corresponding to individual particles will overlap with each other. At this point, they interact and we need to take the spin of the particles into account, i.e., we have to use the appropriate (Bose or Fermi) statistics. If we have N particles, then the number density is N 1 n = ∼ 3 , V r0 −1/3 where λ ∼ r0. Thus, r0 ∼ n , and so

√~ ∼ n−1/3. mT This gives us the degeneracy temperature 2n2/3 T ∼ ~ , deg m

below which we need to use quantum statistics. For solid materials, like metals, Tdeg is so high that in all practical circumstances, we need to treat them with quantum statistics. For neutral atoms (same number of electrons and protons), if the nucleus has an even number of neutrons, the atom is a boson. If it has an odd number of neutrons, it is a fermion. For example, 4He is a bosonic gas. It becomes liquid at ∼ 4 K and a superﬂuid at ∼ 2.2 K. On the other hand, 3He is a fermionic gas and behaves diﬀerently. Consider a gas of bosons. At low temperatures, all particles want to be in the lowest energy state, resulting in what is called a Bose condensate. When that happens, the overall wave function of the system has the form

ψ0(a)ψ0(b) ··· ψ0(last). This assumes the particles don’t interact with each other. The fraction of particles of a real 4He gas, which are in the condensate (i.e. in the ground state with zero momentum) is ∼ 6-10%. Consider a two-particle system (with particles a and b) containing two single-particle states 1 and 2. Then the possible states of the overall system are:

ψ1(a)ψ1(b)

ψ1(a)ψ2(b)

ψ2(a)ψ1(b)

ψ2(a)ψ2(b) Notice that the ﬁrst and last states are symmetric. The other two are neither symmetric nor antisymmetric, so we must construct symmetric or antisymmetric states from those two by linear combinations: ψ (a)ψ (b) + ψ (b)ψ (a) 1 2 √ 1 2 2 ψ (a)ψ (b) − ψ (b)ψ (a) 1 2 √ 1 2 2 10.4. Bosons and Fermions 157

Here, the first is symmetric, and the second is antisymmetric. Of these overall states, only the antisymmetric state can be used by fermions. The three symmetric states can be used by bosons. This difference in the number of available states is the cause of the differences between bosons and fermions in nature. Note, the antisymmetric state is often written as the Slater determinant

ψ (a)ψ (b) − ψ (b)ψ (a) 1 ψ (a) ψ (b) 1 2 1 2 1 1 √ = √ . 2 2 ψ2(a) ψ2(b) If we have a system of N particles, then the Slater determinant takes the form

ψ (a)ψ (b) ··· ψ (z) 1 1 1

1 ψ2(a)ψ2(b) ··· ψ2(z)

√ . . N! .

ψz(a)ψz(b) ··· ψz(z)

Example 10.4.1: A Pionic Mesoatom

A negatively-charged pion (π−) can be captured by another atom. The pion and the nucleus form a bound state called a mesoatom. Consider the case in which a deuteron (i.e. the nucleus of deuterium) captures a pion to form a pionic mesoatom with ` = 0, which results in nuclear dissociation into two neutrons π− + d −→ n + n. The product—two neutrons—is a pair of identical fermions. Therefore, the overall wave function of the product is required to be antisymmetric. The pion has spin 0 and the deuteron has spin 1, so the total angular momentum of the initial system is J = 1. By conservation of angular momentum, the ﬁnal system must also have J = 1. The orbital angular momentum of the 2-neutron system can have values `2n = 0, 1, 2,.... The total spin of the 2-neutron system must be S = 0 or S = 1 since both neutrons are spin-1/2 particles. The overall wave function of the 2-neutron system,has the form

Y`m(~r)χspin(s1, s2), where ~r is the relative coordinate. The parity of the 2-neutron state is (−1)`2n , which implies that the states with `2n even have positive parity (and thus symmetric spatial wave functions) and the state with `2n odd have negative parity (and thus antisymmetric spatial wave function). When adding the spins of two spin-1/2 fermions, we know that S = 0 corresponds to an antisymmetric spin wave function, and S = 1 corresponds to the triplet of symmetric spin wave functions. We have to combine the spatial and spin parts properly so that the ﬁnal overall wave function is antisymmetric. Thus, we have to combine the symmetric spatial wave functions (`2n = 0, 2) with the antisymmetric spin wave function (S = 0) and the antisymmetric spatial wave function (`2n = 1) with the symmetric spin wave functions (S = 1). However, we can’t combine `2n = 0, 2 with S = 0, since there is no way to get the required total angular momentum J = 1. The only possible state is then `2n = 1 with S = 1. This state has negative parity. The deuteron, it turns out, has positive parity, so parity conservation implies that the pion must have negative parity. 158 Many-body Quantum Mechanics

For two particles, the symmetry for coordinates, i.e., the spatial part of the wave Tip function, is just the parity. That is,

The symmetry of the spa- even parity =⇒ symmetric spatial part tial part of the wave function is just the parity. odd parity =⇒ antisymmetric spatial part.

Example 10.4.2: The Hydrogen Gas Molecule Tip In the H2 molecule, we have two protons and two electrons, where the protons (separated by some distance a) are covalently bonded to each other by the Remember, for two parti- electrons. The H system has two rotational degrees of freedom and a single cles, parity is equivalent to 2 vibrational degree of freedom. spatial inversion. This is The wave function of the 2-electron system, should be such that the protons not generally true for wave are bound as tightly as possible. The Pauli exclusion principle implies that the functions of more than two electrons cannot have the same wave function. They have the same orbital wave particles. function, but with opposite spin. We know that the energy of the electrons, the vibrational energy of the protons, and the rotational energy of the system are given by

2 ε ∼ ~ ele ma2 r m ε ∼ ε vib ele M m ε ∼ ε . rot ele M For the 2-electron wave function, we know the it must be overall antisymmetric, since they are fermions. There is inversion (~r → −~r) symmetry which implies parity conservation. Recall that parity is given by (−1)`. If ` = 0, the system is symmetric in space, and so it requires antisymmetric spin wave function. There are two diﬀerent allotropes of hydrogen gas. The parahydrogen allotrope occurs when the spatial wave function is symmetric with ` = 0, 2, 4,..., and the spin part of the wave function is antisymmetric with S = 0. The orthohydrogen allotrope occurs when the spatial wave function is antisymmetric with ` = 1, 3, 5,..., and the spin part of the wave function is symmetric with S = 1. At high temperatures, many rotational degrees of freedom are excited, and the orthohydrogen allotrope dominates. This is especially true since the lowest orthohydrogen allotrope (S = 1) has three projections (Sz = −1, 0, 1) versus the single projection (Sz = 0) for the lowest state (S = 0) of the parahydrogen allotrope. So the orthohydrogen allotrope dominates simply by virtue of it having more available states provided that the temperature of the gas is suﬃciently high. At room temperature, the most common allotrope is parahydrogen. The orthohydrogen states become excited when

2`(` + 1) T ∼ ε = ~ . rot 2ma2

10.5 Second Quantization

Recall that Hilbert space is the space of all possible states for a ﬁxed number of particles. Fock space is a generalization in that the number of particles N is now just another variable {|ni} −→ {|n, Ni} . 10.5. Second Quantization 159

In the “second quantization” of quantum mechanics, we use

aˆ† ← creation operator aˆ ← annihilation operator Nˆ =a ˆ†aˆ ← number operator |0i ← vacuum state, where

aˆ |0i = 0 n aˆ† |ni = √ |0i . n! Previously, we used this formalism for the harmonic oscillator. Now we generalize it to all systems. Recall that the algebra of the system is determined by the commutator

a,ˆ aˆ† = 1.

This is all we need if we have a single kind of particle. But we don’t. We have two kinds of particles—bosons and fermions.

Bosons Let |λ) be the state of a single particle. The parameter λ encodes all the quantum numbers for a single particle. Now, we write our creation and annihilation operators as

† aˆλ, aˆλ.

The algebra is now h † i aˆλ, aˆλ0 = δλλ0 , and the number operator is ˆ † Nλ =a ˆλaˆλ. The general wave function of a multi-particle system is now

nλ aˆ† Y λ |ψi = √ |0i . n ! λ λ

This only works for bosons where, nλ = 0, 1, 2,..., i.e. the number of particles in state λ, can be any positive number.

Fermions For fermions, we again have |λ), but now due to the Pauli exclusion princple, we can only have nλ = 0, 1. Fermi statistics disallows nλ > 1 particles in the same state. Since we cannot generate more than one particle of the same kind in the same state, this implies that aˆ†aˆ† ≡ 0. So we have a strange algebra where products of two or more of the same creation operators are zero. This is called Grassman algebra. Likewise,

aˆaˆ ≡ 0. 160 Many-body Quantum Mechanics

Now, instead of commutators, we have anticommutators

h † i † † aˆλ, aˆλ0 =a ˆλaˆλ0 +a ˆλ0 aˆλ = δλλ0 . +

The full algebra is deﬁned by the three anticommutators

h † i aˆλ, aˆλ0 = δλλ0 + h † † i aˆλ, aˆλ0 = 0 +

[ˆaλ, aˆλ0 ]+ = 0.

For fermions, the general wave function takes the form

n Y † λ |ψi = aˆλ |0i , λ

since nλ can only be 0 or 1. Example 10.5.1

Simplify the commutator h † †i aˆ1, aˆ2aˆ3 , keeping in mind that the behavior is diﬀerent for bosons and fermions. For bosons, we know that the creation and annihilation operators for diﬀerent states λ and λ0 have the commutation relation

h † i † † aˆλ, aˆλ0 =a ˆλaˆλ0 − aˆλ0 aˆλ = δλλ0 .

This implies that

† † aˆλaˆλ0 = δλλ0 +a ˆλ0 aˆλ † † aˆλ0 aˆλ = −δλλ0 +a ˆλaˆλ0 . Thus, for bosons,

h † †i † † † † aˆ1, aˆ2aˆ3 =a ˆ1aˆ2aˆ3 − aˆ2aˆ3aˆ1

† † † † = δ12 +a ˆ2aˆ1 aˆ3 − aˆ2 −δ13 +a ˆ1aˆ3 † † † † † † = δ12aˆ3 +a ˆ2aˆ1aˆ3 +a ˆ2δ13 − aˆ2aˆ1aˆ3 † † = δ12aˆ3 + δ13aˆ2. For fermions, we have the anticommutation relation

h † i † † aˆλ, aˆλ0 =a ˆλaˆλ0 +a ˆλ0 aˆλ = δλλ0 . + This implies that

† † aˆλaˆλ0 = δλλ0 − aˆλ0 aˆλ † † aˆλ0 aˆλ = δλλ0 − aˆλaˆλ0 . 10.6. Nuclear Conﬁgurations 161

Thus, for fermions,

h † †i † † † † aˆ1, aˆ2aˆ3 =a ˆ1aˆ2aˆ3 − aˆ2aˆ3aˆ1

† † † † = δ12 − aˆ2aˆ1 aˆ3 − aˆ2 δ13 − aˆ1aˆ3 † † † † † † = δ12aˆ3 − aˆ2aˆ1aˆ3 − aˆ2δ13 +a ˆ2aˆ1aˆ3 † † = δ12aˆ3 − δ13aˆ2.

When calculating vacuum expectation values like † h0|aˆ1aˆ2|0i , there are two approaches. One can use the commutation (or anticommutation) relation for bosons (or fermions) to rewrite as

† † † Bosons: h0|aˆ1aˆ2|0i = h0|δ12 +a ˆ2aˆ1|0i = h0|δ12|0i + h0|aˆ2aˆ1|0i = δ12 † † † Fermions: h0|aˆ1aˆ2|0i = h0|δ12 − aˆ2aˆ1|0i = h0|δ12|0i − h0|aˆ2aˆ1|0i = δ12 using the fact that any annihilation operator acting on the vacuum state (e.g.a ˆ1 |0i) is zero. If we reorder all operators in this way so that annihilation operators are on the right side of all creation operators in each term in which they appear, we say that the operators are in normal order. The other approach is to note that we have a vacuum state on the right side of the braket and a vacuum state on the left side. So if we leave the operators in the original anti-normal order, then for every creation operator, the only nonzero terms are those which include the corresponding annihilation operator (since the state must be returned to the vacuum state). Thus, we can say right from the beginning that † h0|aˆ1aˆ2|0i = δ12.

10.6 Nuclear Conﬁgurations

0 We have a set of single-particle levels {|λ)}, with orthogonality such that (λ |λ) = δλλ0 . † We also have a particle creation operator aλ. Then our commutation relation is

h † i aλ0 , aλ = δλ0λ, ∓ where the subscript “−” indicates the commutator, and “+” indicates anticommutator. We have some other basis |µ), then we can write

X λ0∗ λ δλλ0 = Cµ0 Cµ . µ The basis we use doesn’t matter, but we should be consistent. We have that † λh0|aλ0 aλ|0i = δλλ0 † µh0|aµ0 aµ|0i = δµµ0 . 162 Many-body Quantum Mechanics

We have some total number Ω of single-particle states. This is ﬁxed in this space. Now we put in N particles. How many diﬀerent ways can those N particles be distributed into the Ω states? For fermions, the Pauli principle allows no more than one particle per single-particle state, thus, we necessarily have the condition that Ω ≥ N. The number of ways to distribute N fermions into Ω states is Ω · (Ω − 1) ··· (Ω − (N − 1)) , N! which can also be written as

Ω! Fermions: . N!(Ω − N)!

For bosons, we ﬁnd that (Ω + N − 1)! Bosons: . N!(Ω − 1)!

This is the number of allowed conﬁgurations for N bosons in Ω states. Example 10.6.1

Suppose we have three spin-1 bosons in a potential box. Find the allowed values of total spin S of the three-particle system. Our goal is to add the three spins together to obtain the possible values of the total spin, while taking boson statistics into account. Since we have N = 3 particles and Ω = 3 possible states for each (since ms = −1, 0, 1), we expect to ﬁnd a total of (Ω − 1 + N)! = 10 (Ω − 1)!N! states. For a single spin-1 boson, we have spin magnitude s = 1 and projections ms = −1, 0, 1. Bosons are not limited by the Pauli exclusion principle, so all three of them can be in the same single-particle state. The maximum spin magnitude, then, occurs when all three spins are aligned. Since

MS = ms,1 + ms,2 + ms,3,

this maximum is MS = 3.

m1 m2 m3 MS

1 1 1 3

Since this is the largest possible MS, it must be the largest projection of a multiplet with spin magnitude S = 3. However, this is only one state, and S = 3 has seven projections MS = −3, −2, −1, 0, 1, 2, 3, therefore, we expect to find six more states to finish filling this multiplet. Now we drop down a projection to MS = 2 by mentally applying the lowering operator to the above state. We find that there are three possibilities: m1m2m3 = 110, 101, 011. However, bosonic statistics implies that these three are equivalent— they just represent permutations of the creation operators. So we ignore two of them. 10.6. Nuclear Configurations 163

m1 m2 m3 MS

1 1 0 2

Recall that our multiplet with S = 3 has a spot for a state with MS = 2. Therefore, this state is part of the previous multiplet, and does not open a new multiplet. So far, we’ve ﬁlled two states of the S = 3 multiplet

(S = 3) : MS = −3, −2, −1, 0, 1, 2, 3 .

Now, we drop down to MS = 1. Now we have the following non-equivalent possibilities:

m1 m2 m3 MS

1 1 −1 1

1 0 0 1

The S = 3 multiplet has an opening for a single state with MS = 1. The other one of these must open a new multiplet with total spin S = 1. So now we have ﬁlled:

(S = 3) : MS = −3, −2, −1, 0, 1, 2, 3

(S = 1) : MS = −1, 0, 1

Next, we drop down to MS = 0. We have the following non-equivalent possibilities:

m1 m2 m3 MS

1 −1 0 0

0 0 0 0

This is two states, which is precisely the number of MS = 0 states we need for the already opened multiplets. Now, we’ve ﬁlled

(S = 3) : MS = −3, −2, −1, 0, 1, 2, 3

(S = 1) : MS = −1, 0, 1

We keep going like this down through MS = −1, −2, −3, to ﬁll the remaining states in these two multiplets

(S = 3) : MS = −3, −2, −1, 0, 1, 2, 3

(S = 1) : MS = −1, 0, 1

No new multiplets will have to be opened since we have now a total of 10 states, which is the number we expected to ﬁnd. So our ﬁnal result for the possible values of the total spin is S = 3 or S = 1. 164 Many-body Quantum Mechanics

10.7 Atomic Conﬁgurations

Quantum Numbers

Each electron in an atom is characterized by four quantum numbers—n, `, m`, and ms. Electron shell: An electron shell, also called a principal energy level is characterized by the quantum number n = 1, 2, 3, ··· . In X-ray notation, these are labeled K, L, M, . . .. Each shell can be thought of as an “orbit” followed by the electrons in that shell. In general, at higher n shells, the electrons are farther from the nucleus. Each electron shell is limited in the number of electrons it can hold. In general, each shell can hold up to 2n2 electrons. Subshell: Each electron shell is composed of one or more subshells. So each electron in some shell is also in one of that shell’s subshells. The diﬀerent subshells are characterized by the angular quantum number, which gives the angular momentum of the electrons in the subshell. The possible values are ` = 0, 1, 2, . . . n − 1. They are often labeled with the letters s, p, d, f, . . .. Where the principal quantum number indicates the overall size of an electron’s orbit, the angular quantum number gives it shape. The s subshell, which has ` = 0, is a spherical orbit. The p subshell, which has ` = 1, is a polar orbit. The d subshell, which has ` = 2, is an orbit shaped kind of like a clover leaf. In general, each subshell can hold a maximum of 2(2` + 1) electrons. The 2` + 1 factor is the number of projections m`. This is multiplied by two for the number of ms values. Atomic Orbital: Each subshell is composed of one of more atomic orbitals. These are characterized by the magnetic quantum number m` which can have the values m` = −`, . . . , `. The magnetic quantum number speciﬁes the orientation of the subshell. For example, the ` = 0 subshell is spherical, and so it has only a single unique orientation m` = 0. The ` = 1 subshell has three orientations— m` = −1, 0, 1. In the context of quantum mechanics, it is helpful to think of the values of m` as the possible projections of the orbital angular momentum `.

Spin: The ﬁnal quantum number is the spin quantum number ms which has possible values ms = ±1/2, indicating whether the electron is spin up or spin down. Ac- cording to the Pauli principle, an atomic orbital (which has a given n, `, and m`) can hold a maximum of two electrons and then only if they have opposite spin.

Electron configuration notation is used to indicate the electron shell and subshell configuration for an atom. For example, hydrogen has a single electron in the n = 1 shell in the ` = 0 subshell corresponding to the letter s, so its electron configuration is 1s1. For neon, the electron configuration is 1s22s22p6.

LS Coupling In light atoms, the spin-orbit coupling is weak. Recall that spin-orbit coupling is when an electron’s spin couples to its own orbital angular momentum. What happens instead, for the valence electrons in light atoms is LS coupling. This is when the individual spins s of the electrons interact with each other, coupling together to form a total spin S X S~ = ~si. i Similarly, the orbital momenta ` of the different electrons tend to couple together to form a total orbital angular momentum X L~ = ~ì. i 10.7. Atomic Configurations 165

The total spin and total orbital angular momentum then couple together to form the total angular momentum J~ = L~ + S~.

Since LS coupling is a good model for light atoms, L and S make good quantum numbers for these atoms. Note, this is only for the valence electrons. We can ignore non-valence electrons, since full shells and subshells do not contribute to the overall momentum. Note: For heavier atoms, a diﬀerent kind of coupling occurs—the jj-coupling where a given electron’s spin s couples to its own orbital angular momentum ` forming a total angular momentum j for that speciﬁc electron. The j’s of all the valence electrons then couple together to form a total momentum J.

Term Symbols Given LS-coupling of the valence electrons in an atom, we often want to know what the possible values are of the total orbital angular momentum L and total spin S of the atom. We need to know that in order to write the allowed atomic configurations in the term symbol notation (2S+1) (L)J . Let’s consider the simple case of the nitrogen atom. We will take it step-by-step. Step 1 (Valence electrons): Nitrogen has 7 electrons. The first two electrons go into the 1s orbital. The next two go into the 2s orbital. That leaves N = 3 valence electrons to partially fill the 2p orbital. Step 2 (Number of states): The 2p orbital corresponds to ` = 1, and it has three projections m` = −1, 0, 1. Each of these can hold two electons (↑↓) without violating the Pauli exclusion principle. So the 2p orbital can hold a maximum of Ω = 6 electrons. Then the number of possible states that can be formed with N = 3 electrons in Ω = 6 states is Ω! = 20. N!(Ω − N)! Our goal is to classify these twenty states. Step 3 (Single electron states): First, let’s enumerate and label the possible single-electron states. The electrons in the 2p orbital can have m` = −1, 0, 1, and they can have spin projection ms = ±1/2. So there are a total of six possible single-particle states. We label all six of them in the form (m`, ms) as follows:

1 1 a := 1, , a0 := 1, − 2 2 1 1 b := 0, , b0 := 0, − 2 2 1 1 c := −1, , c0 := −1, − . 2 2

Step 3 (Multi-electron states): Now we need to combine these single-particle states into three-particle states (since that’s how many valence electrons are in 2p) without violating the Pauli principle. We start with the largest possible orbital angular momentum projection. We do this by adding the m` from three diﬀerent single-particle states. We see that the largest possible value is ML = 2, which comes from adding together both m` = 1 particles and one m` = 0 particle. The possible three-particle states with ML = m`,1 + 0 m`,2 +m`,3 = 2 are the two states listed in the table. Note, (aa b) means the three-electron 0 state containing the electrons a, a , and b. The total spin is MS = ms,1 + ms,2 + ms,3. 166 Many-body Quantum Mechanics

(m`, ms) ML MS

0 1 (aa b) 2 2

0 0 1 (aa b ) 2 − 2

Now, we go down a level, and enumerate the states with ML = 1.

(m`, ms) ML MS

0 1 (abb ) 1 2

0 1 (aa c) 1 2

0 0 1 (a bb ) 1 − 2

0 0 1 (aa c ) 1 − 2

Next, we go down to ML = 0.

(m`, ms) ML MS

3 (abc) 0 2

0 1 (abc ) 0 2

0 1 (ab c) 0 2

0 1 (a bc) 0 2

0 0 1 (ab c ) 0 − 2

0 0 1 (a bc ) 0 − 2

0 0 1 (a b c) 0 − 2

0 0 0 3 (a b c ) 0 − 2

Next, we go down to ML = −1.

(m`, ms) ML MS

0 1 (cbb ) −1 2

0 1 (acc ) −1 2

0 0 1 (c bb ) −1 − 2

0 0 1 (a cc ) −1 − 2

Finally, we drop to the lowest possible total angular momentum projection ML = −2. 10.7. Atomic Conﬁgurations 167

(m`, ms) ML MS

0 1 (bcc ) −2 2

0 0 1 (b cc ) −2 − 2

If we count the number of rows in these five tables, we find that all twenty three-particle states are accounted for. Step 4 (Term Symbols): Now we want to list the different term symbols that are possible. Since we started with the largest possible projection ML = 2, we know this corresponds to the maximum orbital momentum L. That is, we have a state with total orbital momentum L = 2. Since MS = ±1/2 for these two states, we know that 1 S = 2 . Then the possible values of the total angular momentum are J = L+S = 5/2 and J = L − S = 3/2. Thus, our first table implies that we have opened the two multiplets

2 2 D3/2, D5/2. The tricky thing is, that both of these states correspond to multiple of the states we listed 2 2 above. For example, D3/2 denotes a total of 4 states, and D5/2 denotes a total of 6 states. In general, for total momentum J, there are 2J + 1 states. In the next table, we have four states with ML = 1. However, two of these belong to the two multiplets we already opened, since ML = 1 is an allowed projection of L = 2. So we can add two of the states to the previous two multiplets, leaving the other two states to open two new multiplets. These multiplets have maximum projection ML = 1, so they correspond to L = 1. Then, since ms = ±1, we have S = 1/2. Thus, J = L + S = 3/2 or J = L − S = 1/2. In spectroscopic notation, these multiplets are

2 2 P1/2, P3/2.

Next, we move to the table with ML = 0. This table contains eight states. Since ML = 0 is an allowed projection of multiplets with L = 2 and L = 1, four of these states are part of the previously opened multiplets. What do we do with the other four? We open a new multiplet with maximum angular momentum L = 0. The spin projections MS = ±1/2 and MS = ±3/2 are all allowed projections of S = 3/2. Then the total momentum of this multiplet is J = S ± L = 3/2. In spectroscopic notation, the multiplet is 4 S3/2. So far, we have opened the multiplets

4 2 2 2 2 S3/2, P1/2, P3/2, D3/2, D5/2.

(2S+1) Note that for LJ , the total number of states is 2J + 1, since it has projections Jz = −J, . . . J. So if we enumerate all the states in these five multiplets, we get a total of 20, which is precisely the number of states we needed to classify. So we are done. These five multiplets are five different possibilities that we predict for nitrogen.

Hund’s Rules Hund’s rule tells us that the state with maximum S is the ground state. This is due to the Coulomb interaction between the electrons. Thus, we predict the ground state of nitrogen to be 4 S3/2. If there were multiple states with the same maximum spin S, then the ground state is the one of those with the largest L. 168 Many-body Quantum Mechanics

For a partially-ﬁlled subshell (i.e. for a given n and `), the maximum number of electrons that can occupy that subshell is N = 2(2` + 1). If the number of electrons actually occupying the subshell is n, then the total number of available microstates is

N N! Ω = = . n n!(N − n)!

