UChicago Graduate Diagnostic Exam Prep - Quantum Mechanics

July 30, 2024. Before the start of my PhD program at UChicago this fall, there is a diagnostic exam to test knowledge/ability in the core subjects of classical mechanics, electromagnetism, quantum mechanics, and statistical mechanics. Here, I share some interesting problems and solutions I came across while preparing for this exam. This is the first post in this series, where I go through quantum mechanics problems.

Superconducting Microwave Resonator
Measuring a Two-Spin System
Particle on a Ring
Emancipation from a Distance
Temperature and Adiabatic Changes of the 3-D Harmonic Oscillator
Quantum Zeno Effect
Positronium Stern-Gerlach
Quantum Single-Slit Diffraction
Gravitational Potential
Rigid Rotor in Magnetic Field
Symmetries and Transition Amplitudes
Thermal States
Scattering of Identical Particles
Spin-Orbit Coupling in the 3-D Harmonic Oscillator
Identical Particles in an Infinite Cubic Well
Time-Varying Force
Two Interacting Fermions
A Quantum Spin Chain
Anomalous Magnetic Moment of the Electron
A Heisenberg Ferromagnet
The Supersymmetric Method
Spin-1/2 in Time-Dependent Magnetic Field
Particle on a 1-D Lattice
Spin-1/2 and Stern-Gerlach
Harmonic Oscillator and Translation

Superconducting Microwave Resonator

Source: UBC Spring 2024 Physics Qualifying Exam Q1

Problem Statement: A superconducting microwave resonator is a small loop with an inductance of $L$ and a capacitance of $C$ . It can be modelled quantum-mechanically. Its quantum state can be described by the charge $Q$ on the capacitor and the magnetic flux $ϕ$ through the loop. The operators for these quantities satisfy an uncertainty relation $[\hat{Q}, \hat{ϕ}] = i ℏ$ . Write down the Hamiltonian for this system, and determine its ground state energy.

Solution: We start by analyzing the system using knowing what we know about circuits in a classical setting, then promote the classical variables by classical quantization to obtain the quantum Hamiltonian and ground state energy.

The potential drop across a capacitor with charge $d Q$ and capacitance $C$ is:

V_{cap} = \frac{Q}{C}

and hence the energy stored by adding infinitesimal charge $d Q$ is:

d E_{cap} = V d Q = \frac{Q}{C} d Q .

Thus if we charge up the capacitor from 0 charge to final charge $Q$ we can integrate the above to obtain the total stored energy in the capacitor:

E_{cap} = \frac{1}{2 C} Q^{2} .

Next, the potential drop across an inductor is:

V_{ind} = L \dot{I} = \dot{ϕ}

where the second equality follows via Faraday’s law. Integrating both sides of the second equality from zero current/flux at $t = 0$ to a final current/flux of $I$ and $ϕ$ , we obtain:

L I = ϕ

Furthermore, the energy stored in the inductor per unit time (using $P = I V$ ) is:

{\dot{E}}_{ind} = I V_{ind} = L I \dot{I}

Therefore the energy stored per infinitesimal current $d I$ is:

d E_{ind} = L I d I

Thus if we go from 0 current to final current $I$ the inductor stores energy:

E_{ind} = \frac{1}{2} L I^{2} .

Rewriting this in terms of $ϕ$ :

E_{ind} = \frac{1}{2 L} ϕ^{2}

Thus we have determined the energy of the system in terms of $L, C, Q, ϕ$ to be:

E = \frac{1}{2 C} Q^{2} + \frac{1}{2 L} ϕ^{2} .

Via canonical quantization we promote the classical variables $Q, ϕ$ to quantum operators $\hat{Q}, \hat{ϕ}$ and thus the energy $E$ to the Hamiltonian $\hat{H}$ , thus obtaining:

\hat{H} = \frac{1}{2 C} {\hat{Q}}^{2} + \frac{1}{2 L} {\hat{ϕ}}^{2} .

To find the ground state energy, we note the similarities of the above Hamiltonian with that of the quantum harmonic oscillator:

{\hat{H}}_{HO} = \frac{1}{2} k {\hat{x}}^{2} + \frac{1}{2 m} {\hat{p}}^{2}

which has eigenenergies $E_{n} = ℏ ω (n + \frac{1}{2})$ with $ω = \sqrt{k / m}$ . Since $\hat{Q}, \hat{ϕ}$ obey the same (canonical) commutation relations as $\hat{x}, \hat{p}$ and this is the only algebraic data necessary to determine the eigenenergies, $\hat{H}$ has the same eigenenergies (just replacing $k = \frac{1}{C}, m = L$ ) ans so the ground state energy of the system is:

E_{0} = \frac{1}{2} ℏ \sqrt{\frac{1}{L C}} .

Measuring a Two-Spin System

Source: UBC Spring 2024 Physics Qualifying Exam Q6

Problem Statement: Many copies of a system of two spin-1/2 particles are prepared in an identical spin state given by: $\begin{matrix} (1) & | ψ ⟩ = a | ↑↑ ⟩ + b | ↑↓ ⟩ + c | ↓↑ ⟩ + d | ↓↓ ⟩ \end{matrix}$ An experimenter performs many measurements of these identical systems, and finds the following results:

If only the first spin is measured, the result is spin up with 30% probability.
If only the second spin is measured, the result is spin up with 40% probability.
If both spins are measured, the result is both spins up with 10% probability. Derive the upper and lower limits on the probability that a measurement of the total spin of this state will yield zero.

Solution: From the experimental data (and the Born rule), we know that $| a |^{2} + | b |^{2} = 3 / 10$ , $| a |^{2} + | c |^{2} = 4 / 10$ , and $| a |^{2} = 1 / 10$ . For completeness’ sake let us go through the full calculation for the first of these. The probability of measuring the first spin to be spin up is given by the Born rule to be:

p (s_{1} =↑) = ⟩ ψ | Π_{1}^{↑} \otimes I_{2} | ψ ⟩

Where $Π_{↑}^{(1)} = | ↑ ⟩_{1} ⟨ ↑ |_{1}$ is the projector onto the $↑$ state for the first spin and $I_{2}$ is the identity matrix on the second spin. Thus using that $⟨ ↑ | ↑ ⟩ = 1$ and $⟨ ↑ | ↓ ⟩ = 0$ we find:

p (s_{1} =↑) = (a^{*} ⟨ ↑ |_{2} + b^{*} ⟨ ↓ |_{2}) (a | ↑ ⟩_{2} + b | ↓ ⟩_{2}) = | a |^{2} + | b |^{2}

which from the experimental data is equal to $3 / 10$ . Analogous calculations follow for the other cases.

Now, substituting the third expression into the first/second we find that $| b |^{2} = 2 / 10$ and $| c |^{2} = 3 / 10$ , and from the normalization of the state it must hold that:

| a |^{2} + | b |^{2} + | c |^{2} + | d |^{2} = 1

from which we find that $| d |^{2} = 4 / 10$ . The question asks for the probability that we measure the total spin of the state to be zero. To this end we rewrite the state in the $| s m ⟩$ basis, with: $| 11 ⟩ = | ↑↑ ⟩$ , $| 1 - 1 ⟩ = | ↓↓ ⟩$ , $| 10 ⟩ = \frac{1}{\sqrt{2}} (| ↑↓ ⟩ + | ↓↑ ⟩)$ and $| 00 ⟩ = \frac{1}{\sqrt{2}} (| ↑↓ ⟩ - | ↓↑ ⟩)$ . Thus:

| ψ ⟩ = a | 11 ⟩ + \frac{b + c}{\sqrt{2}} | 10 ⟩ + \frac{b - c}{\sqrt{2}} | 00 ⟩ + d | 1 - 1 ⟩

We are interested in the probability of measuring $s = 0$ , and hence:

p (s = 0) = | \frac{b - c}{\sqrt{2}} |^{2} = \frac{1}{2} (| b |^{2} + | c |^{2} - b^{*} c - b c^{*}) = \frac{1}{2} (\frac{1}{2} - b^{*} c - b c^{*})

where in the last equality we use the known values of $| b |^{2}$ and $| c |^{2}$ . $b, c$ have undetermined phases with $b = e^{i ϕ_{b}} \sqrt{\frac{2}{10}}$ and $c = e^{i ϕ_{c}} \sqrt{\frac{3}{10}}$ . We thus determine the upper and lower bounds on:

$- (b^{*} c + b c^{*}) = - \frac{\sqrt{6}}{10} (e^{i (- ϕ_{b} + ϕ_{c})} + e^{i (ϕ_{b} - ϕ_{c})}) = - \frac{\sqrt{6}}{5} \cos (ϕ)$ where we defined $ϕ = ϕ_{b} - ϕ_{c}$ and applied Euler’s identity. This quantity maximized when $ϕ = π$ for which we have:

p (s = 0)_{max} = \frac{1}{4} + \frac{\sqrt{6}}{10} .

Conversely, we have the minimum when $ϕ = 0$ for which:

p (s = 0)_{min} = \frac{1}{4} - \frac{\sqrt{6}}{10} .

Particle on a Ring

Source: UBC Summer 2022 Physics Qualifying Exam Q3, UBC Summer 2013 Physics Qualifying Exam Q4

Problem Statement: An electron is constrained to move in a ring with radius $R$ in the $x - y$ plane.

(a) Find the eigenstates and eigenvalues for the electron.

(b) In organic chemistry, aromatic compounds are cyclic compounds that satisfy Hückel’s rule, which says that the compound must have $4 n + 2$ valence electrons. Such compounds are known for their stability. Explain how Hückel’s rule can be derived from the simple model of an electron on a ring.

(c) Now an electric field of strength $E$ is applied along the $x$ -direction. Find the leading (non-zero) correction to the ground state energy.

(d) Calculate the polarizability of the system (the ratio of the electric dipole moment to the electric field strength)

(e) Consider again the case with no electric field. If the system is heated to some temperature $T$ , calculate the expectation value of the magnitude of the system’s angular momentum, $⟨ | \vec{L} | ⟩$ , in the high- $T$ limit.

Solution: (a) In the absence of a potential, we have the 1-D Shrodinger equation:

\frac{d^{2}}{d x^{2}} ψ (x) = - \frac{2 m E}{ℏ^{2}} ψ (x)

since we are confined to a ring, it is convenient to work with the angle $ϕ = x / R$ :

\frac{d^{2}}{d ϕ^{2}} ψ (ϕ) = - \frac{2 m R^{2} E}{ℏ^{2}} ψ (ϕ)

with the boundary conditions of $ψ (0) = ψ (2 π), ψ^{'} (0) = ψ^{'} (2 π)$ . The solutions can be found via inspection:

ψ_{n} (ϕ) = \frac{1}{\sqrt{2 π}} e^{i n ϕ}, n \in Z

with associated energies:

E_{n} = \frac{ℏ^{2} n^{2}}{2 m R^{2}} .

where we note that the energies (excepting $E_{0} = 0$ ) are two-fold degenerate (with $E_{n} = E_{- n}$ ).

(b) Filling up an orbital (all states of a given energy) gives rise to a more stable molecule (as it has a full valence shell and is less likely to react). Accounting for the two spin states $s_{z} = \pm \frac{ℏ}{2}$ , each of the energy levels $E_{n}$ as found in (a) can furnish up to two electrons - with our prior observation that the energy levels are two-fold degenerate, with the additional spin degeneracy to fill up all orbitals up to energy $E_{n}$ requires $2 \cdot (2 n) + 2 = 4 n + 2$ electrons, which is precisely Hückel’s rule.

V = E \hat{x} = E \cos (ϕ) = \frac{E}{2} (e^{i ϕ} + e^{- i ϕ})

Looking at the matrix elements of the peturbation:

⟨ ψ_{m} | V | ψ_{n} ⟩ = \frac{E}{2} \frac{1}{2 π} \int_{0}^{2 π} e^{- i m ϕ} (e^{i ϕ} + e^{- i ϕ}) e^{i n ϕ} = \frac{E}{2} (δ_{m, n + 1} + δ m, n - 1)

where in the last equality we use the orthogonality of imaginary exponentials. Thus we find that the first-order energy correction to the ground state (note: we can use non-degenerate perturbation theory here, as the ground state is non-degenerate ) vanishes:

E_{0}^{(1)} = ⟨ ψ_{0} | V | ψ_{0} ⟩ = 0

and we must go to second order to obtain the non-vanishing energy corrections.

E_{0}^{(2)} = \sum_{n \neq 0} \frac{| ⟨ ψ_{n} | V | ψ_{0} ⟩ |^{2}}{E_{0}^{(0)} - E_{n}^{(0)}} = \frac{E^{2}}{4} (\frac{1}{E_{0} - E_{1}} + \frac{1}{E_{0} - E_{- 1}}) = \frac{E^{2}}{4} \cdot 2 \frac{- ℏ^{2} (1)^{2}}{2 m R^{2}}

thus:

E_{0}^{(2)} = - \frac{E^{2} ℏ^{2}}{4 m R^{2}}

(d) We proceed by computing the first order corrections to the ground state:

| ψ_{0}^{(1)} ⟩ = \sum_{n \neq 0} \frac{| ⟨ ψ_{n} | V | ψ_{0} ⟩ |^{2}}{E_{0}^{(0)} - E_{n}^{(0)}} | ψ_{n} ⟩ = - \frac{E ℏ^{2}}{4 m R^{2}} (| ψ_{1} ⟩ + | ψ_{- 1} ⟩)

Thus computing the dipole moment:

⟨ d ⟩ = (⟨ ψ_{0}^{(1)} | + ⟨ ψ_{0}^{(0)} |) x (| ψ_{0}^{(1)} ⟩ + | ψ_{0}^{(0)} ⟩)

Recalling the matrix elements, only the off-diagonal elements between $n = \pm 1$ and $0$ are nonvanishing, and thus we obtain:

⟨ d ⟩ = - \frac{E ℏ^{2}}{4 m R^{2}} - \frac{E ℏ^{2}}{4 m R^{2}} = - \frac{E ℏ^{2}}{2 m R^{2}}

Thus the polarizability is:

α = \frac{| ⟨ d ⟩ |}{E} = \frac{ℏ^{2}}{2 m R^{2}}

(e) The $| ψ_{m} ⟩$ are also eigenstates of $L_{z} = - i ℏ \frac{d}{d ϕ}$ with eigenvalue $ℏ m$ . Since the electron is confined to the $x - y$ plane, it can only have a $z$ -component of angular momentum and hence $| L | = | L_{z} | = ℏ | n |$ .

At finite temperature, we can compute the expectation value using the machinery of statistical mechanics. The partition function is computed as the sum of Boltzmann factors:

Z = \sum_{n = - \infty}^{\infty} e^{- β E_{n}} = \sum_{n = - \infty}^{\infty} e^{- \frac{ℏ^{2}}{2 m R^{2} k_{B} T} n^{2}}

and the expectation value is then:

⟨ | L | ⟩ = \frac{\sum_{n = - \infty}^{\infty} | L | (n) e^{- β E_{n}}}{Z} = \frac{\sum_{n = - \infty}^{\infty} ℏ | n | e^{- \frac{ℏ^{2}}{2 m R^{2} k_{B} T} n^{2}}}{\sum_{n = - \infty}^{\infty} e^{- \frac{ℏ^{2}}{2 m R^{2} k_{B} T} n^{2}}} .

These sums cannot be computed analytically, but in the high temperature limit we can approximate them as integrals:

Z \approx \int_{- \infty}^{\infty} e^{- \frac{ℏ^{2}}{2 m R^{2} k_{B} T} n^{2}} d n = \sqrt{\frac{2 π m R^{2} k_{B} T}{ℏ^{2}}}

\sum_{n = - \infty}^{\infty} ℏ | n | e^{- \frac{ℏ^{2}}{2 m R^{2} k_{B} T} n^{2}} \approx 2 ℏ \int_{0}^{\infty} n e^{- \frac{ℏ^{2}}{2 m R^{2} k_{B} T} n^{2}} d n = 2 ℏ \frac{1}{2} \frac{2 m R^{2} k_{B} T}{ℏ^{2}} = \frac{2 m R^{2} k_{B} T}{ℏ}

Thus:

⟨ | L | ⟩ \approx \sqrt{\frac{2 m R^{2} k_{B} T}{π}}

Emancipation from a Distance

Source: UBC Summer 2022 Physics Qualifying Exam Q5

Problem Statement: Consider a 1-D quantum mechanical system with an attractive short-range potential which we approximate by a delta function such that the initial potential energy is given by:

$\begin{matrix} (2) & U_{0} (x) = - \frac{ℏ^{2} κ}{m} δ (x) . \end{matrix}$

We assume that we set the system up such that the particle occupies the unique bound state of this potential. This bound state has energy $E_{0} = - \frac{ℏ^{2} κ^{2}}{2 m}$ . Then, we modify the problem by placing an impenetrable wall at position $x = - L$ so that the resulting potential energy function is well approximated by:

$\begin{matrix} (3) & U (x) = {\begin{cases} - \frac{ℏ^{2} κ}{m} δ (x) & - L < x < \infty \infty & - \infty < x < - L \end{cases} \end{matrix}$

with $L$ initially very large. Indeed, if $L$ is very large, the system should be undisturbed by the presence of the wall. Now, let us imagine that we adiabatically decrease $L$ . We expect that the binding energy of the bound state begins to depend on $L$ ; is there a value of $L$ at which the binding energy goes to zero and the particle is liberated? If so, what is that value?

Solution: It is clear that there is some value of $L$ for which the binding energy must go to zero. This conclusion can be reached by considering the limiting case of $L \to 0$ where we have the infinite step potential which admits no bound states, only scattering states.

Finding the liberating value of $L$ requires a little more work. We solve the Schrödinger equation in the left region $- L \leq x \leq 0$ and the right region $0 \leq x \leq \infty$ . For each of those two regions, the potential is zero, and the SE thus reads:

\frac{d^{2} ψ}{d x^{2}} = k^{2} ψ

with $k = \frac{\sqrt{2 m (- E)}}{ℏ^{2}}$ . Note that here $E$ is negative (as we assume a bound state) and thus $k$ is positive. In the left region we have:

ψ_{L} (x) = A e^{k x} + B e^{- k x}

with the boundary condition of $ψ (- L) = 0$ yielding:

A e^{- k L} + B e^{k L} = 0 ⟹ A = - B e^{2 k L} .

In the right region we have:

ψ_{R} (x) = C e^{k x} + D e^{- k x}

with the boundary condition of $ψ (x \to \infty) = 0$ enforcing that $C = 0$ . The continuity of the wavefunction at $x = 0$ gives:

ψ_{L} (0) = ψ_{R} (0) ⟹ A + B = D .

