Quantum Mechanics Revision Notes

Home , Ehrenfest theorem, Exchange operator

C.R.D. Guetta

April 7, 2008 2 Contents

1 Failure of Classical Physics 7 1.1 Waves are particles ...... 7 1.2 Particles are waves ...... 8

2 Basics 11 2.1 Basic postulates ...... 11 2.2 Wavepackets ...... 11 2.2.1 Introduction ...... 11 2.2.2 Momentum Representation of a Wavepacket ...... 11 2.2.3 The Dispersion Relation for a Wavepacket ...... 12 2.2.4 Time Evolution of a general Wavepacket ...... 12 2.2.5 Time evolution of a wavepacket representing a particle ...... 12 2.3 The Heisenberg Uncertainty Principle ...... 12 2.4 Probability current ...... 12 2.5 Beams of particles ...... 13 2.5.1 Position representation ...... 13 2.5.2 Momentum representation ...... 13 2.6 Practical stuﬀ ...... 13

3 The Schr¨odingerEquation 15 3.1 Derivation ...... 15 3.2 Stationary states ...... 15 3.3 Boundary Conditions ...... 15 3.4 Solutions for Constant V ...... 16 3.4.1 Plane wave solution ...... 16 3.4.2 Non-plane wave solutions ...... 16

4 The Wave Approach to QM 17 4.1 Unbound states – scattering ...... 17 4.1.1 Applications of tunnelling ...... 18 4.2 Bound states ...... 18 4.2.1 The Inﬁnite Well ...... 19 4.2.2 The δ-function potential ...... 19 4.2.3 Finite square well ...... 19 4.2.4 The 1D Harmonic Oscillator ...... 19 4.2.5 Three Dimensions ...... 19

5 The Basic Postulates & Operators 21

3 4 CONTENTS

5.1 The Basic Postulates of QM ...... 21 5.2 Operators ...... 21 5.2.1 Dirac notation ...... 21 5.2.2 Properties of Observables ...... 21 5.3 Expectation values ...... 22 5.4 The Generalised Uncertainty Principle ...... 22 5.4.1 Compatible Observables ...... 22 5.4.2 Incompatible Observables ...... 23 5.4.3 Conjugate Observables ...... 23 5.4.4 Minimum Uncertainty States ...... 24 5.5 Examples of operators ...... 24 5.6 Ladder Operators ...... 25 5.6.1 The Harmonic Oscillator ...... 25

6 Time Dependence 27 6.1 Measurements ...... 27 6.2 Ehrenfest’s Theorem ...... 27 6.2.1 Classical limit ...... 28 6.3 Conserved Quantities ...... 28 6.3.1 Parity ...... 28 6.4 The Time-Energy Uncertainty Relation ...... 28

7 3 Dimensions 29 7.1 Angular Momentum ...... 29 7.2 Commutation relations ...... 29

8 Angular Momentum 31 8.1 Orbital Angular Momentum Eigenvalues ...... 31 8.2 Eigenstates ...... 32 8.3 Bits and bobs ...... 33 8.4 The Stern-Gerlach Experiment ...... 34

9 Spin 35 9.1 Experimental Evidence ...... 35 9.2 Eigenvalues and Eigenvectors ...... 35 9.3 Uncertainties ...... 36 9.4 Spin in any direction ...... 36 9.5 Combining Orbital & Spin Angular Momentum ...... 36 9.5.1 Introduction ...... 36 9.5.2 The values of J ...... 36 9.5.3 Combined wavefunctions ...... 37 9.6 Conservation of total angular momentum ...... 37

10 Central Potentials 39 10.1 Conservation of angular momentum ...... 39 10.2 Quantum numbers ...... 39 10.3 Separation of Variables ...... 39 10.4 Normalisation & Probabilities ...... 40 10.5 Practical tips ...... 40 CONTENTS 5

11 Two-Particle Systems 41 11.1 Position Probabilities ...... 41 11.2 Observables ...... 41 11.3 The Hamiltonian ...... 41 11.4 Conservation of Total Momentum ...... 41 11.5 Centre of Mass Motion ...... 42 11.6 Separation of CoM Motion and Relative Motion ...... 42 11.7 Combining spins ...... 43

12 Examples of real systems 45 12.1 The rigid rotor ...... 45 12.2 The Harmonic Oscillator ...... 45 12.3 The Hydrogenic Atom ...... 45

13 Identical Particles 47 13.1 Exchange Symmetry ...... 47 13.2 The Spin-Statistics Theorem ...... 48 13.3 Non-interacting particles ...... 48 13.4 N particles ...... 48 13.4.1 Distinguishable ...... 48 13.4.2 Indistinguishable ...... 48 13.5 N =2...... 49 13.5.1 Diatomic molecules ...... 49 13.6 Correlation and Exchange Forces ...... 50 13.7 Interacting Particles ...... 50 13.7.1 Exchange energy ...... 50 13.7.2 Energy Spectra ...... 51 13.8 Practical tips ...... 51

14 Matrix Methods 53 14.1 Introduction ...... 53 14.2 Orbital angular momentum ...... 53 14.3 Spin angular momentum ...... 53

15 Heat capacities, etc. . . 55 15.1 Vibrational Speciﬁc Heat of Diatomic molecules ...... 55 15.2 Rotational Speciﬁc Heat of Diatomic molecules ...... 55 15.3 Identical Particles and Rotational Heat Capacities ...... 56 6 CONTENTS Chapter 1

Failure of Classical Physics

1.1 Waves are particles – Feeding this into the formula |k| = π|p|/L, taking into account both polar- 3 There is a plethora of evidence that waves can be- isation of EM waves, and dividing by L have as particles, with the following properties: to find the density, we obtain the relation above. h E = hν p = • Classically, the average energy of each har- λ monic mode (2 degrees of freedom) is kBT – so the energy density (ie: energy per unit vol- Black body radiation ume) is given by the Rayleigh-Jeans Law: 8πν2 Classically: ρ(ν, T ) = kBT c3 • For a cavity, the number of EM modes per unit This, however, predicts that ρ increases with- volume with frequency between ν and ν + dν out bound at high ν – the ultraviolet catastro- is given by phe. 8πν2 • Planck hypothesised that the energy of each N(ν) dν = dν c3 mode is not just kT , but is quantised in units of hν – in other words, the average energy per Derive as follows: mode is given by (note the major assumption – Let k be the wavevector of our wave. that there is no zero-point energy – we say that Boundary conditions dictate that it must this is the ‘energy of the vacuum’) be quantised as |k| = π|p|/L, where p Boltzmann is a vector with integer compoenents in n=∞ z }| { P nhν e−nhν/kB T hν 3D space, and L is the dimension of the ¯ = n=0 = Pn=∞ −nhv/kB T hν/kB T (square) cavity. n=0 e e − 1 | {z } – In p-space, the volume density of states Partition func is, by definition, 1 (because p must This expression can easily be obtained by not- have integer components). Therefore, the ing that the denominator is an infinite geomet- number of states in the range p to p + dp ric series (Pn=∞ anr = a/(1−r)) and that the is given by n=0 top is the negative differential of the bottom, 1 with respect to β = k T . dN(p) = N(p) dp = 4πp2 dp B 8 • We then get (Given that we are only interested in p- 8πν2 hν vectors with positive components. ρ(ν, T ) = c3 ehν/kB T − 1

7 8 CHAPTER 1. FAILURE OF CLASSICAL PHYSICS

This fits data, and reduces to Rayleigh-Jeans The X-rays are made out of photons as ν → 0. with well-defined momentum given by Conclusion: h p = The energy of each mode of λ electromagnetic radiation is quantised in units of hν Interference of light at low intensity The Photoelectric Effect In a Young’s Double Slit experiment at a very low When light falls on a metal, electrons emerge with intensity of light, we detect the light as discrete different energies, which can be measured by mak- photons. However, the resulting pattern is what is ing the metal the positive-end of an electrode, and expected of a double-slit pattern – each quantum varying V , the voltage across the electrode. The particle interacts with both slits, and its trajectory stopping voltage, Vstop is the voltage above which is indeterminate – trying to identify which slit the no electrons reach the cathode, and is proportional electron passes through destroys the pattern. to the maximum kinetic energy of electrons that leave the metal (Emax). The phenomenon is characterised by the following 1.2 Particles are waves properties

• Emax = hν −W , where W is the work function Evidence exists that suggests particles with mo- of the metal (the minimum energy required to mentum p have wave-like properties, with wave- release an electron from the metal). Classi- length given by cally, one would expect the energy in the wave h λ = to vary with intensity, not frequency. p • The number of electrons emitted (the satura- and tion current) is proportional to the intensity of light. Higher intensity means more packets. angular momentum quantized in units of • Electron emission is instantaneous – classi- ~ cally, one would expect a time delay to absorb enough energy to emit an electron. Atomic structure & the de Broglie Hypoth- These can be explained by the following conclusion: esis

Light exists as quantised packets, each Incandescent gasses have characteristic line spec- of energy hν tra. The positive charge in the atom is known to be concentration in the nucleus (α-scattering ex- The Compton Effect periments), and the electrons orbit the nucleus. To explain why the electrons do not radiate and spiral in, Bohr suggested that In Compton Scattering, X − Rays are fired onto a metal sample and scattered by electrons. It is The electron’s angular momentum L is found that the wavelength of the scattered X-Rays, quantized in units of ~ λ0 varies with φ, the angle between the new beam and the undeflected beam. Starting from this assumption, he was able to recover the correct form for the line spectrum of hy- The correct relation between λ0 and φ can be de- drogen. rived directly from conservation of momentum and energy as long as we assume that De Broglie hypothesised that 1.2. PARTICLES ARE WAVES 9

Particles behave with wave-like properties, with wavelength h λ = p

This nearly justiﬁed one of Bohr’s assumptions – for a stable orbit (a ‘standing wave’ around the nucleus), the circumference should be an integer number of de Broglie wavelengths: nh 2πr = nλ = p And angular momentum is therefore:

L = pr = n~ As hypothesised by Bohr.

Electron & Molecular Diﬀraction

The Davidson-Germer experiment diffracted electrons through a crystalline array of atoms. The diffracted beam directions depended on the incident energy (and momentum). Diffraction has since been shown to occur with many other particles, as heavy as fluorofullerene. It is experimental support for the de Broglie hypothesis.

