Quantum Mechanics Exercises Notes

Home , Delta potential, Finite potential well

Elijan Mastnak

Winter Semester 2020-2021

About These Notes

These are my notes from the Exercises portion of the class Kvanta Mehanika (Quantum Mechanics), a required course for third-year physics students at the Faculty of Math and Physics in Ljubljana, Slovenia. The exact problem sets herein are specific to the physics program at the University of Ljubljana, but the content is fairly standard for a late- undergraduate quantum mechanics course. I am making the notes publically available in the hope that they might help others learning similar material—the most recent version can be found on GitHub. Navigation: For easier document navigation, the table of contents is “clickable”, meaning you can jump directly to a section by clicking the colored section names in the table of contents. Unfortunately, the clickable links do not work in most online or mobile PDF viewers; you have to download the file first. On Authorship: The exercises are led by Asst. Prof. TomaˇzRejec, who has assembled the problem sets and guides us through the solutions. Accordingly, credit for the problems in these notes goes to Prof. Rejec. I have merely typeset the problems and provided an additional explanation or two where I saw fit. Disclaimer: Mistakes—both trivial typos and legitimate errors—are likely. Keep in mind that these are the notes of an undergraduate student in the process of learning the material himself—take what you read with a grain of salt. If you find mistakes and feel like telling me, by Github pull request, email or some other means, I’ll be happy to hear from you, even for the most trivial of errors.

1 Contents

1 First Exercise Set4 1.1 Theory: Basic Concepts in Quantum Mechanics...... 4 1.2 Bound States in a Finite Potential Well...... 5 1.3 Theory: Wavefunctions of a Particle in an Even Potential...... 7 1.4 Finite Potential Well, Second Approach...... 8

2 Second Exercise Set 10 2.1 Bound States in a Delta Potential Well...... 10 2.2 Bound States in a Delta Potential, Second Approach...... 11 2.3 Theory: Scattering...... 12 2.4 Scattering From a Delta Potential...... 15

3 Third Exercise Set 17 3.1 Theory: Uncertainty Product of Observable Quantities...... 17 3.2 Wave Function Minimizing the Uncertainty Product...... 19

4 Fourth Exercise Set 20 4.1 Wave Function Minimizing the Uncertainty Product (continued)...... 20 4.2 Theory: Time Evolution of a Wave Function...... 21 4.3 Theory: Time-Dependent Expectation Values and Operators...... 22 4.4 Time Evolution of the Gaussian Wave Packet...... 24

5 Fifth Exercise Set 28 5.1 Time Evolution of a Gaussian Wave Packet (continued)...... 28 5.2 Theory: Quantum Harmonic Oscillator...... 29 5.3 Time Evolution of a Particle in a Harmonic Oscillator...... 30

6 Sixth Exercise Set 33 6.1 Theory: Coherent States of the Quantum Harmonic Oscillator...... 33 6.2 Harmonic Oscillator in an Electric Field...... 39

7 Seventh Exercise Set 41 7.1 Theory: Two-Dimensional Harmonic Oscillator...... 41 7.2 Two-Dimensional Isotropic Harmonic Oscillator...... 43

8 Eight Exercise Set 45 8.1 Two-Dimensional Isotropic Harmonic Oscillator (continued)...... 45 8.2 Larmor Precession...... 49

9 Ninth Exercise Set 53 9.1 Larmor Precession (continued)...... 53 9.2 Rashba Coupling...... 55

10 Tenth Exercise Set 61 10.1 Rashba Coupling with Pauli Spin Matrices...... 61 10.2 Theory: Time Reversal Symmetry...... 62 10.3 Time-Reversal Symmetry and Rashba Coupling...... 63 10.4 Scattering on a Spin-Dependent Delta Function...... 65

2 11 Eleventh Exercise Set 66 11.1 Scattering on a Spin-Dependent Delta Function (continued)...... 66

12 Twelfth Exercise Set 74 12.1 Theory: Non-Degenerate Perturbation Theory...... 74 12.2 QHO with a Quartic Perturbation...... 75 12.3 QHO with a Cubic Perturbation...... 76

13 Thirteenth Exercise Set 79 13.1 Theoretical Background for the Coming Problem...... 79 13.1.1 Degenerate Perturbation Theory...... 79 13.1.2 Review of Hydrogen Atom Eigenvalues and Eigenfunctions..... 80 13.1.3 Parity Symmetry...... 81 13.2 Hydrogen Atom in a Homogeneous Electric Field...... 82

14 Fourteenth Exercise Set 87 14.1 Theory: First-Order Time-Dependent Perturbations...... 87 14.2 Hydrogen Atom Exposed to an Electric Field Pulse...... 88 14.3 Spin 1/2 Particle in a Time-Varying Magnetic Field...... 91

15 Fifteenth Exercise Set 95 15.1 Rabi Oscillations...... 95

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1 First Exercise Set

1.1 Theory: Basic Concepts in Quantum Mechanics • The fundamental quantity in quantum mechanics is the wave function, which in one dimension reads ψ(x, t). All measurable quantities are fundamentally derived from the wave function. As an example, we consider probability density dp ρ(x, t) = |ψ(x, t)|2 = , dx where dp represents the probability of ﬁnding a particle with the wave function ψ(x, t) in the x interval dx at time t.

• For the majority of this course, we will work with a one-dimensional, time-independent Hamiltonian operator of the form

pˆ2 ∂ Hˆ = + V (x) wherep ˆ = −i . 2m ~∂x Notation: By convention, we usually write operators without the hat symbol and distinguish between operators and scalar quantities based on context.

• We will often solve the stationary Schr¨odingerequation, which reads

Hψn(x) = Enψn(x).

Note that ψ is time-independent, and that the stationary Schr¨odingerequation is an eigenvalue equation, where En are the eigenvalues of the Hamiltonian operator and ψn are the corresponding eigenfunctions.

We then use the solutions ψn(x) of the stationary Schr¨odingerequation to ﬁnd the time-dependent wave function ψ(x, t). This process is called time evolution.

• The time evolution procedure goes as follows: Start with an initial wavefunction ψ(x, 0) at time t = 0. Expand ψ(x, 0) over a basis of the eigenfunctions ψn(x) to get X ψ(x, 0) = cnψn(x). n The time-dependent wave function is then

X − iEn t ψ(x, t) = cne ~ ψn(x). n

We have essentially replaced the constant coefficients cn in the stationary expansion − iEn t with the time-dependent coefficientsc ñ = cne ~ . • Next, we consider a particle’s energy eigenvalues. A particle or quantum system typically has as many eigenvalues as the dimension of the Hilbert space containing the system’s wave functions. Often, such as for a free particle, the Hilbert space is infinite. The set of all energy eigenvalues is {En} and is often called the spectrum. The number of linearly independent eigenfunctions a system has at a given energy eigenvalue En is called the eigenvalue’s degeneracy.

4 1.2 Bound States in a Finite Potential Well github.com/ejmastnak/fmf

• Next, we consider a quantum particle’s eigenfunctions. We start with a non-degenerate energy level with eigenfunction ψn with the eigenvalue relation

Hψn(x) = Enψn(x).

If ψn solves the eigenvalue problem, then mathematically λψn also solves the eigenvalue problem for all λ ∈ C. But physically, there is a restriction: the probability density must be normalized to unity, i.e. Z ∞ 2 |ψn(x)| dx = 1. −∞ The normalization condition requires |λ| = 1, which is often written λ = eiα where α ∈ R. This result is important: essentially, any eigenfunction ψn is undetermined up to a constant phase factor λ = eiα of magnitude |λ| = 1. In other words, the two wavefunctions

iα ψn(x) and e ψn(x) are physically identical. We can’t distinguish between the two in experiments, because the phase factor eiα is lost during the process of ﬁnding the probability density when squaring the absolute value. All wave functions are determined only up to a constant phase factor of magnitude one. As an example, consider an inﬁnite potential well with x ∈ [0, a]. The energy eigenvalues and eigenfunctions are

2π2n2 r2 nπx E = ~ and ψ (x) = sin , n ∈ +. n 2ma2 n a a N q ˜ 2 nπx + However, e.g. ψn(x) = i a sin a , n ∈ N represents the same physical information as ψn.

1.2 Bound States in a Finite Potential Well Find the energy eigenvalues of the bound states of a particle with energy E in a ﬁnite potential well of depth V0 > 0 and width a. Assume |E| < V0; note that bound states have negative energies. • Split the x axis into three regions: to the left of the well, inside the well, and to the right of the well, e.g. regions 1, 2 and 3, respectively. The potential energy is  0, x ∈ region 1  V (x) = −V0, x ∈ region 2 . 0, x ∈ region 3

• The Schr¨odinger equation for region 1 is 2 ∂2ψ − ~ + 0 · ψ = Eψ, 2m ∂x2 where we have sbustituted in V = 0 in region 1. The general solution is a linear combination of exponential functions of the form

κx −κx ψ1(x) = Ae + Be .

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To find κ, insert ψ1 into the Schrödingerequation for region 1. Differentiating, canceling the wave function terms from both sides, and rearranging gives

2 2mE − ~ κ2 Aeiκx + Be−iκx = E Aeiκx + Be−iκx =⇒ κ2 = 0 , 2m ~2

where E0 = −E is a positive quantity. • The Schr¨odinger equation for region 2 is

2 ∂2ψ − ~ − V · ψ = Eψ. 2m ∂x2 0 We use a plane wave ansatz of the form

ikx −ikx ψ2(x) = Ce + De .

To find k, insert ψ2 into the Schrödingerequation for region 2. The result after differentiation is 2 + ~ k2 Ceikx + De−ikx − V Ceikx + De−ikx = E Ceikx + De−ikx . 2m 0

The wave function terms in parentheses cancel, leading to k2 = 2m(V0−E0) where ~2 E0 = −E is a positive quantity. • The Schr¨odingerequation for region 3 is analogous to the equation for region 1. Following a similar procedure, we would get

κx −κx 2 2mE0 ψ3 = F e + Ge where κ = . ~2 • Next, to find energy, we apply boundary conditions, namely that the wave function is continuous and continuously differentiable at the boundaries between regions and that the functions vanishes at ±∞, meaning there is no probability of finding the particle infinitely far away from the well. The conditions are:

Region Condition 1 Condition 2 ∂ψ ∂ψ 1/2 ψ (0) = ψ (0) 1 (0) = 2 (0) 1 2 ∂x ∂x ∂ψ ∂ψ 23 ψ (a) = ψ (a) 2 (a) = 3 (a) 2 3 ∂x ∂x ±∞ ψ1(−∞) = 0 ψ3(∞) = 0

The last two conditions at ±∞ require that B = 0 and F = 0, respectively; otherwise ψ1 and ψ3 would diverge. The continuity conditions (in the Condition 1 column) require A = C + D and Ceika + De−ika = Ge−κa. The continuous diﬀerentiability conditions require

Aκ = ik(C − D) and ik(Ceika − De−ika) = −κGe−κa.

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• We could then write the four equations in a 4 × 4 coeﬃcient matrix

··· A 0 ··· B 0     =   . ··· C 0 ··· D 0

To proceed, we would require the matrix’s determinant be zero for a unique solution. The determinant of the coefficient matrix is a function of κ and k, which are in turn functions of energy. We then view the matrix’s determinant as a function of energy and try to find its zeros. Each zero is an energy eigenvalue of the particle in a finite potential well. However, this involves solving the above system of four equations for the four unknowns A, B, C and D. That’s tedious! We will stop here and solve the same problem more elegantly in the next exercise. But first, we’ll need some theoretical machinery.

1.3 Theory: Wavefunctions of a Particle in an Even Potential Theorem: The eigenfunctions of a particle in an even potential can always be chosen such that the eigenfunctions are either even or odd.

• First, recall the superposition principle: if two eigenfunctions ψ1 and ψ2 have the same eigenvalue E, than any linear combination of ψ1 and ψ2 is also an eigenfunction with eigenvalue E. In equation form, if Hψ1(x) = Eψ1(x) and Hψ2(x) = Eψ2(x) then H (αψ1(x) + βψ2(x)) = E (αψ1(x) + βψ2(x)) . We’ll use the superposition principle later in the proof when considering degenerate states.

• To prove the even/odd theorem for particles in an even potential V (x) = V (−x), we start with the stationary Schr¨odingerequation

Hψ(x) = Eψ(x).

We make the parity transformation x → −x and note the Hamiltonian is invariant ∂ ∂ ∂2 ∂2 under parity transformation if V is even, since ∂(−x) = − ∂x , and ∂(−x)2 = ∂x2 . Under the transformation x → −x, the Schr¨odingerequation reads

Hψ(−x) = Eψ(−x).

This means that if ψ(x) is an eigenfunction of H for the energy eigenvalue E, then ψ(−x) is also an eigenfunction for the same energy eigenvalue.

• Next, we have two options: either the eigenvalue E is degenerate or it is not. If E is not degenerate, then ψ(−x) and ψ(x) are equal up to a constant phase factor. In other words, if E is not degenerate and V is even, then ψ(x) and ψ(−x) are linearly dependent. In equation form,

ψ(x) = eiαψ(−x) = e2iαψ(x) =⇒ e2iα = 1 =⇒ eiα = ±1.

If eiα = 1, then ψ is even, and if eiα = −1, then ψ is odd. There are no other options. In other words, the eigenfunctions corresponding to the non-degenerate states of a particle in an even potential are either even or odd, as we wanted to show.

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• Next, if E is degenerate, then ψ(x) and ψ(−x) are by deﬁnition linearly independent. By the superposition principle, we can create two more linear combinations of ψ(x) and ψ(−x): a sum and a diﬀerence, e.g.

ψ+(x) = ψ(x) + ψ(−x) and ψ−(x) = ψ(x) − ψ(−x).

Note that, by construction, ψ+ is even and ψ− is odd. In other words, even though ψ1 and ψ2 may not be even or odd themselves, we can always combine the two into a valid eigenfunction that is either even or odd. • The general takeaway is: if our problem’s potential has reﬂection symmetry about the origin, we can a priori search for only even or odd functions, because we know there exists a basis of the surrounding Hilbert space formed of only odd and even eigenfunctions.

1.4 Finite Potential Well, Second Approach Continued from the previous problem: find the energy eigenvalues of the bound states of a particle with energy E in a finite potential well of depth V0 > 0 and width a. • Here’s how to proceed: First, note the problem has reflection symmetry if we place a a the center of the well at the origin. The well then spans from x = − 2 to x = 2 . In this case V (x) = V (−x), i.e. the potential is an even function of x. Why is this useful? It reduces the number of wave functions we have to find by one. Namely, if the eigenfunctions are even, then ψ3 is a copy of ψ1. If the eigenfunctions are odd, then ψ3 = −ψ1. In both cases, we only need to find ψ1. We then only need to solve the problem on half of the real axis. • First, we assume the solution is an even function. Reusing results from the earlier treatment of the finite potential well, the wavefunctions in region 3 read

−κx 2 2mE0 ψ3 = Ge where κ = , ~2 and for region 2, inside the well, the even wavefunctions read

2 2m(V0 − E0) ψ2 = A cos(kx) where k = . ~2 A few notes: we switched from plane waves to sinusoidal functions, where are better suited to odd/even symmetry. And we dropped the B sin(kx) term and kept only the cosine because we’re solving for an even function a priori. For region 1, because we’re searching for even functions, the result is the same as for region 3, i.e. κx ψ1 = Ge . • If we assume an odd solution, the wavefunctions are

−κx κx ψ3(x) = Ge ψ2(x) = B sin(kx) ψ1(x) = −Ge .

• On to boundary conditions. Because an even or odd function will have the same behavior (up to a minus sign for odd functions) at both region boundaries, we only a need to consider one region; we’ll consider x = 2 . The boundary conditions are a a ∂ψ a ∂ψ a ψ = ψ and 2 = 3 . 2 2 3 2 ∂x 2 ∂x 2

8 1.4 Finite Potential Well, Second Approach github.com/ejmastnak/fmf

• For an even solution, the boundary conditions read:

a − κa a − κa A cos k = Ge 2 and − kA sin k = −κGe 2 . 2 2 Next, a trick: divide the equations and cancel like terms to get a k tan k = κ. 2

• For an odd solution, the boundary conditions read:

a − κa a − κa B sin k = Ge 2 and kB cos k = −κGe 2 . 2 2 Dividing the equations and cancel like terms gives a k cot k = −κ. 2

• The solutions of the two equations

ka κ ka κ tan = and cot = − 2 k 2 k

will give the energy eigenvalues of the even and odd states, respectively. These are transcendental equations. They don’t have analytic solutions, and we’ll have to solve them graphically. First, we introduce the dimensionless variable u = ka. We then express κ in terms of u using

u2 2mV κ2 + k2 = κ2 + = 0 . a2 ~2 2 Alternatively, in dimensionless form, we deﬁne u2 = 2mV0a , which leads to 0 ~2 u2 − u2 κ2 = 0 . a2

• In terms of u and u0, the two transcendental equations become r r u pu2 − u2 u2 u u2 tan = 0 = 0 − 1 and cot = − 0 − 1. 2 u u2 2 u2 We then plot both sides of the equations and look for values of u where the left side equals the right side. These values of u give the energy eigenvalues, which are the solution to our problem. Finally, some notes:

– The right sides of the equations are deﬁned only for u ≤ u0 and diverge as u → 0. – As u increases, k increases, E becomes more positive, and states become less bound. The ground state occurs at the smallest value of u.

– The number of bound states increases as u0 increases. This is achieved by in- creasing V0 (well depth), a (well width), or m (particle mass). A ﬁnite potential well always has at least one bound state, near u = 0.

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2 Second Exercise Set

2.1 Bound States in a Delta Potential Well Find the energies and wave functions of the bound states of a quantum particle of mass m in a delta function potential well by reusing the results from a ﬁnite potential well.

• We will reuse our results from the previous problem set by interpreting a delta potential as a ﬁnite potential well in the limit

a → 0,V0 → ∞ and aV0 = constant ≡ λ.

We then write the potential as V (x) = −λδ(x).

• Recall from the previous exercise set that even bound states in a ﬁnite potential well of width a and depth V0 obey

r 2 2 u u0 2 2mV0a tan = − 1 where u0 = and u = ak. 2 u ~2 The wave numbers k and κ inside and outside the well are r r 2m(V − E ) 2mE k = 0 0 and κ = 0 . ~2 ~2

• The limit a → 0 forces u0 → 0. In the limit u0 → 0, we expect at most one, even bound state. For small u and u0, we expect u and u0 to be very close, so we write

u = u0 − where 1.

We continue with a Taylor expansion approximation of the even bound state equation. To ﬁrst order, the tangent function is u u + u + tan = tan 0 ≈ 0 + ..., 2 2 2 while the square root is

r s 2 s −2 u 2 u r 0 − 1 = 0 − 1 = 1 − − 1 ≈ 2 . u u0 + u0 u0

• With the approximations, the original even bound state equation becomes

u + r 0 = 2 . 2 u0

2 2 We square both sides and take only the leading u0 term from (u0 + ) to get

(u + )2 u2 u 3 0 ≈ 0 = 2 =⇒ = 0 . 4 4 u0 2

Next, we solve for u in terms of u0 to get u 3 u = u − = u − 0 . 0 0 2

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• We then use the equation for k to solve for E in terms of u0:

r 2 2 2 2 u 2m(V0 − E0) ~ u ~ u0 3 k = = =⇒ E0 = V0 − = V0 − u0 − . a ~2 2ma2 2ma2 2

6 Multiplying out and dropping the small terms of order u0 gives 2 u4 E = V − ~ u2 − 0 . 0 0 2ma2 0 4

2 We then substitute in the expression u2 = 2mV0a , which simpliﬁes things to 0 ~2 2 2 2 ma V0 mλ E0 = = , 2~2 2~2

where the last equality uses λ = aV0. This is the result for the bound state’s energy. • We ﬁnd the corresponding wave function with the plane-wave ansatz ( r Aeκx x < 0 2mE mλ ψ(x) = where κ = 0 = . Be−κx x > 0 ~2 ~2

Because the wave function is even, A = B, allowing us to write ψ = Ae−κ|x|. we ﬁnd the coeﬃcient A from the normalization condition Z Z ∞ 1 ≡ |ψ(x)|2 dx = A2 e−2κ|x| dx. −∞ Because ψ is even, we can integrate over only half the real line and double the result:

Z ∞ 2 ∞ 2 2 −2κx A −2κx A 2 1 = 2A e dx = − e = =⇒ A = κ. 0 κ 0 κ The wave function is thus √ mλ ψ(x) = κe−κ|x| where κ = . ~2 2.2 Bound States in a Delta Potential, Second Approach Find the energies and wave functions of the bound states of a quantum particle of mass m in a delta function potential well using a plane wave ansatz and appropriate boundary conditions.

• Write the potential well in the form V (x) = −λδ(x). The stationary Schr¨odinger equation for the potential reads

2 ∂2ψ − ~ − λδ(x)ψ = Eψ. 2m ∂x2 We ﬁnd the boundary condition on the derivative ψ0 by integrating the Schr¨odinger equation over a small region [−, ].

2 Z Z ~ ∂ψ − − λ δ(x)ψ(x) dx = E ψ(x) dx. 2m ∂x − − −

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In the limit → 0, this becomes

2 − ~ ψ0(0 ) − ψ0(0 ) − λψ(0) = 0, 2m + − 0 0 where ψ (0+) and ψ (0−) denote ψ’s derivatives from the right and left, respectively. The appropriate boundary condition ψ0(x) are thus

0 0 2mλψ(0) ψ (0+) − ψ (0−) = − . ~2 • We construct the wavefunction with a plane-wave ansatz of the form ( Aeκx x < 0 ψ(x) = . Be−κx x > 0

As boundary conditions, we require that ψ is continuous and that ψ0 obey the earlier 0 0 2mλψ(0) condition ψ (0+) − ψ (0−) = 2 . Applying the continuity condition at x = 0 ~ gives Aeκ·0 = Be−κ·0 =⇒ A = B. Applying the condition on the derivatives (and using A = B) gives 2mλ 2mλ mλ −κA−κ·0 − κAκ·0 = − ψ(0) = − Aeκ·0 =⇒ κ = . ~2 ~2 ~2 The wave function is thus mλ ψ(x) = Ae−κ|x| where κ = . ~2 • We can ﬁnd the bound state’s energy from

r 2 2mE0 mλ κ = =⇒ E0 = . ~2 2~2 which matches the result from the previous problem. To ﬁnd the wave function, we just need the value of A, which we ﬁnd with the usual normalization condition Z Z ∞ Z ∞ 1 ≡ |ψ(x)|2 dx = A2 e−2κ|x| dx = 2A2 e−2κx dx =⇒ A2 = κ. −∞ 0 The wave function is thus √ mλ ψ(x) = κe−κ|x| where κ = , ~2 in agreement with the result from the previous problem.

2.3 Theory: Scattering • The basic scattering problem involves a particle with energy E and two regions of the x axis with potentials V1 and V3 respectively, where E > V1,V3. Both energies and potentials are positive quantities. Because E > V1,V3, the particle is free in the regions of V1 and V2. Near the origin, between regions 1 and 3, the particle encounters a small potential barrier where V (x) > E; the exact form of V (x) is not important for our purposes. We call this region 3.

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• The general ansatz solutions in regions 1 and 3 are

ik1x −ik1x −ik3x ik3x ψ1 = A1e + B1e and ψ3 = A3e + B3e . The terms with A coeﬃcients represent movement toward the potential barrier

• In region 2, we don’t know V (x), so we have to be more general. Because ψ2 comes from a second-order linear diﬀerential equation (i.e. the Schr¨odingerequation), its general form is a linear combination of two linearly independent functions, e.g.

ψ2 = Cf(x) + Dg(x). where f and g are linearly independent solutions of the Schr¨odingerequation in region 2.

• Because E > V1,V3, the states in regions 1 and 3 are free, meaning the particle’s Hamiltonian has a continuous spectrum of energy eigenvalues. As a result, energy eigenvalues aren’t particularly useful for characterizing the problem. Instead, for unbound situations with a potential barrier, we describe the problem in terms of transmissivity T and reﬂectivity R, which describe the probabilities of the scattered particle passing through and reﬂecting from the barrier, respectively. Probability Current • For scattering problems with T and R, we work in terms of probability current

h 0∗ i j(x) = ~ ψ∗(x)ψ0(x) − ψ (x)ψ(x) = ~ Im ψ∗ψ0(x) , 2mi m where ψ∗ and ψ0 denote the complex conjugate and derivative of ψ, respectively.

ik x −ik x • Substituting in the region 1 ansatz ψ1 = A1e 1 + B1e 1 gives the probability current in region 1: n o j = ~ Im A A∗ik − B B∗ik + B∗A ik e2ik1x − B A∗ik e−2ik1x 1 m 1 1 1 1 1 1 1 1 1 1 1 1 k = ~ 1 (A A∗ − B B∗) ≡ v (A A∗ − B B∗) = v (|A |2 − |B |2). m 1 1 1 1 1 1 1 1 1 1 1 1

~k1 ~k1 Noting that m has units of velocity, we’ve deﬁned the constant v1 ≡ m . The procedure for j3 in region 3 is analogous. The result is k j = v (B B∗ − A A∗) = v (|B |3 − |A |3) where v ≡ ~ 3 . 3 3 3 3 3 3 3 3 3 3 m • Next, we run into a slight problem: the plane waves in regions 1 and 3 can’t be normalized. They oscillate! Instead, we normalize ψ1 and ψ3 per unit “probability velocity”. We redeﬁne A B A B 1 ik1x 1 −ik1x 3 −ik3x 3 ik3x ψ˜1 = √ e + √ e and ψ˜3 = √ e + √ e . v1 v1 v3 v3

The corresponding probability currents, using the normalized ψ˜1 and ψ˜3 are ˜ ∗ ∗ 2 2 j1(x) = A1A1 − B1 B1 = |A1| − |B1| ˜ ∗ ∗ 2 2 j3(x) = B3 B3 − A3A3 = |B3| − |A3| .

√1 As a general rule, we always normalize plane waves with the probability velocity v hk where v = m .

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• The next step in the scattering problem is formulating four boundary conditions on ψ requiring continuity and continuous differentiability at the two boundary regions between regions 1 and 2 and between regions 2 and 3. We end up with 4 linearly independent equations for six unknown coefficients. Our plan is to eliminate C and D and express B1 and B3 in terms of A1 and A3. The amplitudes A1 and A3 of the waves incident on the potential barrier thus parameterize our problem. Informally, all this means is that if we know the wavefunction of the incident particle (by specifying the incident amplitudes A1,A3), we can solve the scattering problem by finding the reflected amplitudes B1,B3. In general, it is best practice to parameterize a scattering problem with the incident amplitudes A1 and A3. In any case, we end up with a matrix equation

B A 1 = S 1 or more concisely B = SA, B3 A3 where S is called the scattering matrix, in our case a 2 × 2 matrix. By convention, we parameterize a 2 × 2 scattering matrix in terms of the parameters r,r ˜, t and t˜:

r t˜ S = . t r˜

Probability Conservation • Some identities:

2 2 2 2 T = |t| R = |r| and T˜ = t˜ R˜ = |r˜| .

The identities T + R = 1 and T˜ + R˜ = 1 imply probability conservation, i.e. the total incident and reﬂected probabilities sum to one.