Finally, we have the boundary condition on the derivative of the wavefunction at $x = 0$ , modified accordingly via the presence of the delta well at $x = 0$ (which makes the derivative discontinuous):

ψ_{L}^{'} (0) - ψ_{R}^{'} (0) = - \frac{2 m}{ℏ^{2}} (- \frac{ℏ^{2} κ}{m}) ψ (0) ⟹ A - B = (\frac{2 κ}{k} - 1) D

Dividing this equation by the continuity condition:

\frac{A - B}{A + B} = \frac{2 κ}{k} - 1

and now using the boundary condition at $x = - L$ :

\frac{- B e^{2 k L} - B}{- B e^{2 k L} + B} = \frac{2 κ}{k} - 1.

Rearranging this expression we obtain the transcendental equation for $k$ :

\frac{k}{1 - e^{2 k L}} = κ .

Since we are interested in the $E \to 0$ limit, $k \to 0$ and thus $e^{2 k L} \approx 1 - 2 k L$ :

\frac{k}{1 - (1 - 2 k L)} = \frac{k}{2 k L} = κ

thus rearranging for $L$ , we find the length at which the particle is liberated:

L = \frac{1}{2 κ} .

Temperature and Adiabatic Changes of the 3-D Harmonic Oscillator

Source: UBC Spring 2022 Physics Qualifying Exam Q5, UBC Fall 2020 Physics Qualifying Exam Q2

Problem Statement: Consider an atom of mass $m$ trapped in a 3D harmonic potential $V_{c} (\vec{r}) = \frac{1}{2} k r^{2}$ . In modern laboratories, such confining potentials can be easily induced by lasers.

(a) Negating the internal energy of the atom, what are the first three eigenenergies of the atom in the central potential $E_{0}, E_{1}, E_{2}$ ?

(b) At what temperature $T$ will the probability of finding the atom to have energy $E_{0}$ be the same as finding it to have energy $E_{2}$ ?

(c) Suppose we load the trap with atoms at an initial temperature of 1K. Afterwards, the trap is adiabatically turned into a much stepper trap with $k$ ten times bigger, i.e. $k \to 10 k$ . Estimate the final temperature of the atoms.

(d) Consider again the zero- $T$ single atom case, where now we turn on an additional potential such that $V_{a} (\vec{r}) = \infty$ for the half space of $z > 0$ . How much work is done on the atom during this adiabatic switching on?

(e) After keeping $V_{t o t} = V_{c} + V_{a}$ for some time, one decides to suddenly switch off all the potentials, freeing the atom into free space. How much work is done on the atoms in this process?

Solution: (a) The 3D harmonic potential is simply the sum of independent harmonic potentials in the $x, y, z$ directions and thus the eigenenergies are just the respective sum of the 1-D harmonic oscillators:

E_{n_{x}, n_{y}, n_{z}} = E_{n_{x}} + E_{n_{y}} + E_{n_{z}} = ℏ ω (n_{x} + n_{y} + n_{z} + \frac{3}{2})

with $ω = \sqrt{\frac{k}{m}}$ and $n_{i} = 0, 1, 2, \dots$ . From the above we can read off the three lowest energies as $E_{0} = \frac{3}{2} ℏ ω, E_{1} = \frac{5}{2} ℏ ω, \frac{5}{2} ℏ ω .$ Note that $E_{0}$ is non-degenerate, $E_{1}$ is three-fold degenerate (choose one of three $n_{i}$ s to be 1) and $E_{2}$ is six-fold degenerate (choose one of three $n_{i}$ s to be 2, or choose two of three $n_{i}$ s to be 1.)

(b) The probabilities are equal when the Boltzmann weights (with degeneracy) are equal, so:

\frac{P (E_{0})}{P (E_{6})} = 1 ⟹ \frac{e^{- β E_{0}}}{6 e^{- β E_{2}}} = \frac{1}{6} e^{\frac{2 ℏ ω}{k_{B} T}} = 1

therefore:

T = \frac{2 ℏ ω}{k_{B} \ln (6)} .

(c) Since the modification to the trap is done adiabatically, the atoms stay in the $n_{x}, n_{y}, n_{z}$ eigenstate. Their energy is modified via the frequency modification $ω \to ω^{'} = \sqrt{\frac{10 k}{m}} = \sqrt{10} ω$ :

E_{n_{x}, n_{y}, n_{z}}^{'} = ℏ ω^{'} (n_{x} + n_{y} + n_{z} + \frac{3}{2}) = \sqrt{10} ℏ ω (n_{x} + n_{y} + n_{z} + \frac{3}{2}) = \sqrt{10} E_{n_{x}, n_{y}, n_{z}} .

The atoms are cooled in such a way that the Boltzmann factors/probabilities of occupying a given state are preserved:

e^{- β E_{n_{x}, n_{y}, n_{z}}} = e^{- β^{'} E_{n_{x}, n_{y}, n_{z}}^{'}} ⟹ \frac{E_{n_{x}, n_{y}, n_{z}}}{T} = \frac{\sqrt{10} E_{n_{x}, n_{y}, n_{z}}}{T^{'}}

and thus:

T^{'} = \sqrt{10} T .

(d) At zero temperature, the atom is in the ground state with $E_{0} = \frac{3}{2} ℏ ω$ . The $x$ and $y$ components of the state (and thus the energy contributions) are unchanged by the turning on of $V_{a}$ , but the $z$ component is; namely, we note that only odd parity eigenstates (i.e. odd $n_{z}$ ) are allowed with the infinite potential in half space (as there must be a node in the wavefunction at $x = 0$ . These odd parity eigenstates form a complete basis of functions for the half-space of $z < 0$ .) Thus the lowest energy eigenstate after $V_{a}$ is fully switched on is that with $n_{x} = n_{y} = 0$ and $n_{z} = 1$ (the lowest energy odd parity eigenstate in $z$ ) and thus the ground state energy becomes $E_{0} = \frac{5}{2} ℏ ω$ . Thus work of $W = ℏ ω$ is done on the atom.

(e) If the potential is suddenly switched off, the quantum state of the atom has no time to react to the changing Hamiltonian, and thus stays in the $| n_{x} = 0, n_{y} = 0, n_{z} = 1 ⟩$ state with energy $\frac{5}{2} ℏ ω$ . Thus, no work is done.

Quantum Zeno Effect

Source: UBC Summer 2018 Physics Qualifying Exam Q8

Problem Statement: An electron spin is oriented upward along $+ \hat{z}$ , and subject to a magnetic field of strength $B$ along $+ \hat{x}$ . The precession period of the spin around the magnetic field direction is $T$ . The operator $S_{z}$ is measured repeatedly, $N$ times within a time interval $T / 2$ (assume the measurements are evenly spaced in time). What is the probability that all $N$ measurements returns $+ ℏ / 2$ , in the limit of large $N$ ?

Solution The quantum spin state is given by:

| ψ (t) ⟩ = \cos (\frac{2 π}{T} t) | ↑ ⟩ + i \sin (\frac{2 π}{T} t) | ↓ ⟩ .

To see this, consider that the Hamiltonian of the system is $H = - \frac{ℏ}{2} μ B σ_{x}$ with $μ$ the magnetic moment, and thus the time evolution operator is:

\exp (- \frac{i H t}{ℏ}) = \cos (\frac{μ B}{2} t) I + i \sin (\frac{μ B}{2} t) σ_{x}

(obtained via power series expansion of the exponential) and then identifying the period as $T = \frac{4 π}{μ B}$ and applying it to the initial state of $| ψ (0) ⟩ = | ↑ ⟩$ we obtain the claimed result. The Born rule tells us that the probability of measuring $S_{z} = - \frac{ℏ}{2}$ is the modulus squared of the coefficient:

P_{↓} (t) = | i \sin (\frac{2 π}{T} t) |^{2} = \sin^{2} (\frac{2 π}{T} t)

If we measure at $t = \frac{T}{2 N}$ we have:

P_{↓} (t = \frac{T}{2 N}) = \sin^{2} (\frac{π}{N}) \approx \frac{π^{2}}{N^{2}}

where we use the small-angle approximation in the large- $N$ limit. Thus the probability that all $N$ measurements of $S_{z}$ would all yield $+ \frac{ℏ}{2}$ is given by:

P_{all ↑} = {(1 - P_{↓} (t = \frac{T}{2 N}))}^{N} \approx {(1 - \frac{π^{2}}{N^{2}})}^{N} \approx (e^{- \frac{π^{2}}{N^{2}}})^{N} = e^{- \frac{π^{2}}{N}}

where we use the first-order approximation for the exponential. Thus we conclude:

P_{all ↑} = e^{- \frac{π^{2}}{N}}

which yields $1$ as $N \to \infty$ . This is the quantum Zeno effect, where measuring the system at high frequency stabilizes it.

Positronium Stern-Gerlach

Source: UBC Summer 2013 Physics Qualifying Exam Q3

Problem Statement: A beam of ground state positronium particles (an ``atom’’ composed of an electron and it’s anti-particle, a positron) enters a double Stern-Gerlach apparatus. At the input, the particles are an equal mixture of the two spin configurations, that is 50% para-positronium with total spin $s = 0$ and ortho-positronium with total spin $s = 1$ . Assume also that the beam is completely unpolarized. The positronium first encounters a magnetic field along the $z$ -axis with a gradient increasing in the $z$ -direction and the beam splits into three. The un-deflected beam then flies through a hole in the beam-stop and encounters a magnetic field directed perpendicular to the first one and along the $x$ -axis, causing the beam to split in three again.

(a) What fraction of the total beam flux passes through the first beamstop? What fraction of the total beam flux will pass through to each of the final outputs (right deflected, undeflected, left deflected)? Write down the spin state for the positronium through each of the three outputs in the $z$ basis.

(b) Suppose the positronium is disassociated at the three outputs that the spin of the electron and positron can be measured independently and the electron and positron spin projections along the z axis are measured. What is the chance of finding both the electron and the positron in the same spin state for each of the outputs?

Solution: The joint total spin and joint z-spin eigenstates of positronium we recall from the addition of angular momenta of two spin-1/2 particles. For para-positronium we have the spin singlet:

| s = 0, m_{z} = 0 ⟩ = \frac{| ↑↓ ⟩ - | ↓↑ ⟩}{\sqrt{2}}

and for ortho-positronium we have the three (spin triplet) states:

| s = 1, m_{z} ⟩ = {\begin{cases} | ↑↑ ⟩ & m_{z} = 1 \\ \frac{| ↑↓ ⟩ + | ↓↑ ⟩}{\sqrt{2}} & m_{z} = 0 \\ | ↓↓ ⟩ & m_{z} = - 1 \end{cases}

Assuming we have a 50/50 mixture of para and ortho positronium and the initial beam is unpolarized, to begin the composition of the beam is 1/2 $| s = 0, m_{z} = 0 ⟩$ and 1/6 for each of the $| s = 1 ⟩$ states. Since only $m_{z} = 0$ positronium atoms pass through the first beamstop, we conclude that $2 / 3$ of the initial beam flux passes through.

Let us denote the single-qubit $S_{x}$ eigenstates as $| \pm ⟩ = \frac{| ↑ ⟩ \pm | ↓ ⟩}{\sqrt{2}}$ . The joint total spin and joint x-spin eigenstates are obtained by replacement of $↑ \to +$ , $↓ \to -$ . We have the spin singlet:

| s = 0, m_{x} = 0 ⟩ = \frac{| + - ⟩ - | - + ⟩}{\sqrt{2}}

and the triplet states:

| s = 1, m_{x} ⟩ = {\begin{cases} | + + ⟩ & m_{x} = 1 \\ \frac{| + - ⟩ + | - + ⟩}{\sqrt{2}} & m_{x} = 0 \\ | - - ⟩ & m_{x} = - 1 \end{cases}

The singlet state is basis invariant, so $| s = 0, m_{z} = 0 ⟩ = | s = 0, m_{x} = 0 ⟩$ and hence all the singlet states that pass through the first stop are part of the undeflected output. For the spin triplet $| s = 1, m_{z} = 0 ⟩$ state we observe that:

| s = 1, m_{z} = 0 ⟩ = \frac{(\frac{| + ⟩ + | - ⟩}{\sqrt{2}}) (\frac{| + ⟩ - | - ⟩}{\sqrt{2}}) + (\frac{| + ⟩ - | - ⟩}{\sqrt{2}}) (\frac{| + ⟩ + | - ⟩}{\sqrt{2}})}{\sqrt{2}} = \frac{| + + ⟩ + | - - ⟩}{\sqrt{2}}

Thus:

| s = 1, m_{z} = 0 ⟩ = \frac{1}{\sqrt{2}} (| s = 1, m_{x} = 1 ⟩ + | s = 1, m_{x} = - 1 ⟩)

Therefore of the $| s = 1, m_{z} = 0 ⟩$ states that pass through the first beamstop, 1/2 will be right deflected and 1/2 will be left deflected.

In conclusion - the right-deflected beam contains $\frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}$ of the original beam, in state:

| s = 1, m_{x} = 1 ⟩ = | + + ⟩ = \frac{| ↑↑ ⟩ + | ↑↓ ⟩ + | ↓↑ ⟩ + | ↓↓ ⟩}{2} .

The left deflected beam contains $\frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}$ of the original beam, in state:

| s = 1, m_{x} = - 1 ⟩ = | - - ⟩ = \frac{| ↑↑ ⟩ - | ↑↓ ⟩ - | ↓↑ ⟩ + | ↓↓ ⟩}{2} .

The undeflected beam contains $\frac{1}{2} \cdot 1 = \frac{1}{2}$ of the original beam, in state:

| s = 0, m_{x} = 0 ⟩ = \frac{| + - ⟩ - | - + ⟩}{\sqrt{2}} = \frac{| ↑↓ ⟩ - | ↓↑ ⟩}{\sqrt{2}} .

(b) This can be simply read off from the expressions of the output states (in the $z$ -basis) for each of the three outputs given in (a). For the right and left deflected beams, the probability of finding the electron and positron to be in the same spin state is $\frac{1}{2}$ . For the undeflected beam, the probability is $0$ .

Quantum Single-Slit Diffraction

Source: UBC Fall 2011 Physics Qualifying Exam Q14

Problem Statement: A beam of atoms of mass $m$ and energy $E$ is passed through a hole of diameter $d$ in an opaque plate normal to the beam. The atoms are then detected by a second plate a distance $L$ away. Under the most ideal circumstances, what is the lower bound for the diameter $D$ of the spot that the beam forms on the detector plate? $L$ and $E$ are fixed, but $d$ may be varied. Take the beam cross section to be always much wider than the hole. Notice that as $d \to 0$ , the uncertainty in momentum leads to $D \to \infty$ , while, as $d \to \infty$ , clearly $D$ also becomes very large. You are asked to find the optimal hole size $d$ for which $D$ is the smallest.

Solution: Assuming the hole size $d$ is smaller than the characteristic length scale of an atom (which we take to twice the Bohr radius $a_{0}$ ), when the atom passes through the hole we can approximate their spatial wavefunction as uniform across the hole:

ψ (x) = {\begin{cases} \frac{1}{\sqrt{d}} & - \frac{d}{2} \leq x \leq \frac{d}{2} \\ 0 & otherwise \end{cases} .

Here we consider the $x$ direction as normal to the beam axis. By taking the Fourier transform, we obtain the momentum wavefunction:

\tilde{ψ} (p) = \frac{1}{\sqrt{2 π ℏ}} \int_{- \infty}^{\infty} e^{- i p x / ℏ} ψ (x) d x = \frac{1}{\sqrt{2 π ℏ d}} \int_{- d / 2}^{d / 2} e^{- i p x / ℏ}

Carrying out the integral we obtain:

\tilde{ψ} (p) = \frac{1}{\sqrt{2 π ℏ d}} \frac{ℏ}{- i p} (e^{- i p d / 2 ℏ} - e^{i p d / 2 ℏ}) = \sqrt{\frac{2 ℏ}{π d}} \frac{1}{p} \sin (\frac{p d}{2}) .

If the atoms have energy $E$ , they have total momentum $p = \sqrt{2 m E}$ and the component of their momenta normal to the flight axis is given by $p_{x} = p \sin θ = \sqrt{2 m E} \sin θ$ where $θ$ is the angle between the beam axis and the vertical position on the final detector plate. For small $θ$ we approximate $\sin θ \approx \tan θ = \frac{x}{L}$ and so $p_{x} \approx \frac{\sqrt{2 m E} x}{L}$ . Substituting $p_{x}$ into $\tilde{ψ}$ and taking the modulus squared, we obtain the intensity of atoms on the final detector as a function of $x$ :

I (x) = | \tilde{ψ} (p_{x}) |^{2} = \frac{2 ℏ}{π d} \frac{1}{p_{x}^{2}} \sin^{2} (\frac{p_{x} d}{2}) = \frac{ℏ L^{2}}{π m E d x^{2}} \sin^{2} (\frac{\sqrt{m E} d x}{\sqrt{2} L}) .

The diameter of the spot on the detector is simply the width of the central fringe of $I (x)$ , which has zeroes at $x = \frac{n \sqrt{2} π L}{\sqrt{m E} d}$ for all nonzero $n \in Z$ and so:

D = \frac{2 \sqrt{2} π L}{\sqrt{m E} d}

$D \propto \frac{1}{d}$ and so to minimize the size of the spot we take $d$ as large as possible. The constraint is with our initial assumption of $ψ (x)$ uniform across the hole only applies if $d \leq 2 a_{0}$ , so $D$ is minimized at $d = 2 a_{0}$ where it has the minimum value:

D = \frac{\sqrt{2} π L}{\sqrt{m E} a_{0}} .

Gravitational Potential

Source: UBC Fall 2005 Physics Qualifying Exam Q6

Problem Statement: A particle of mass $m$ is restricted to move in the vertical direction in the Earth’s gravitational field. Assume tht the surface of the Earth reflects this particle elastically (like a steel ball falling on a surface of glass) and go through the following steps to quantize the energy of the motion using the momentum representation instead of the position representation. Set $ℏ = 1$ throughout for convenience.

(a) Set up the Hamiltonian, assuming that the gravitational potential is zero on the surface of the Earth.

(b) Set up the Schrödinger equation in the momentum representation, and then solve it. Do not use any boundary conditions yet, but use the abbreviations:

$\begin{matrix} (4) & 2 m^{2} g = \frac{1}{l^{3}} 2 m E = \frac{λ}{l^{2}} \end{matrix}$

(c) From the momentum space eigenfunctions, obtain the position space eigenfunctions $ψ (x)$ (Do not worry about normalization). Express $ψ (x)$ in terms of the Airy function, whose integral representation is given by:

$\begin{matrix} (5) & Ai (z) = \frac{1}{2 π} \int_{- \infty}^{\infty} \exp (i (\frac{u^{3}}{3} + u z)) d u . \end{matrix}$

(d) Elastic reflection means that $ψ (x = 0) = 0$ - use this to find the transcendental equation that quantizes the energy. How would you use it to find the energy levels?