For example, for carbon, we have the electron configuration 1s22s22p2. Since the s subshells can only hold two electrons each, we see that both the 1s and 2s subshells are filled. We ignore filled subshells because they don’t contribute to the overall angular momentum. The 2p subshell can hold a maximum of N = 6 electrons, but for carbon, it has n = 2 electrons. So the total number of electronic microstates available to the atom is Ω = 6!/2!4! = 15. To identify the ground state term symbol for an atom, follow this procedure: 1. Identify the number of valence electrons and the subshell they are occupying. The subshell gives you the quantum number ` for the valence electrons. Ignore full subshells because they do not contribute to the overall angular momentum. 2. For the subshell ` identified above, note the available orbitals m` = −`, . . . , `. Distribute the valence electrons into these orbitals being sure to observe the Pauli principle (each orbital can have no more than two electrons and then only if they have opposite spin). Fill the orbitals going from highest m` to lowest with spin-up (ms = +1/2). Only after all orbitals contain a spin-up electron, go back through (in the same order), and add spin-down electrons to the orbitals. 3. The total spin S is just the sum of all the ms. Hund’s first rule tells us that for the ground state, all unpaired electrons are spin-up, and S is just 1/2 times the number of unpaired electrons in the orbital. Any paired electrons don’t contribute, since their spins cancel. 4. The total orbital angular momentum L is just the sum of all the m`. For example, if there are two electrons in an orbital with m` = −1, then this orbital would contribute −2 to L. 5. The total angular momentum J is calculated as follows: • If the subshell is less than half-filled, then J = |L − S| • If the subshell is exactly half-filled, then J = S (since L = 0 for a half-filled subshell) • If the subshell is more than half-filled, then J = L + S Note, the above procedure is for obtaining S, L, and J for the ground state term. For general terms, it’s more complicated. 10.8. Summary: Many-body Quantum Mechanics 169

10.8 Summary: Many-body Quantum Mechanics

Skills to Master • Taking particle statistics into account, apply angular momentum and parity conservation and symmetry requirements to two-particle systems • Using parity and symmetry requirements, identify characteristics of the diﬀerent allotropes of a system of identical particles • Be able to simplify expressions containing creation/annihilation operators for bosons (commutation relations) and fermions (anticommutation relations) • Given the number of single-particle states Ω, and the number of particles N, calculate the number of allowed conﬁgurations for a system of bosons or fermions

Bosons and Fermions • What are the possible values of the relative orbital angular momentum of the final system? In There are only two classes of allowed wave functions general, ` = 0, 1, 2,... of identical particles. The wave functions are either • What are the possible values of the total spin bosons (integer spin) with completely symmetric wave of the final system? For two spin-1/2 particles, functions or fermions (half-integer spin) with com- S = 0, 1 pletely antisymmetric wave functions. This has phys- • What is the form of the overall wave function ical consequences. For example, for a bosonic gas at (spatial and spin parts) of the final system tak- low temperature all particles want to be in the lowest ing into account whether the particles are bosons energy state, resulting in a Bose condensate. The Pauli or fermions exclusion principle prohibits a fermionic gas from be- • What is the parity of the final state? This is having in this way. Bosons and fermions also tend to typically (−1)`, where ` are the possible val- have a different number of available states. For exam- ues of the relative orbital momentum of the sys- ple, for a two particle spin-1/2 system, there is only one tem. For example, for ` = 1, the wave function antisymmetric state available for fermions, but there must have odd parity, so if the two particles are are three symmetric states available if the particles are fermions and therefore require an antisymmet- bosons. ric wave function, then the spin part must be Note, for neutral atoms (same number of electrons symmetric. For two spin-1/2 particles recall that and protons), if the nucleus has an even number of neu- S = 0 is antisymmetric spin function, and S = 1 trons, the atom is a boson. If it has an odd number of is a triplet of symmetric spin functions. neutrons, it is a fermion. As the temperature of a gas of particles is reduced, at some point the particles begin to interact with each other. I.e., their wave functions begin to overlap. At this point, we need to take the spin of the particles into account, i.e., we have to use the appropriate (Bose or Fermi) statistics. The temperature at which this occurs For two particles, parity is equivalent to spatial 2 2/3 inversion. So the parity of the wave function is equiv- is the degeneracy temperture T ∼ ~ n , where n deg m alent to the symmetry of the spatial part of the wave is the number density N/V . function

Symmetries and Wave Functions When determining the possible ﬁnal states for a two- even parity =⇒ symmetric spatial part particle system, e.g., the product of a nuclear reaction, use symmetries and particle statistics odd parity =⇒ antisymmetric spatial part.

• Total angular momentum J must be conserved before and after the reaction. Given the known initial state, what is the total angular momentum If the system has inversion symmetry (~r → −~r), then of the ﬁnal state? parity is conserved. 170 Many-body Quantum Mechanics

Second Quantization ways to distribute the particles. Suppose you have N spin-s identical particles in In the “second quantization” of quantum mechanics, a box. What are the possible values of the total spin we use S of the system? Alternatively, you have N spin-less aˆ† ← creation operator particles with orbital momentum `. Then what are the possible values of the total L? aˆ ← annihilation operator 1. Calculate the number of possible states using one Nˆ =a ˆ†aˆ ← number operator of the two formulas given above with N being the |0i ← vacuum state, number of particles and Ω being the number of spin projections ms = −s, . . . , s where 2. Start with the state(s) with largest S. For bosons, we can have any number of particles in aˆ |0i = 0 the same state, so the state with largest S will †n aˆ be the one where all particles have ms = +s. |ni = √ |0i . n! The sum of them gives the total projection MS. For fermions, we have to take the Pauli principle For bosons, the algebra is determined by the com- into account. This largest state opens a multiplet mutator with total spin S = MS. h † i aˆλ, aˆλ0 = δλλ0 . 3. Next, mentally apply the lowering operator to the above state to drop down to the states with one Fermions obey Grassman algebra since we cannot lower value of M . How many possible states generate more than one particle of the same kind in S are there with this value of M ? For bosons, the same state, S don’t repeat states which are just permutations aâˆ =a ˆ†aˆ† ≡ 0. of each other. For example, for bosons, the state Their algebra is determined by the anti-commutators (m1m2) = (01) is equivalent to the state (10). 4. Keep repeating this process of dropping down one h † i h † † i aˆλ, aˆλ0 = δλλ0 , aˆλ, aˆλ0 = 0, [âλ, aˆλ0 ]+ = 0. projection and listing all possible states with this + + value of MS. Keep track of the multiplets being When calculating vacuum expectation values like opened and how many projections each one has. † h0|aˆ1aˆ2|0i, there are two approaches. One is to use the For example, if you start by opening the mul- commutation relation (or anticommutation if fermions) tiplet with S = 3, then you know that it has to reorder all operators so the annihilation operators the projections MS = −3, −2, −1, 0, 1, 2, 3, so as are on the right and then use the fact thata î |0i = 0. you drop down in MS, you will keep back-filling We can also leave them in the original order and note these available slots. Only when there are no pre- that for every creation operator, the only nonzero viously opened slots at the given MS left unfilled terms will be those which include the corresponding do you open a new multiplet. annihilation operator since the state must be returned 5. In the end, the number of possible values for the to the vacuum state. So we can immediately write total spin S is equal to the number of multiplets † down h0|aˆ1aˆ2|0i = δ12. you opened.

Counting States Atomic Conﬁgurations Suppose we have Ω single-particle states and N identi- Each electron in an atom is characterized by four quan- cal particles to be distributed into these states. Then tum numbers—n, `, m`, and ms. if the particles are fermions, they must obey the Pauli principle and there are Shell: An electron shell or energy level is characterized by the quantum number n = 1, 2, 3, ··· . In Ω! Fermions: . general, each shell can hold up to 2n2 electrons. N!(Ω − N)! Subshell: Each electron shell is composed of one or number of ways to distribute the particles. If they are more subshells characterized by angular momen- bosons, then there is no Pauli principle, and there are tum quantum numbers ` = 0, 1, 2, . . . n−1. They (Ω + N − 1)! are often labeled with the letters s, p, d, f, . . .. In Bosons: . N!(Ω − 1)! general, each subshell can hold a maximum of 10.8. Summary: Many-body Quantum Mechanics 171

2(2` + 1) electrons. The 2` + 1 factor is the num- the Ω calculated in the previous step. Write down ber of projections m`. This is multiplied by two all possible pairs of m` and ms and label them for the number of ms values. as the single-particle states. 4. Combine the single particle states into the multi- Orbital: Each subshell is composed of one of more electron states. If we have N valence electrons, atomic orbitals characterized by the orbital mo- then each multi-electron state contains N of the mentum projections m` = −`, . . . , `. single-electron states identified in the previous state. Start with the multi-electron state with Spin: The final quantum number is the the electron P the largest possible ML = m`. List all possi- spin ms = ±1/2. According to the Pauli principle, an atomic orbital (which has a given n, `, ble multi-electron states with this ML. 5. Repeat, each time going down one level in ML and m`) can hold a maximum of two electrons and then only if they have opposite spin. and listing all possible multi-electron states with that value. You don’t have to go to negative val- jj-coupling occurs in heavier atoms, and it means ues since we know these will mirror the positive that J makes a good quantum number. It occurs when values. a given valence electron’s spin s couples to its own or- 6. Using the tables of multi-electron states identi- bital angular momentum ` forming a total angular mo- fied in the previous step, identify all the term mentum j for that specific electron. The j’s of all the symbols. Start again with the highest ML since valence electrons then couple together to form a total this corresponds to a term symbol with L = ML. momentum J. Remember that j = ` ± s. Keep adding term symbols until all states are LS coupling occurs in light atoms, and it means accounted for. Remember that a single term (2S+1) that L and S make good quantum numbers for these symbol (L)J groups 2J + 1 different multi- atoms, and we can specify atomic states by term sym- electron states. bols of the form If we only want to identify the term symbol of the (2S+1) (L)J . ground state then follow this procedure: This is when the individual spins s of the valence elec- 1. Identify the number of valence electrons and the trons interact with each other, coupling together to subshell they are occupying. The subshell gives ~ P you the quantum number ` for the valence elec- form a total spin S, S = i ~si, and the orbital momenta ` of the different valence electrons tend to cou- trons. Ignore full subshells because they do not ple together to form a total orbital angular momentum contribute to the overall angular momentum. ~ P ~ 2. For the subshell ` identified above, note the avail- L = i ì. Finally, the total spin and total orbital angular momentum then couple together to form the able orbitals m` = −`, . . . , `. Distribute the va- total angular momentum J~ = L~ + S~. lence electrons into these orbitals being sure to Given an atom, use this procedure to predict the observe the Pauli principle (each orbital can have possible term symbols: no more than two electrons and then only if they 1. Write down the electron configuration and count have opposite spin). Fill the orbitals going from the number of valence electrons. Only the valence highest m` to lowest with spin-up (ms = +1/2). electrons matter, and writing down the electron Only after all orbitals contain a spin-up electron, configuration will tell you which atomic subshell go back through (in the same order), and add these electrons must be placed into. spin-down electrons to the orbitals. 2. Calculate the number of states. What is the ` of 3. The total spin S is just the sum of all the ms. the subshell identified above? Count the number Hund’s first rule tells us that for the ground state, all unpaired electrons are spin-up, and S is just of projections m` = −`, . . . , ` then multiply by two since each orbital (i.e. projection) can hold 1/2 times the number of unpaired electrons in the two electrons. This number is your Ω, and N is orbital. Any paired electrons don’t contribute, the number of valence electrons. Then the total since their spins cancel. number of states is 4. The total orbital angular momentum L is just the sum of all the m`. For example, if there are two Ω! electrons in an orbital with m = −1, then this . ` N!(Ω − N)! orbital would contribute −2 to L. 5. The total angular momentum J is calculated as 3. Enumerate and label the single electron states. follows: Each electron can have m` = −`, . . . , ` and ms = ±1/2. The number of combinations are equal to • If the subshell is less than half-filled, then 172 Many-body Quantum Mechanics

J = |L − S| If we have a list of possible term symbols for an • If the subshell is exactly half-filled, then atom, then Hund’s rule tells us that the one corre- J = S (since L = 0 for a half-filled subshell) sponding to the ground state is the one with the largest • If the subshell is more than half-filled, then S. If there are multiple ones with this value of S, then J = L + S the ground state is the one of those with largest L. Chapter 11

Approximation Methods

11.1 Variational Method

Suppose we want to estimate the ground state energy E0 of a hamiltonian Hˆ with a discrete spectrum. We know the exact Hamiltonian Hˆ , and we know the boundary conditions of the system. We can take an arbitrary function ψ, called a “variational” or “trial” function, which satisﬁes the boundary conditions and then calculate the quantity

hψ|Hˆ |ψi .

Keep in mind that ψ is an arbitrary function—not an eigenfunction of Hˆ , so this quantity is not hEi. We can in principle expand ψ in the actual eigenstates ψn of Hˆ as X ψ = cnψn, n where Hψˆ n = Enψn. Then we can write

We know that En > E0 for all n > 0. That is, all excited state energies are higher than the ground state. So

X 2 X 2 X 2 |cn| En ≥ |cn| E0 = E0 |cn| = E0. n n n Thus, for any ψ,

hψ|Hˆ |ψi ≥ E0.

So to approximate E0, we calculate hψ|Hˆ |ψi with different ψ and try to minimize it. We can try an entire class of arbitrary functions ψ by including a variational parameter, and then minimizing hψ|Hˆ |ψi with respect to that parameter. When using the variational method, one thing you can check is that the virial theorem is satisfied when you’re dealing with simple potentials. For example, for U ∝ rs, the virial theorem tells us that 2 s hKi ≡ − ~ hψ|∇2|ψi = hUi . 2m 2 On the other hand, for the harmonic oscillator potential (in both 1D and 3D), the virial theorem tells us that hKi = hUi . If the virial theorem is not satisfied, then you made a mistake somewhere. When choosing a trial function, we want it to 174 Approximation Methods

• Be normalizable • Satisfy the boundary conditions implied by the potential • Be orthogonal to any lower states (if we are trying to estimate excited state energies) In general, we can write hψ|Hˆ |ψi E = . hψ|ψi Notice that E is a functional of ψ. We want to minimize E, so we want δE = 0. We get

Thus, the condition for minimum is

11.2 The Two-State System

In general, a 2-state system can be described by " # c1 |ψi = c1 |ψ1i + c2 |ψ2i = . c2

A Hamiltonian must be Hermitian, so the most general Hamiltonian of a 2-state system is " # ε V Hˆ = 1 . (11.1) ∗ V ε2 11.3. Diagonalization on Restricted Basis 175

The eigenvalue problem is " #" # " # ε V c c 1 1 = E 1 . ∗ V ε2 c2 c2

The eigenvalues, i.e. the energies of the stationary states of our system, are obtained by solving

ε − EV 1 = 0. ∗ V ε2 − E We ﬁnd that ε + ε 1 E = 1 2 ± p(ε − ε )2 + 4|V |2. ± 2 2 1 2 What does this have to do with perturbation theory? Suppose you have a 2-state system with energy levels ε1 and ε2. Note, this is a simpler system than the one described above. It is the same as the system above only when V = V ∗ = 0. Now, the Hamiltonian can be written " # ε 0 Hˆ = 1 , 0 ε2 with eigenvalues ε1 and ε2 and eigenvectors " # " # 1 0 , and . 0 1

Now suppose we add a perturbation which mixes the two levels. The interaction elements are V ≡ H12 = h1|Hˆ |2i. Now our system is the more general 2-level system with energies E±.

11.3 Diagonalization on Restricted Basis

Diagonalization on a restricted basis is one form of the variational method that is used a lot in practice. The exact solution (i.e. with an unrestricted basis) is given by

Hˆ |ψi = E |ψi .

We can write X |ψi = cn |ni , n where the |ni form an arbitrary orthonormal basis. Then X X Hˆ cn |ni = E cn |ni . n n If we multiply both sides by hm| from the left, we get X X cnHmn = E cnδmn = Ecm, n n where m = 1, . . . , mmax, and mmax could be inﬁnity. Note that Hmn ≡ hm|Hˆ |ni. Now we have a set of linear algebraic equations X cn (Hmn − Eδnm) = 0. n 176 Approximation Methods

This is still exact (in any usable basis). Nontrivial solutions occur when det Hˆ − E1 = 0.

Typically, if mmax is large, we end up with a determinant with many roots. In practice, and here is where the approximation comes in, we select, based on physical arguments, only a ﬁnite number of terms—the ones which contribute the most to E0. We will consider an example now where we keep only two terms. Then we can use the 2-level formalism introduced in the previous section. We can write the energy levels as ε + ε 1 E = 1 2 ± p∆2 + 4|V |2, ± 2 2

where ∆ = |ε1 − ε2| is the original spacing between the levels. Notice that

p 2 2 s = E+ − E− = ∆ + 4|V | ≥ |∆|.

That is, s ≥ ∆, so the spacing s between the mixed levels can only be greater than the spacing ∆ between the unmixed levels. This phenomenon is called level repulsion, and it is caused by the interaction |V | between the two levels.

What we’ve shown here is that adding an arbitrary perturbation V to the Hamiltonian of a two-state system will cause the two energy levels to be pushed away from each other. Recall that originally our eigenvalues were c1 and c2, then we changed to a1 and a2. I.e., if the unperturbed basis states are |1i and |2i, then

iα/2 −iα/2 |ψi = c1 |1i + c2 |2i = a1e |1i + a2e |2i .

Let us now solve for these eigenvalues for the lower state with energy E−. We will denote − − − them a1 and a2 . For example, for a2 , we can rearrange an earlier equation to write

− V − a2 = a1 . E− − ε2 But we have that ε − ε 1 ∆ s s + ∆ E − ε = 1 2 − s = − − = − . − 2 2 2 2 2 2 Therefore, 2V a− = − a−. 2 s + ∆ 1 Squaring both sides yields 2 2 4V 2 a− = a− . 2 (s + ∆)2 1

But recall that a1 and a2 are the magnitudes of the coeﬃcients on the eigenstates. In 2 2 other words, a1 + a2 = 1. So −2 −2 a2 = 1 − a1 . 11.3. Diagonalization on Restricted Basis 177

Equating with the earlier equation, we get

2 4V 2 2 a− = 1 − a− (s + ∆)2 1 1 2 2 4V a− 1 + = 1 1 (s + ∆)2 2 2 2 2 s + ∆ + 2∆s + 4V a− = 1 1 (s + ∆)2 2 2 2s + 2∆s a− = 1 1 (s + ∆)2

2 2s a− = 1. 1 s + ∆)

Thus, r s + ∆ a− = . 1 2s Similarly, we can ﬁnd the orthogonal eigenstate

r s − ∆ a− = − . 2 2s

If we write, − − a1 = cos θ, a2 = sin θ, then we can think of this as a rotation for some angle θ. If ∆ = 0, i.e., the level is degenerate, then s = 2|V |. This is the maximum mixing, and it happens when θ → 45◦. − − We found the eigenstates a1 and a2 for the energy level E−. Similarly, we can ﬁnd + + the eigenstates a1 and a2 for the energy level E+. What is the probability for the level spacing to have some value s? We can write s = px2 + y2, where x = ∆ and y = 2|V |. Then ∆ and V have their own probability distributions. That is, we have some probability distributions Px(x) and Py(y) for the probabilities of x and y. Then

p P (s) = dx dy P (x)P (y)δ s − x2 + y2 . ˆ x y

We want to know what happens when s → 0.

P (s) = lim s ds dπ Px(s cos φ)Py(s sin φ) s→0 ˆ

= lim s dφ Px(s cos φ)Py(s sin φ) s→0 ˆ = s dφ P (0) P (0) ˆ x y

= 2πs Px(0) Py(0).

So as s → 0, the probability goes to zero. That is, the levels do not cross. Incidentally, for a time-reversal invariant system, P (s → 0) ∼ s. For a time-reversal non-invariant system, P (s → 0) ∼ s2. What we’ve found is that the energy levels in terms of some variational parameter λ of the interaction, do not cross each other. 178 Approximation Methods

11.4 Stationary Perturbation Theory

We will be using Rayleigh-Schrodinger perturbation theory. There’s another form called Brillouin-Wigner perturbation theory, which is less useful for us. In perturbation theory, we have a known Hamiltonian Hˆ , and unknown energy eigenstates and energy spectrum |ψi and E

Hˆ |ψi = E |ψi .

We write the given Hamiltonian Hˆ as the sum of a simple Hamiltonian Hˆ 0 with known energy eigenstates |ni and a perturbation Hˆ 0

Hˆ = Hˆ 0 + Hˆ 0.

The perturbation Hˆ 0 should be small in some sense. We know the eigenstates and energy spectrum of Hˆ 0, and ˆ 0 0 H |ni = En |ni . The fundamental assumption we make with perturbation theory is that the general structure of the eigenstates and energy spectrum of Hˆ 0 remains after the perturbation Hˆ 0 is added. Formally, we assume that all the functions are smooth and analytic when we go from Hˆ 0 −→ Hˆ 0 + Hˆ 0. We don’t know the eigenstates |ψi of Hˆ , but we can expand them as X |ψi = cn |ni . n Then to do a Taylor expansion, we temporarily make use of a small parameter g

0 0 X X Hˆ + gHˆ cn |ni = E cn |ni n n X 0 X ˆ 0 cn E − En |ni = g cnH |ni . n n Let |mi be a member of {|ni}, then multiply the above equation from the left by hm|, and we get 0 X 0 cm E − Em = g cnHmn. n Note that m = 0, 1,..., ∞ runs over the whole basis. So far, this is still exact. Now we make an approximation. At this point, the unknowns are E and the set of coeﬃcients {cn}. We expand E in terms of small g as

0 (1) 2 (2) E = Ek + gEk + g Ek + ··· . 11.4. Stationary Perturbation Theory 179

Note, if the small parameter g is part of H0, then it will not explicitly appear as it does here. That is, if we deﬁne H = H0 +H0 such that H0 is small as opposed to H = H0 +gH0 0 (1) (2) where g is small, then we would have E = Ek + Ek + Ek + ··· . Remember that the zeroth order correction to the energy is just the known eigenvalue of Hˆ 0 0 0 ˆ 0 Ek = Hkk ≡ hk|H |ki . We do the same with the coeﬃcients

(1) 2 (2) cm = δmk + gcm + g cm + ··· . Plugging these expansions into the equation above gives us

(1) 2 (2) 0 (1) 2 (2) 0 δmk + gcm + g cm + ··· Ek + gEk + g Ek + · · · − Em =

X (1) 2 (2) 0 (11.2) = g δnk + gcn + g cn + ··· Hmn. n If m = k, this simpliﬁes to

(1) 2 (2) (1) (2) X (1) 2 (2) 0 1 + gck + g ck + ··· Ek + gEk + ··· = δnk + gcn + g cn + ··· Hkn. n If we expand both sides and equate the orders, we ﬁnd the ﬁrst order energy correction

(1) 0 ˆ 0 Ek = Hkk ≡ hk|H |ki .

Similarly, equating second-order terms, we ﬁnd

(2) (1) (1) X (1) 0 (1) 0 X (1) 0 Ek + ck Ek = cn Hkn = ck Hkk + cn Hkn. n n6=k

Notice that we treat H0 as a ﬁrst-order term, so its product with another ﬁrst-order term 0 (1) yields a term of overall order two. Since Hkk = Ek , we can cancel those terms from both sides to get our second-order correction for the energy

(2) X (1) 0 Ek = cn Hkn. n6=k

Going back to Eq. (11.2), if we treat m 6= k, expand both sides, and equate terms of the same order, then we ﬁnd for the ﬁrst order,

(1) 0 0 0 gcm Ek − Em = gHmk.

0 0 If |ki is not degenerate, then Ek − Em 6= 0, then we are safe to divide both sides by 0 0 Ek − Em to get 0 (1) Hmk cm = 0 0 . Ek − Em We can plug this back into the equation for the second-order correction of the energy, to get 0 0 (2) X H H E = nk kn . k E0 − E0 n6=k k n We can also write this as 0 2 (2) X |H | E = kn . k E0 − E0 n6=k k n 180 Approximation Methods

Notice that for the ground state, i.e. when k = 0, the right-hand side is always negative. So the eﬀect of a second-order perturbative correction to the ground state energy is always to lower the ground state energy. We can go higher. The third-order energy correction is

(3) X (2) 0 Ek = cm Hkm, m6=k where 1 X H0 H0 H0 H0 c(2) = mn nk − kk mk , for m 6= k. m E0 − E0 E0 − E0 E0 − E0 k m n6=k k n k m We’ve looked at the energy expansion and the corrections up to third order. What about the eigenstates? The eigenstate, including ﬁrst-order correction turns out to be

X H0 |ψi = |ki + g nk |ni . (0) (0) n6=k Ek − En

To second order,

(1) (2) and the coeﬃcients cn for n 6= k and cn for n 6= k were given earlier. Note, this perturbative approach is only valid if (i.e. it provably converges if)

0 |Hmk| << 1. (0) (0) Ek − Em

Keep in mind that the above formulas only work for non-degenerate states. If they are degenerate, we must diagonalize on the degenerate subspace. Example 11.4.1

Consider the anharmonic oscillator with Hamiltonian pˆ2 1 xˆ3 Hˆ = + mω2xˆ2 + g . 2m 2 3 Here, we have Hˆ = Hˆ 0 + gHˆ 0, where the known Hamiltonian is pˆ2 1 Hˆ 0 = + mω2xˆ2, 2m 2 and the perturbation is xˆ3 Hˆ 0 = . 3 We will assume that g > 0. Our total potential is now 1 1 U = mω2x2 + gx3. 2 3 11.4. Stationary Perturbation Theory 181

In the plot below, we show the normal harmonic oscillator potential (black curve), and the new potential (blue curve).