The Stern-Gerlach Experiment

• In the Stern-Gerlach experiment, a beam of neutral particles with magnetic moment µ are passed through a non-uniform electric ﬁeld, that varies most rapidly in the positive z- direction (upwards). • Classically, the orientation of the magnetic moments should be random, and the net force on any atom should very continuously be- ∂Bz tween ±µ ∂z – the emerging beam should be broadened in the z-direction to reﬂect this. • In fact, the beam is split into a small number of discrete beams, indication that the orientation of magnetic moments is quantized. This is experimental support for Bohr’s Hypothesis. 10 CHAPTER 1. FAILURE OF CLASSICAL PHYSICS Chapter 2

Basics

2.1 Basic postulates 2.2 Wavepackets

Some of the basic postulates of Quantum Mechan- 2.2.1 Introduction ics can be expressed as follows: 1. A particle is represented by a ‘wavefunction’, The problem with this representation of a parti- ψ(x, t), which contains all possible information cle is that it P (x, t) is completely uniform over all about the particle. space – there is no information about the particle’s position. 2. The probability density is given by If we want to know something about the particle’s ρ(x, t) = |ψ(x, t)|2 position, ψ must to some degree be localised, such that ψ(x, t) → 0 as x → ±∞. This is achieved The probability that the position of the parti- by the superposition of planes waves of diﬀerent cle is between x and x + dx is given by wavenumbers and frequencies P (x, t) dx = ρ(x, t) dx 1 Z ∞ ψ(x, t) = √ g(k)ei(kx−ωt) dk (2.2) 2π The wavefunction needs to be normalised such −∞ that the probability of ﬁnding the particle in For t = 0, this becomes a Fourier Integral. all space is 1. As such Z ∞ ρ(x, t) dx = 1 2.2.2 Momentum Representation of −∞ a Wavepacket

3. A free particle of mass m, momentum p and Let us deﬁne energy E is represented by a plane wave of the 1 p form Φ(p, t) = √ = g(k, t) = g , t ~ ~ i(kx−ωt) i( p x− E t) Ψ(x, t) = Ae = Ae ~ ~ (2.1) We then have, in accordance with Equation 2.2: The form of each term in the wavefunction arises from experimental evidence: Z ∞ 1 ( ix p) Ψ(x, t) = √ Φ(p, t)e ~ dp • The De Broglie Hypothesis, supported by 2π~ −∞ the Compton and Davidson-Germer experiments, gives us k = p/~ and Z ∞ • The photoelectric eﬀect gives rise to ω = 1 (− ip x) Φ(x, t) = √ Ψ(x, t)e ~ dx E/ ~ 2π~ −∞

11 12 CHAPTER 2. BASICS

And, as above, the probability of finding the parti- 2.3 The Heisenberg Uncer- cle with momenta p to p + dp is tainty Principle P (p) = |φ(p)| dp Consider Ψ – the wavefunction itself has intrinsic Ψ and Φ are two different ways of representing the positional uncertainty (since Ψ extends over a range wavefunction fully – the in position space and mo- of values of x), but there is also uncertainty in the mentum space respectively. momentum of the particle, because the wavefunction is made up of a number of plane waves, each with a different momentum. 2.2.3 The Dispersion Relation for a Wavepacket These uncertainties are reciprocal, because of the properties of Fourier Transforms – the larger the In general, ω in Equation 2.2 is related to k by a one, the smaller the other. dispersion relation, ω = ω(k). Mathematically, the uncertainty in x is given by For a wavepacket representing a particle, we know p 2 2 the following ∆(x) = hx i − hxi Where • E = ~ω Z ∞ 2 • p = k hxi = x|Ψ(x)| dx ~ −∞ • E = p2/2m For a Gaussian Wavepacket: The Dispersion Relation is therefore: ∆x∆p = ~ k2 2 ω = ~ (2.3) 2m This is the least possible uncertainty attainable by any wavepacket (see Section 5.4.4). The group velocity of a wavepacket is therefore: dω k p v = = ~ = (2.4) g dk m m 2.4 Probability current

2.2.4 Time Evolution of a general For the probability density, ρ(r, t) to remain nor- Wavepacket malised for all t, the probability current j (r, t) must satisfy the continuity equation Consider the following wavepacket: Z ∂ Z j (r, t) · dS = − ρ(r, t) dV ∞ Z S ∂t V Ψ(x, t) ∝ g(k)ei(kx−ω(k)t) dk −∞ Where S is the surface bounding some V . Where ω(k) is the dispersion relation We can simplify as follows: • Express |Ψ| as Ψ∗Ψ, and evaluate the deriva- 2.2.5 Time evolution of a tive. wavepacket representing a • Get rid of the derivatives ∂t by using the time- particle dependent Schrodinger Equation (see section 3). Because the dispersion relation is not linear, differ- • Use Green’s Second Theorem ent components of the wavepacket (with different Z Z momenta) will travel at different speeds. As such, [g(∇f)−f(∇g)]·dS = [g∇2f −f∇2g]dV the wavepacket will change shape over time. S V 2.6. PRACTICAL STUFF 13

We then obtain the following expression: 2.5.1 Position representation

In the position representation of a beam j (r, t) = ~ [Ψ∗(r, t)∇Ψ(r, t) − Ψ(r, t)∇Ψ∗(r, t)] 2mi Z |Ψ|2 dx = |A|2 For stationary states: unit length So |A|2 is the number of particles per unit length j (r) = ~ [ψ∗(r)∇ψ(r) − ψ(r)∇ψ∗(r)] (or volume in 3D) – the particle number density. 2mi The particle ﬂux (Equation 2.5) is then given by Re-arranging: p j(x) = |A|2 m ~ ∗ ∗ ( ∗ j (r) = [ψ (r)∇ψ(r) − (ψ (r)∇ψ r)) ] (In other words, the product of the number density 2mi and the velocity – as expected).

p j = < ψ∗ ~ ∇ψ = < ψ∗ b ψ (2.5) im m 2.5.2 Momentum representation

Feeding the expression for Ψ into the momentum We note two important points: representation, we find that the time-independent • For a stationary state, the probability current momentum-representation of the beam is j is time-independent. φ(p) ∝ δ(p − p0) • From the equation above, we have ∇ · j = ·ρ. The momentum is known exactly for the beam. However, we saw above that for a stationary state, ·ρ = 0, and so ∇·j = 0. In 1-Dimension, this means that 2.6 Practical stuff dj = 0 dx • When trying to find the minimum uncertainty in a position measurement as a result of a Which means that the flux j is independent wavepacket spreading, find an expression for of position as well. In more than 1 dimension, the uncertainty in terms of the original uncer- however, this is no longer the case. tainty ∆x0 and differentiate. • Alternatively, for the harmonic oscillator, write the expectation value of the energy as 2.5 Beams of particles hpi2 hEi = k x2 + 2m The plane wave And note that 2 2 Ψ(x, t) = Aei(kx−ωt) – x = (∆x) , because hxi = 0, since the potential is symmetric. is readily normalised if it only extends over a finite – p2 = (∆p)2, because hpi = 0, since the volume. Otherwise, it represents a particle of well- particle isn’t drifting away. defined momentum but completely unknown position. Then, use the uncertainty relation to obtain ∆p in terms of ∆x, feed it into the expression By choosing A properly, Ψ can be made to repre- for hEi above, and hence obtain an expression sent a beam of particles with momentum ~k. for ∆x. 14 CHAPTER 2. BASICS Chapter 3

The Schr¨odinger Equation

3.1 Derivation 3.3 Boundary Conditions

By considering the form of the wavepacket in Equa- tion 2.2, we can deduce that There are a number of Boundary Conditions that Quantum Mechanics imposes on ψ – our general 2 2 ∂Ψ ~ ∂ Ψ method for finding ψ is simply to use a number of i~ = − ∂t 2m ∂x2 trial solutions, and to fit the boundary conditions. If a potential V (x, t) is present, though, and we move to 3D, this becomes: • Boundary conditions on ψ 2 ∂Ψ(r, t) ~ 2 i~ = − ∇ Ψ(r, t) + V (r, t)Ψ(r, t) – ψ must be finite everywhere. Otherwise, ∂t 2m |ψ|2 → ∞, unphysically. (3.1) This can also be written using the Hamiltonian Op- – ψ must be continuous. Otherwise, the erator (see below - Equation 3.4) as follows particle flux (Equation 2.5) would pro- ∂Ψ(r, t) duce a singular, unphysical result (since i = HcΨ(r, t) it involves ψ0). ~ ∂t • Boundary conditions on ψ0 3.2 Stationary states – ψ0 must be finite, because ψ is finite.

If V depends on r only and does not change with – If V is finite, and since ψ is finite the time, then Equation 3.1 becomes separable, and Schrödinger Equation in 1D implies that this leads to a solution of the form ψ00 must also be finite. This means that 0 00 Ψ(r, t) = ψ(r)e−iEt/~ (3.2) ψ must be continuous, or else ψ would have an infinite singularity. Where the positional wavefunction ψ(r) satisfies – HOWEVER, if V is not finite, and Hcψ(r) = Eψ(r) (3.3) tends to ∞ discontinuously, then ψ00 → ∞, so ψ0 is discontinuous. Where E is the energy of the particle and Hc is the Hamiltonian Operator, given by To summarise KE PE ψ is finite and continuous at all points, z 2}| { ~ 2 z }| { and differentiable at every point where Hc = − ∇ + V (r) (3.4) 2m V does not discontinuously tend to ∞.

15 16 CHAPTER 3. THE SCHRODINGER¨ EQUATION

3.4 Solutions for Constant V

3.4.1 Plane wave solution

When dealing with unbound particles, we seek a plane wave solution. Feeding the trial solution into the Schr¨odinger Equation gives

ψ(x, t) = Aei(kx−ωt)

Where r 2m[E − V (x)] k = ± ~2 • For E > V (ie: the particle’s has real kinetic energy), ψ oscillates. • For E < V , the kinetic energy is imaginary, and ψ decays exponentially. In that case, we write κ = −ik. A general approach to solving the Schr¨odinger Equation for any potential V is simply to solve it in each region of constant V , and to match boundary conditions.

3.4.2 Non-plane wave solutions

For bound states, we need non-plane wave solutions, and we therefore use the most general solution of the equation (involving both the positive and negative exponential). The boundary conditions on the wave can then often we used to elimi- nate one of the exponentials. Chapter 4

The Wave Approach to QM

4.1 Unbound states – scatter- • In cases of, for example, a potential barrier, ing one could consider waves infinitely bouncing back and forth within the barrier – but it turns out to be just as general to consider one We first consider a beam of particles, represented forward-going wave and one backward going by a plane wave, travelling through some potential wave. and being reflected/transmitted from an obstruc- tion. Possible cases include a potential step, a po- • For a potential barrier, it is found that T = 1 tential barrier or a potential well. when k2a = nπ, in other words, when a = In each case the strategy is the same: nλ/2 – at that point, the wave reflected from one side of the barrier and from the other side • Find a plane-wave solution in each case. interfere destructively, resulting in R = 0. Note that reflected waves will have negative wavenumbers • For a potential barrier, V > E, tunnelling oc- • Match boundary conditions, bearing in mind curs. If κ2a is very large (very high or thick that the wave at any point is given by the su- barrier), we are in the weak tunnelling limit, perposition of all the waves present there. where we ignore all e−κ2a terms, and we have A few notes: 16k2κ2e−2κ2a • Two kinds of reflection coefficients exist: 1 2 T ≈ 2 2 2 (k1 + κ2) – Amplitude reflection coefficients,r ˜ and t˜, are simply given by dividing the reflected amplitude over the incident amplitude. Note that – Flux reflection coefficients, R and T are – This is a strong function of a, m and V0 − found by diving the reflected flux (Equa- E. tion 2.5) over the incident flux. If the wavenumber is the same in the incident and reflected regions, T = t˜2, and R =r ˜2. – A single evanescent wave carries no particle current, but a superposition of two In all these problems, it is simplest to assume opposite evanescent waves does. This is that the incident wave has an amplitude of 1, how tunnelling occurs. and it is useful to remember, in that case, that T + R = 1. • Note, in all cases, that the energy of the beam • Negative reflection coefficients indicate a phase is not specified – a continuous range of energies change of π. is possible.