• In general, the incident current is the sum of the two squares of the incidence amplitudes A1 and A3, while the reﬂected current is the sum of the reﬂected amplitudes B1 and B3, i.e.

2 2 2 2 jin = |A1| + |A3| and jout = |B1| + |B3| .

In terms of the coeﬃcient vector A, the incident probability current is ∗ ∗ ∗ ∗ A1 † jin = A1A1 + A3A3 = A1,A3 = A A. A3 An analogous procedure with the reﬂected amplitudes gives ∗ ∗ ∗ ∗ B1 † jout = B1B1 + B3B3 = B1 ,B3 = B B. B3

† † • Conservation of probability requires jin = jout or A A = B B. Combining the scattering matrix equation B = SA with the requirement A†A = B†B implies

A†A = B†B = (A†S†)(SA) =⇒ S†S = I.

In other words, the scattering matrix S must be unitary to conserve probability in scattering problems.

14 2.4 Scattering From a Delta Potential github.com/ejmastnak/fmf

2.4 Scattering From a Delta Potential Analyze a quantum particle scattering oﬀ of a delta function potential. • We consider a particle of energy E scattering oﬀ the potential V (x) = λδ(x). We’ll divide the problem into two separate cases. First, we’ll assume the particle is incident on the potential barrier only from the left, so the incident amplitude vector reads A = (1, 0). In this case, the scattering matrix equation reads r t˜ 1 r B = SA = = . t r˜ 0 t

• Similarly, if we assume the particle is incident only from the right, i.e. A = (0, 1), the scattering equation reads r t˜ 0 t˜ B = SA = = . t r˜ 1 r˜

• First, for A = (1, 0) and B = (r, t), the wave functions in region 1 and 3 are A B eikx r ψ (x) = √1 eikx + √1 e−ikx = √ + √ e−ikx, 1 v v v v A B t ψ = √3 e−ikx + √3 eikx = 0 + √ eikx, 3 v v v where k2 = 2mE . Note that because the potential is the same on both sides of the ~2 delta function, k1 = k3 ≡ k and v1 = v3 ≡ v. • On to boundary conditions. Continuity between regions 1 and 3 requires

ψ1(0) = ψ3(0) =⇒ 1 + r = t. For the the boundary conditions on the derivative ψ0, we reuse the earlier diﬀeren- 0 0 2mλψ(0) tiability condition ψ (0+) − ψ (0−) = 2 from the previous problem involving ~ bound states in a delta potential. Changing the sign of of λ because the delta 0 function in a scattering problem points upward, and changing notation from ψ± to 0 ψ1,3the diﬀerentiability condition becomes

0 0 2mλ 2mλ ψ3(0) − ψ1(0) = ψ(0) =⇒ ik(t + r − 1) = t. ~2 ~2 Substituting in the earlier continuity requirement t = 1 + r gives 2mλ iα mλ 2ikr = (1 + r) =⇒ r = − where α = . ~2 k + iα ~2 The expression for the parameter t is then iα k t = 1 + r = 1 − = . k + iα k + iα • We would then ﬁndr ˜ and t˜ using an analogous procedure with the incidence vector A = (0, 1). It turns out (exercise left to the reader) that the results are the same: iα k r˜ = − and t˜= . k + iα k + iα The probability matrix can then be written 1 −iα k mλ S = where α = . k + iα k −iα ~2

15 2.4 Scattering From a Delta Potential github.com/ejmastnak/fmf

On the Problem’s Symmetry: A few notes on why the scattering matrix for the delta potential has only two independent parameters, i.e. why t˜= t andr ˜ = r.

• We begin by noting that the wave functions ψ1,3 in regions 1 and 3 solve the stationary Schr¨odingerequation

2 ∂2 Hψ = Eψ where H = −~ + V (x). m ∂x2 We then take the Schr¨odingerequation’s complex conjugate to get

Hψ∗ = Eψ∗.

In other words, if ψ is a wave function for the energy eigenvalue E, then so is ψ∗. This holds in general for any real Hamiltonian.

• In our speciﬁc problem in regions 1 and 3, the conjugate wave function reads A∗ B∗ A∗ B∗ ψ∗ = √1 e−ikxe + √1 eikx and ψ∗ = √3 eikxe + √3 eikx. 1 v v 3 v v

Recall k1 = k3 = k and v1 = v3 = v. • Conjugation effectively switches the role of the incident and reflected waves: the waves with B coefficients are now incident and waves with A coefficients reflected. The equation linking the incident and reflected waves becomes

∗ ∗ A1 B1 ∗ ∗ ∗ = S ∗ or more concisely A = SB . A3 B3 Recall the original equation, before conjugation, was B = SA and note that the scattering matrix S is the same, since the potential is invariant under conjugation.

• Remember probability conservation requires that S be unitary, i.e. S†S = I. With this in mind, multiply the original equation B = SA from the left by S†, and then take the complex conjugate of the equation, which gives

S†B = A =⇒ A∗ = ST B∗.

Substituting the result A∗ = ST B∗ into the conjugate equation A∗ = SB∗ gives

ST B∗ = SB∗,

meaning that S is symmetric for scattering oﬀ a delta potential. This symmetry is the reason why t = t˜ in the delta function scattering matrix. Formally, this is a consequence of the problem’s Hamiltonian being invariant under time reversal.

• Next, why r =r ˜. This is a consequence of the problem’s reﬂection symmetry. Start again with the Schr¨odingerequation:

Hψ(x) = Eψ(x).

The parity transformation x → −x (recall H is invariant under parity) gives:

Hψ(−x) = Eψ(−x).

16 github.com/ejmastnak/fmf

The parity-transformed wave functions for regions 1 and 3 are A B A B ψ (−x) = √1 e−ikx + √1 eikx and ψ (−x) = √3 eikx + √3 e−ikx. 1 v v 3 v v

The roles of the incident and reﬂected waves are mixed up under parity: ψ1(−x) now refers to the region of x > 0 (right of the delta function), and ψ3(−x) to the region of x < 0. The equation linking the incident and reﬂected waves becomes

B A 3 = S 3 (after reﬂection). B1 A1 Recall the original equation, before parity transformation, was

B A 1 = S 1 (before reﬂection). B3 A3

• We can get the reflected equation from the original one by multiplying the original 0 1 equation by the first Pauli spin matrix σ = . The reflected equation is x 1 0

σxB = SσxA =⇒ B = σxSσxA.

Substituting the expression for B into the original equation B = SA gives

σxSσxA = SA =⇒ S = σxSσx.

The relationship S = σxSσx holds in general for any even Hamiltonian. Applied to our parameterization of S, we have

r t˜ r t˜ r˜ t S = σ Sσ ⇐⇒ = σ σ = . x x t r˜ x t r˜ x t˜ r

The result is r =r ˜ and t = t˜. This accounts for both of the symmetries in our delta function scattering matrix! So invariance under parity transformation is a powerful symmetry: it reduces the number of independent elements in the scattering matrix by two, even under in the absence of time reversal symmetry.

3 Third Exercise Set

3.1 Theory: Uncertainty Product of Observable Quantities • We are interested in ﬁnding an expression for the uncertainty in the product ∆A∆B of two arbitrary observable quantities A and B. We start by assigning both A and B a corresponding Hermitian operator (Hermitian because A and B are observable). The operators are A = A† and B = B†.

• By deﬁnition, the uncertainty in A is

(∆A)2 = A2 − hAi2 = h(A − hAi)i ≡ A˜2 ,

where we’ve deﬁned A˜ = A − hAi I. Here hAi is a scalar and I is the identity operator.

17 3.1 Theory: Uncertainty Product of Observable . . . github.com/ejmastnak/fmf

• First, the deﬁnition of an operator’s expectation value, in both coordinate and braket notation, is Z hAi = ψ∗Aψ dx ≡ hψ| A |ψi = hψ|Aψi .

Because A is Hermitian, hψ|Aψi = hAψ|ψi. This is a special case of the more general adjoint relationship hAψ|ψi = hψ|A∗ψi with A = A∗ because A is Hermitian.

• The expectation value of a Hermitian operator is real. The classic proof reads

∗ ∗ ∗ hAi ≡ hψ|Aψi = hA ψ|ψi = hAψ|ψi = hψ|Aψi ≡ hAi =⇒ hAi ∈ R.

• In general, the sum of two Hermitian operators A = A† and B = B† is

(A + B)† = A† + B† = A + B.

So the sum of two Hermitian operators is a Hermitian operator. If we return to A˜ = A − hAi I it follows that A˜† = A˜ since both A and I are Hermitian.

• We return to the product ∆A∆B. First, we square everything, since uncertainty is expressed as (∆A)2.

2 2 2 2 2 (∆A∆B) = A˜ B˜ = hψ| A˜ |ψi hψ| B˜ |ψi = Aψ˜ Aψ˜ Bψ˜ Bψ˜ 2 2 ˜ ˜ = Aψ Bψ . The last equality occurs in the Cauchy-Schwartz inequality: 2 2 2 2 ˜ ˜ ˜ ˜ ˜ ˜ Aψ Bψ ≥ Aψ Bψ = ψ ABψ .

• Next, some more manipulations of the last result: 2 2 hψ| A˜B˜ |ψi = hψ| 1 A˜B˜ − B˜A˜ + A˜B˜ + B˜A˜ |ψi . 2 We introduce two quantities: the commutator and anti-commutator. They’re written

A,˜ B˜ = A˜B˜ − B˜A˜ and A,˜ B˜ = A˜B˜ + B˜A.˜

Intermezzo: Some Properties of Commutators • The anti-commutator of two Hermitian operators is a Hermitian operator

A, B † = (AB + BA)† = (AB)† + (BA)† = B†A† + A†B† = AB + BA = A, B .

• The commutator of two Hermitian operators is anti-Hermitian

A, B† = (AB − BA)† = (AB)† − (BA)† = B†A† − A†B† = −(AB − BA) = −A, B.

• Finally, the expectation value of an anti-Hermitian operator A† = −A is

The result implies hAi is a purely imaginary number if A is anti-Hermitian.

18 3.2 Wave Function Minimizing the Uncertainty Product github.com/ejmastnak/fmf

Back to the Uncertainty Principle • Speciﬁcally, back to where we left oﬀ with 2 2 2 2 ˜ ˜ ˜ ˜ (∆A∆B) = ··· = Aψ Bψ ≥ Aψ Bψ = ··· 2 D E 2 = hψ| 1 A,˜ B˜ + 1 A,˜ B˜ |ψi = 1 A,˜ B˜ + 1 A,˜ B˜ 2 2 2 2 1 2 2 = A,˜ B˜ + A,˜ B˜ . 4 In last parentheses we have a sum of two positive quantities, so we can drop the anti-commutator if we write 2 1 2 2 1 2 1 A,˜ B˜ + A,˜ B˜ ≥ A,˜ B˜ = [A, B] . 4 4 2

The last equality uses the identity A,˜ B˜ = [A, B], which follows from some basic algebra and the deﬁnitions A˜ = A − hAiI and B˜ = B − hBiI:

A,˜ B˜ = A − hAiI,B − hBiI = ... = AB − BA = [A, B].

• Here’s the entire derivation so far in one place. Note the two inequalities.

∂ • Now we apply the result speciﬁcally to x and p, where x = x, p = −i~ ∂x and [x, p] = i~. The result is

2 (∆x∆p)2 ≥ 1 |h[x, p]i|2 = 1 |hi i|2 = 1 |i |2 = ~ =⇒ ∆x∆p ≥ ~. 2 2 ~ 2 ~ 2 2 3.2 Wave Function Minimizing the Uncertainty Product Find all wave functions with the minimum position-momentum uncertainty product allowed ~ by the Heisenberg uncertainty principle, i.e. ﬁnd all wave functions for which ∆x∆p = 2 . • To solve this problem, we need a wavefunction for which both inequalities in the above theory section are strict equalities. First, the Cauchy-Schwartz inequality is an equality for two linearly dependent vectors. 2 2 2 ˜ ˜ ˜ ˜ ˜ ˜ Aψ Bψ = Aψ Bψ if Aψ = αBψ, α ∈ C. The second inequality is an equality if

A,˜ B˜ = 0.

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• We have two conditions: Aψ˜ = α Bψ˜ and A,˜ B˜ = 0. We start with the second condition and perform some manipulations... 0 ≡ A,˜ B˜ = hψ| A˜B˜ + B˜A˜ |ψi = hψ| A˜B˜ |ψi + hψ| B˜A˜ |ψi

= Aψ˜ Bψ˜ + Bψ˜ Aψ˜ = 0. We then substitute in the Cauchy-Schwartz condition to get

0 = Aψ˜ Bψ˜ + Bψ˜ Aψ˜ = αBψ˜ Bψ˜ + Bψ˜ αBψ˜ ∗ ∗ = α Bψ˜ Bψ˜ + α Bψ˜ Bψ˜ = (α + α) Bψ˜ Bψ˜ = 0. Mathematically, the equality holds if either of the two product terms equal zero. 2 2 But the bra-ket term is Bψ˜ Bψ˜ = B˜ = (∆B) , and this term being zero would imply zero uncertainty in B, which implies inﬁnite uncertainty in A. We reject this option as non-physical, so we’re left with (α∗ + α) = 0, meaning α is a purely imaginary number. To make this requirement explicit, we write α = ic where c ∈ R. The Cauchy-Schwartz condition then reads

Aψ˜ = ic Bψ˜ , c ∈ R. Substituting in the deﬁnition A˜ = A − hAi I gives |Aψi − hAi |ψi = ic |Bψi − ichBi |ψi .

• Any solutions ψ to this equation will satisfy the equality 1 ∆A∆B = [A, B] . 2 ∂ For the speciﬁc case A → x and B → p where p → −i~ ∂x and |ψi = ψ(x), the equation reads ∂ψ xψ − hxi ψ = c − ic hpi ψ. ~ ∂x 4 Fourth Exercise Set

4.1 Wave Function Minimizing the Uncertainty Product (continued) Continued from the previous exercise set... • Review of last exercise set: We started with two Hermitian operators A and B and showed 1 ∆A∆B ≥ |h[A, B]i|. 2 An equality = replaces the inequality ≥ if {A,˜ B˜} = 0 and A˜ |ψi = αB˜ |ψi ,

where A˜ = A − hAi, B˜ = B − hBi and α = ic where c ∈ R is real constant. In the concrete momentum-position case, the minimum uncertainty condition reads

0 xψ(x) − hxi ψ = ~cψ (x) − ic hpi ψ. The wave functions ψ solving this equation will minimize the product ∆x∆p.

20 4.2 Theory: Time Evolution of a Wave Function github.com/ejmastnak/fmf

• Rearranging the equation for ψ0 gives x − hxi i hpi ψ0(x) = + ψ(x). c~ ~ Next, we separate variables and integrate: dψ x − hxi i hpi (x − hxi)2 i hpi = + =⇒ ln ψ = + x + λ. ψ c~ ~ 2c~ ~ Solving for x gives 2 λ (x − hxi) −i hpi x ψ(x) = e exp e ~ . 2c~ • Now we just have to normalize the wavefunction:

Z ∞ 2 Z ∞ (x − hxi)2 2 λ |ψ(x)| dx ≡ 1 = e exp dx. −∞ −∞ 2c~ Note that c must be negative for the integral to converge. To make this requirement explicit, we’ll write

2 Z ∞ (x − hxi)2 1 ≡ eλ exp − dx, 2 −∞ 2σ where we introduce σ to match the form of a Gauss bell curve, which then implies 2 1 eλ = √ . 2πσ2 The ﬁnal result for ψ(x) is

2 1 (x − hxi) −i hpi x ψ(x) = √ exp e ~ . 4 2πσ2 4σ2 These functions are called Gaussian wave packets. This concludes the problem: the functions minimizing the product ∆x∆p are Gaussian wave packets. −i hpi x As a final note, the e ~ term is a plane wave, with wavelength hpi h λ = 2π =⇒ λ = . ~ hpi 4.2 Theory: Time Evolution of a Wave Function • We find the time evolution of a wave function by solving the Schrödingerequation ∂ Hψ(x, t) = i ψ(x, t). ~∂t One option is to solve the stationary Schrödinger equation

Hψn(x) = Enψn(x),

and expand the initial state ψ(x, 0) in terms of the basis functions ψn solving the stationary Schr¨odingerequation, i.e. writing ψ(x, 0) as X ψ(x, 0) = cnψn(x). n

21 4.3 Theory: Time-Dependent Expectation Values and . . . github.com/ejmastnak/fmf

• We then ﬁnd the time evolution by adding a time-dependent phase factor to the eigenfunction expansion

X −i En t ψ(x, t) = cnψn(x)e ~ . n In Dirac bra-ket notation, the stationary Schr¨odingerequation and eigenfunction expansion of the initial state read X H |ni = En |ni and |ψ, 0i = cn |ni , n and the corresponding time evolution reads

X −i En t |ψ, ti = cne ~ |ni . n The time evolution depends on the potential V (x) we’re working in.

4.3 Theory: Time-Dependent Expectation Values and Operators • Consider an operator A. A function of the operator is deﬁned in terms of the function’s Taylor series:

X 1 f(A) = √ f (n)(A)AN . n N!

• Next, consider an arbitrary observable A = A†. We then ask what is the expectation value hA, ti of A at time t? We can ﬁnd the wave function |ψ, ti with

X −i En t |ψ, ti = cne ~ |ni . n We’ll now show that we get the same result by acting on the initial state |ψ, 0i with the time evolution operator, i.e. we will prove the equality

X −i En t −i H t |ψ, ti ≡ cne ~ |ni = e ~ |ψ, 0i . n

• To show this, we ﬁrst expand |ψ, 0i in terms of its eigenstates and bring the operator inside the sum.

−i H t −i H t X X −i H t e ~ |ψ, 0i = e ~ cn |ni = cne ~ |ni . n n We write the exponent of the Hamiltonian operator in terms of the exponential function’s Taylor series:

" m # −i H t X X 1 it m e ~ |ψ, 0i = c − H |ni . n m! n m ~

• First, we’ll tackle the term Hm |ni, recalling that |ni are eigenfunctions of H:

m m−1 m−1 m H |ni = H (H |ni) = H (En |ni) = ··· = En |ni .

22 4.3 Theory: Time-Dependent Expectation Values and . . . github.com/ejmastnak/fmf

This step is important—it converts the operator H to the scalar values En. We then have " m # −i H t X X 1 it m X −i En t e ~ |ψ, 0i = c − E |ni = c e ~ |ni . n m! n n n m ~ n This completes the proof that

X −i En t −i H t |ψ, ti = cne ~ |ni = e ~ |ψ, 0i . n In other words, instead of ﬁnding a time evolution by expanding an initial state |ψ, 0i in terms of basis functions, we can act directly on the initial state with the time evolution operator.

Time-Dependent Expectation Values

• Next, we return to the expectation value hA, ti:

hA, ti ≡ hψ, t| A |ψ, ti .

We’ve just shown the ket term can be found with the time evolution operator

−i H t |ψ, ti = e ~ |ψ, 0i .

We ﬁnd the bra term by taking the complex conjugate of the ket term:

† −i H t i H t hψ, t| = hψ, 0| e ~ = hψ, 0| e ~ .

Note that the |ψ, 0i ket term is reversed under complex conjugation! We then have

i H t −i H t hA, ti = hψ, 0| e ~ Ae ~ |ψ, 0i ≡ hψ, 0| A(t) |ψ, 0i ,

i H t −i H t where we’ve deﬁned A(t) = e ~ Ae ~ for shorthand.

• The above result is useful: we have a new way to find a time-dependent expectation value. If we are given an initial state |ψ, 0i, we can use the time evolution operator to get i H t −i H t hA, ti = hψ, 0| e ~ Ae ~ |ψ, 0i ≡ hψ, 0| . In other words, we don’t need to find the time evolution of the wave function |ψ, ti. This is convenient, since finding |ψ, ti is often tedious.

• Finally, some vocabulary. Finding |A, ti via time evolution of the wave function, i.e.

hA, ti = hψ, t| A |ψ, ti ,

Theory: Time Evolution of an Operator

23 4.4 Time Evolution of the Gaussian Wave Packet github.com/ejmastnak/fmf

• Next, we ask how to ﬁnd the time evolution A(t) of an operator A? To do this, we solve the diﬀerential equation

d d h i H t −i H ti i i H t −i H t i H t H −i H t A(t) = e ~ Ae ~ = H e ~ Ae ~ + e ~ A −i e ~ . dt dt ~ ~ This is found with the Taylor series definition of an operator A. In both terms, we have the product of the Hamiltonian H with the time evolution operator. These operators commute, so we can switch their order of multiplication and factor to get d i H t i i −i H t i H t i −i H t i A(t) = e ~ HA − AH e ~ = e ~ [H,A]e ~ ≡ [H,A](t). dt ~ ~ ~ ~ We thus find A(t) by solving the differential equation d i A(t) = [H,A](t). dt ~ • One more useful identity: consider two operators A and B. We’re interested in the time evolution of their product AB:

i H t −i H t i H t −i H t (AB)(t) = e ~ ABe ~ = e ~ AIBe ~ i H t −i H t i H t −i H t = e ~ Ae ~ e ~ Be ~ = A(t)B(t).

4.4 Time Evolution of the Gaussian Wave Packet p2 Given a Gaussian wave function ψ(x, 0) and the time-independent Hamiltonian H = 2m + V (x), ﬁnd the initial wavefunction’s time evolution ψ(x, t).

• We’ll work in the potential V (x) = 0. Because the basis functions of a free particle are continuous, and n is traditionally used to index discrete quantities, we’ll index the basis functions with k. For a particle in a constant potential, the eigenfunctions are plane waves. We have

ikx 2 2 e ~ k ψk(x) = √ and Ek = . 2π 2m √ Note the normalization of the plane wave with 2π. The factor √1 is chosen so 2π that a product of two basis functions produces a delta function:

D E Z 1 Z ˜ k k˜ ≡ ψ∗(x)ψ (x) dx = ei(k−k)x dx ≡ δ(k˜ − k). k k˜ 2π

• First, an overview: To perform the time evolution, we ﬁrst expand the initial state ψ(x, 0) in terms of the basis functions ψk Z eikx ψ(x, 0) = c(k)√ dk. 2π Note that we use integration instead of summation because the basis functions are continuous and not discrete. The above expansion is a Fourier transform of c(k) from the k to the x domain, so c(k) is found with the inverse Fourier transform

Z e−ikx c(k) = ψ(x, 0) √ dx. 2π

24 4.4 Time Evolution of the Gaussian Wave Packet github.com/ejmastnak/fmf

We would then ﬁnd the time evolution with

Z E ikx −i k t e ψ(x, t) = c(k)e ~ √ dk. 2π Again, integration replaces summation.

• In the end, we’ll be more interested in the observables x and p than the actual wave function. For example the expectation values of x and p at time t, written

hx, ti ≡ hψ, t| x |ψ, ti and hp, ti ≡ hψ, t| p |ψ, ti .

Likewise, we’ll be interested in the uncertainties q q ∆x(t) = hx2, ti − hx, ti2 and ∆p(t) = hp2, ti − hp, ti2.

Note: Directly performing the above eigenfunction expansion and time evolution by solving the integrals is a tedious mathematical exercise. We’ll take a slight detour, and introduce some machinery that solves the problem in a more physically elegant way.

Theory: Commutator Properties • First, a distributive property: [AB, C] = A[B,C] + [A, C]B.

• And second: [λA, B] = λ[A, B].

• And ﬁnally: [B,A] = −[A, B]. Back to the Time Evolution of the Gaussian Wave Packet • We now have the formalism we need to easily ﬁnd the expectation values

hx, ti ≡ hψ, t| x |ψ, ti and hp, ti ≡ hψ, t| p |ψ, ti

p2 for a Gaussian wave packet with the simple Hamiltonian H = 2m . Using the Heisen- berg approach, we’ll first find the time evolution of the operators x and p, then find the expectation values with

hA, ti = hψ, 0| A(t) |ψ, 0i .

• First, the equation for the operator x(t):

d i i p2 x(t) = [H, x](t) = , x (t). dt ~ ~ 2m Writing p2 = pp, applying basic commutator properties and recalling the result [x, p] = i~ gives d i i p(t) x(t) = (p[p, x] + [p, x]p)(t) = (−pi~ − i~p)(t) = . dt 2m~ 2m~ m Second, the diﬀerential for the operator p(t), noting that [p2, p] = 0, is

d i i p2 (t) = [H, p](t) = , p (t) = 0. dp ~ ~ 2m

25 4.4 Time Evolution of the Gaussian Wave Packet github.com/ejmastnak/fmf

The general solution is a constant: p(t) = p0. We need an initial condition to ﬁnd a solution speciﬁc to our problem, which we get from general expression

i H ·0 −i H ·0 A(t) t=0 = e ~ Ae ~ = A. In our case, p(0) = p, which implies p(t) = p. In other words, the momentum operator stays the same with time; it is always the usual ∂ p = −i ~∂x for all time t. Next, with p(t) known, we solve the equation for x(t):

d p(t) p p = = =⇒ x(t) = t + x . dx m m m 0 With the initial condition x(0) = x, we now have the expression for the time evolution of the operator x(t) p ∂ x(t) = t + x → −i ~ t + x. m m ∂x • With x(t) and p(t) known, we can ﬁnd the expectation values using

hA, ti = hψ, 0| A(t) |ψ, 0i .

We start with momentum:

hp, ti = hψ, 0| p(t) |ψ, 0i = hψ, 0| p |ψ, 0i ≡ hp, 0i = hpi .

In other words, because p(t) = p is constant, we can just ﬁnd the initial expectation value hp, 0i using the initial wave function. This is the same hpi term in the exponent of the plane wave term in the Gaussian wave packet. Next, the expectation value of position. Inserting x(t) and splitting the bra-ket into two parts gives p hx, ti = hψ, 0| x(t) |ψ, 0i = hψ, 0| t + x |ψ, 0i m t t = hψ, 0| p |ψ, 0i + hψ, 0| x |ψ, 0i = hp, 0i + hx, 0i . m m These are the same hxi and hpi term in the exponents of the initial the Gaussian wave packet.

• Next, we want to ﬁnd the uncertainties q q ∆x(t) = hx2, ti − hx, ti2 and ∆p(t) = hp2, ti − hp, ti2.

It remains to ﬁnd the the time evolution of the operators p2 and x2 and then the expectation values p2, t and x2, t . We start with the operator p2(t).

p2, t = hψ, 0| p2(t) |ψ, 0i = hψ, 0| p(t)2 |ψ, 0i = hψ, 0| p2 |ψ, 0i .

In other words, p2, t is the expectation value of the time-independent operator p2.