If g is small, then the ridge at

mω2 x = − , max g is high, and there can be “bound” states. The energy of this local maximum is

1 m3ω6 U = U(x ) = . 1 max 6 g2

If E < U1, then we have “bound” states. I put this in quotation marks, because there is nonzero probability that the particle will tunnel out toward the left. Thus, the particle will eventually tunnel out. In that sense, there are no truly bound states. The probability of tunneling per second, is

2 1 p mω2 m3ω5 T ∼ exp − p dx ∼ exp − mU1 C ∼ exp −C , ~ ˆ ~ g ~g2 for some constant C. So when g is small, T ∼ 0. When g is not small, there is still tunneling probability, so technically, there are no bound states. But practically, their lifetime is so long that they can be treated as bound states. Now consider a symmetric anharmonicity

xˆ4 Hˆ 0 = g , g > 0. 4

In this case, perturbation theory is still not valid. That is, it doesn’t converge to the true solution. However, it is good enough if to approximate the lower order corrections if g is very small. 182 Approximation Methods

11.5 Time-dependent Perturbation Theory

where t0 is some time at which the system is unperturbed. Adding time-dependence,

i X − Ent |Ψ(t)i = ane ~ |ni . n

2 Note that the probabilities |an| are independent of time. So far, we have not added the perturbation. We’ve seen the above before. Now we apply the time-dependent perturbation to the system. The coefficients, and therefore the probabilities, will now be time-dependent. New wave components can now appear since the coefficients, rather than just the wave components, also now depend on time. Suppose the perturbation lasts only for finite time. Then we want to know the transition probabilities for ending up in different final states given some initial state. The Schrodinger equation is ∂ i |Ψ(t)i = Hˆ 0 + Hˆ 0(t) |Ψ(t)i , ~∂t where the general wave function is now

i X − Ent |Ψ(t)i = an(t)e ~ |ni . n Notice the new time-dependence on the coeﬃcients. This approach is called the interaction picture (as opposed to the Schrodinger or Heisenberg pictures). The |ni are still the eigenstates of Hˆ 0. The full Schrodinger equation is now " # " # ∂ i i X − Ent 0 0 X − Ent i a (t)e ~ |ni = Hˆ + Hˆ (t) a (t)e ~ |ni . ~∂t n n n n Taking the derivative on the left and applying the product rule, this simpliﬁes to

i i X − Ent ˆ 0 X − Ent i~ a˙ ne ~ |ni = H (t) an(t)e ~ |ni . n n

Next, we multiply from the left by hm|, then on the left-hand side, hm|ni = δmn, and only one term remains

i i − Emt X − Ent ˆ 0 i~a˙ me ~ = an(t)e ~ hm| H (t) |ni . n 11.5. Time-dependent Perturbation Theory 183

So i X iωmnt 0 a˙ m = − an(t)e (H (t))mn , ~ n

0 ˆ 0 where (H (t))mn ≡ hm| H (t) |ni, and ~ωmn ≡ Em − En. Suppose at t → −∞, we start with some initial state |ii. At this time,

an(t) = an(−∞) = δni.

2 Assume the transition probabilities are small, i.e. |an6=i| << 1. Then at a later time, the dominant coeﬃcient is still that of the state |ii i iωmit 0 a˙ m ≈ − e (H (t))mi . ~ Thus, t i 0 0 iωmit 0 0 am(t) = − dt e (H (t ))mi . ~ ˆ−∞ Suppose we are interested in the ﬁnal state |fi at t → ∞. Then

∞ i 0 0 iωfit 0 0 af (∞) = − dt e (H (t ))fi . ~ ˆ−∞ Then the probability of |ii → |fi is

∞ 2 1 0 2 0 iωfit 0 0 wfi = |af (∞)| = 2 dt e (H (t ))fi . ~ ˆ−∞

So we need to know the spectrum of the unperturbed system (to write down ωfi) as well 0 0 as the matrix element (H (t ))fi of the perturbation. Notice that this is really a Fourier transformation. Example 11.5.1: Harmonic Oscillator

Consider a harmonic oscillator with frequency ω. Suppose |ii = |0i is the ground state. Then we apply an electric ﬁeld, such that the perturbation is

0 −t2/τ 2 H (t) = −eE~(t)x = −eE0e x. Then the transition probability is

2 2 ∞ 2 e E 2 2 0 −t /τ iωfit wfi = 2 dt e xfie . ~ ˆ−∞ For a harmonic oscillator, r x = ~ aˆ +a ˆ† . 2mω So to go from the ground state to the ﬁrst excited state, r x → x = ~ . fi 10 2mω So the probability of jumping from the ground state to the ﬁrst excited state is

2 2 ∞ 2 e E0 ~ −t2/τ 2+iωt wfi = w10 = 2 dt e . ~ 2mω ˆ−∞ 184 Approximation Methods

We can complete the square in the exponential, then integrate to get

2 2 2 2 2 e E0 ~ 2 −ω2τ 2/2 πτ e E0 −ω2τ 2/2 w10 = 2πτ e = e . ~2 2mω 2m~ω What is the eﬀect of diﬀerent pulses? We can express everything in terms of momentum p = dt eE(t), ˆ then 2 p 1 −ω2τ 2/2 w10 = e , 2m ~ω so the energy change is

2 p 2 2 ∆E = ω · w = e−ω τ /2. ~ 10 2m If we have a sudden perturbation (ωτ << 1) where the pulse is very narrow and nearly instantaneous, then

p2 ∆E ' . 2m On the other hand, if we have an adiabatic perturbation (ωτ >> 1) where the width of the signal is much larger than the period of oscillation, then

p2 ∆E << . 2m

11.6 Periodic Perturbations

Now, instead of Hˆ 0(t) being the Hamiltonian of a pulse, it is the Hamiltonian of a periodic perturbation. This is our entrance to the quantum-mechanical treatment of radiation. The most general periodic perturbation is Hˆ 0(t) = Hˆ 0e−iωt + H0†eiωt, where ω is the frequency of the external perturbation. Then i T/2 h i 0 −i(ω−ωfi)t 0† i(ω+ωfi)t af (T ) = − dt Hfie + H fie . ~ ˆ−T/2

Note that ωfi are frequencies of the system. We assume T is very large, so we can write i ∞ ∞ 0 −i(ω−ωfi)t 0† i(ω+ωfi)t af (T ) = − Hfi dt e + H fi dt e ~ ˆ−∞ ˆ−∞ i h 0 0† i = − Hfi2πδ(ω − ωfi) + H fi2πδ(ω + ωfi) . ~ When Ef − Ei ω = ωfi = , ~ we have excitation. On the other hand, when

Ei − Ef ω = −ωfi = , ~ we have radiation. 11.6. Periodic Perturbations 185

Case: Excitation For excitation, we have 2πi 0 af = − Hfi δ (ω − ωfi) . ~ Then the probability of |ii → |fi, in this case, is

T/2 2π 2 1 2 0 −i(ω−ωfi)t wfi = |af | = 2 Hfi δ(ω − ωfi) dt e . ~ 2π ˆ−T/2 Because we don’t know otherwise know how to interpret a delta-function squared, we wrote one of the delta functions in integral form above. Now, because of the ﬁrst delta- function, we know that wfi = 0 unless ω = ωfi. So we can go ahead and set ω = ωfi in the integral. When we do that, we get T/2π from the integral, then

2π 0 2 wfi = T Hfi δ(ω − ωfi), ~2 where T is the duration of the process. The rate of the process, i.e., the transition probability per unit time is

2π 0 2 rate =w ˙ fi = Hfi δ(ω − ωfi). ~2 This is the Fermi golden rule.

Case: Radiation For radiation, we get 2 2π 0† wfi = T H fi δ(ω + ωfi). ~2 Density of States We need to know what the density of states is in the continuum. We can write

δ(ω ∓ ωfi) = δ (~ω ∓ [Ef − Ei]) ~. Then for excitation, 2π 0 2 w˙ fi = Hfi δ(Ef − Ei ± ~ω) · dνf , ~ where dνf is the number of states possible at that energy. This depends on the resolution of the experiment. Consider a big box with volume V . The phase volume is

3 V d ρ V 2 dpf 3 = 3 pf dΩf dEf , (2π~) (2π~) Ef where the subscript f refers to the “ﬁnal” state, and dΩ is the angular element. For a non-relativistic particle,

2 pf pf dpf m Ef = =⇒ dEf = dpf =⇒ = . 2m m dEf pf 186 Approximation Methods

So the phase volume is V d3ρ V = mpf dΩf dEf , (2π~)3 (2π~)3 then 2π 0 2 V w˙ fi = Hfi mpf dΩf . ~ (2π~)3 11.7. Summary: Approximation Methods 187

11.7 Summary: Approximation Methods

Skills to Master • Be comfortable calculating the eigenvalues of matrices • Know the unperturbed energy spectrum of the harmonic oscillator, inﬁnite square well, and hydrogen atom

Variational Method We write the given Hamiltonian Hˆ as the sum of a simple Hamiltonian Hˆ 0 with known energy eigenstates |ni If we know the exact Hamiltonian Hˆ with a discrete and a perturbation Hˆ 0 spectrum, and we know the boundary conditions of the system, then Hˆ = Hˆ 0 + Hˆ 0, hψ|Hˆ |ψi ≥ E0, ˆ for any ψ. So to approximate E0, we calculate hψ|H|ψi where the perturbation Hˆ 0 is small in some sense. We with different ψ and try to minimize it. We can try an know the eigenstates and energy spectrum of Hˆ 0, and entire class of arbitrary functions ψ by including a vari- ˆ ˆ 0 0 ational parameter, and then minimizing hψ|H|ψi with H |ni = En |ni . respect to that parameter. You can also use the variational method to es- Then the energy spectrum of the whole Hamiltonian timate excited state energies. If an arbitrary func- Hˆ is tion ψ is orthogonal to the ground state ψgs, i.e., if (0) (1) (2) E = Ek + Ek + Ek + ··· , hψ|ψgsi = 0, then where the first several corrections are hψ|Hˆ |ψi ≥ E1, (0) 0 ˆ 0 is an (upper bound) estimate of E1. Ek = Hkk ≡ hk|H |ki When choosing a trial function, we want it to (1) 0 ˆ 0 Ek = Hkk ≡ hk|H |ki • Be normalizable 0 2 (2) X |H | • Satisfy the boundary conditions implied by the E = kn . k E0 − E0 potential n6=k k n • Be orthogonal to any lower states (if we are trying to estimate excited state energies) The eigenstate, including first-order correction is

When using the variational method, one thing you 0 X Hnk can check is that the virial theorem is satisfied when |ψi = |ki + |ni . E(0) − E(0) you’re dealing with simple potentials. For example, for n6=k k n U ∝ rs, the virial theorem tells us that Note, this perturbative approach is only valid if 2 s hKi ≡ − ~ hψ|∇2|ψi = hUi . 2m 2 |H0 | mk << 1. On the other hand, for the harmonic oscillator poten- (0) (0) E − Em tial (in both 1D and 3D), the virial theorem tells us k that Keep in mind that the above formulas only work hKi = hUi . for non-degenerate states. If they are degenerate, we If the virial theorem is not satisfied, then you made a must diagonalize on the degenerate subspace. mistake somewhere. If given Hˆ 0 with multiple terms and asked for the leading-order perturbative corrections, be sure to cal- Stationary Perturbation Theory culate the leading order corrections for all terms. For example, this might mean only having to calculate first- In perturbation theory, we have a known Hamiltonian order corrections for some terms but second-order cor- Hˆ , and unknown energy eigenstates and energy spec- rections for other terms. trum |ψi and E The general procedure for solving stationary per- Hˆ |ψi = E |ψi . turbation theory problems is as follows: 188 Approximation Methods

1. Write down the perturbation Hamiltonion Hˆ 0. Be comfortable working with the ladder operators. For example, In particular, know • For electric field r xˆ = ~ aˆ +a ˆ† , Hˆ 0 = −E~ · d~ = −eEx, 2mω if the field is in the x direction. and how they act on the unperturbed harmonic oscil- • For magnetic field lator states 0 √ Hˆ = −~µ · B~ , ~µ = g~J~. aˆ |ni = n |n − 1i √ If the particle interacts with the magnetic aˆ† |ni = n + 1 |n + 1i . field via spin, then ~µ = gs~~s. If it’s a charged particle and interacts via orbital For a particle constrained to a circle, ~ motion through the field, then ~µ = g`~`. 2Lˆ2 1 Hˆ = ~ z , ψ = √ eimφ, m = 0, ±1, ±2,... 2. When calculating the perturbative correction for 2I 2π a given energy level, if there is only a single state at this energy level, then proceed as normal using You should also know the hydrogen ground state the formulas given above 1 |100i = e−r/a0 , 3. If there are multiple states at this level, then you p 3 πa0 must use degenerate perturbation theory a) Identify and label the degenerate states in and its energy spectrum this level. For example, for n = 2 in hydro- 1 E = − Ry, n = 1, 2, 3,... gen, we have the degenerate states |n`mi = n n2 |200i , |210i , |211i , |21 − 1i all at the same For a 3D spherical harmonic oscillator, the energy energy. It is convenient to relabel these as levels are n = 0, 1, 2,..., and the angular momentum |1i , |2i , |3i , |4i for now is b) Write down the perturbation matrix. If ( 0, 2, 4, . . . , n if n is even there are N degenerate states at this energy ` = . level, then the perturbation matrix is N ×N 1, 3, 5, . . . , n if n is odd Each ` has projections m = −`, . . . , `, and then there  H0 H0  ` 11 12 are spin projections m = −s, . . . , s. The parity of a  .  s  H0 .. . given state is (−1)`.  21  0 HNN Time-dependent Perturbation Theory c) Calculate the matrix elements The transition probability for |ii at t = −∞ to go to H0 = hm|H0|ni . mn |fi at t = +∞ is For example, 1 ∞ 2 • For the harmonic oscillator, x and y (if iωfit 0 wfi = 2 dt e Hfi(t) , they exist in Hˆ 0) are written in terms ~ ˆ−∞ of the ladder operators, and then it’s where ~ωfi = Ef − Ei. To calculate this transition relatively easy to calculate the matrix probability, we have to calculate the matrix element elements 0 0 • For the hydrogen atom, x, y, z should be Hfi(t) = hf|H (t)|ii . written in spherical coordinates, then you calculate the matrix elements by Miscellaneous performing the integration in spherical coordinates Need to know the lowest several spherical harmonics: d) Calculate the eigenvalues of the perturba- 1 Y00 = √ tion matrix. These are the first-order cor- 4π rections to the energies of these states r 3 You should know the (unperturbed) states and en- Y10 = cos θ 4π ergy spectrum of the infinite square well r r 3 ±iφ 2 nπx n2π2 2 Y1±1 = ∓ sin θe . |ni = sin ,E = ~ , n = 1, 2,... 8π a a n 2ma2 Chapter 12

Atoms

12.1 Lamb Shift

Recall the basic hydrogen energy levels shown below. In particular, consider the ﬁrst excited state which consists of a mixture of levels including 2s1/2 and 2p1/2 with diﬀerent parity but the same energy.

Then if we include relativistic eﬀects, there occurs ﬁne structure splitting which splits this level.

Then in 1947, Lamb and Retherford discovered an additional splitting that separates the 2s1/2 and 2p1/2 levels into levels with slightly diﬀerent energies. 190 Atoms

This splitting is of the order 4 × 10−6 eV. The corresponding wavelength is λ ∼ 2.74 m, and the frequency range is ω ∼ 1057.8 MHz ∼ 1 GHz. At the time it was discovered, this Lamb splitting was not predicted by any theory. Hans Bethe was the first to explain it, and we will follow his approach. Around the hydrogen atom, there is the electromagnetic vacuum. This vacuum fluctuates as virtual photons are created and annihilated. So the electron in the hydrogen atom feels these random E~ and B~ fields. These random virtual photons popping in and out of existence cause the electron to randomly fluctuate in position. The electron feels a potential U(~r), and we can encode the fluctuation as ~r → ~r + ζ~. Then 1 ∂2 U(~r + ζ~) = U(~r) + ζ~ · ∇~ U + ζiζj U(~r) + ··· 2 ∂xi∂xj Note that i and j give the components (i.e. x, y, and z), and are summed over. Taking the average, we know that ζ~ = 0,

since the fluctuations are random. Since ζx, ζy, and ζz are uncorrelated and equivalent, 1 ζ ζ = ζ2 = ζ~ 2. i j x 3 So 1 U(~r) = U(~r + ζ~) = U(~r) + ζ~ 2∇2U. 6 We want to estimate ζ~ 2. We have some E~ and B~ in vacuum. They can have different frequency ω components. ~ Then for the ω component of the fluctuation, ζω, we have the equation of motion ~¨ ~ mζω = eEω. We are neglecting B~ , which we can for small atoms in which the electrons are non- relativistic. This leads us to the solution 2~ ~ −mω ζω = eEω, and so e2 ζ~ 2 = E~ 2. ω m2ω4 ω Then the energy for the ω-component of the fluctuation is

E~ 2 + B~ 2 E~ 2 1 E = V ω ω = V ω = ω. ω 8π 4π 2~ Thus, 2π ω E~ 2 = ~ , ω V 12.1. Lamb Shift 191 and so 2π e2 ζ~ 2 = ~ . ω V m2ω3 This is for a given frequency ω. To get ζ~ 2 for all frequencies, we simply add all the contributions separately since each component is independent. We do this by integrating

ζ~ 2 = dω ρ(ω)ζ~ 2, ˆ ω where ρ(ω) is the density of the oscillations, or the number of modes in a given interval. We ﬁnd that V d3k 2 V ρ(ω) dω = · 2 = 4πV k2 dk = ω2 dω, (2π)3 (2π)3 π2c3 where the factor of 2 in the ﬁrst term comes from the polarization modes of light. Plugging everything into the integral gives us

2π e2 V 2 e2 dω ζ~ 2 = dω ρ(ω)ζ~ 2 = ~ ω2 dω = ~ . ˆ ω ˆ V m2ω3 π2c3 πm2c3 ˆ ω

This integral is divergent in both directions (in the limit ω → 0 and in the limit ω → ∞). QED takes care of the inﬁnities rigorously. We will just use simple arguments. For example, by assuming minimum and maximum frequencies. Then

2 e2 ωmax dω 2 e2 ω ~ 2 ~ ~ max ζ = 2 3 = 2 3 ln . πm c ˆωmin ω πm c ωmin We estimate the rest mass of the electron to be

2 mc ∼ ~ωmax, and the diﬀerence between subsequent hydrogenic energy levels to be

2 2 (Zα) mc ∼ ~ωmin. So ωmax f ln ∼ ln 2 , ωmin (Zα) where f ∼ 1. So 2 e2 ζ~ 2 = ~ Λ, πm2c3 where f Λ ≡ ln . (Zα)2 For any potential U, the change in the energy of the electron is 1 δE = ζ~ 2 ∇2U . 6 Now we apply this to the hydrogenic atom with

Ze2 U = − . r Then 1 ∇2U = −Ze2∇2 = 4πZe2δ(~r). r 192 Atoms

Then 4πZe2 δE = ζ~ 2 hψ |δ(~r)|ψ i , 6 s s 2 where the ψs indicate s-waves. Note that hψs|δ(~r)|ψsi is |ψs| at the origin, so

Z3 hψ |δ(~r)|ψ i = |ψ (0)|2 = , s s s πa3n3 since for s-waves, which have ` = 0, we know that r Z3 ψ (0) = . s πa3n3 Plugging this in, we get 4 Z4 e4 1 δE = ~ Λ. 3π2 m2c3a3 n3 For hydrogen with Z = 1, the energy levels are

me4 e2 En = − = − . 2~2n2 2an2 Then 2 δE ~e 1 3 1 ∼ 2 3 2 Λ = α Λ. |En| m c a n n Recall that the fine structure energy shift is ∝ α2. Here we have a ∝ α3 effect. However, with the logarithm in Λ, the effect of this Lamb shift is of order 1/10 that of fine structure splitting.

12.2 Polarizability

We will now use perturbation theory to analyze systems in a static electric ﬁeld. In a static and uniform electric ﬁeld E~ = −∇~ φ, the system (e.g. atom, molecule, or nucleus) experiences the potential

φ = −E~ · ~r.

We assume the ﬁeld is small and can be treated perturbatively. If our system consists of charges ea in this ﬁeld, then the perturbation Hamiltonian of the system is

0 X Hˆ = − ea E~ · ~ra . a We can write this as Hˆ 0 = −E~ · d~, where X d~ = ea~ra, a is the dipole operator. In our case, E~ is small, so we can apply perturbation theory. Then

~ ~ X H0 X E · dnk |ψi = |ki + nk |ni = |ki + |ni . (12.1) (0) (0) (0) (0) n6=k Ek − En n6=k Ek − En 12.2. Polarizability 193

Note, if the original (unperturbed) system had a deﬁnite parity, then it will no longer have a deﬁnite parity. This is because the dipole operator changes parity. For example, suppose we have a state ψ+ with positive parity, then hψ+|d~|ψ+i −→ hψ+|ψ−i. So if our original system had positive or negative parity, the d~ in Eq. (12.1), will mix it all up. The induced dipole moment is the expectation value of the dipole operator

hd~i = hψ|d~|ψi .

Plugging in Eq. (12.1) gives us " # X d~H0 H0 d~ hd~i = hk| nk |ki + hn| kn |ki (0) (0) (0) (0) n6=k Ek − En Ek − En ~ ~ ~ ~ ~ ~ X −E · dkn dnk − E · dnk dkn = (0) (0) n6=k Ek − En ~ ~ ~ ~ ~ ~ X dnk E · dkn + dkn E · dnk = . (0) (0) n6=k En − Ek

0 ∗ 0 Note, in the ﬁrst line, we used Hnk = Hkn. Note that hd~i is a vector with x, y, and z components. We can write the expectation value for a single component as

µ µν ν hd ik = α E , (12.2) where the µ and ν can be any of {x, y, z}. The subscript k indicates that this is the dipole moment component in state k. That is, hdik ≡ hk|d|ki. Note, we are also using Einstein summation notation here. The αµν are the elements of the polarizability tensor

µ µ X d dν + dν d αµν = kn nk kn nk . k (0) (0) n6=k En − Ek

This quantity has dimensions of volume. The change in the energy due to the applied electric ﬁeld is

1 ∆E = − αµν EµEν , (12.3) k 2 where we are again using Einstein notation. Then

∂ ∆E = −αλν Eν = −dλ, ∂Eλ where λ ∈ {x, y, z}. Example 12.2.1: Harmonic Oscillator in Electric Field

Consider a harmonic oscillator in a weak electric ﬁeld. Its potential is 1 V = mω2x2 − eEx. 2 194 Atoms

Completing the square, we get

1 eE 2 e2E2 V = mω2 x − − . 2 mω2 2mω2

The electric ﬁeld shifts the oscillator to the right by eE ∆x = . mω2 This corresponds to a dipole

e2E d = e∆x = . mω2 Comparing with Eq. (12.2), we see that the oscillator’s polarizability is

e2 α = . mω2 From V , we also see that the electric ﬁeld shifts the oscillator energy

e2E2 ∆E = − . 2mω2 This matches what we get from Eq. (12.3) 1 ∆E = − αE2. 2

Suppose state |ki has angular momentum J~. Thus, this is a multiplet of 2J + 1 states |JMi. Then we can apply the electric field along different directions. Consider a field Ez in the z-direction. For any state |JMi, αµν should be a function of the angular momentum components. The polarizability tensor is symmetric, meaning αµν = ανµ. The most general second-rank symmetric tensor made of the components of angular momentum we can write down is 2 αˆµν = χδµν + β JˆµJˆν + Jˆν Jˆµ − δµν J~ 2 . 3 For example,

2 2 αzz(M) = χ + β 2J 2 − J~ 2 −→ χ + β 2M 2 − J(J + 1) . z 3 3

If the system has zero angular momentum, i.e. it is spherically symmetric, then there will only be αµν = χδµν , since all the J will be zero. Keep in mind, that the α, χ, and β all depend on k.

TRK Sum Rule “TRK” comes from “Thomas-Reiche-Kuhn.” Consider a general (unperturbed) many-particle Hamiltonian

X ~p 2 Hˆ = a + U({~r }). 2m a a a 12.2. Polarizability 195

Then we can write down a sum rule 1 h h i i S = hk| z,ˆ Hˆ , zˆ |ki k 2 1 = hk| zˆHˆ zˆ − Hˆ zˆzˆ − zˆzˆHˆ +z ˆHˆ zˆ |ki 2

Next, insert a complete set of states P |ni hn| between each of the above operators, then we get 1 S = z E z − E z z − z z E + z E z k 2 kn n nk k kn nk kn nk k kn n nk 1 X = (E − E ) z z 2 n k kn nk n 1 X 2 = (E − E ) |z | . 2 n k kn n We could also have calculated the commutator explicitly. The inner commutator gives us h i p2 pˆ z,ˆ Hˆ = z, z = i z . 2m ~ m Then the overall commutator is h h i i h pˆ i i 2 z,ˆ Hˆ , zˆ = i z , zˆ = ~ [ˆp , zˆ] = ~ . ~ m m z m Equating the two results gives us the sum rule

2 1 X 2 S = (E − E ) |z | = ~ . k 2 n k kn m n Any non-relativistic single-particle system will obey this. The above is for a single component. We now apply it to all components of the dipole moment. Without loss of generality, we assume the dipole moment points in the z direction, then N X d~ = ezˆ, where the sum goes up to the number of particles N. Then we ﬁnd that

2 2 X 2 e |d | (E − E ) = ~ N. n0 n 0 2m n

This is the TRK sum rule. Thus,

2m X 2 |d | (E − E ) = N 2e2 n0 n 0 ~ n gives us the number of charged particles in the system. The oscillator strength of a given transition is deﬁned as

2m 2 2mωn0 2 fn0 = (En − E0) |dn0| = |dn0| , e2~2 e2~ where ~ωn0 ≡ En − E0. So X N = fn0. n 196 Atoms

µν Recall that if there is no angular momentum dependence, then α = χδµν . Thus,

2 2 X |dn0| X e χ = 2 = f . ω mω2 n0 n ~ n0 n n0

2 2 We can think of the coeﬃcient e /mωn0 as the polarizability of one oscillator, and the whole thing, χ, as the polarizability of the entire system of oscillators. Recall the hydrogen energy spectrum:

If we apply an electric field, then we get a dipole moment. We will assume there are no spin effects, i.e., we assume the electric field is strong enough to drown out all relativistic effects due to spin. We know that d~n0 has ∆` = 1. That is, the dipole operator changes the angular momentum by ±1. It also changes the parity. So if the electron was initially in the 1s ground state, it is now pushed to the 2p first excited state. Generally, this splitting due to an electric field is called the Stark effect. What if the electron was originally in the first excited state? The first excited state consists of the 2s and 2p levels—a total of four degenerate states. If the applied electric field is in the zˆ direction, then our selection rules are ∆` = ±1 and ∆m = 0. If the electron was initially in one of the two 2p states with m = ±1, then it will be excited to the 3d state. For the 2s (with m = 0) and the remaining 2p (with m = 0) states, it’s a little more complicated since there is coupling (h2s|d|2pi) between the two states. Recall that ε + ε 1q E = 1 2 ± (ε − ε )2 + 4|V |2. ± 2 2 1 2

Since the levels are degenerate with ε1 = ε2, we have

E± = ε ± |V |.