17 18 CHAPTER 4. THE WAVE APPROACH TO QM

4.1.1 Applications of tunnelling • However, in the radioactive decay of U238, the emitted α-particles have energies less than Field emission from a metal this. • The radioactive decay process occurs by tun- • Electrons do not usually spontaneously escape nelling. from metals because of the potential step at the surface – the ‘work function’. • However, if a strong applied electric field is 4.2 Bound states applied, the potential barrier becomes V = −eEx, where x is the distance from the metal Unbound particles can have any energy – but (the field cannot penetrate into the metal). for bound particles, wavefunctions satisfying the • This results in a barrier of finite width, and Schrödinger Equation can only have particular en- tunnelling occurs – field emission. ergies (eigenvalues of the Hamiltonian). A few general points: The Scanning Tunnelling Microscope • A particle bound in a symmetric potential V must be equally probably found at ±x – this • Strong field occur between a metal surface and means that

a very sharp metal tip, suitably biased – the 2 2 iα resulting tunnelling current, I, can be moni- |ψ(−x)| = |ψ(x)| ⇒ ψ(−x) = e ψ(x) tored. By repeating the whole process the other way • The tip is scanned horizontally over the sur- round, we obtain face, and I is kept constant by moving the tip iα up and down with very sensitive piezo-electric ψ(x) = e ψ(−x) actuators. Combining both • Because I is constant, the tunnelling barrier 2 must be constant, sot he top follows a trace ψ(x) = eiα ψ(x) corresponding to a particular value of the electron wavefunction. This means that

• The voltages controlling the tip are recorded, eiα = ±1 and allow a surface contour map to be gen- erated, imaging the wavefunction density over Going back to our original equation, we get the surface. that • Extremely high spatial resolution can be ψ(−x) = Pb ψ(x) = ±ψ(x) achieved ( 0.1 nm), imaging individual In other words, all the eigenstates have either atoms. even or odd parity. • The ground state of a symmetric well always Radioactive α-Decay has even parity, because it results in less ‘cur- vature’ in the wavefunction, which in turn • In Rutherford scattering of α-particles from means that the Hamiltonian (involving ∇2) is the U238 nucleus, 8.8 MeV particles are kept as small as possible. strongly back-scattered. • The correspondence principle can be shown to • This means that the Coulomb potential around hold in all cases, both in position and in mo- the nucleus is at least 8.8 MeV high. mentum space. 4.2. BOUND STATES 19

4.2.1 The Inﬁnite Well 4.2.4 The 1D Harmonic Oscillator

The energy of the states is The energies of the oscillator are 2n2π2 1 E = ~ E = n + ~ω 2ma2 2 Where a is the width of the well, and n is an integer, and their wavefunctions are n ≥ 1. −q2/2 ψn = AnHn(q)e p 4.2.2 The δ-function potential Where q = x mω/~ and the H are the Hermite Polynomials. When faced with a δ-function potential: The constant ω is deﬁned as • Integrate the Schr¨odinger Equation over the r α ω = range x0 + to x0 − . m

• Any contribution on the RHS not involving the Where the potential is given by δ-function will vanish as → 0. 1 2 • The result will be an expression for the discon- V (x) = αx 0 2 tinuity in ψ at x0, in terms of ψ(x0). Thereafter, the tactic is to ﬁnd trial solutions 4.2.5 Three Dimensions around the δ-function and then to match the boundaries using continuity of ψ and the discon- The Harmonic Oscillator tinuity in ψ0 found above. In that case, each potential is harmonic, and the overall wavefuntion is a product of three 1D wave- 4.2.3 Finite square well functions. The energy is characterised by three quantum numbers, and is given by For the ﬁnite square well, the ‘trial solution — boundary conditions’ approach can be used, and E = (nx + ny + nz + 3/2)~ω this leads to the following two equations:

Y = X tan X The inﬁnite well

Y = −X cot X If the inﬁnite well has sides a, b and c, the total energy is given by Where X = ka/2 and Y = κa/2, and k and κ " # are the wavenumbers in the classically allowed and π2 2 n2 n2 n2 E = ~ x + y + z forbidden regions respectively. 2m a2 b2 c2 From the form of κ and k, we can deduce that Again, the wavefunction is a product of the 1D mV a2 ones. X2 + Y 2 = 0 2~2 Finding the intersection of this circle with the functions above gives the bound states, the energy of which can be recovered by using either the expression for k or κ. 20 CHAPTER 4. THE WAVE APPROACH TO QM Chapter 5

The Basic Postulates & Operators

5.1 The Basic Postulates of 5.2.1 Dirac notation QM When dealing with Hermitian operators, it is easi- est to use Dirac Notation, in which The basic postulates of QM are as follows: • |ψi denotes “the state with wavefunction 1. The most complete knowledge that can be had ψ(x)”. of a system is represented by the state vector |χi. • hψ| denotes “the state with the wavefunction ψ∗(x)”. 2. To every observable A, there corresponds a Hermitian operator Ab . The results of the mea- • The ‘inner product’ or ‘overlap integral’ is surement of A must be one of the eigenvalues written Z ∞ of Ab . hφ|ψi = φ∗ψ dx −∞ 3. If the eigenvalue ai corresponds to eigenstate |ψii, then the probability of obtaining the result ai when the system is in state |χi is 2 5.2.2 Properties of Observables |hψ1|χi| . 4. As a result of a measurement of A in which the A few key properties of operators corresponding to value ai is obtained, the state of the system is observables: changed to the corresponding eigenstate |ψ i. i • They are linear, such that 5. Between measurements, the state |χi evolves with time according to the time-dependent Ab (αΨ + βΦ) = α(Ab Ψ) + β(Ab Ψ) Schr¨odinger equation

∂ • Operators commute over addition, but not i |χi = Hc |χi over multiplication. The commutator of two ~∂t operators as deﬁned as h i 5.2 Operators Ab , Bb = Ab Bb − Bb Ab

A key part of these postulates is that observables • Operators corresponding to observables are are represented as operators. When measuring the Hermitian – this means that value of the observable, the result is always an † eigenvalue of the operator. Ab = Ab

21 22 CHAPTER 5. THE BASIC POSTULATES & OPERATORS

† Where Ab , the Hermitian conjugate of Ab is • The set of eigenfunctions of a Hermitian oper- deﬁned by ator is complete. In other words, if the φi are the orthogonal and normalised eigenfunctions, D E∗ D † E φ|Ab |ψ = ψ|Ab |φ and |ψi is a wavefunction spanning the same space, we can write For all wavefunctions Φ and Ψ. X • Some properties of Hermitian operators: |ψi = ci |φii i †† – Cb = Cb where

† † ci = hφi|ψi – (Ab + Bb )† = Ab + Bb † † † † • Note that this idea can, and indeed must, be † – (Ab Bb Cb ... Zb ) = Zb ... Cb Bb Ab extended to operators with an inﬁnite number – If Ab and Bb are Hermitian, then Ab Bb is of eigenvalues and eigenvectors – for example, not necessarily Hermitian, but the position operator. Cb = Ab Bb + Bb Ab is. 5.3 Expectation values – If Ab and Bb are Hermitian, then Ab Bb − h i Consider a wavefunction ψ, and an operator A. ψ Bb Ab = Ab , Bb is not necessarily Hermi- b can clearly be expressed in terms of the eigenfunc- tian, but P h i tions, φi of Ab – ψ = i ciφi. i Ab , Bb It then stands to reason that if we make a mea- is. surement of A when the particle is in state ψ, the † expected value the measurement will be equal to the – If Ab |ψi = |φi, then hψ| Ab = hφ| product of each eigenvalue with the probability of • Operators corresponding to observables have obtaining it. This is given by real eigenvalues, because the observables are real quantities. This follows from the hermitic- D E D E ity of the operators. Ab = ψ|Ab |ψ • Operators corresponding to observables have orthogonal eigenfunctions, provided that the 5.4 The Generalised Uncer- eigenfunctions are not degenerate. We also normalise the eigenfunctions φi such that tainty Principle

hφi|φji = δij 5.4.1 Compatible Observables If the eigenfunctions are degenerate, we can make them orthogonal using the Gram- Consider two observables A and B that commute Schmidt orthogonalisation process. Consider h i (ie: Ab , Bb = 0). Let the orthogonal eigenfunc- eigenfunctions φ1 and φ2, both with the same eigenvalue a, but not necessarily orthogonal. tions of Ab be φi, with corresponding eigenvalues We can construct an eigenvector φ that will ai. Degeneracy or not, these eigenfunctions can be be orthogonal to φ1 as follows: chosen to be orthogonal.