26 4.4 Time Evolution of the Gaussian Wave Packet github.com/ejmastnak/fmf

Next, the evolution of x(t). Using x2(t) = x(t)2, substituting in the definition of x(t), and splitting up the expectation value into three terms gives p 2 x2, t = hψ, 0| x2(t) |ψ, 0i = hψ, 0| x(t)2 |ψ, 0i = hψ, 0| t + x |ψ, 0i m p2t2 t = hψ, 0| + (px + xp) + x2 |ψ, 0i m2 m t2 t = hψ, 0| p2 |ψ, 0i + hψ, 0| px + xp |ψ, 0i + hψ, 0| x2 |ψ, 0i . m2 m We’ve now succeeded in expressing the time-dependent expectation values of x, x2, p and p2 in terms of quantities that appear in the initial state, namely the exponent parameters hxi and hpi. • From the beginning of this problem, recall the initial wave function is 2 1 (x − hxi) −i hpi x ψ(x, 0) = √ exp e ~ . 4 2πσ2 4σ2 The associated probability density is 1 (x − hxi)2 ρ(x, t) = √ exp . 2πσ2 2σ2 The expectation values of x and p are simply hψ, 0| x |ψ, 0i = hxi and hψ, 0| p |ψ, 0i = hpi . By definition, the initial uncertainty in x is q ∆x(0) = hx2, 0i − hx, 0i2. There is a shortcut: for a Gaussian wave packet, the uncertainty is simply the wave packet’s standard deviation σ. We can take advantage of this relationship to easily find x2, 0 : q ∆x(0) = hx2, 0i − hx, 0i2 = σ =⇒ x2, 0 = σ2 + hx, 0i2 .

• Another shortcut: instead of the deﬁnition, we ﬁnd the uncertainty in p using the uncertainty principle, which for a Gaussian wave packet is the equality

∆x∆p = ~ =⇒ ∆p(0) = ~ = ~ . 2 2∆x(0) 2σ With ∆p(0) in hand, we can find p2, 0 with q 2 ∆p(0) = hp2, 0i − hp, 0i2 = ~ =⇒ p2, 0 = ~ + hp, 0i2 . 2σ 4σ2 • The last piece needed to find x2, t is the quantity hψ, 0| px + xp |ψ, 0i. To find this, we’ll use the relationship px = p†x† = (xp)†. In general, † † ∗ ∗ A = ψ A ψ = hAψ|ψi = hψ|Aψi = hAi , which implies A + A† = hAi+hAi∗ = 2 Re hAi. The expression hψ, 0| px+xp |ψ, 0i then simplifies to hψ, 0| px + xp |ψ, 0i = 2 Re hψ, 0| xp |ψ, 0i . We ran out of time at this point. The problem is completed in the next exercise set.

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5 Fifth Exercise Set

5.1 Time Evolution of a Gaussian Wave Packet (continued) • In the last exercise set, we wanted to ﬁnd the uncertainties q q ∆x(t) = hx2, ti − hx, ti2 and ∆p(t) = hp2, ti − hp, ti2.

We approached ﬁnding the time evolution of the operators x(t) and p(t) using the Heisenberg approach. We left oﬀ with the following operator expressions: pt p(t) = p and x(t) = + x. m For expected momentum values, we found

hp, ti = hp, 0i and p2, t = p2, 0 .

hp, 0i = hpi hx, 0i = hxi 2 x2, 0 = σ2 + hxi2 p2, 0 = ~ + hpi2 . 2σ

The only remaining expectation value is hpx + xp, 0i, which we showed to be

hxp + px, 0i = 2 Re hxp, 0i .

We ﬁnd the expectation value by deﬁnition, using the wave function: Z ∞ ∗ d hxp + px, 0i = 2 Re hxp, 0i = 2 Re ψ (x, 0)x −i~ ψ(x, 0) dx −∞ dx ∞ 2 Z d 1 (x−hxi) hpi ∗ − 2 i x 4σ ~ = 2 Re ψ (x, 0) · x · (−i~) √4 e e dx −∞ dx 2πσ2 Z ∞ = 2 ψ∗(x, 0)x hpi ψ(x, 0) dx = 2 hpi hxi . −∞ Note that the imaginary portion of the derivative in the middle line disappears when taking the real component.

• We now have everything we need to ﬁnd the uncertainties ∆x and ∆p. Starting with ∆p, we have:

2 2 ∆p(t)2 = p2, t − hp, ti2 = ~ + hpi2 − hpi2 = ~ . 2σ 2σ

28 5.2 Theory: Quantum Harmonic Oscillator github.com/ejmastnak/fmf

For position, the uncertainty ∆x is

t2 t t 2 ∆x(t)2 = x2, t − hx, ti2 = p2, 0 + hpx + xp, 0i + x2, 0 − hp, 0i + hx, 0i m2 m m " # t2 2 t t2 t = hpi2 + ~ + 2 hpi hxi + σ2 + hxi2 − hpi2 − 2 hpi hxi − hxi2 m2 2σ m m2 m t 2 = σ2 + ~ . 2mσ

The product in the two uncertainties is ! s 2 t 2 t 2 ∆x(t)∆p(t) = ~ σ2 + ~ = ~ 1 + ~ . 2σ 2mσ 2 2mσ2

~ Note that the product in uncertainties starts at the initial value 2 and increases with time. This immediately implies that the time evolution ψ(x, t > 0) of the initial Gaussian wave packet ψ(x, 0) is no longer a Gaussian wave packet. It couldn’t be—it ~ has an uncertainty product greater ∆x∆p > 2 , while Gaussian wave packets are by ~ deﬁnition constructed to have ∆x∆p = 2 .

5.2 Theory: Quantum Harmonic Oscillator • The Hamiltonian of a particle of mass m in a harmonic oscillator reads r p2 kx2 p2 mω k H = + = + x2, ω = . 2m 2 2m 2 m We analyze the harmonic oscillator using the annihilation and creation operators. The annihilation operator a is r 1 x p ~ ~ a = √ + i where x0 = , p0 = , 2 x0 p0 mω x0 while the creation operator, equal to the adjoint of a, is

1 x p a† = √ − i . 2 x0 p0 The annihilation and creation operators don’t commute—their commutator is

a, a† = 1.

Adding and subtracting the two operators give expressions for x and p x p x = √0 (a + a†) and p = √0 (a − a†). 2 2i

• In terms of a and a†, the harmonic oscillator’s Hamiltonian reads

1 H = ω a†a + . ~ 2

29 5.3 Time Evolution of a Particle in a Harmonic . . . github.com/ejmastnak/fmf

The Hamiltonian’s eigenstates are indexed by n = 0, 1, 2,... and read

1 H |ni = ω n + |ni , n = 0, 1, 2,.... ~ 2

The annihilation and creation operators act on the oscillator’s eigenstates as follows: √ √ a |ni = n |n − 1i and a† |ni = n + 1 |n + 1i .

Note the operator equality a†a |ni = n |ni—in other words, acting on an eigenstate |ni with the operator a†a reveals the state’s index n.

Theory: Ehrenfest Theorem In our case (for a harmonic oscillator), the theorem reads

d hp, ti d d hx, ti = and hp, ti = − V (x) = −k hx, ti . dt m dt dx

5.3 Time Evolution of a Particle in a Harmonic Oscillator Consider a particle in a harmonic potential with the initial eigenfunction 1 1 |ψ, 0i = √ |0i + √ |1i . 2 2 Compute the wavefunction’s time evolution |ψ, ti the time-dependent expectation values hx, ti and hp, ti and use this to determine the validity of the Ehrenfest theorem. Finally, determine the product of uncertainties ∆x(t)∆p(t).

• First, we ﬁnd the wavefunction’s time evolution. This is relatively simple, since the initial wavefunction is a linear combination of the Hamiltonian’s eigenstates. We just −i En t have to add the time-dependent phase factors e ~ . Substituting in the energies for a particle in a harmonic oscillator, the result is

1 −i E0 t 1 −i E1 t 1 −i ω t 1 −i 3ω t |ψ, ti = √ e ~ |0i + √ e ~ |1i = √ e 2 |0i + √ e 2 |1i . 2 2 2 2

• Next, working with x in terms of the annihilation and creation operators, we find the expectation value hxi as √ x0 D †E x0 ∗ hxi = √ a + a = √ (hai + hai ) = 2x0 Re hai . 2 2 Following similar lines for hpi, we have √ p0 D †E p0 ∗ hpi = √ a − a = √ (hai − hai ) = 2p0 Im hai . 2i 2i These two results are useful: they show that by finding the expectation value hai we also determine the values of hxi and hpi. Naturally, the next step is to find hai (we’ll find the time-dependent form) 1 −i ω t 1 −i 3ω t ha, ti = hψ, t| a |ψ, ti = hψ, t| √ e 2 a |0i + √ e 2 a |1i . 2 2

30 5.3 Time Evolution of a Particle in a Harmonic . . . github.com/ejmastnak/fmf

The annihilation operator eliminates the ground state, i.e. a |0i = 0 and turns the ﬁrst state into the ground state, i.e. a |1i = |0i. The expression for ha, ti becomes 1 −i 3ω t 1 i ω t 1 i 3ω t 1 −i 3ω t ha, ti = hψ, t| √ e 2 |0i = √ e 2 h0| + √ e 2 h1| √ e 2 |0i 2 2 2 2 1 = e−iωt, 2 where the last equality makes use of the eigenstates’ orthogonality. With ha, ti known, we can now ﬁnd the expectation values of position and momentum These are

√ x0 √ p0 hx, ti = 2x0 Re ha, ti = √ cos ωt and hp, ti = 2p0 Im ha, ti = −√ sin ωt. 2 2

• Next, we conﬁrm the validity of the Ehrenfest theorem.

d d x ωx hp, ti p hx, ti = √0 cos ωt = − √ 0 sin ω =? = −√0 sin ωt. dt dt 2 2 m 2

Comparing the equalities, canceling like terms and inserting the definition of p0 and ω, we have ? p0 ~ ? h ~ mω ωx0 = = =⇒ ω = 2 = = ω, m x0m mx0 m ~ which satisfies the first part of the Ehrenfest theorem. To confirm the second part of the theorem, we compute

d p p ω −√0 sin ωt = − √0 cos ωt =? −k hx, ti . dt 2 2 The equality reduces to

? ~ 2 2 ? ~ p0ω = kx0 =⇒ ω = mω x0 =⇒ x0 = . x0 mω The last equality leads to ω = ω, already shown above when conﬁrming the ﬁrst part of the Ehrenfest theorem.

• Finally, we compute the product ∆x(t)∆p(t). By deﬁnition, q q ∆x(t) = hx2, ti − hx, ti2 and ∆p(t) = hp2, ti − hp, ti2.

First, we express hxi in terms of the annihilation operator a:

* 2+ 2 x x D 2 E x2 = √0 (a + a†) = 0 a2 + aa† + a†a + a† 2 2 2 x D 2 E = 0 a2 + 2a†a + a† + 1 . 2 Where the last equality uses the identity a, a† = 1 =⇒ aa† = 1 + a†a. Continuing on, we have

2 x 2 x2 = 0 2 a†a + a2 + a† + 1 = x2 Re a2 + a†a + 1 . 2 0 2

31 5.3 Time Evolution of a Particle in a Harmonic . . . github.com/ejmastnak/fmf

Next, we express p2 in terms of a with a similar procedure: 2 p D E p D 2 E p2 = √0 (a − a†)2 = − 0 a2 − aa† − a†a + a† 2i 2 p D 2 E = − 0 a2 − 2a†a − 1 + a† = p2 − Re a2 + a†a + 1 . 2 0 2 • Next, we ﬁnd the expectation value a2, t : 2 1 −i ω t 2 1 −i 3ω t 2 a , t = hψ, t| √ e 2 a |0i + √ e 2 a |1i . 2 2 We make a brief intermezzo to note that a2 |0i = a (a |0i) = a · 0 = 0 and a2 |1i = a (a |1i) = a |0i = 0. With these identities in hand, we have a2, t = hψ, t| · 0 = 0.

• One more expectation value to go: a†a, t † 1 −i ω t † 1 −i 3ω t † 1 −i 3ω t a a, t = hψ, t| √ e 2 a a |0i + √ e 2 a a |1i = hψ, t| √ e 2 |1i 2 2 2 1 i ω t 1 i 3ω t 1 −i 3ω t 1 = √ e 2 h0| + √ e 2 h1| √ e 2 |1i = . 2 2 2 2 The derivation uses the identity a†a |ni = n |ni and the eigenstates’ orthogonality. • We now have everything we need to ﬁnd the product of uncertainties in x and p. Putting the pieces together, we have

2 2 2 † 1 2 1 1 2 x , t = x0 Re a + a a + 2 = x0 0 + 2 + 2 = x0 2 2 2 † 1 2 D † E 1 2 p , t = p0 − Re a + a a + 2 = p0 0 + a a + 2 = p0 √ x0 hx, ti = 2x0 Re ha, ti = √ cos ωt 2 √ p0 hp, ti = 2p0 Im ha, ti = −√ sin ωt. 2 The uncertainties ∆x(t) and ∆p are then x2 cos2 ωt ∆x(t)2 = x2, t − hxi2 = x2 − 0 cos2 ωt = x2 1 − 0 2 0 2 p2 sin2 ωt ∆p(t)2 = p2, t − hpi2 = p2 − 0 sin2 ωt = p2 1 − . 0 2 0 2 The product in uncertainties is s cos2 ωt sin2 ωt ∆x(t)∆p(t) = x p 1 − 1 − 0 0 2 2 r1 1 r1 1 = + sin2 ωt cos2 ωt = + sin2 2ωt. ~ 2 4 ~ 2 16 In other words, the product of uncertainties oscillates in time between a minimum value of √~ to a maximum value of 3~ . 2 4

32 github.com/ejmastnak/fmf

Take Two, Using the Heisenberg Approach • We start with an expression for hx, ti, using the Heisenberg approach √ hx, ti = 2x0 Re hψ, 0| a(t) |ψ, 0i . We start with the equation for the operator a(t): i a˙(t) = [H, a](t). ~ Making use of a, a† = 1 and a†, a = −1, the commutator [H, a] is

† 1 † † † [H, a] = [~ω(a a + 2 ), a] = ~ω a a, a = ~ω [a , a]a + a [a, a] = −~ωa, and the equation for a(t) reads da a˙(t) = −iωa(t) =⇒ = −iω dt =⇒ ln a(t) = −iωt + C a a(t) = De−iωt. We find the constant D with the initial condition a(0) = a, which leads to the final solution a(t) = ae−iωt. √ Inserting a(t) into the expression hx, ti = 2x0 Re hψ, 0| a(t) |ψ, 0i gives √ √ −iωt −iωt hx, ti = 2x0 Re hψ, 0| ae |ψ, 0i = 2x0 Re e hψ, 0| a |ψ, 0i . This general result holds for any initial state |ψ, 0i of a particle in a harmonic potential. For our specific initial state, the expression reads 1 1 1 1 hψ, 0| a |ψ, 0i = hψ, 0| √ a |0i + √ a |1i |ψ, 0i = hψ, 0| √ |0i = . 2 2 2 2 The result for hx, ti then reads √ √ 1 hx, ti = 2x Re e−iωt hψ, 0| a |ψ, 0i = 2x Re e−iωt 0 0 2 x = √0 cos ωt, 2 which matches the result from the earlier Schrödingerapproach.

6 Sixth Exercise Set

6.1 Theory: Coherent States of the Quantum Harmonic Oscillator • The annihilation operator’s eigenstates are called coherent states of the quantum harmonic oscillator’s Hamiltonian. Coherent states |ψi of a QHO satisfy the eigenvalue equation a |ψi = λ |ψi , λ ∈ C. Recall that a is not Hermitian, so the eigenvalue λ is in general complex. By convention, we make the potentially complex nature of λ explicit by writing a |zi = z |zi , where |zi is just a new notation for the coherent eigenstate and z is the corresponding eigenvalue.

33 6.1 Theory: Coherent States of the Quantum . . . github.com/ejmastnak/fmf

This is a recursive relation between the coeﬃcients cn of the eigenstate expansion of the coherent states. We can recognize the general form from the pattern in the ﬁrst few terms:

n = 0 =⇒ c1 = c0z z z2 n = 1 =⇒ c2 = c1 √ = c0 √ . 2 1 · 2 . zn cn = √ c0. n!

As long as we know c0, we can ﬁnd all cn with the above expression. Returning to the expansion of the coherent |zi, we have X X zn |zi = cn |ni = c0 √ |ni . n n n!

• We ﬁnd c0 by requiring the coherent states are normalized: ∗n ! m ! ∗ X z X z 1 ≡ hz|zi = c0 √ hn| c0 √ |mi . n n! m m! We only have to evaluate one sum, since orthogonality of the QHO eigenstates means hn|mi = δnm. We then have ∗n n 2n X z z X |z| 2 1 ≡ hz|zi = |c |2 = |c |2 = |c |2e|z| ≡ 1. 0 n! 0 n! 0 n n

2 − |z| Solving the last equality for c0 gives c0 = e 2 . We now have everything we need for the eigenfunction expansion of the coherent states. Putting the pieces together gives n 2 n X X z − |z| X z |zi = cn |ni = c0 √ |ni = e 2 √ |ni . n n n! n n!

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• We now consider another way to write the coherent states |zi using creation operator’s action on the QHO eigenstates:

a† |0i = |1i =⇒ |1i = a† |0i 2 √ a† |1i a† a†|1i = 2 |2i =⇒ |2i = √ = √ |0i . 2 1 · 2 . n a† |ni = √ |0i . n! Substituting in the expression for the QHO eigenstate |ni into the expansion of the coherent states |zi gives

n †n 2 n †n 2 X X z c0 a − |z| X z a − |z| za† |zi = c |ni = √ √ |0i = e 2 |0i = e 2 e |0i . n n! n n n! n! n In other words, the coherent states |zi and ground state |0i are related by the exponential operator eza† via

2 − |z| za† |zi = e 2 e |0i .

• Next, we will ﬁnd the time evolution |z, ti of the coherent states. This is

1 where we have substituted in the QHO’s energy eigenvalues En = ~ω n + 2 . Some rearranging of |z, ti leads to

2 n − |z| X z |zi = e 2 √ |ni , n n! with z replaced by ze−iωt. We use this relationship to write

−i ω t −iωt |z, ti = e 2 ze .

• Finally, an approach to ﬁnding the time evolution |z, ti in the Heisenberg picture. We write |z, ti in terms of the time evolution operator and use the earlier relationship between |zi and |0i via the creation operator a† to get

2 2 −i H t −i H t − |z| za† − |z| −i H t za† i H t −i H t |z, ti = e ~ |zi = e ~ e 2 e |0i = e 2 e ~ e e ~ e ~ |0i .

35 6.1 Theory: Coherent States of the Quantum . . . github.com/ejmastnak/fmf

−i H t za† i H t The three terms e ~ e e ~ , with the exponent written as a Taylor series, are

† n ! n −i H t za† i H t −i H t X (za ) i H t X z −i H t † n i H t e ~ e e ~ = e ~ e ~ = e ~ (a ) e ~ n! n! n n n X z h −i H t † i H t −i H t † i H t −i H t † i H ti = e ~ a e ~ e ~ a e ~ ··· e ~ a e ~ n! n n n X z −i H t † i H t = e ~ a e ~ . n! n Recall that in the Heisenberg approach the time evolution of an operator O reads i H t −i H t O(t) = e ~ Oe ~ . Applied to our above expression (note the reversed roles of the time evolution operators), we see n −i H t za† i H t X z † n za†(−t) e ~ e e ~ = a (−t) = e . n! n The initial expression for |z, ti in the Heisenberg picture then simpliﬁes to

2 2 − |z| za†(−t) −i H t − |z| za†(−t) −i ω t |z, ti = e 2 e e ~ |0i = e 2 e e 2 |0i .

• Next, we ﬁnd an expression for a†(−t). In the Heisenberg approach, this is

† † −i H t † i H t h −i H t i H ti † a (−t) = e ~ a e ~ = e ~ ae ~ = a(−t) .

Using the relationship a(t) = ae−iωt from the previous exercise set, we ﬁnally have

† a†(−t) = a(−t)† = aeiωt = a†e−iωt.

Substituting the expression for a†(−t) into the time evolution |z, ti to get

2 2 |ze−iωt| − |z| ze−iωta† −i ω t −i ω t − ze−iωta† |z, ti = e 2 e e 2 |0i = e 2 e 2 e |0i .

2 −i H t − |z| za† |z, ti = e ~ e 2 e |0i .

−i ω t Aside from the factor e 2 , the expressions are the same, just with z shifted to ze−iωt. This allows us to write

2 |ze−iωt| −i ω t − ze−iωta† −i ω t −iωt |z, ti = e 2 e 2 e |0i = e 2 ze , in agreement with the earlier result using the Schr¨odingerapproach. • Next, we want to get a better sense for the behavior of the wave function |z, ti. From the last exercise set, we know the following expressions for any state of quantum harmonic oscillator: √ √ hxi = 2x0 Re hai hpi = 2p0 Im hai 2 2 † 2 1 2 2 † 2 1 x = x0 a a + Re a 2 p = p0 a a − Re a 2 .

36 6.1 Theory: Coherent States of the Quantum . . . github.com/ejmastnak/fmf

Next, using the eigenvalue equation a |zi = z |zi for coherent states, we have

n hz| (a†)nam |zi = hanz|amzi = hznz|zmzi = z∗ zm.

We will use this expression to evaluate the expectation values of x and p above. Note the diﬀerence between z as an eigenvalue and |zi as a coherent state.

• Next, using the just-derived identity hz| (a†)nam |zi = z∗n zm, the expectation value hai is 0 hai = h1 · ai = (a∗)0a = z∗ z1 = 1 · z1 = z. Using hai = z, the expectation values of x and p are √ √ hxi = 2x0 Re z and hpi = 2p0 Im z.

To find x2 and p2 , we first have to find a†a and a2 . These are

a†a = |z|2 and a2 = z2.

The expectation values x2 and p2 are then

2 2 2 2 1 2 2 2 2 1 x = x0 |z| + Re z + 2 and p = p0 |z| − Re z + 2

2 2 z2+z∗ • Next, using the general complex number identity Re z = 2 , we write

z2 + z∗2 1 1 |z|2 + Re z2 = |z|2 + = z + z∗2 = (2 Re z)2 = 2 Re2 z, 2 2 2 which, substituted into x2 , gives

2 2 2 2 1 2 2 1 x = x0 |z| + Re z + 2 = x0 2 Re z + 2 .

A similar procedure for p2 gives

2 2 1 ∗ 1 2 2 1 p = p0 − 2 z − z + 2 = p0 2 Im z + 2 .

• Next, the uncertainties ∆x and ∆p are

√ 2 x2 ∆x2 = x2 − hxi2 = x2 2 Re2 z + 1 − 2x Re z = 0 0 2 0 2 √ 2 p2 ∆p2 = p2 − hpi2 = p2 1 + 2 Im2 z − 2p Im z = 0 . 0 2 0 2 The product of uncertainties, recalling the deﬁnition p = ~ , is 0 x0 r x2 p2 x p ∆x∆p = 0 0 = 0 0 = ~. 2 2 2 2

~ Note that the product of uncertainties takes the minimum possible value 2 , meaning a coherent state |zi of the QHO must be a Gaussian wave packet.

37 6.1 Theory: Coherent States of the Quantum . . . github.com/ejmastnak/fmf

• Recall the general form of a Gaussian wave packet is

2 1 (x − hxi) i hpi x ψ(x) = √ exp − e ~ . 4 2πσ2 4σ2

For the coherent state |zi, substituting in the values of hxi, hpi and σ2 ≡ x2 , the Gaussian wave packet reads √ " 2 # √ 1 x − 2x0 Re z i 2p0 Im z x ψz(x) = exp − e ~ . p4 2 2x2 πx0 0

• Finally, we will examine two more properties of the coherent states of a QHO: the expectation values of the Hamiltonian H. First, recall the Hamiltonian can be written in terms of a and a† as

† 1 H = ~ω a a + 2 .

We evaluate this with the help of the earlier identity hz| (a†)nam |zi = z∗n zm. Ap- plied to a†a, we have ω ω ω hHi = ~ + ω a†a = ~ + ωz∗z = ~ + ω|z|2. 2 ~ 2 ~ 2 ~ For the expectation value H2 we ﬁrst write H in terms of a and a†:

2 2 2 † † † 1 H = ~ ω a aa a + a a + 4 .

To simplify a†aa†a we use the commutator identity

a†, a = −1 =⇒ aa† = 1 + a†a.

Substituting aa† = 1 + a†a into H2 and an intermediate step of algebra gives

2 2 2 †2 2 † 1 H = ~ ω a a + 2a a + 4 .

We again use the earlier identity hz| (a†)nam |zi = z∗n zm to get

2 2 2 ∗2 2 ∗ 1 2 2 4 2 1 H = ~ ω z z + 2z z + 4 = ~ ω |z| + 2|z| + 4 .

The energy uncertainty is then

ω 2 ∆H2 = H2 − hHi2 = 2ω2 |z|4 + 2|z|2 + 1 − ~ + ω|z|2 ~ 4 2 ~ 2 2 2 = ~ ω |z| =⇒ ∆H = ~ω|z|.

• Next, we consider the ratio

∆H ω|z| 1 ∆H = ~ ≈ , |z| 1 =⇒ lim = 0. hHi 2 1 |z| |z|→∞ hHi ~ω |z| + 2

∆H In other words, the relative uncertainty in energy hHi approaches zero for large |z|.

38 6.2 Harmonic Oscillator in an Electric Field github.com/ejmastnak/fmf

We want a better understanding of what |z| 1 means. To do this, we ﬁrst examine the earlier expressions for a QHO coherent state: √ √ −iωt hxi = 2x0 Re z and hx, ti = 2x0 Re ze .

If we write the complex number z in the polar form z = |z|eiδ, hx, ti becomes √ √ n iδ−iωto hx, ti = 2x0 Re ze = 2x0|z| cos(ωt − δ).

In other words, large |z| means a large amplitude of oscillation in the time-dependent expectation value hx, ti.

6.2 Harmonic Oscillator in an Electric Field

Consider a particle of mass m and charge q at the equilibrium position x0 of a spring with spring constant k constrained to oscillate along the x axis. At time t = 0, we turn on an external electric ﬁeld E in the positive x direction. Solve for the particle’s motion using quantum mechanics.

• First, we ﬁnd the classical solution with Newton’s law to get a feel for the solution. Newton’s law reads: mx¨ = −kx + qE. Turning the ﬁeld exerts a force on the particle in the x direction, moving the equilibrium position tox ˜0 = x0 + x1 > x0. The particle starts at the far left amplitude x0 and oscillates about the new equilibrium positionx ˜0 with amplitude x1.

Placing the origin at x0 ≡ 0, the initial conditions are x(0) = 0 andx ˙(0) = 0. The solution is r k qE x(t) = x (1 − cos ωt) where ω = , x = . 1 m 1 k

• Next, the quantum mechanical picture. The particle’s Hamiltonian is

( 2 p + 1 kx2 t < 0 H(t) = 2m 2 . p2 1 2 2m + 2 kx − qEx t > 0 Note that turning on the ﬁeld adds the potential energy term −qEx to the initial Hamiltonian. Notation: all quantities for t > 0, after the ﬁeld is turned on, will be written with a tilde.

• In the classical picture, the particle takes the initial position x(0) = 0. This doesn’t carry over in quantum mechanics—specifying an exact position violates the uncertainty principle. Instead, we place the particle in the harmonic oscillator’s ground 1 state: |ψ, 0i = |0i where H |0i = 2 ~ω |0i. Our goal will be to ﬁnd the time-dependent wave function |ψ, ti and then compute the position expectation value hx, ti.