The perturbation is Hˆ = −eEz, then the matrix element is

h2s|d|2pi = h2s, m = 0|z|2p, m = 0i = 3a.

12.3 Van der Waals Forces

Consider two atoms separated by a large distance R >> a0, where a0 is the Bohr radius. Some fluctuation causes a dipole in one atom whose field fluctuation induces a dipole in 12.3. Van der Waals Forces 197 the second atom. Now, we have a dipole-dipole interaction between the two atoms. The Hamiltonian of this interaction is 1 h i Wˆ Hˆ = d~ · d~ − 3 d~ · nˆ d~ · nˆ ≡ , dip R3 1 2 1 2 R3 where nˆ = R~ /R is the unit vector pointing from one atom to the other. Note that if d~1 k d~2, then Hˆdip > 0. On the other hand, if d~1 ⊥ d~2, then Hˆdip = 0. If both atoms are in their ground state, then the perturbative change in the energy is 2 ˆ 1 X h00| W |00i C ∆E00 = 6 = − 6 , R E00 − En1n2 R n1n2 where C is the positive constant 2 ˆ X h00| W |00i C = > 0. E00 − En1n2 n1n2 We can approximate this infinite sum using a closure approximation 1 X C ≈ h00| Wˆ |n1n2i hn1n2| Wˆ |00i ∆E n1n2 1 = h00| Wˆ 2 |00i ∆E 1 = e4 h00| (x x + y y + z z − 3z z )2 |00i . ∆E 1 2 1 2 1 2 1 2 Only the squared terms will be nonzero after the averaging, so 1 C = e4 h00| x2x2 + y2y2 + 4z2z2 |00i . ∆E 1 2 1 2 1 2 2 2 2 2 For the hydrogen ground state, hx i = hy i = hzi = a0, so 1 C = 6e4a4. ∆E 0 Excitation is typically to the closest p-level, so typically, e2 3 3 e2 ∆E ' 2 = , 2a0 4 4 a0 where the factor of 2 in front is due to there being two atoms, and the factor of 3/4 is for the distance between the 1s and 2p levels. This will be a lower bound, so we can say that e2 ∆E = ζ , a0 where ζ > 3/4 is some constant. Then 4 4 4 2 6e 4 6e a0 6 e 6 C = a0 = 2 ∼ a0. ∆E ζe /a0 ζ a0 So the expected energy for the Van der Waals force is

2 6 0 6 e a0 hHV dW i ≈ − . ζ a0 R Note, this analysis of Van der Waals forces is only valid for a certain range of distances (between the atoms). It is valid only in the far range—but not in the very far range where retardation becomes non-negligible. At the very far range, E ∼ 1/R7 instead of E ∼ 1/R6. 198 Atoms

12.4 Summary: Atoms

Skills to Master •

Lamb Shift Note that hd~i is a vector with x, y, and z components. We can write the expectation value for a single The first excited state of hydrogen consists of a mix- component as ture of levels including 2s1/2 and 2p1/2 with different µ µν ν parity but the same energy. If we include relativis- hd ik = α E , tic effects, there occurs fine structure splitting which splits this level. Then in 1947, Lamb and Retherford where the µ and ν can be any of {x, y, z}. The sub- discovered an additional splitting that separates the script k indicates that this is the dipole moment com- 2s1/2 and 2p1/2 levels into levels with slightly different ponent in state k. That is, hdik ≡ hk|d|ki. Note, we energies. are also using Einstein summation notation here. This splitting is of the order 4 × 10−6 eV. The The αµν are the elements of the polarizability ten- corresponding wavelength is λ ∼ 2.74 m, and the fre- sor µ µ quency range is ω ∼ 1057.8 MHz ∼ 1 GHz. X d dν + dν d αµν = kn nk kn nk . The Lamb shift is ultimately due to the virtual k (0) (0) E − E photons popping in and out of existence around the n6=k n k electron in the hydrogen atom, causing the electron to The change in the energy due to the applied elec- randomly fluctuate in position. tric field is

1 µν µ ν Polarizability ∆Ek = − α E E , 2 In a static and uniform electric field E~ = −∇~ φ, the system experiences the potential where we are again using Einstein notation. Then φ = −E~ · ~r. ∂ ∆E = −αλν Eν = −dλ, Assuming a small field strength, we can treat the elec- ∂Eλ tric field as a perturbative effect. The perturbation can be written as where λ ∈ {x, y, z}. Hˆ 0 = −E~ · d~, For example, to calculate the polarizability of a system due to a perturbing electric field where X d~ = e ~r , a a Hˆ 0 = −E~ · d~, a is the dipole operator. Keep in mind that the dipole operator changes the parity of a state it acts on, so calculate the energy change ∆E between perturbed if the system initially had a definite parity, it will not and unperturbed states. For example, calculate the 1 2 after this electric field is applied. second order perturbative correction ∆E = − 2 αE , The induced dipole moment is the expectation which implies the polarizability is value of the dipole operator 2∆E ~ ~ α = − . hdi = hψ|d|ψi . E2 Chapter 13

Photons and the Electromagnetic Field

13.1 Hamiltonian of EM Field

There are two ways to quantize fields: • path integrals • canonical quantization (commutators) Consider the free electromagnetic field in a vacuum. There are no charges, so div E~ = 0. We have the scalar and vector potentials φ and A~. The electromagnetic fields are

1 ∂A~ E~ = −∇~ φ − c ∂t B~ = ∇~ × A~.

We will use the gauge φ = 0, A~ 6= 0, which is convenient for free ﬁeld in a vacuum. We will consider some box with volume V and periodic boundary conditions. The wave equation is 1 ∂2A~ − ∇2A~ = 0. c2 ∂t2 Any plane wave i(~k·~r)−iωkt e , ωk = ck, k = |~k|, ωk > 0, is a solution to the wave equation. The general solution is a superposition of all plane waves X h ~ ~ i A~(~r, t) = ~b ei(k·~r−ωkt) +~b ∗ e−i(k·~r−ωkt) . ~k ~k ~k The complex conjugate is included to ensure that A~ ends up being real. The ﬁelds corresponding to this are

1 ∂A~ i X h ~ ~ i E~(~r, t) = − = ω ~b ei(k·~r−ωkt) −~b ∗ e−i(k·~r−ωkt) c ∂t c k ~k ~k ~k X h ~ ~ i B~ (~r, t) = ∇~ × A~ = i ~k × ~b ei(k·~r−ωkt) −~b ∗ e−i(k·~r−ωkt) . ~k ~k ~k The energy of the ﬁeld is

E~ 2 + B~ 2 E = dV = E + E . ˆ 8π electric magnetic 200 Photons and the Electromagnetic Field

There’s no cross term E~ · B~ because that would violate parity conservation. The energy of the electric ﬁeld is

2 i h i h 0 0 i X i(~k·~r−ωkt) ∗ −i(~k·~r−ωkt) i(~k ·~r−ω 0 t) ∗ −i(~k ·~r−ω 0 t) E = ω ω 0 ~b e −~b e ~b e k −~b e k . ele 8πc2 k k ~k ~k ~k 0 ~k 0 ~k,~k

We have four integrals of the type

0 dV ei(~k±~k )·~r = V δ , ˆ ~k 0,∓~k

where V is the volume. So we get

V X h i E = − w2 ~b ·~b e−2iωkt + ~b ∗ ·~b ∗ e2iωkt − ~b ·~b ∗ − ~b ∗ ·~b . ele 8πc2 k ~k −~k ~k −~k ~k ~k ~k ~k ~k So we have time dependence in the electric field energy. The energy is exchanged between the electric and magnetic fields, but the total energy is constant. If we repeat the above procedure for the magnetic field energy, and then add the two together, we get

V X h i E = E + E = ω2 ~b ·~b ∗ + ~b ∗ ·~b . ele mag 4πc2 k ~k ~k ~k ~k ~k

Note, for every ~k, there should be two polarizations. Since ∇~ · E~ = 0 and ∇~ · B~ = 0, we have for a plane wave,

ei~k·~r −→ ~k · E~ = ~k · B~ = 0. ~ For linear polarization, for every k, we introduce two unit vectors ~eλ~k where λ = 1, 2 for the two polarizations. These are orthogonal to the wave direction of motion such that ~ ~eλ~k · k = 0, and we have the orthogonality condition

~eλ~k · ~eλ0~k = δλλ0 .

So the {~eλ~k} form an orthonormal basis and we can write ~ X b~k = ~eλkbλk, λ

where the bλk are coeﬃcients. We can also have circular polarization. We deﬁne

(±) 1 ~e~ = ∓√ (~ex ± i~ey) , k 2 then we have the orthogonality condition

(±)∗ (∓) ~e~k · ~e~k = 0. Plugging these into the energy,

V X h i E = ω2 b · b∗ + b∗ · b . 4πc2 k λ~k λ~k λ~k λ~k λ,~k 13.2. Quantization of the EM Field 201

The λ and ~k are the diﬀerent modes. We deﬁne the “coordinate” of the photon mode as r V Q = b + b∗ , λ~k 4πc2 λ~k λ~k and the “momentum” as r V P = −iω b − b∗ . λ~k k 4πc2 λ~k λ~k Then r r 1 4πc2 i 4πc2 bλ~k = Qλ~k + Pλ~k 2 V 2ωk V r r 1 4πc2 i 4πc2 b∗ = Q − P . λ~k λ~k λ~k 2 V 2ωk V The time derivatives of Q and P are

˙ Qλ~k = Pλ~k ˙ 2 Pλ~k = −ωkQλ~k. Thus, we get the equation of motion

¨ 2 Qλ~k + ωkQλ~k = 0. This is the equation of motion for a harmonic oscillator. Thus, we can write the energy as 1 E = H = ω2 Q2 + P 2 = P 2 + ω2Q2 . k λ~k 2 λ~k λ~k k λ~k ωk This is the eﬀective Hamiltonian for every mode characterized by ~k and λ.

13.2 Quantization of the EM Field

The next step in the quantization procedure is to deﬁne Q and P to be operators and to ﬁnd the commutation relations. Each mode is an independent degree of freedom, so when we go to operator notation, we can write the Hamiltonian as

ˆ X ˆ E −→ H = Hλ,~k, λ,~k where every mode is characterized by the polarization λ and wave vector ~k 1 Hˆ = Pˆ2 + ω2Qˆ2 . λ~k 2 λ~k k λ~k

Recall that ωk = ck. Then for every mode (λ~k),

˙ ˙ ¨ 2 Q = P, P = Q = −ωkQ.

For the “coordinate” and “momentum” operators, we now have the commutation relation h ˆ ˆ i Qλ~k, Pλ0~k 0 = i~δλλ0 δ~k~k 0 . 202 Photons and the Electromagnetic Field

The operators are r V Qˆ = ˆb + ˆb† λ~k 4πc2 λ~k λ~k r V ˆ ˆ† Pˆ = −iωk b − b . λ~k 4πc2 λ~k λ~k

The ˆb and ˆb† operators are interpreted as annihilation and creation operators, respectively, and we can solve the above pair of equations for them

r 2 ˆ πc ˆ i ˆ bλ~k = Qλ~k + Pλ~k V ωk r 2 ˆ† πc ˆ i ˆ b ~ = Qλ~k − Pλ~k . λk V ωk These have the commutation relation

2 2 hˆ ˆ† i πc i i 2π~c bλ~k, b ~ = − i~ + (−i~) = . λk V ωk ωk ωkV We can rescale the creation and annihilation operators as s s 2 2 ˆ 2π~c ˆ† 2π~c † bλ~k = aˆλ~k, b ~ = aˆ ~ . ωkV λk ωkV λk

Then the commutation relation for the rescaled operators is h i aˆ , aˆ† = 1, λ~k λ~k

or more generally, h † i aˆ , aˆ = δλλ0 δ 0 . λ~k λ0~k 0 ~k~k We can express Qˆ and Pˆ in terms ofa ˆ anda ˆ†, and then plug these into the Hamiltonian to get ~ωk † † ~ωk † 1 Hˆ = aˆ aˆ +a ˆ aˆ = 2ˆa aˆ + 1 = ωk Nˆ + . λ~k 2 λ~k λ~k λ~k λ~k 2 λ~k λ~k ~ λ~k 2

This is the form of the Hamiltonian for a harmonic oscillator. Now we have inﬁnitely many of them—one for each possible mode of the electromagnetic ﬁeld. Now, s 2 2π c h ~ ~ i ~ X ~ i(k·~r−ωkt) ∗ † −i(k·~r+ωkt) A(~r, t) = ~eλ~kaˆλ~ke + (~eλ~k) aˆ ~ e . ωkV λk λ,~k

Then the electric and magnetic ﬁelds are

~ r i ∂A X 2π ωk h ~ ~ i E~(~r, t) = − = i ~ ~e aˆ ei(k·~r−ωkt) − (~e )∗aˆ† e−i(k·~r+ωkt) c ∂t V λ~k λ~k λ~k λ~k λ,~k s 2 2π c n h ~ ~ io ~ ~ ~ X ~ ~ i(k·~r−ωkt) ∗ † −i(k·~r+ωkt) B(~r, t) = ∇ × A = i k × ~eλ~kaˆλ~ke − (~eλ~k) aˆ ~ e . ωkV λk λ,~k 13.3. Radiation 203

Now E~ and B~ are operators which can create and annihilate modes of energy. For example,a ˆ† creates a photon with polarization λ and wave vector ~k. The wave vector λ~k determines its frequency and direction of motion. This is the correct way to quantize the electromagnetic ﬁeld. We do this so we can correctly calculate electromagnetic processes in quantum mechanics. We denote the vacuum state as |vaci = Nλ~k = 0 .

That is, it is the state in which every element of the set of Nλ~k is zero. Then each mode contributes energy 1 1 Hˆ = ω Nˆ + = ω . λ~k ~ k λ~k 2 2~ k So if we add the contributions of all infinite modes, we find that the vacuum state has infinite energy. This is just a basic example of the many infinities that occur in QED. To resolve this, we don’t consider the absolute energy of the fields—we just consider changes in the energy. Note that changing the boundary conditions of an experiment changes the vacuum energy in that region. This is the Casimir effect. The electric and magnetic fields also have momenta. The Poynting vector is c h i S~ = E~ × B~ . 4π Then the momentum of the electromagnetic field is

1 X 1 P~ = d3r S~ = ~k aˆ† aˆ + . c2 ˆ ~ λ~k λ~k 2 λ~k

For every ~k, there is a −~k, so when we add all of these for the vacuum, we find that P~ vac = 0. Note, the field operators E~ and B~ at different points in spacetime do not in general commute. That is, they are not simultaneously measurable.

13.3 Radiation

We will now look more closely at radiation which occurs in a transition |ii → |fi, where Ei > Ef . In radiation, the electron drops from a higher energy state |ii to a lower energy state |fi, and radiates a photon with energy

~ω = Ei − Ef .

The photon is characterized by a wave vector ~k and a polarization ~e.

We can apply the Fermi Golden rule, then

2π 0 2 dw˙ fi = Hfi δ (Ei − Ef − ~ω) dνf . ~ 204 Photons and the Electromagnetic Field

0 Our main problem is to calculate the matrix elements Hfi. The unperturbed Hamil- tonian is X ~p 2 Hˆ = a + U ({~r }) . 2m a a a We assume the system is non-relativistic. Then for every particle a, we make the substitution e ~p −→ ~p − a A~(~r ). a a c a Adding this perturbation gives us

2 X 1 h ea i Hˆ = ~p − A~(~r ) + U ({~r }) 2m a c a a a a 2 2 X 1 ea e = ~p 2 − ~p · A~(~r ) + A~(~r ) · ~p + a A~ 2(~r ) + U. 2m a c a a a a c2 a a a

This is the Hamiltonian in the presence of the EM ﬁeld. Since this is an approximation, we only keep the linear terms in A~. Then our perturbation is

X ea Hˆ 0 = − ~p · A~(~r ) + A~(~r ) · ~p . 2m c a a a a a a

Recall that s 2 2π c ~ ~ ~ X ~ i(k·~r)−iωkt ∗ −i(k·~r)+iωkt † A(~r, t) = ~eλ~ke aˆλ~k + ~eλ~ke aˆ ~ . ωkV λk λ,~k

ˆ 0 ~ ~ ~ ~ Plugging this into H , we use the fact that ~pa and A(~ra) commute: [~p, A] = −i~∇·A = 0 to write s 2 ea 2π c h ~ ~ i ˆ 0 X X ~ i(k·~ra)−iωkt ∗ −i(k·~ra)+iωkt † H = 2 ~pa · ~eλ~k e aˆλ~k + ~pa · ~eλ~k e aˆ ~ . 2mac ωkV λk a λ,~k

In radiation, a photon is created, so we take only the term containing the creation operator aˆ† s 2 ea 2π c h ~ i ˆ 0 X X ~ ∗ −i(k·~ra)+iωkt † Hrad = 2 ~pa · ~eλ~k e aˆ ~ . 2mac ωkV λk a λ,~k

So for a given λ and ~k, r X ea 2π ~ Hˆ 0 = ~ ~p · ~e∗ e−i(k·~ra)eiωktaˆ† . rad,λ,~k m ω V a λ~k λ~k a a k

Then our matrix elements are r X ea 2π ~ H0 = ~ eiωkt hf| ~p · ~e∗ e−i(k·~ra)aˆ† |ii . fi m ω V a λ~k λ~k a a k

Then the radiation rate is 2 2π e r 2π X a ~ iωkt ∗ −i(~k·~ra) † dw˙ fi = e hf| ~p · ~e e aˆ |ii δ (Ei − Ef − ω) dνf . m ω V a λ~k λ~k ~ ~ a a k 13.4. Electric Dipole Transitions 205

We can pull the creation operator outside of the matrix elements, Tip 2 hf| aˆ† |ii = n + 1 . λ~k λ~k Electric dipole transitions are the dominant radia-

The nλ~k gives the induced photons radiated, and the +1 gives the spontaneous radiation. tion for the interaction of Simplifying, gives us an atomic electron with an electromagnetic ﬁeld. 2 2 4π X ea ~ dw˙ = hf| ~p · ~e∗ e−i(k·~ra) |ii n + 1 δ (E − E − ω) dν . fi ω V m a λ~k λ~k i f ~ f k a a

Recall that for photons, we can write

2 V 3 V 2 dEf V ω dνf = 3 d p = 3 p dp dΩp = 3 3 dΩp dEf , (2π~) (2π~) dEf (2π) ~c

2 since Ef = ~ω = pc, so dp = dEf /c = (~ω/c) . Integrating the energy over the delta- function gives us V ω2 dνf = dΩf . (2π)3~c3 Plugging this back in,

2 ω X ea ~ dw˙ = hf| ~p · ~e∗ e−i(k·~ra) |ii n + 1 dΩ . (13.1) fi 2π c3 m a λ~k λ~k f ~ a a where ω = (Ei − Ef )/~. This is valid for non-relativistic radiation. We got to this result by following this procedure: 1. We assumed minimal coupling. I.e. we neglected spin, and for every particle, we substituted e ~p −→ ~p − A~(~r). c 2. We quantized the ﬁeld 3. We applied the Golden Rule 4. We used the density of states in the continuum In the next section, we make additional assumptions. The intensity of the radiation is given by

dI = ~ω dw˙ fi.

13.4 Electric Dipole Transitions

If the size of the system is R, and if kR << 1, then we can use the dipole approximation and set ~ e−i(k·~ra) = 1. If the wavelength λ of the radiation is much larger than the size of the system R, then

ω ~ω ~ωR ~ω R ~k · ~ra < kR = R = R = = . c ~c 2 × 10−5 eV · cm 2 × 10−5 eV cm This is a rough estimate. For an atom, kR ∼ 0.5 × 10−3. For a nucleus, kR ∼ 1/20. When dipole transitions are allowed by the selection rules, then they are preferred by the system. 206 Photons and the Electromagnetic Field

Applying the dipole approximation in Eq. (13.1) gives us

2 ω X ea dw˙ = ~e∗ · (~p ) n + 1 dΩ , fi 2π c3 m λ~k a fi λ~k f ~ a a

where

˙ 1 h ˆ i im ~pfi ≡ hf| ~p |ii = m~rfi = m ~r, H = − ~rfi (Ei − Ef ) = −imω~rfi. i~ fi ~ Therefore, 3 2 ω ∗ ~ dw˙ fi = ~e ~ · dfi n ~ + 1 dΩf , 2π~c3 λk λk where the dipole operator is X X d~fi = ea~rfi = ea hf| ~ra |ii . a a

The rate of spontaneous (i.e. we drop the nλ~k term) dipole radiation is given by

3 2 ω ∗ ~ dw˙ fi = ~e ~ · dfi dΩf . (13.2) 2π~c3 λk To get this result, in addition to the assumptions made in the previous section, we assumed 1. That the radiation wavelength is much larger than the size of the system. This allowed us to set ~ e−i(k·~ra) = 1. 2. We assumed that

~pfi = −imω~rfi. This is true for non-relativistic systems where the interaction does not depend on velocities. These assumptions led us to the electric dipole (i.e. E1) approximation. We still need to sum over the polarizations and integrate over the emission angles. A transverse plane wave with wave vector ~k, can have two polarizations ~e1 and ~e2 in the plane perpendicular to ~k. Since these polarization unit vectors are perpendicular to each other and to ~k, we know that

~e1 · ~e2 = 0

~e1 · ~k = 0

~e2 · ~k = 0.

Let the direction ~k of the emitted radiation be in the z-direction. Then we can choose the two polarization vectors ~e1 and ~e2 to be real, and we can choose them such that one of them, ~e2, is orthogonal to d~fi. Then ~e2 · d~fi = 0, and

~e1 · d~fi = |d~fi| sin θ,

where θ is the angle between ~k and d~fi. Alternatively, we could write

~e1 · d~fi = |nˆ × d~fi|,

where nˆ = (sin θ cos φ, sin θ sin φ, cos θ) is the unit vector in the direction of ~k. 13.4. Electric Dipole Transitions 207

Then, summing Eq. (13.2) over the polarizations gives us Tip 3 2 ω ~ 2 dw˙ fi = dfi dΩf sin θ. To determine if a given 2π c3 ~ dipole transition is allowed, Then we integrate over the angles check the selection rules.

3 2 ω 2 w˙ fi = d~fi dΩf sin θ, 2π~c3 ˆ which gives us the radiation rate

4 ω3 2 w˙ fi = d~fi . 3 ~c3 The intensity radiated per second is

4 ω4 2 I = ω w˙ = d~ . ~ fi 3 c3 fi

Compare this with classical EM theory,

2 1 ¨ 2 I = d~ . classical 3 c3 Recall from classical EM theory that if the total charge of a system is not zero, then the dipole moment is not independent of the choice of origin. In our case, if the system is not charge neutral, then the dipole moment is not well-defined. Recall that the dipole operator is X d~ = ea~ra. a If we translate it by R~ , then X X X d~ = ea~ra −→ ea ~ra − R~ = d~− R~ ea. a a a P So if the total charge, a ea, is not zero, then the dipole operator is not translationally invariant. To fix this, we define the dipole operator at the center of mass of the system.

Selection Rules (Dipole Transitions) When is dipole radiation allowed to cause the transition from some given initial state |ii to some given ﬁnal state |fi? This is what the selection rules answer. The dipole operator changes the parity Π. So one selection rule is that the parity of the ﬁnal state must be opposite the parity of the initial state

Πf = −Πi.

The parity of an atomic orbital is P Π = (−1) `, where the sum is over the electrons. The sum P ` is not always equal to the total orbital momentum L that characterizes a given orbital. If there’s only a single electron, like in hydrogen, then Π = (−1)L. Given an initial state |JiMii and ﬁnal state |Jf Mf i, our matrix elements are

d~fi = hJf Mf | d~|JiMii . 208 Photons and the Electromagnetic Field

~ 2 2 2 2 We can write d in Cartesian form, then d = dx + dy + dz, where the components are dx = ex, dy = ey, and dz = ez. Note that for E1, we can ignore spin, then J = L. Recall that Lz = xpy − ypx and Lz |LMi = M |LMi. It can be shown that [Lz, z] = 0. This implies that

0 = hLf Mf | [Lz, z] |LiMii

= hLf Mf | Lzz − zLz |LiMii

= hLf Mf | Lzz |LiMii − hLf Mf | zLz |LiMii

= M hLf Mf | z |LiMii − hLf Mf | z |LiMii Mi

= (Mf − Mi) hLf Mf | z |LiMii .

This implies that hLf Mf | z |LiMii = 0, or ∆M = 0. In other words, z dfi = 0, unless ∆M = 0.

If we deﬁne x± ≡ x ± iy, then it can be shown that [Lz, x±] = ±~x±. Using the same procedure as above, we ﬁnd that

x y dfi = dfi = 0, unless ∆M = ±1. The only possibilities are ∆M = 0, ±1.

x y What we showed above is that for the case ∆M = 0, we know that dfi = dfi = 0, and for z the case ∆M = ±1, we know that dfi = 0. Remember that our z-axis is in the direction ~k in which the radiated photon propagates. If we take into account angular momentum conservation, then Jf = Ji,Ji ± 1, or

∆J = 0, ±1 (except 0 → 0).

The special case Ji = 0 going to Jf = 0 is forbidden. On the other hand, if the state is characterized by L and S instead of J, then Lf = Li,Li ± 1 and ∆S = 0 since d~ does not operate on spin. So

∆S = 0, ∆L = 0, ±1 (except 0 → 0).