φ = αφ1 + βφ2 Now, consider

where Ab (Bb φi) = Bb (Ab φi) = ai(Bb φi) α hφ |φ i = − 1 2 β hφ |φ i 1 1 So we see that Bb φi is also an eigenfunction of Ab . 5.4. THE GENERALISED UNCERTAINTY PRINCIPLE 23

2 However, this means that Bb φi must be propor- • Likewise, (∆B) must be real and positive. tional to φi (or at least to one of the φi that has an h i the eigenvalue ai. • We also showed above that i Ab , Bb is Hermi- Dh iE In other words, Ab and Bb have simultaneous eigen- tian, so Ab , Bb is also real. states. This means that Since all the coeﬃcients of Quadratic 5.1 are real, If two observables commute, they but we also require the quadratic to be ≥ 0, we share eigenfunctions, and one such must have b2 − 4ac ≤ 0. This means that eigenfunction has deﬁnite, precise values of the two operators D h iE2 i A , B − 4(∆A)2(∆B)2 ≤ 0 simultaneously. The two operators are b d b d compatible. Finally, we note that h i h i 5.4.2 Incompatible Observables Ab d, Bb d = Ab , Bb

Now, consider two operators Ab and Bb that do not As such, we have that commute. We deﬁne the operator Ab d, which de- ﬁnes the deviation of A from its mean value A¯: 1 D h iE

∆A · ∆B ≥ i Ab , Bb ¯ 2 Ab d = Ab − A 2 ¯ 2 2 ¯ ¯2 This is the generalised uncertainty principle. Ab d = (Ab − A) = Ab − 2AAb + A So: =2A¯A¯ z }| { 5.4.3 Conjugate Observables D 2E D 2E ¯ D E Ab d = Ab − 2A Ab Conjugate pairs of Observables, pi and qi are de- D 2E D 2E ¯2 2 Ab d = Ab − A = (∆A) termined from the Lagrangian

A similar operator Bb d can be constructed for Bb . L = T − V Now, consider the state vector |φi obtained from any state vector |ψi as follows where ∂L pi = |φi = (Ab d + iλBb d) |ψi ∂qi Where λ is an arbitrary real number. In QM, it is a fundamental assumption that these And consider the inner product hφ|φi, which must are replaced by operators, according to be positive: ~ ∂ qi → q i and pi = pi = D E b b i ∂qi hφ|φi = ψ|(Ab d − iλBb d)(Ab d + iλBb d)|ψ Examples of conjugate observables are posi- 2 2 2 D h iE tion/momentum and angle/angular momentum. = (∆A) + λ (∆B) + λ i Ab d, Bb d(5.1) ≥ 0 For conjugate observables A and B, it is always the case that h i Now: Ab , Bb = i~ 2 • Ab and Ab are Hermitian, which implies that and we then get the familiar uncertainty principle 2 Ab d is Hermitian. This means that hAid = ∆A must be real, and that (∆A)2 must be real and 1 ∆p ∆q ≤ positive. i i 2~ 24 CHAPTER 5. THE BASIC POSTULATES & OPERATORS

5.4.4 Minimum Uncertainty States The Momentum Operator

It is reasonably clear from the analysis above that Classically, the relation between kinetic energy and if we want the uncertainty to be kept to a mini- momentum is given by T = p2/2m. Given the form mum, we require the inner product hφ|φi to vanish. of T in Quantum Mechanics, we can conjecture that Therefore the equivalent momentum operator in QM is |φi = Ab d + iλBb d |ψi = 0 pb = −i~∇ This gives a diﬀerential equation for ψ: The Parity Operator ∂ (x − x¯) + iλ ~ − p¯ ψ(x) = 0 i ∂x The Parity Operator simply inverts the wavefunction through the origin ∂ψ x − x¯ p¯ = − + i ψ(x) ∂x ~λ ~ Pb ψ = ψ(−x) This has, as a solution, the standard Gaussian wavepacket The Translation and Rotation Operators 2 ψ(x) = Ce−(x−x¯) /2~λei¯px/~

q q The Translation Operator, Db , produces a shift ~λ ~ For which ∆x = 2 ∆p = 2λ along Ox:

Db f(x) = f(x + ) 5.5 Examples of operators For small : pbx Db = 1 + i ~ The Hamiltonian Operator

The Rotation Operator, Rb , produces a rotation We saw above that the Hamiltonian Operator is about Oz: given by Rb f(φ) = f(φ + ) 2 ~ 2 Hc = − ∇ + V (r) 2m The Position Operator We also interpreted the two parts of the operator to represent the kinetic energy and potential energy of the particle, so we obtain two new operators: The position operator, xb, is simply a number, x – the position of the particle. The eigenfunctions of • The Kinetic Energy operator, Tb , is given by the operator are Dirac Delta functions, at the point x. This means that, if we want, for example, to ﬁnd 2 2 ~ 2 p the probability that the particle is at a position x , Tb = − ∇ = b 0 2m 2m we simply use Postulate 3:

2 2 Probability = | hδ(x − x0)|ψi | = |ψ(x0)| • The Potential Energy operator, Vc, is simply a number Which is consistent with our previous interpreta- Vc = V (r) tion of the wavefunction. 5.6. LADDER OPERATORS 25

5.6 Ladder Operators 5.6.1 The Harmonic Oscillator

Suppose that two operators, X and N, have the We can put this all to good use to work out the en- commutation relation ergy levels of the Harmonic Oscillator, with Hamil- tonian [N,X] = cX p2 mω2x 2 Hc = Tb + Vc = b + b We can then derive a number of useful results 2m 2

Ladder action The Operators

Let |φi be an eigenstate of N with eigenvalue α, Consider and consider N acting on X |φi: r 1 p p NX |φi = (XN + [N,X]) |φi ab = Vc + i Tb ω~ = (XN + cX) |φi rmω p = Xα |φi + cX |φi a = x + i√ b b 2 b = (α + c) X |φi ~ 2m~ω Now In other words, the operator X has changed our h i eigenstate to another eigenstate with eigenvalue c Hc, ab = −~ω higher than the previous one. † ab therefore acts as a lowering operator, and a as a raising operator. Lowering operator

Now, consider X† The Energy Spectrum

† † N,X† = NX† − X†N = XN † − N †X The ‘extra bit of information’ we have here is that the energy of the oscillator must be positive, be- Now, if N is a Hermitian operator, this gives D 2E D 2E cause pb and xb must be positive, which †† D E N,X = XN − NX = −cX means that Hc ≥ 0. As such, there must be Taking the Hermitian Conjugate of both sides a lowest energy ground state satisfying

† † N,X = −cX ab |φ0i = 0 And this operator will therefore lower the eigen- Otherwise, value by C. ab |φ0i In other words, if X is a raising operator for the would be a lower energy state. Hermitian operator N, then X† is a lowering operator for N. Now, we note that we can prove We can use these as follows: h †i ab, ab = 1 • The ladder operators themselves reveal the spacing between the energy levels. 1 † † † 1 Hc = 2 ~ω abab + ab ab = ~ω ab ab + 2 • Extra information about the states (eg: they must have a minimum value) allows us to ﬁnd As such the eigenstates. 1 Hc |φ0i = 2 ~ω |φ0i 26 CHAPTER 5. THE BASIC POSTULATES & OPERATORS

The excited states are obtained by operating re- peatedly with the raising operator – we found that each is ~ω higher than the previous one. This gives the energy spectrum

1 En = (n + 2 )~ω

The Wavefunctions

To find the ground state wavefunction, simply write the equation ab |φ0i in full, and solve the resulting differential equation. To find higher wavefunctions, remember that

Time Dependence

6.1 Measurements 6.2 Ehrenfest’s Theorem

We know that the expectation value for an observ- We mentioned above that a measurement of a given able A for a system in state Ψ(x, t) is given by observable A produces a value ai, one of the eigenvalues of Ab , and collapses the wavefunction into D E hAi = Ψ|A|Ψ the corresponding eigenfunction ψi. b

This eigenfunction ψi can be expressed in terms of We can ﬁnd an expression that describes how this the eigenfunctions of the Hamiltonian (the energy changes with time eigenfunctions): dt dx z}|{ z }| { X d hAi d D E ψi = biφi = Ψ|Ab |Ψ dt dt i " # Z ∂Ψ∗ ∂Ab ∂Ψ = Ab Ψ + Ψ∗ Ψ + Ψ∗Ab dx And we know that between measurements, the ∂t ∂t ∂t wavefunction evolves according to the time- * + Z ∂Ψ∗ ∂Ψ ∂Ab dependent Schrodinger equation. This means that = Ab Ψ + Ψ∗Ab dx + ∂t ∂t ∂t each of the eigenfunctions of the Hamiltonian | {z } (which, we know, are time-independent stationary Use time dependent S.E. states of the Schr¨odinger equation – see section 3.2) have a time dependence e−iEit/~. * + d hAi 1 D h iE ∂Ab = i Hc, Ab + In other words: dt ~ ∂t

X −iEit/~ Ψi(t) = biΦi(0)e This last result is Ehrenfest’s Theorem – a QM i Equation of Motion.

For observables, Ab has no explicit time-dependence, and so So we see that, in general, an eigenfunction of an observable is not necessarily a stationary state. The state of the system remains a superposition of en- d hAi 1 D h iE ergy eigenstates with the same relative amplitudes = i Hc, Ab but diﬀerent relative phases. dt ~

27 28 CHAPTER 6. TIME DEPENDENCE

6.2.1 Classical limit 6.3 Conserved Quantities

We consider the classical limit of Ehrenfest’s The- dhAi We note that if [Hc, Ab ] = 0, = 0. orem in the case of two operators: dt This means that any operator that commutes with the Hamiltonian is a constant of the motion. Position – xb Similarly, if Ψ is a stationary state, all observables have time-independent values. Since Vc = V and xb = x commute (they’re just numbers), we have that1

1 2 6.3.1 Parity [Hc, x] = [p , x] = −2i p b 2m b b ~ Therefore: We note that the parity operator, Pb , has two eigen- d hxi hpi values – +1 and −1 (clearly, those are the only ones, b = b 2 dt m since Pb = Ib). To be compared with the classical result We also note that Pb always commutes with pb, and dx p commutes with V if V (x) = V (−x). Therefore, Pb = dt m commutes with Hc, and is a conserved quantity, if the potential itself is symmetric.

Momentum – pb This means that a particle with a given energy in a symmetric potential will always retain the same 2 In this case pb obviously commutes with pb , but not parity as time passes. with V . So: More generally, this is because Hc is invariant under dV Pb . In fact, for every symmetry property of Hc, there [Hc, pb] = [Vc, pb] = i~ dx exists a symmetric operator which commutes with H . (Using the deﬁnition of pb, and assuming it acts on c an arbitrary function f(x), in the last step). An example of this is translational invariance – if As such dV/dx = 0, then [Hc, pbx] = 0 (this can be shown in dp dV b = − = hF (x)i a rather roundabout way using the Taylor expan- dt dx sion of the translation operator above).

Classically, we would have dp 6.4 The Time-Energy Uncer- b = F (hxi) dt tainty Relation The two expressions are not generally equivalent. However, if we perform a Taylor expansion of F (x) This is very different from the general uncertainty relations we have discussed thus far – time is not a about hxbi and find its expected value, we find that they are the same as long as we can ignore the dynamical variable, represented by an operator – it quantum uncertainty ∆x. This is true in, for exam- cannot be “measured”. Also, it does not appear in ple, particle accelerators, where ∆x is much smaller the Lagrangian formulations. It is simply a param- than any scale over which we would expect the eter identifying the instant at which the property fields to change, and we would expect to recover of the system is measured. the classical equations.