• Let’s begin! First, we expand the electric ﬁeld Hamiltonian to a perfect square

p2 k eE 2 1 qE 2 p2 k 1 H˜ = + x − − k = + (x − x )2 − kx2, 2m 2 k 2 k 2m 2 1 2 1

39 6.2 Harmonic Oscillator in an Electric Field github.com/ejmastnak/fmf

qE where we have substituted in the amplitude x1 = k from the classical solution. Next, we denote the post-field potential k 1 V˜ (x) = (x − x )2 − kx2. 2 1 2 1 1 2 Note that while the pre-field potential V (x) = 2 kx has its minimum at (0, 0), the ˜ 1 2 post-field potential V (x)’s minimum is shifted to x1, − 2 kx1 . With this shift of the equilibrium position in mind, we introduce new coordinates d d x˜ = x − x andp ˜ = −i = −i = p. 1 ~d˜x ~dx The Hamiltonian H˜ then reads p˜2 1 1 H˜ = + kx˜2 − kx2. 2m 2 2 1 • The Hamiltonian’s shifted annihilation operator is 1 x˜ p˜ a˜ = √ + i , 2 x0 p0 q where x = ~ and p = ~ . The difference betweena ˜ and the unshifted a is 0 mω 0 x0 x x a˜ − a = −√ 1 =⇒ a˜ = a − √ 1 . 2x0 2x0 • Recall the annihilation operator actions on the quantum harmonic oscillator’s eigen- √ states as a |ni = n |n − 1i and a |0i = 0. In our case, the shifted annihilation operatora ˜ acts on our problem’s initial state |ψ, 0i = |0i as x x x a˜ |ψ, 0i = a − √ 1 |0i = 0 − √ 1 |0i =⇒ a˜ |ψ, 0i = −√ 1 |ψ, 0i . 2x0 2x0 2x0 In other words, our initial state |ψ, 0i is an eigenstate of the shifted annihilation operatora ˜ with the eigenvalue − √x1 . This means the initial state is a coherent 2x0 state of the annihilation operatora ˜, so we can use the results from the theory section on coherent states. • From the theory section, an coherent state |ψ, 0i satisfying a |ψ, 0i = z |ψ, 0i with eigenvalue z evolves in time as − ω t −iωt |ψ, ti = e 2 ze , where the eigenvalue z changes with time as z(t) = ze−iωt. Meanwhile, the position expectation value is √ hxi = 2x0 Re z. In our case, the time-dependent expectation value of the shifted positionx ˜ is thus √ √ x1 −iωt hx,˜ ti = 2x0 Re z(t) = 2x0 Re −√ e = −x1 cos ωt. 2x0

The unshifted position is x = x1 +x ˜, which implies

hx, ti = x1 − x1 cos ωt = x1(1 − cos ωt),

in agreement with the classical result x(t) = x1(1 − cos ωt).

40 github.com/ejmastnak/fmf

7 Seventh Exercise Set

7.1 Theory: Two-Dimensional Harmonic Oscillator • We generalize the harmonic oscillator to two dimensions by introducing the vector operators ∂ ∂ r = (x, y) and p = (p , p ) = −i , −i = −i ∇. x y ~∂x ~∂y ~ We write the oscillator’s Hamiltonian as p2 1 1 p2 1 p2 1 H = + k x2 + k y2 = x + k x2 + y + k y2 ≡ H + H . 2m 2 x 2 y 2m 2 x 2m 2 y x y In other words, we can write the two-dimensional Hamiltonian as the sum of two one-dimensional Hamiltonians corresponding to motion in the x and y directions. • We want to solve the stationary Schr¨odingerequation

Hψ(x, y) = Eψ(x, y).

The equation can be solved with separation of variables because the two-dimensional Hamiltonian can be decomposed into the sum of one-dimensional x and y Hamilto- nians. We make this separation explicit by deﬁning r k H ψ = E ψ (x) where E = ω n + 1 , ω = x x nx nx nx nx ~ x x 2 x m r k H ψ = E ψ (y) where E = ω n + 1 , ω = y y ny ny ny ny ~ y y 2 y m

for nx, ny = 0, 1, 2,... and then writing

Hψ(x, y) = Eψ(x, y) → Hψnx (x)ψny (y) = (Enx + Eny )ψnx (x)ψny (y).

• In terms of the annihilation and creation operators, the Hamiltonian reads

† 1 † 1 H = Hx + Hy = ~ωx axax + 2 + ~ωy ayay + 2 , where the annihilation and creation operators are deﬁned as r 1 x px ~ ~ ax = √ + i where x0 = , p0x = 2 x0 p0x mωx x0 s 1 y py ~ ~ ay = √ + i where y0 = , p0y = . 2 y0 p0y mωy y0

• With the annihilation and creation operators deﬁned, we write

1 1 Hx |nxi = ~ωx nx + 2 |nxi and Hy |nyi = ~ωy ny + 2 |nyi . For the two-dimensional Hamiltonian we write separation of variables as

1 1 H |nxi |nyi = ~ωx nx + 2 + ~ωy ny + 2 |nxi |nyi . A slightly shorter notation for the same expression reads

1 1 H |nxnyi = ~ωx nx + 2 + ~ωy ny + 2 |nxnyi .

41 7.1 Theory: Two-Dimensional Harmonic Oscillator github.com/ejmastnak/fmf

A Few Special Cases

• Consider the degenerate case with kx > 0 and ky = 0. In this case the y Hamiltonian has only a kinetic term:

p2 1 p2 H = x + k x2 and H = y . x 2m 2 x y 2m In this case the y Hamiltonian’s eigenfunctions are simply plane waves:

2 2 2 2 ~ κy ~ κy H eiκyy = eiκyy or H |κ i = |κ i . y 2m y y 2m y We would then solve the eigenvalue equation

" 2 2 # ~ κy H |n κ i = ω n + 1 + |n κ i . x y ~ 2 x 2 2m x y

• Next, if both spring constants are zero with kx = ky = 0 (i.e. a free particle in two p2 dimensions) we have the purely kinetic Hamiltonian H = 2m . Both the x and y eigenfunctions are plane waves

2 2 2 2 κ ~ κy H eiκxx = ~ x eiκxx and H eiκyy = eiκyy, x 2m y 2m and the Schr¨odingerequation reads

" 2 2 2 2 # 2 2 κ ~ κy κ H |κ κ i = ~ x + |κ κ i or H |κi = ~ = |κi , x y 2m 2m x y 2m

where κ = (κx, κy).

• Finally, we consider the case kx = ky ≡ k > 0 when both spring constants are equal. The potential then reads 1 1 V (r) = k(x2 + y2) = kr2. 2 2 Note that the potential depends only on the distance r from the origin; such an oscillator is called isotropic.

q k Because kx = ky ≡ k we have ωx = ωy ≡ ω = m . The stationary Schr¨odinger then reads H |nxnyi = ~ω(nx + ny + 1) |nxnyi .

The system’s lowest possible energy occurs for nx = ny = 0, where the stationary Schr¨odingerequation reads H |00i = ~ω |00i. This is the ground state, and the energy E00 = ~ω is called the zero-point energy. The ﬁrst two excited states are

H |10i = ~ω |10i and H |01i = ~ω |01i . Note that the ﬁrst excited states is doubly degenerate—two linearly independent eigenfunctions |10i and |01i have the same energy. In general, the nth excited state of a 2D isotropic harmonic oscillator has degeneracy n + 1.

42 7.2 Two-Dimensional Isotropic Harmonic Oscillator github.com/ejmastnak/fmf

Theory: The z Component of Angular Momentum

• Eigenstates |ψi of the z component of angular momentum operator Lz satisfy

Lz |ψi = λ |ψi .

The operator Lz can be written ∂ L = xp − yp = − , z y x ~∂φ where φ is the azimuthal angle in the spherical coordinate system. Using the def- ∂ inition Lz = −~ ∂φ in spherical coordinates, the eigenvalue equation in coordinate notation reads 0 −i~ψ (φ) = λψ(φ). Separating variables and integrating the equation gives

iλ i λ φ ln ψ = φ + C =⇒ ψ(φ) = Ce˜ ~ . ~ To satisfy periodicity over a full rotation of 2π, we require ψ(φ + 2π) = ψ(φ), which implies the quantization

λ imφ = m; m ∈ Z =⇒ ψ(φ) = Ce˜ . ~ We ﬁnd the integration constant C˜ with the normalization condition

Z 2π 2 1 ∗ ˜ ˜ 1 ≡ ψ (φ)ψ(φ) dφ = 2π C =⇒ C = √ . 0 2π

The eigenfunctions of the angular momentum operator Lz are thus

1 imφ ψm(φ) = √ e , m ∈ Z. 2π In Dirac notation, the eigenfunctions are written simply |mi, where the reference to the angular momentum operator is usually clear from context. Using the relationship λ = hm, the eigenvalue equation in Dirac notation reads

Lz |ψi = m~ |ψi .

7.2 Two-Dimensional Isotropic Harmonic Oscillator Consider a two-dimensional isotropic harmonic oscillator. Within the subspace of energy eigenstates with energy E = 2~ω, ﬁnd those states that are also eigenstates of the z- component of angular momentum Lz.

• The energy E = 2~ω corresponds to the isotropic oscillator’s ﬁrst two excited states |10i and |01i. These states have degeneracy two, meaning the subspace is spanned by two linearly independent states. The subspace is formed of all normalized linear combinations of |01i and |10i, i.e. all |ψi for which

2 2 |ψi = α |10i + β |01i , α, β ∈ C, |α| + |β| = 1.

43 7.2 Two-Dimensional Isotropic Harmonic Oscillator github.com/ejmastnak/fmf

The Hamiltonian acts on |ψi according to

H |ψi = αH |10i + βH |01i = α2~ω |10i + β2~ω |01i = 2~ω α |10i + β |01i = 2~ω |ψi .

Evidently, ψ is an eigenfunction of H with energy eigenvalue 2~ω. We are interested in ψ that are also eigenfunctions of Lz, which must satisfy the eigenvalue equation

Lz |ψi = m~ |ψi .

• It is possible to ﬁnd functions that are simultaneously eigenfunctions of two operators if the two operators commute. Applied to our problem, if we show that H and Lz commute, we can be sure there exist functions that are eigenfunctions of both H and Lz. In other words, we want to show our problem is solvable by proving H,Lz = 0. We will ﬁrst write the Hamiltonian in polar coordinates:

p2 1 2 1 2 1 ∂ ∂ 1 ∂2 1 H = + kr2 = − ~ ∇2 + kr2 = − ~ r + + kr2. 2m 2 2m 2 2m r ∂r ∂r r2 ∂φ2 2

Using the deﬁnition of Lz in polar coordinates, the commutator reads 2 1 ∂ ∂ 1 ∂2 1 ∂ H,L = − ~ r + + kr2, −i z 2m r ∂r ∂r r2 ∂φ2 2 ~∂φ 2 ∂2 ∂ = − ~ , −i = 0. 2mr2 ∂φ2 ~∂φ

The entire proof rests on the commutativity of second derivatives, i.e.

∂2 ∂2 ∂2 ∂ ∂ ∂2 = and = . ∂φ∂r ∂r∂φ ∂φ2 ∂φ ∂φ ∂φ2 The derivatives with respect to φ then cancel the r-dependent terms in the Hamil- tonian. All we did here is show that eigenfunctions solving our problem exist in the ﬁrst place by showing H and Lz commute. • To actually solve the problem, we will take both a heuristic and a formal approach. We start with a heuristic approach. From the introductory theory section, the eigenfunctions solving H |10i = 2~ω |10i and H |01i = 2~ω |01i can be written in the separated form

ψ10(x, y) = ψ1(x)ψ0(y) and ψ01(x, y) = ψ0(x)ψ1(y).

Next, recall the annihilation operator action on the ground state produces

a |0i = 0 |0i .

In other words, |0i is an eigenstate of the annihilation operator (i.e. a coherent state) with eigenvalue z = 0. From the previous exercise set, we know that the eigenfunctions of coherent states are Gaussian wave packets. If z = 0, then hxi = 0 and hpi = 0, and the wavefunction would read

2 − x 1 2x2 ψ0(x) = e 0 . p4 2 πx0

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• We can get an expression for the ﬁrst excited state |1i with the creation operator a† |0i = |1i. In the coordinate representation of the operator, this reads † 1 x p 1 x ~ d ψ1(x) = a ψ0(x) = √ − i ψ0(x) = √ − ψ0(x) 2 x0 p0 2 x0 p0 dx 1 x d = √ − x0 ψ0(x). 2 x0 dx

Applied to the Gaussian wave packet ψ0(x), we have

" 2 # − x 1 x x0 d 1 2x2 1 x x0 x ψ1(x) = √ ψ0(x) − √ e 0 = √ ψ0(x) + √ ψ0(x) p4 2 2 2 x0 2 dx πx 2 x0 2 x0 √ 0 2 = xψ0(x). x0

We can then write the states ψ10 and ψ01 as

√ √ 2 2 − x − y 2 2 x 2x2 2x2 ψ10(x, y) = ψ1(x)ψ0(y) = xψ0(x)ψ0(y) = p e 0 e 0 x0 x0 πx2 √ √ 0 r2 r2 2 x − 2 2r cos φ − 2 = e 2x0 = e 2x0 , x p 2 p 2 0 πx0 x0 πx0 and, analogously, √ r2 2r sin φ − 2 ψ (x, y) = ψ (x)ψ (y) = e 2x0 . 01 0 1 p 2 x0 πx0 We ran out of time at this point; the problem is continued in the eight exercise set.

8 Eight Exercise Set

8.1 Two-Dimensional Isotropic Harmonic Oscillator (continued) We left oﬀ in the previous exercise set having found the harmonic oscillator’s states ψ10(x, y) and ψ01(x, y); these were √ √ r2 r2 2r cos φ − 2 2r sin φ − 2 ψ (x, y) = e 2x0 and ψ (x, y) = e 2x0 . 10 p 2 01 p 2 x0 πx0 x0 πx0 These are the two linearly independent states spanning the isotropic harmonic oscillator’s subspace of eigenstates with energy E = 2~ω. Our next step is to ﬁnd the linear combinations of ψ10 and ψ01 that are also eigenstates of the operator Lz.

• Recall that the eigenfunctions of Lz are

1 imφ ψm(φ) = √ e , m ∈ Z. 2π

Recall also the identities cos φ + i sin φ = eiφ and cos φ − i sin φ = e−iφ.

45 8.1 Two-Dimensional Isotropic Harmonic Oscillator . . . github.com/ejmastnak/fmf

• Comparing these identities to the deﬁnitions of ψ10, ψ01 and ψm(φ), we can construct the Lz eigenstate with m = 1 via 1 |m = 1i = √ |10i + i |01i , 2

where the √1 term ensures the state is normalized. To conﬁrm the state |m = 1i is 2 normalized, note that 1 hm = 1|m = 1i = h10| − i h01| |10i + i |01i 2 1 = h10|10i − i h01|10i + i h10|01i + h01|01i 2 1 = 1 − i + i + 1 = 1. 2 Finally, we construct the state with m = −1 with 1 |m = −1i = √ |10i − i |01i . 2

Formal Approach

• Consider an eigenstate |ψi that is both an eigenstate of H with E = 2~ω and an eigenstate of Lz. This state must satisfy

|ψi = α |10i + β |01i and Lz |ψi = λ |ψi .. First, we write the 2D Hamiltonian H in terms of the creation and annihilation operators:

† 1 † 1 H = ~ωx axax + 2 + ~ωy ayay + 2 ≡ Hx + Hy.

• For the isotropic oscillator with ωx = ωy, this simpliﬁes to † † H = ~ω axax + ayay + 1 .

More so, both the constants x0, y0 and p0x , p0y are equal for the x and y annihilation and creation operators; we have 1 x px 1 y py ax = √ + i and ay = √ + i , 2 x0 p0 2 x0 p0 q where x = ~ and p = ~ . 0 mω 0 x0

• The angular momentum operator is Lz = xpy − ypx; in terms of the annihilation and creation operators this becomes

x0 † p0 † x0 † p0 † Lz = √ ax + a · √ ay − a − √ ay + a · √ ax − a 2 x 2i y 2 y 2i x x p h i = 0 0 a + a† a − a† − a + a† a − a† . 2i x x y y y y x x

Applying x0p0 = ~ and multiplying out the terms in parentheses and combining like † † † terms (recall ax, ay = ax, ay = ax, ay = 0) simpliﬁes the expression for Lz to

L = ~ 2a† a − 2a† a = ~a† a − a† a . z 2i x y y x i x y y x

46 8.1 Two-Dimensional Isotropic Harmonic Oscillator . . . github.com/ejmastnak/fmf

† • Next, using the just-derived expression for Lz in terms of ax,y and ax,y, we investigate the action of Lz on our desired state |ψi; we have

L |ψi = −~αa |1i a† |0i + ~βa† |0i a |1i . z i x y i x y

Remember ax and ay act only on their respective components of |01i and |10i, which † can be factored√ into |0i |1i and |1i |0i. Finally, using the identities a |0i = |1i and † a |1i = 2 |2i, Lz |ψi simpliﬁes further to

L |ψi = −~α |0i |1i + ~β |1i |0i = ~ − α |01i + β |10i . z i i i

• To be an eigenstate of Lz, |ψi must satisfy the eigenvalue equation Lz |ψi = λ |ψi. Using our just-derived expression for Lz |ψi, the eigenvalue equation reads

L |ψi = λ |ψi =⇒ ~ − α |01i + β |10i ≡ λα |10i + β |01i . z i Multiplying (scalar product on the Hilbert space) the equation from the left by h10| and applying the orthonormality of |00i and |10i leaves

~β = λα, i while multiplying the equation from the left by h01| produces

−~α = λβ. i In matrix form, these two equations form a simple eigenvalue problem:

0 ~ α α i = λ . ~ − i 0 β β We solve the problem by ﬁnding the zeros of the characteristic polynomial:

−λ ~ det i ≡ 0 =⇒ λ2 − 2 = 0 =⇒ λ = ± . ~ ~ ~ − i −λ

Recall Lz’s eigenvalues are λ = m~, so we have λ = ~m = ±~ =⇒ m = ±1 (we had found the same solutions m = ±1 in the earlier heuristic approach).

• Substituting the m = 1 =⇒ λ = ~ solution into the matrix equation gives − ~ α ~ i = 0 =⇒ β = iα. ~ − i −~ β

47 8.1 Two-Dimensional Isotropic Harmonic Oscillator . . . github.com/ejmastnak/fmf

Substituting the m = 1 result β = iα into the general linear combination |ψi = α |10i + β |01i produces

|m = 1i = α |10i + (iα) |01i = α |10i + i |01i .

The normalization condition requires α = √1 , which gives 2 1 |m = 1i = √ |10i + i |01i , 2 which agrees with the m = 1 solution in the heuristic approach.

• Substituting the m = −1 =⇒ λ = −~ solution into the matrix equation gives + ~ α ~ i = 0 =⇒ β = −iα. ~ − i +~ β Analogously to the m = 1 case, substituting the m = −1 result β = −iα into the general linear combination |ψi = α |10i + β |01i produces

|m = −1i = α |10i + (−iα) |01i = α |10i − i |01i .

The normalization condition requires α = √1 , which gives 2 1 |m = −1i = √ |10i − i |01i , 2 which agrees with the m = −1 solution in the heuristic approach.

• The ﬁnal solution for the eigenstates of H with energy 2~ω that are also eigenstates of Lz is thus 1 |m = 1i = √ |10i + i |01i 2 1 |m = −1i = √ |10i − i |01i . 2

Why This Result is Useful

• Next, we ask why we would be interested in ﬁnding states that are simultaneously eigenstates of two operators (e.g. both H and Lz). As an example, consider a 2D isotropic harmonic oscillator in an external magnetic ﬁeld B. Its Hamiltonian is

H = H0 − µ · B,

where H0 is the Hamiltonian of a 2D isotropic harmonic oscillator without a magnetic ﬁeld and µ is the particle’s magnetic moment. The magnetic moment is proportional to the angular momentum, i.e. µ = γL, where γ is a constant proportionality factor.

• Assume the magnetic ﬁeld points in the z direction (i.e. B = Bzˆ) in which case the Hamiltonian simpliﬁes to H = H0 − γBLz.

48 8.2 Larmor Precession github.com/ejmastnak/fmf

In this case, the states that are simultaneously eigenstates of both H0 and Lz will also be eigenstates of the Hamiltonian H = H0 −γBLz, which is useful for analyzing particles in a external magnetic ﬁeld. For example, using our earlier solutions |m = ±1i, we have the eigenvalue equation

H |m = ±1i = H0 |m = ±1i − γBLz |m = ±1i = 2~ω |m = ±1i − γB(±1 · ~) |m = ±1i = (2~ω ± γB~) |m = ±1i .

In other words, |m = ±1i are indeed eigenstates of H, with eigenvalues (2~ω±γB~).

8.2 Larmor Precession Analyze the motion of a quantum-mechanical particle with magnetic moment µ and angular momentum quantum number l = 1 in an external magnetic ﬁeld B = Bzˆ, where µ forms an initial angle θ with B at t = 0.

• We consider a particle in an external magnetic ﬁeld B = Bzˆ whose magnetic moment µ makes an initial angle θ with the magnetic ﬁeld. The particle’s Hamiltonian is

H = −µ · B = −γL · B = −γBLz.

Recall the values of angular momentum are quantized. In general, a quantum particle with angular momentum quantum number l = 1 exists in a 2l+1 dimensional Hilbert space with basis |lmi, where m ∈ {−l, −l+1, . . . , l−1, l}. In our case l = 1, meaning m ∈ {−1, 0, 1}.

• In Cartesian coordinates, the angular momentum operator is L = (Lx,Ly,Lz). The 2 2 2 operator L acts on the state |lmi as L |lmi = h l(l + 1) |lmi. The operator Lz acts on |lmi as Lz |lmi = ~m |lmi. In other words, the basis |lmi is the basis of states that are eigenstates of the operator L2 with eigenvalue h2l(l + 1) and eigenstates of the operator Lz with eigenvalue ~m. In the coordinate wave function representation, the states |lmi are the spherical harmonics |lmi = Ylm(φ, θ).

iml The φ component of Y is proportional to √e . lm 2π • We return to the Hamiltonian of a particle with a magnetic moment in an external magnetic ﬁeld: H = −γBLz. For a particle with l = 1 (and thus magnitude of 2 2 2 angular momentum L = ~ l(l + 1) = 2~ ), the Hamiltonian’s eigenstates |ψi occurs in a 2l + 1 = 3 dimensional Hilbert space. In the |lmi basis, the general expression for |ψi is thus the linear combination

|ψi = a |11i + b |10i + c |1 − 1i ,

where l = 1 and m = 1, 0, −1.

• At t = 0, the particle’s magnetic moment forms a polar angle θ with the magnetic field B = Bzˆ. Our first step will be finding an appropriate expression for this initial condition in the language of quantum mechanics.

49 8.2 Larmor Precession github.com/ejmastnak/fmf

Finding the Initial State • In general, if some state |ψi is an eigenstate of the operator L·eˆ, (i.e. the projection of angular momentum L in the direction eˆ), and the operator acts on ψ as

L · eˆ|ψi = ~l |ψi , then we say the state |ψi’s angular momentum points in the direction eˆ. Back to the initial condition. We write L · eˆ|ψ, 0i = ~ |ψ, 0i where eˆ points in the initial direction of the magnetic moment µ. If we assume the magnitude moment initially lies in the x, z plane with φ = 0, the unit vector eˆ, in spherical coordinates, reads eˆ = (sin θ, 0, cos θ). To summarize, the quantum mechanical way of saying the particle’s magnetic moment forms an initial angle θ with the magnetic ﬁeld B = Bzˆ is

L · eˆ|ψ, 0i = ~ |ψ, 0i , where |ψ, 0i is the particle’s initial wave function and eˆ = (sin θ, 0, cos θ). • To ﬁnd |ψ, 0i, we need to ﬁnd the constants a, b, c in the linear combination |ψi = a |11i + b |10i + c |1 − 1i. Inserting eˆ into the condition L · eˆ|ψ, 0i = ~ |ψ, 0i gives

(Lx sin θ + Lz cos θ) |ψ, 0i = ~ |ψ, 0i . Substituting in the linear combination |ψi = a |11i+b |10i+c |1 − 1i and rearranging gives (Lx sin θ + Lz cos θ − h) a |11i + b |10i + c |1 − 1i = 0.

Theoretical Interlude: Action of Lx,y on |mli

• Recall the eigenvalue equation Lz |lmi = ~m |lmi for the operator Lz. We now ask how Lx and Ly act on the states |lmi. To do this, we introduce the new operators † L+ ≡ Lx + iLy and L− ≡ L+ = Lx − iLy.

In terms of L+ and L−, Lx and Ly read L + L L − L L = + − and L = + − . x 2 y 2i

The plan is to ﬁnd out how L+ and L− act on |lmi, and then reconstruct the action of Lx and Ly on |lmi using the relationship between L+,− and Lx,y.

• In our case (working with a particle with l = 1 and thus m ∈ {−1, 0, 1}), we have L+ |11i = 0 (nominally this would create a state with m = 2, outside the range of allowed m). In general, L+ |l, m = li = 0, since m cannot increase beyond l. √ √ Next, L+ |10i = 2~ |11i and L+ |1 − 1i = 2~ |10i. √ √ Meanwhile, L− produces L− |11i = 2~ |10i, L− |10i = 2~ |1 − 1i and L− |1 − 1i = 0, since m cannot decrease below l = 01. In general, L+ |l, m = −li = 0, since m cannot decrease below −l.

50 8.2 Larmor Precession github.com/ejmastnak/fmf

Back to Finding the Initial State

• We left oﬀ with (Lx sin θ + Lz cos θ − h) a |11i + b |10i + c |1 − 1i = 0.

Writing Lx,y in terms of L+,− and substituting the just-derived action of L+,− on |11i , |10i and |1 − 1i gives

L + L 0 = + − sin θ + L cos θ − h a |11i + b |10i + c |1 − 1i 2 z √ ! 2 = a ~ |10i sin θ + cos θ |11i − h |11i 2 ~ √ √ ! 2 2 + b ~ |11i sin θ + ~ sin θ |1 − 1i − |10i 2 2 ~ √ ! 2 + c ~ |10i sin θ − cos θ |1 − 1i − |1 − 1i . 2 ~ ~

Notice that this equation now involves only scalar values; all operators are gone.