The special case Li = 0 going to Lf = 0 is forbidden. If atomic and nuclear angular momenta are both conserved, then

F~ = J~ + I~,

where I~ is the nuclear angular momentum, and J~ is the total angular momentum of the electrons. Then we have the selection rule Ff = Fi,Fi ± 1, or

∆F = 0, ±1.

Example 13.4.1: Lyman Series

The Lyman series is the set of electron transitions from initial state n0 ≥ 2 to ﬁnal state n = 1 in the hydrogen atom. This series results in ultraviolet (UV) emission lines. Identify the allowed transitions from n0 = 2. 0 At n = 2, we have the states 2s1/2, 2p1/2, and 2p3/2. The ﬁnal state must 13.4. Electric Dipole Transitions 209

be 1s1/2 since this is the only state at n = 1. For atomic orbitals, the parity is Π = (−1)L. So the parity of the ﬁnal state which has L = 0 is +1. Since the parity must change in the transition, the initial states must be the p states with L = 1. The 2p1/2 and 2p3/2 states are split by ﬁne structure. They are coupled so that J~ = L~ + S~, so they have no L and S quantum numbers. If we neglect the Lamb shift, then 2s1/2 and 2p1/2 have the same energy.

Example 13.4.2: Balmer Series

The Balmer series is the set of electron transitions from initial state n0 ≥ 3 to ﬁnal state n = 2 in the hydrogen atom. This series results in visible emission lines. Identify the allowed transitions from n0 = 3. 0 At n = 3, we have the states 3s1/2, 3p1/2, 3p3/2, 3d3/2, and 3d5/2. We know that 3s −→ 2s and 3p −→ 2p transitions are not allowed because the parity does not change. We cannot have 3d −→ 2s because then ∆L = 2. We also cannot have 3d5/2 −→ 2p1/2 because then we would have ∆J = 2. The other seven possibilities are allowed and illustrated below.

Transition Lifetime The lifetime is given by 1 τfi = . w˙ fi

For hydrogen, a rough estimate of the dipole is d = ea0. Then

3 2 ω 2 1 ω 2 2 2 w˙ fi ∼ (ea) ∼ ω e a0 ∼ ωα(ka0) , ~c3 ~c c2 where k = ω/c. We can estimate

ω ~ω me4 1 ~2 e2 ka0 ∼ a0 ∼ a0 ∼ ∼ ∼ α. c ~c ~2 ~c me2 ~c 210 Photons and the Electromagnetic Field

So 3 w˙ fi ∼ ωα . So the lifetime is approximately 1 1 τ ∼ . fi ω α3 Then τ 1 fi ∼ ∼ 106, T α3 where T is the period of the electron’s orbital motion. So we need to wait about 106 periods for radiation to occur.

13.5 Multipole Transitions

Now instead of considering dipole transitions, we consider higher multipole transitions. ~ Previously, we approximated e−i(k·~ra) = 1. Now, we go to the next higher order

−i(~k·~ra) e ≈ 1 − i(~k · ~ra).

The ‘1’ gave us the electric dipole (E1) piece. The second term, i(~k · ~ra), will give us two new pieces. This higher order term preserves parity, so quadrupole transitions can occur between initial and ﬁnal states with the same parity.

Now Eq. (13.1) for spontaneous radiation (i.e. nλ~k = 0) gives the new piece 2 ω X ea dw˙ = hf| ~p · ~e∗ ~k · ~r |ii dΩ . fi 2π c3 m a λ~k a f ~ a a We can write X ea Fˆ = ~p · ~e∗ ~k · ~r = Fˆ+ + Fˆ−, m a λ~k a a a where

X ea h i Fˆ± = ~p · ~e∗ ~k · ~r ± ~r · ~e∗ ~k · ~p . 2m a λ~k a a λ~k a a a ˙ Then using ~pa = ma~ra, we can write

X ea h i Fˆ+ = ~r˙ · ~e∗ ~k · ~r + ~r · ~e∗ ~k · ~r˙ 2 a λ~k a a λ~k a a 1 X d h i = e ~r · ~e∗ ~k · ~r 2 a dt a λ~k a a 1 X µ d = e ~e∗ kν (xµxν ) . 2 a λ~k dt a a a Quadrupole moments are deﬁned as

µν X µ ν 2 µν Q = ea 3xa xa − raδ . a This is a second-rank, symmetric, and traceless tensor. So we can write

1 X µ d 1 X µ h i Fˆ+ = ~e∗ kν Qˆµν = ~e∗ kν Qˆµν , Hˆ . 6 λ~k dt 6i λ~k a ~ a 13.5. Multipole Transitions 211

The matrix elements of this operator are

+ 1 ∗ µ ν h ˆµν ˆ i 1 ∗ µ ν µν ω ∗ µ ν µν Ffi = ~e ~ k Q , H = ~e ~ k (Ei − Ef ) Qfi = ~e ~ k Qfi . 6i~ λk fi 6i~ λk 6i λk Then the electric quadrupole (E2) transition rate is ω3 2 E2 ∗ µ ν µν dw˙ fi = ~e ~ k Qfi dΩ. 72π~c3 λk Now, consider F −. We can write (~p · ~e) ~k · ~r − (~r · ~e) ~k · ~p = ~k × ~e∗ · (~r × ~p) . To see this, we write ~ ∗ ∗ k × ~e · (~r × ~p) = εij`kje` εimnxmpn ∗ = (δjmδ`n − δjnδ`m) kje` xmpn ∗ ∗ = kjenxjpn − kje` x`pj = (~p · ~e) ~k · ~r − (~r · ~e) ~k · ~p . Also, we can write ∗ ∗ ~k × ~e · (~r × ~p) = ~k × ~e · ~~`. Then X ea Fˆ− = hf| ~k × ~e∗ · ~` |ii . fi 2m λ~k ~ a a ~ ω ~ Using k = c nˆ, where nˆ is the unit vector in the k direction, this becomes

X ea Fˆ− = ω hf| ~ ~` · nˆ × ~e∗ |ii fi 2m c λ~k a a X (`)~ ∗ = ω hf| ga ` · nˆ × ~eλ~k |ii a ∗ = ω hf| ~µ` · nˆ × ~eλ~k |ii , (`) where ga is the gyromagnetic ratio, and ~µ` is the orbital magnetic moment. If we wanted to include spin, we would replace this with ~µ` + ~µs. We can also write ∗ ˆ nˆ × ~eλ~k = B, where Bˆ = B~ /|B~ | is the unit vector along the magnetic ﬁeld. Then the magnetic dipole (M1) radiation is determined by 3 M1 ω ∗ 2 dw˙ fi = nˆ × ~e ~ · ~µfi dΩf . 2π~c3 λk We integrate this, as we did for the electric dipole radiation, then

3 M1 4 ω 2 w˙ fi = ~µfi . 3 ~c3 The selection rules for M1 radiation are: • Parity does not change • The same vector rules hold as for E1 transitions. E.g. ∆J = 0, ±1 except the transition 0 → 0 is forbidden

In the hydrogen atom, the M1 transition 2s1/2 → 1s1/2 is forbidden by the selection rules. However, relativistic eﬀects conspire to allow this transition, although it has a very long lifetime of ∼ 1/7 s. A similar eﬀect occurs in orthohelium with S = 1 and parahelium with S = 0. 212 Photons and the Electromagnetic Field

13.6 Photoabsorption

Now we look at the opposite of radiation—when photons are absorbed rather than emitted. The transition rate is 2 2π 2π e 2 δ(ω − ω ) ~ X a i(~k·~ra) fi dw˙ fi = hf| ~e ~ · ~p e |ii hn ~ − 1|aˆ|n ~ i . ωV m λk a λk λk ~ a a ~ √ Simplifying the ladder operator acting on the number states, hnλ~k − 1|aˆ|nλ~ki = nλ~k, so photoabsorption is proportional to the number of available photons

2 2 4π X ea ~ dw˙ = hf| ~e · ~p ei(k·~ra) |ii n δ(ω − ω ). fi ω V m λ~k a λ~k fi ~ a a Now we can make the same approximations as we did for radiation. The cross-section is given by dw˙ dw˙ σ = = n . ﬂux V c It has units of area. We see that 2 2 4π X ea ~ σ = hf| ~e · ~p ei(k·~ra) |ii δ(ω − ω ). fi ωc m λ~k a fi ~ a a In reality, the energy width is not inﬁnitesimal as implied by the delta function. Due to uncertainty, there is some width. So we replace the delta-function by a Lorentzian γ 1 δ(ω − ωfi) −→ 2 2 , 2π (ω − ωfi) + γ /4 where Γ + Γ γ = i f . ~ The γ is the full width at half maximum (FWHM).

The lifetime is 1 τ = . γ For an atom, 2π2e2 dω σ = Z. ˆ mc For a nucleus, 2π2e2 NZ dω σ = . ˆ mc A 13.6. Photoabsorption 213

In many books, this is written as

m c2 π 2π2e2~ NZ 0.06 MeV · barn NZ dE σ = = −24 2 , ˆ0 mc A 10 cm A where the integral only goes up to the energy of pion production. This excludes the center-of-mass motion. If we include COM motion, then

2 mπ c 2π2e2 NZ Z2 2π2e2 dE σ = ~ + = ~Z. ˆ0 mc A A mc 214 Photons and the Electromagnetic Field

13.7 Summary: Photons and the EM Field

Skills to Master •

Quantizing the EM Field The lifetime is given by The procedure to quantize the EM field is: 1 τfi = . 1. Write down the Hamiltonian in terms of a ‘co- w˙ fi ordinate’ Qˆ and its conjugate ‘momentum’ Pˆ, where Qˆ and Pˆ satisfy the standard commuta- Electric Dipole (E1) Transitions tion relation Keeping only the first term in the multipole expansion a) Write down A~ as a linear combination of allows us to simplify and integrate out the angles to plane waves get the radiation rate b) Write down E~ and B~ from A~ E2+B2 3 2 c) Write down the energy E = dV E1 4 ω ~ 8π w˙ fi = dfi , d) Treat everything as operators´ and derive Qˆ 3 ~c3 ˆ and P where d~fi = hf|d~|ii is the dipole operator evaluated e) Determine their commutation relation between the final and initial states. 2. Write down E~ and B~ now in terms of operators Remember, the dipole operator is defined as 3. Interpret the coefficients as creation/annihilation ~ X operators d = ea~ra. a Radiation So, for example, if there is a single particle and it moves only in the x direction, then d = ex. Then you would Radiation occurs when an electron transitions |ii → |fi have to calculate a single element dfi = e hf|x|ii. In where Ei > Ef . The energy drop corresponds to a pho- general, you have to calculate three elements—one for ton with energy ~ω = Ei − Ef leaving the atom. This each Cartesian component of d~. These can also be ex- photon is characterized by some wave vector ~k and po- pressed in spherical coordinates. larization ~e. Only certain E1 transitions are allowed. To deter- The radiation can be induced by an external per- mine if an electric dipole transition is allowed, check the turbation such as an applied electric field or it can be selection rules. To be an allowed transition |fi → |ii, spontaneous. We only consider spontaneous emission • The parity of initial and final states must be op- here. The rate at which spontaneous radiation occurs, posite. This is because d~ changes parity. The in the non-relativistic approximation, is given by parity of an atomic orbital is P 2 Π = (−1) `, ω X ea ~ dw˙ = hf| ~p · ~e∗ e−i(k·~ra) |ii dΩ . fi 2π c3 m a λ~k f ~ a a where the sum is over the electrons. For example, for a hydrogen atom with a single electron, the p If the system (e.g. atom) is of size R, then if kR << 1, orbital, which has ` = 1, has negative parity. An we can do a multipole expansion of the exponential in E1 transition like from one p orbital to another is dw˙ fi not allowed because the parity does not change. −i(~k·~ra) e = 1 − i(~k · ~ra) + ··· . • If the total angular momentum J is a good quantum number, then a selection rule is The first term gives us electric dipole (E1) radiation. The second term gives us magnetic dipole (M1) and ∆J = 0, ±1, (except 0 → 0). electric quadrupole (E2) radiation. E1 transitions For example, in the hydrogen atom, the transi- are the dominant radiation from the interaction of an tion d5/2 → s1/2 is not allowed because ∆J = 2. atomic electron with an electromagnetic field. • If the state is characterized by L and S instead The intensity radiated per second is of J, then a selection rule is

I = ~ω w˙ fi. ∆S = 0, ∆L = 0, ±1 (except 0 → 0). 13.7. Summary: Photons and the EM Field 215

• If atomic and nuclear angular momenta are both The selection rules for M1 radiation are: conserved, then F~ = J~+I~, where I~ is the nuclear angular momentum, and J~ is the total angular • Parity does not change momentum of the electrons. Then we have the • The same vector rules hold as for E1 transitions. selection rule E.g. ∆J = 0, ±1 except the transition 0 → 0 is ∆F = 0, ±1. forbidden

Magnetic Dipole (M1) Transitions Misc. The second term in the multipole expansion gives us The following may be helpful to know: two pieces—the electric quadrupole (E2) and the mag- −22 netic dipole (M1). Here we consider only the latter. ~ ' 6.58 × 10 MeV · s For the magnetic dipole, the radiation rate is 2 ~ ˚ a0 = 2 ' 0.529 A (Bohr radius) 3 mee M1 4 ω 2 2 w˙ fi = ~µfi , e 1 3 ~c3 α = ' (Fine structure constant) ~c 137 where 1 2 2 X ~ α mec = 1 Ry = 13.6 eV (Atomic energy unit) ~µfi = hf| ga` |ii , 2 is the magnetic dipole operator evaluated between the ~c ' 197 MeV · fm 2 ﬁnal and initial states. me ' 0.511 MeV/c Chapter 14

Relativistic Quantum Mechanics

14.1 Klein-Gordon Equation

In free motion, any particle should satisfy the Klein-Gordon equation. In the non-relativistic case, a free particle has energy

p2 E = . 2m To get the quantum results, we substitute

∂ E → Hˆ = i ~∂t ~p → −i~∇~ .

Doing so gives us the Schrodinger equation. In relativity, the energy of a free particle has the form

E = p(mc2)2 + ~p 2c2.

We know the quantity E2 − c2~p 2 = m2c4, is invariant. We can write this as

(E + mc2)(E − mc2) = c2~p 2.

In the non-relativistic limit, this gives us

~p 2 2mc2(E − mc2) ≈ c2~p 2 =⇒ E = mc2 + . NR 2m Now we substitute in the operators and apply everything to a wave function Ψ(~r, t). So E2 − c2~p 2 = m2c4, becomes ∂ 2 2 i Ψ − c2 −i ∇~ Ψ = m2c4Ψ. ~∂t ~ This simpliﬁes to 1 ∂2 m2c2 − ∇2 + Ψ = 0. c2 ∂t2 ~2 14.1. Klein-Gordon Equation 217

This is Klein-Gordon equation, which is the equation of motion for any free particle with mass m. If m = 0, then we get the equation of motion for a photon. In relativistic units, the Klein-Gordon equation is written as ∂2 − ∇2 + m2 Ψ = 0. ∂t2 An important case is that of a “static source”, i.e., when the particle doesn’t move. Then the time derivative is zero, and we add a source term gδ(~r) on the right-hand side.

m2c2 ∇2Ψ − Ψ = gδ(~r). ~2 If m = 0, this just gives us the Coulomb potential. Recall the Compton wavelength λ = ~/mc. Then the solution is e−r/λ Ψ = const · . r

This is the Yukawa potential. For example, if m ≡ mπ, then this is the pion ﬁeld. In the static case, Ψ is not a wave function—it’s just a spatial distribution. The uncertainty of the particle’s position is

∆x ≤ ~ = λ. mc

Then ∆p ∼ mc, ∆E ∼ mc2, ∆t ∼ ~/mc2, and ∆p · ∆t ∼ ~/c. These are the physical applicability limits of our equation. E.g., if ∆E > mc2, then we could have multiple particles, and we would then need the full formalism of QFT. The Klein-Gordon equation has plane wave solutions

− i Et+ i ~p·~r Ψ(~r, t) = Ae ~ ~ .

If we take the square root of the invariant quantity E2 = c2~p 2 + m2c4, then we get a negative energy solution p E = ± m2c4 + ~p 2c2. We cannot sweep away the negative energies with the excuse of unphysicality. Complete- ness of the set of solutions requires that we keep them. Instead, we interpret a solution with E < 0 as the complex conjugate of the solution for an antiparticle. For an electron with charge e and an external potential φ with minimal coupling, 2 e 2 Hˆ − eφ Ψ = c2 ~p − A~ Ψ + m2c4Ψ. c Putting in the operators for Hˆ and ~p gives us

∂ 2 e 2 i − eφ Ψ = c2 −i ∇~ − A~ Ψ + m2c4Ψ. ~∂t ~ c The complex conjugate is

∂ 2 e 2 −i − eφ Ψ∗ = c2 i ∇~ − A~ Ψ∗ + m2c4Ψ∗. ~∂t ~ c We can factor out a negative from each set of parentheses to get

∂ 2 e 2 i + eφ Ψ∗ = c2 −i ∇~ + A~ Ψ∗ + m2c4Ψ∗. ~∂t ~ c 218 Relativistic Quantum Mechanics

So we see that this is a solution for a particle with charge −e instead of +e. The charge conjugation operator Cˆ gives us

CˆΨ = Ψ∗.

For E > 0 solutions, take Ψ(+). If E < 0, take Ψ(−) ∗. We are still in the regime of quantum mechanics, as opposed to quantum ﬁeld theory. In quantum ﬁeld theory, there are creation and annihilation operators allowing particles to be created and annihilated. In quantum mechanics, however, particle number is conserved. This implies a continuity equation

∂ρ + ∇~ ·~j = 0. ∂t

Incidentally, this continuity equation is relativistically invariant. The “current” is

h i ~j = ~ Ψ∗∇~ Ψ − Ψ∇~ ψ∗ . (14.1) 2mi

Taking the divergence of the current gives us

h i ∇~ ·~j = ~ ∇~ Ψ∗ · ∇~ Ψ + Ψ∗∇2Ψ − ∇~ Ψ · ∇~ Ψ∗ − Ψ∇2Ψ∗ 2mi = ~ Ψ∗∇2Ψ − Ψ∇2Ψ∗ 2mi 1 ∂2Ψ m2c2 1 ∂2Ψ∗ m2c2 = ~ Ψ∗ + Ψ − Ψ + Ψ∗ 2mi c2 ∂t2 ~2 c2 ∂t2 ~2 ∂2Ψ ∂2Ψ∗ = ~ Ψ∗ − Ψ 2mc2i ∂t2 ∂t2 ∂ ∂Ψ ∂Ψ∗ = ~ Ψ∗ − Ψ . 2mc2i ∂t ∂t ∂t

Comparing with the continuity equation, we see that

i ∂Ψ ∂Ψ∗ ρ = ~ Ψ∗ − Ψ . (14.2) 2mc2 ∂t ∂t

So if Ψ is a solution of the Klein-Gordon equation, then the continuity equation is satisfied given the definitions Eq. (14.1) and Eq. (14.2). For the plane wave solution noted earlier, we find that the current is

i i ~p ~j = ~ |A|2 ~p − − ~p = |A|2 , 2mi ~ ~ m and the density is i i i E ρ = ~ |A|2 − E − E = |A|2. 2mc2 ~ ~ mc2 This is the density for a relativistic wave, as opposed to the usual |A|2 for a non-relativistic wave. The extra factor pf E/mc2 is due to Lorentz contraction. Let E = mc2 + ε. Then the non-relativistic limit is ε << mc2. We want to subtract out the rest energy, so we write the wave as

− i mc2t Ψ(~r, t) = e ~ Ψ(˜ ~r, t). 14.1. Klein-Gordon Equation 219

Plugging this back into the Klein-Gordon equation and simplifying the time derivative Tip gives us 2im ∂Ψ˜ 1 ∂2Ψ˜ − + − ∇2Ψ˜ = 0. Keep in mind that the ~ ∂t c2 ∂t2 Klein-Gordon equation re- Then we can write ally only works for spinless − i εt particles. Ψ˜ = e ~ ψ(~r), where ψ(~r) is independent of time. Plugging this in and simplifying gives us

2mε ε2 − ψ − ψ − ∇2ψ = 0. ~2 ~2c2 We can rearrange this as 2 ε2 − ~ ∇2ψ = εψ + ψ. 2m 2mc2 In the non-relativistic limit, the second term on the right is zero, and we recover the ordinary (time-independent) Schrodinger equation for a free particle. The energy used in the ordinary Schrodinger equation does not include the rest energy of the particle, so we see ε on the right-hand side rather than E = mc2 + ε. With minimal coupling, the Klein-Gordon equation for a particle with charge e and mass m in an electromagnetic ﬁeld with scalar potential φ and vector potential A~ is

" # 1 ∂ 2 e 2 i − eφ − ~p − A~ − m2c2 Ψ(~r, t) = 0. c2 ~∂t c or in relativistic units, " # ∂ 2 2 i − eφ − ~p − eA~ − m2 Ψ(~r, t) = 0. ∂t

In relativity, E2 = ~p 2 + m2. Thus, the stationary states ψ(~r) with energy E satisfy the eigenvalue problem ~p 2 + m2 ψ(~r) = E2ψ(~r). But for a charged particle in EM ﬁelds, we need to make the minimal coupling substitutions ~p → ~p − eA~ and E → E − eφ. So the stationary states satisfy the eigenvalue problem

2 ~p − eA~ + m2 ψ(~r) = (E − eφ)2 ψ(~r).

For example, for an electron (NOTE: I’m only using the electron here for illustrative purposes. The Klein-Gordon equation is not actually valid for particles with spin.) with charge e and mass m in the static electric ﬁeld of a nucleus with charge Ze, the scalar potential is φ = Ze/r and the vector potential is A~ = 0. Then the Klein-Gordon equation becomes " # ∂ Ze2 2 i − − ~p 2 − m2 Ψ = 0. ∂t r

The stationary states ψ(~r) with energy E satisfy the eigenvalue problem

Ze2 2 ~p 2 + m2 ψ(~r) = E − ψ(~r). r 220 Relativistic Quantum Mechanics

14.2 Four-vectors and Minkowski Space

In Euclidean space R3, the main objects are vectors

~v = (vx, vy, vz), and their deﬁning transformation is the rotation. The length squared of a vector is

3 2 2 2 2 X 2 |~v| = vx + vy + vz = vi . i=1 This quantity is invariant under the deﬁning transformation. For two vectors, ~u and ~v, the inner product is ~u · ~v = u1v1 + u2v2 + u3v3. In Minkowski space, the main objects are 4-vectors, which in contravariant form are written aµ = a0, a1, a2, a3 ≡ a0,~a .

The covariant vector is then

0 1 2 3 0 aµ = a , −a , −a , −a ≡ a , −~a .

Their deﬁning transformation is the Lorentz transformation. The “length squared” of a 4-vector is 2 µ 0 2 1 2 2 2 3 2 0 2 2 a ≡ a aµ = (a ) − (a ) + (a ) + (a ) = (a ) − ~a . This quantity is invariant under the deﬁning transformation. In general, the scalar product of two 4-vectors is µ 0 0 a bµ = a b − ~a ·~b.

We will often use relativistic units c = ~ = 1. Then E, m, and p all have units of energy and length and time both have units of length. With these units, E2 = m2 + ~p 2 and the invariant quantity is E2 − ~p 2 = m2. The relativistic velocity is d~x ∂E ~p ~v = = = . dt ∂~p E With full units, ~v = c2~p/E. We can write

E~v 2 E2~v 2 E2 = m2c4 + c2 = m2c4 + . c2 c2 Then mc2 E = ≥ mc2. p1 − ~v 2/c2 In relativistic units, m E = √ = γm, 1 − ~v 2 where the Lorentz factor is 1 γ = √ . 1 − ~v 2 The contravariant spacetime 4-vector is

xµ = x0, ~x , 14.2. Four-vectors and Minkowski Space 221 where µ = 0, 1, 2, 3, and x0 = t. The covariant version is

0 xµ = x , −~x . The quantity µ 0 2 2 x xµ = (x ) − ~x , is a scalar and is invariant. The four-momentum is P ≡ pµ = (E, ~p) . Note the capital ‘P’ without indices will refer to this four-momentum. The invariant quantity is 2 µ 2 2 2 P = p pµ = E − ~p = m . For a photon, P 2 = 0. The covariant gradient operator is ∂ ∂ = . µ ∂xµ

Example 14.2.1: Compton Eﬀect

Consider a photon with initial frequency ω that is scattered oﬀ of an electron at an angle θ, and leaves with a frequency ω0.

Initially, the energy of the electron is just the rest mass E = m. After the scattering event, the electron has some energy E0. The photon has initial 4- momentum K and ﬁnal 4-momentum K0, and the electron has initial 4-momentum P and ﬁnal 4-momentum P 0. Conservation of energy-momentum requires that

K + P = K0 + P 0.

We are interested in the scattering angle θ, so we rearrange to put the vectors K and K0 involving this angle on one side

K − K0 = P 0 − P.

Squaring both sides gives us

K2 + K0 2 − 2K · K0 = P 0 2 + P 2 − 2P · P 0.

Recall that in general P 2 = E2 − ~p 2 = m2. For the photon, K2 = K0 2 = 0. For the electron, P 2 = P 0 2 = m2. This implies

K · K0 = P · P 0 − m2.

This expression is still invariant. So far, we have not speciﬁed any frame of reference. 222 Relativistic Quantum Mechanics

Tip Now, consider the problem in the lab frame. The initial 4-momentum of the electron is P = (m,~0), and its ﬁnal 4-momentum is P 0 = (E0, ~p 0). This implies The Dirac equation is for that particles with spin-1/2. It P · P 0 = E0m. is not valid for particles with arbitrary spin. The photon, in terms of angular frequency, has initial 4-momentum K = (ω, ~ω) = (ω, ω, 0, 0) and ﬁnal 4-momentum K0 = (ω0, ~ω 0) = (ω0, ω0 cos θ, ω0 sin θ, 0). This implies that K · K0 = ωω0 − ωω0 cos θ = ωω0(1 − cos θ). Plugging these into K · K0 = P · P 0 − m2 gives us

ωω0(1 − cos θ) = E0m − m2.