1 2 The trick to simplify [pb , xb] is to write it as pbpbxb + pbxbpb − pbxbpb − xbpbpb. Chapter 7

3 Dimensions

h i 7.1 Angular Momentum Lbx, yb = i~xb h i Angular momentum in QM is deﬁned, as in classi- Lbx, pby = i~pbz cal mechanics, by Lb = r × p. Its components are b b h 2i h 2i therefore Lbx, rb = Lbx, pb = 0 Lbi = bj pbk − kbpbj Where i, j and k are any cyclic permutation of x, y and z. The operator for the square of the total orbital angular momentum Lb is 2 2 2 2 Lb = Lbx + Lby + Lbz

7.2 Commutation relations

The following can be proved: xbα, pbβ = i~δαβ h i Lbi, Lbj = i~Lbk h 2 i Lb , Lbα = 0 Where α and β are any of x, y and z, and i, j and k are any cyclic permutation of x, y and z. It follows that the magnitude of L and one of Lx, Ly and Lz can be known simultaneously (ie: are compatible observables) – no two components of the angular momentum can be precisely known known simultaneously. By convention, we choose to accurately specify the z-component. Other commutation relations, which might or might not be useful: h i h i Lbx, xb = Lbx, pbx = 0

29 30 CHAPTER 7. 3 DIMENSIONS Chapter 8

Angular Momentum

8.1 Orbital Angular Momen- This means that all the |φm` i have the same eigen- 2 tum Eigenvalues value of Lb – call it Λ~2. We will now label our states Ladder operators

|φΛ,m` i Consider the following operator Restraining condition Lb+ = Lbx + iLby

We note that Obviously, Lz must be smaller or equal to L, so h i D 2E D 2E Lbz, Lb+ = ~Lb+ Lbz ≤ Lb

And so, we must have As such, Lb+ is a ladder operator for angular momentum 2 m` ≤ Λ √ Separation of Eigenvalues Furthermore, if λ < Λ is the largest possible value of m`, then, by symmetry, the smallest value must be −λ. We therefore know that the eigenvalues of Lbz are separated by ~. Let’s call these eigenvalues m`~, and the eigenstates |φm i. ` The Eigenvalues m` of Lbz

Total angular momentum Using the ladder operators, this means that m` must have values 2 Now, Lb and Lbz commute, so they will have simul- m` = λ, λ − 1,..., 1 − λ, −λ taneous eigenstates. 2 The number of eigenvalues is 2n + 1 and must be However, Lb also commutes with Lby and Lbz, and an integer. For this to happen, we must have therefore also commutes with Lb±. This means that 2 λ = ` or ` + 1 Lb and Lb± have simultaneous eigenstates. As such, 2 2 L |φ i has the same eigenvalue of L as |φ i.1 b± m` b m` Where ` is an integer.

1 2 2 To prove, let |φm` i have eigenvalue Λ~ of Lb , and note For orbital angular momentum, it will be shown 2 2 2 that Lb Lb± |φm` i = Lb±Lb |φm` i = Λ~ Lb± |φm` i that only the λ = ` case is possible.

31 32 CHAPTER 8. ANGULAR MOMENTUM

2 The Eigenvalues Λ of Lb 8.2 Eigenstates

2 We can write Lb in terms of the raising and lowering Effect of Raising and Lowering Operators operators We have, so far, derived that + 2 2 Lb−Lb = Lb − Lbz − ~Lbz Lb+ |`, mì = D`,m` |`, m` + 1i 2 Lb = Lb−Lb+ + ~Lbz + Lbz Lb− |`, mì = C`,m` |`, m` − 1i

Now, consider the state |φΛ,m =ì – Lb+ acting on ` We can work out what C and D are as follows: that state should give 0, so • Take the Hermitian Conjugate of both the 2 2 2 Λ~ = Lb |φΛ,m`=ì = 0 + ~`~ + (`~) equations above, and then apply to the original equations, from the left. And therefore Λ = `(` + 1) • Simplify using the following identity 2 2 To summarise: Lb = Lb+Lb− − ~Lbz + Lbz 2 (Prove by expanding L L ). The eigenstates of both Lbz and Lb are b+ b− denoted • Assume |`, mì and |`, m` − 1i are normalised, and that C and D are real. |`, mì This gives The corresponding eigenvalues are given by p C`,m` = `(` + 1) − m`(m` − 1) 2 ~ L |`, m i = `(` + 1) 2 |`, m i b ` ~ ` p D`,m` = ~ `(` + 1) − m`(m` + 1) Lbz |`, mì = m`~ |`, mì

Where The Operators in Spherical Polars

m` = −`, −`+1,..., −1, 0, 1, . . . , `−1, ` In 3D spherical polars ` is 0 or an integer pb r pb θ pb φ z }| { z }| { z }| { ~ ~ ∂ 1 ~ ∂ ~ 1 ∂ Notes: pb = ∇ = r + θ + φ 2 i i ∂r r i ∂θ i r sin θ ∂φ • We can think of this as the eigenvalues of Lb restricting the length of the L-vector in space, We can use these to explicitly ﬁnd the forms of and eigenvalues of Lbz restricting its direction. each component of Lb, in terms of the angular pb • Note that Lx and Ly are still completely un- operators, the form of which we can obtain from known. the Hamiltonian above: • The states with ` = 0, 1, 2, 3, 4,... are called ∂ Lbz = −i~ s-, p-, d-, f-, g-, ... states respectively. ∂φ

• For the s state, all the components of angular ∂ ∂ Lbx = i sin φ + cot θ cos φ momentum are simultaneously 0. ~ ∂θ ∂φ 8.3. BITS AND BOBS 33

∂ ∂ The resulting eigenfunctions must be normalised to Lby = i~ − cos φ + cot θ sin φ ∂θ ∂φ unity when integrated over all angles

The ladder operators then become Z 2π Z π |Y |2 sin θdθdφ = 1 `,m` ±iφ ∂ ∂ 0 0 Lb± = e ± + i cot θ ~ ∂θ ∂φ To summarise All these operators depend on θ and φ only – so all the angular momentum eigenfunctions |`, m`i will The eigenfunctions of Lbz, the Spherical be functions of these angles only – say Y (θ, φ). `,m` Harmonics, Y`,m` have the form

im`φ Y`,m` = F`,m` (θ)e The Eigenfunctions

Where the θ-part F`,m` is given by

We ﬁnd the eigenfunctions in two parts ` F`,` ∝ (sin θ) The φ part The eigenvalue equation for Lbz is dF F ∝ `,m` + m cot θF (θ) `,m`−1 dθ ` `,m` LbzY`,m` (θ, φ) = ~m`Y`,m` (θ, φ)

Using the operator derived above ∂ 8.3 Bits and bobs −i = m Y (θ, φ) ~∂φ ~ ` `,m`

Which means that Note the following properties of the spherical harmonics

im`φ Y`,m` (θ, φ) = F`,m` (θ)e Y = (−1)m` Y ∗ (θ, φ) = Y (θ, π − φ) `,−m` `,m` `,m` Note that this is what forces the m` to be in- tegers for orbital angular momentum – we re- ` Y`,m` (π − θ, π + φ) = (−1) Y`,m` (θ, φ) quire a rotation of 2π about the z axis to leave the wavefunction unchanged. Parity The θ part As earlier, Lb+ has to give 0 when it acts on the highest eigenstate of Lbz (the one In 3D, the parity operator inverts the wavefunction with m` = `. So through the origin: ∂ ∂ eiφ + i cot θ |`, `i = 0 ~ ∂θ ∂φ Pb ψ(r = ψ(−r)

This gives Pb ψ(r, θ, φ) = ψ(r, π − θ, π + φ) dF sin θ `,` = ` cos θF From the second property above, therefore dθ `,`

Which gives ` Pb Y`,m` = (−1) Y`,m` ` F`,`(θ) ∝ (sin θ) 2 The eigenstates of Lb and Lbz are all To obtain lower eigenfunctions, we successively either even or odd, depending on apply the lowering operator. whether ` is even or odd... 34 CHAPTER 8. ANGULAR MOMENTUM

Conservation of Angular Momentum and angular momentum L = mevr. It is effectively a loop of area A = πr2 carrying a current I = As we showed above, the rotation operator, for a −e/T , and so it has a magnetic moment of small rotation about the z-axis is given by eL ∂ψ µ = − Rb ψ(r, θ, φ) = ψ(r, θ, φ + ) = ψ + 2me ∂φ But we now know that Lz is quantified, as Lz = But using the form of Lbz: m`~. So i e~ Rb = 1 + Lbz µz = − m` ~ 2me Thus, if the system is rotationally invariant, then So this means that the z-component of the mag- Rb will commute with the Hamiltonian and so will netic field is quantized in units of Lbz – angular momentum is then a conserved quantity. e~ µB = – the Bohr magneton 2me Particle Flux It is now clear, in the Stern-Gerlach experiment, why a discrete number of beams emerge – taking Using the expression for ∇ in spherical polars the z-axis to lie along the B-field gradient, there is ∂ 1 ∂ 1 ∂ a quantized number of values the magnetic moment ∇ = r + θ + φ was able to have. ∂r r ∂θ r sin θ ∂φ

We can work out the particle ﬂux for the spherical harmonics. We ﬁnd that

• For those spherical harmonics with m` = 0 (ie: zero z-component of angular momentum), there is no particle ﬂux, even though the total momentum is non-zero. This is because the momentum is only in the x and y direction, but is completely indeterminate, and so cancels to 0. • For other spherical harmonics, we ﬁnd a net current around the z-axis, as indeed expected.

Energy

The energy of the particular egeinstates depends on the system under consideration – we will consider the example of the rigid rotor and of the hydrogen atom later on.

8.4 The Stern-Gerlach Exper- iment

Classically, an electron moving in a circular orbit of radius r with speed v has orbital period T = 2πr/v Chapter 9

Spin

9.1 Experimental Evidence ‘bolting on’ a spin wavefunction, |Si, to the spatial wavefunction. We saw above that the splitting of beams in the Stern-Gerlach experiment was due to quantisation Spatial operators (like L) only act on the spatial of angular momentum. There are 2` + 1 possible wavefunction, whereas spin operators (denoted S) values of m`, and since ` must be in integer for only acts on the spin wavefunction. orbital angular momentum, this should produce an ODD number of beams. To summarise: However, there are some experiments where an even number of beams is produced – there seems to 2 be a non-orbital magnetic moment in some atoms. The eigenstates of both Sb z and Sb are We call this SPIN. denoted

Spin has no classical counterpart. If we try and |Si view spin as a rotation of the electron about its axis, not only do we end up with nonosensical num- The corresponding eigenvalues are bers, but we also defeat the purpose of introducing given by spin, because we would still require quantisation in 2 2 integer units of ~. Sb |Si = s(s + 1)~ |Si

Sb z |Si = ms~ |Si 9.2 Eigenvalues and Eigenvec- Where tors ms = −s, −s+1,..., −1, 0, 1, . . . , s−1, s

Every statement made above about angular mo- s is 0, an integer or a half-integer mentum also applies to spin. The eigenfunctions, however, require a somewhat diﬀerent treatment, because they do not involve 1 1 For s = 2 , the ms = ± 2 states are usually denoted any spatial coordinates – spin cannot be described |χ+i and |χ−i. They are called ‘up’ and ‘down’ spin in terms of matter moving through space. states, even though in truth, they point at ≈ 55o to the horizontal. It turns out that it emerges naturally from the Dirac Equation – a relativistic wave equation, and it turns out that spin can be represented by Ladder operators work as before.