Solution: (a) The potential energy (assuming zero potential at $z = 0$ ) is given by $V (z) = m g z$ , and thus the Hamiltonian has the form:

\hat{H} = \frac{{\hat{p}}^{2}}{2 m} + m g \hat{z} .

with $\hat{p}$ the operator corresponding to $z$ -momentum.

(b) In the momentum representation $\hat{p} \to p$ and $\hat{z} \to i ℏ \frac{\partial}{\partial p}$ so the Schrodinger equation $H | ψ ⟩ = E | ψ ⟩$ reads:

(\frac{p^{2}}{2 m} + m g i \frac{\partial}{\partial p}) \tilde{ψ} (p) = E \tilde{ψ} (p)

with $\tilde{ψ} (p)$ the momentum eigenfunctions. Using the provided abbreviations, we write the above as:

\frac{\partial}{\partial p} \tilde{ψ} (p) = - i (l λ - l^{3} p^{2}) \tilde{ψ} (p)

This can be solved by inspection:

\tilde{ψ} (p) = \exp (- i (l λ p - \frac{1}{3} l^{3} p^{3})) .

ψ (x) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{i p x} \tilde{ψ} (p) d p = \frac{1}{2 π} \int_{- \infty}^{\infty} \exp (i (- (l λ - x) p + \frac{1}{3} l^{3} p^{3}))

Substituting $u = l p$ the above can be written in terms of the Airy function:

ψ (x) = \frac{\sqrt{2 π}}{l} Ai (\frac{x}{l} - λ) .

(d) Using the boundary condition of $ψ (0) = 0$ , we find the transcendental equation:

Ai (- λ) = 0

Thus based on the zeros $- λ_{n}$ of the Airy function (which can be found numerically), we can invert the defining relation of $λ$ to find the energies:

E_{n} = - \sqrt[3]{\frac{m g^{2}}{2}} λ_{n} .

Note that the energies will be positive as the roots of the Airy function are negative.

Rigid Rotor in Magnetic Field

Source: UBC PHYS 402 2022 Final Q4

Problem Statement: The Hamiltonian of a rigid rotor placed in a magnetic field perpendicular to its axis takes the approximate: $\begin{matrix} (6) & H = α L^{2} + β L_{z} + γ L_{x} \end{matrix}$ with $α, β, γ$ real and $L_{i}$ the angular momentum operators.

(a) Show that the exact eigenenergies are given by: $\begin{matrix} (7) & E_{l, m} = α ℏ^{2} l (l + 1) + \sqrt{β^{2} + γ^{2}} ℏ m . \end{matrix}$

(b) Assuming $γ ≪ β$ , use perturbation theory to lowest non-vanishing order to independently obtain approximate energy eigenvalues.

(c) Check consistency between the results in (a) and (b).

Solution: (a) We can exactly solve the problem via a coordinate transformation. In particular, we rotate the system about the $y$ -axis by angle $θ$ such that $\tan θ = γ / β$ . This gives us the transformed Hamiltonian:

H = α L^{2} + \sqrt{β^{2} + γ^{2}} L_{z}^{'}

where we note that $L^{2}$ is invariant under rotations and $L_{z}^{'}$ is the $z$ -angular momentum operator in the transformed coordiante system. Since $[L^{2}, L_{z}^{'}] = 0$ , they share a simultaneous eigenbasis, and labelling the eigenstates as $| l, m^{'} ⟩$ we find the eigenergies to be:

E_{l, m^{'}} = α ℏ^{2} l (l + 1) + \sqrt{β^{2} + γ^{2}} ℏ m^{'}

as claimed.

(b) The perturbation theory, we take the base Hamiltonian to be $H_{0} = α L^{2} + β L_{z}$ and the perturbation to be $H^{'} = γ L_{x} = γ \frac{L_{+} + L_{-}}{2}$ where we write $L_{x}$ in terms of the raising/lowering operators of angular momentum. The eigenbasis of $H_{0}$ is $| l, m ⟩$ . Thus, the zeroth order energy eigenvalues are:

E_{l, m}^{(0)} = α ℏ^{2} l (l + 1) + β ℏ m .

Computing the first order corrections, we find:

E_{l, m}^{(1)} = ⟨ l, m | H^{'} | l, m ⟩ = \frac{γ}{2} ⟨ l, m | (L_{+} + L_{-}) | l, m ⟩ = 0.

These vanish and we must go to second order:

E_{l, m}^{(2)} = \sum_{l^{'}, m^{'} \neq l, m} \frac{| ⟨ l, m | H^{'} | l^{'}, m^{'} ⟩ |^{2}}{E_{l, m}^{(0)} - E_{l^{'}, m^{'}}^{(0)}} = \frac{γ^{2} ℏ^{2}}{4} [\frac{l (l + 1) - m (m - 1)}{E_{l, m}^{(0)} - E_{l, m - 1}^{(0)}} + \frac{l (l + 1) - m (m + 1)}{E_{l, m}^{(0)} - E_{l, m + 1}^{(0)}}]

Substituing in the zeroth order energies we find:

E_{l, m}^{(2)} = \frac{γ^{2} ℏ m}{2 β} .

E_{l, m} = α ℏ^{2} l (l + 1) + β \sqrt{1 + {(\frac{γ}{β})}^{2}} ℏ m = α ℏ^{2} l (l + 1) + β (1 + \frac{1}{2} {(\frac{γ}{β})}^{2}) ℏ m + O ({(\frac{γ}{β})}^{3})

thus:

E_{l, m} = α ℏ^{2} l (l + 1) + β ℏ m + \frac{γ^{2} ℏ m}{2 β} + O ({(\frac{γ}{β})}^{3})

and the exact result agrees to perturbation theory to second order.

Symmetries and Transition Amplitudes

Source: UBC PHYS 500 2023 HW2 Q2

Problem Statement: One of the most important applications of symmetries in quantum mechanics is that they can be used to constrain the amplitudes for many dynamical processes without doing any calculation.

(a) Consider a time-independent Hamiltonian $H$ and a symmetry operation implemented by a unitary operator $S$ that commutes with $H$ , $[S, H] = 0$ . Show that the amplitude $⟨ ψ_{f} | U (t) | ψ_{i} ⟩$ for a state $| ψ_{i} ⟩$ to transition to a different state $| ψ_{f} ⟩$ vanishes if $| ψ_{i, f} ⟩$ are eigenstates of $S$ with different eigenvalues, i.e. if $S | ψ_{i, f} ⟩ = e^{i ϕ_{i, f}} | ψ_{i, f} ⟩$ with $ϕ_{i} \neq ϕ_{f}$ . Here $U (t) = \exp (- i H t)$ is the time evolution unitary.

(b) The quadratic/harmonic potential is symmetric under parity, $Π : x \to - x$ and $Π : p \to - p$ . Denote the unitary opeartor that implements parity by the same symbol, $Π$ . What are the possible eigenvalues of $Π$ ? What is the parity of the $n$ th energy eigenstate $| n ⟩ = \frac{(a^{†})^{n}}{\sqrt{n!}} | 0 ⟩$ ? Suppose we start with an initial state equal to the ground state $| 0 ⟩$ of the Harmonic oscillator, and turn on a quartic potential $V_{4} = α x^{4}$ . What is the amplitude to find the particle in the first excited state of the harmonic oscillator at time $t$ later?

Solution: (a) We evaluate the expression $⟨ ψ_{f} | S^{†} U (t) S - U (t) | ψ_{i} ⟩$ in two different ways. First, since $[S, H] = 0$ then $[S, U (t)] = 0$ and so:

⟨ ψ_{f} | S^{†} U (t) S - U (t) | ψ_{i} ⟩ = ⟨ ψ_{f} | S^{†} S U (t) - U (t) | ψ_{i} ⟩ = ⟨ ψ_{f} | U (t) - U (t) | ψ_{i} ⟩ = 0

For the other way we evaluate the two terms:

⟨ ψ_{f} | S^{†} U (t) S - U (t) | ψ_{i} ⟩ = ⟨ ψ_{f} | S^{†} U (t) S | ψ_{i} ⟩ - ⟨ ψ_{f} | U (t) | ψ_{i} ⟩ = (e^{i (ϕ_{i} - ϕ_{f})} - 1) ⟨ ψ_{f} | U (t) | ψ_{i} ⟩

Thus:

0 = (e^{i (ϕ_{i} - ϕ_{f})} - 1) ⟨ ψ_{f} | U (t) | ψ_{i} ⟩

so unless $ϕ_{i} = ϕ_{f}$ then $⟨ ψ_{f} | U (t) | ψ_{i} ⟩ = 0.$

(b) Note that applying spatial inversion twice amounts to doing nothing; $P^{2} (x) = P \circ P (x) = P (- x) = x$ (and analogously for $p$ ) and so $P^{2} = I$ . Hence the square of the eigenvalues of $P$ are $λ^{2} = 1$ and hence $λ = \pm 1$ . Next, note the definition of the raising operator $a^{†}$ :

a^{†} = \sqrt{\frac{m ω}{2}} (x - i \frac{p}{m ω}) .

$a^{†}$ is linear in $x, p$ and so $P a^{†} = - a^{†}$ , and more generally $P (a^{†})^{n} = (- 1)^{n} (a^{†})^{n}$ . Noting that $| 0 ⟩$ has even parity (its wavefunction is a Gaussian symmetric about $x = 0$ ), the parity of $| n ⟩ = \frac{1}{\sqrt{(} n + 1)!} (a^{†})^{n} | 0 ⟩$ is therefore $(- 1)^{n}$ .

Next, note that $V_{4} = α x^{4}$ commutes with $P$ as it is even in $x$ . Therefore, $[P, H_{SHO} + V_{4}] = 0$ and since the ground state $| 0 ⟩$ and first excited state $| 1 ⟩$ of the harmonic oscillator have different parities, by our result in (a) the transition amplitude vanishes.

Thermal States

Source: UBC PHYS 500 2023 HW3 Q1

Problem Statement: Consider a single electron spin (spin-1/2) in a magnetic field with Hamiltonian:

H = - g μ_{B} B S_{z}

where $g$ is the g-factor of the spin, $μ_{B}$ is the Bohr magneton, $B$ is the magnetic field (here in the $z$ -direction) and $S_{z} = \frac{1}{2} σ_{z}$ is the $z$ -component of spin (with $ℏ = 1$ ).

(a) What is the the expected $z$ -magnetization, $m = ⟨ S_{z} ⟩ = ⟨ \frac{1}{2} σ_{z} ⟩$ for a thermal equilibrium state of the spin at temperature $T$ ?

(b) Compute the spin suceptibility $χ = \frac{\partial m}{\partial B}$ as a function of temperature. How does this quantity behave asymptotically in the extreme limits of $T \to 0, \infty$ ?

(c) Compute the entropy as a function of temperature. Compute and explain the limiting value of the entropy in the extreme limits of $T \to 0, \infty$ ?

Solution: The thermal equilibrium state is given by (working in units with $k_{B} = 1$ ):

ρ_{T} = \frac{e^{- H / T}}{Z} = \frac{1}{e^{- \frac{g μ_{B} B}{T}} + e^{\frac{g μ_{B} B}{T}}} (\begin{matrix} e^{\frac{g μ_{B} B}{T}} & 0 \\ 0 & e^{- \frac{g μ_{B} B}{T}} \end{matrix}) = \frac{1}{2 \cosh (\frac{g μ_{B} B}{T})} (\begin{matrix} e^{\frac{g μ_{B} B}{T}} & 0 \\ 0 & e^{- \frac{g μ_{B} B}{T}} \end{matrix}) .

From this we can easily compute the $z$ -magnetization:

m = ⟨ \frac{1}{2} σ_{z} ⟩ = Tr (\frac{1}{2} σ_{z} ρ_{T}) = \frac{1}{4 \cosh (\frac{g μ_{B} B}{T})} Tr ((\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) (\begin{matrix} e^{\frac{g μ_{B} B}{T}} & 0 \\ 0 & e^{- \frac{g μ_{B} B}{T}} \end{matrix})) = \frac{e^{\frac{g μ_{B} B}{T}} - e^{- \frac{g μ_{B} B}{T}}}{4 \cosh (\frac{g μ_{B} B}{T})}

so we conclude:

m = \frac{1}{2} \tanh (\frac{g μ_{B} B}{T}) .

(b) Computing the susceptibility we have:

χ = \frac{\partial m}{\partial B} = \frac{g μ_{B}}{2 T} {sech}^{2} (\frac{g μ_{B} B}{T})

where we use that $\partial_{x} \tanh (x) = {sech}^{2} (x)$ and the chain rule. In the $T \to \infty$ limit, we note first that $lim_{T \to \infty} sech (\frac{g μ_{B} B}{T}) \sim \frac{1}{4}$ and so:

lim_{T \to \infty} χ \sim \frac{g μ_{B}}{32 T}

and the magnetization drops to 0 as $\sim 1 / T$ . in the $T \to 0$ limit, $lim_{T \to 0} sech (\frac{g μ_{B} B}{T}) \sim \frac{e^{- \frac{g μ_{B} B}{T}}}{2}$ and so:

lim_{T \to 0} χ \sim \frac{g μ_{B}}{8 T} e^{- \frac{2 g μ_{B} B}{T}}

so the magnetization drops exponentially quickly to 0.

(c) We recall the entropy of a quantum state is given by $S (ρ) = - Tr (ρ \log ρ) = - \sum_{i} λ_{i} \log λ_{i}$ with $λ_{i}$ the eigenvalues of $ρ$ . Therefore:

S (ρ_{T}) = - \frac{e^{\frac{g μ_{B} B}{T}}}{Z} \log (\frac{e^{\frac{g μ_{B} B}{T}}}{Z}) - \frac{e^{- \frac{g μ_{B} B}{T}}}{Z} \log (\frac{e^{- \frac{g μ_{B} B}{T}}}{Z}) = \frac{\log (Z)}{Z} 2 \cosh (\frac{g μ_{B} B}{T}) - \frac{1}{Z} \frac{g μ_{B} B}{T} 2 \sinh (\frac{g μ_{B} B}{T})

which with $Z = 2 \cosh (\frac{g μ_{B} B}{T})$ becomes:

S (ρ_{T}) = \log (2) + \log (\cosh (\frac{g μ_{B} B}{T})) - \frac{g μ_{B} B}{T} \tanh (\frac{g μ_{B} B}{T})

In the $T \to \infty$ limit the second/third terms go to zero and we have:

lim_{T \to \infty} S (ρ_{T}) = \log (2) .

This makes sense as at infinite temperature both of the spin states are equally favoured and hence the quantum state is maximally mixed, thus yielding the maximal 1-spin entropy of $\log (2)$ . The $T \to 0$ limit must be taken more carefully as both terms diverge; letting $x = \frac{T}{g μ_{B} B}$ we find:

lim_{x \to 0} \log (\cosh (\frac{1}{x})) - \frac{1}{x} \tanh (\frac{1}{x}) = lim_{x \to 0} \log (\frac{e^{\frac{1}{x}}}{2}) - \frac{1}{x} \frac{e^{\frac{1}{x}}}{e^{\frac{1}{x}}} = lim_{x \to 0} \log (e^{\frac{1}{x}}) - \log (2) - \frac{1}{x} = - \log (2)

Thus:

lim_{T \to 0} S (ρ_{T}) = 0.

This also makes sense, as at $T = 0$ the thermal equilibrium is simply the pure ground state $| ↑ ⟩$ which has no entropy.

Scattering of Identical Particles

Source: UBC PHYS 500 2023 HW6 Q4

Problem Statement: Consider two particles of mass $m$ , interacting by the Yukawa potential: $\begin{matrix} (8) & V (r) = \frac{e^{- κ r}}{r} \end{matrix}$ where $r = | {\vec{r}}_{1} - {\vec{r}}_{2} |^{2}$ is the distance between the two particles. Denote the initial momenta of the particles as $\pm k \hat{z}$ where $\hat{z}$ is the unit vector in the $z$ -direction. Find the differential cross section $\frac{d σ}{d Ω}$ for particles to scatter into a thin shell of solid angle $d Ω = \sin θ d θ$ around the final momenta $\pm {\vec{k}}_{f}$ , where ${\vec{k}}_{f}$ has angle $θ$ from the $z$ -axis. Express your final answer in terms of $k, m θ$ . Compute separately for the case of (spin-less or spin-polarized) bosons and fermions. Comment on the effect of identical particle statistics for the case $θ = π / 2$ . The Fourier transform of the Yukawa potential may be useful: $\begin{matrix} (9) & \tilde{V} (\vec{q}) = \int d^{3} r e^{- i \vec{q} \cdot \vec{r}} V (r) = \frac{4 π}{q^{2} + κ^{2}} . \end{matrix}$

Solution: The state corresponding to the particles with momenta ${\vec{k}}_{1}, {\vec{k}}_{2}$ is given by:

| ψ ({\vec{k}}_{1}, {\vec{k}}_{2}) ⟩ = | {\vec{k}}_{1}, {\vec{k}}_{2} ⟩ + ζ | {\vec{k}}_{2}, {\vec{k}}_{1} ⟩

where $ζ = \pm 1$ accounts for the particle exchange statistics (1 for bosons, -1 for fermions). Expanding out the above in the position basis:

| ψ ({\vec{k}}_{1}, {\vec{k}}_{2}) ⟩ = \int d^{3} r_{1} d^{3} r_{2} \frac{1}{L^{3} \sqrt{2}} (e^{i {\vec{k}}_{1} ({\vec{r}}_{1} - {\vec{r}}_{2})} + ζ e^{- i {\vec{k}}_{2} ({\vec{r}}_{1} - {\vec{r}}_{2})}) | r_{1}, r_{2} ⟩

The initial state is the above with ${\vec{k}}_{1} = {\vec{k}}_{i} = k \hat{z}, {\vec{k}}_{2} = - {\vec{k}}_{i} = - k \hat{z}$ and the final state is the above with ${\vec{k}}_{1} = {\vec{k}}_{f}, {\vec{k}}_{2} = - {\vec{k}}_{f}$ . The scattering matrix element is:

M ({\vec{k}}_{f}, {\vec{k}}_{i}) = ⟨ ψ ({\vec{k}}_{f}, - {\vec{k}}_{f}) | V ({\vec{r}}_{1} - {\vec{r}}_{2}) | ψ ({\vec{k}}_{i}, - {\vec{k}}_{i}) ⟩

as an integral this takes the form:

M ({\vec{k}}_{f}, {\vec{k}}_{i}) = \int d^{3} r_{1} d^{3} r_{2} \frac{1}{2 L^{6}} [e^{i ({\vec{k}}_{f} - {\vec{k}}_{i}) \cdot ({\vec{r}}_{1} - {\vec{r}}_{2})} + c . c . + ζ e^{i ({\vec{k}}_{f} + {\vec{k}}_{i}) \cdot ({\vec{r}}_{1} - {\vec{r}}_{2})} + c . c .] V ({\vec{r}}_{1} - {\vec{r}}_{2})

with c.c. denoting the complex conjugate. If we convert to center of mass $\vec{R} = \frac{{\vec{r}}_{1} + {\vec{r}}_{2}}{2}$ and relative coordinates $\vec{r} = {\vec{r}}_{1} - {\vec{r}}_{2}$ , the above simplifies significantly:

M ({\vec{k}}_{f}, {\vec{k}}_{i}) = \int \frac{d^{3} R}{L^{3}} \frac{d^{3} r}{2 L^{3}} [e^{i ({\vec{k}}_{f} - {\vec{k}}_{i}) \cdot \vec{r}} + c . c . + ζ e^{i ({\vec{k}}_{f} + {\vec{k}}_{i}) \cdot \vec{r}} + c . c .] V (\vec{r}) .