Energy conservation tells us that

ω + m = ω0 + E0.

Using this to eliminate E0 gives us

0 ω ω = ω . 1 + m (1 − cos θ) Adding units gives us ω ω0 = . ~ω 1 + mc2 (1 − cos θ)

14.3 Dirac Equation

Recall that for relativistic particles,

Hˆ 2 =p ˆ2 + m2. ˆ ∂ ~ This should hold true for all spin-1/2 free particles. Plugging in H = i ∂t and ~p = −i∇ gives us the Klein-Gordon equation

∂2Ψ 2 − = −∇~ Ψ + m2Ψ. ∂t2 This is valid for free and spinless particles. Since electrons have spin, we want an equation for free particles with spin. Instead of using the approach that leads to the Klein-Gordon equation, we now try to generalize the Schrodinger equation by writing

∂ i Ψ = βm + c~α · ~p Ψ, ~∂t

where ~α = (αx, αy, αz). This is the Dirac equation. The αi and β are unknown. We know they cannot be numbers, but they could be operators acting on the internal degrees of freedom of the particle. Our new Hamiltonian, at least for a free particle, is

Hˆ = βm + c~α · ~p = βm + cαxpˆx + cαypˆy + cαzpˆz.

We know that Hˆ 2 =p ˆ2 + m2, so let’s square our new Hamiltonian. Keep in mind that we don’t know if β and αi commute, but we do know that pi and pj commute. We ﬁnd 14.3. Dirac Equation 223 that

ˆ 2 2 2 2 2 2 2 2 2 H = β m + αxpˆx + αypˆy + αzpˆz

+ m {(βαx + αxβ)p ˆx + (βαy + αyβ)p ˆy + (βαz + αzβ)p ˆz}

+p ˆxpˆy [αxαy + αyαx] +p ˆypˆz [αyαz + αzαy] +p ˆzpˆx [αzαx + αxαz] .

By comparison with Hˆ 2 =p ˆ2 + m2, we see that

2 2 2 2 β = 1, αx = αy = αz = 1.

We also see that β and the αi must anticommute

[αi, β]+ = 0, so that terms like βαx + αxβ are zero. Finally, we must have

[αi, αj]+ = 0 for i 6= j, so that terms like αyαz + αzαy are zero. Assuming β and αi are matrices, the lowest dimension of matrices which can satisfy all of these conditions is four. So we use 4 × 4 matrices to represent these operators. There are many possible choices but the standard representation is " # " # 1 0 0 σi β = , αi = , 0 −1 σi 0 where each element of these matrices is itself a 2 × 2 matrix. In particular, the σi are the Pauli matrices. Importantly, 2 2 β = αi = 1.

The eigenvalues of β and all three αi are ±1. 2 2 Note, the notation αi is ambiguous. Some people use the terrible notation Ai ≡ AiAi then the doubled index implies a sum over i. That is not the interpretation here. In this 2 2 2 2 example, αi means αx, αy, or αz but never the sum of all three. The β and αi in the Dirac equation are 4 × 4 matrices, so the wave functions are 4-component vectors   Ψ1    Ψ2  Ψ(~r, t) =  .  Ψ   3  Ψ4

The 4 × 4 matrices β and αi are related to the Dirac gamma matrices

γ0 = β, γi = βαi, where i = 1, 2, 3.

Free Particle The Dirac equation and Hamiltonian given above are for a free Dirac particle (i.e. spin- 1/2). What is the velocity of the particle? The Heisenberg operator is

˙ h i i~~r = ~r, Hˆ = [~r, c~α · ~p] . 224 Relativistic Quantum Mechanics

Here we think of ~r as ~r1 since we are dealing with 4 × 4 matrices. We know that ~r commutes with β and ~α. In components,

i~r˙i = [ˆri, cαjpj] = cαji~δij = i~cαi. Thus, the velocity components are given by

vˆi = cαˆi. Apparently, for free motion, ~v does not depend on ~p. Example 14.3.1: Particle at Rest

Consider a particle at rest with ~p = 0 and some energy E. The Hamiltonian eigenvalue equation is

Hˆ Ψ = βmΨ.

Recall that β is a diagonal matrix, β = diag(1, 1, −1, −1). Thus,

E = βm.

This implies four solutions—two positive solutions and two negative solutions. With full units, E = ±mc2.

We will use a two-component notation to write the bispinor   Ψ1 " #   φ  Ψ2  Ψ = ≡  , χ  Ψ   3  Ψ4 in terms of a pair of spinors " # " # Ψ Ψ φ ≡ 1 , χ ≡ 3 . Ψ2 Ψ4

In general, we have a moving particle ~p 6= 0, with some energy satisfying E2 = p2 + m2. Now, ( " # " # ) 1 0 0 ~σ EΨ = m + · ~p Ψ. 0 −1 ~σ 0 Note that we use the abuse of notation  " #  0 σx    σx 0  " #  " #  0 ~σ  0 σ   y  ~α = ≡  . ~σ 0  σy 0     " #   0 σz  σz 0 Using the 2-component form of Ψ, we have " # " #" # " #" # φ 1 0 φ 0 ~σ φ E = m + ~p · . χ 0 −1 χ ~σ 0 χ 14.3. Dirac Equation 225

This gives us a pair of coupled equations

Eφ = mφ + (~σ · ~p) χ Eχ = −mχ + (~σ · ~p) φ.

We can rearrange these as ~σ · ~p φ = χ E − m ~σ · ~p χ = φ. E + m In the non-relativistic limit, if E > M, then φ >> χ. Otherwise, if E < −m, then χ >> φ. So in the non-relativistic limit, for E > 0, it is convenient to choose the solution φ. For E < 0, choose χ. We can write ~σ · ~p ~σ · ~p ~σ · ~p φ = χ = φ. E − m E − m E + m Since (~σ · ~p)(~σ · ~p) = ~p 2, this implies

~σ · ~p ~σ · ~p ~p 2 1 = = , E − m E + m (E − m)(E + m) which gives us E2 = p2 + m2 as expected. Our solution is " # " # φ φ Ψ = N = N ~σ·~p , χ E+m φ where N is some normalization constant. Assuming that the 2-component spinor φ is already normalized so that φ†φ = 1, we ﬁnd that " # h i φ 1 = hΨ|Ψi = φ† χ† N ∗N χ " # h i φ = |N|2 φ† χ† χ = |N|2 φ†φ + χ†χ ! (~σ · ~p)2 = |N|2 1 + (E + m)2 ~p 2 = |N|2 1 + . (E + m)2

Using ~p 2 = E2 − m2, we get r E + m N = . 2E So the time-independent part of our solution is r " # E + m φ Ψ = . 2E ~σ·~p E+m φ In general, for a particle in free motion, we can separate the space and time parts of the wave function as Ψ(~r, t) = ψ(~r)e−iEt, 226 Relativistic Quantum Mechanics

where ψ(~r) satisﬁes Hψˆ = Eψ. We can take the solution to be a plane wave, so

ψ(~r) = ei~p·~ru(~p).

The full solution is Ψ(~r, t) = e−iEt+i~p·~ru(~p), with energy p 2 2 E~p = m + ~p , and bispinor " # r " # φ E + m φ u(~p) = = . 2E ~σ·~p χ E+m φ If we go to a diﬀerent frame, then we’ll have some diﬀerent E and ~p. Recall that the velocity of the particle is ~v = ~α in units where c = 1. What is the real velocity of the particle? The expectation value of the particle in a stationary state is " #" # E + m ~σ · ~p 0 ~σ φ h~vi = u† ~αu = φ† φ† 2E E + m ~σ·~p ~σ 0 E+m φ " # E + m ~σ · ~p ~σ(~σ·~p φ = φ† φ† E+m 2E E + m ~σφ E + m ~σ(~σ · ~p) (~σ · ~p)~σ = φ† φ + φ† φ 2E E + m E + m 1 = φ†~σ(~σ · ~p)φ + φ†(~σ · ~p)~σφ . 2E In components, 1 p hv i = φ†σ (σ p )φ + φ†(σ p )σ φ = j φ† σ σ + σ σ φ. i 2E i j j j j i 2E i j j i

Using σiσj = δij + iεijkσk, this becomes p p p hv i = j φ† 2δ φ = i φ†φ = i . i 2E ij E E So (in full units) c2~p h~vi = . E This is the correct relativistic velocity for the case E > 0. A moving wave function is a superposition of particles which would have positive and negative energies at rest. One interpretation is that a moving particle is a superposition of the bare particle plus virtual pairs. Consider the graphic below which depicts a moving electron. As it moves along, virtual e+ and e− pairs pop into existence, the e+ annihilating with the initial bare particle, leaving the virtual e− to become the real particle. This idea is called “zitterbewegung” which means “trembling motion” in German. So-called since it looks like a single electron jittering along. 14.3. Dirac Equation 227

Angular Momentum The spin operator of a Dirac particle is a vector of 4 × 4 matrices

" # 1 σi 0 ~s = Σ~ , Σi = . 2 0 σi

The Heisenberg equation of motion of the spin operator is h i 1 h i 1 h i i~s˙ = ~s, Hˆ = Σ~ , Hˆ = Σ~ , ~α · ~p . 2 2 In terms of components, 1 is˙ = [Σ , α ] p i 2 i j j "" # " ## 1 σi 0 0 σj = , pj 2 0 σi σj 0 " # 1 0 σiσj − σjσi = pj 2 σiσj − σjσi 0 " # 1 0 [σi, σj] = pj 2 [σi, σj] 0 " # 0 σk = iεijk pj σk 0

= iεijkpjαk.

Thus, ~s˙ = ~p × ~α. We see that spin is conserved only for stationary particles with ~p = 0. For moving particles, it is not conserved due to the relativistic eﬀects of spin-orbit interaction. The helicity operator Σ~ · ~p hˆ = , |~p| is the projection of the spin onto the direction of motion ~p. Spin is not conserved in the Dirac equation, but helicity is. In other words, the projection of the spin onto the momentum does not change in time d (~s · ~p) = 0. dt

The helicity operator hˆ has eigenvalues ±1. The eigenvalue +1 corresponds to a particle with right-handed longitudinal polarization (or just “right polarization”), and the eigenvalue −1 corresponds to a particle with left polarization. Helicity is conserved, but it is not relativistically invariant for massive particles. For a massive particle, one can always ﬁnd a frame in which the helicity is reversed. However, helicity is invariant for massless particles since they move at the speed of light—so there exists no frame in which the helicity is reversed. The Heisenberg equation for the orbital momentum of the particle is ˙ h i i~` = ~`, Hˆ = [~r × ~p, ~α · ~p] . 228 Relativistic Quantum Mechanics

In components, ˙ i`i = [εijkxjpk, αnpn] = εijkpkαniδjn = iεijkpkαj = i (~α × ~p)i . So ˙ ~` = ~α × ~p. So the orbital momentum is also not conserved. The total momentum ~j = ~` + ~s, however, is conserved.

Parity Does the Dirac equation conserve parity Pˆ? This is equivalent to asking if it commutes with the Hamiltonian PˆHˆ =? Hˆ Pˆ. Since Pˆ is an idempotent operator, this is equivalent to PˆHˆ Pˆ =? H.ˆ We can evaluate the left-hand side PˆHˆ Pˆ = Pˆ (βm + ~α · ~p) Pˆ = βm + ~α · Pˆ~pPˆ = βm − ~α · ~pPˆPˆ = βm − ~α · ~p. So clearly, PˆHˆ Pˆ = 6 H.ˆ However, if we replace Pˆ with βPˆ, then βPˆHˆ Pˆβ = β (βm − ~α · ~p) β = βm + ~α · ~p = H.ˆ So Pˆ does not commute with Hˆ but βPˆ does. This is a new deﬁnition of the parity operator.

Continuity Equation Recall the continuity equation ∂ρ + ∇~ ·~j = 0. ∂t For the free Dirac particle, the conserved charge is ρ = Ψ†Ψ, and the conserved current is ~j = Ψ† ~αΨ. Notice the similarity with the classical ~j = ~vρ. Now, we have ~v = ~α. The charge density ρ and the current density ~j can be considered the components of the 4-current density jµ = ρ,~j , if we transform Ψ† → Ψ. That is, we transform the Hermitian conjugate bispinor to the Dirac conjugate bispinor which is deﬁned as † † Ψ = Ψ β = Ψ γ0. Then the 4-current density is jµ = ΨγµΨ, with components ρ = Ψγ0Ψ = Ψ†Ψ, ji = ΨγiΨ = ΨαiΨ. 14.3. Dirac Equation 229

The Massless Limit For a massive particle (m 6= 0), the energy spectrum has a gap.

If the mass is zero, then there’s no gap in the energy. The massless limit is relevant for extremely low mass particles like neutrinos, and in general, for any ultrarelativistic particle with p so large that E >> m and the mass can be neglected. If m = 0, then " # 0 ~σ Hˆ = ~α · ~p = · ~p. ~σ 0 Then the stationary solutions " # φ u(~p) = , χ imply the energies E = ±c|~p|. Our coupled equations become Eφ = (~σ · ~p) χ Eχ = (~σ · ~p) φ. We can rearrange these as ~σ · ~p φ = χ E ~σ · ~p χ = φ. E We can write ~σ · ~p φ + χ = (φ + χ) E ~σ · ~p φ − χ = − (φ − χ). E Now we can say that our solution is truly a 2-component spinor instead of a 4-component bispinor. For m = 0, ~p ~p ~p ~n = = = sign(E) E |E| sign(E) |~p| is a unit vector. Then the operator ~σ · ~p ~σ · ~n = , |E| 230 Relativistic Quantum Mechanics

corresponds to the helicity of the particle. We can use the basis 1 ψ(±) = √ (φ ± χ) . 2 These give us the disconnected Weyl equations

ψ(±) = ± (~σ · ~n) ψ(±).

If the particle has E > 0, then ~n = ~p/|~p|, and we get two solutions

~p ψ = + ~σ · ψ ← right-polarized particle |~p| ~p ψ = − ~σ · ψ ← left-polarized particle |~p|

If the particle has E < 0, i.e. it is an antiparticle, then ~n = −~p/|~p|, and we get two solutions ~p ψ = − ~σ · ψ ← left-polarized antiparticle |~p| ~p ψ = + ~σ · ψ ← right-polarized antiparticle |~p|

The Weyl equations imply that we have four types of particle that are permanently polarized either right-handed or left-handed.

Neutrinos in the Massless Limit Experiments demonstrate the existence of right-handed antineutrinos and left-handed neutrinos, but not right-handed neutrinos or left-handed antineutrinos. Consider the neutron β-decay

− n → p + e + νe.

The electron e− has lepton number +1, so the neutrino must have lepton number −1. By convention, this neutrino is called an antineutrino since it has negative lepton charge. This electron antineutrino is always right-polarized with ~σ · ~n = +1. We can also have the β− decay of a nucleus

A A − (Z)N −→ (Z + 1)N−1 + e + νe,

or the β+ decay of a nucleus

A A + (Z)N −→ (Z − 1)N+1 + e + νe.

The electron neutronino νe is always left-polarized. The Wu experiment performed in 1956 using the β-decay 60Co → 60Ni showed that parity is not conserved β-decays. Consider the table below which shows the four possible states in terms of particle/antiparticle and left/right polarization for massless particles. A parity transformation P moves us horizontally on this table. For example, performing a parity transformation on a right-handed particle gives us the left-handed particle. Charge conjugation C moves us vertically on the table. For example, charge conjugating the right-handed particle gives us the right-handed antiparticle. Although there are four possibilities, only two of them are found in nature. The right-handed particle and the left-handed antiparticle are apparently non-physical. 14.3. Dirac Equation 231

right particle left particle

right antiparticle left antiparticle

Polarization is less meaningful for massive particles since for those, one can always ﬁnd a frame in which the polarization (aka helicity) is reversed. Separately, charge conjugation C and parity P are violated, but the combined transformation CP is conserved for the most part. CP is modestly violated in some meson decays, but its conservation is still a pretty good assumption.

Charge Conjugation We need some external field, e.g. the EM field, to distinguish particles and antiparticles. To include the EM field, we use minimal coupling as before. Now, Hˆ = βm + ~α · ~p −→ Hˆ − eA0 = βm + ~α · ~p − eA~ ,

µ where A0 is the ﬁrst component of A . I.e., it is the scalar potential of the EM ﬁeld. Now, the Dirac equation is ∂Ψ h i i = βm + eA + ~α · ~p − eA~ Ψ. ∂t 0

Keep in mind that the four component bispinor Ψ is a function of ~r and t as are A0 and A~. Taking the complex conjugate of the Dirac equation gives us ∂Ψ∗ h i −i = β∗m + eA + ~α∗ · i∇~ − eA~ Ψ∗ ∂t 0 ∂Ψ∗ h i i = −β∗m − eA + ~α∗ · ~p + eA~ Ψ∗. ∂t 0 Compare this with the Dirac equation. Suppose we multiply this result with some 4 × 4 matrix C. Then ∂(CΨ∗) h i i = C(−β∗)C−1Cm − eA C + C~α∗C−1C · ~p + eA~ Ψ∗. ∂t 0 Next, deﬁne Ψ˜ = CΨ∗, then

∂Ψ˜ h i i = C(−β∗)C−1m − eA + C~α∗C−1 · ~p + eA~ Ψ˜ . ∂t 0 We choose the matrix C such that

C(−β∗)C−1 = β, C~α∗C−1 = ~α.

Then ∂Ψ˜ h i i = βm − eA + ~α · ~p + eA~ Ψ˜ , ∂t 0 which is the Dirac equation (with minimal coupling) for a particle with charge −e. We call the matrix C the charge conjugation operator. An example that satisﬁes the conditions given above is C = iβα2. The imaginary unit is included to make it Hermitian (C = C†). Note that

∗ −1 −1 2 2 C(−β )C = −CβC = −iβα2βiβα2 = βα2β α2 = βα2 = β, 232 Relativistic Quantum Mechanics

as desired. Similarly, this choice for C satisﬁes C~αC−1 = ~α. The physical solutions for the Dirac equation (with minimal coupling) are ( Ψ if E > 0 . CΨ∗ if E < 0

Time Reversal Consider the Schrodinger equation

∂Ψ(t) i = Hˆ Ψ(t). ∂t

Assuming Hˆ is independent of t, time reversal gives us

∂Ψ(−t) ∂Ψ(−t) i = Hˆ Ψ(−t) =⇒ −i = Hˆ Ψ(−t). ∂(−t) ∂t

Complex conjugating gives us

∂Ψ∗(−t) = Hˆ ∗Ψ∗(−t). ∂t

Now we come to a new Schrodinger equation with Hˆ → Hˆ ∗ and Ψ(t) → Ψ∗(−t). For the Schrodinger equation, Hˆ = p2/2m + U = Hˆ ∗. Thus, we have time reversal invariance since we now have a Schrodinger equation

∂Ψ∗(−t) = Hˆ Ψ∗(−t), ∂t for a particle with t → −t. For example, for a plane wave,

e−iEt+i~p·~r −→ eiE(−t)−i~p·~r.

That is, we get the particle with same energy but going in the opposite direction. Under time reversal, the orbital momentum

~` −→ −~`,

and the spin ~s −→ −~s. In general, the time reversal operator has the form

Tˆ = UˆT Kˆ Oˆt,

where Oˆt is the operator that explicitly takes t → −t, Kˆ is the complex conjugation operator, and UˆT is some additional unitary operator that ensures the correct physical result. For example, UˆT corrects for things like spin which are not explicitly related to time. For an operator Aˆ, we can write the similarity transform

TˆAˆTˆ −1 = Aˆ0TˆTˆ −1 = Aˆ0,

where Aˆ0 is the time-reversed version of Aˆ. Since Tˆ −1Tˆ = 1, we can write Tˆ Aψˆ = TˆAˆTˆ −1Tˆψ = TˆAˆTˆ −1 Tˆψ . 14.3. Dirac Equation 233

Any angular momentum operator is T -odd. For example,

~p 0 = T ~pT −1 = −~p ~` 0 = T ~`T −1 = −~`.

The T -oddness of ~` follows from the deﬁnition of ~` and the fact that ~p is T -odd. However, for spin ~s, we need some explicit operator UT to perform the time-reversal transformation. Since ~s = ~σ/2, we need a T such that

T ~σT −1 = −~σ.

Consider the rotation operator

−iαS −i α σ α α Rˆ (α) = e y = e 2 y = cos − iσ sin . y 2 y 2 Importantly, Rˆy(π) = −iσy.

In general, Tˆ = UˆT Kˆ Oˆt. However, the Pauli matrices contain no explicit t, so we don’t need the Oˆt. We now deﬁne the time-reversal operator T for the spin matrices to be

T = UˆT Kˆ = Rˆy(π)Kˆ = −iσyK,ˆ

−1 −1 where Kˆ is the complex conjugation operator. Then T = iK σy. Let’s see if it works.

−1 −1 −1 T σxT = −iσyKσˆ xKˆ σyi = σyKσˆ xKˆ σy.

−1 Since σx contains only real numbers, Kσˆ xKˆ = σx. Thus,

−1 2 T σxT = σyσxσy = −σxσy = −σx.

So T reverses the sign of σx as it should. For σy,

−1 −1 −1 T σyT = −iσyKσˆ yKˆ σyi = σyKσˆ yKˆ σy.

−1 Since σy is imaginary, Kσˆ yKˆ = −σy, so we get

−1 T σyT = −σyσyσy = −σy.

Similarly, we can conﬁrm that −1 T σzT = −σz, so all three σi change sign under time reversal as they should. Suppose Hˆ is T -invariant and Ψ is not degenerate. Applying T to Ψ gives us some phase T Ψ = eiαΨ. If we apply it twice, and remembering that T also performs complex conjugation,

T 2Ψ = T (T Ψ) = T eiαΨ = e−iαT Ψ = e−iαeiαΨ = Ψ.

So any stationary non-degenerate state, if time-reversal is applied twice, one returns to the original state. Suppose we have N spin-1/2 particles. We apply T 2 to this state. Now, our time- reversal operator has the form

y y y ˆ T = (−iσ )1 (−iσ )2 ··· (−iσ )N K, 234 Relativistic Quantum Mechanics

y where, for example, (−iσ )1 operates only on particle 1. Then 2 N y y ˆ N y y ˆ T = (−i) σ1 ··· σN K(−i) σ1 ··· σN K N y y N ˆ y y ˆ = (−i) σ1 ··· σN (i) Kσ1 ··· σN K N y y N N y y ˆ ˆ = (−i) σ1 ··· σN (i) (−1) σ1 ··· σN KK.

Here we used the fact that each σy gives a factor of −1 when complex conjugated. So

2 N y 2 y 2 N T = (−1) (σ1 ) ··· (σN ) = (−1) ,

2 since σi = 1 for any Pauli matrix. Recall that the eigenvalue of T 2 is 1. This is true for non-degenerate states, regardless of the number of particles. Now we showed that for N spin-1/2 particles, T 2 has eigenvalue (−1)N . This implies that a system with an odd number of spin-1/2 particles cannot have a non-degenerate stationary state. This is Kramer’s theorem. For example, for a single particle (N = 1), adding a magnetic ﬁeld removes the degeneracy. This is not a violation of Kramer’s theorem, it’s just a sign that the magnetic ﬁeld is not time-reversal invariant. For a single spin-1/2 particle, " # " # 0 −i 0 −1 UT = −iσy = −i = . i 0 1 0

In the standard basis " # " # 1 0 χ+ = , χ− = , 0 1 we ﬁnd that " #" # " # 0 −1 1 0 UT χ+ = = = χ− 1 0 0 1 " #" # " # 0 −1 0 −1 UT χ− = = = −χ+. 1 0 1 0

So for a single particle, time reversal changes the spin direction and adds a phase. For spin J, consider instead 2J particles each with spin-1/2. For a single particle we can write 1 −m UT χm = (−1) 2 χ−m. For all 2J particles, Y 1 −m UT |JMi = (−1) 2 |J − Mi , where the product is over all 2J particles. This simpliﬁes to

J−M UT |JMi = (−1) |J − Mi .

What happens with the Dirac equation ∂Ψ i = (βm + ~α · ~p)Ψ. ∂t under time-reversal? We only consider the Dirac equation without (minimal coupling) ﬁelds here. Doing the explicit time-reversal and complex conjugation (remember ~p∗ = −~p) gives us ∂Ψ∗(−t) i = β∗m − ~α∗ · ~p Ψ∗(−t). ∂t 14.4. Relativistic Corrections 235

Applying the time reversal operator should return us to the original Dirac equation. We get ∂ i T Ψ∗(−t) = T β∗m − T ~α∗ · ~p Ψ∗(−t). ∂t We can insert T −1T , then

∂ i T Ψ∗(−t) = T β∗m − T ~α∗ · ~p T −1T Ψ∗(−t) ∂t = T β∗T −1m − T ~α∗T −1 · ~p T Ψ∗(−t).

Thus to obtain the original Dirac equation, we require

T β∗T −1 = β, T ~α∗T −1 = −~α, and by deﬁnition, T Ψ∗(−t) = Ψ(t). We try

T = −iα1α3. Then since β is real,

∗ −1 T β T = −iα1α3βα3α1i = α1α3βα3α1 = −α1α3α3βα1 = −α1βα1 = α1α1β = β.

Similarly, we can show that T ~α∗T −1 = −~α.

14.4 Relativistic Corrections

Pauli Equation Recall again the wave function

Ψ(~r, t) = e−iEtψ(~r), where " # φ(~r) ψ(~r) = , χ(~r) is a 4-component bispinor. Consider the ﬁeld (A0, A~) in stationary states. The eigenvalue equation is " # " # φ h i φ Hˆ = βm + eA0 + ~a · ~p − eA~ . χ χ

This is minimal coupling. This gives us the set of coupled diﬀerential equations h i Eφ = m + eA0 + ~σ · ~p − eA~ χ h i Eχ = −m + eA0 + ~σ · ~p − eA~ φ.