35 36 CHAPTER 9. SPIN

9.3 Uncertainties Because the spin operators do not aﬀect the spatial ones, and vice versa, it doesn’t matter what order one applies Lb and Sb in, so Lb and Sb commute. Measurement of Sb z when the spin part of the system is in one of the eigenstates of Sb z gives the Using this fact, we can show that the usual com- corresponding eigenvalue with 0 uncertainty. mutation relations apply to the components of Jb, When the system is in one of the x or y eigenstates, and this means that everything we’ve done so far applies. So: the expected value of Sb z is 0, but the uncertainty ~ is 2 (use ladder operators to prove). 2 The eigenstates of both Jb z and Jb (which are compatible observables) are 9.4 Spin in any direction denoted

|j, mji Spin in any direction can be expressed as a linear combination of up and down spin: The corresponding eigenvalues are given by |αi = c1 |χ+i + c2 |χ−i 2 2 Jb |j, mji = j(j + 1)~ |j, mji And the vector spin operator is given by Jb z |j, mji = js~ |j, mji Sb = (Sb x, Sb y, Sb z) D E D E We can ﬁnd Sb by explicitly evaluating α|Sb|α , Where and we obtain mj = −j, −j+1,..., −1, 0, 1, . . . , j−1, j D E ~ ∗ ∗ ∗ ∗ ∗ ∗ Sb = (c c1 + c c2)i + i(c c1 − c c2)j + (c c1 + c c2)k 2 2 1 2 1 1 2 j is an integer or a half-integer

Let the spin be pointing in the n direction. • The spin |αi will be an eigenvector of a mea- 9.5.2 The values of J surement of spin that direction (ie: n.S) with b Spins add vectorially, so taking the z component, eigenvalue ± 1 (assuming s = 1 ). 2 2 we clearly have • To ﬁnd an expression for an arbitrary spin |χi m = m + m in terms of |χ+i and |χ−i, we can use the j l s above, or simply solve the equation n.Sb |χi = 1 Therefore, the maximum and minimum values of ± ~ |χi. 2 mj are given by

mj,max = ml,max + ms,max = ` + s 9.5 Combining Orbital & Spin m = |` − s| Angular Momentum j,min (In the ﬁrst case, the spins are ‘maximally aligned’, and in the second, they are ‘maximally anti- 9.5.1 Introduction aligned).

The total angular momentum, J, consists of both Therefore, orbital and spin angular momentum. The operator is therefore given by j = (`+s), (`+s−1),..., |`−s|+1, |`−s| Jb = Lb + Sb 9.6. CONSERVATION OF TOTAL ANGULAR MOMENTUM 37

9.5.3 Combined wavefunctions Which means that Jb z will be a constant of the motion – angular momentum is conserved. If mj and j are definitely known for a particular particle, then the orbital part of its wavefunction will be a linear combination of all the possible individual orbital and spin angular momentum eigenfunctions that would lead to those values of j and mj. To find them: • First, construct one definite wavefunction, for which there is no linear combination – just one possibility for both angular and orbital angular momentum eigenfunctions.

• Use the lowering operator, Jb − = Lb− + Sb −, to ﬁnd the eigenfunction which is the next mj down, as follows: – Apply the operator in its Jb form on the LHS. – Apply it in its Lb and Sb form on the RHS. – Deduce the next wavefunction down. • For other values of j, use the fact that all wavefunctions must be orthogonal, and apply that condition with an already-determined wavefunction that contains the same constituent parts.

9.6 Conservation of total angular momentum

The rotation operator for a small angle around the z-axis is given by i Rb ,L = 1 + Lbz ~ Assuming a similar expression for spin, and ignoring terms in 2, then the operator for overall rotation in both spin and spatial coordinates is i Rb = 1 + Jb z ~ If Hc is invariant under such rotations, then it obviously commutes with Rb , so [Hc, Rb ] = 0, which implies that [Hc, Jb z] = 0 38 CHAPTER 9. SPIN Chapter 10

Central Potentials

10.1 Conservation of angular • The r-motion, however, will also be quantized, momentum and will contribute to determining the energy of the state – let the quantum number n label the radial motion. If we consider a Hamiltonian in which the potential, • For a central potential, E cannot depend on V , is a function of r only, we find, as before m`, because its value depends on the choice of h i z-axis, which does not affect the Hamiltonian Hc, Lbα = 0 (which depends only on r). So E depends on n and `. h 2i Hc, Lb = 0 • The radial part affects the energy, so we can label it Rn,`(r). This follows from the fact that the angular momentum operators are functions of angle only, and therefore commute with any function of r, and the 10.3 Separation of Variables fact they commute with the pb operators. This means that angular momentum is a conserved ∇2, in the Schrödinger Equation, contains both ra- quantity. dial and angular parts, and to separate these requires tedious calculations. The result of these calculations gives

10.2 Quantum numbers 2 2 ~ 1 ∂ 2 ∂ψ Lb ψ − 2 r + 2 + V (r)ψ = Eψ 2 2m r ∂r ∂r 2mr It also means, however, that Lb , Lbz and Hc must have a common set of eigenstates, and their eigen- Feeding in ψ(r, θ, φ) = R(r)Y (θ, ψ) into this, we values can be speciﬁed simultaneously. Further- get more, since the angular momentum operators depend only an angle, and the hamiltonian operator depends only on r (for central potentials), the com- 2 d2U 2`(` + 1) mon eigenstates will be of the form − ~ + ~ + V (r) U = EU 2m dr2 2mr2 ψ(r, θ, φ) = R(r)Y (θ, φ) `,m` Where U(r) = rR(r) Notes: This is just a 1D “radial” Schr¨odinger Equation in • The angular motion is quantized by ` and m`. r, with an extra contribution to the potential of

39 40 CHAPTER 10. CENTRAL POTENTIALS the form L2 /2mr2 – which corresponds to the “centrifugal potential” term in the radial equation of a classical orbit. For small r, and as long as V (r) doesn’t diverge as fast as r−2, the centrifugal potential dominates, and the wavefunction looks like R ∝ r`. We see that only for ` = 0 (s-state) can the wavefunction be nonzero at the origin.

10.4 Normalisation & Proba- bilities

The normalisation of any wavefunction will be such that Z ∞ Z 2π Z π |ψ(r, θ, φ)|2r2 sin θ dθ dφ dr = 1 0 0 0

In the case where the wavefunction is separated into an angular and a radial part (ψ = RY ), the probability of ﬁnding the particle between spherical shells of radius r and r + dr is simply

P (r) dr = r2|R(r)|2 dr

Note that this is no factor of 4π, because the spherical harmonics are already normalised over all angular integrations.

10.5 Practical tips

When considering a spherically symmetric potential in 3D therefore, simply use the ‘1D radial equation’ above to reduce the problem to a 1- dimensional one. But note • If considering the ground state, remember to set ` = 0. • Remember that the 1D equation solves for rR, not R. As such, R = U/r. Therefore, at r = 0, we must have U = 0, or else the limit is unde- ﬁned or the wavefunction unphysically tends to inﬁnity. Chapter 11

Two-Particle Systems

Many systems involve two particles – for example, will aﬀect the coordinates of b, or vice-versa, so the hydrogenic atom involves the orbiting electron h i h i and the nucleus. r , p = 0 r , p a b b a

11.1 Position Probabilities 11.3 The Hamiltonian

If we consider both particles, Ψ must be written in In the presence of a potential, the Hamiltonian is terms of both the spatial coordinates p2 p2 a b Hc = + + V (r a, r b) |Ψi = Ψ(r a, r b, t) 2ma 2mb

(ignoring spin for the time being). V includes any external potential as well as any The probability of ﬁnding particle a in a volume arising from their mutual interaction. Often, V is 3 2 a function only of the separation between the two d r a around r a and particle b in a volume d r b particles around r b is 2 3 3 V (r , r ) = V (|r − r |) = V (r) |Ψ(r a, r b, t)| d r a d r b a b a b

The probability for particle a without regard to position b is obtain by integrating the wavefunction 11.4 Conservation of Total over all possible values positions of particle b: Momentum Z 3 2 3 3 P (r a, t) d r a = |Ψ(r a, r b, t)| d r b d r a The total momentum operator is

P = p + p 11.2 Observables a b We can show that (α and β are any of x, y and z) The observables relating to each particle corre- spond to operators acting only on the relevant r- • [Pb α, rbβ] = 0, where r is the separation be- dependence of |Ψi. tween the two particles. • [P , p2 ] = [P , p2 ] = 0, because each of The usual commutation relations apply for observ- b α ba,β b α bb,β ables for any one particle. Furthermore, it is ob- p and b commute with p2 and p2, so P does ba bb ba bb b vious that no momentum operator pertaining to a also.

41 42 CHAPTER 11. TWO-PARTICLE SYSTEMS

Together, these two facts tell us that 11.6 Separation of CoM Mo- h i tion and Relative Motion Hcα, Pb β We will attempt to re-write the Hamiltonian in In other words, Pb is a conserved quantity – just as terms of new variables – R, representing the posi- expected from classical physics, since there is only tion of the centre of mass and deﬁned above, and r, the “internal force”, and no “external force”. representing the separation between particles (also However, note that p and p do not individually deﬁned above). ba bb commute with r, and therefore not with V (r), and Now therefore not with H . The momenta are not con- c ∂ ∂R ∂ ∂r ∂ served. = x + x ∂x ∂x ∂R ∂x ∂r a xb a xb x a xb x m ∂ ∂ = a − 11.5 Centre of Mass Motion M ∂Rx ∂rx So we can say that

ma The position of the centre of mass for a two-particle p = Pb − p system is ba M br m r + m r R = a a b b And likewise for b. ma + mb 2 2 By using these expressions to write pba and pbb in terms of Pb and p , we can write the Hamiltonian Consider the commutator of R with the two- br b as follows particle Hamiltonian. It can be shown that 2 2 ! h i h i Pb pbr r , p2 + r , p2 Hc = + + V (r) = HcCoM + Hcr h i ba ba bb bb 2M 2µ Rb , Hc = 2(ma + mb) −1 −1 −1 Where µ is the reduced mass – µ = ma + mb . Expanding the vectors out, we find that Substituting ψ(R, r) = U(R)u(r), and feeding h i into the Schrödinger Equation, we obtain r , p2 = 2i p ba ba ~ba 1 1 HcCoMU + Hcru = E and that a similar expression applies for b. U u Therefore, we have In other words, the equation becomes separable, with each part equal to a constant, and we get the following two equations h i Pb Rb , Hc = i ~M 2 Pb U(R) = ECoMU(R) Where P is the total angular momentum, and M 2M is the total mass. ! p2 br + V (|r|) u(r) = E u(r) Using the Ehrenfest Theorem, we therefore have 2µ r D E D E With a total energy of d Rb Pb = dt M E = ECoM + Er Which is, indeed, as expected, since the velocity of • The first equation describes the motion of a the centre of mass is P/M. free particle of mass M. 11.7. COMBINING SPINS 43

• The second equation describes the motion of a single particle, mass µ, moving in a potential V (r) (a problem which we have already solved). • The energy eigenvalues are therefore

P 2 E = + E 2M n`

Where P is the eigenvalue of Pb . The overall wavefunction then has separated form

i PR ψ(r a, r b) = ψ(R, r) = e ~ ψn,`,m` (r)

11.7 Combining spins

Spins for two particles can be combined, using precisely the same results as those derived earlier for the combination of orbital and spin angular momentum.