• Next, we ﬁnd the constants a, b and c. First, divide through by ~, which cancels out from each equation. Next, multiplying the above equations by h11| and applying orthonormality of the states |11i, |10i, |1 − 1i, we have

b a cos θ − a + √ sin θ = 0. 2 Next, multiplying the equation by |10i gives a c √ sin θ − b + √ sin θ = 0. 2 2 Finally, multiplying the equation by |1 − 1i gives

b √ sin θ − c cos θ − c = 0. 2

• We’ve found a system of three equations for the three unknowns a, b and c. Recall from e.g. introductory linear algebra that because the system of equations is homogeneous (the right side of each equation is zero), the system has a solution only up to a free parameter. Solving the ﬁrst equation for a in terms of b gives b b sin θ a(cos θ − 1) + √ sin θ = 0 =⇒ a = √ . 2 2(1 − cos θ)

From the third equation, we get c in terms of b: b b sin θ √ sin θ − c(1 + cos θ) = 0 =⇒ c = √ . 2 2(1 + cos θ)

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• The solution is easier to find with sin θ and cos θ written in terms of double angle identities, which produce b θ b θ a = √ cot and c = √ tan . 2 2 2 2 • With a and c known in terms of the free parameter b, the initial condition on |ψ, 0i reads 1 θ 1 θ |ψ, 0i = b √ cot |11i + |10i + √ tan |1 − 1i . 2 2 2 2 We find b from the normalization condition, which, with the help of some trigonometric identities, reads ! 1 θ 1 θ cos4 θ + 2 cos2 θ sin2 θ + sin4 θ hψ, 0|ψ, 0i ≡ 1 = |b|2 cot2 + 1 + tan2 = |b|2 2 2 2 2 2 2 2 2 2 θ 2 θ 2 sin 2 cos 2 2 cos2 θ + sin2 θ |b|2 = |b|2 2 2 = ≡ 1. 2 θ 2 θ 2 θ 2 θ 2 sin 2 cos 2 2 sin 2 cos 2 Because of the |b|2, b is mathematically determined only up to a constant phase factor eiφ of magnitude 1; if we make the simple choice eiφ = 1, the normalization condition gives √ θ θ b = 2 sin cos . 2 2 √ θ θ • Substituting b = 2 sin 2 cos 2 into the expression for |ψ, 0i gives the desired result θ √ θ θ θ |ψ, 0i = cos2 |11i + 2 sin cos |10i + sin2 |1 − 1i . 2 2 2 2 To summarize: precisely this state |ψ, 0i satisfies the initial condition that a particle’s magnitude moment µ makes an angle θ with an external magnetic field B = Bzˆ. Time Evolution of the Initial State • With the initial state |ψ, 0i known, the next step is to find the time-dependent wave function |ψ, ti for a particle with magnetic moment µ in an external magnetic field. We’ll find the time evolution in the Schrödingerpicture using the particle’s Hamiltonian H = −γBLz. • We’ll first solve the stationary Schrödingerequation H |ψi = E |ψi for the basis functions |11i, |10i and |1 − 1i; this gives

H |11i = −γBLz |11i = −γB(1 · ~) |11i = −γB~ |11i H |10i = −γBLz |10i = −γB(0 · ~) |10i = 0 |10i H |1 − 1i = −γBLz |1 − 1i = −γB(−1 · ~) |1 − 1i = γB~ |1 − 1i .

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where in the last line we’ve deﬁned the Larmor frequency ωL ≡ γB. The time evolution is relatively simple to ﬁnd because the initial state |ψ, 0i was already written in the basis |11i, |10i, |1 − 1i.

• Next, using the wave function |ψ, ti, we will ﬁnd the expectation value hµ, ti of the particle’s magnetic moment at time t, which is related to angular momentum L via   hLx, ti hµ, ti = γ hL, ti = γ hLy, ti . hLz, ti

where the operator Lz acts on the basis states |11i, |10i and |1 − 1i to produce the eigenvalues ~, 0 and −~. Multiplying out the terms and applying the orthonormality of the basis states, followed by some trigonometric identities, leads to the considerably simpler expression θ θ θ θ θ θ hL , ti = cos4 − sin4 = cos2 + sin2 cos2 − sin2 z ~ 2 ~ 2 ~ 2 2 2 2 = ~ · (1) · (cos θ) = ~ cos θ.

In other words, the z component of angular momentum Lz is constant! We ran out of time at this point and continued in the ninth exercise set

9 Ninth Exercise Set

9.1 Larmor Precession (continued) Analyze the motion of a quantum-mechanical particle with magnetic moment µ and angular momentum quantum number l = 1 in an external magnetic ﬁeld B = Bzˆ, where µ forms an initial angle θ with B at t = 0. • In the last exercise set we found the particle’s time-dependent wave function θ √ θ θ θ |ψ, ti = cos2 eiωLt |11i + 2 sin cos |10i + sin2 e−ωLt |1 − 1i , 2 2 2 2

where ωL ≡ γB is the Larmor frequency. We left off with using |ψ, ti to find the expectation value hLz, ti, which turned out to be the constant quantity |Lz, ti = ~ cos θ. Next, we will use |ψ, ti to find Lx and Ly, which we will use to find the particle’s magnetic moment via |µ, ti = γ hLx, ti , hLy, ti , hLz, ti .

• It will be easiest to ﬁnd hLx, ti and hLy, ti indirectly, in terms of the operators L+ and L−; recall these are related by

L+ = Lx + iLy and L− = Lx − iLy L + L L − iL L = + − and L = + − . x 2 y 2i

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It follows that hLx, ti and hLy, ti can be found via 1 1 hL , ti = hL , ti + hL , ti and hL , ti = hL , ti − hL , ti . x 2 + − y 2i + − ∗ In fact, hL−, ti = hL+, ti is the complex conjugate of hL+, ti, so we only need to † ∗ ﬁnd hL+, ti, from which we can ﬁnd hL−, ti using the general identity O = hOi . We then have 1 hL , ti = hL , ti + hL , ti∗ = Re hL , ti x 2 + + + 1 hL , ti = hL , ti − hL , ti∗ = Im hL , ti . y 2i + + +

The angular momentum components hLx, ti and hLy, ti are then

hLx, ti = Re hL+, ti = ~ sin θ cos(ωLt) hLy, ti = Im hL+, ti = ~ sin θ sin(ωLt). In one place, the components of angular momentum are

hLx, ti = ~ sin θ cos(ωLt) hLy, ti = ~ sin θ sin(ωLt) hLz, ti = ~ cos θ.

In other words, analogously to the classical picture, the expectation values hLx, ti and hLy, ti precess around the direction of the external magnetic field B = Bzˆ with the Larmor frequency ωL = γB. • Recall that an physical measurement (observation) generally disturbs quantum mechanical systems. With this issue in mind, we’ll find what happens with an observation of Lz at the time t. Theoretical Interlude: Observations in Quantum Mechanics • We make an observation of a physical quantity in quantum mechanics as follows. First, expand the relevant system’s wave function in the basis consisting of the eigenfunctions of the operator corresponding to the quantity being measured (e.g. to measure a system’s Lz, expand the system wave function in the Lz basis |lmi.) • The possible outcomes of the observation are the discrete eigenvalues of the operator corresponding to the quantity being measured. When the wavefunction is a linear combination of multiple eigenstates, each of the states’ corresponding eigenvalues is a possible outcome, and the probability of each eigenvalue being the outcome is the squared absolute value of the corresponding eigenstate’s coefficient.

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Eigenstate Outcome Probability

2 θ −iωLt 4 θ |11i ~ cos 2 e = cos 2 2 θ 2 θ |10i 0 2 cos 2 sin 2 2 θ −iωLt 4 θ |1 − 1i −~ sin 2 e = sin 2 Table 1: Ingredients to consider in a measurement of an l = 1 Larmor-precessing particle’s angular momentum Lz.

• Finally, immediately after an observation yielding a given eigenvalue, the system’s wavefunction assumes (“collapses into”) the eigenfunction corresponding to the measured eigenvalue.

Back to the Observation of Lz

• Next, we consider the possible outcome of the observation of Lz. Our wavefunction is a combination of three states |11i, |10i and |1 − 1i; the corresponding Lz eigenvalues of these states are ~, 0 and −~, and the corresponding probabilities are the squared absolute values of the corresponding eigenstate’s coeﬃcients. Table1 summarizes the eigenstates, possible outcomes, and corresponding probabilities. • Because of wavefunction collapse to the measured eigenvalue’s corresponding eigenfunction, any measurement of Lz would cause the particle’s eigenfunction to stop precessing at a polar angle θ about the magnetic ﬁeld B = Bzˆ in agreement with classical Larmor precession.

For example, measuring Lz = ~ would cause the particle to assume the eigenfunction |11i, which directly points in the direction of the magnetic ﬁeld and zˆ axis. The lesson here is that even though the classical and quantum-mechanical results for a particle with magnetic moment µ in an external magnetic ﬁeld both display Lar- mor precession about the Bˆ direction, any observation of the quantum-mechanical observation of the system breaks this agreement because of wave function collapse.

9.2 Rashba Coupling Consider a particle in the two-dimensional x, y plane with magnitude of spin s = 1/2 and Hamiltonian p2 H = + λ(p S − p S ), 2m x y y x where p = (px, py, 0) and S = (Sx,Sy,Sz) are the momentum and spin operators and λ is a constant.1 Find the Hamiltonian H’s eigenfunctions and eigenvalues. Spin and Momentum States

• The Hamiltonian H acts on states in the combined Hilbert space H = Hp ⊗Hs where Hp is the momentum Hilbert space and Hs is the spin Hilbert space. Analogously, the states in H are |ψi = |ψpi ⊗ |ψsi.

1 The term λ(pxSy − pySx) is called the Rashba coupling term, which gives this problem its name.

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1 • Since we’re working with states with spin magnitude s = 2 and thus z component 1 1 ms ∈ − 2 , 2 , the spin wavefunction ψs is the linear combination 1 1 1 1 |ψsi = α + β − , 2 2 2 2 2 where |smsi is the basis of the spin operator S —s is the spin quantum number and ms is the quantum number corresponding to the projection of spin on the z axis. The spin operator S2 acts on the spin basis states to produce 2 1 1 2 1 1 1 1 2 3 1 1 S = ~ + 1 = ~ 2 2 2 2 2 2 4 2 2 2 1 1 2 1 1 1 1 2 3 1 1 S − = ~ + 1 − = ~ − . 2 2 2 2 2 2 4 2 2

• Meanwhile, the operator Sz acts on the basis states to produce 1 1 ~ 1 1 1 1 ~ 1 1 Sz = and Sz − = − − , 2 2 2 2 2 2 2 2 2 2 1 1 When working with particles with spin s = 1/2, we usually denote the states 2 2 1 1 and 2 − 2 with |↑i and |↓i, respectively. In this notation, our spin wavefunction is

|ψsi = α |↑i + β |↓i .

In our case, the spin Hilbert space Hs is two-dimensional, since it consists of all linear combinations of the two linearly independent states |↑i and |↓i.

• Next, on to the momentum states |ψpi, which correspond to the kinetic energy operator. The eigenstates of the kinetic energy operator are plane waves of the form eik·r. Because the wave vectors k are continuously distributed and k generates a unique set of eigenvalues, the position space Hp is infinitely dimensional. Finding Eigenfunctions and and Eigenvalues • To find the Hamiltonian’s eigenfunctions and eigenvalues, we need to solve the stationary Schrödingerequation H |ψi = E |ψi . First, to make the rest of the problem easier, we will show that H commutes with the operator p, which reduces to showing H commutes with px, py and pz. First: " # p2 + p2 [H, p ] = x y + λ(p S − p S ), p . x 2m x y y x x

2 2 The operators px and px commute, as do py and px, which takes care of the ﬁrst term. Next, we consider the Rashba coupling term (pxSy − pySx). Because the Sy,x operators act on the spin Hilbert space Hs and px acts on the independent momentum Hilbert space Hp, and because px commutes with both px and py, the operator px also commutes with the Rashba term (pxSy − pySx). It follows that

[H, px] = 0.

We can show [H, py] = 0 with an analogous procedure. Finally, since pz = 0, the identity [H, pz] = 0 is trivial—it isn’t even strictly necessary, since our particle’s momentum is conﬁned to the xy plane.

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• Next, recall that if two operators commute, it is possible to ﬁnd eigenfunctions that are simultaneously eigenfunctions of both operators (see e.g. the 2D harmonic oscillator problem in Subsection 7.2). In our case, because p = (px, py, 0) and H commute, p’s eigenfunctions will also be H’s eigenfunctions, and instead of ﬁnding H’s eigenfunctions, we can search for p’s eigenfunctions instead, which is easier.

• To ﬁnd p’s eigenfunctions, we would solve the equation

p |ψi = a |ψi or p |ψpi ⊗ |psi = a |ψpi ⊗ |psi .

I’ve used a instead of the conventional λ to denote the eigenvalue to avoid confusion with the Rashba coupling parameter λ. Since p acts only on the momentum Hilbert space Hp (and thus only on |ψpi and not |psi), we can immediately simplify to the purely momentum equation

p |ψpi = a |ψpi where p = −i~∇. Recall the momentum eigenfunctions are plane waves of the form eik·r, for which the eigenvalue equation in the coordinate wavefunction representation reads

ik·r ik·r ik·r ik·r pe = −i~∇e = −i~(ik)e = ~ke . In Dirac notation, the equation reads

p |ki = ~k |ki , where the eigenstate |ki corresponds to the wavefunction eik·r and the eigenvalue is ~k = ~(kx, ky).

• Returning to the stationary Schr¨odingerequation and using |ψpi = |ki we have H |ψi = H |ψpi ⊗ |ψsi = H |ki ⊗ |ψsi = E |ki ⊗ |ψsi .

Again, the replacement of |ψpi with k in the Schrödinger equation is possible only because H and p commute. We have thus reduced our problem to finding spin wavefunctions |ψsi. Notation: I practice, we drop the tensor product symbol ⊗; in this convention the above stationary Schrödingerreads H |ki |ψsi = E |ki |ψsi .

p2 H |ki |ψ i = + λ(p S − p S ) |ki |ψ i s 2m x y y x s p2 = |ki |ψ i + λp |ki S |ψ i − p |ki S |ψ i 2m s x y s y x s 2k2 = ~ |ki |ψ i + λ k |ki S |ψ i − λ k |ki S |ψ i 2m s ~ x y s ~ y x s ≡ E |ki |ψsi .

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We’re now left with an equation containing only the spin wavefunction |ψsi. Theoretical Interlude: Spin Ladder Operators

• Recall that for a particle with spin s = 1/2, the function |ψsi is an element of the two-dimensional spin Hilbert space Hs and is written as the linear combination

|ψsi = α |↑i + β |↓i .

• Just like it is more convenient to write the orbital angular momentum components Lx and Ly in terms of the ladder operators L+ and L−, it is best to analyze Sx and Sy in terms of the analogously deﬁned operators S+ and S−, which read

S+ = Sx + iSy and S− = Sx − iSy S + S S − S S = + − and S = + − . x 2 y 2i

Back to Our Problem • First, some useful intermediate results: 1 1 S+ |↑i ≡ S+ 2 2 = 0 1 1 S+ |↓i ≡ S+ 2 − 2 = ~ |↑i 1 1 S− |↑i ≡ S− 2 2 = ~ |↓i 1 1 S− |↓i ≡ S− 2 − 2 = 0. The ﬁrst and last identities hold from a physical interpretation since m cannot increase beyond s = 1/2 or decrease below s = −1/2, respectively.

• In terms of S+ and S− and using the above intermediate results, Sx and Sy act on |ψsi to produce S + S S |ψ i = + − α |↑i + β |↓i = ~α |↓i + β |↑i x s 2 2 S − S S |ψ i = + − α |↑i + β |↓i = ~ − α |↓i + β |↑i . y s 2i 2i

• We can now evaluate the earlier eigenvalue equation for |ψsi, i.e. 2k2 ~ |ψ i + λ (k S − k S ) |ψ i = E |ψ i . 2m s ~ x y y x s s

Using the just-derived action of Sx and Sy act on |ψsi, we have 2k2 λk 2 λk 2 ~ α |↑i+β |↓i + x~ −α |↓i+β |↑i − y~ α |↓i+β |↑i = Eα |↑i+β |↓i . 2m 2i 2 Note that this equation contains only scalar values and the two linearly independent states |↓i and |↑i—all operators are gone from the equation.

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• Multiplying the equation from the left by h↑| and applying the orthonormality of |↓i and |↑i produces 2k2 λ 2k λ 2k ~ α + ~ x β − ~ y β = Eα, 2m 2i 2 while multiplying the equation from the left by h↓| produces 2k2 λ 2k λ 2k ~ β − ~ x α − ~ y α = Eβ. 2m 2i 2 We have a system of two equations for the two unknown coeﬃcients α and β; in matrix form this system reads

" 2 2 2 # ~ k λ~ 2m 2i (kx − iky) α α 2 2 2 = E , λ~ ~ k β β − 2i (kx + iky) 2m which is an (2 × 2) eigenvalue problem.

• Before tackling the eigenvalue problem, we’ll write the 2D wave vector k = (kx, ky) in polar coordinates:

iφ −iφ kx + iky = ke and kx − iky = ke . With k written in polar coordinates, the eigenvalue equation reads

2 2 2 " ~ k λ~ −iφ# 2m 2i ke α α 2 2 2 = E . λ~ iφ ~ k β β − 2i ke 2m We ﬁnd the energy eigenvalues from the zeros of the characteristic polynomial

2 2 2 " ~ k λ~ −iφ# 2 2 2 2 2 2m − E 2i ke ~ k λ~ k p(E) = det 2 2 2 = − E − . λ~ iφ ~ k 2m 2 − 2i ke 2m − E We want to ﬁnd the E such that p(E) = 0, i.e. solving

2k2 2 λ 2k 2 ~ − E = ~ . 2m 2 Squaring both sides and solving for E gives 2k2 λ 2k E = ~ ± ~ . ± 2m 2 Note that the case λ = 0, which corresponds to zero coupling, recovers the free- 2 2 ~ k particle energy E = 2m . In this case we have energy degeneracy 2 for each k, since the two linearly independent eigenfunctions corresponding ±k have the same energy 2 2 ~ k eigenvalue E = 2m . For λ 6= 0, in the presence of Rashba coupling, the degeneracy is broken, since the states with ±k have diﬀerent energies E±, shifted up or down from the free particle 2 λ~ k energy by the quantity 2 . • Next, we consider the eigenstates. In the degenerate case λ = 0, we have 2k2 H |ki α |↑i + β |↓i = ~ α |↑i + β |↓i . 2m

for each α and β. In other words, when λ = 0, any linear combination |ψsi = 2 2 ~ k α |↑i + β |↓i is an eigenstate of the Hamiltonian with the energy eigenvalue 2m .

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• When λ 6= 0, only two unique linear combination of |↑i and |↓i will be eigenstates of the Hamiltonian; we ﬁnd these states by ﬁnding the eigenvectors of the earlier eigenvalue problem 2 2 2 " ~ k λ~ −iφ# 2m − E 2i ke α 2 2 2 = 0. λ~ iφ ~ k β − 2i ke 2m − E 2 2 2 ~ k λ~ k Substituting in the two energy eigenvalues E± = 2m ± 2 gives

2 2 " λ~ k λ~ −iφ# ∓ 2 2i ke α 2 2 = 0. λ~ iφ λ~ k β 2i ke ∓ 2 Multiplying the matrices leads to the equation

λ 2k λ 2k ∓ ~ α + ~ e−iφβ = 0 =⇒ β = ±αieiφ. 2 2i The two eigenvectors are thus

α 1 α 1 1 = α or, when normalized, = √ . β ±ieiφ β 2 ±ieiφ

The corresponding eigenfunctions for the eigenvalues E± are thus

|↑i + ieiφ |↓i |↑i − ieiφ |↓i |ψ+i = |ki √ and |ψ−i = |ki √ . 2 2 Next, we will analyze the just-derived wavefunctions.

Theoretical Interlude: Orientation of Spin in Space

• If the projection S · eˆ of a particle’s spin S in the direction eˆ acts on the spin ~ particle’s wavefunction |ψsi to produce S · eˆ|ψsi = 2 |ψsi, then we say the particle’s spin points in the direction eˆ.

• If we write eˆ in spherical coordinates as

eˆ = cos ϕ sin θ, sin ϕ sin θ, cos θ,

In other words, if we know a particle’s spin wavefunction |ψsi and can deduce the values of ϕ and θ by comparing the particle’s wavefunction to the above equation, we can determine the spatial orientation eˆ = cos ϕ sin θ, sin ϕ sin θ, cos θ of the particle’s spin.

Back to Our Problem

• In our case, the particle has the spin wavefunctions

1 1 iφ 1 1 i(φ± π ) ψs = √ |↑i ± √ ie |↓i = √ |↑i + √ e 2 |↓i . ± 2 2 2 2

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By comparing ψs± to the above general expression for a spin wave function, i.e. θ θ |ψ i = cos |↑i + sin eiϕ |↓i , s 2 2 we see that our particle has polar angle θ = π (since θ = π =⇒ cos θ = cos π = √1 2 2 2 4 2 reproduces the |↑i coeﬃcient √1 ) and azimuthal angle ϕ = φ ± π , where φ is the 2 2 angle in the phase term eiφ (φ was the azimuthal angle in the polar-coordinate representation of the wave vector k = (k, φ)). In both cases, the particle’s spin is perpendicular to wave vector k.

10 Tenth Exercise Set

10.1 Rashba Coupling with Pauli Spin Matrices Use the Pauli spin matrices to ﬁnd the energy eigenvalues and eigenfunctions of a particle in the two-dimensional x, y plane with magnitude of spin s = 1/2 and Hamiltonian p2 H = + λ(p S − p S ), 2m x y y x where p = (px, py) and S = (Sx,Sy,Sz) are the momentum and spin operators and λ is a constant. Quick Review of Previous Solution • In the ninth exercise set, we solved this same problem without using spin matrices. For a particle with spin s = 1/2, the spin wavefunctions occur in a 2D spin Hilbert space and take the general form

|ψsi = α |↑i + β |↓i .

We then showed the commutation relation [H, p] = 0 which allowed us to make the wavefunction factorization |ψi = |ki |ψsi . We substituted this factorization in the Schr¨odingerequation H |ψi = E |ψi and ended up with the (2 × 2) eigenvalue problem

" 2 2 2 # ~ k λ~ 2m 2i (kx − iky) α α 2 2 2 = E , λ~ ~ k β β − 2i (kx + iky) 2m which we solved for the Hamiltonian’s energy eigenvalues and eigenfunctions. Theory: The Pauli Spin Matrices • Today we will use the Pauli matrix representation of the spin operator, which is

S = ~σ, 2

where σ = (σx, σy, σz); σx, σy, and σz are called the Pauli spin matrices. In the standard |↑i |↓i basis for particles with spin s = 1/2, the spin matrices are

0 1 0 −i 1 0 σ = σ = σ = . x 1 0 y i 0 z 0 −i

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Recall our problem’s spin wavefunctions took the general form |ψsi = α |↑i + β |↓i. When working with Pauli matrices, we represent the wavefunction with the vector [α, β]T , which we call a spinor. Solution Using Spin Matrices • In terms of the spin matrices, our problem’s Rashba-coupled Hamiltonian reads p2 H = + λ~(p σ − p σ ), 2m 2 x x y x

and the wavefunction |ψi = |ki |ψsi is, in the coordinate representation, α ψ = eik·r . β The stationary Schr¨odingerequation Hψ = Eψ reads 2k2 α λ α α Hψ = ~ eik·r + ~ ( k σ − k σ ) eik·r ≡ Eψ = Eeik·r . 2m β 2 ~ x y ~ y x β β If we cancel eik·r from both sides of the equation, we end up with 2k2 α λ 2 α α ~ + ~ [k σ − k σ ] = E . 2m β 2 x y y x β β

• If we insert the deﬁnitions of the Pauli matrices σy and σx and multiply out the T matrices σy and σx with the spinor α, β , we end up with the matrix equation

" 2 2 2 # ~ k λ~ 2m 2i (kx − iky) α α 2 2 2 = E . λ~ ~ k β β − 2i (kx + iky) 2m This is exactly the same 2×2 eigenvalue problem as in the previous exercise set. The procedure from here forward is same as in the previous problem—we would solve the eigenvalue problem for the Hamiltonian’s energy eigenvalues and eigenfunctions as before. The lesson is that when working with particles with s = 1/2, we are able to use the Pauli spin matrix representation of the spin operator to derive the same eigenvalue problem as in the previous Hilbert space approach.

10.2 Theory: Time Reversal Symmetry • A problem is symmetric under time reversal if the problem’s Hamiltonian H obeys [H, T ] = 0 where T is time reversal operator. The time reversal operator is anti- unitary, i.e. ∗ hT ψ1|T ψ2i = hψ1|ψ2i , where ∗ denotes complex conjugation. • From linear algebra, recall that unitary operators preserve the inner product, i.e.

hUψ1|Uψ2i = hψ1|ψ2i . If we move U from the ket to the bra term, the equality reads † U Uψ1 ψ2 = hψ1|ψ2i , which implies a unitary operator must satisfy U †U = I.

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• Next, we introduce the complex conjugation operator K : ψ 7→ ψ∗. We will now show that if T is written as the product of a unitary operator U and the conjugation operator K, then T is anti-unitary:

which results in the anti-unitary identity

∗ hT ψ1|T ψ2i = hψ1|ψ2i .

• Finally a theorem: for spin 1/2 particles, the time-reversal operator takes the form

T = iσyK.

In this case (comparing to the general form T = UK) we have U = ıσy. We can show that T is anti-unitary by proving that U = iσy is unitary. The proof reads:

† 2 U†U = σyiσy = σy(−i)iσy = σy = I.

Since U = iσy is indeed unitary, T = UK is anti-unitary.

10.3 Time-Reversal Symmetry and Rashba Coupling Analyze the Rashba-coupled Hamiltonian from the previous two problems, i.e.

p2 λ H = + ~(p σ − p σ ), 2m 2 x y y x using the formalism of time-reversal symmetry for a particle with spin s = 1/2.

• First, we will show the problem’s Hamiltonian is invariant under time reversal, i.e. [H, T ] = 0. We can rewrite the commutation condition in the form HT = T H and multiply by T −1 from the left to get

H = T HT −1.

We will show H and T commute by proving H = T HT −1.

• Recall from the theory section that for particles with spin s = 1/2, the time reversal operator is T = iσyK, where σy is the second Pauli spin matrix and K is the complex conjugation operator. Substituting T = iσyK into the commutation condition gives

−1 −1 −1 −1 T HT = (iσyK)H(K σy i ).

2 −1 −1 We simplify this expres sion by noting that K = I =⇒ K = K; σy = σy ; and i−1 = −i, which results in

−1 ∗ T HT = (iσyK)H(Kσy(−i)) = iσyH K(Kσy(−i)) † ∗ = iσyH KKσy(−i) = iσyH σy(−i) ∗ = σyH σy.

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Substituting in the Rashba-coupled Hamiltonian H gives

p2 λ ∗ T HT −1 = σ + ~(p σ − p σ ) σ y 2m 2 x y y x y p2 = σ + λ~ ((−p )(−σ ) − (−p )(σ )) σ . y 2m 2 x y y x y

2 ∗ 2 ∗ ∗ ∗ where we have used (p ) = p ; px,y = −px,y; σy = −σy and σx = σx. Multiplying 2 out the above expression and applying σy = I and σxσy = −σyσx gives

p2 λ p2 λ T HT −1 = σ2 + ~ p σ3 + p σ σ σ = + ~ (p σ + p (−σ σ )σ ) 2m y 2 x y y y x y 2m 2 x y y x y y p2 λ = + ~ (p σ − p σ ) = H. 2m 2 x y y x Because T HT −1 = T , H is invariant under time reversal.

HT |ψi = T H |ψi = T E |ψi = ET |ψi .

In other words, both |ψi and T |ψi are eigenfunctions of H.