The integral over $R$ yields one as the integrand is independent of $R$ , and the leftover integrals are simply the fourier transform of $V$ :

M ({\vec{k}}_{f}, {\vec{k}}_{i}) = \frac{1}{2} (\tilde{V} ({\vec{k}}_{f} - {\vec{k}}_{i}) + \tilde{V} ({\vec{k}}_{f} - {\vec{k}}_{i}) + 2 ζ \tilde{V} ({\vec{k}}_{f} + {\vec{k}}_{i}))

Up until this point the discussion has been for a totally general potential $V / \tilde{V}$ . However, using the Hint we see that $\tilde{V} (\vec{q})$ for the Yukawa potential only depends on the magnitude of the momenta, and so the above simplifies even further:

M ({\vec{k}}_{f}, {\vec{k}}_{i}) = \tilde{V} (| {\vec{k}}_{f} - {\vec{k}}_{i} |) + ζ \tilde{V} (| {\vec{k}}_{f} + {\vec{k}}_{i} |)

Since the scattering is elastic, $| {\vec{k}}_{f} | = | {\vec{k}}_{i} |$ and hence:

| {\vec{k}}_{f} \pm {\vec{k}}_{i} | = k \sqrt{2 (1 \pm \cos θ)} = {\begin{cases} 2 k \cos (θ / 2) & (+) \\ 2 k \sin (\frac{θ}{2}) & (-) \end{cases}

where $θ$ is the angle between the initial and final momenta (and here also the polar angle, as ${\vec{k}}_{i} = k \hat{z}$ ). Thus:

M ({\vec{k}}_{f}, {\vec{k}}_{i}) = M (k, θ) = \tilde{V} (2 k \sin (\frac{θ}{2})) + ζ \tilde{V} (2 k \cos (\frac{θ}{2}))

Then computing the differential cross section as the square of the scattering matrix element/amplitude:

\frac{d σ}{d Ω} = | \frac{m}{2 π} M (k, θ) |^{2} = \frac{m^{2}}{4 π^{2}} | \tilde{V} (2 k \sin (\frac{θ}{2})) + ζ \tilde{V} (2 k \cos (\frac{θ}{2})) |^{2}

using the form of $\tilde{V} (\vec{q})$ provided we conclude:

\frac{d σ}{d Ω} = 4 m^{2} {(\frac{1}{4 k^{2} \sin^{2} (\frac{θ}{2}) + κ^{2}} + \frac{ζ}{4 k^{2} \sin^{2} (\frac{θ}{2}) + κ^{2}})}^{2}

for $θ = π / 2$ , for bosons we have:

\frac{d σ}{d Ω} (θ = π / 2) = 4 m^{2} {(\frac{2}{2 k^{2} \sin^{2} + κ^{2}})}^{2} .

which is twice the result for distinguishable particles. For fermions we have:

\frac{d σ}{d Ω} (θ = π / 2) = 0.

arising from destructive interference between the two possible scattering processes ( $k_{i} \hat{z} \to k_{f} \hat{x}, - k_{i} \hat{z} \to - k_{f} \hat{x}$ and $k_{i} \hat{z} \to - k_{f} \hat{x}, - k_{i} \hat{z} \to k_{f} \hat{x}$ ).

Spin-Orbit Coupling in the 3-D Harmonic Oscillator

Source: UChicago 2014 Quantum Mechanics Graduate Diagnostic Exam Q2

Problem Statement: Consider a spin-1/2 particle of mass $m$ moving in a three-dimensional harmonic oscillator potential: $\begin{matrix} (10) & V (r) = \frac{1}{2} m ω^{2} r^{2} \end{matrix}$ with $r^{2} = x^{2} + y^{2} + z^{2}$ .

(a) What are the energy and degeneracy (including spin) of the ground state(s) and the first excited state(s) of this system?

(b) The general form of the spin-orbit coupling for a particle of mass m and spin \vec{S} moving in a radial potential $V (r)$ is: $\begin{matrix} (11) & H_{s - o} = \frac{1}{2 m^{2} c^{2}} \vec{S} \cdot \vec{L} \frac{1}{r} \frac{d V (r)}{d r} \end{matrix}$ What are the corrections to the ground state and first excited state energy eigenvalues for a spin-1/2 particle in the three-dimensional isotropic harmonic oscillator potential above when this spin-orbit interaction is included?

Solution: (a) We can write the 3-D harmonic oscillator Hamiltonian as the sum of 1-D harmonic oscillator Hamiltonians:

H_{3 D} = \frac{{\vec{p}}^{2}}{2 m} + V (r) = \frac{p_{x}^{2} + p_{y}^{2} + p_{z}^{2}}{2 m} + \frac{1}{2} m ω^{2} (x^{2} + y^{2} + z^{2}) = H_{x} + H_{y} + H_{z}

Thus the eigenstates of $H_{3 D}$ are simply the tensor product of the eigenstates of the 1D harmonic oscillators in each direction, denoted by the four (including spin) quantum numbers to be $| n_{x}, n_{y}, n_{z}, s_{z} ⟩$ , and with energies:

E_{n_{x}, n_{y}, n_{z}, s_{z}} = E_{n_{x}} + E_{n_{y}} + E_{n_{z}} = ℏ ω (n_{x} + \frac{1}{2}) + ℏ ω (n_{y} + \frac{1}{2}) + ℏ ω (n_{z} + \frac{1}{2}) = ℏ ω (n_{x} + n_{y} + n_{z} + \frac{3}{2}) .

Thus the ground state energy is obtained by taking $n_{x} = n_{y} = n_{z} = 0$ :

E_{g} = \frac{3}{2} ℏ ω

which is 2-fold degenerate due to the spin (which can be $\pm 1 / 2$ ). The first excited state has one of the $n_{i}$ equal to one and the rest to 0, and has energy:

E_{1 - e x} = \frac{5}{2} ℏ ω .

There are three options for which of the $n_{i}$ are one and two choices for the spin state, and thus this state is 6-fold degenerate.

(b) We first convert the eigenstates of the previous section to be eigenstates of the angular momentum operators $L^{2}$ and $L_{z}$ . First, we note that $L_{i} = ϵ_{i j k} r_{j} p_{k}$ (with $ϵ_{i j k}$ the Levi-Civita symbol and using the Einstein summation notation), and that we can express the position/momentum operators in terms of the raising/lowering operators $a_{i}^{†} / a_{i}$ as:

r_{i} = \sqrt{\frac{ℏ}{2 m ω}} (a_{i}^{†} + a_{i})

p_{i} = i \sqrt{\frac{ℏ m ω}{2}} (a_{i}^{†} - a_{i}) .

Thus we can express the angular momentum operators in terms of the raising/lowering operators:

L_{i} = i ℏ ϵ_{i j k} a_{j} a_{k}^{†}

$L^{2} = L_{i} L_{i} = ℏ^{2} (N (N + 1) - a_{k}^{†} a_{k}^{†} a_{j} a_{j})$ where repeated indices are summed over and $N = a_{i}^{†} a_{i}$ is the (total) number operator.

Using these expressions, we find that the states $| n_{x} = 0, n_{y} = 0, n_{z} = 0, s_{z} = \pm 1 / 2 ⟩$ of the last section are eigenstates of $L^{2}$ and $L_{z}$ with eigenvalue zero (using that $a_{j} | n_{j} = 0 ⟩ = 0$ ), and thus we may write them as angular momentum eigenstates of $| l = 0, m = 0, s = 1 / 2, s_{z} = \pm 1 / 2 ⟩$ .

The first excited states with one quanta are eigenstates of $L^{2}$ - the $a_{j} a_{j}$ term vanishes and we are left with $L^{2} = ℏ^{2} l (l + 1) = ℏ^{2} (1) (1 + 1) = 2 ℏ^{2}$ . The $| n_{x} = 0, n_{y} = 0, n_{z} = 1, s_{z} = \pm 1 / 2 ⟩$ states are also an eigenstate of $L_{z}$ (with eigenvalue 0) and can be identified with $| l = 1, m = 0, s = 1 / 2, s_{z} = \pm 1 / 2 ⟩$ . The other two states are not immediately eigenstates of $L_{z}$ but by taking linear combinations we obtain:

| l = 1, m = \pm 1, s = 1 / 2, s_{z} = \pm 1 / 2 ⟩ = \frac{| n_{x} = 1, n_{y} = 0, n_{z} = 0, s_{z} = \pm 1 / 2 ⟩ \pm | n_{x} = 0, n_{y} = 1, n_{z} = 0, s_{z} = \pm 1 / 2 ⟩}{\sqrt{2}} .

We now have a basis more convenient for thinking about spin-orbit coupling. We treat the spin-orbit coupling as a perturbation, with the form:

H_{s - o} = \frac{1}{2 m^{2} c^{2}} \vec{S} \cdot \vec{L} \frac{1}{r} \frac{d V (r)}{d r} = \frac{1}{2 m^{2} c^{2}} \vec{S} \cdot \vec{L} \frac{1}{r} (\frac{1}{2} m ω^{2} 2 r) = \frac{ω^{2}}{m c^{2}} \vec{S} \cdot \vec{L}

The $| l, m, s, s_{z} ⟩$ basis we use is inconvenient for calculating the energy corrections as these states are not eigenstates of $\vec{S} \cdot \vec{L}$ . However, defining the joint angular momentum $J_{i} = L_{i} + S_{i}$ we find that $J^{2} = L^{2} + S^{2} + 2 \vec{S} \cdot \vec{L}$ and so $\vec{S} \cdot \vec{L} = \frac{J^{2} - L^{2} - S^{2}}{2}$ . Thus we will find it convenient to work in the $| j, j_{z}, l, s ⟩$ basis. To this end we note:

| j = 1 / 2, j_{z} = \pm 1 / 2, l = 0, s = 1 / 2 ⟩ = | l = 0, m = 0, s = 1 / 2, s_{z} = \pm 1 / 2 ⟩

| j = 3 / 2, j_{z} = \pm 3 / 2, l = 1, s = 1 / 2 ⟩ = | l = 1, m = \pm 1, s = 1 / 2, s_{z} = \pm 1 / 2 ⟩

\begin{aligned} | j = 3 / 2, j_{z} = 1 / 2, l = 1, s = 1 / 2 ⟩ & = \sqrt{\frac{1}{3}} | l = 1, m = 1, s = 1 / 2, s_{z} = - 1 / 2 ⟩ \\ + \sqrt{\frac{2}{3}} | l = 1, m = 0, s = 1 / 2, s_{z} = 1 / 2 ⟩ \end{aligned}

\begin{aligned} | j = 1 / 2, j_{z} = 1 / 2, l = 1, s = 1 / 2 ⟩ & = \sqrt{\frac{2}{3}} | l = 1, m = 1, s = 1 / 2, s_{z} = - 1 / 2 ⟩ \\ - \sqrt{\frac{1}{3}} | l = 1, m = 0, s = 1 / 2, s_{z} = 1 / 2 ⟩ \end{aligned}

\begin{aligned} | j = 3 / 2, j_{z} = - 1 / 2, l = 1, s = 1 / 2 ⟩ & = \sqrt{\frac{1}{3}} | l = 1, m = - 1, s = 1 / 2, s_{z} = 1 / 2 ⟩ \\ + \sqrt{\frac{2}{3}} | l = 1, m = 0, s = 1 / 2, s_{z} = - 1 / 2 ⟩ \end{aligned}

\begin{aligned} | j = 1 / 2, j_{z} = - 1 / 2, l = 1, s = 1 / 2 ⟩ & = - \sqrt{\frac{2}{3}} | l = 1, m = - 1, s = 1 / 2, s_{z} = 1 / 2 ⟩ \\ + \sqrt{\frac{1}{3}} | l = 1, m = 0, s = 1 / 2, s_{z} = - 1 / 2 ⟩ . \end{aligned}

These equations can be obtained by comparing the quantum numbers on each side, with the coefficients in the last four equations being determined with the help of a Clebsh-Gordon table (Note: to do this part of the question properly, we could apply the raising/lowering operators of angular momentum to find the coefficients (and indeed this is how the CG-table is derived). However, the exact coefficients presented here are not strictly relevant for solving the problem, and so we omit this (slightly tedious) step of the calculation). With this basis transformation established, we can compute the energy corrections using first-order perturbation theory:

\begin{aligned} E_{j, j_{z}, l, s}^{(1)} = ⟨ j, j_{z}, l, s | H_{s - 0} | j, j_{z}, l, s ⟩ & = \frac{ω^{2}}{m c^{2}} ⟨ j, j_{z}, l, s | \frac{J^{2} - L^{2} - S^{2}}{2} | j, j_{z}, l, s ⟩ \\ = \frac{ℏ^{2} ω^{2}}{2 m c^{2}} (j (j + 1) - l (l + 1) - s (s + 1)) \end{aligned}

Thus for the ground state(s) with $l = 0$ , $s = 1 / 2$ and $j = 1 / 2$ we find:

E_{j = 1, j_{z} = \pm 1 / 2, l = 0, s = 1 / 2}^{(1)} = \frac{ℏ^{2} ω^{2}}{2 m c^{2}} (\frac{1}{2} (\frac{1}{2} + 1) - 0 (0 + 1) - \frac{1}{2} (\frac{1}{2} + 1)) = 0.

Note that we could have deduced this without any calculation, as the ground state(s) are spherically symmetric, and spin-orbit coupling does not impact such states. For the first excited state(s), the energy correction depends on the total angular momenta, as there are $j = 3 / 2$ and $j = 1 / 2$ sectors:

E_{j = 3 / 2, j_{z} = \pm 3 / 2 \pm 1 / 2, l = 1, s = 1 / 2}^{(1)} = \frac{ℏ^{2} ω^{2}}{2 m c^{2}} (\frac{3}{2} (\frac{3}{2} + 1) - 1 (1 + 1) - \frac{1}{2} (\frac{1}{2} + 1)) = \frac{ℏ^{2} ω^{2}}{2 m c^{2}} .

E_{j = 1 / 2, j_{z} = \pm 1 / 2, l = 1, s = 1 / 2}^{(1)} = \frac{ℏ^{2} ω^{2}}{2 m c^{2}} (\frac{1}{2} (\frac{1}{2} + 1) - 1 (1 + 1) - \frac{1}{2} (\frac{1}{2} + 1)) = - \frac{ℏ^{2} ω^{2}}{m c^{2}} .

Identical Particles in an Infinite Cubic Well

Source: UChicago 2014 Quantum Mechanics Graduate Diagnostic Exam Q3

Problem Statement: (a) First, consider a single particle of mass $m$ in a three-dimensional box of dimensions $L \times L \times L / 4$ :

$\begin{matrix} (12) & V (x, y, z) = {\begin{cases} 0 & if 0 \leq x \leq L, 0 \leq y \leq L, 0 \leq z \leq L / 4 \infty & otherwise. \end{cases} \end{matrix}$

Find the three lowest energy eigenvalues and all the corresponding eigenfunctions of the particle in the box. Express all energies in units of $ϵ_{0} = \frac{ℏ^{2} π^{2}}{2 m L^{2}}$ .

(b) Now suppose that we place three identical spin-0 particles in the box. For the combined system of three particles, find the two lowest eigenvalues of the total energy. Find the degeneracy associated with each of these eigenvalues (i.e. find the total number of distinct three-particle states corresponding to each eigenvalue).

(c) Repeat part (b) for spin-1/2 particles.

(d) Repeat part (b) for spin-1 particles.

Solution: We can solve the problem independently in $x, y, z$ and then combine the eigenvalues/eigenfunctions to obtain the 3D solution. The eigenvalues and eigenfunctions of the 1-D infinite square well of length $L$ are $E_{n}, ψ_{n} (x)$ where:

E_{n} = \frac{ℏ^{2} π^{2} n^{2}}{2 m L^{2}} = ϵ_{0} n^{2}

ψ_{n} (x) = \sqrt{\frac{2}{L}} \sin (\frac{n π x}{L})

Thus the eigenvalues/eigenfunctions of the 3D cubic infinite square well are labelled by quantum numbers $n_{x}, n_{y}, n_{z}$ , with:

E_{n_{x}, n_{y}, n_{z}} = ϵ_{0} n_{x}^{2} + ϵ_{0} n_{y}^{2} + 16 ϵ_{0} n_{z}^{2}

ψ_{n_{x}, n_{y}, n_{z}} (x, y, z) = ψ_{n_{x}} (x) ψ_{n_{y}} (y) ψ_{n_{z}} (4 z) = 2 {(\frac{2}{L})}^{3 / 2} \sin (\frac{n_{x} π x}{L}) \sin (\frac{n_{y} π y}{L}) \sin (\frac{n_{z} 4 π z}{L})

where $n_{i} \in N$ and we note that the length is $L / 4$ in the $z$ -direction. From this, we can easily read off the three lowest eigenenergies to be $E_{1, 1, 1} = 18 ϵ_{0}, E_{2, 1, 1} = E_{1, 2, 1} = 19 ϵ_{0}, E_{2, 2, 1} = 20 ϵ_{0}$ . With the respective eigenfunctions given in the equation above with the corresponding $n_{x}, n_{y}, n_{z}$ .

(b) We first note the celebrated spin-statistics theorem, which tells us that bosons have integer spin and fermions have half-integer spin. Thus the spin-0 particles here are bosons, which can occupy the same quantum state. They each only have a single $s = 0$ spin state, and thus this does not enter into the degeneracy.

The lowest eigenvalue is with all three particles in the $(n_{x}, n_{y}, n_{z}) = (1, 1, 1)$ state, with a total energy of $3 \cdot 18 ϵ_{0} = 54 ϵ_{0}$ with no degeneracy. The next lowest eigenvalue is with one particle in either $(2, 1, 1)$ or $(1, 2, 1)$ with the other two particles in the $(1, 1, 1)$ state, giving a total energy of $2 \cdot 18 ϵ_{0} + 19 ϵ_{0} = 55 ϵ_{0}$ . Since the particles are indistinguishable, the only choice is whether the more energetic particle is in $(2, 1, 1)$ or $(1, 2, 1)$ and thus this eigenvalue is 2-fold degenerate.