Remember that ~p = −i∇~ . We can rearrange them as h i (E − m − eA0)φ = ~σ · ~p − eA~ χ h i (E + m − eA0)χ = ~σ · ~p − eA~ φ. 236 Relativistic Quantum Mechanics

Solving the second equation for χ and plugging that back into the ﬁrst gives us

1 h i2 (E − m − eA0)φ = ~σ · ~p − eA~ φ. E + m − eA0

If we write E = m + ε,

then E − m = ε and E + m = 2m + ε, and the equation becomes

1 h i2 (ε − eA0)φ = ~σ · ~p − eA~ φ ε + 2m − eA0 h i2 (ε − eA0)(ε + 2m − eA0)φ = ~σ · ~p − eA~ φ.

Now we look at the non-relativistic limit ε << m and the weak-ﬁeld limit eA0 << m. Then h i2 (ε − eA0)(2m)φ = ~σ · ~p − eA~ φ.

We can write this as

1 2 εφ = eA + ~σ · ~p − eA~ φ. 0 2m

This is the Pauli equation. Notice that this is essentially the Schrodinger equation with 2 2 U(~r) = eA0. If there is no ﬁeld, then A0 = 0 and A~ = 0 and (~σ · ~p) = ~p , then this gives us the regular Schrodinger equation for a free particle. We can write

2 ~σ · ~p − eA~ = σiσj (pi − eAi)(pj − eAj)

= (δij + iεijkσk)(pi − eAi)(pj − eAj) 2 = ~p − eA~ + iεijkσk (pi − eAi)(pj − eAj) 2 2 = ~p − eA~ + iεijkσk pipj + e AiAj − e(piAj + Aipj) .

When simplifying something like this, keep in mind that

εijkfij = 0,

for any function fij that is symmetric under exchange of i and j. This allows us to drop 2 the pipj and e AiAj terms. Then

2 2 ~σ · ~p − eA~ = ~p − eA~ − ieεijkσk [piAj + Aipj] .

Adding and subtracting Ajpi gives us

2 2 ~σ · ~p − eA~ = ~p − eA~ − ieεijkσk [piAj − Ajpi + Ajpi + Aipj] . 14.4. Relativistic Corrections 237

But Ajpi + Aipj is symmetric, so the εijk out front means we can drop this term. Then

2 2 ~σ · ~p − eA~ = ~p − eA~ − ieεijkσk [piAj − Ajpi] 2 = ~p − eA~ − ieεijkσk [pi,Aj]

2 ~ ∂Aj = ~p − eA − ieεijkσk −i~ ∂xi 2 ~ ∂Aj = ~p − eA − e~εijk σk ∂xi 2 ~ ~ ~ = ~p − eA − e~ ∇ × A σk k 2 = ~p − eA~ − e~ ~σ · B~ .

Notice that the first term is just the regular kinetic energy with minimal coupling. The second term gives us the new (first-order) relativistic effect. The Pauli equation is now

1 2 εφ = eA + ~p − eA~ − e ~σ · B~ φ. 0 2m ~

Recall that 1 ~s = ~σ. 2 So the new term can be written as

e e − ~ ~σ · B~ = − ~ ~s · B~ = −g ~s · B~ . 2mc mc s

We’ve added the c back into the denominator to get the proper units. What we see is that the Pauli equation correctly predicts the spin gyromagnetic ratio

e g = ~ , s mc which happens to be twice as big as the orbital one g`.

Second-order Eﬀects Kinetic Correction

The Pauli equation is the ﬁrst-order relativistic correction. We now look at second-order corrections. The simplest second order correction is

~p 2 ~p 2 ~p 4 −→ − . 2m 2m 8m3c4

This is the kinetic correction, and it comes from keeping the next higher order term in the expansion of p E = ~p 2c2 + m2c4. 238 Relativistic Quantum Mechanics

Spin-orbit Correction In the rest frame of the electron, there is a magnetic ﬁeld due to the relative motion of the charged nucleus. This magnetic ﬁeld is 1 B~ = − ~v × E~ c 1 = − ~p × E~ mc 1 = − ~p × (∇~ U) mce 1 1 ∂U = (~p × ~r) mce r ∂r 1 ∂U = − ~ ~`. mce r ∂r Here we used ∂U ∂U ~r ∇~ U = rˆ = , ∂r ∂r r and then ~p × ~r = −~~`. Since B~ is proportional to ~` and the Pauli equation contains ~s, we now have spin-orbit coupling in the form

1 ∂U 1 ∂U −g ~s · B~ = −g ~s · − ~ ~` = ~ ~s · ~` . s s mce r ∂r m2c2 r ∂r

Plug in the atomic potential (for lowest n)

Ze2 ∂U Ze2 U = − =⇒ = . r ∂r r2 Then Ze2 −g ~s · B~ = ~ ~s · ~` . s m2c2 r3 This is the spin-orbit correction, and it gives us the ﬁne structure of hydrogen. Notice that this term only contributes when ~` 6= 0.

Darwin Term The third second-order relativistic correction is called the “Darwin term”. It comes from considering ﬂuctuations ξ~ in position (also known as “Zitterbewegung”). We replace

U(~r) −→ U(~r + ξ~).

We can expand

1 ∂2U U(~r + ξ~) = U(~r) + ξ~∇U + ξiξj + ··· 2 ∂xi∂xj 2 1 2 ∂ U = U(~r) + ξ~∇U + δijξ~ + ··· 6 ∂xi∂xj 1 = U(~r) + ξ~ 2 ∇2U + ··· 6 So 1 1 2 δU = ξ~ 2 ∇2U ∼ ~ ∇2U. 6 6 mc 14.4. Relativistic Corrections 239

This correction acts only on the s-wave of the hydrogen atom since

∇2U ∼ δ(~r).

Recall that the spin-orbit correction acts only when ~` 6= 0. The Darwin term ﬁxes that by correcting only when ~` = 0. 240 Relativistic Quantum Mechanics

14.5 Summary: Relativistic Quantum Mechanics

Skills to Master •

Klein-Gordon Equation 4-Vectors A non-relativistic free particle has energy E = p2/2m. In Minkowski space, the main objects are 4-vectors, A relativistic free particle’s energy satisﬁes which in contravariant form are written

E2 − c2~p 2 = m2c4. aµ = a0, a1, a2, a3 ≡ a0,~a .

If we put in quantum operators, E → Hˆ and p → pˆ, The covariant vector is then and act everything on a wave function, we are led to 0 1 2 3 0 the Klein-Gordon equation aµ = a , −a , −a , −a ≡ a , −~a .

1 ∂2 m2c2 Their defining transformation is the Lorentz transfor- − ∇2 + Ψ = 0. c2 ∂t2 ~2 mation. In general, the scalar product of two 4-vectors is Any relativistic, spinless, free particle of mass m should µ 0 0 satisfy this equation of motion. If m = 0, we get the a bµ = a b − ~a ·~b, equation of motion of a photon. The Klein-Gordon equation has plane wave solu- and it is invariant under Lorentz transformation. tions The four-momentum in relativistic units is − i Et+ i ~p·~r Ψ(~r, t) = Ae ~ ~ . P ≡ pµ = (E, ~p) . A Klein-Gordon particle can have positive or negative energy E = ±p~p 2c2 + m2c4. We interpret the The invariant quantity is solutions with E < 0 as the complex conjugate of the 2 µ 2 2 2 solutions for antiparticles (with opposite charge). P = p pµ = E − ~p = m . The Klein-Gordon equation satisfies a continuity For a photon, P 2 = 0. equation ∂ρ When solving relativistic particle collision prob- + ∇~ ·~j = 0, lems, use energy-momentum conservation and the in- ∂t variance of P 2 = m2. where the “current” is h i ~j = ~ Ψ∗∇~ Ψ − Ψ∇~ ψ∗ , Dirac Equation 2mi The Dirac equation and the conserved charge is ∂ i ∂Ψ ∂Ψ∗ i Ψ = βm + c~α · ~p Ψ, ρ = ~ Ψ∗ − Ψ . ~∂t 2mc2 ∂t ∂t is the equation of motion for a relativistic free particle For a charged particle in EM fields, we need to with spin-1/2. The eigenvalue equation si make the minimal coupling substitutions ~p → ~p − eA~ and E → E − eφ. Then the Klein-Gordon equation Hˆ Ψ = βm + c~α · ~p Ψ. becomes " 2 # The β and α are 4 × 4 matrices 1 ∂ e 2 i i − eφ − ~p − A~ − m2c2 Ψ(~r, t) = 0. c2 ~∂t c " # " # 1 0 0 σi β = , αi = . The stationary states ψ(~r) satisfy the eigenvalue prob- 0 −1 σi 0 lem They are related to the Dirac gamma matrices 2 ~p − eA~ + m2 ψ(~r) = (E − eφ)2 ψ(~r). γ0 = β, γi = βαi, 14.5. Summary: Relativistic Quantum Mechanics 241 for i = 1, 2, 3. Important properties include helicity is. The helicity operator hˆ has eigenvalues ±1, corresponding to right and left-handed polarizations. 2 2 2 2 β = 1, αx = αy = αz = 1, In addition to being conserved, for massless particles, the helicity is relativistically invariant since one can and the anticommutation relations find no frame in which the helicity of a massless particle is reversed. For a massless particle, the helicity [αi, β]+ = [αi, αj]+ = 0 for i 6= j. is ~σ · ~p The β and αi matrices all have eigenvalues ±1. hˆ = . The components of the velocity of a Dirac particle |E| are The parity operator Pˆ transforms ~r → −~r. Parity vî = cαî. is not conserved by the Dirac equation, but the quan- This can be shown by calculating the Heisenberg oper- tity βPˆ is conserved. This is a new definition of the ator of ~r˙. parity operator. In general, the bispinor Ψ contains four compo- To include EM fields (minimal coupling), substi- nents. We can write it in the form of tute ~ " # H → H − eA0, ~p → ~p − eA. φ Ψ = , Then the Dirac equation becomes χ ∂ h i i Ψ = βm + eA + c~α · ~p − eA~ Ψ. where φ and χ are spinors. Then the eigenvalue prob- ~∂t 0 lem Hu(~p) = (βm+~α·~p)u(~p) gives us a pair of coupled The charge conjugation operator for a Dirac par- equations for the spinors. For a free particle, the plane ticle is Cˆ = iβα2. This operator converts the minimal wave solution is coupling Dirac equation into a minimal coupling Dirac Ψ(~r, t) = e−iEt+i~p·~ru(~p), equation for a particle with the opposite charge. We have to use the Dirac equation with minimal coupling p 2 2 with energy E~p = m + ~p , and bispinor since we need a field to distinguish the particles and antiparticles. Then for E < 0, the physical solution " # r " # ˆ ∗ φ E + m φ for the Dirac equation with minimal coupling is CΨ , u(~p) = = . whereas the solution for E > 0 is simply Ψ. When χ 2E ~σ·~p φ E+m taking the complex conjugate of the Dirac equation, ~ ∗ Note, for E < 0, it is more convenient to write the remember that ~p = −i∇ so ~p = −~p. solution in terms of χ rather than φ. The time-reversal operator for the Dirac equation ˆ The real velocity of the Dirac particle is without minimal coupling is T = −iα1α3. c2~p h~vi = u† ~αu = . Relativistic Corrections E For hydrogen, there are a number of relativistic correc- The spin operator of a Dirac particle is a vector of tions that we can make. 4 × 4 matrices The Pauli equation " # σ 0 2 1 ~ i 1 ~ ~ ~s = Σ, Σi = . εφ = eA0 + ~p − eA − e~ ~σ · B φ, 2 0 σi 2m

˙ is the non-relativistic limit of the minimal coupling Its Heisenberg equation of motion is ~s = ~p× ~α, so spin Dirac equation. It can be thought of as a Schrodinger is conserved only for stationary particles with ~p = 0. equation for spin-1/2 particles in an electromagnetic For moving particles, spin is not conserved due to the field. It correctly predicts the spin gyromagnetic ratio relativistic effects of spin-orbit interaction, but s2 is gs = e~/mc, which is twice the orbital gyromagnetic conserved. The helicity ratio. The Pauli equation is a first-order relativistic Σ~ · ~p correction to the Schrodinger equation with minimal hˆ = , coupling. |~p| The kinetic correction is is the projection of the spin onto the direction of mo- ~p 2 ~p 2 ~p 4 −→ − . tion ~p. Spin is not conserved in the Dirac equation, but 2m 2m 8m3c4 242 Relativistic Quantum Mechanics

It is a second-order relativistic correction that derives U = −gsS~ · B~ in the Hamiltonian, which gives us the from keeping the next higher order term in the expan- fine structure of hydrogen. It only contributes when sion E = p~p 2c2 + m2c4. ` 6= 0. The spin-orbit correction is a second-order rela- The Darwin term is a second-order correction that tivistic correction due to the magnetic field of the nu- comes from considering the random fluctuations in the cleus. This magnetic field exists because in the elec- position of the electron. This correction only con- tron’s frame, the charged nucleus appears to be mov- tributes when ` = 0. ing. The spin-orbit correction is an additional term Chapter 15

Scattering

15.1 Perturbation Theory

Suppose we have in incident beam with energy Ei, and which is scattered by the target through an angle angle θ. The scattered beam has energy Ef . In this process, some momentum is transferred to the target. Suppose its initial energy is Ei and its ﬁnal energy is Ef .

In general, we will work in the center-of-mass frame. We use m to mean the reduced mass. Energy conservation implies that

Ef + Ef = Ei + Ei. We can write down the golden rule

2π 0 2 dw˙ fi = Hfi δ (Ef + Ef − Ei − Ei) dνf , ~ where 3 V d pf V 2 dνf = = pf dpf dΩf , (2π~)3 (2π~)3 is the density of ﬁnal states in the continuum. In general, we should also sum over the spin states here. For non-relativistic particles which have dispersion relation ε = p2/2m, we can write dpf m dpf = dεf = dεf . dεf pf Then V 0 2 dw˙ fi = Hfi mpf dΩf , 4π2~4 The diﬀerential cross-section is N dw˙ fi dσfi = N p , V m 244 Scattering

where N is the number of incident particles, N/V is the number density, and the incoming flux is this number density times p/m. Thus, for an initial state prepared by the experimenter and a specified final state,

2 V m pf 0 2 dσfi = 2 Hfi dΩf . 2π~ pi

0 Hfi is the piece that contains the initial and ﬁnal wave functions and the interaction Hamiltonian. The perturbation is

0 0 Hfi = hf; ~pf | H (~r) |i; ~pii ,

where ~r is the position of the projectile. Suppose the initial and ﬁnal states are plane waves, then

0 3 1 − i ~p ·~r 0 1 i ~p ·~r H = d r √ e ~ f H (~r)√ e ~ i fi ˆ V fi V

1 3 − i (~p −~p )·~r 0 = d r e ~ f i hf| H (~r) |ii V ˆ 1 = d3r e−i~q·~r hf| H0(~r) |ii , V ˆ where ~pf − ~pi ~q = = ~kf − ~ki, ~ is the momentum transfer. Now the diﬀerential cross-section is

2 2 dσfi m pf 3 −i~q·~r 0 = 2 d r e hf| H (~r) |ii . dΩf 2π~ pi ˆ The incident beam can be scattered at any point ~r inside the scattering region. When it’s scattered from anywhere but the origin, there is a phase diﬀerence ~q · ~r coming from the exponential.

The scattering amplitude is

m 3 −i~q·~r 0 ffi = − d r e hf| H (~r) |ii . 2π~2 ˆ

0 We can think of this as the Fourier component of Hfi as a function of ~r. Now,

dσfi 2 pf = |ffi| . dΩ pi 15.2. Potential Scattering 245

For elastic scattering, pf = pi, then

dσ fi = |f |2 . dΩ fi The momentum transfer-squared is

2 2 ~ ~ 2 2 ~ ~ 2 2 q = kf − ki = kf + ki − 2ki · kf = kf + ki − 2kikf cos θ.

For elastic scattering, |~kf | = |~ki|. I.e., only their angles diﬀer. Then

θ q2 = 2k2 (1 − cos θ) = 4k2 sin2 . 2

Thus, θ q = 2k sin . 2

This is the only place in the cross-section where there is angular dependence. We can estimate e−i~q·~r ∼ eiqR.

iqR If qR << 1, then e ∼ 1 and there is no more dependence on angles in ffi or in the cross-section. Then the scattering is isotropic. This is what we expect for small q. On the other hand, if qR >> 1, then e−i~q·~r becomes highly oscillatory, and we expect the cross0section to go down. If kR >> 1, then small angles are selected, and we get a scattering “cone” in the forward direction.

• low energies: scattering is roughly isotropic • higher energies: scattering is cone-like

15.2 Potential Scattering

For a system of interacting particles,

0 X Hˆ (~r) = Ua(~r − ~ra), a where the sum goes over the particles. Then

−i~q·~ra 0 Here we multiplied and divided by e then deﬁned ~r ≡ ~r − ~ra. We can write the Fourier component 0 U (~q) = d3r0 e−i~q·~r U (~r 0), a ˆ a then X d3r e−i~q·~r hf| Hˆ 0(~r) |ii = hf| e−i~q·~ra U (~q) |ii . ˆ a a 246 Scattering

Then the scattering amplitude is

m X f (~q) = − hf| e−i~q·~ra U (~q) |ii . fi 2π 2 a ~ a

Consider the Coulomb potential

Ua(~r) = eaφ(~r),

where the electric potential φ satisﬁes the Poisson equation

2 ∇ φ = −4πρch(~r),

where ρch is the charge density. As a function of ~q, the electric potential is 4π φ = ρ (~q). ~q q2 ch

2 For a point charge at the origin, ρch = δ(~r), then φq = 4π/q and 4πe U (~q) = a . a q2

Thus, for a particle of charge ea scattering from a Coulomb potential,

4πmea X f (~q) = − hf| e e−i~q·~ra |ii . fi 2π 2q2 a ~ a

The form factor is deﬁned as 1 X F (~q) = hf| e e−i~q·~ra |ii . fi Ze a a

Notice that if q → 0, then Ffi → δfi. If qR << 1, then we can expand the form factor as

1 X 1 F (~q) = hf| e 1 − i~q · ~r + (−i~q · ~r )2 + ··· |ii . fi Ze a a 2 a a

For the diagonal elements,

1 X 1 F ≈ hi| e 1 − i~q · ~r − (~q · ~r )2 |ii ii Ze a a 2 a a q2 X = 1 − hi| r2 |ii . 6 a a

The second term contains the mean-square radius of the system. The procedure that we followed so far is: 1. We started with the golden rule plus the density of states to write down the cross section 2 dσfi 2 m pf 0 2 = V 2 4 hf; ~pf | H |i; ~pii . dΩ 4π ~ pi 2. Using normalized states of the projectile,

2 2 dσfi m pf i i 3 − ~pf ·~r 0 ~pi·~r = 2 d r e ~ hf|H (~r)|ii e ~ . dΩ 2π~ pi ˆ 15.2. Potential Scattering 247

3. Then we deﬁned the momentum transfer ~q = ~kf − ~ki and the scattering angle θ. Assuming no spin eﬀects, the scattering is independent of the azimuthal angle ϕ. For elastic scattering, |~kf | = |~ki|. Then

θ q = 2k sin . 2

4. Next we consider the potential interaction

0 X H (~r) = Ua(~r − ~ra), a which gives us the cross-section

2 2 dσfi m pf X = hf| U (~q) |ii . dΩ 2π 2 p a ~ i a

5. For elastic scattering, |fi = |ii,

2 2 dσ m X 2 = hi| U (~q) |ii = |f(~q)| . dΩ 2π 2 a ~ a 6. The scattering amplitude is m X f = − hi| U (~q) |ii . 2π 2 a ~ a 7. Then we considered Coulomb scattering where e e 4πe e 0 a 0 a −i~q·~ra Ua = =⇒ Ua(~q) = 2 e , |~r − ~ra| q

where e0 is the charge of the projectile, and ea are the charges in the scattering region. Then the scattering amplitude is * + 2me0 X f(~q) = − e e−i~q·~ra . 2q2 a ~ a 8. The charge form factor, which is the Fourier image of the charge distribution, is * + 1 X F (~q) = e e−i~q·~ra . Ze a a

Can expand F (~q) in spherical multipoles. We are interested in the charge distribution of the nucleus. We can extract information by varying q and examining the form factor via analysis of the cross-section dσ/dΩ. Assuming a point-like nucleus at the center of the scattering region, * + 1 X F (~q) = 1 + e e−i~q·~ra . Ze a a The ﬁrst term is from the point-like nucleus, and the second term is from the electrons. If qRa << 1, where Ra is the atomic radius, then F → 1 − 1 = 0. If qRa >> 1, then qtypical ∼ 1/Ra. Then F → 1. I.e., only the nucleus term contributes. 248 Scattering

9. The Coulomb cross-section is

dσ 2me Ze2 = 0 |F (~q)|2 . dΩ ~2q2 10. The classical Rutherford cross-section is 2 2 dσ 2me0Ze Ze0e 1 = 2 2 = 4 , dΩ Ruth ~ q 4ε sin (θ/2)

2 where q2 = 4k2 sin (θ/2) and the kinetic energy is ε = p2/2m = ~2k2/2m. The total cross-section dσ σ = dΩ , ˆ dΩ diverges at small θ (i.e. for weak interactions). For an electron with relativistic (but not ultra-relativistic) energy, we have Mott scattering 2 dσ dσ v 2 θ = 1 − 2 sin . dΩ Mott dΩ Ruth c 2 Backscattering (θ = π) goes to zero as v/c → 1. I.e. as v/c → 1, scattering is compressed into the forward direction. Physically, this preserves helicity conservation for relativistic particles. An important theorem is that

Eσ(E) grows with E.

We can prove it by starting with

dσ = |f(~q)|2 , dΩ

and q = 2k sin(θ/2) and E = ~2k2/2m, where m is the reduced mass. Let µ = q2, then

σ(E) = dϕ sin θdθ |f(~q)|2 ˆ π = 2π sin θdθ |f(~q)|2 ˆ0 4k2 dµ 2 = 2π 2 |f(µ)| ˆ0 2k 2 π 2 8mE/~ = ~ dµ |f(µ)|2 . 2mE ˆ0 Thus, 2 π 2 8mE/~ Eσ(E) = ~ dµ |f(µ)|2 . 2m ˆ0 Every quantity on the right-hand side is positive, therefore, Eσ(E) increases monotoni- cally with E.

15.3 Born (Series) Approximation

Consider elastic scattering at a localized potential U(~r). I.e., we have a free particle before and after, but not in the middle. 15.3. Born (Series) Approximation 249

We consider scattering in the stationary formalism, where the states satisfy the eigenvalue problem 2 − ~ ∇2 + U(~r) ψ(~r) = Eψ(~r). 2m We can write this as 2mU(~r) ∇2 + k2 ψ(~r) = ψ(~r), ~2 where k2 = 2mE/~2. The solution is

ψ(~r) = ei~k·~r + ψ˜(~r), where the ﬁrst term solves (∇2 +k2)ψ = 0, and ψ˜ solves the equation with the interaction. What can we say about ψ˜(~r)? Since we are only considering elastic scattering, we know that ψ˜(~r) consists of waves with total energy conserved. In asymptotics,

eikr ψ˜(~r) = f(~k 0~k), r since the surface area of the wave fronts grows as r2. It turns out that f(~k 0,~k) is the same scattering amplitude f as before. The cross-section is the ratio of the scattered flux j = ~ ψ˜∗∇ψ˜ − ψ˜∇ψ˜∗ = ~ 2iIm ψ˜∗∇ψ˜ , scatt 2mi 2mi to the incident flux k j = v = ~ . inc m The r-component of the scattered flux is

∂ eikr e−ikr k j = ~ Im ψ˜∗ ψ˜ = ~ Im ik f · f ∗ = ~ |f|2 . scatt m ∂r m r r m r2 Then dσ ( k/m)|f|2 = ~ r2 = |f|2. dΩ r2~k/m This is the same f as before. Keep in mind that this is valid in the asymptotics. Green’s function G(~r, ~r 0) is a solution to the Schrodinger equation for a point source

2 2 0 0 ∇r + k G(~r, ~r ) = −4πδ(~r − ~r ). Then we take the solution 0 eik|~r−~r | G(~r, ~r 0) = . |~r − ~r 0| Note, this is not a unique solution, but we choose it because it has the right asymptotic behavior. Then we can write the formal solution to 2mU(~r) ∇2 + k2 ψ˜(~r) = ψ(~r), ~2 as 1 2m ψ˜(~r) = − d3r0 G(~r, ~r 0) U(~r 0)ψ(~r 0). 4π ˆ ~2 The full solution is

m ψ(~r) = ei~k·~r − d3r0 G(~r, ~r 0)U(~r 0)ψ(~r 0), 2π~2 ˆ 250 Scattering

where the ﬁrst term is the original plane wave. Notice that we have ψ on the left and also on the right under the integral. So we can solve this by iteration (i.e. by plugging ψ back into itself repeatedly. m 0 m ψ(~r) = ei~k·~r − d3r0 G(~r, ~r 0)U(~r 0) ei~k·~r − d3r00 G(~r 0, ~r 00)U(~r 00) {· · · } 2π~2 ˆ 2π~2 ˆ m 0 = ei~k·~r − d3r0 G(~r, ~r 0)U(~r 0)ei~k·~r 2π~2 ˆ m 2 00 + d3r0 G(~r, ~r 0)U(~r 0) d3r00 G(~r 0, ~r 00)U(~r 00)ei~k·~r 2π~2 ˆ ˆ . .

We can think of this as giving the multiple scattering from points ~r 0, ~r 00,... inside the scattering region. As such, this is an example of a path integral—an integral over all possible paths (i.e. from all possible points in the scattering region) to the detector at ~r. At any level, this perturbative expansion can be cut oﬀ by replacing the ψ(~r) under the integral with ei~k·~r. For example, the ﬁrst Born approximation is

m 0 ψ(~r) = ei~k·~r − d3r0 G(~r, ~r 0)U(~r 0)e−i~k·~r . 2π~2 ˆ The second Born approximation is m 0 m 00 ψ(~r) = ei~k·~r − d3r0 G(~r, ~r 0)U(~r 0) ei~k·~r − d3r00 G(~r 0, ~r 00)U(~r 00)ei~k·~r . 2π~2 ˆ 2π~2 ˆ Consider the asymptotics (r >> r0).