• Next, we consider the projection of |ψi onto |T ψi, i.e. hψ|T ψi. Since T is anti- unitarity, we have

∗ 2 ∗ hψ|T ψi = hT ψ|T (T ψ)i = T ψ T ψ .

2 We ﬁnd T using the spin 1/2 identity T = iσyK:

2 2 T = iσyKiσyK = iσy(−i)(−σy)K = −σyI = −I.

Substituting T 2 = −I into the inner product hψ|T ψi gives

2 ∗ ∗ ∗ hψ|T ψi = T ψ T ψ = hT ψ|−ψi = − hT ψ|ψi = − hψ|T ψi ,

• To summarize: For a spin 1/2 particle (which has T 2 = −I) whose Hamiltonian is invariant under time reversal, if |ψi is a solution of the stationary Schr¨odinger equation H |ψi = E |ψi , then the wavefunction T |ψi also solves the stationary Schr¨odinger equation with the same energy eigenvalue, and |ψi is orthogonal to T |ψi. This means every eigenstate of a T -invariant spin 1/2 Hamiltonian has degeneracy of at least two, since we have two linearly independent solutions |ψi and T |ψi for each energy eigenvalue. This result is called Kramers degeneracy, and (|ψi , T |ψi) are called a Kramers pair.

64 10.4 Scattering on a Spin-Dependent Delta Function github.com/ejmastnak/fmf

• Next, referring to the solutions from the previous two problem, we will identify the Kramer’s pairs in the energy eigenstates of the Rasbha-coupled Hamiltonian. Recall that for a given k, the Hamiltonian has energy eigenvalues

2k2 λ 2k E = ~ ± ~ , ± 2m 2 with corresponding eigenfunctions

|ki iφ |ψ±i = √ |↑i ± ie |↓i , 2 where we wrote the wave vector k in polar coordinates as k = (k, φ).

• We consider the solution ψ+. Theory predicts that the Kramer’s pair T |ψ+i is also an energy eigenstate with the same eigenvalue and is orthogonal to |ψ+i—we’ll see if this is the case. Writing |ki in the coordinate wavefunction representation and working in terms of spinors, the Kramer’s pair T |ψ+i is

∗ |ki iφ ik·r α 0 −i −ik·r α T |ψ+i = T √ |↑i + ie |↓i = (iσyK)e = i e 2 β i 0 β∗ 0 1 α∗ β∗ = e−ik·r = e−ik·r . −1 0 β∗ −α∗

iφ Returning to Dirac notation and noting that α = √1 and β = ie√ , we have 2 2

−iφ ∗ ∗ −ie 1 T |ψ+i = |−ki (β |↑i − α |↓i) = |−ki √ |↑i − √ |↓i 2 2 |↑i − ieiφ |↓i = −ie−iφ |−ki √ . 2

The phase factor −ie−iφ has magnitude one; since a wavefunction is determined only up to a constant phase factor of magnitude one, the above is equivalent to

|↑i − ieiφ |↓i T |ψ+i = |−ki √ 2 |↑i + iei(φ+π) |↓i = |−ki √ , 2

iπ where the second line uses the identity −i = ie . Recall the original eigenstate |ψ+i has a phase angle φ, corresponding to the direction of k and spin up. It’s Kramer pair T |ψ+i has phase angle φ+π, corresponding to the direction −k and spin down.

10.4 Scattering on a Spin-Dependent Delta Function

Consider a spin-dependent scattering problem with two particles: particle 1 with spin s1 = 1/2 in the state |↑i and particle 2 with spin s2 = 1 in the state |10i. The particles have the Hamiltonian 2 p1 λ H = − S1 · S2δ(x). 2m ~2 Particle 2 is ﬁxed at the origin at x = 0 and particle 1 approaches and scatters oﬀ particle 2 from the region x → −∞. Determine the particle’s spins after the collision.

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• First, we’ll establish the problem’s Hilbert space and choose an appropriate basis. The total Hilbert space is H = Hp ⊗H1 ⊗H2. Hp is the momentum space of the first particle and H1,2 are the spin spaces of the first and second particles, respectively. Note that the second particle’s Hilbert space doesn’t have a momentum component because particle 2 is fixed at the origin and doesn’t move.

The ﬁrst particle has spin s = 1/2, so its spin Hilbert space H1 is two-dimensional, and we will use the standard basis |↑i and |↓i. The second particle has spin s = 1 and thus a three-dimensional space; we’ll use the basis functions |11i, |10i and |1 − 1i.

• Next, we determine a basis for the combined spin space H1 ⊗ H2. There are two standard options:

2 2 1. A product basis, which uses commutativity of S1 , S1z, S2 , and S2z, which implies the existence of eigenfunctions that are simultaneously eigenfunctions of all four operators. The product basis consists of the states

|↑i |11i , |↑i |10i , |↑i |1 − 1i , |↓i |11i , |↓i |10i , |↓i |1 − 1i .

This set is just all possible combinations of the products of the basis functions of the individual spin spaces H1 and H2. 2. The other possible basis is the total angular momentum basis, which works in terms of the total spin operator S = S1 + S2. This basis uses commutativity 2 2 2 of S , Sz, S1 , and S2 . In general, the total spin operator obeys |s1 − s2| ≤ s ≤ s1 + s2. In our case 1 3 with s1 = 1/2 and s2 = 1, the above relationship implies s ∈ 2 , 2 . In our case the total angular momentum basis would be

3 3 1 3 1 1 3 1 1 3 3 1 1 1 1 1 1 1 2 , 2 2 1 , 2 , 2 2 1 , 2 , − 2 2 1 2 , − 2 2 1 , 2 , 2 2 1 , 2 , − 2 2 1 .

By convention, because s1 = 1/2 and s2 = 1 remain ﬁxed throughout the problem, we drop the last two terms 1/2 and 1 from each basis state and write the basis functions as

3 3 3 1 3 1 3 3 1 1 1 1 2 , 2 , 2 , 2 , 2 , − 2 2 , − 2 , 2 , 2 , 2 , − 2 . We ran out of time at this point and continue the problem in the next exercise set.

11 Eleventh Exercise Set

11.1 Scattering on a Spin-Dependent Delta Function (continued)

Consider a spin-dependent scattering problem with two particles: particle 1 with spin s1 = 1 1 1/2 in the state |↑i ≡ 2 , 2 and particle 2 with spin s2 = 1 in the state |10i. The particles have the Hamiltonian 2 p1 λ H = − S1 · S2δ(x). 2m ~2 Particle 2 is ﬁxed at the origin at x = 0 and particle 1 approaches and hits particle 2 from the region x → −∞. Determine the particle’s spins after the collision.

• Review from the last exercise set: we had just begun the problem, and established the two possible bases to describe the problem—the product basis and the total angular momentum basis.

66 11.1 Scattering on a Spin-Dependent Delta Function . . . github.com/ejmastnak/fmf

• Next we’ll review the problem of scattering in a spinless delta potential (from the second exercise set) where the Hamiltonian is

p2 H = + λδ(x). 2m We describe free scattering states with energy E > 0 with the scattering matrix

r t0 1 −iκ k S = = , t r0 k + iκ k −iκ

where r 2mE mλ k = and κ = . ~2 ~2 If we send a particle in towards the delta potential from the left, with incidence vector A = (1, 0), the particle is partly reflected and partly transmitted, according to ψ(x < 0) = eikx + re−ikx and ψ(x > 0) = teikx. Recall that the scattering matrix S is unitary and thus obeys S†S = I. Finally, recall the coefficient parameters obey |t|2 + |r|2 = T + R = 1, where T and R are the probability of transmission and reflection, respectively.

• From the last exercise set, recall the system’s Hilbert space is

H = Hp ⊗ H1 ⊗ H2,

where Hp is the first particle’s momentum space, and H1 and H2 are the first and second particle’s two and three dimensional spin Hilbert spaces, respectively. The combined spin Hilbert space Hs = H1 ⊗ H2 is six-dimensional. We identified two possible bases for use with our problem: The first is a product 2 2 basis using the commutativity of S1 , Sz1 , S2 and Sz2 .

Nominally, the states in this basis are written |s1m1s2m2i, but because the particle spins s1 and s2 are ﬁxed at 1/2 and 1, respectively, we adopt a shorthand convention and write the states simply as |m1m2i, with s1 = 1/2 and s2 = 1 implicit. • The second basis, called the total angular momentum basis, uses the commutativity 2 2 of S2, Sz, S1 and S2 . Recall total spin is the vector sum S = S1 + S2. We write the states in this basis as

|ss1s2mi or, in shorthand, |smi .

Again, by convention, we leave out the s1 and s2 for conciseness because in this problem their values are ﬁxed at s1 = 1/2 and s2 = 1, respectively. We transform between the bases using the Clebsch-Gordan coeﬃcients.

End of Review, Start of New Material

• Recall the Hamiltonian for our spin-dependent scattering problem is

2 p1 λ H = − S1 · S2δ(x1). 2m ~2

67 11.1 Scattering on a Spin-Dependent Delta Function . . . github.com/ejmastnak/fmf

First, we’ll rewrite the dot product in terms of total spin S:

2 2 2 S = S1 + S2 =⇒ S = S1 + 2S1S2 + S2 ,,

which implies S2 − S2 − S2 S · S = 1 2 . 1 2 2

Substituting this expression for S1 · S2 into the Hamiltonian gives

2 p1 λ 2 2 2 H = − (S − S1 − S2 )δ(x). 2m 2~2 • Next note that the Hamiltonian H commutes with the spin operators:

2 2 2 [H,S ] = [H,S1 ] = [H,S2 ] = [H,Sz] = 0.

Because H commutes with all of the spin basis operators, we can ﬁnd eigenfunctions that are eigenfunctions of both H and all of the spin operators simultaneously.

• Next, we solve the stationary Schr¨odingerequation:

H |ψi = E |ψi .

Because H and the spin operators commute, we can factor the system’s wavefunction into a momentum and spin portion of the form |ψi = |ψ1i ⊗ |smi, which gives

2 p1 λ 2 2 2 H |ψ1i ⊗ |smsi = |ψ1i ⊗ |smsi − δ(x1) |ψ1i ⊗ (S − S1 − S2 ) |smsi 2m 2~2 = E |ψ1i ⊗ |smsi .

We stress that the factorization |ψi = |ψ1i ⊗ |smsi is possible only because H commutes with each spin operator.

• The spin operators S, S1 and S2 then act on their corresponding eigenstates to give

= E |ψ1i ⊗ |smsi .

Conveniently, |smsi cancels from both sides of the equation. Substituting in s1 = 1/2 and s2 = 1 produces the purely momentum problem

p2 λδ(x) 3 1 − s(s + 1) − − 2 |ψ i = E |ψ i , 2m 2 4 1 1

where |ψ1i is the ﬁrst particle’s momentum wavefunction. Note that this equation depends only on the size s of the total spin, not on the z components ms. For our problem with s1 = 1/2 and s2 = 1, the identity |s1 − s2| ≤ s ≤ s1 + s2 means there are only two possible values for s; these are s = 1/2 and s = 3/2.

68 11.1 Scattering on a Spin-Dependent Delta Function . . . github.com/ejmastnak/fmf

• For s = 3/2 we have p2 λδ(x) 1 − |ψ i = E |ψ i . 2m 2 1 1 For s = 1/2 we have p2 1 + λδ(x) |ψ i = E |ψ i . 2m 1 1 To review: using the commutation of H and the spin operators, we have transformed our problem to a problem in only the momentum Hilbert space. All spin dependence is gone; the eigenvalues depend only on the delta function parameter λ. Bound States • A delta function potential has a bound state only for a downward-pointing delta function, which occurs if the delta function’s coefficient is negative. Looking at the above equations for s = 1/2 and s = 3/2, we see we have a bound state only for s = 3/2, where the δ(x) term has a negative coefficient. For s = 1/2, the δ function has a positive coefficient, and there is no bound state. • For conciseness, we define the constants λ λ = and λ = −λ. 3/2 2 1/2 The Hamiltonians for s = 3/2 and s = 1/2 then read p2 p2 H = 1 − λ δ(x) and H = 1 − λ δ(x). 3/2 2m 3/2 1/2 2m 1/2 • Recall (e.g. from the second exercise set) that a bound state in a delta potential V (x) = −λδ(x) has the wavefunction

√ −κ|x| mλ ψ0(x) = κe , where κ = , ~2 mλ2 and the corresponding energy eigenvalue E0 = 2 . 2~ • For the s = 3/2 problem, which has a bound state, the quantity κ is

mλ3/2 mλ κ3/2 = = , ~2 2~2 and the corresponding bound eigenfunctions are

ms p −κ3/2|x| 3 |ψ0 i = κ3/2e 2 ms .

This energy eigenvalue E0 is four-times degenerate because four linearly independent ms 3 1 1 3 eigenfunctions |ψ0 i (one each for ms ∈ − 2 , − 2 , 2 , 2 ) all have the same energy 2 mλ3/2 eigenvalue E0 = 2 . 2~ • Side Note: Technically, we are guilty of a slight abuse of notation. Namely, we’ve used both coordinate and Dirac notation when writing the wavefunction

69 11.1 Scattering on a Spin-Dependent Delta Function . . . github.com/ejmastnak/fmf

Free Scattering States

• Having discussed the single bound state for s = 3/2, we now consider the free scattering states. For our problem (reusing the results from the delta function scattering problem in the second exercise set), the scattering matrix is rs ts 1 −iκs k Ss = = , ts rs k + iκs k −iκs

where we have introduced a subscript s to make the dependence of r, t and κ on the value of total spin s explicit. The possible values of κs for s ∈ {1/2, 3/2} are

mλ3/2 mλ mλ1/2 mλ κ3/2 ≡ = and κ1/2 ≡ = − . ~2 2~2 ~2 ~2 • For a particle incident on the origin from the left, the scattering wavefunction on x ∈ (−∞, 0) (to the left of the delta potential) takes the general form:

− ikx −ikx ψb = e + rse |smsi ,.

The subscript b indicates this is a basis function for scattering states on the region x < 0. Note that the ansatz contains the incident wave eikx and the reflected wave re−ikx. Meanwhile, the wavefunction on x ∈ (0, ∞) (to the right of the delta potential) is + ikx ψb = tse |smsi , ikx where tse is the transmitted wave. Note the explicit dependence of t and r on the particle’s spin s. The above two wavefunctions, which are written in the total angular momentum basis |smsi, form a complete basis of the Hilbert space of free scattering states. In the remainder of the problem, we will use these basis states to construct the initial and final scattering states in our specific problem.

• Just to review what the problem asks for: particle 1, in state |↑i, is incident from the left on particle 2, which is ﬁxed at the origin and in state |0i, and we are interested in the scattered spin states of the two particles. The incident part of the wavefunction takes the form

ikx |ψini = e |↑i |0i .

Note that this wavefunction is written in the product basis—it is the product of the ﬁrst particle’s spin state |↑i and the second particle’s spin state |0i. However, the − + above eigenfunctions ψb and ψb spanning the Hilbert space of free scattering are written in the total angular momentum basis. At this point, we need a way to transform the initial state from the product basis to the total angular momentum basis—we make the transformation using the Clebsch- Gordan (CG) coeﬃcients.

Using the Clebsch-Gordan Coeﬃcients

• First, using a table of CG coeﬃcients, ﬁnd the sub-table whose heading has the correct combination s1 × s2 of spin numbers for each particle. In our case, our particles have spins s1 = 1/2 and s2 = 1, so we use the 1 × 1/2 table, shown in Figure1 (note that the order of s1 and s2 is not important in this step.

70 11.1 Scattering on a Spin-Dependent Delta Function . . . github.com/ejmastnak/fmf

Figure 1: The Clebsch-Gordan coeﬃcients for two particles with spins s1 = 1/2 and s2 = 1.

• We start with the initial incident spin state |↑i |0i ≡ |1/2i |0i, which is written in the product basis |m1m2i. In the CG table, we find the row containing 0 and +1/2. The row immediately to the right contains the numbers 2/3 and −1/3—these are coefficients. Each coefficient has an implicit square root—for example −1/3 means −p1/3—but the square roots are not written for conciseness. For each coefficient, in our case 2/3 and −1/3, we find the column in the box immediately above. For the coefficient 2/3, the column above is 3/2, +1/2, and for the coefficient −1/3, the column above is 1/2, +1/2. The columns represent states in the total angular momentum basis—the top row, in this case 3/2 and 1/2, represents values of s, and the bottom row, in this case +1/2 and +1/2, represents values of ms. Put together, each column is a basis state |smsi. We assemble the transformation between the product and total angular momentum basis as follows: q q 1 2 3 1 1 1 1 |↑i |0i ≡ 2 |0i → 3 2 2 − 3 2 2 . The transformation of the initial state |↑i |0i from the product to the total angular momentum basis is then written q q ikx ikx 2 3 1 1 1 1 |ψini = e |↑i |0i = e 3 2 2 − 3 2 2 .

• Recall the basis scattering states on the region x < 0 are

− ikx −ikx ψb = e + rse |smsi ,

while the second set of basis states, for the region x > 0, are

+ ikx ψb = tse |smsi .

Both scattering state basis functions are written the basis |smsi. • Our next step is to construct our speciﬁc scattering problem’s wavefunctions on the regions x < 0 and x > 0 using the using the above basis functions.

71 11.1 Scattering on a Spin-Dependent Delta Function . . . github.com/ejmastnak/fmf

3 1 To do this, we first note that the incident state |ψini contains the state 2 2 with q 2 1 1 q 1 coefficient 3 , and the state 2 2 with coefficient − 3 . We thus construct the wavefunction |ψ−i, which describes the particle on the region x < 0, from a linear − 3 1 q 2 combination of the basis state ψb with |smsi = 2 2 weighted by 3 , and the − 1 1 q 1 basis state ψb with |smsi = 2 2 weighted by − 3 . The result is r r 2 ikx −ikx 3 1 1 ikx −ikx 1 1 |ψ−i = e + r3/2e − e + r1/2e . 3 2 2 3 2 2

Similarly, we construct |ψ+i, which describes our scattering problem on the region x > 0, via r r 2 ikx 3 1 1 ikx 1 1 |ψ+i = t3/2e − t1/2e . 3 2 2 3 2 2

• Next, now that we have properly constructed the basis states |ψ−i and |ψ+i, we will transform them from the total angular momentum basis back into the product basis |m1m2i. Again, we make the transformation with the CG coefficients. The transformation of |ψini within |ψ−i is already known, i.e. ikx |ψini = e |↑i |0i . 3 1 Next, we transform the state 2 2 to the product basis. In the CG table, we find the column containing 3/2 and +1/2, which correspond to s and ms. The column below, in this case 1/3 and 2/3, contains the coefficients needed for the transformation to the product basis. The row to the left of each coefficient (+1, −1/2 for the coefficient 1/3 and 0, +1/2 for the coefficient 2/3) represents the corresponding states in the product basis |m1m2i; each coefficient applies to the |m1m2i state in the corresponding row. 3 1 The transformation of 2 2 to the product basis reads r r 3 1 1 2 → |↓i |1i + |↑i |0i . 2 2 3 3 1 1 Analogously, the transformation of 2 2 to the product basis reads r r 1 1 2 1 → |↓i |1i − |↑i |0i . 2 2 3 3

3 1 1 1 • Substituting in the product basis representations of |ψini, 2 2 , and 2 2 into the state |ψ−i gives " ! r2 r1 r2 |ψ i = eikx |↑i |0i + e−ikx r |↓i |1i + |↑i |0i − 3 3/2 3 3 !# r1 r2 r1 − r |↓i |1i − |↑i |0i . 3 1/2 3 3

72 11.1 Scattering on a Spin-Dependent Delta Function . . . github.com/ejmastnak/fmf

The state |ψ−i is now fully transformed back to the product basis. • Following an analogous transformation procedure, the product basis representation of |ψ+i is " ! r2 r1 r2 |ψ i = eikx t |↓i |1i + |↑i |0i + 3 3/2 3 3 !# r1 r2 r1 − t |↓i |1i − |↑i |0i . 3 1/2 3 3

• Next, we combine like terms in |ψ−i to get " √ √ ! # 2 2 2 1 |ψ i = eikx |↑i |0i + e−ikx r − r |↓i |1i + r + r |↑i |0i − 3 3/2 3 1/2 3 3/2 3 1/2 ikx −ikx ≡ e |↑i |0i + e r↓1 |↓i |1i + r↑0 |↑i |0i ,

where we have defined the probability amplitudes r↓1 and r↑0, which represent, respectively, the probability of the first particle reflecting in the state |↓i and the second particle reflecting in the state |1i; and the probability of the first particle reflecting in the state |↑i and the second particle reflecting in the state |0i.

Similarly, combining like terms in |ψ+i gives " √ √ ! # 2 2 2 1 |ψ i = eikx t − t |↓i |1i + t + t |↑i |0i + 3 3/2 3 1/2 3 3/2 3 1/2 ikx ≡ e t↓1 |↓i |1i + t↑0 |↑i |0i .

Here t↓1 and t↑0 represent, respectively, the probability of the ﬁrst particle transmitting in the state |↓i and the second particle transmitting in the state |1i; and the probability of the ﬁrst particle transmitting in the state |↑i and the second particle transmitting in the state |0i.

• Evidently, there are four possible outcomes: the incident particle can either reflect or transmit; and the system’s spin configuration can either flip (from |↑i |0i to |↓i |1i) or remain the same. The probability for a given process comes from the squared amplitude of the corresponding r or t coefficient. For example, the probability for reflection with a flipped spin configuration

√ √ 2 2 2 2 R↓1 = |r↓1| = r3/2 − r1/2 . 3 3

• Finally, we will show the problem obeys conservation of total probability, i.e.

Ptot ≡ R↓1 + R↑0 + T↓1 + R↑0 = 1.

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We substitute in the values of the probability amplitudes to get √ √ 2 √ √ 2 2 2 2 1 2 2 2 2 1 2

Ptot = r3/2 − r1/2 + r3/2 + r1/2 + t3/2 − t1/2 + t3/2 + t1/2 3 3 3 3 3 3 3 3 2 2 2 2 2 ∗ 4 2 1 2 2 ∗ = r3/2 + r1/2 − 2 Re r r1/2 + r3/2 + r1/2 + 2 Re r r1/2 9 9 9 3/2 9 9 9 3/2 2 2 2 2 2 ∗ 4 2 1 2 2 ∗ + t3/2 + t1/2 − 2 Re t t1/2 + t3/2 + t1/2 + 2 Re t t1/2 9 9 9 3/2 9 9 9 3/2 2 2 2 1 2 2 = r3/2 + t3/2 + r1/2 + t1/2 . 3 3 Next, note that the r and t terms are precisely the elements of the scattering matrices r1/2 t1/2 r3/2 t3/2 S1/2 = and S3/2 = . t1/2 r1/2 t3/2 r3/2 Since every scattering matrix is unitary, i.e. S†S = I, it follows that 2 2 2 2 r3/2 + t3/2 = 1 and r1/2 + t1/2 = 1.

Substituting these results into the expression for Ptot gives

2 2 2 1 2 2 2 1 Ptot = r3/2 + t3/2 + r1/2 + t1/2 = · 1 + · 1 = 1. 3 3 3 3

In other words, because Ptot = 1, our scattering problem obeys conservation of probability, as it must.

12 Twelfth Exercise Set

12.1 Theory: Non-Degenerate Perturbation Theory

• Consider a Hamiltonian H0 with known eigenfunctions |n0i and non-degenerate en- (0) ergy eigenvalues En satisfying the stationary Schr¨odingerequation (0) H0 |n0i = En |n0i . We then add a small perturbation term H0 to the original Hamiltonian to get the perturbed Hamiltonian 0 H = H0 + H . • In this case, to second order in perturbation theory, the energy eigenvalues of the perturbed Hamiltonian are

0 2 X | hm0|H |n0i| E = E(0) + hn |H0|n i + , n n 0 0 (0) (0) m6=n En − Em while the corrected eigenfunction |ni are

0 2 X |hm0| H |n0i| |ni = |n i + |m i . 0 (0) (0) 0 m6=n En − Em For both eigenvalues and eigenfunctions, we generally take only the first nonzero (non-trivial) term in the perturbation expansion (e.g. if the first-order correction is non-zero, we would stop there, without finding the second-order correction).

74 12.2 QHO with a Quartic Perturbation github.com/ejmastnak/fmf

12.2 QHO with a Quartic Perturbation Use perturbation theory to estimate the energy eigenvalues of the perturbed Hamiltonian

p2 1 H = + kx2 + λx4, λ > 0, 2m 2 which contains the quartic perturbation term λx4.

• First, we identify the unperturbed Hamiltonian: the familiar harmonic oscillator

p2 1 H = + kx2, 0 2m 2

(0) with eigenstates |n0i and eigenvalues En obeying

(0) (0) 1 H0 |n0i = En |n0i ,En = ~ω n + 2 .

• The perturbation term is H0 = λx4. To ﬁrst order, the corrected energy of the perturbed Hamiltonian H is

(0) 4 (0) 4 En = En + hn0|λx |n0i = En + λ hn0|x |n0i .

4 Next, we evaluate the expectation value matrix element n0 x n0 . We will work in terms of the annihilation and creation operators. Nominally, x4 takes the form

x4 x4 = 0 (a + a†)4. 4 However, since a and a† don’t commute, (a + a†)4 would have 16 terms if multiplied out. To avoid dealing with 16 terms, we make two x act on the hn| term and two on the |ni term as follows:

4 2 2 hn0|x |n0i = x n0 x n0 .

4 We can factor x and move it unchanged between hn0| and |n0i because x is Hermi- tian.

• In terms of a and a†, the operator x2 reads

x2 x2 x2 = 0 (a + a†)2 = 0 (a2 + 2a†a + (a†)2 + 1), 2 2 where we have used the commutator identity a, a† = 1. Using this expression for 2 2 2 x , we now assemble the matrix element x n0 x n0 in pieces. First, (temporarily dropping the subscript 0 for conciseness and referring to the unperturbed |n0i as |ni), we have

75 12.3 QHO with a Cubic Perturbation github.com/ejmastnak/fmf

Since the coeﬃcients in x2 |ni are real, making complex conjugation redundant, we 2 2 have x n = x |ni. Using orthonormality of the states |ni produces

4 2 2 x0 2 x n x n = n(n − 1) + (2n + 1) + (n + 1)(n + 2) 4 x4 = 0 6n2 + 6n + 3. 4

2 2 4 • Using the just-calculated matrix element x n0 x n0 = hn0|x |n0i, the perturbed Hamiltonian’s energy to ﬁrst order in perturbation theory is 3x4 E = E(0) + λ 0 (2n2 + 2n + 1). n n 4 We end our approximation here, since the ﬁrst-order correction is nontrivial. Written out in full, the corrected energy eigenvalues are 3x4 E = ωn + 1 + λ 0 (2n2 + 2n + 1). n ~ 2 4 (0) Note that, unlike En , the energies En are no longer equally spaced by ~ω.

12.3 QHO with a Cubic Perturbation Use perturbation theory to estimate the energy eigenvalues of the perturbed Hamiltonian p2 1 H = + kx2 + λx3, λ > 0, 2m 2 which contains the cubic perturbation term λx3.

• As in the previous problem, the unperturbed Hamiltonian H0 is the familiar quantum harmonic oscillator p2 1 H = + kx2, 0 2m 2 with the eigenvalue relation

(0) (0) 1 H0 |n0i = En |n0i ,En = ~ω n + 2 .