(c) The spin-1/2 particles are fermions, which obey Pauli exclusion - hence any two particles cannot be in the same state. We also note that there is an additional quantum number for spin with $s_{z} = \pm 1 / 2$ - this does not affect the energy of the single-particle states, but will enter into our considerations of degeneracy.

The lowest eigenvalue is with one particle in $(n_{x}, n_{y}, n_{z}, s_{z}) = (1, 1, 1, 1 / 2)$ , another in $(1, 1, 1, - 1 / 2)$ , and the last particle in one of $(2, 1, 1, \pm 1 / 2)$ or $(1, 2, 1, \pm 1 / 2)$ . Hence the eigenvalue is $55 ϵ_{0}$ with 4-fold degeneracy. The second lowest eigenvalue is $56 ϵ_{0}$ , which has combinations of one particle in $(1, 1, 1, \pm 1 / 2)$ , and the other two particles being in $(2, 1, 1, \pm 1 / 2)$ or $(1, 2, 1, \pm 1 / 2)$ states (so long as they are not in the same state). There are $2 \cdot 6 = 12$ possible choices in this way. Another possibility is for one particle in $(1, 1, 1, 1 / 2)$ , the second in $(1, 1, 1, - 1 / 2)$ , and the last in $(2, 2, 1, \pm 1 / 2)$ . There are 2 possible choices in this configuration. Thus in total this eigenvalue is 14-fold degenerate.

(d) The spin-1 particles are bosons. They can occupy the same state, and have the spin quantum number $s_{z} = - 1, 0, 1$ . This again does not affect the energy of the single particle states, but will again enter in degeneracy counting.

The lowest eigenvalue is with all particles in $(n_{x}, n_{y}, n_{z}, s_{z}) = (1, 1, 1, s_{z})$ states, with eigenvalue $54 ϵ_{0}$ . The possibilities for the spins are all $+ 1$ , all $0$ , all $- 1$ , two $+ 1$ /one $0$ or $- 1$ , two $0$ /one $+ 1$ or $- 1$ , two $- 1$ /one $+ 1$ or $0$ . Thus this eigenvalue is 9-fold degenerate. The next lowest eigenvalue is $55 ϵ_{0}$ . Here two particles are in $(1, 1, 1, s_{z})$ and one particle is in $(2, 1, 1, s_{z})$ or $(1, 2, 1, s_{z})$ . There are 6 choices for the excited particle (3 for the spin state, 2 for which of $n_{x}, n_{y}$ is 2) and 6 choices for the spin states of the ground state particles (both $+ 1$ , both $0$ , both $- 1$ , one $+ 1$ /one $0$ , one $+ 1$ /one $- 1$ , one $0$ /one $- 1$ ) and by multiplying we find 36-fold degeneracy.

Time-Varying Force

Source: UChicago 2014 Quantum Mechanics Graduate Diagnostic Exam Q4

Problem Statement: A one-dimensional harmonic oscillator of mass $m$ and frequency $ω$ is in its ground state for $t < 0$ . For $t \geq 0$ it is subject to a time-dependent but spatially constant force in the $x$ -direction of $F (t) = F_{0} \exp (- t / τ)$ .

(a) Calculate the probability of finding the oscillator in the first excited state for $t > 0$ as a function of time $t$ to lowest order in $F_{0}$ . You may need the creation and annihilation operators in terms of $x$ and $p$ :

$\begin{matrix} (13) & a = \sqrt{\frac{m ω}{2 ℏ}} x + i \frac{1}{\sqrt{2 m ω ℏ}} p \end{matrix}$ $\begin{matrix} (14) & a^{†} = \sqrt{\frac{m ω}{2 ℏ}} x - i \frac{1}{\sqrt{2 m ω ℏ}} p . \end{matrix}$

(b) The probability of finding the oscillator in the $n$ th excited state for $t > 0$ is proportional to a power of $F_{0}$ for small $F_{0}$ . What is that power?

Solution: (a) First recalling that $F = - \frac{\partial V}{\partial x}$ for a conservative force, we have that a time-dependent potential of $V (t) = - F_{0} x \exp (- t / τ)$ can give rise to $F (t)$ . To first order in perturbation theory, the transition probability to the first excited state as a function of time $t$ is given by:

p_{0 \to 1} (t) = | \int_{0}^{t} ⟨ 1 | V (t^{'}) | 0 ⟩ e^{- i ω_{0, 1} t^{'}} d t^{'} |^{2}

The difference in energy between the ground state and first excited state is $E_{0, 1} = \frac{3}{2} ℏ ω - \frac{1}{2} ℏ ω = ℏ ω$ and so the characteristic frequency between the two states is just the frequency of the oscillator $ω 0, 1 = \frac{E_{0, 1}}{ℏ} = ω$ . To evaluate the matrix element, we first note that (inverting the definitions of the creation/annihilation operators):

x = \sqrt{\frac{ℏ}{2 m ω}} (a + a^{†})

and so:

⟨ 1 | V (t^{'}) | 0 ⟩ = - F_{0} \exp (- t^{'} / τ) ⟨ 1 | x | 0 ⟩ = - F_{0} \exp (- t^{'} / τ) \sqrt{\frac{ℏ}{2 m ω}} ⟨ 1 | (a + a^{†}) | 0 ⟩ = - F_{0} \exp (- t^{'} / τ) \sqrt{\frac{ℏ}{2 m ω}}

Thus the transition probability becomes:

p_{0 \to 1} (t) = \frac{F_{0}^{2} ℏ^{2}}{2 m ω} | \int_{0}^{t} \exp (- (i ω + \frac{1}{τ}) t^{'}) d t^{'} |^{2}

Carrying out the integral:

p_{0 \to 1} (t) = \frac{F_{0}^{2} ℏ^{2}}{2 m ω} | \frac{\exp (- (i ω + \frac{1}{τ}) t) - 1}{i ω + \frac{1}{τ}} |^{2}

Computing the modulus squared we conclude:

p_{0 \to 1} (t) = \frac{F_{0}^{2} ℏ^{2} τ^{2}}{2 m ω} \frac{1 - 2 \exp (- t / τ) \cos (ω t) + \exp (- 2 t / τ)}{τ^{2} ω^{2} + 1}

(b) Now considering higher-order perturbation theory, the transition amplitude to the $| n ⟩$ state only becomes nonvanishing at $n$ th order (as each order can raise the energy by one level). Each order picks up a factor of $F_{0}$ in the amplitude and thus the dependency of the probability is $F_{0}^{2 n} .$

Two Interacting Fermions

Source: MIT Fall 2015 Doctoral General Examination Quantum Q2

Problem Statement: Consider two identical fermions of mass $m$ interacting with each other through an attractive harmonic potential. The Hamiltonian is

$\begin{matrix} (15) & H = \frac{{\vec{p}}_{1}^{2}}{2 m} + \frac{{\vec{p}}_{2}^{2}}{2 m} + \frac{k}{2} ({\vec{x}}_{1} - {\vec{x}}_{2})^{2} \end{matrix}$

${\vec{x}}_{1, 2}$ are the coordinates of the two fermions, and ${\vec{p}}_{1, 2}$ are the conjugate momenta. For simplicity assume that the spins of both fermions are polarized in the same direction (e.g, in the $↑$ direction) so that we may ignore the spin degree of freedom.

(a) State the restriction imposed by Fermi statistics on acceptable wave functions $ψ ({\vec{x}}_{1}, {\vec{x}}_{2})$ of this system.

(b) Rewrite $H$ in terms of center-of-mass and relative coordinates.

(c) Ignoring the restriction imposed by Fermi statistics what is the ground state energy? What is the energy of the first excited bound state?

(d) Including the effects of Fermi statistics what is the ground state energy? What is the degeneracy of the ground state?

Solution: (a) Fermi statistics dictate that the wavefunction must have odd parity/be antisymmetric under exchange of the particles, and therefore:

ψ ({\vec{x}}_{1}, {\vec{x}}_{2}) = - ψ ({\vec{x}}_{2}, {\vec{x}}_{1})

(b) We define the center of mass and relative coordinates $\vec{R} = \frac{m {\vec{r}}_{1} + m {\vec{r}}_{2}}{m + m} = \frac{{\vec{r}}_{1} + {\vec{r}}_{2}}{2}$ and $\vec{r} = {\vec{r}}_{1} - {\vec{r}}_{2}$ . We also have the conjugate center of mass and relative momenta of $\vec{P} = {\vec{p}}_{1} + {\vec{p}}_{2}$ and $\vec{p} = \frac{{\vec{p}}_{1} - {\vec{p}}_{2}}{2}$ , which inverting we obtain ${\vec{p}}_{1} = \frac{\vec{P} + 2 \vec{p}}{2}$ and ${\vec{p}}_{2} = \frac{\vec{P} - 2 \vec{p}}{2}$ , and thus we can rewrite $H$ as:

H = \frac{{\vec{P}}^{2}}{4 m} + \frac{{\vec{p}}^{2}}{m} + \frac{k}{2} {\vec{r}}^{2}

(c) The Hamiltonian decouples into $H = H_{c m} + H_{r e l}$ where $H_{c m}$ is the Hamiltonian of a free particle with mass $2 m$ and $H_{r e l}$ is the Hamiltonian of a particle of mass $\frac{m}{2}$ in a 3-D harmonic oscillator potential with spring constant $k$ . The former has unbounded spectrum $E_{c m} =\in [0, \infty)$ and for simplicity we assume that $E_{c m} = 0$ for the rest of the problem. The latter has quantized spectrum $E_{n_{x}, n_{y}, n_{z}} = \frac{3}{2} ℏ ω (n_{x} + n_{y} + n_{z} + \frac{3}{2})$ (as $H_{r e l}$ can be decomposed into 3 1-D simple harmonic oscillators, each with energy $E_{n} = \frac{ℏ}{ω} (n + \frac{1}{2})$ ), with $ω = \sqrt{\frac{2 k}{m}}$ . Ignoring the restriction from Fermi statistics, the ground state is then the state with $n_{x} = n_{y} = n_{z} = 0$ and has energy $\frac{3}{2} ℏ ω$ . The excited state has one $n_{i} = 1$ and the other two zero, with energy $\frac{5}{2} ℏ ω$ .

(d) Including the effects of fermionic exchange statistics, the answer to (c) changes. To this end, we note that the eigenstates of the SHO are eigenstates of parity (this follows as $H$ is quadratic in the $x, p$ quadrature operators and thus $[Π, H] = 0$ ), and in particular (recalling the parity of the Hermite polynomials/wavefunctions, or thinking about the parity of raising operators as is done in Symmetries and Transition Amplitudes ) $Π | n ⟩ = (- 1)^{n} | n ⟩$ . In 3-D this generalizes $Π | n_{x}, n_{y}, n_{z} ⟩ = (- 1)^{n_{x} + n_{y} + n_{z}} | n_{x}, n_{y}, n_{z} ⟩$ . Since fermionic exchange statistics requires that the acceptable states of this system have odd parity, the $n_{x} = n_{y} = n_{z}$ state from (c) with even parity is forbidden. The ground states are instead those with one $n_{i} = 1$ and the other two zero (these have odd parity), with energy $\frac{5}{2} ℏ ω$ . The ground space is thus 3-fold degenerate.

Note: We could reach the same conclusion by observing that the eigenstates of the 3-D harmonic oscillator are eigenstates of angular momentum (see Spin-Orbit Coupling in the 3-D Harmonic Oscillator for more discussion) and can be labelled by quantum numbers $l, m$ . We also recall that the angular momentum eigenstates are eigenstates of parity $Π | l, m ⟩ = (- 1)^{l} | l, m ⟩$ , thus the fermionic exchange statistics/antisymmetric wavefunction forbids $l = 0$ and the ground state(s) consist of the 3-fold degenerate $l = 1$ subspace with $E = \frac{5}{2} ℏ ω$ .

A Quantum Spin Chain

Source: MIT Fall 2012 Doctoral General Examination Quantum Q1

Problem Statement: Consider a one-dimensional chain of $N$ spin-1/2 particles coupled through the Hamiltonian

\begin{equation}\label{eq:heisenbergchain} H = J\sum_{i=1}^{N-1}\vec{S}{i} \cdot \vec{S}{i+1} \end{equation}

where $\vec{S} = (S_{x}, S_{y}, S_{z})$ are the usual spin operators for a spin-1/2 particle, $J > 0$ is a positive constant, and $N ≫ 1$ .

In a famous 1931 paper, Hans Bethe showed that for this Hamiltonian, the ground state energy per particle $E_{G S} / N \equiv E_{0}$ was equal to $ℏ^{2} J (\log 2 - 1 / 4) \approx - 0.433 ℏ^{2} J$ . You will not be required to reproduce this result. Instead, you will determine upper and lower bounds for the ground state energy per particle.

(a) If the spin operators are treated as classical spin vectors with $| \vec{S} | = ℏ / 2$ , what is the ground state spin configuration and what is the ground state energy per particle $E_{0}$ ?

(b) Consider the trial wave function

$\begin{matrix} (16) & | Ψ ⟩ = ⨂_{i = odd} | i, i + 1 ⟩_{0} = | 1, 2 ⟩_{0} \otimes | 3, 4 ⟩_{0} \otimes \dots \otimes | N - 1, N ⟩_{0} \end{matrix}$

where $| i, j ⟩_{0}$ is the spin singlet state formed from the spins on sites $i$ and $j$ . Use this state to find an upper bound on $E_{0}$ .

(c) Prove the following lower bound on E_0:

$\begin{matrix} (17) & - \frac{3}{4} ℏ^{2} J \leq E_{0} \end{matrix}$

Solution: (a) If the spin operators are classical spin vectors, then the most energetically optimal configuration is to make ${\vec{S}}_{i} \cdot {\vec{S}}_{i + 1} = - | {\vec{S}}_{i} | | {\vec{S}}_{i + 1} | = - \frac{ℏ^{2}}{4}$ for each term in the Hamiltonian. The spin configuration that accomplishes this is ${\vec{S}}_{i} = + \frac{ℏ}{2} \hat{z}$ for odd sites, ${\vec{S}}_{i + 1} = - \frac{ℏ}{2} \hat{z}$ for even sites (or vise versa). Since each term in the Hamiltonian contributes $- \frac{1}{4} ℏ^{2} J$ , the ground state energy is thus $E_{G S} = - (N - 1) \frac{1}{4} ℏ^{2} J$ and thus the energy per particle (in the $N ≫ 1$ limit in which $N - 1 \sim N$ ) is $E_{0} = E_{G S} / N = - \frac{1}{4} ℏ^{2} J .$

(b) We first prove the variational theorem, that says that for a Hamiltonian $H$ with ground state energy $E_{G S}$ that $⟨ ψ | H | ψ ⟩ \geq E_{G S}$ for any normalized $| ψ ⟩$ . To begin, we expand $| ψ ⟩$ in the eigenbasis of $H$ , so $| ψ ⟩ = \sum_{n} c_{n} | n ⟩$ where $H | n ⟩ = E_{n} | n ⟩$ , with $E_{0} = E_{G S} \leq E_{n}$ for all $n$ . It then follows that:

⟨ ψ | H | ψ ⟩ = (\sum_{n} c_{n}^{*} ⟨ n |) H (\sum_{n^{'}} c_{n^{'}} | n^{'} ⟩) = \sum_{n} | c_{n} |^{2} E_{n} \geq \sum_{n} | c_{n} |^{2} E_{G S} = E_{G S} \sum_{n} | c_{n} |^{2} = E_{G S} .

In the second equality we use the eigenvalue equation and the orthogonality of the eigenstates, and in the last equality we use the normalization of $| ψ ⟩$ that ensures that $\sum_{n} | c_{n} |^{2} = 1$ . We now apply the variational theorem with the provided trial state. We will find it convenient to split the Hamiltonian into two parts:

⟨ Ψ | H | Ψ ⟩ = J \sum_{i odd} ⟨ Ψ | {\vec{S}}_{i} \cdot {\vec{S}}_{i + 1} | Ψ ⟩ + \sum_{i even} ⟨ Ψ | {\vec{S}}_{i} \cdot {\vec{S}}_{i + 1} | Ψ ⟩

Studying the first sum, we have $(N - 1) / 2$ terms of the form: $⟨ i, i + 1 |_{0} {\vec{S}}_{i} \cdot {\vec{S}}_{i + 1} | i, i + 1 ⟩_{0}$ . Defining $\vec{S} = {\vec{S}}_{i} + {\vec{S}}_{i + 1}$ , we note that ${\vec{S}}_{i} \cdot {\vec{S}}_{i + 1} = \frac{1}{2} (\vec{S} - {\vec{S}}_{1} - {\vec{S}}_{2})$ and so joint eigenstates of $\vec{S}, {\vec{S}}_{1}, {\vec{S}}_{2}$ have eigenvalue $\frac{ℏ^{2}}{2} (s (s + 1) - s_{1} (s_{1} + 1) - s_{2} (s_{1} + 1)) = \frac{ℏ^{2}}{2} (s (s + 1) - \frac{3}{2})$ where $s$ is the total joint spin. For a singlet state $s = 0$ and so:

| i, i + 1 |_{0} {\vec{S}}_{i} \cdot {\vec{S}}_{i + 1} | i, i + 1 ⟩_{0} = \frac{ℏ^{2}}{2} (0 - \frac{3}{2}) = - \frac{3}{4} ℏ^{2}

thus we compute the energy from the odd terms:

J \sum_{i odd} ⟨ Ψ | {\vec{S}}_{i} \cdot {\vec{S}}_{i + 1} | Ψ ⟩ = - \frac{N - 1}{2} \frac{3}{4} ℏ^{2} J = - (N - 1) \frac{3}{8} .

Now studying the second sum, we have $(N-1)/2$$ terms of the form:

⟨ i, i + 1 |_{0} ⟨ i + 2, i + 3 |_{0} {\vec{S}}_{i + 1} \cdot {\vec{S}}_{i + 2} | i, i + 1 ⟩_{0} | i + 2, i + 3 ⟩_{0}

which we evaluate to be:

⟨ i, i + 1 |_{0} {\vec{S}}_{i + 1} | i, i + 1 ⟩_{0} ⟨ i + 2, i + 3 |_{0} {\vec{S}}_{i + 2} | i + 2, i + 3 ⟩_{0} = 0 \cdot 0 = 0

where we use that the ${\vec{S}}_{i + 1} \cdot {\vec{S}}_{i + 2}$ acts on the subspaces independently (allowing us to factor the expectation value) and we use that the singlet state is spherically symmeric to conclude that $⟨ i, i + 1 |_{0} {\vec{S}}_{i + 1} | i, i + 1 ⟩_{0} = 0$ . Thus:

\sum_{i even} ⟨ Ψ | {\vec{S}}_{i} \cdot {\vec{S}}_{i + 1} | Ψ ⟩ = 0

Thus, using the variational theorem we conclude:

E_{G S} \leq ⟨ Ψ | H | Ψ ⟩ \leq - \frac{N - 1}{2} \frac{3}{4} ℏ^{2} J

from which we can bound $E_{0} = E_{G S} / N$ to be:

E_{0} \leq - \frac{3}{8} ℏ^{2} J .