For two particles of spin s1 and s2, the total spin, S, will, once again, range as follows:

S = (s1 +s2), (s1 +s2)−1,..., |s1 −s2|+1, |s1 −s2|

Once again, the eigenfunctions are found by ﬁnding all the possible combinations, and using the ladder operators. 44 CHAPTER 11. TWO-PARTICLE SYSTEMS Chapter 12

Examples of real systems

12.1 The rigid rotor We can solve this as follows 1. Re-write in dimensionless form, by substitut- A rigid rotor is a particle of mass m ﬁxed to the ing origin O by a rigid light rod of length a. A diatomic 2 molecule can be considered as two rigid rotors ro- 2m Ze 2 2mEn,` A = 2 κ = − 2 tating about the common centre of mass. ~ 4π0 ~ Now, the kinetic energy of the rigid rotor is so that E = 1 Iω2 2 2 d Un,` `(` + 1) A 2 2 − 2 − + κ Un,` = 0 (No potential energy term for the free rotation). In dr r r each case, L = Iω, so 2. Consider the solution for r → ∞. L2 E = 2I 3. Consider the solution for r → 0. With the known angular momentum eigenstates, 4. Try a solution consisting of the product of the this gives the allowed energy levels: last two multiplied by a polynomial

`(` + 1) 2 −κr `+1 E = ~ degeneracy of 2` + 1 Un,`(r) = e r G(r) ` 2I (The degeneracy is obtained from the number of This results in the associated Laguerre Equa- tion. possible m` for every `). 5. Substitute a polynomial solutions. 6. Require the series to terminate so that it can 12.2 The Harmonic Oscillator be physically acceptable. Do this by ﬁnding cp+1/cp as p → ∞, and not that it diverges as 12.3 The Hydrogenic Atom e2κr. 7. The termination requires that In the hydrogenic atom A = n Ze2 2κ V (r) = − 4π0r and And the radial equation becomes: p + ` + 1 = n 2 2 2 2 ~ d Un,` ~ `(` + 1) Ze where n is the principal quantum number, an − 2 + 2 − Un,` = En,`Un,` 2m dr 2mr 4π0r integer larger than 0.

45 46 CHAPTER 12. EXAMPLES OF REAL SYSTEMS

8. The fact that p must be larger than 0 implies lar momentum for each of these orbitals will that ` ≤ n − 1. be 0. 9. We can then ﬁnd an expression for the energy Such wavefunctions are useful in certain situations of each state. • Isolated atoms involve electrons moving in the To summarise: central Coulomb potential of the nucleus – angular momentum is conserved, and we can use The stationary states are labelled by the standard orbitals. the principal (n), orbital (`) and • In a molecule or solid, the potential is no longer magnetic (m`) quantum numbers, which specify three compatible central – it is perturbed by the potential of 2 neighbouring atoms. observables Hc, Lb and Lbz. • Solving Schr¨odinger ’s Equation for these is n ≥ 1 hard, though a natural starting point to dis- cuss the true wavefunctions is to use wavefunc- ` = 0, 1, 2, ··· , n − 1 tions that conform to the local cartesian sym-

m` = −`, −`+1, ··· , −1, 0, 1, ··· `−1, ` metry of the potential – hybridised orbitals. All the quantum numbers are integer. A number of symmetries are possible. The energy of each state is determined by n alone

Z2e4m 1 ~2 Z2 1 En = − 2 2 2 = − 2 2 2(4π0) ~ n 2m a0 n And the degeneracy of each state is therefore

n−1 X 2 (2` + 1) = 2n2 `=0

(The additional factor 2 arises because of spin).

Hybridisation

• All the wavefunctions corresponding to a given n have degenerate energies, any linear combination of them is also an eigenstate of Hc. • We can therefore combine our existing orbitals into any number of new ones, pointing in any number of arbitrary directions. • These are called hybridised orbitals.

• The results are still eigenstates of Hc (and of 2 Lb , if only orbitals of equal ` are combined), but are no longer eigenstates of Lbz. In fact, the expectation value of each component of angu- Chapter 13

Identical Particles

13.1 Exchange Symmetry • For identical particles, particle exchange cannot aﬀect the overall particle density – so 2 2 • For an N-particle system, the Hamiltonian and |Ψ| = |Pb ijΨ| N-particle wavefunction would take the forms And so iφ H = H (r , p ; ... ; r , p ) |Ψi = e P ij |Ψi cN c 1 1 N N b But since i and j are identical, we can imag- ψ = ψ(r ,..., r ) N 1 N ine starting from a system with i and j inter- HcN |ψN i = EN |ΨN i changed, and we would also expect iφ Where EN is the eigenenergy of the entire N- Pb ij |Ψi = e |Ψi particle system. Therefore 2iφ • The particle exchange operator, Pb ij has the |Ψi = e |Ψi eﬀect of swapping the two particles i and j. Which means that eiφ = ±1. Therefore • We prove that Pb ij is Hermitian as follows The wavefunction ΨN , including two † identical particles i and j, must be an – Pb ij |ΨN i = hΨN | Pb ij = ΨN,exchanged eigenstate of Pb ij with eigenvalues D † E η = ±1 – ΨN |Pb ijPb ij|ΨN =

hΨN,exchanged|ΨN,exchangedi = 1 • If two particles are indistinguishable, no measurement of an operator Ab can distinguish – Since hΨN |ΨN i = 1, this implies that † between the them. Therefore, Ab |ΨN i = Pb Pb ij = Ib. ij Ab Pb ij |ΨN i. So

– But applying Pb ij twice has no eﬀect, so h i 2 Ab , Pb ij = 0 Pb = Ib † Exchange-symmetry is compatible with any – Therefore, Pb ij = Pb ij observable. η is a conserved quantity. • If two particles are identical, the Hamiltonian Even if measurements are made, this is the cannot depend on which particle is labelled case. Every eigenfunction of an operator that what – therefore, HcN is invariant under par- forms part of a state |ψi must have the same ticle exchange: exchange symmetry eigenvalue η as |ψi. We can see this by writing the expansion of |ψi Pb ijHcN = HcN and applying Pb ij separately to each eigenfunc- This can be see as the deﬁnition of indistin- tion and to |ψi itself, and writing both results guishable particles... as a series expansion.

47 48 CHAPTER 13. IDENTICAL PARTICLES

13.2 The Spin-Statistics The- eigenenergies, will be different. orem The overall wavefunction if the particles are in states α, β, etc. . . is then The spin-statistics Theorem comes from relativistic quantum field theory ΨN = u1α(1)u2β(2) ··· uNκ(N) If identical particles have integer spin And this has eigenenergy quantum number s, their overall wavefunctions must be symmetric EN = E1α + E2β + ··· + ENκ under exchange (η = 1) – these are called bosons. We can split each of the u wavefunctions into their spatial and spin parts If identical particles have half-integer spin quantum number s, the uiγ (i) = φig(r i) |χii wavefunctions must be antisymmetric under exchange (η = −1 – these are called fermions. Now, the probability of finding particle i at any point r i is then Z ∗ 13.3 Non-interacting particles P (r i) = (φ1a(r 1)φ2b(r 2) ··· ) (φ1a(r 1)φ2b(r 2) ··· ) dr 1dr 2 dr i For an N-particle system, the potential of each par- 2 ticle i in general depends on the position of ev- = |φik(r i)| ery other particle j, and on any external potential. In other words, the particle behaves as if the other Let’s denote the hamiltonian for particle i by h , bi ones weren’t there – the motion of the particles is and that for the whole system by H . cN uncorrelated. We have X hbi = Tb i + Vext(r i) + NVi|r i − r j| 13.4.2 Indistinguishable j6=i However, the term if we assume that the particles For identical particles, we can drop the numerical do not interact, and that the term in red above can eigenfunction subscripts above, because the parti- be ignored, we then have cles are all identical, and so are their eigenstates.

N However, because the particles are identical,ΦN X must be symmetric or antisymmetric under parti- HcN = Tb i + Vext(r i) j cle exchange. The simple product of appropriate eigenstates, And the whole wavefunction, ΨN can bee con- however, does not satisfy this condition – Φ must structed from the individual solutions of hbiui = N therefore be formed from linear combinations of Eiui. various products of the eigenstates

13.4 N particles Fermions – antisymmetric ΨN

13.4.1 Distinguishable The most general linear combination of products of N single-particle wavefunctions uκ(i) which is If the particles are distinguishable and diﬀer- antisymmetric under exchange of any two particles ent, then their Hamiltonians, and eigenstates and is given by the Slater Determinant 13.5. N = 2 49

– If they are in the same spatial states, the Ψ(1, 2,...,N) = spatial part of the wavefunction will be symmetric, so this forces the spin part to uα(1) uα(2) uα(3) ··· uα(N) be antisymmetric. 1 uβ(1) uβ(2) uβ(3) ··· uβ(N) √ – If they are in different spatial states, the N! ··············· spatial part can be either symmetric or uκ(1) uκ(2) uκ(3) ··· uκ(N) antisymmetric, and the spin part must This does, indeed, have the required property that have opposite exchange symmetry. interchanging two particle labels interchanges two – The way to find the wavefunctions in this columns, and therefore changes the sign of the de- case is to list every possible eigenstate terminant. (spatial & spin parts included) and find If any two of the particles are chosen to be in the linear combinations of then by calculating same state (ie: α = β), then two rwos of the deter- Slater Determinants (or by inspection). minant are the same, and ΨN immediately goes to – Sometimes, Salter Determinants cannot 0. be separated into a spatial and a spin part, but linear combinations of these This means that problematic states usually can be. No two identical Fermions can be in • Bosonic case the same single-particle state – they cannot have the same set of quantum – Very similar considerations apply. numbers. – For example, if the we have spin-0 bosons, the spin-part of the wavefunction must be This is the Pauli Exclusion symmetric (because must particles must Principle have the same spin – 0), and the spatial part therefore must also be symmetric.