• The perturbation term is H0 = λx3. Finally, we deﬁne the perturbed potential 1 V (x) ≡ kx2 + λx3. 2

• To ﬁrst order, the perturbed Hamiltonian’s energy eigenvalues are

(0) 3 En = En + hn0|λx |n0i . In the coordinate wavefunction representation the expectation value reads Z ∞ 3 ∗ 3 hn0|λx |n0i = ψnλx ψn dx = 0. −∞ This expectation value turns out to be zero—a short explanation follows. Because 1 2 V (x) = 2 kx is an even potential, the Hamiltonian H0’s eigenfunctions are either symmetric or asymmetric (see e.g. Subsection 1.3).

76 12.3 QHO with a Cubic Perturbation github.com/ejmastnak/fmf

∗ 3 If ψn(x) is even, the integrand ψnλx ψn, being the product of an even, odd, and even function, is odd. Similarly, if ψn(x) is odd, the integrand is a product of three odd functions, and is thus still odd. In both cases we take the integral of an odd function over −∞ to ∞—result is zero.

• Because the ﬁrst-order correction to the perturbed energies En gives a trivial result, we move on to the second order perturbation approximation

0 2 X | hm0|H |n0i| E = E(0) + . n n (0) (0) m6=n En − Em

Temporarily dropping the subscript 0 for conciseness and referring to the unperturbed eigenfunctions |n0i as |ni, the expectation value can be simpliﬁed to

hm|H0|ni = hm|λx3|ni = λ hxm|x2|ni .

As in the previous problem, we assemble the expectation value piece by piece. Reusing the previous problem’s result of x2 |ni, the ket term is

3 2 x0 h√ p √ hxm|x |ni = √ m n(n − 1)δm−1,n−2 + m(2n + 1)δm−1,n 8 √ p √ p + m (n + 1)(n + 2)δm−1,n+2 + m + 1 n(n − 1)δm+1,n−2 √ √ p i + m + 1(2n + 1)δm+1,n + m + 1 (n + 1)(n + 2)δm+1,n+2 .

Re-shifting indexes on the Kronecker deltas (just for clarity, e.g. δm−1,n−2 ≡ δm,n−1 or δm+1,n−2 ≡ δm,n−3) gives

3 2 x0 h√ p √ hxm|x |ni = √ m n(n − 1)δm,n−1 + m(2n + 1)δm,n+1 8 √ p √ p + m (n + 1)(n + 2)δm,n+3 + m + 1 n(n − 1)δm,n−3 √ √ p i + m + 1(2n + 1)δm,n−1 + m + 1 (n + 1)(n + 2)δm,n+1 .

Writing m in terms of n with reference to the δmn (e.g. m → n − 1 for δm,n−1) gives 3 √ 2 x0 h√ hxm|x |ni = √ n(n − 1)δm,n−1 + n + 1(2n + 1)δm,n+1 8 p p + (n + 1)(n + 2)(n + 3)δm,n+3 + n(n − 1)(n − 2)δm,n−3 √ √ i + n(2n + 1)δm,n−1 + (n + 2) n + 1δm,n+1 .

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Finally, combining like terms and putting the δmn in order produces 3 2 x0 hp √ hxm|x |ni = √ n(n − 1)(n − 2)δm,n−3 + 3n nδm,n−1 8 √ p i + 3(n + 1) n + 1δm,n+1 + (n + 1)(n + 2)(n + 3)δm,n+3 .

Switching back to subscript notation for the unperturbed eigenstates, with hxm|x2|ni ≡ 2 hxm0|x |n0i known, we just multiply by λ to ﬁnd matrix element 0 2 hm0|H |n0i = λ hxm0|x |n0i ,

needed to ﬁnd the perturbed energy En.

0 • Using the just-derived hm0|H |n0i, the perturbed Hamiltonian’s energy En is thus 0 2 2 6 3 X | hm0|H |n0i| λ x n(n − 1)(n − 2) 9n E = E(0) + = E(0) + 0 + n n (0) (0) n 8 (0) (0) (0) (0) m6=n En − Em En − En−3 En − En−1 9(n + 1)3 (n + 1)(n + 2)(n + 3) + + . (0) (0) (0) (0) En − En+1 En − En+3

Note that, nominally, we must sum over all m 6= n to ﬁnd En. However, because of 0 the Kronecker deltas in the matrix element hm0|H |n0i, there are only four m that are not equal to n at a given value of n. (0) (0) Next, we substitute in the energy diﬀerences (e.g. En − En−3 = 3~ω) in the denominators, which are known from the unperturbed Hamiltonian H0, where the energies are equally spaced by ~ω. Substituting in energies and factoring out ~ω gives

2 6 (0) λ x0 3 3 En = En + n(n − 1)(n − 2) + 27n − 27(n + 1) − (n + 1)(n + 2)(n + 3) 24~ω 2 6 (0) λ x0 3 2 3 3 2 3 2 = En + n − 3n + 2n + 27n − 27n − 81n − 81n − 27 − n − 6n − 11n − 6 24~ω 2 6 (0) λ x0 2 = En − 90n + 90n + 33 24~ω 2 6 (0) λ x0 2 = En − 30n + 30n + 11 . 8~ω Note that the energy correction is negative—the perturbed energies En are lower (0) than the standard QHO energies En . • Next, we ﬁnd the perturbed eigenfunctions |ni. From the theory section, the |ni are deﬁned as 0 2 X |hm0| H |n0i| |ni = |n i + |m i . 0 (0) (0) 0 m6=n En − Em 0 which in our case, using the matrix element hm0|H |n0i, is

3 "p √ λx0 n(n1)(n − 2) 3n n |ni = |n0i + √ |n0 − 3i + |n0 − 1i 8 3~ω ~ω √ # 3(n + 1) n + 1 p(n + 1)(n + 2)(n + 3) − |n0 + 1i − |n0 + 3i . ~ω 3~ω

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The perturbed ground state wavefunction |0i is

3 h √ i ˜ λx0 0 = |0i0 + √ −9 |1i0 − 6 |3i0 , 3 8~ω

3 √ x0 † λx0 x |0i = √ (a + a ) |0i0 + √ −9 |1i0 − 6 |3i0 2 3 8~ω 4 √ √ √ √ √ x0 λx0 h i = √ |1i0 + −9 |0i0 − 6 3 |2i0 + |1i0 − 9 2 |2i0 − 6 4 |4i0 . 2 12~ω

Next, using the orthonormality of the |n0i states, the full expectation value is

λx4 3λx4 h0|x|0i = − 0 (9 + 9) = − 0 . 12~ω 2~ω In other words, according to perturbation theory, the position expectation value in the perturbed state |0i shifts left from the origin to the region of negative x.

13 Thirteenth Exercise Set

13.1 Theoretical Background for the Coming Problem 13.1.1 Degenerate Perturbation Theory

0 0 • Consider the Hamiltonian H = H0 + H where H0 H has the known energy eigenvalue relation (0) H0 |ψi = E |ψi and the energy E(0) is N-times degenerate, meaning the wavefunction |ψi can be written as the general linear combination

N X (0) |ψi = cn |ni where H |ni = E |ni for each n. n=1

• In this course, we will consider only the ﬁrst-order correction; the procedure to ﬁnd this correction proceeds as follows:

1. Create a matrix of perturbation elements in a chosen basis of the subspace of degenerate states, i.e. {|ni}. The matrix reads

 h1|H0|1i h1|H0|2i · · · h1|H0|Ni   h2|H0|1i h2|H0|2i · · · h2|H0|Ni    .  . .. .   . ··· . .  hN|H0|1i hN|H0|2i · · · hN|H0|Ni

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2. Next, diagonalize the matrix and ﬁnd its eigenvalues λj and and eigenvectors uj. To ﬁrst order in perturbation theory, the corrected energies are then

(0) Ej = E + λj.

In general the degeneracy is broken if all of the eigenvalues λj are diﬀerent, i.e. λj 6= λi for all i 6= j. Note that if certain eigenvalues λj are equal, then a degree of degeneracy remains in the corrected energy eigenvalues. 3. Finally, again to ﬁrst order, the corrected eigenstates are

N X |ψji = ujk |ki , k=1

where uj is the perturbation matrix’s eigenvector corresponding to λj, and ujk is uj’s k-th component.

13.1.2 Review of Hydrogen Atom Eigenvalues and Eigenfunctions

• We now quickly review the unperturbed hydrogen states for H0. In Dirac notation:

H0 |nlmli |msi = En |nlmli |msi , where n is principle quantum number, l is magnitude of orbital angular momentum. The energy eigenvalue depends only on n; it is Ry E = − , n n2 where Ry = 13.6 eV is the Rydberg energy unit. The quantum numbers take the values n = 1, 2,...; l = 0, . . . , n − 1; ml = −l, . . . , l and ms ∈ {−1/2, 1/2}; we write the ms states as |↓i and |↓i. • The coming problem involves the first excited state with n = 2, so we give n = 2 special consideration here. The energy eigenvalues of the first excited state is Ry E = − . 2 4 We now conisder the degeneracy of the first excited state. There are 8 possible linearly independent eigenfunctions with energy E2; these are |200i |↑i , |200i |↓i |211i |↑i , |211i |↓i . |210i |↑i , |210i |↓i |21 − 1i |↑i , |21 − 1i |↓i

• Next, we consider the radially-dependent wavefunction Rnl(r) and spherical harmonics Ylm(θ, φ) in the expression

ψnlm(r) = Rnl(r)Ylm(θ, φ). For our problem we will need the following radial functions: 2 r − r R (r) = 1 − e 2rB 20 3/2 (2rB) 2rB 1 r − r R (r) = √ e 2rB , 21 3/2 3(2rB) rB

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where rB is the Bohr radius, and the following spherical harmonics 1 Y00(θ, φ) = √ 4π r 3 Y (φ, θ) = − sin θeiφ 11 8π r 3 Y (φ, θ) = cos θ 10 4π r 3 Y (φ, θ) = sin θe−iφ. 1−1 8π

Note the presence of m ∈ −1, 0, 1 in the exponents of the spherical harmonics Y11, Y10 and Y1−1.

13.1.3 Parity Symmetry • The parity operator P is deﬁned by the relation

Pψ(r) = ψ(−r).

In other words, the parity operator reverses a wavefunction’s spatial coordinates.

• Next, we assume the generic eigenvalue relation

Pψ(r) = pψ(r)

and apply the parity operator to get

PPψ(r) = P2ψ(r) = p2ψ(r) = ψ(r).

which implies p = ±1. The eigenvalues of P are thus ±1. We say that a wavefunction ψ has even parity if p = 1 and odd parity of p = −1.

Parity and the Hydrogen Eigenfunctions

• We now consider the action of the parity operator on the hydrogen eigenfunctions: Pψnlm(r) = P Rml(r)Ylm(θ, φ) = Rml(r)PYlm(θ, φ),

where we note that Rml(r) is unaﬀected because the distance r is unchanged by parity inversion.

• Next, we will show that [P, L] = [P, (Lx,Ly,Lz)] = 0. The derivation reads [P,Lz] = PLz − LzP = P(xpy − ypz) − LzP = (−x)(−py) − (−y)(−px) P − LzP

= (xpy − ypx)P − LzP = LzP − LzP = 0.

The procedure for Lx and Ly is analogous, and the end result is [P, L] = 0. • Since [P, L] = 0, both P and L have mutual eigenfunctions, meaning the spherical harmonics, which are nominally eigenfunctions of L, are also eigenfunctions of P .

Without derivation, the corresponding eigenvalue between P and Ylm is

l PYlm(θ, φ) = (−1) Ylm(θ, φ).

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13.2 Hydrogen Atom in a Homogeneous Electric Field Consider a hydrogen atom placed in a homogeneous electric ﬁeld of magnitude E. Analyze the resulting energy splitting of the ﬁrst excited state.

• The system’s Hamiltonian is

p2 q2 H = − − qEz, 2m 4π0r

where r is the distance between electron and proton and q = −e0 it the electron’s 0 charge. The unperturbed Hamiltonian H0 and perturbation term H are

2 2 p q 0 H0 = − and H = −qEz. 2m 4π0r Note that neither Hamiltonian contains a spin-dependent term, which corresponds to our problem’s rotation symmetry about the z axis. This symmetry will greatly simply our problem when we calculate matrix elements.

• On to our perturbation matrix, which has dimensions 8 × 8 because of ﬁrst excited state’s N = 8 degeneracy. A general matrix element takes the form

0 0 0 0 2lmlms H 2l mlms .

In principle there are 8 × 8 = 64 matrix elements, but we can take care of a lot of these by considering the problem’s symmetry.

• First, note that H0 = −qEz acts only in the position Hilbert space. We can thus factor the spin term |msi out of the main matrix element to get

0 0 0 0 0 0 0 0 0 0 0 0 2lmlms H 2l mlms = 2lml H 2l ml ms ms = 2lml H 2l ml δms,ms .

0 We then write the 8 × 8 matrix in four blocks of size 4 × 4 the value of hms|msi. 0 The oﬀ-diagonal blocks are zero because of δms,ms , while the two diagonal blocks are 0 0 equal because of the symmetry of δms,ms = δms,ms . The perturbation matrix is thus a block diagonal matrix with two identical 4 × 4 diagonal blocks, which we call D. Because these D sub-matrices are identical, each of the perturbation matrix’s eigenvalues is at least doubly degenerate. We now focus only the 4 × 4 sub-matrix D.

• Next, we note that the Hamiltonian H0 commutes with the angular momentum 0 operator Lz: 0 H ,Lz = − qEz, xpy, ypx = 0..

This is because both of the ∂y and ∂x terms in py and pz commute with z, as do the 0 0 x and y coordinates. Because H ,Lz = 0, the Hamiltonian H is invariant with respect to rotation in the xy plane about the z axis. 0 Note: going forward, since we have already used the spin term hms|msi to introduce the sub-matrix D, we only need to consider the position Hilbert space |nlmli ≡ |nlmi, where we’ll write ml = m since there is no longer any ambiguity with ms.

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• We now consider the basis in which we ﬁnd the matrix elements. The quantum 2 2 number l corresponds to the operator L and m to Lz, and in the basis oc L and Lz, the relevant eigenvalue relations read

2 2 L |nlmi = ~ l(l + 1) |nlmi and Lz |nlmi = ~m |nlmi , or, in the coordinate representation in terms of the spherical harmonics,

2 2 L Ylm(θ, φ) = ~ l(l + 1)Ylm(θ, φ) and LzYlm(θ, φ) = ~mYlm(θ, φ).

• Next, we consider the matrix element

0 0 0 0 0 0 0 MHLz ≡ 2lm [H ,Lz] 2l m = 2lm (H Lz − LzH ) 2l m 0 0 0 0 0 0 = 2lm H Lz 2l m − 2lm LzH 2l m .

† Next, since Lz is Hermitian, which means Lz = Lz and ~m ∈ R we have 0 0 0 0 0 0 0 0 0 2lm LzH 2l m = Lz2lm H 2l m = (~m)2lm H 2l m ∗ 0 0 0 0 0 0 = (~m) 2lm H 2l m = ~m 2lm H 2l m .

The matrix element MHLz is then

0 0 0 0 0 0 MHLz = 2lm H Lz 2l m − 2lm LzH 2l m 0 0 0 0 0 0 0 = hm 2lm H 2l m − ~m 2lm H 2l m 0 0 0 0 = ~(m − m) 2lm H 2l m ≡ 0.

0 0 0 0 The last equality holds from [H ,Lz] = 0 =⇒ MHLz ≡ h2lm|[H ,Lz]|2l m i = 0. 0 0 0 0 The equality ~(m − m) h2lm|H |2l m i ≡ 0 implies 0 0 0 0 2lm H 2l m = 0 if m 6= m.

• We now return to the 4 × 4 sub-matrix D and choose a basis so that states with the same m are grouped together and sub-divide D into blocks in which the m terms in the bra and ket terms are equal. We choose columns to run in the order |200i , |210i , |211i , |21 − 1i and rows to run in the order h200| , h210| , h211| , h21 − 1|. In terms of |lmi (dropping n = 2 from |2lmi for conciseness) the matrix D is then

 h00|H0|00i h00|H0|10i h00|H0|11i h00|H0|1 − 1i   h10|H0|00i h10|H0|10i h10|H0|11i h10|H0|1 − 1i  D =   .  h11|H0|00i h11|H0|10i h11|H0|11i h11|H0|1 − 1i  h1 − 1|H0|00i h1 − 1|H0|10i h1 − 1|H0|11i h1 − 1|H0|1 − 1i

We then apply the earlier result h2lm|H0|2l0m0i = 0 if m0 6= m, which simpliﬁes the matrix to  h00|H0|00i h00|H0|10i 0 0   h10|H0|00i h10|H0|10i 0 0  D =   .  0 0 h11|H0|11i 0  0 0 0 h1 − 1|H0|1 − 1i

Next, we consider parity symmetry, which will simplify the matrix even further. Parity Symmetry

83 13.2 Hydrogen Atom in a Homogeneous Electric Field github.com/ejmastnak/fmf

• Recall from the theory section beginning this exercise set that P and L commute, so the spherical harmonics Ylm are eigenfunctions of P. The relevant eigenvalue relation is

l l PYlm(θ, φ) = (−1) Ylm(θ, φ) or, in Dirac notation, P |lmi = (−1) |lmi .

Next, we show that H0 and P anti-commute, i.e.

{H0, P} = H0P + PH0 = H0P + P(−qEz) = H0P + (−qE)(−z)P = H0P − H0P = 0.

We then apply {H0, P} = 0 and the eigenvalue relation P |lmi = (−1)l |lmi to the matrix elements in the D matrix to get

0 0 0 0 0 0 0 0 0 MHP = 2lm {H , P} 2l m = 2lm H P 2l m + 2lm PH 2l m 0 0 0 l0 0 0 0 l = 2lm H 2l m (−1) + 2lm H 2l m (−1) l0 l 0 0 0 = (−1) + (−1) 2lm H 2l m ≡ 0,

0 0 0 0 where the last equality MHP = h2lm|{H , P}|2l m i = 0 holds because {H , P} = 0. The end result is

l0 l 0 0 0 (−1) + (−1) 6= 0 =⇒ 2lm H 2l m = 0.

In other words, if l and l0 are either both even or both odd, then h2lm|H0|2l0m0i = 0.

• We now return to matrix D, which read (using only |lmi and not |2lmi for shorthand)

 h00|H0|00i h00|H0|10i 0 0   h10|H0|00i h10|H0|10i 0 0  D =   .  0 0 h11|H0|11i 0  0 0 0 h1 − 1|H0|1 − 1i

We now apply the result (−1)l0 + (−1)l 6= 0 =⇒ h2lm|H0|2l0m0i = 0, which means that the diagonal matrix elements in the diagonal blocks are zero. The matrix simpliﬁes further to

 0 h200|H0|210i 0 0  h210|H0|200i 0 0 0 D =   ,  0 0 0 0 0 0 0 0

where we have returned to the longer |nlmi notation, since the matrix has cleared up considerably.

• Recall that D is a diagonal sub-matrix corresponding to the Hamiltonian H0, which is a Hermitian operator. Since Hermitian operators correspond to Hermitian matrices, the matrix D must also be Hermitian, which implies

h200|H0|210i = h210|H0|200i∗ .

We are conveniently left with a single non-zero matrix element from the 64 elements in the original perturbation matrix.

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• We ﬁnd the matrix element h200|H0|210i by direct calculation in the coordinate representation. Substituting in the relevant radial eigenfunctions and spherical harmonics, which were given in the theory section opening this exercise set, and converting to spherical coordinates (e.g. z = r cos θ), the calculation reads Z 0 3 ∗ h200|H |210i = d rψ200(r)(−qEz)ψ210(r)

Z ∞ Z 1 Z 2π r 2 2 r − 1 = −qE r dr d[cos θ] dφ 1 − e 2rB √ r cos θ 3/2 0 −1 0 (2rB) 2rB 4π " r # 1 1 r − r 3 · √ e rB cos θ 3/2 3 (2rB) rB 4π √ Z ∞ 4 r Z 1 Z 2π 3 r r − 2 2 rB √ = −qE dr 3 1 − e · d[cos θ] cos θ dφ 4π (2rB) rB 2rB 3 √0 −1 0 2 3 r Z ∞ u 2 = −qE √ · B u4 1 − e−u du · · (2π) , 3 4π 8 0 2 3 where we have introduced the new variable u = r . We ﬁnd the integral over u rB using the gamma function, and the result is

1 1 h200|H0|210i = −qEr 4! − 5! = 3qEr . B 12 2 B

0 0 ∗ ∗ • Since h210|H |200i = h200|H |210i = (3qErB) = 3qErB, the matrix D is   0 3qErB 0 0 3qErB 0 0 0 D =   .  0 0 0 0 0 0 0 0

Because D has two zero diagonal elements, it has two zero eigenvalues

λ1 = λ2 = 0.

The matrix D’s two non-zero eigenvalues come from D’s 2 × 2 upper-left sub-matrix and are the zeroes of the second-degree characteristic polynomial

2 2 (−λ) − (3qErB) = 0.

The results are

λ3 = 3qErB = −3e0ErB and λ4 = −3qErB = 3e0ErB.

• Finally, recall the original 8 × 8 perturbation matrix contains two identical 4 × 4 sub-matrices D along its diagonal, so the degeneracy of D’s eigenvalues λ1, λ2, λ3 and λ4 are doubled when considering the original 8 × 8 perturbation matrix.

As a result, the eigenvalues λ3 and λ4, which decrease and increase linearly with E, respectively, have twofold degeneracy two.

Meanwhile, the energy eigenvalues λ1 and λ2, which are unchanged in the presence of an electric ﬁeld, have fourfold degeneracy.

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• The unchanged eigenvalues λ1 and λ2 with fourfold degeneracy correspond to the four-dimensional linear combination

|ψ1,2i = a |211i |↑i + b |211i |↓i + c |21 − 1i |↑i + d |21 − 1i |↓i ,

where we require a2 + b2 + c2 + d2 = 1 for normalization.

• To ﬁnd the |nlmi |msi states corresponding to the lowered energy λ3, we solve the 2 × 2 eigenvector problem r 3e Er −3e Er u u 1 1 0 B 0 B 3,200 = 0 =⇒ 3,200 = . −3e0ErB 3e0ErB u3,210 u3,210 2 1

The state associated the eigenvalue λ3 is thus the two-dimensional linear combination 1 · |200i + 1 · |210i |ψ3i = √ . 2

• To ﬁnd the |nlmi |msi states corresponding to the raised energy λ4, we solve the 2 × 2 eigenvector problem r −3e Er −3e Er u u 1 1 0 B 0 B 4,200 = 0 =⇒ 4,200 = . −3e0ErB −3e0ErB u4,210 u4,210 2 −1

The state associated the eigenvalue λ4 is thus the two-dimensional linear combination 1 · |200i − 1 · |210i |ψ4i = √ . 2

Quick Summary:

• In the presence of the basic hydrogen atom Hamiltonian H0, i.e.

p2 q2 H0 = − , 2m 4π0r a hydrogen atom’s ﬁrst excited state has energy degeneracy N = 8, and the corresponding energy eigenfunctions are any functions in the 8-dimensional Hilbert space spanned by the 8 basis functions

|200i |↑i , |200i |↓i |211i |↑i , |211i |↓i . |210i |↑i , |210i |↓i |21 − 1i |↑i , |21 − 1i |↓i

In other words, any linear combination of the 8 basis functions is an eigenfunction (0) Ry of H0 with the same energy E2 = − 4 . • We then turn on a homogeneous electric ﬁeld E, which adds a perturbation H0 = −qzE to the original Hamiltonian H0. As a result, the originally eight-times degen- (0) erate energy E2 splits into

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(0) – A four-times degenerate energy, unchanged relative to E2 , corresponding to λ1 and λ2. The corresponding eigenfunctions come from the four-dimensional linear combination

|ψ1,2i = a |211i |↑i + b |211i |↓i + c |21 − 1i |↑i + d |21 − 1i |↓i ,

where a2 + b2 + c2 + d2 = 1; – a two-times degenerate energy level

(0) (0) E2 + λ3 = E2 − 3e0ErB,

(0) lowered relative to E2 in proportion to the electric ﬁeld E. The corresponding eigenfunctions come from the two-dimensional linear combination

1 · |200i + 1 · |210i |ψ3i = √ ; 2

– and a two-times degenerate energy level

(0) (0) E2 + λ4 = E2 + 3e0ErB,

(0) raised relative to E2 in proportion to the electric ﬁeld E. The corresponding eigenfunctions come from the two-dimensional linear combination

1 · |200i − 1 · |210i |ψ4i = √ . 2

14 Fourteenth Exercise Set

14.1 Theory: First-Order Time-Dependent Perturbations • Consider a Hamiltonian of the form

0 H(t) = H0 + H (t),

where H0 is analytically solvable and time-independent, while the time-dependent 0 perturbation H (t) is small relative to H0.

• Assume the eigenstates and eigenvalues of H0 obey

H0 |ni = En |ni ,

where |ni denotes H0’s (and not H’s) energy eigenstates. • We them aim to solve the time-dependent Schr¨odingerequation ∂ i |ψ, ti = H(t) |ψ, ti . ∂t

First, we expand the wavefunction in the basis of H0 eigenstates {|ni}:

X −i En t |ψ, ti = cn(t) |n, ti where |n, ti = |ni e ~ . n

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We then substitute the ansatz for |ψ, ti into the time-dependent Schrödingerequa- tion, which results in a system of coupled integral equations for the coefficients cn(t) of the form Z t i 0 0 0 0 0 cn(t) = cn(t0) − dt n, t H (t ) m, t cm(t ). ~ t0 Note that up to this step we have not made any approximations—the above system of integral equations is equivalent to the time-dependent Schrödingerequation for |ψ, ti.

• To proceed, we make the following approximation:

Assume H0(t) is weak enough that it doesn’t drastically disturb |ψ, ti from the initial state |ψ, t0i, so that the coeﬃcients cn(t) don’t change much from their initial values cn(t0)

Under the assumption cm(t) ≈ cm(t0), the coeﬃcients cn(t) become

Z t i 0 0 0 0 0 cn(t) ≈ cn(t0) − dt n, t H (t ) m, t cm(t0) ~ t0 Z t i 0 0 0 0 0 = cn(t0) − cm(t0) dt n, t H (t ) m, t . ~ t0

• To solve the problem, we solve for the coeﬃcients cn(t), and then ﬁnd the perturbed eigenfunction with

X X −i En t |ψ, ti = cn(t) |n, ti = cn(t) |ni e ~ , n n

where |ni and En are the energy eigenfunctions and eigenvalues of the static Hamil- tonian H0. It is then best practice to retrospectively check that the coeﬃcients cm didn’t actually change much—if they did, the approximation cm(t) ≈ cm(t0) would not hold and we would need a higher-order approximation, which is beyond the scope of this course.

14.2 Hydrogen Atom Exposed to an Electric Field Pulse Consider an electron in a hydrogen atom in an electric ﬁeld E(t), with Hamiltonian

2 2 p q E0 H = − − qE(t)z, E(t) = 2 . 2m 4π0r t 1 + τ Before the electric ﬁeld is turned on (at t → −∞), the electron is in the ground state, i.e.

|ψ, −∞i = |100i .