(c) Let $H$ be any Hamiltonian of the form $H = \sum_{k} H_{k}$ such that $H$ has ground state energy $| ψ_{G S} ⟩$ and $E_{G S}$ and the $H_{k}$ s have ground state energy $E_{G S, k}$ . Then it follows that:

E_{G S} = ⟨ ψ_{G S} | H | ψ_{G S} ⟩ = \sum_{k} ⟨ ψ_{G S} | H_{k} | ψ_{k} ⟩ \geq \sum_{k} E_{G S, k}

where in the last inequality we use the variational theorem on the individual $H_{k}$ terms. Since the ground state energy of ${\vec{S}}_{i} \cdot {\vec{S}}_{i + 1}$ is $- \frac{3}{4} ℏ^{2} J$ , using this we can lower bound $E_{G S} \geq - (N - 1) \frac{3}{4} ℏ^{2} J$ and thus we obtain the upper-bound on $E_{0}$ :

E_{0} \geq - \frac{3}{4} ℏ^{2} J .

Combining the above two results we have an upper and lower bound on $E_{0}$ :

- \frac{3}{4} ℏ^{2} J \leq E_{0} \leq - \frac{3}{8} ℏ^{2} J

which (as must be the case) the exact result falls within.

Anomalous Magnetic Moment of the Electron

Source: MIT Fall 2012 Doctoral General Examination Quantum Q2

Problem Statement: *The gyromagnetic factor of the electron $g$ determines the relationship between the electron magnetic moment $\vec{μ}$ and the electron spin $\vec{S}$ ,

$\begin{matrix} (18) & \vec{μ} = g \frac{e}{2 m} \vec{S}, \end{matrix}$

where $e$ is the electron charge and $m$ is the electron mass. Famously, the Dirac equation predicts $g = 2$ , but in quantum electrodynamics, the electron picks up na anomalous magnetic moment $g = 2 (1 + a)$ , where the current experimental value is $a = 0.00115965218076 (27)$ .

One way to experimentally measure $a$ is the allow a beam of electrons to interact with a constant magnetic field $\vec{B} = B \hat{z}$ via the Hamiltonian:

$\begin{matrix} (19) & H = \frac{1}{2 m} {(\vec{p} - e \vec{A})}^{2} - \vec{μ} \cdot \vec{B} \end{matrix}$

where $\vec{A}$ is the vector potential. The electrons are confined to the $x - y$ plane, and you can ignore any electron-electron interactions. The electrons will exhibit cyclotron motion with frequency $ω = e B / m$ , but will also exhibit spin precession with a slightly different frequency. In this problem, you will show how to use this phenomenon to extract $a$ .

(a) Verify the commutation relations

$\begin{matrix} (20) & [v_{x}, H] = i ℏ ω v_{y}, [v_{y}, H] = - i ℏ ω v_{x}, \end{matrix}$

where $\vec{v} = (\vec{p} - e \vec{A}) / m$ is the gauge-invariant velocity operator. Hint: $\vec{v}$ is gauge-invariant and thus this problem can be solved in any choice of gauge.

(b) Consider the two expectation values

$\begin{matrix} (21) & C_{1} (t) = ⟨ S_{x} v_{x} + S_{y} v_{y} ⟩, C_{2} (t) = ⟨ S_{x} v_{y} - S_{y} v_{x} ⟩ . \end{matrix}$

Derive a set of coupled differential equations that describe the time evolution of $C_{1} (t)$ and $C_{2} (t)$ . In the special case that $a = 0$ (i.e. $g = 2$ ), verify that $C_{1} (t)$ and $C_{2} (t)$ do not change with time.

(c) A beam of electrons of velocity $\vec{v}$ is prepared at time $t = 0$ in a spin state with known values of $C_{1} (0)$ and $C_{2} (0)$ . The beam interacts with a magnetic field $\vec{B} = B \hat{z}$ between $t = 0$ and $t = T$ . The expectation value $C_{1} (T)$ is experimentally measured to be periodic with period $2 π / Ω$ , i.e. $C_{1} (T) = C_{1} (T + 2 π / Ω)$ . Use this information to determine the value of $a$ in terms of $Ω$ and other physical parameters.

Solution: (a) We follow the hint and work in the Landau gauge $\vec{A} = (0, B x, 0)$ (one can easily verify that $\vec{\nabla} \times \vec{A} = B \hat{z} = \vec{B}$ ) to make the calculations as simple as possible. We then note that:

\vec{v} = \frac{\vec{p} - e \vec{A}}{m} = \frac{1}{m} (p_{x}, p_{y} - e B x, p_{z})

we then observe:

[v_{x}, v_{x}] = [v_{y}, v_{y}] = 0

[v_{x}, v_{y}] = - [v_{y}, v_{x}] = - \frac{1}{m^{2}} [p_{y} - e B x, p_{x}] = \frac{e B}{m^{2}} [x, p_{x}] = i \frac{ℏ e B}{m^{2}} = i ℏ \frac{ω}{m}

Thus if we write:

H = \frac{m}{2} (v_{x}^{2} + v_{y}^{2}) - (1 + a) \frac{e}{m} B S_{z}

then we can compute:

[v_{x}, H] = \frac{m}{2} [v_{x}, v_{y}^{2}] = \frac{m}{2} (i ℏ \frac{ω}{m} v_{y}) = i ℏ ω v_{y}

[v_{y}, H] = \frac{m}{2} [v_{y}, v_{x}^{2}] = \frac{m}{2} (- i ℏ \frac{ω}{m} v_{x}) = - i ℏ ω v_{x}

Thus we conclude:

[v_{x}, H] = i ℏ ω v_{y}, [v_{y}, H] = - i ℏ ω v_{x} .

(b) We recall Ehrenfest’s theorem:

\frac{d ⟨ A ⟩}{d t} = \frac{1}{i ℏ} ⟨ [A, H] ⟩ .

to this end we compute the commutators:

\begin{aligned} = \frac{m}{2} [S_{x} v_{x}, v_{y}^{2}] - (1 + a) \frac{e}{m} B [S_{x} v_{x}, S_{z}] + \frac{m}{2} [S_{y} v_{y}, v_{x}^{2}] - (1 + a) \frac{e}{m} B [S_{y} v_{y}, S_{z}] \\ = \frac{m}{2} S_{x} (i ℏ \frac{ω}{m} v_{y}) - (1 + a) \frac{e}{m} B (- i ℏ S_{y}) v_{x} + \frac{m}{2} S_{y} (- i ℏ \frac{ω}{m} v_{x}) - (1 + a) \frac{e}{m} B (i ℏ S_{z}) v_{y} \\ = - i ℏ ω a [S_{x} v_{y} - S_{y} v_{x}] \end{aligned}

\begin{aligned} = \frac{m}{2} [S_{x} v_{y}, v_{x}^{2}] - (1 + a) \frac{e}{m} B [S_{x} v_{y}, S_{z}] - \frac{m}{2} [S_{y} v_{x}, v_{y}^{2}] + (1 + a) \frac{e}{m} B [S_{y} v_{x}, S_{z}] \\ = \frac{m}{2} S_{x} (- i ℏ \frac{ω}{m} v_{x}) - (1 + a) \frac{e}{m} B (- i ℏ S_{y}) v_{y} + \frac{m}{2} S_{y} (i ℏ \frac{ω}{m} v_{y}) - (1 + a) \frac{e}{m} B (i ℏ S_{x}) v_{x} \\ = - i ℏ ω a [S_{x} v_{x} - S_{y} v_{y}] . \end{aligned}

Thus we obtain the two (coupled) differential equations:

\frac{d C_{1}}{d t} = ⟨ [S_{x} v_{x} + S_{y} v_{y}, H] ⟩ = - ω a C_{2}

\frac{d C_{2}}{d t} = ⟨ [S_{x} v_{y} - S_{y} v_{x}, H] ⟩ = ω a C_{1}

if $a = 0$ then $\frac{d C_{1}}{d t} = \frac{d C_{2}}{d t} = 0$ so $C_{1} (t), C_{2} (t)$ are constant in time.

{\ddot{C}}_{1} = - ω a {\dot{C}}_{2} = - ω a (ω a C_{1}) = - ω^{2} a^{2} C_{1}

from which we find:

C_{1} (t) = A \cos (ω a t) + B \sin (ω a t)

with $A, B$ determined by the known initial conditions on $C_{1}$ and $C_{2}$ . Thus we find $C_{1}$ to have period $\frac{2 π}{ω a}$ . If we experimentally measure $C_{1}$ to have a period of $\frac{2 π}{Ω}$ , equating the two expressions we can experimentally measure $a$ as the ratio:

a = \frac{Ω}{ω} .

A Heisenberg Ferromagnet

Source: MIT Spring 2012 Doctoral General Examination Quantum Q1

Problem Statement: In a ferromagnetic material the electron spins are aligned, suggesting that an interaction of the form:

$\begin{matrix} (22) & δ H = κ {\vec{S}}_{1} \cdot {\vec{S}}_{2} \end{matrix}$

is present between each pair of electrons, where ${\vec{S}}_{1}, {\vec{S}}_{2}$ are the operators corresponding to the spins of the two electrons. While Eq. $(22)$ does not appear explicitly in the Hamiltonian for a ferromagnet, Heisenberg realized that Eq. $(22)$ could appear as an effective interaction, arising from Coulomb repulsion and fermionic properties of electrons.

In this problem you will derive a Heisenberg ferromagnetism for two spin-1/2 electrons in a common potential $V (\vec{r})$ , with Hamiltonian:

$\begin{matrix} (23) & H = \frac{{\vec{p}}_{1}^{2}}{2 m} + \frac{{\vec{p}}_{2}^{2}}{2 m} + V ({\vec{r}}_{1}) + V ({\vec{r}}_{2}) + \frac{e^{2}}{| {\vec{r}}_{1} - {\vec{r}}_{2} |} \end{matrix}$

Note that $δ H$ does not appear in the above, so there is no explicit spin dependence. The single-particle Schrödinger equation with potential $V (\vec{r})$ has eigenstates with energies $E_{i}$ and wavefunctions $ψ_{i} (\vec{r})$ .

(a) The total wave function $Ψ ({\vec{r}}_{1}, s_{1} {\vec{r}}_{2}, s_{2})$ depends on the set of spin variables $s_{1}, s_{2}$ (use whichever notation for spin that you prefer, including bra/ket notation). Show that the eigenstates of the Hamiltonian can be written as a separable form $ψ ({\vec{r}}_{1}, {\vec{r}}_{2}) χ (s_{1}, s_{2})$ . Construct eigenstates of total spin, and describe the symmetry properties of $ψ ({\vec{r}}_{1}, {\vec{r}}_{2})$ under particle exchange for each of the spin states.

(b) In the absence of Coulomb repulsion, consider the two-particle configurations where one electron is in state $ψ_{a} (\vec{r})$ and the other is in state $ψ_{b} (\vec{r})$ . What are the corresponding two-particle wavefunctions, including spin?

(c) The degeneracy of the states in part (b) is broken by Coulomb repulsion, yielding two distinct energy levels. Identify the two-particle wavefunctions associated with these two levels, and find an expression for the energy splitting $δ E$ to first order in the Coulomb interaction in terms of $ψ_{a}$ and $ψ_{b}$ .

(d) Show that the energy splitting in part (c) can be mimicked (at first order) by turning off the Coloumb interaction in Eq. $(23)$ and replacing it with Eq. $(22)$ . Find an expression for $κ$ in terms of $δ E$ .

(e) Determine the sign of $κ$ . Hint: Judicious use of Fourier transforms may be helpful. Recall that

$\begin{matrix} (24) & \int d^{3} r e^{i \vec{q} \cdot \vec{r}} (2 π)^{3} δ^{(3)} (\vec{q}), \int d^{3} r e^{i \vec{q} \vec{r}} \frac{1}{| \vec{r} |} = \frac{4 π}{{\vec{q}}^{2}} . \end{matrix}$

Solution: (a) Since Eq. $(23)$ is independent of spin, its eigenstates are the position eigenstates $| ψ ({\vec{r}}_{1}, {\vec{r}}_{2}) ⟩$ multiplied by an arbitrary spin state $| χ (s_{1}, s_{2}) ⟩$ . Thus the wavefunctions have the claimed separable form of $Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s_{1}, s_{2}) = ψ ({\vec{r}}_{1}, {\vec{r}}_{2}) χ (s_{1}, s_{2})$ . The eigenstates of total spin $\vec{S} = {\vec{S}}_{1} + {\vec{S}}_{2}$ are the triple states:

| s = 1, m ⟩ = {\begin{cases} | ↑↑ ⟩ & m = 1 \\ \frac{| ↑↓ ⟩ + | ↓↑ ⟩}{\sqrt{2}} & m = 0 \\ | ↓↓ ⟩ & m = - 1 \end{cases}

and the singlet state:

| s = 0, m = 0 ⟩ = \frac{| ↑↓ ⟩ - | ↓↑ ⟩}{\sqrt{2}} .

Since electrons are fermions, the total wavefunction $Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s_{1}, s_{2})$ must be antisymmetric under particle exchange $1 \leftrightarrow 2$ . For the triplet states (for which the spins are symmetric under exchange) we therefore must have the position wavefunction be antisymmetric and vise versa for the singlet state (for which the spins are antisymmetric under exchange), so we conclude:

ψ ({\vec{r}}_{1}, {\vec{r}}_{2}) = {\begin{cases} - ψ ({\vec{r}}_{2}, {\vec{r}}_{1}) & s = 1 \\ ψ ({\vec{r}}_{2}, {\vec{r}}_{1}) & s = 0 \end{cases} .

(b) In the absence of the coloumb repulsion, $H$ splits into the sum of $H_{1} + H_{2}$ for the individual electrons. Thus we can find the individual eigenfunctions $ψ_{a} (\vec{r}), ψ_{b} (\vec{r})$ and antisymmetrize (for triplet states)

Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 1, m) = \frac{ψ_{a} ({\vec{r}}_{1}) ψ_{b} ({\vec{r}}_{2}) + ψ_{b} ({\vec{r}}_{1}) ψ_{a} ({\vec{r}}_{2})}{\sqrt{2}} χ (s = 1, m)

or symmetrize (for the singlet states):

Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 0, m = 0) = \frac{ψ_{a} ({\vec{r}}_{1}) ψ_{b} ({\vec{r}}_{2}) + ψ_{b} ({\vec{r}}_{1}) ψ_{a} ({\vec{r}}_{2})}{\sqrt{2}} χ (s = 0, m = 0) .

(c) The distinct particle-wavefunctions are those in the triplet states $Ψ^{'} ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 1, m)$ and those in the singlet state $Ψ^{'} ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 0, m)$ (note that due to the Coloumb repulsion, we can no longer write $H = H_{1} + H_{2}$ and thus the decomposition into individual one-particle wavefunctions $ψ_{a}, ψ_{b}$ from (b) no longer holds - thus the eigenfunctions $Ψ^{'}$ here are distinct from the $Ψ$ s in (b)). We can find the energy splitting $δ E$ to first order using perturbation theory, treating $δ H = \frac{e^{2}}{| {\vec{r}}_{1} - {\vec{r}}_{2} |}$ as a perturbation:

δ E = E_{s = 1}^{(1)} - E_{s = 0}^{(1)} = ⟨ Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 1, m) | δ H | Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 1, m) ⟩ - ⟨ Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 0, m) | δ H | Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 0, m) ⟩ .

The degeneracy of $| Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 1, m) ⟩, | Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 0, m = 0) ⟩$ is a priori a cause for concern, as the perturbation must be diagonal in the above basis for degenerate perturbation theory to give a valid result for the energy splitting. However, since $δ H = \frac{e^{2}}{| {\vec{r}}_{1} - {\vec{r}}_{2} |}$ is symmetric under particle interchange, the off diagonal terms indeed vanish, as can be seen via an insertion of the particle exchange operator $Π$ :

\begin{aligned} ⟨ Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 1, m) | δ H | Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 0, m = 0) ⟩ \\ = ⟨ Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 1, m) | Π^{†} δ H Π | Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 0, m = 0) ⟩ \\ = - ⟨ Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 1, m) | δ H | Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 0, m = 0) ⟩ . \end{aligned}

Thus the expression for the energy splitting using perturbation theory is valid. We can evaluate the (diagonal) matrix elements via carrying out integrals by taking integrals in position space and sums in spin space - the sum over spin states drops out as $δ H$ is independent of spin, and thus we have:

δ E = \int d^{3} r_{1} d^{3} r_{2} \frac{e^{2}}{| {\vec{r}}_{1} - {\vec{r}}_{2} |} (\frac{ψ_{a} ({\vec{r}}_{1}) ψ_{b} ({\vec{r}}_{2}) - ψ_{b} ({\vec{r}}_{1}) ψ_{a} ({\vec{r}}_{2})}{\sqrt{2}} \cdot c . c . - \frac{ψ_{a} ({\vec{r}}_{1}) ψ_{b} ({\vec{r}}_{2}) + ψ_{b} ({\vec{r}}_{1}) ψ_{a} ({\vec{r}}_{2})}{\sqrt{2}} \cdot c . c .)

with c.c. denoting the complex conjugate. After the cancellation of terms, we conclude:

δ E = - 2 \int d^{3} r_{1} d^{3} r_{2} \frac{e^{2}}{| {\vec{r}}_{1} - {\vec{r}}_{2} |} ψ_{a}^{*} ({\vec{r}}_{1}) ψ_{b}^{*} ({\vec{r}}_{2}) ψ_{b} ({\vec{r}}_{1}) ψ_{a} ({\vec{r}}_{2}) .

(d) Turning off the Coloumb interaction and turning on the spin interactions, if we again study:

δ E = E_{s = 1}^{(1)} - E_{s = 0}^{(1)} = ⟨ Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 1, m) | δ H | Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 1, m) ⟩ - ⟨ Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 0, m) | δ H | Ψ ({\vec{r}}_{1}, {\vec{r}}_{2}, s = 0, m) ⟩

it is now the integral over position space that drops out as $δ H = κ {\vec{S}}_{1} \cdot {\vec{S}}_{2}$ is independent of position, and we are left with:

δ E = κ (⟨ s = 1, m | {\vec{S}}_{1} \cdot {\vec{S}}_{2} | s = 1, m ⟩ - ⟨ s = 0, m = 0 | {\vec{S}}_{1} \cdot {\vec{S}}_{2} | s = 0, m = 0 ⟩)

We write ${\vec{S}}_{1} \cdot {\vec{S}}_{2} = \frac{1}{2} (S^{2} - S_{1}^{2} - S_{2}^{2})$ , which has eigenvalues $\frac{ℏ^{2}}{2} (s (s + 1) - s_{1} (s_{1} + 1) - s_{2} (s_{2} + 1))$ . Here $s_{1} = s_{2} = \frac{1}{2}$ , and for the triplet ( $s = 1$ ) states evaluates to $- \frac{ℏ^{2}}{4}$ and for the singlet ( $s = 0$ )states evaluates to $- \frac{3 ℏ^{2}}{4}$ . Thus:

δ E = κ (\frac{ℏ^{2}}{4} - (- \frac{3 ℏ^{2}}{4})) = κ ℏ^{2}

Thus the energy splitting in (c) can be mimicked via the spin interaction. Specifically, we tune $$\kappa$ to:

κ = \frac{δ E}{ℏ^{2}} .