When expanding G(~r, ~r 0), we only expand the denominator—leave the exponential in the numerator alone since this adds phase information to the solution. For the denominator, p p |~r − ~r 0| = r2 + r0 2 − 2~r · ~r 0 ' r2 − 2~r · ~r 0 r 2~r · ~r 0 ~r · ~r 0 = r 1 − ' r 1 − r2 r2 ~r · ~r 0 = r − = r − ~r 0 · ~n, r where ~n = ~r/r. We discarded r0 2 since it is second order in the small parameter r0. Now,

0 0 eikr−i~k ·~r G(~r, ~r 0) ≈ , r where ~k 0 = k~n. So in the asymptotics,

ikr m e 0 0 ψ˜(~r) = − d3r0 e−i~k ·~r U(~r 0)ψ(~r 0). 2π~2 r ˆ So eikr ψ~ ≈ ei~k·~r + f(~k,~k 0) . r 15.3. Born (Series) Approximation 251

The ﬁrst term is a plane wave, and the second term is a spherical wave, and the scattering amplitude is

m 0 f(~k 0,~k) = − d3r e−i~k ·~rU(~r)ψ(~r). 2π~2 ˆ This is the exact deﬁnition of the scattering amplitude. In the ﬁrst Born approximation, we take ψ(~r) = ei~k·~r. Then

m 0 m 0 f(~k 0,~k) = − d3r e−i~k ·~rU(~r)ei~k·~r = − d3r e−i(~k −~k)·~rU(~r). 2π~2 ˆ 2π~2 ˆ Then since ~q = ~k 0 − ~k, we get

m m f B1(~q) = − d3r e−i~q·~rU(~r) = − d3r e−iqr cos θU(~r). 2π~2 ˆ 2π~2 ˆ If U(r) is spherically symmetric, then in the ﬁrst Born approximation, we can integrate out the angle ∞ B1 m 2 −iqr cos θ 2m 1 f (q) = − 2 r dr U(r) dΩ e = − 2 dr r sin(qr)U(r). 2π~ ˆ ˆ ~ q ˆ0 Remember, if the scattering is elastic, then q = 2k sin(θ/2). But don’t plug this in until you are calculating the cross-section (leave f as a function of q until then)

dσ = |f(q)|2, σ = dΩ |f(q)|2. dΩ ˆ

What is the condition of validity for Born approximation? Consider a spherically- symmetric potential U(r) of extent R. Then the condition of validity is

m 3 0 0 0 0 i~k·~r d r G(~r, ~r )U(~r )ψ(~r ) << e 2π~2 ˆ ik|~r−~r 0| m e 0 d3r0 U(r0)ei~k·~r << ei~k·~r 2 0 2π~ ˆ |~r − ~r | ikr0 2π π m e 0 dr0 r0 2 U(r0) dφ dθ sin θeikr cos θ << 1 2 0 2π~ ˆ r ˆ0 ˆ0 π m 0 0 ikr0 0 ikr0 cos θ 2 dr r e U(r ) dθ sin θe << 1 ~ ˆ ˆ0

m 0 0 2ikr0 dr U(r ) e − 1 << 1. ~2k ˆ The integral is over the scattering region. We can approximate kr = kR. There are two cases: Low energy: For low energy scattering, kR << 1 (R << λ) High energy: For high energy scattering, kR >> 1 (R >> λ) For low energy scattering, kr << 1, then e2ikr − 1 = cos(2kr) + i sin(2kr) − 1 ≈ 1 + i2kr − 1 = 2ikr. Then the validity condition becomes

2m 0 0 0 dr r U(r ) << 1. ~2 ˆ 252 Scattering

If U is the average potential, then

2 2m 0 0 0 2m R U dr r U(r ) ∼ · U = << 1, ~2 ˆ ~2 2 K

where K = ~2/mR2 is a kinetic energy. So the validity condition is

U << 1, (low energy). K

For high energy, e2ikr oscillates wildly, and we can neglect it. Then

2 m 0 0 m mR 1 U 1 dr U(r ) ∼ RU = U = << 1. ~2k ˆ ~2k ~2 kR K kR So the validity condition is

U 1 << 1, (high energy). K kR

The Born approximation is much better for higher energy scattering.

15.4 The Method of Partial Waves

For a scattering problem, the asymptotic wave function is composed of the incident plane wave plus the spherical scattered wave

eikr ψ(~r) ∝ ei(~k·~r) + f(~k,~k 0) . r Then the diﬀerential cross-section is dσ = |f|2. dΩ If the incident beam is in the z-direction, then

eikr ψ(~r) ∝ eikz + f(θ) . r Then dσ = |f(θ)|2. dΩ Let ρ ≡ kr, then we can write

eiρ ψ(~r) ∝ eiρ cos θ + kf(θ) . ρ

We want to expand in a basis of spherical functions. For a central scattering potential, the orbital angular momentum ` is conserved, so we can expand in terms of “partial waves” each with a speciﬁc `. Recall the radial Schrodinger equation for a radial function R`(r) with orbital angular momentum ` in a spherically-symmetric potential U(r)

1 d dR `(` + 1) r2 ` + k2R = R , r2 dr dr ` r2 ` 15.4. The Method of Partial Waves 253 where 2m k2 = (E − U(r)) . ~2 For free motion, U(r) = 0, with ρ ≡ kr, the two independent solutions are the spherical Bessel functions j`(ρ) and the spherical Neumann functions n`(ρ)

1 d ` sin ρ j (ρ) = (−ρ)` ` ρ dρ ρ 1 d ` cos ρ n (ρ) = −(−ρ)` . ` ρ dρ ρ Their behavior near the origin (ρ → 0) is

ρ` j (ρ) → ` 1 · 3 · 5 ··· (2` + 1) 1 · 3 · 5 ··· (2` + 1) n (ρ) → . ` ρ`

Notice that the behavior of the j`(ρ) is ﬁne in the region containing ρ = 0, but the n`(ρ) are singular at ρ = 0. Their asymptotic behavior (ρ >> `) is

sin ρ − π ` j (ρ) → 2 ` ρ cos ρ − π ` n (ρ) → 2 . ` ρ

We require a “regular” solution, e.g. j` at the origin. In the asymptotics, we can have a superposition of the j` and n`. We can expand a plane wave over spherical waves as

i~k·~r X ` ∗ ~ e = 4π i Y`m(k)Y`m(~r)j`(kr). `,m For a plane wave traveling along the z axis,

∞ ikz iρ cos θ X ` e = e = i (2` + 1)P`(cos θ)j`(ρ). `=0 Recall that the Legendre polynomials are related to the spherical harmonics as

r 4π P (cos θ) = Y , ` 2` + 1 `0 and their orthogonality condition is

1 2 dx P`(x)P`0 (x) = δ``0 . ˆ−1 2` + 1 We can expand the scattering amplitude in spherical functions as

∞ X f(θ) = 2(` + 1)P`(cos θ)f`, `=0 where the f` are some coeﬃcients. Then the cross-section is

dσ 2 X 0 ∗ = |f(θ)| = 2(` + 1)2(` + 1)f f 0 P (cos θ)P 0 (cos θ). dΩ ` ` ` ` `,`0 254 Scattering

If we integrate over the angles and apply the orthogonality condition for the Legendre polynomials, we get dσ X σ = dΩ = 4π (2` + 1)|f |2. ˆ dΩ ` `

So to calculate the cross-section σ, it is suﬃcient to calculate the set of f` coeﬃcients. The asymptotic wave function can now be written as

X eip X ψ(~r) ∝ i`(2` + 1)P j + k (2` + 1)P f ` ` ρ ` ` ` ` X eip = (2` + 1)P i`j + kf . ` ` ρ ` ` In asymptotics,

sin ρ − π ` eiρ−i`π/2 − e−iρ+i`π/2 (−i)` j (ρ) → 2 = = eiρ − (−1)è−iρ . ` ρ 2iρ 2iρ Then X (−i)` eip ψ(~r) ∝ (2` + 1)P (cos θ) i` eiρ − (−1)è−iρ + kf ` 2iρ ρ ` ` i X = (2` + 1)P (cos θ) (−1)è−iρ − (1 + 2ikf ) eiρ . 2ρ ` ` ` The incoming wave e−iρ = e−ikr is not changed, but the outgoing wave eiρ = eikr now has a factor of (1 + 2ikf`). These are the elements

S` = 1 + 2ikf`, of the scattering matrix S (in the `-representation) which transforms the incoming wave into the scattered wave. For a central potential like ours, S is diagonal. For elastic scattering, the ﬂux is conserved which implies

2 |S`| = 1.

2 This implies |1 + 2ikf`| = 1, which gives us

∗ 2 2 1 = 1 + 2ikf` − 2ikf` + 4k |f`| ∗ 2 2 −2ik(f` − f` ) = 4k |f`| 2 2 4k Im(f`) = 4k |f`| .

Thus, the condition on f` for ﬂux conservation is

2 Im(f`) = k|f`| .

This is related to the unitarity of the S-matrix

S†S = 1.

We can write X † X ∗ SfnSni = Snf Sni = δfi. n n For f = i, X ∗ X 2 SniSni = 1 = |Sni| . n n 15.4. The Method of Partial Waves 255

The total cross section is

X 4π X σ = 4π (2` + 1)|f |2 = (2` + 1)Im f . ` k ` ` `

The scattering amplitude is X f(θ) = (2` + 1)f`P`(cos θ). `

For forward scattered particles, i.e. for θ = 0, X f(0) = (2` + 1)f`. `

So 4π σ = Im f(θ = 0). k This is the optical theorem. We derived this for elastic scattering, but σ will satisfy the optical theorem even for inelastic scattering. Since S` = 1 + 2ikf` for elastic scattering, we can write the total cross-section for elastic scattering as

2 X S` − 1 π X 2 σel = 4π (2` + 1) = (2` + 1)|S` − 1| . 2ik k2 ` `

For inelastic scattering,

π X σ = (2` + 1) 1 − |S |2 . inel k2 ` `

Inelastic scattering is always accompanied by elastic scattering. That is, we can have elastic scattering without inelastic scattering, but not vice versa. Consider classical scattering depicted below. Classical scattering is parametrized by an impact parameter b, which can be related to the orbital angular momentum ` of the incident particle as 1 ` = mvb =⇒ b = ~ ` = ~` = `. ~ ` ` mv p k Similarly, 1 b = (` + 1). `+1 k 256 Scattering

So we can write the classical cross-section for particles with orbital angular momentum near ` as π π σ = π b2 − b2 = (` + 1)2 − `2 = (2` + 1). `,cl `+1 ` k2 k2

If we compare this with the σ` for quantum scattering, we see that the only diﬀerence is that in the quantum case, we multiply each of these terms by a factor involving the scattering matrix S`. We return now to quantum scattering. For elastic scattering. |S`| = 1, so we can 2iδ choose a phase and write S` = e ` . Then the scattering amplitude is

S − 1 e2iδ` − 1 f = ` = , ` 2ik 2ik and the cross-section is

2 2iδ` X e − 1 4π X 2 σ = 4π (2` + 1) = (2` + 1) sin δ`. 2ik k2 ` ` Notice that the maximum of the cross-section for a given `-term is 4π (σ ) = (2` + 1). max ` k2

Example 15.4.1

Consider scattering from the potential well of radius R shown below. Consider only s-wave scattering, i.e. ` = 0.

For a spherically symmetric potential U(r), we know that

u(r) ψ(~r) = Y , r `m where u(r) is the solution to the diﬀerential equation

2 d2 `(` + 1) − ~ + + U(r) u(r) = Eu(r). 2m dr2 r2

In our case, ` = 0, so the wave function is

u(r) u(r) 1 ψ(~r) = Y00 = √ , r r 4π and u(r) is the solution to the ODE 2m u00 = − (E − U)u. ~2 15.4. The Method of Partial Waves 257

The ﬁrst step is to get the solution in each region. In the region r < R, we 00 0 2 have U(r) = −U0. So the ODE becomes u = −k u, where 2m(E + U ) k0 2 = 0 . ~2 We have the boundary condition u(0) = 0, so our solution for this region is u(r) = A sin(k0r). In the region r > R, we have U(r) = 0, so the ODE becomes u00 = −k2u, where 2mE k2 = . ~2 For this region we can write the solution as u(r) = sin(kr + δ). We only need a single normalization constant, which we’ve tacked onto the solution for the other region. At r = R, we stitch the two solutions and their ﬁrst derivatives together A sin(k0R) = sin(kR + δ) Ak0 cos(k0R) = k cos(kR + δ). Dividing one line by the other, we get k tan (kR + δ) = tan (k0R) , k0 which implies k δ = tan−1 tan (k0R) − kR. k0 √ 0 At low energy, k ≡ k0 ≈ 2mU0/~, then −1 k δ = tan tan (k0R) − kR. k0

When k0R = π/2 a resonance appears. In this way, scattering can you information about the energy levels in the well. For a shallow well, k0R << 1, there is no level (i.e. no bound state). We need to go to a higher order expansion 1 tan x ≈ x + x3, 3 for x << 1. Then 1 δ ≈ (kR)3, 3 and 4π 4π (kR)6 σ = sin2 δ ' . 0 k2 k2 9 The scattering amplitude is

X iδ` f = (2` + 1)P`e sin δ`. `

Take ` = 0 and δ` << 1, then 1 f = (kR)3 ∼ volume. 0 3 258 Scattering

We now analyze the same problem as in the example above, but for arbitrary `. We go back to the Schrodinger equation

2 − ~ ∇2 + U(r) − E ψ(~r) = 0, 2m where u(r) ψ(~r) = Y . r `m This implies

2 d2 2`(` + 1) − ~ + U + ~ − E u(r) = 0 2m dr2 2mr2 d2 2mU `(` + 1) 2mE − − + u(r) = 0 dr2 ~2 r2 ~2 d2 2mU `(` + 1) − − + 1 u(r) = 0. dρ2 ~2k2 ρ2

Here we substituted k2 = 2mE/~2 and ρ = kr. So the equation to solve is d2u U `(` + 1) ` − u − u + u = 0, (15.1) dρ2 E ` ρ2 ` `

where u` = u`(ρ). Compare this with the equation for free motion with solution v` d2v `(` + 1) ` − v + v = 0, (15.2) dρ2 ρ2 ` `

If we multiply Eq. (15.1) by v` and Eq. (15.2) by u` and subtract one from the other, we get U u v00 − v u00 + u v = 0. ` ` ` ` E ` ` We can write this was d U W (u , v ) = − u v , dρ ` ` ` E ` ` where 0 0 W`(u`, v`) = u`v` − v`u`, is the Wronskian. So we have a ﬁrst-order ODE for the Wronskian. This implies

ρ0 0 U(ρ) W`(ρ ) = − dρ u`v` + W`(0). ˆ0 E Then ∞ U(ρ) W`(∞) = W`(0) − dρ u`(ρ)v`(ρ). ˆ0 E In asymptotics, we know that π π v = sin ρ − ` , u = sin ρ − ` + δ . ` 2 ` 2 ` So we can write

0 0 W` = u`v` − v`u` π π π π W (∞) = sin ρ − ` + δ cos ρ − ` − cos ρ − ` + δ sin ρ − ` ` 2 ` 2 2 ` 2 = sin δ`. 15.5. Scattering of Particles with Spin 259

At the origin, u` ∝ v`, which implies W (0) = 0. Then

∞ U(ρ) W`(∞) = − dρ u`(ρ)v`(ρ) = sin δ`. ˆ0 E

We can in principle solve this numerically and that gives us the phase δ` which lets us write down the cross-section. For low energies and for ` 6= 0, sin δ` ≈ δ` and we can estimate

∞ U(ρ) 2 δ` ≈ − dρ v` (ρ). ˆ0 E

2 2 2`+2 Since ρ = kr, E ∼ k , and v` ∼ k , we can estimate 1 δ ∼ k · · k2`+2 ∼ k2`+1. ` k2 This is not valid for the s-wave (` = 0), but is otherwise a good estimate at low energies.

15.5 Scattering of Particles with Spin

Suppose you have a beam of spin-1/2 particles and a target with total momentum J = 0. The scattering is elastic, so J of the target does not change. Suppose the incident beam is unpolarized. In reality, each individual particle is polarized since it has spin. But if you have many particles with random polarization, then the resulting beam is unpolarized. The polarization per particle is hS~i P~ = = h~σi. 1/2 Here the overline signiﬁes the statistical average over the multiple particles and the braket signiﬁes the quantum expectation value. For an unpolarized incident beam

P~ = h~σi = 0.

Given the assumption that the incident beam is unpolarized, is it possible for the scattered beam to be polarized? If so, what will be the direction of polarization?

We can assume the incident particles have momentum

~k = k~n, where ~n = (0, 0, 1), is a unit vector along the z-direction. Then since the scattering is elastic, the scattered particles have momentum ~k 0 = k~n 0, 260 Scattering

where ~n 0 = (sin θ cos ϕ, sin θ sin ϕ, cos θ), is the direction in which the particles are scattered. We will assume that the scattering interaction preserves parity and time-reversal. What does this tell us about the possibility of the scattered particles being polarized? The system is characterized with the two vectors ~n and ~n 0. These two are regular polar vectors. But remember that angular momenta like ~`, ~s, and ~j are all axial vectors, and axial vectors do not change sign under parity reversal. So if our setup can possibly result in polarization, it must give us an axial polarization vector. The only axial vector we can construct with the pair of polar vectors ~n and ~n 0 is ~n × ~n 0 = − sin θ sin ϕ, sin θ cos ϕ, 0 = sin θ − sin ϕ, cos ϕ, 0 .

So if the scattered beam is polarized, to preserve parity, the polarization vector must be proportional to ~n × ~n 0. What about time-reversal? Naively, it appears that a polarization proportional to ~n× ~n 0 would violate time-reversal since neither of these vectors is time-dependent. However, under time-reversal, the process is reversed. The incident beam has direction −~n 0 and the scattered beam has direction ~n. So under time-reversal, ~n × ~n 0 −→ (−~n 0) × (−~n) = ~n 0 × ~n = −~n × ~n 0. So polarization along ~n × ~n 0 satisﬁes time-reversal symmetry. Let ~ν be the unit vector in the direction of ~n × ~n 0, so ~n × ~n 0 ~n × ~n 0 ~ν = = = − sin ϕ, cos ϕ, 0 . |~n × ~n 0| sin θ Recall that without spin, we would have a scattering amplitude 1 X f(θ) = (2` + 1)(S − 1)P (cos θ). 2ik ` ` ` What do we get with spin? For a single particle, in the decoupled scheme, we can write

i~k·~r ψinitial = e χ. For spin-1/2, 1 j = ` ± ≡ j(±). 2 In the scattering process, `, j, jz, and parity are all conserved. (+) 1 We now introduce two projection operators for convenience. Λ selects j = ` + 2 (−) 1 and Λ selects j = ` − 2 when they are applied to any function with orbital or spin momentum. We require Λ(+) + Λ(−) = 1. We want to construct these operators. Since S~ = ~σ/2, we can write 3 ~` · ~σ = 2 ~` · S~ = ~j 2 − ~` 2 − S~ 2 = j(j + 1) − `(` + 1) − . 4 Then 1 3 3 ~` · ~σ = ` + ` + − `(` + 1) − = ` j=`+1/2 2 2 4 1 1 3 ~` · ~σ = ` − ` + − `(` + 1) − = −(` + 1). j=`−1/2 2 2 4 15.5. Scattering of Particles with Spin 261

So we can write the projection operators as

` + 1 + ~` · ~σ Λ(+) = 2` + 1 ` − ~` · ~σ Λ(−) = . 2` + 1

(+) (−) Now instead of just S`, we have S` for j = ` + 1/2 and S` for j = ` − 1/2. We can write the scattering phases in the form

(±) (±) 2iδ` S` = e . Now the scattering amplitude is

1 X h 2iδ(+) (+) 2iδ(−) (−)i f(θ) = (2` + 1) e ` − 1 Λ + e ` − 1 Λ P (cos θ) 2ik ` ` ` ` 1 X 2iδ(+) ` + 1 2iδ(−) ` = (2` + 1) e ` − 1 + e ` − 1 2ik 2` + 1 2` + 1 ` ~ ~ # 2iδ(+) ` · ~σ 2iδ(−) ` · ~σ + e ` − 1 + e ` − 1 P (cos θ). 2` + 1 2` + 1 `

We need to see how ~` · ~σ acts on the Legendre polynomials P`. We can write

~` · ~σ = `xσx + `yσy + `zσz.

The last term does not contribute since the P` are independent of ϕ. Then ∂ ` = −i sin ϕ + terms that don’t contribute x ∂θ ∂ ` = i cos ϕ + terms that don’t contribute. y ∂θ Then ∂ ∂ ~` · ~σ = ` σ + ` σ = i (− sin ϕσ + cos ϕσ ) = i (~ν · ~σ) . x x y y x y ∂θ ∂θ This is precisely the same vector ~ν that we noted earlier as the only possible direction in which the scattered beam can be polarized. So our new scattering amplitude is

fˆ(θ, ϕ) = A(θ) + B(θ)(~ν · ~σ) . where the ﬁrst term corresponds to normal scattering and the second corresponds to scattering resulting in polarization. The ϕ-dependence is in the (~ν · ~σ) operator. The coeﬃcients are

1 X h 2iδ(+) 2iδ(−) i A(θ) = (` + 1) e ` − 1 + ` e ` − 1 P (cos θ) 2ik ` `

1 X h 2iδ(+) 2iδ(−) i dP` B(θ) = e ` − e ` . 2k dθ `

The derivative dP`/dθ gives an associated Legendre polynomial. The A-term gives regular scattering. The B-term gives us polarized scattering since it depends on ϕ via ~ν. If there is no spin-orbit coupling then B(θ) = 0. Given the spin-orbit coupling, we can solve the Schrodinger equation to get the phases δ`, and then we can write down the scattering amplitude f(δ, ϕ). 262 Scattering

The amplitude of scattering in the direction ~n 0(θ, ϕ) with a change from initial spin 0 0 projection µ = Sz to final spin projection µ = Sz is given by the matrix element ˆ fµ0µ = hχµ0 |f|χµi . The differential cross-section is dσ † = fµ0µfµ0µ. dΩ µ0µ If the final spin projections µ0 are not measured, then we have to sum over them dσ X dσ = . dΩ dΩ µ µ0 µ0µ We can calculate the average

fˆ†fˆ = (A∗ + B∗ (~σ · ~ν)) (A + B (~σ · ~ν)) = |A|2 + |B|2(~σ · ~ν)2 + (A∗B + AB∗) (~σ · ~ν).

For any vector ~a,(~σ · ~a)2 = ~a 2. Here, ~ν is a unit vector. Using, P~ = h~σi, we can write D E fˆ†fˆ = |A|2 + |B|2 + (A∗B + AB∗) ~ν · P~ = |A|2 + |B|2 1 + α ~ν · P~

where A∗B + AB∗ 2Re(A∗B) α = = , |A|2 + |B|2 |A|2 + |B|2 is the coefficient of azimuthal asymmetry. Note that |α| ≤ 1. Now we can write the differential cross-section is dσ dσ = 1 + α ~ν · P~ . dΩ dΩ unpolarized dσ We measure first for unpolarized initial state to get dΩ unpolarized. Then we introduce polarization of the initial state and remeasure to get α. If the initial beam is unpolarized, can the scattered beam come out polarized? We can write the final polarization as f †~σf P~ final = . f †f We can expand the numerator using f †~σf = (A∗ + B∗ (~σ · ~ν)) ~σ (A + B (~σ · ~ν)) . We can use the algebra of Pauli matrices,

σiσj = δij + iεijkσk, to reduce any powers of (~σ · ~ν) to end up with

P~ final = α~ν, where α is the polarization. It turns out that this is the same as the coeﬃcient of azimuthal asymmetry α identiﬁed above. 15.6. Scattering of Identical Particles 263

15.6 Scattering of Identical Particles

When two identical particles are scattered from each other, there are two diﬀerent processes that give the same result.

The classical cross-section is dσ dσ dσ = + . dΩ classical dΩ θ dΩ π−θ If θ = 90◦, then dσ dσ = 2 . dΩ classical dΩ 90◦ 264 Scattering

15.7 Summary: Scattering

Skills to Master •

For low energy scattering (kR << 1), use partial For the first Born approximation, ψ(~r) = ei~k·~r, then waves. At low energies k → 0, and the scattering becomes dominated by the ` = 0 term. m f = − d3r e−i~q·~rU(~r) For high energy scattering (kR >> 1), use Born 2π~2 ˆ approximation. 2π π ∞ m 2 −iqr cos θ For identical particles, the scattering wave func- = − 2 dϕ sin θ dθ r dr e U(r, θ, ϕ), 2π~ ˆ0 ˆ0 ˆ0 tion must be symmetric or antisymmetric depending on whether the particles are bosons or fermions. where ~q = ~k 0 − ~k is the momentum transfer. If the Particles with spin bring the following complica- potential is spherically-symmetric, then tions: ∞ • The spin of a particle can interact with an ex- 2m 2 sin(qr) f = − 2 dr r U(r) . ternal field, so even a neutral particle (e.g. a ~ ˆ0 qr neutron) will scatter from a charged particle • The spins of two spin-full particles will interact The differential cross-section is with each other as in the scattering of a neutron dσ from a proton. = |f(q)|2 . dΩ Born Approximation If the scattering is elastic, then

In general, θ q = 2k sin . eikr 2 ψ ≈ ei~k·~r + f(~k,~k 0) , r The total cross-section is where

m 0 f(~k,~k 0) = − d3r e−i~k ·~rU(~r)ψ(~r). σ = dΩ |f|2. 2π~2 ˆ ˆ