Determine the probability of finding the electron in the first excited state when the electric field pulse ends at t = ∞.

Note: Formally, a complete hydrogen atom eigenstate reads |nlmlmsi, but since spin terms don’t occur in the Hamiltonian, spin eﬀects don’t play a role in our problem and we’ll leave out |msi for conciseness and write the eigenstates as |nlmi.

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• The first excited state has principle quantum number n = 2, and the possible |nlmi eigenfunctions in the first excited state are |200i, |211i, |210i and |21 − 1i. We define the probability 2 P2lm(∞) = |c2lm(∞)| , which is the probability of finding the electron in the state |2lmi at time t → ∞. Our goal is to find this P2lm for the four possible |2lmi states, and then find the total probability of finding the electron in the first excited state with

P2 = P200 + P211 + P210 + P21−1.

• In the electron’s initial ground state |ψ, −∞i = |100i, the corresponding coefficient cnlm is c100(−∞) = 1, while cnlm(−∞) = 0 for all other n, l, and m. The final coefficients c2lm(t) at t → ∞ are thus Z ∞ i 0 0 0 0 0 0 0 c2lm(∞) = c2lm(−∞) − dt 2lm, t H (t) n l m , t cn0l0m0 (−∞) ~ −∞ Z ∞ i 0 0 0 0 = c2lm(−∞) + dt 2lm, t −qE(t )z 100, t c100(−∞) ~ −∞ Z ∞ i 0 0 0 0 = dt 2lm, t −qE(t )z 100, t , ~ −∞

where we have used cnlm(∞) = 0 for all (nlm) 6= (100) to simplify the system of equations in the ﬁrst line to the single equation in the second line, and then c2lm(−∞) = 0 and c100(−∞) = 1 to get the third line.

−i En t • Next, we substitute in |nlm, ti = |nlmi e ~ to get

Z ∞ E i 0 i 2lm t0 0 −i E100 t0 c2lm(∞) = − dt e ~ h2lm|−qE(t )z|100i e ~ . ~ −∞ Note that, as usual, we must take the complex conjugate of the exponent in the bra Ry term. We then substitute in the Rydberg energy formula Enlm = En = n2 to get Z ∞ i 0 i (E2−E1) t0 −qE0 c (∞) = − dt e ~ h2lm|z|100i . 2lm 0 2 −∞ t ~ 1 + τ We have two integrals: one over time and one over position (implicitly contained in the matrix element h2lm|z|100i).

• We begin by ﬁnding the matrix element h2lm|z|100i. We reuse a result from the previous exercise set: since [z, Lz] = 0, the matrix elements h2lm|z|100i are diagonal with respect to Lz’s quantum number m. In other words, the matrix element is non- zero only if the m quantum numbers are equal on both sides of the braket, i.e.

h2lm|z|100i = h2lm|z|100i δm,0.

Next we note that {z, P} = 0, which, again reusing a result from the previous exercise set, means that

0 0 0 l0 l nlm z n l m = 0 if (−1) + (−1) 6= 0.

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In our case the matrix element h2lm|z|100i is no zero only for l satisfying

(−1)l + (−1)0 = 0 =⇒ l = 1.

Because of the above to results, i.e. l = 1 and h2lm|z|100i = h2lm|z|100i δm,0, it follows that c200(∞) = c211(∞) = c21−1(∞) = 0. In other words, the probability of ﬁnding the electron in the ﬁrst excited state at t → ∞ is simply

2 P2 = P200 + P211 + P210 + P21−1 = 0 + 0 + P210 + 0 = |c210| ,

and we only need to calculate the single matrix element h210|z|100i.

• In terms of spherical coordinates and the radial and spherical harmonic eigenfunctions, which are given in the theory section opening the previous exercise set, the matrix element h210|z|100i ≡ h210|r cos θ|100i is Z ∗ ∗ h210|z|100i = drR21Y10(θ, φ)r cos θR10(r)Y00(θ, φ)

Z ∞ Z 1 Z 2π r 2 1 1 r − = r dr d[cos θ] dφ √ e 2a0 · 3/2 0 −1 0 3 (2rB) rB r ! ! 3 2 − r 1 · cos θ r cos θ e rB √ 4π 3/2 4π rB Z ∞ 4 3 r Z 1 r − 2 = e 2 rB dr d[cos θ] cos θ 3/2 4 0 2 rB −1 1 Z ∞ 2r 5 cos3 θ 1 = B u4e−u du · 3/2 4 2 rB 0 3 3 cos θ=−1 7/2 Z ∞ 7/2 2 2 4 −u 2 2 = rB 5 · u e du = rB 5 · · 4! 3 3 0 3 3 215/2 = r , B 35 where we have introduced the new variables u = 3r and evaluated the integral over 2rB u with the gamma function.

• After substituting in the just-derived matrix element h210|z|100i, the coeﬃcient c210(∞) simpliﬁes to

Z ∞ 15/2 ! i 0 i (E2−E1) t0 −qE0 2 c (∞) = − dt e ~ · r . 210 0 2 B 5 −∞ t 3 ~ 1 − τ We now ﬁnd the integral over time, which we will do with complex integration and the residue theorem. First, to make the complex poles more obvious, we rewrite the integral as (E −E ) Z ∞ 2 i 2 1 t0 τ e ~ 0 c210(∞) ∝ 0 0 dt , −∞ (t + iτ)(t − iτ) where we have dropped constant factors for conciseness. We then interpret t0 as a complex variable with Re t0 and Im t0 spanning the complex. We will close our complex integration contour as a semicircle in the upper complex plane, so that the

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exponential factor in the integrand’s numerator of converges, since Im t0 > 0 in the upper complex plane. We then deﬁne the complex function

(E −E ) 2 i 2 1 z τ e ~ f(z) = (z + iτ)(z − iτ)

and apply the residue theorem at the complex pole t0 = iτ. As the semicircular integration contour’s radius approaches inﬁnity, the integral approaches

(E −E ) I Z ∞ 2 i 2 1 t0 τ e ~ 0 f(z) dz → 0 + 0 0 dt = 2πi Res(f(z); iτ) −∞ (t + iτ)(t − iτ) E −E 2 − 2 1 τ τ e ~ − E2−E1 τ = 2πi = πτe ~ . 2iτ

Using the above integral, the coeﬃcient c210(∞) is then

(E −E ) 15/2 ! Z ∞ 2 i 2 1 t0 i 2 τ e ~ 0 c210(∞) = − rB 5 (−qE0) 0 0 dt ~ 3 −∞ (t + iτ)(t − iτ) 15/2 i 2 − E2−E1 τ = rB qE0πτe ~ . ~ 35

• With c210(∞) known, the probability P210 of the electron transition to the ﬁrst excited state |210i, and thus the probability of transitioning to the ﬁrst excited state in general, is

15 2 2 2 2 π qE0rB 2 − 2(E2−E1) τ P2(∞) = P210(∞) = |c210(∞)| = τ e ~ . 310 ~

Note that P210(∞) ∝ E0. This concludes the main portion of this problem.

• As a final note, we ask what value of τ at fixed electric field amplitude E0 will maximize the probability P2(∞). We do this by finding the extrema of P2(∞) with respect to τ, as in elementary single variable calculus. Ignoring the constant coefficients, the calculation reads d − 2(E2−E1) τ 2 E2 − E1 − 2(E2−E1) τ 0 ≡ P2(∞) = 2τe ~ + τ −2 e ~ dτ ~ E − E = 1 − τ 2 1 ~ τ = ~ . E2 − E1 In other words, the pulse length maximizing the transition amplitude from the ground state to the first excited state is directly related to the energy difference between the two states.

14.3 Spin 1/2 Particle in a Time-Varying Magnetic Field Consider a particle with spin s = 1/2 in the initial state |ψ, 0i = |↓i and Hamiltonian

H = µBσ · B(t),

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where σ is written in terms of the Pauli spin matrices as σ = (σx, σy, σz), and the magnetic ﬁeld is t B(t) = B , 0,B ,B B . x τ z x z Find the particle’s wavefunction |ψ, τi at t = τ, and analyze the limiting cases of large and small τ.

• The terms of the magnetic ﬁeld B’s components, the Hamiltonian reads t H(t) = µ σ · B(t) = µ σ B + µ σ B . B B z z B x x τ

Assuming Bx Bz, the static Hamiltonian is H0 = µBσzBz, while the time- 0 t dependent perturbation term is H (t) = µBσxBx τ . The magnetic ﬁeld’s x component Bx increase from 0 to Bx as t increases from 0 to τ. In the initial state |ψ, 0i = |↓i, the particle’s spin points in the opposite direction of the magnetic ﬁeld. Since the particle has spin s = 1/2, the problem’s eigenfunctions basis is spanned by the spin eigenfunctions |↑i and |↓i.

while the corresponding eigenvalues are E↑ = µBBz and E↓ = −µBBz. Note that the initial state |ψ, 0i = |↓i is a ground state of the initial Hamiltonian H(t = 0) = H0.

−i En t • Next, using the general expression |n, ti = |ni e ~ together with the energy eigenvalues E↑ = µBBz and E↓ = −µBBz, it follows that the basis functions |↑i and |↓i evolve in time as

−i µB Bz t i µB Bz t |↑, ti = |↑i e ~ and |↓, ti = |↓i e ~ .

The particle’s wavefunction at t = τ is thus the linear combination

|ψ, τi = C↑(τ) |↑, τi + C↓(τ) |↓, τi .

• We find the first coefficient C↑(τ) using the general approximation

Z t i 0 0 0 0 0 cn(t) ≈ cn(t0) − n, t H (t ) m, t cm(t0) dt . ~ t0 In our case, with the two dimensional eigenbasis {|↑i , |↓i}, the expression reads Z τ Z τ i 0 i 0 C↑(τ) = C↑(0) − dt h↑, t|H (t)|↓, ti C↓(0) − dt h↑, t|H (t)|↑, ti C↑(0). ~ 0 ~ 0

Since C↑(0) = 0 and C↓(0) = 1 (because |ψ, 0i = |↓i), the expression simpliﬁes to Z τ i 0 C↑(τ) = − dt h↑, t|H (t)|↓, ti . ~ 0

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• We ﬁnd the second coeﬃcient C↑(τ) with Z τ Z τ i 0 i 0 C↓(τ) = C↓(0) − dt h↓, t|H (t)|↓, ti C↓(0) − dt h↓, t|H (t)|↑, ti C↑(0). ~ 0 ~ 0

Substituting in C↑(0) = 0 and C↓(0) = 1 simpliﬁes the expression to Z τ i 0 C↓(τ) = 1 − dt h↓, t|H (t)|↓, ti . ~ 0

0 • Next, we substitute |↑, ti, |↓, ti and H (t) into the expressions for C↑(τ) and C↓(τ) to get Z τ i i µB Bz t t i µB Bz t C↑(τ) = − dte ~ h↑|µBBx σx|↓i e ~ ~ 0 τ Z τ i −i µB Bz t t i µB Bz t C↓(τ) = − dte ~ h↓|µBBx σx|↓i e ~ . ~ 0 τ Note the complex conjugations when moving the exponential factor out of the bra term. To ﬁnd C↑(τ) and C↓(τ), we have to calculate the two matrix elements h↑|σx|↓i and h↓|σx|↓i, which we’ll ﬁnd in spin basis {|↑i , |↓i} using spinors.

We start by ﬁnding the ket term σx |↓i, which reads

0 1 0 1 σ |↓i → = → |↑i . x 1 0 1 0

Using σx |↓i = |↑i and applying the orthonormality of |↑i and |↓i gives

h↑|σx|↓i = h↑|↑i = 1 and h↓|σx|↓i = h↓|↑i = 0.

Because h↓|σx|↓i = 0, we have C↓(τ) = 0.

• We ﬁnd the remaining coeﬃcient C↑(τ) bt direct calculation: Z τ i i µB Bz t t i µB Bz t C↑(0) = − e ~ h↑|µBBx σx|↓i e ~ dt ~ 0 τ Z τ i 2i µB Bz t t = − e ~ µBBx dt, ~ 0 τ

where we’ve used h↑|σ |↓i = 1. In terms of the new variable u = 2µB Bz t, the x ~ resulting integral, using integration by parts, evaluates to

Z 2µB Bz τ i µBBx ~ ~ iu ~ C↑(τ) = − du e u ~ τ 0 2µBBz 2µBBz 2 2µB Bz τ i µ B Z ~ = − B x ~ ueiu du ~ τ 2µBBz 0 i µ B 2 2µB Bz τ B x ~ iu ~ = − (−iu + 1) e ~ τ 2µBBz 0 iBx ~ 2µBBzτ i 2µB Bzτ = − 1 − i e ~ − 1 . 2Bz 2µBBzτ ~

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• In terms of C↑(τ) the first-order approximation to the wavefunction |ψ, τi is thus iBx ~ 2µBBzτ i 2µB Bzτ |ψ, τi = 1 · |↓, τi − 1 − i e ~ − 1 |↑, τi . 2Bz 2µBBzτ ~ Limit Case for Large τ • First, for shorthand, we define 2µ B τ ∆E α ≡ B z = τ, ~ ~ where ∆E is the energy difference between the energies E↓,↑ = ±µBBz. We start with the limit case α 1, which corresponds to large τ, i.e. turning on the x component of the magnetic field slowly. In the limit α 1, we have (1 − iα)eiα − 1 ≈ −ieiα. α The complete wavefunction |ψ, τi is then

S · nˆ |ψi = ~ |ψi , 2 where nˆ = cos φ sin θ, sin φ sin θ, cos θ is a unit direction vector, the corresponding solution for |ψi is θ θ |ψi = cos |↑i + sin eiφ |↓i . 2 2 We can then ﬁnd the spatial orientation (φ, θ) of the wavefunction’s spin by matching the wavefunction to the above expression in terms of φ and θ. • Recall that a wavefunction is determined only up to a constant phase factor. With θ + this in mind, we will ﬁrst re-write the wavefunction |ψ, τi so that cos 2 ∈ R , which will make it possible to compare |ψ, τi to the general form θ θ |ψi = cos |↑i + sin eiφ |↓i , 2 2 and thus determine the directions φ and θ of our particle’s spin at t = τ. We do this i µB Bzτ by factoring out and then neglecting a constant phase factor e ~ from |ψ, τi, which gives i µB Bzτ 1 Bx 1 Bx |ψ, τi = −e ~ |↑i − |↓i → |↑i − |↓i . 2 Bz 2 Bz

θ θ iφ We then compare |ψ, τi to |ψi = cos 2 |↑i + sin 2 e |↓i to get θ 1 B θ cos = x and sin eiφ = −1. 2 2 Bz 2

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• Because Bx Bz, we can write θ = π − for 1 and thus θ π B B cos = cos − = sin ≈ = x =⇒ = x . 2 2 2 2 2 2Bz Bz Next, we consider

θ π − sin = sin = cos ≈ 1 + O(2) =⇒ eiφ = −1 =⇒ φ = π. 2 2 2

Geometrically, the angles θ = π − where = Bx and φ = π mean the particle’s spin Bz points in the third quadrant and makes an angle with the −z axis. Physically, this means the particle’s spin points opposite the total magnetic ﬁeld B(t) = (Bx(t), 0,Bz for the course of the perturbation for t ∈ [0, τ]. This orientation of time-dependent spin is analogous to the spin orientation in the initial ground state |↓i (spin down), which pointed opposite the initial magnetic ﬁeld B = (0, 0,Bz). Limit Case for Small τ

• We now consider the opposite case, in which we turn on the ﬁeld rapidly relative ~ ~ to the characteristic time ∆E . In this case we have τ ∆E , which corresponds to α 1.

• To make the next step clearer, we write the time-evolved wavefunction in terms of α as iα i α Bx (1 − iα)e − 1 i α |ψ, τi = e 2 |↓i − i e 2 |↓i . 2Bz α We then expand the second term in the limit α 1 to get

(1 − iα)eiα − 1 (1 − iα)(1 + iα + ··· ) − 1 1 + α2 + · · · − 1 ≈ ≈ = O(α) 1. α α α This result, in terms of the time-dependent wavefunction |ψ, τi, implies

i α |ψ, τi = e 2 |↓i .

In other words, for small τ, the particle’s spin points in the same direction as the initial state |ψ, 0i = |{i ↓} throughout the entire perturbation. As a general rule, ~ we expect τ & ∆E for any notable change in |ψi from the initial state during the course of the perturbation.

15 Fifteenth Exercise Set

15.1 Rabi Oscillations Consider a particle with spin s = 1/2 in the initial state |ψ, 0i = |↓i in the presence of the Hamiltonian ∆E H = σ + λ cos(ωt)σ . 2 z x Find the particle’s time-dependent wavefunciton |ψ, ti.

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• First, we note that this Hamiltonian is in fact nearly identical to the Hamiltonian in the previous problem, where the Hamiltonian was t ∆E t H(t) = µ σ B + µ σ B ≡ σ + λσ , B z z B x x τ 2 z x τ

where we have deﬁned ∆E = 2µBBz and λ = µBBx to highlight the similarity. The only diﬀerence between our current problem and the previous problem is the time dependence—this was linear in the previous problem, but is sinusoidal in the current problem.

• When solving the present problem, because of the time-dependent cosine term, we won’t assume a priori that the probability for transition from the initial ground state to higher excited states is small. As a result, we cannot use the time-dependent perturbation approach from the previous section, which assumes transition probability is small. Instead, we will begin with the time-dependent Schr¨odingerequation ∂ H(t) |ψ, ti = i |ψ, ti . ~∂t Note that we cannot use the usual time-evolution approach because the Hamiltonian is time-dependent—we must solve the Schr¨odingerequation directly.

Attempt 1 (won’t work, but useful as an intermediate step)

• We begin our ﬁrst attempt by writing the time-dependent wavefunction as the linear combination |ψ, ti = c↑(t) |↑i + c↓(t) |↓i .

Next, recall that in Dirac notation, the spin matrices σx and σz are written

σz = |↑i h↑| − |↓i h↓| and σx = |↑i h↓| + |↓i h↑| .

• We then subsitute the time-dependent wavefunction into the Schrödingerequation and apply the Dirac-notation definitions of the spin matrices to get ∆E ∆E c |↑i − c |↓i + λ cos(ωt)c |↓i + λ cos(ωt)c |↑i = i c˙ |↑i + i c˙ |↓i , 2 ↑ 2 ↓ ↑ ↓ ~ ↑ ~ ↓ wherec ˙ denotes the time-derivative of the coefficients c.

• Next, we apply the orthonormality of the basis states |↓i and |↑i to get the system of equations ∆E ∆E c + λ cos(ωt)c = i c˙ and − c + λ cos(ωt)c = i c˙ . 2 ↑ ↓ ~ ↑ 2 ↓ ↑ ~ ↓

Unfortunately, this system of coupled differential equations for c↓ and c↑ is not solveable because of the time-dependent coefficient cos(ωt). Evidently, we need to make an approximation. Since we have ruled out first-order time-dependent perturbation theory, we will need a novel approach.

Second Approach

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• We begin by moving some of the time dependence of the coefficients c(t) into the basis states, similar to the first steps in first-order perturbation-theory. This gives

−iω0t iω0t |ψ, ti = c↑(t)e |↑i + c↓(t)e |↓i ,

0 where ~ω = ∆E. We then substitute |ψ, ti into the Schr¨odingerequation to get

• As before, we apply the orthonormality of the basis states |↓i and |↑i to get the system of equations

∆E 0 0 0 0 c e−iω t + λ cos(ωt)c eiω t = i c˙ e−iω t + ω0c e−iω t 2 ↑ ↓ ~ ↑ ~ ↑ ∆E 0 0 0 0 − c eiω t + λ cos(ωt)c e−iω t = i c˙ eiω t − ω0c eiω t. 2 ↓ ↑ ~ ↓ ~ ↓

Next, we multiply the ﬁrst equation by eiω0t and the second by e−iω0t and then rearrange to get ∆E 0 − ω0 c + λ cos(ωt)e2iω tc = i c˙ 2 ~ ↑ ↓ ~ ↑ ∆E 0 − + ω0 c + λ cos(ωt)e−2iω tc = i c˙ . 2 ~ ↓ ↑ ~ ↓

Again, we have a system of coupled diﬀerential equations, but this one will be approximately solveable. Our trick is to write

0 1 h 0 0 i cos(ωt)e2iω t = ei(ω+2ω )t + ei(−ω+2ω )t . 2 Note that ω0 doesn’t have a priori physical meaning—it is related to the energy diﬀerence ∆E, which is a free parameter of our system. We now choose ω0 such that 2ω0 = ω, i.e. ω0 is half the oscillation frequency of the x-dependent perturbation term. With 2ω0 = ω, the expression cos(ωt)e2iω0t simpliﬁes to

0 1 cos(ωt)e2iω t = (e2iωt + 1). 2 Similarly, the analogous expression with a negative exponent reads

0 1 cos(ωt)e−2iω t = (e−2iωt + 1). 2

In terms of these expressions, the system of equations for c↓ and c↑ reads

∆E λ − ω0 c + (e2iωt + 1)c = i c˙ 2 ~ ↑ 2 ↓ ~ ↑ ∆E λ − + ω0 c + (e−2iωt + 1)c = i c˙ . 2 ~ ↓ 2 ↑ ~ ↓

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• We now make an assumption: we assume any possible oscillation between the ground and excited states in the eventual solution will be small relative to the frequency ω of the x-dependent perturbation term. In other words, we assume c↑(t) and c↓(t) change slowly relative to 2ω. We will then check back to ensure the assumption is valid when we get our ﬁnal solution.

Under the assumption that c↑(t) and c↓(t) change slowly relative to ω, we can neglect ±2iωt the e terms in the coefficients of c↓ and cua, which simplifies the system of equations for c↑(t) and c↓(t) to ∆E ω λ − ~ c + c = i c˙ 2 2 ↑ 2 ↓ ~ ↑ ∆E ω λ − + ~ c + c = i c˙ , 2 2 ↓ 2 ↑ ~ ↓ where we have substituted in ω0 = ω/2. This system of equations is now analytically solvable, since the coefficients are constant. • Recall the initial state is |ψ, 0i = |↓i, which corresponds to

c↑(0) = 0 and c↓(0) = 1.

With these initial coeﬃcients in mind, we solve the coupled equations for c↓ and c↑ with the ansatz −iΩt −iΩt c↑(t) = αe and c↓(t) = βe . Substituting this ansatz into the system of equations leads to α λ −( ω − ∆E) + β = + Ωα ~ 2 2 ~ λ β α + ( ω − ∆E) = Ωβ, 2 ~ 2 ~ where we have canceled out the common exponential term. • In matrix form, the system of equations reads

ω−∆E λ ! − ~ − ~Ω α 2 2 = 0. λ ~ω−∆E β 2 2 − ~Ω This (2×2) system will have two linearly independent solutions, and the ﬁnal solution is a linear combination of the two. To solve the system for the eigenvalues Ω, we require the matrix’s determinant is zero, which gives

ω − ∆E ω − ∆E λ2 −~ − Ω ~ − Ω − ≡ 0 2 ~ 2 ~ 2 ω − ∆E 2 λ2 ( Ω)2 − ~ − = 0 ~ 2 2 s ω − ∆E 2 λ2 Ω = ± ~ + ≡ ± Ω , ~ 2 2 ~ 0 where we have deﬁned s 2 2 1 ~ω − ∆E λ Ω0 = + . ~ 2 2

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• With the eigenvalues Ω = ±Ω0 known, the corresponding eigenvectors are ω − ∆E λ −~ ∓ Ω α + β = 0. 2 ~ 0 2

The coeﬃcients α and β in the ansatzes for c↓ and c↑ are thus related by ! α λ ∝ 2 . β ~ω−∆E 2 ± ~Ω0 We leave oﬀ normalization for now—it turns out we won’t need it.

• The solutions for c↓ and c↑ are thus the linear combinations λ λ c (t) = A e−iΩ0t + B eiΩ0t ↑ 2 2 ω − ∆E ω − ∆E c (t) = A ~ + Ω e−iΩ0t + B ~ − Ω eiΩ0t. ↓ 2 ~ 0 2 ~ 0

We ﬁnd the coeﬃcients A and B from the initial conditions λ λ 0 ≡ c (0) = A + B ↑ 2 2 ω − ∆E ω − ∆E 1 ≡ c (0) = A ~ + Ω + B ~ − Ω , ↓ 2 ~ 0 2 ~ 0

which leads to 1 2A~Ω0 = 1 =⇒ A = −B = . 2~Ω0

• In terms of c↓ and c↑, the wavefunction |ψ, ti is then

Discussion of Results

∆E • Interpretation: our system is a two-state system with energies ± 2 . We then consider a hypothetical measurement at some time t > 0 to determine our particle’s state. In principle, we could measure either energy or z component of spin, since spin and Hamiltonian commute. Since the perturbation term λσx cos ωt complicates an energy measurement, we choose to measure the z component of the particle’s spin.

The state |ψ, ti is already written in terms of Sz’s basis functions |↓i and |↑i, and the possible measurement outcomes for Sz are

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Result Probability ~ 2 P↑(t) ~ − 2 P↓(t)

• The probability P↑ for spin up is 2 2 iλ −i ω t λ 2 P↑(t) = − sin Ω0te 2 = sin Ω0t. 2~Ω0 2~Ω0

• Meanwhile, the probability P↓ for spin up is 2 2 ~ω − ∆E i ω t ωω − ∆E 2 2 P↓(t) = −i sin Ω0t + cos Ω0t e 2 = sin Ω0t+cos Ω0t. 2~Ω0 2~Ω0

We then rearrange P↓ to get

" 2# 2 2 2 ~ω − ∆E 2 (2~Ω0) − (~ω − ∆E) P↓(t) = 1 − sin Ω0t 1 − = 1 − sin (Ω0t) 2 . 2~Ω0 (2~Ω0)

Finally, we substitute in deﬁnition of Ω0 to get 2 2 λ P↓(t) = 1 − sin (Ω0t) . 2~Ω0

Note that the sum of probabilities P↑(t) + P↓(t) is one, as it must be.

• The initial probability for detecting the particle with spin-up is zero; P↑ then oscil- 2 lates with time from 0 to λ . Meanwhile, the initial probability for detecting 2~Ω0 2 λ the particle with spin down is one, and P↓ then oscillates with time from 1 to . 2~Ω0

• The frequency Ω0 is called the Rabi frequency—recall that it is deﬁned as s ω − ∆E 2 λ2 Ω = ~ + . ~ 0 2 2

In the resonant case ~ω = ∆E, the Rabi frequency simpliﬁes to λ λ 2 ~Ω0 = and = 1. 2 2~Ω0 2 In this case the amplitude of probability oscillations λ is exactly one. 2~Ω0 • Finally, we check the validity of the earlier assumption/approximation ±2iωt e ≈ 0 ⇐⇒ Ω0 2ω. λ At resonance, where ~Ω0 = 2 , this condition reads λ 2ω = 2∆E =⇒ λ ∆E. 2 The requirement λ ∆E is consistent with the perturbation theory examples we have studied so far. Note, however, that we could not solve the problem with perturbation theory—time-dependent perturbation theory assumes a low probability of transition between states, but the probability for transition between states is large—even equal to one at resonance.

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