(e) We determine the sign of $κ$ by reasoning about $δ E$ from the expression in (c). Defining $F (\vec{r}) = ψ_{a}^{*} ({\vec{r}}_{1}) ψ_{b} ({\vec{r}}_{1})$ and $G ({\vec{r}}_{2} - {\vec{r}}_{1}) = \frac{1}{| {\vec{r}}_{2} - {\vec{r}}_{1} |}$ , the expression becomes:

δ E = - 2 e^{2} \int d^{3} r_{1} d^{3} r_{2} F ({\vec{r}}_{1}) G ({\vec{r}}_{2} - {\vec{r}}_{1}) F^{*} ({\vec{r}}_{2})

Now using the expression in the hint, we can write $G ({\vec{r}}_{2} - {\vec{r}}_{1})$ in terms of its Fourier transform:

G ({\vec{r}}_{2} - {\vec{r}}_{1}) = \frac{1}{(2 π)^{3}} \int d^{3} q \tilde{G} (\vec{q}) e^{i \vec{q} ({\vec{r}}_{2} - {\vec{r}}_{1})}

Thus:

δ E = - 2 e^{2} \frac{1}{(2 π)^{3}} \int d^{3} r_{1} d^{3} r_{2} d^{3} q F ({\vec{r}}_{1}) e^{- i \vec{q} \cdot {\vec{r}}_{1}} \tilde{G} (\vec{q}) F^{*} ({\vec{r}}_{2}) e^{i \vec{q} \cdot {\vec{r}}_{1}}

The integrals over $r_{1}, r_{2}$ simply give the Fourier transforms of $F ({\vec{r}}_{1}) / F^{*} ({\vec{r}}_{2})$ and thus:

δ E = - 2 e^{2} \frac{1}{(2 π)^{3}} \int d^{3} q \tilde{F} (\vec{q}) \tilde{G} (\vec{q}) {\tilde{F}}^{*} (\vec{q}) = - 2 e^{2} \frac{1}{(2 π)^{3}} \int d^{3} | \tilde{F} (\vec{q}) |^{2} \tilde{G} (\vec{q})

Now evaluating substituting the expression for $\tilde{G} (\vec{q})$ from the hint, we obtain:

δ E = - 2 e^{2} \frac{1}{(2 π)^{3}} \int d^{3} q | \tilde{F} (\vec{q}) |^{2} \frac{4 π}{{\vec{q}}^{2}} = - \frac{e^{2}}{π^{2}} \int d^{3} q \frac{| \tilde{F} (\vec{q}) |^{2}}{{\vec{q}}^{2}} .

Since the integrand is positive definite, we conclude $δ E < 0$ . Since $κ = δ E / ℏ^{2}$ we conclude that $κ < 0$ . The spins being aligned is favoured, as we would expect for a ferromagnet.

The Supersymmetric Method

Source: MIT Spring 2012 Doctoral General Examination Quantum Q2

Problem Statement: In this problem you will solve for the energy spectrum of a particle of mass $m$ confined to a potential

$\begin{matrix} (25) & V (x) = V_{0} [\sec^{2} \frac{x}{x_{0}} + \tan^{2} \frac{x}{x_{0}}] = V_{0} [1 + 2 \tan^{2} \frac{x}{x_{0}}] \end{matrix}$

where $- \frac{π}{2} x_{0} \leq x \leq \frac{π}{2} x_{0}$ (recall that $\sec x = 1 / \cos x$ , and the equality of the two expressions above follows from trigonometric identities). Amazingly, this system is exactly solvable for the especial value:

$\begin{matrix} (26) & V_{0} = \frac{ℏ^{2}}{2 m} \frac{1}{x_{0}^{2}}, \end{matrix}$

and you will derive the spectrum using the supersymmetric method. (Supersymmetry is a possible symmetry between bosons and fermions, but that fact will not be relevant for this problem).

(a) Consider two Hamiltonians

$\begin{matrix} (27) & H = A^{†} A, \tilde{H} = A A^{†}, \end{matrix}$

Where $A$ is an unspecified operator. Assume that $H$ has (normalized) eigenstates $| n ⟩$ with

$\begin{matrix} (28) & H | n ⟩ = E_{n} | n ⟩ \end{matrix}$

for $n \geq 1$ . For every $n$ with $E_{n} \neq 0$ , show that $A | n ⟩$ is an unnormalized eigenstate of $\tilde{H}$ . Find the normalized eigenstates $| \tilde{n} ⟩$ and their eigenvalues ${\tilde{E}}_{n}$ under $\tilde{H}$ . What goes wrong with this argument if $E_{n} = 0$ ? Can $E_{n}$ or ${\tilde{E}}_{n}$ ever be negative? Note: For the remainder of this problem, you can assume that $| \tilde{n} ⟩$ forms a complete basis for $\tilde{H}$ , up to possible zero energy states.

(b) Now consider a specific operator $A$ of the form

$\begin{matrix} (29) & A = \frac{\partial}{\partial x} + W (x) \end{matrix}$

where $W (x)$ is real. Show that $H$ and $\tilde{H}$ each describe a particle moving in a potential, in units where $ℏ^{2} / 2 m = 1$ . Find the two potential energy functions, $V (x)$ and $\tilde{V} (x)$ , corresponding to $H$ and $\tilde{H}$ , respectively.

(c) We will be considering Hamiltonians defined on a finite range $- \frac{π}{2} x_{0} \leq x \leq \frac{π}{2} x_{0}$ . This means that the wave functions will have Dirichlet boundary conditions at $x = \frac{π}{2} x_{0}$ , i.e. $ψ (\pm \frac{π}{2} x_{0}) = 0$ . Assume that $H$ has a zero energy ground state consisten with these boundary conditions. Find the unnormalized ground state wave function $ψ_{0} (x)$ for $H$ in terms of $W (x)$ . Show that $\tilde{H}$ cannot have a zero energy ground state consistent with these boundary conditions.

(d) The potential in Eq. $(25)$ is dual to a constant potential. That is, there is a $W (x)$ such that for $- \frac{π}{2} x_{0} \leq x \leq \frac{π}{2} x_{0}$ ,

$\begin{matrix} (30) & V (x) = a, \tilde{V} (x) = b [\sec^{2} \frac{x}{x_{0}} + \tan^{2} \frac{x}{x_{0}}] = b [1 + 2 \tan^{2} \frac{x}{x_{0}}] \end{matrix}$

where $a, b$ are constants. What is $W (x)$ , and what are $a$ and $b$ ? You may find the following formulas helpful:

$\begin{matrix} (31) & \int \frac{d x}{1 + x^{2}} = \arctan x, \int d θ \sec^{2} θ = \tan θ, \int \tan^{2} θ = - θ + \tan θ \end{matrix}$

(e) Find the energy spectrum and energy eigenstates for the potential in Eq. $(25)$ . Does this system have a zero energy ground state? You should assume that all wave functions vanish at $x = \frac{π}{2} x_{0}$ (i.e. Dirichlet boundary conditions), and you should restore all factors of $ℏ$ and $m$ . You do not need to normalize the states.

Solution: (a) We apply $\tilde{H}$ to $A | n ⟩$ :

\tilde{H} A | n ⟩ = A A^{†} A | n ⟩ = A H | n ⟩ = A E_{n} | n ⟩ = E_{n} A | n ⟩ .

Thus $A | n ⟩$ is an eigenstate of $\tilde{H}$ . Since $⟨ n | A^{†} A | n ⟩ = ⟨ n | H | n ⟩ = E_{n} ⟨ n | n ⟩$ , the $A | n ⟩$ have norm of $\sqrt{E_{n}}$ and thus the normalized eigenstates of $\tilde{H}$ are $| \tilde{n} ⟩ = \frac{1}{\sqrt{E_{n}}} A | n ⟩$ with eigenvalues ${\tilde{E}}_{n} = E_{n}$ . The argument fails if $E_{n} = 0$ as then $A | n ⟩$ has zero norm, i.e. it is the zero vector (which cannot be an eigenstate). Further, since $E_{n}$ is the norm squared of $A | n ⟩$ and norms are positive semidefinite, $E_{n} / {\tilde{E}}_{n}$ cannot be negative.

(b) Note that $A = i p + W (x)$ as $p = - i \frac{\partial}{\partial x}$ . Note that from the canonical commutation relation $[x, p] = i$ it follows that $[p, x^{n}] = - i n x^{n - 1}$ (by induction on $n$ ) and thus Taylor expanding a function of $x$ it follows that $[p, G (x)] = - i G^{'} (x)$ (alternatively, since $p = - i \frac{\partial}{\partial x}$ is given, the commutator can be evaluated by acting on a test function). With this result, we compute:

H = A^{†} A = (- i p + W (x)) (i p + W (x)) = p^{2} - i [p, W (x)] + W^{2} (x) = - \frac{\partial^{2}}{\partial x^{2}} - W^{'} (x) + W^{2} (x)

\tilde{H} = A A^{†} = (i p + W (x)) (- i p + W (x)) = p^{2} + i [p, W (x)] + W^{2} (x) = - \frac{\partial^{2}}{\partial x^{2}} + W^{'} (x) + W^{2} (x)

Thus $H, \tilde{H}$ both describe a particle in a potential with $V (x) = - W^{'} (x) + W^{2} (x)$ and $\tilde{V} (x) = W^{'} (x) + W^{2} (x) .$

(c) From (a), we know that for eigenstates $| n ⟩$ with energy $E_{n} = 0$ it follows that $A | n ⟩ = 0$ . Thus we obtain the following ODE for the wavefunction $ψ_{0} (x)$ :

(\frac{\partial}{\partial x} + W (x)) ψ_{0} (x) = 0 ⟹ \frac{\partial}{\partial x} ψ_{0} (x) = - W (x) ψ_{0} (x) .

Which can be solved via inspection:

ψ_{0} (x) = \exp (- \int_{0}^{x} W (x^{'}) d x^{'}) .

Now consider $\tilde{H}$ ; by an anlogous argument, if an eigenstate $| \tilde{n} ⟩$ exists with energy ${\tilde{E}}_{n} = 0$ then $A^{†} | n ⟩ = 0$ and thus:

(- \frac{\partial}{\partial x} + W (x)) {\tilde{ψ}}_{0} (x) = 0 ⟹ \frac{\partial}{\partial x} {\tilde{ψ}}_{0} (x) = W (x) {\tilde{ψ}}_{0} (x) .

This has solution:

{\tilde{ψ}}_{0} (x) = \exp (\int_{0}^{x} W (x^{'}) d x^{'}) = \frac{1}{ψ_{0} (x)}

But $ψ_{0} (\pm \frac{π}{2} x_{0}) = 0$ means that ${\tilde{ψ}}_{0} (\pm \frac{π}{2} x_{0}) = \infty$ , so it is impossible for both $ψ, \tilde{ψ}$ to have a zero-energy eigenstate that satisfies the Dirchlet boundary conditions.

(d) Since $\tilde{V} (x) = b [\sec^{2} \frac{x}{x_{0}} + \tan^{2} \frac{x}{x_{0}}] = W^{'} (x) + W^{2} (x)$ , recalling that $\frac{\partial}{\partial x} \tan x = \sec^{2} x$ we observe $W (x) = \frac{1}{x_{0}} \tan (\frac{x}{x_{0}})$ works with with $b = \frac{1}{x_{0}^{2}}$ . Then since $V (x) = a = W^{'} (x) - W^{2} (x)$ we find that $a = - \frac{1}{x_{0}^{2}} .$ .

(e) We find the spectrum and eigenstates for the dual potential and use this to construct the spectrum and eigenstates for the potential of Eq. $(30)$ . The Dirchlet boundary conditions of $ψ (\pm \frac{π}{2} x_{0}) = 0$ means that the dual problem is simply that of the infinite square well with walls at $x = \pm \frac{π}{2} x_{0}$ and with the bottom of the potential shifted by $-V_0 = \frac{\hbar^2}{2m}\frac{1}{x_0^2}$$. These we recall to be:

ψ_{n} (x) = \sin (n (\frac{x}{x_{0}} - \frac{π}{2}))

E_{n} = \frac{ℏ^{2} n^{2}}{2 m x_{0}^{2}} - V_{0} = V_{0} (n^{2} - 1) n \geq 1.

The eigenvalues of $\tilde{H}$ are identical:

{\tilde{E}}_{n} = V_{0} (n^{2} - 1) n \geq 2

noting that since the result of (c) tells us that $H, \tilde{H}$ cannot share states with zero eigenvalue and hence $E_{1} = 0$ means that $\tilde{H}$ cannot have the $n = 1$ state. The eigenstates of $\tilde{H}$ are obtained by applying $A$ to the above eigenstates and so:

{\tilde{ψ}}_{n} (x) = (\frac{\partial}{\partial x} + W (x)) ψ_{n} (x) = (\frac{\partial}{\partial x} + \frac{1}{x_{0}} \tan (\frac{x}{x_{0}})) \sin (n (\frac{x}{x_{0}} - \frac{π}{2}))

thus we conclude:

{\tilde{ψ}}_{n} (x) = \frac{1}{x_{0}} (n \cos (n (\frac{x}{x_{0}} - \frac{π}{2})) + \sin (n (\frac{x}{x_{0}} - \frac{π}{2})) \tan (\frac{x}{x_{0}})) .

Spin-1/2 in Time-Dependent Magnetic Field

Source: MIT Fall 2002 Doctoral General Examination Quantum Q2

Problem Statement: Consider a spin-1/2 particle interacting with a time-dependent magneitc field:

$\begin{matrix} (32) & H (t) = \vec{B} (t) \cdot \vec{σ} \end{matrix}$

where $\vec{B} (t) = B (\sin θ \cos ω t, \sin θ \sin ω t, \cos θ)$ and $\vec{σ} = (σ_{x}, σ_{y}, σ_{z})$ with $σ_{i}$ the Pauli matrices:

σ_{x} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) σ_{y} = (\begin{matrix} 0 & - i \\ i & 0 \end{matrix}) σ_{z} = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})

(a) Find the instantaneous eigenstates $| +, t ⟩$ and $| -, t ⟩$ of $H (t)$ with eigenvalues $+ B$ anf $- B$ respectively.

(b) Let $| ψ, t ⟩$ obey the Schrödinger equation:

$\begin{matrix} (33) & i ℏ \frac{\partial}{\partial t} | ψ, t ⟩ = H (t) | ψ, t ⟩ \end{matrix}$

with $| ψ, 0 ⟩ = | +, 0 ⟩$ . Calculate $P_{+} (t) | ⟨ +, t | ψ, t ⟩ |^{2}$ exctly. Hint: Expand $| ψ, t ⟩$ in terms of $| +, t ⟩$ and $| -, t ⟩$ and find the equation for the expansion coefficients.

(c) What happens to $P_{+} (t)$ for $ω ≪ B / ℏ$ ? Explain.

Solution: (a) From Spin-1/2 and Stern-Gerlach (taking $θ \to θ$ , $ϕ = ω t$ ) we find:

| +, t ⟩ = (\begin{matrix} \cos (\frac{θ}{2}) \\ e^{i ω t} \sin (\frac{θ}{2}) \end{matrix}) | -, t ⟩ = (\begin{matrix} - \sin (\frac{θ}{2}) \\ e^{i ω t} \cos (\frac{θ}{2}) \end{matrix}) .

(b) We expand:

| ψ, t ⟩ = c_{+} | +, t ⟩ + c_{-} | -, t ⟩

which we substitute into the SE (with $ℏ = 1$ ) to obtain:

i ({\dot{c}}_{+} | +, t ⟩ + c_{+} \dot{| +, t ⟩} + {\dot{c}}_{-} | -, t ⟩ + c_{-} \dot{| -, t ⟩}) = H (t) (c_{+} | +, t ⟩ + c_{-} | -, t ⟩) = B c_{+} | +, t ⟩ - B c_{-} | -, t ⟩ .

We note that:

\dot{| +, t ⟩} = (\begin{matrix} 0 \\ i ω e^{i ω t} \sin (\frac{θ}{2}) \end{matrix}) \dot{| -, t ⟩} = (\begin{matrix} 0 \\ i ω e^{i ω t} \cos (\frac{θ}{2}) \end{matrix})

Now by multiplying both sides by $⟨ +, t |$ we obtain the equation for $c_{+}$ :

i ({\dot{c}}_{+} + c_{+} ⟨ +, t \dot{| +, t ⟩} + c_{-} ⟨ +, t \dot{| -, t ⟩}) = B c_{+}

where we have used the orthogonality of the instantneous eigenstates. An anlogous equation can be derived for ${\dot{c}}_{-}$ , and evaluating the inner products we obtain a set of two coupled differential equations which can be expressed in matrix form:

i (\begin{matrix} {\dot{c}}_{+} \\ {\dot{c}}_{-} \end{matrix}) = (\begin{matrix} B + ω \sin^{2} (\frac{θ}{2}) & ω \sin (\frac{θ}{2}) \cos (\frac{θ}{2}) \\ ω \sin (\frac{θ}{2}) \cos (\frac{θ}{2}) & - B + ω \cos^{2} (\frac{θ}{2}) \end{matrix}) (\begin{matrix} c_{+} \\ c_{-} \end{matrix}) .

This can be solved via diagonalization (the matrix appearing on the RHS is Hermitian). The result (after solving and plugging in the initial condition of $| ψ, 0 ⟩ = | +, 0 ⟩$ ) is:

| ψ, t ⟩ = [\cos (\frac{λ t}{2}) - i \frac{(B - ω \cos θ)}{λ} \sin (\frac{λ t}{2})] | +, t ⟩ - i \frac{B}{λ} \sin θ \sin (\frac{λ t}{2}) | -, t ⟩

with $λ = \sqrt{ω^{2} + B^{2} - 2 ω B \cos θ}$ . We can then calculate $P_{+} (t)$ as:

P_{+} (t) = | \cos (\frac{λ t}{2}) - i \frac{(B - ω \cos θ)}{λ} \sin (\frac{λ t}{2}) |^{2} .