Bosons – symmetric ΨN 13.5.1 Diatomic molecules All we need to do in the Bosonic case is to make every sign in the expression for the Slater Deter- Diatomic molecules are good examples of two ‘ef- minant positive instead of alternately positive or fectively non-interacting particles’. Notes: negative, and the result is symmetric. • Only the rotational parts of the wavefunctions We note that the Pauli Exclusion Principle does must be taken into consideration, because r is not apply to Bosons – in fact, every particle can be ﬁxed. in the same single-particle state, which is indeed what happens in Bose-Einstein Condensation. • The nuclei are well separated, and there is only a negligibly weak interaction between their spins. 13.5 N = 2 • Exchange symmetry, on this case, is equivalent to parity inversion. But we know that the spherical harmonics behave as follows under Things become rather complicated for N > 2 – parity inversion: however, a number of comments can be made about the case N = 2 ` Pb Y`,m` = (−1) Y`,m` • Fermionic case So the spatial wavefunction is exchange- – The particles can be in the same spatial symmetric for even `, and exchange- states, or in diﬀerent spatial states. antisymmetric for odd `. 50 CHAPTER 13. IDENTICAL PARTICLES

Hydrogen However, for identical particles, Φ must be either symmetric or antisymmetric: The nuclei are protons – spin- 1 fermions – so 2 • If Φ is antisymmetric, then the probability of the spin wavefunctions are just the exchange- finding the two particles at the same place is 0. symmetric signlet and the exchange symmetric There is “effective repulsion” between them. triplet states. • If Φ is antisymmetric, then the probability of Since the protons are fermions, the overall wave- finding the two particles at the same place is function must be antisymmetric. larger than it would be in the indistinguishable There are therefore two distinct populations case – the two particles seem to “attract”

Para-H2 consists of the anti-symmetric spin sin- This is a signiﬁcant eﬀect even in quasi-classical glet, has S = 0, and needs to have an angular systems. part with even `. 1 In the case of two spin- 2 particles, either with ‘up’ Ortho-H2 consists of the symmetric spin triplet, or ‘down’ spin: has S = 1, and needs to have an angular part • For the triplet spin state, S = 1, with “paral- with odd `. lel” spins, the spin wavefunction is exchange The latter population is three times more likely symmetric, and so the spatial wavefunction than the former, at high temperatures. must be antisymmetric. Therefore, 1 Parallel 2 -spins “repel” Deuterium • Similarly, for the singlet spin state, S = 0, with “anti-parallel” spins, the spatial wave- The nuclei now consist of a proton and a neutron. function must be symmetric and so The resulting deuteron is a spin-1-boson. This Anti-parallel 1 -spins “attract” means that 2 • There are now six symmetric spin wavefunctions (a sextet) and three anti-symmetric one 13.7 Interacting Particles (a triplet).

• The overall wavefunction now has to be sym- 13.7.1 Exchange energy metric. There are, once again, two distinct populations, If the two particles do not mutually interact, their as above. The ‘para’ population has an anti- correlated motion makes no difference to the ener- symmetric spin wavefunction, and the ‘ortho’ pop- gies of the 2-particle states – if we turn the inter- ulation has symmetric spin wavefunction. action on, things are different. For example, for electrons: • The further apart the electrons are, the lesser 13.6 Correlation and Ex- the Coulomb energy. change Forces • Therefore, the correlated states (anti-parallel spins) is higher in energy than the parallel- For distinguishable particles, the overall wavefunc- spins case. tion is the simple product of the single-particle • The difference in energy is called the exchange wavefunctions, and so the probability of finding the energy. particles in the same place is simply given by the product of the moduli of the two spatial wavefunc- • All other things being equal, the exchange in- tions squared. teractions produces an energetic preference for 13.8. PRACTICAL TIPS 51

the spins to line up – this lies at the root of – One way is to consider the possible spin ferromagnetism in atoms and solids. wavefunctions (eg: the triplet and singlet 1 states for a spin- 2 Fermion), and then separately consider the possible sym- 13.7.2 Energy Spectra metric and anti-symmetric wavefunctions that could go with them. However, the Coulomb Interaction sometimes sig- – Another (usually more sensible) way is nificantly alters the overall Hamiltonian, and the single-particle wavefunctions are no longer true ∗ Consider the spatial degeneracy (ie: eigenstates. the number of ways the spatial wavefunction can be constructed). A good example of this is the He atom. Without interaction, everything we discussed so far applies. For example, for two particles each of which can be in two states spatial, Switching on interactions, however, the spin-state the spatial degeneracy is 3 (see next affects the overall energy, so we have two distinct point). populations: ∗ Consider the spin degeneracy • Para-helium For example, for two spin- 1 particles, – The spatial wavefunction must be sym- 2 the spin degeneracy is 3 (see next metric. point). – The closeness of the electrons consider- and multiply them together. ably raises the energies of the eigenstates. If the actual wavefunctions need to be – The (1s, 2p) state is slightly higher than found, find Slater Determinants for every the (1s, 2s) state, because of “spin-orbit possible combination. interactions” (whatever they might be). • In the second method above, when considering • Ortho-helium spatial combinations, it is important to sepa- – The spatial wavefunction must be anti- rately consider cases in which the spatial states symmetric. are the same and those in which they are different, for two reasons: – (1s, 1s) is not possible. – In the case of Fermions, the spatial – However, (1s, 2s) and (1s, 2p) are slightly states might restrict the permitted spin lower than in Para-helium, because the states. For example, if the spatial states exchange-correlation keeps the electrons of the two particles are identical, then further apart. In the (1s, 2s) cases, the only those non-identical spin states are difference in energy is the exchange en- allowed. ergy. – In all cases, the similarity or otherwise of As a result, their atomic absorption and emission the spatial states will affect the number spectra (which depend on transitions between the of spin states allowed. Consider: levels) are entirely different. ∗ If the two spatial states are identical, then it doesn’t matter whether the spins are ↑↓ or ↓↑ – the overall wave- 13.8 Practical tips function will still end up the same, because the spatial parts are identi- • When finding degeneracies, we can proceed in cal. This therefore removes ones de- two ways: gree of degeneracy. 52 CHAPTER 13. IDENTICAL PARTICLES

∗ However, if the particles were in different spatial states, then this would definitely make a difference, because there would be a number of different products in the wavefunctions. • To determine whether a multiparticle system (eg: an atom) is Fermionic or Bosonic, imagine two identical such systems, and imagine swapping each of the constituent particles one-by- one. The change in the wavefunction at each ‘swap’ is obvious from whether these individual particles are Fermionic or Bosonic. By looking at the final overall change in wavefunction (ie: once all constituent particles – and therefore the whole system – has been swapped), one can easily say whether the system is Fermionic or Bosonic. Chapter 14

Matrix Methods

14.1 Introduction 14.3 Spin angular momentum

We deﬁne the matrix of an operator Ab as follows: When spin is represented as a column vector with respect to |χ+i and |χ−i, these vectors are called D E spinors. Amn = φm|Ab |φn The Pauli matrices are as follows: Where the φ are some complete set of eigenfunc- 1 0 S = ~ tions, but no necessarily those of Ab . z 2 0 −1

Any wavefunction can then be represented as a col- 0 1 S = ~ umn vector – an expansion in terms of the said x 2 1 0 eigenfunctions. 0 −i S = ~ The eigenvalues of the operator are then simply y 2 i 0 given by the eigenvalues of the matrix, and this can be used to ﬁnd the eigenvectors. We usually write this 1 Sb = σ 2~b 14.2 Orbital angular momen- Where the σ are the Pauli matrices. tum And

For example, with ` = 1 0 1 S = + ~ 0 0   1 0 0 L = 0 0 0 0 0 z ~   S− = ~ 0 0 −1 1 0

 √  0 2√ 0 L+ = ~  0 0 2  0 0 0

  1√ 0 0 L− = ~  0 2√ 0  0 0 2

53 54 CHAPTER 14. MATRIX METHODS Chapter 15

Heat capacities, etc. . .

One of the main ways Quantum eﬀects can be ob- • As T → ∞, C → NkB, the classical limit. served at a macroscopic level is through the heat • C → 0 as T → 0 – this is consistent with capacities of various things: vib the Third Law of Thermodynamics.

15.1 Vibrational Speciﬁc Heat 15.2 Rotational Speciﬁc Heat of Diatomic molecules of Diatomic molecules

A diatomic molecule is modelled as two masses, m 1 A diatomic molecule can be modelled as a rigid and m , connected by a spring. The system can be 2 rotor, for which we saw the energy levels were modelled as a single harmonic oscillator oscillating with the reduced mass. `(` + 1)~2 E` = Clasically, each degree of freedom (KE and PE) 2I 1 of the oscillator should have 2 kBT of energy, and each with degeneracy 2` + 1. so the vibrational heat capacity (dE/dT for N molecules should be Nk (a constant). The mean rotational kinetic energy per molecule at B T is therefore Quantum mechanically, we expect the average en- P∞ −βE` `=0 E`(2` + 1)e ergy of each molecule (we’re not working with de- hEi = ∞ P −βE` grees of freedom anymore) to be `=0(2` + 1)e Which we can re-write by expressing temperature 1 ~ω hEi = ~ω + 2 2 e~ω/kB T − 1 in units of θrot = ~ /2IkB: ∞ Which gives a speciﬁc heat capacity of P `(`+1)θrot/T `=0 `(` + 1)(2` + 1)e hEi = θrotkB ∞ 2 P `(`+1)θrot/T d hEi ω e~ω/kB T `=0(2` + 1)e C = N = Nk ~ vib B 2 dT kBT e~ω/kB T − 1 This, sadly, does not simplify as easily as the vi- A plot of C/NkB against kBT/~ω gives brational case. However, a graph looks like

Note a few interesting things: A few notes:

55 56 CHAPTER 15. HEAT CAPACITIES, ETC. . .

• As expected, Crot → 0 as T → 0.

• Crot → nR = NkB as T → ∞, as expected for two degrees of freedom.

• Crot turns on at about θrot/2. In general, rotational modes “turn on” at much lower temperatures than vibrational modes.

15.3 Identical Particles and Rotational Heat Capaci- ties

Whether molecules are in their ‘para’ or ‘ortho’ forms considerably restricts the energies that the molecule can take, and therefore considerably alters the variation of heat capacity with temperature. A few notes:

• At high temperature (T θrot), many rotational levels are populated and so the equilibrium fractions of ortho- and para- reﬂect their spin degeneracies. • As T falls, the population fractions change in true equilibrium, as the relative probabilities of each rotational state changes (the higher the energy, the lesser the probability). • At very low spins, the lowest levels will be occupied (para-H2 and ortho-D2). As T sub- sequently rises fast, the curve corresponds to these pure chemicals.