Thus we conbclude: $P_{+} (t) = \cos^{2} (\frac{λ t}{2}) + \frac{ω^{2} \cos^{2} θ + B^{2} - 2 ω B \cos θ}{λ^{2}} \sin^{2} (\frac{λ t}{2})$

\frac{ω^{2} \cos^{2} θ + B^{2} - 2 ω B \cos θ}{λ^{2}} = \frac{ω^{2} \cos^{2} θ + B^{2} - 2 ω B \cos θ}{ω^{2} + B^{2} - 2 ω B \cos θ} \approx \frac{B^{2}}{B^{2}} = 1

thus:

P_{+} (t) \approx \cos^{2} (\frac{λ t}{2}) + \sin^{2} (\frac{λ t}{2}) = 1.

This is consistent with the adiabatic theorem - because in this limit the magnetic field direction (and hence Hamiltonian) changes slowly, the system remains in the ground state $| +, t ⟩$ throughout the evolution.

Particle on a 1-D Lattice

Source: MIT Spring 2002 Doctoral General Examination Quantum Q1

Problem Statement: Consider a particle confined to move on a 1-D lattice. The Hilbert space is spanned by the orthonormal basis states $| n ⟩$ where $n \in Z$ denotes the lattice site and $⟨ n | m ⟩ = δ_{n m}$ .

(a) Consider the translation operator $T$ defined by:

$\begin{matrix} (34) & T | n ⟩ = | n + 1 ⟩ . \end{matrix}$

What are the eigenstates and eigenvalues of $T$ ?

(b) Consider the Hamiltonian:

$\begin{matrix} (35) & H | n ⟩ = - \frac{1}{2 Δ^{2}} (| n + 1 ⟩ + | n - 1 ⟩ - 2 | n ⟩) . \end{matrix}$

Does $T$ commute with $H$ ? Let $| ψ_{k} ⟩$ be given by:

$\begin{matrix} (36) & ⟨ n | ψ_{k} ⟩ = e^{i k n Δ} . \end{matrix}$

Show that $| ψ_{k} ⟩$ is an eigenstate of $H$ . What is the relationship between the eigenvalue $E_{k}$ and $k$ ? What is $E_{k}$ as $k \to 0$ ?

(c) Add to the Hamiltonian a potential of height $V$ at lattice site $n = 0$ . What is the translation probability through the barrier as a function of $k, V$ , and $Δ$ ?

Solution: (a) Since $T$ translates by a lattice site, the eigenstates must be an equal weight superposition (up to a phase). For the eigenstates we propose:

| θ ⟩ = \sum_{n} e^{i n θ}

with eigenvalue $e^{- i θ} .$ As proof:

T | θ ⟩ = \sum_{n} e^{i n θ} T | n ⟩ = \sum_{n} e^{i n θ} | n + 1 ⟩ = \sum_{n} e^{- i θ} e^{i n θ} | n ⟩ = e^{- i θ} | θ ⟩ .

(b) We can rewrite:

H = - \frac{1}{2 Δ^{2}} (T + T^{†} - 2 I)

and since $T$ commutes with itself, its Hermitian conjugate, and the identity, we conclude that $[T, H] = 0.$ Now, verifying that $| ψ_{k} ⟩$ is an eigenstate:

H | ψ_{k} ⟩ = - \frac{1}{2 Δ^{2}} \sum_{n} e^{i k n Δ} (T + T^{†} - 2 I) | n ⟩ = e^{i k n Δ} (| n + 1 ⟩ + | n - 1 ⟩ - 2 | n ⟩)

Re-indexing the sum to collect terms:

H | ψ_{k} ⟩ = - \frac{1}{2 Δ^{2}} \sum_{n} (e^{i (n - 1) k Δ} + e^{i (n + 1) k Δ} - 2 e^{i n k Δ}) | n ⟩ = - \frac{1}{2 Δ^{2}} (e^{- i k Δ} + e^{i k Δ} - 2) \sum_{n} e^{i n k Δ} | n ⟩

and applying Euler’s identity:

H | ψ_{k} ⟩ = \frac{1 - \cos (k Δ)}{Δ^{2}} | ψ_{k} ⟩ .

Thus the $| ψ_{k} ⟩$ states are eigenstates with eigenvalue $\frac{1 - \cos (k Δ)}{Δ^{2}}$ . The $k$ enters the eigenvalue through the cosine term. As $k \to 0$ , $E_{k} \to 0.$

H = - \frac{1}{2 Δ^{2}} (T + T^{†} - 2 I) + V | 0 ⟩ ⟨ 0 |

And the eigenstates $| ψ_{k} ⟩$ of (b) are modified to the scattering states $| S_{k} ⟩$ with energy $E_{k}$ , where:

⟨ n | S_{k} ⟩ = e^{i k n Δ} + R e^{- i k n Δ} n \leq 0

$$ \langle n \vert S_k \rangle = Te^{ikn\Delta} \quad n \geq 0

⟨ 0 | S_{k} ⟩ = 1 + R = T

The first equation corresponds to a sum of an incoming wave $e^{i k n Δ}$ and a reflected wave $R e^{- i k n Δ}$ with reflection amplitude $R$ . The second equation corresponds to the transmitted wave $T e^{i k n Δ}$ with transmission amplitude $T$ . The third equation has the interpretation of probability conservation through the barrier. Now, we take the eigenvalue equation $H | S_{k} ⟩ = E_{k} | S_{k} ⟩$ and multiply it by $⟨ 0 |$ , which yields:

- \frac{1}{2 Δ^{2}} (⟨ 1 | S_{k} ⟩ + ⟨ - 1 | S_{k} ⟩ - 2 ⟨ 0 | S_{k} ⟩) + V ⟩ 0 | S_{k} ⟩ = E_{k} ⟨ 0 | S_{k} ⟩

Using the previously computed expression for $E_{k}$ in (b) and the coefficients of $| S_{k} ⟩$ as defined above, we obtain an equation relating $T, k, Δ, V$ :

- \frac{1}{2 Δ^{2}} (T e^{i k Δ} + e^{- i k Δ} + (1 - T) e^{i k Δ} - 2 T) + V T = - \frac{1}{2 Δ^{2}} (e^{- i k Δ} + e^{i k Δ} - 2) T

which solving for $T$ yields:

T = \frac{i \sin k Δ}{\sin k Δ + Δ^{2} V}

so we conclude the transmission probability to be:

| T |^{2} = \frac{\sin^{2} k Δ}{\sin^{2} k Δ + Δ^{4} V^{2}}

Spin-1/2 and Stern-Gerlach

Source: MIT Spring 2001 Doctoral General Examination Quantum Q1

Problem Statement: Consider a spin-1/2 particle where $S_{j} = \frac{ℏ}{2} σ_{j}$ with $σ_{j}$ the Pauli matrices: $σ_{x} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) σ_{y} = (\begin{matrix} 0 & - i \\ i & 0 \end{matrix}) σ_{z} = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$

(a) The operator along the $\hat{n} = (\sin θ \cos ϕ, \sin θ \sin ϕ, \cos θ)$ direction is:

$\begin{matrix} (37) & S_{\hat{n}} = n_{x} S_{x} + n_{y} S_{y} + n_{z} S_{z} . \end{matrix}$

Find the eigenvalues of $S_{\hat{n}}$ and the associated eigenvectors.

(b) Is it possible for a spin-1/2 particle to be in a state $| ψ ⟩$ such that:

⟨ ψ | S_{x} | ψ ⟩ = ⟨ ψ | S_{y} | ψ ⟩ = ⟨ ψ | S_{z} | ψ ⟩ = 0 ?

If it is possible, find $| ψ ⟩$ . If it is not, show why.

(c) A beam of spin-1/2 particles enters a Stern-Gerlach filter which allows only particles whose spin is $+ \frac{ℏ}{2}$ along the $z$ direction to pass. the particles then pass through a region where there is a magnetic field $\vec{B} = (0, B, 0)$ and spend time $T$ in this region. Then the particles enter a Stern-Gerlach filter which allows only particles whose spin is $+ \frac{ℏ}{2}$ along the $x$ direction to pass. What fraction of the electrons which exit the first filter exit the second? Assume that the particles interact with the magnetic field via $H = - \sum_{j} B_{j} S_{j}$ .

Solution: (a) We first calculate:

S_{\hat{n}} = \frac{ℏ}{2} (\begin{matrix} \cos θ & \sin θ e^{- i ϕ} \\ \sin θ e^{i ϕ} & - \cos θ \end{matrix}) .

Then if $λ_{+}, λ_{-}$ are the two eigenvalues of $S_{\hat{n}}$ , we have $det S_{\hat{n}} = - \frac{ℏ^{2}}{4} = λ_{+} λ_{-}$ and $Tr S_{\hat{n}} = 0 = λ_{+} + λ_{-}$ from which we conclude that $λ_{\pm} = \pm \frac{ℏ}{2}$ .

We now solve for the eigenvectors, which obey ${\hat{S}}_{\hat{n}} | \pm \hat{n} ⟩ = \pm \frac{ℏ}{2} | \pm \hat{n} ⟩$ and so:

(\begin{matrix} x \cos θ + y \sin θ e^{- i ϕ} \\ x \sin θ e^{i ϕ} - y \cos θ \end{matrix}) = \pm (\begin{matrix} x \\ y \end{matrix})

where $x, y$ are the entries of the eigenvectors. Since the global phase of a state is irrelevant, WLOG we can take $x$ to be real and set $x = \cos α$ . Then by normalization we have $| y | = \sin α$ and so $y = e^{i β} \sin α$ including the relative phase. Thus the above equations become:

$(\begin{matrix} \cos α \cos θ + e^{i β} \sin α \sin θ e^{- i ϕ} \\ \cos α \sin θ e^{i ϕ} - e^{i β} \sin α \cos θ \end{matrix}) = \pm (\begin{matrix} \cos α \\ e^{i β} \sin α \end{matrix})$ The only choice of $β$ that satisfies the above equations is $β = ϕ$ . The above then reduces to:

(\begin{matrix} \cos (θ - α) \\ \sin (θ - α) \end{matrix}) = \pm (\begin{matrix} \cos α \\ \sin α \end{matrix})

Which is satisfied by $α = \frac{θ}{2}$ for the $+$ case and $α = \frac{θ}{2} + π$ for the $-$ case, and hence:

| + \hat{n} ⟩ = (\begin{matrix} \cos (\frac{θ}{2}) \\ e^{i ϕ} \sin (\frac{θ}{2}) \end{matrix}) | - \hat{n} ⟩ = (\begin{matrix} - \sin (\frac{θ}{2}) \\ e^{i ϕ} \cos (\frac{θ}{2}) \end{matrix}) .

(b) Such a state cannot exist. Suppose for the sake of contradiction that such a $| ψ ⟩$ did - since all spin-1/2 pure states can be case in the form of $| + \hat{n} ⟩$ for appropriate choices of $θ, ϕ$ (and possible adjustment of the global phase) it follows that there exists some $\hat{n}$ for which $⟨ ψ | S_{\hat{n}} | ψ ⟩ = \frac{ℏ}{2} ⟨ ψ | ψ ⟩ \frac{ℏ}{2}$ . But then:

\frac{ℏ}{2} = ⟨ ψ | S_{\hat{n}} | ψ ⟩ = n_{x} ⟨ ψ | S_{x} | ψ ⟩ + n_{y} ⟨ ψ | S_{y} | ψ ⟩ + n_{z} ⟨ ψ | S_{z} | ψ ⟩ = n_{x} \cdot 0 + n_{y} \cdot 0 + n_{z} \cdot 0 = 0

which is a contradiction.

(c) The particles coming out of the first apparatus are $| ↑ ⟩ = (\begin{matrix} 1 \\ 0 \end{matrix})$ . In the magnetic field region we have the time-evolution operator (for a constant Hamiltonian):

U (T) = \exp (- \frac{i H T}{ℏ}) = \exp (- \frac{i (- B S_{y}) T}{ℏ}) = \exp (\frac{B T σ_{y}}{2}) = \cos (\frac{B T}{2}) + i \sin (\frac{B T}{2}) σ_{y}

where the last equality can be obtained by expanding out the exponential in a Taylor series, grouping the odd and even terms, and noting that $σ_{y}^{2} = 1$ . Applying this to the initial state we have:

| ψ (T) ⟩ = U (T) | ↑ ⟩ = \cos (\frac{B T}{2}) | ↑ ⟩ + i \sin (\frac{B T}{2}) (i | ↓ ⟩) = (\begin{matrix} \cos (\frac{B T}{2}) \\ - \sin (\frac{B T}{2}) \end{matrix}) .

The probability that we measure $S_{x} = + \frac{ℏ}{2}$ in the second filter is given by the Born rule to be:

P_{+} = | ⟨ + x | ψ (T) ⟩ |^{2} = | \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \end{matrix}) (\begin{matrix} \cos (\frac{B T}{2}) \\ - \sin (\frac{B T}{2}) \end{matrix}) |^{2} = \frac{1 - 2 \cos (\frac{B T}{2}) \sin (\frac{B T}{2})}{2}

where we use that $| + x ⟩ = \frac{| ↑ ⟩ + | ↓ ⟩}{\sqrt{2}}$ . The above is precisely the fraction of particles that passes through the second filter relative to the first, and hence:

P_{+} = \frac{1 - \sin (B T)}{2} .

Harmonic Oscillator and Translation

Source: MIT Spring 2001 Doctoral General Examination Quantum Q2

Problem Statement: A particle with mass $m = 1$ is in the ground state of a harmonic oscillator with Hamiltonian

$\begin{matrix} (38) & H = \frac{1}{2} (p^{2} + ω^{2} x^{2}) \end{matrix}$

The state is translated a distance $d$ while $H$ is left alone.

(a) What is the probability that a measurement of the energy will yield the value $ℏ ω (j + \frac{1}{2})$ ? You may find the following formulas useful:

$\begin{matrix} (39) & a = \frac{1}{\sqrt{2 ℏ ω}} (ω x + i p) | n ⟩ = \frac{(a^{†})^{n}}{\sqrt{n!}} | 0 ⟩ \end{matrix}$

and also if $A, B$ are two operators such that $[A, B]$ is proportional to a unit operator, then

$\begin{matrix} (40) & e^{A + B} = e^{A} e^{B} + e^{- \frac{1}{2} [A, B]} . \end{matrix}$

At $t = 0$ th equantum system is prepared in the translated state described above. It then evolves according to the Schrödinger equation with the Hamiltonian $H$ .

(b) Supose that at the time $t$ the momentum is measured. At what values of $t$ wil the probability distribution for getting any value of the momentum be identical to the $t = 0$ probability distribution?

(c) Repeat (b) with momentum measurement replaced by energy measurement.

Solution: (a) We first note that the translation operator by distance $d$ is given by $T (d) = e^{- \frac{i p d}{ℏ}}$ recalling momentum as the generator of translations. We also can express $p$ in terms of $a, a^{†}$ :

p = i \sqrt{\frac{ℏ ω}{2}} (a^{†} - a)

We also calculate $[a^{†}, a]$ :

[a^{†}, a] = [\frac{1}{\sqrt{2 ℏ ω}} (ω x - i p), \frac{1}{\sqrt{2 ℏ ω}} (ω x + i p)] = \frac{1}{2 ℏ ω} ([ω x, ω x] + [ω x, i p] - [i p, ω x] - [i p, i p]) = - 1

where we have used the canonical commutation relation $[x, p] = i ℏ$ . Thus the operator exponential identity of Eq. $(40)$ applies. Now calculating the probability of measuring $E_{j} = ℏ ω (j + \frac{1}{2})$ using the Born rule, we have:

p (j) = | ⟨ j | e^{- \frac{i p d}{ℏ}} | 0 ⟩ |^{2} = | ⟨ j | e^{\sqrt{\frac{ω d^{2}}{2 ℏ}} (a^{†} - a)} | 0 ⟩ |^{2}

For brevity let us define $α = \sqrt{\frac{ω d^{2}}{2 ℏ}}$ . Then using Eq. $(40)$ with $A = α a^{†}$ and $B = - α a$ (and $[α a^{†}, - α a] = - α^{2} [a^{†}, a] = α^{2}$ ) we obtain:

p (j) = | ⟨ j | e^{α a^{†}} e^{- α a} e^{- \frac{α^{2}}{2}} | 0 ⟩ |^{2}

Now, since $a | 0 ⟩ = 0$ , when expanding $e^{- α a}$ in a power series all term vanish except the $n = 0$ identity term. Therefore the above reduces to:

p (j) = e^{- α^{2}} | ⟨ j | e^{α a^{†}} | 0 ⟩ |^{2}

Since $| j | i ⟩ = δ_{j i}$ the only term of the above that survives when expanding $e^{α a^{†}}$ in a power series is the $n = j$ term, so:

p (j) = e^{- α^{2}} | ⟨ j | \frac{(α a^{†})^{j}}{j!} | 0 ⟩ |^{2} = e^{- α^{2}} | ⟨ j | \frac{α^{j}}{\sqrt{j}!} | j ⟩ |^{2}

Thus we conclude:

p (j) = \frac{e^{- α^{2}} α^{2 j}}{j!}

(b) Let us study the time evolution of the state (which starts in $| ψ (0) ⟩ = e^{- \frac{i p d}{ℏ}} | 0 ⟩$ ):

| ψ (t) ⟩ = \exp (- i H t) | ψ (0) ⟩ = (\sum_{n} e^{- i \frac{ℏ ω (n + \frac{1}{2})}{ℏ} t} | n ⟩ ⟨ n |) | ψ (0) ⟩ = \sum_{n} e^{- i ω n t} ⟨ n | ψ (0) ⟩ | n ⟩

where in the last equality we discard the global phase factor $e^{- i ω t / 2}$ as this does not affect the measurement statistics. Looking at the above, the $t = 0$ measurement probabilities of momentum (and any other observable) is reproduced when $e^{- i ω n t} = 1$ , i.e. when:

t = \frac{2 π m}{ω}, m \in N .

(c) Since $H$ commutes with itself, the energy is conserved through the time evolution and thus the probability for measuring a given energy is constant in time.

Contents

Superconducting Microwave Resonator

Measuring a Two-Spin System

Particle on a Ring

Emancipation from a Distance

Temperature and Adiabatic Changes of the 3-D Harmonic Oscillator

Quantum Zeno Effect

Positronium Stern-Gerlach

Quantum Single-Slit Diffraction

Gravitational Potential

Rigid Rotor in Magnetic Field

Symmetries and Transition Amplitudes

Thermal States

Scattering of Identical Particles

Spin-Orbit Coupling in the 3-D Harmonic Oscillator

Identical Particles in an Infinite Cubic Well

Time-Varying Force

Two Interacting Fermions

A Quantum Spin Chain

Anomalous Magnetic Moment of the Electron

A Heisenberg Ferromagnet

The Supersymmetric Method

Spin-1/2 in Time-Dependent Magnetic Field

Particle on a 1-D Lattice

Spin-1/2 and Stern-Gerlach

Harmonic Oscillator and Translation