Past exam of the quantum theory course of the University of Oxford / 2021 (A11, A13761W1)

Solutions to all three questions of the University of Oxford Part A quantum theory paper, sat on 16 June 2021. Source: original released PDF, bundle pages 145–148; paper pages 1–4. The symbol $\hslash$ denotes the reduced Planck constant. All four measurement experiments in question 1(c) start independently from the wavefunction in question 1(b).

For a free particle, the Hamiltonian operator is $H=P^2/(2m)$, where $P=-i\hslash d/dx$ is the momentum operator. The time-independent Schrödinger equation is thereforeThe full stationary state is $\Psi (x,t)=e^{-iEt/\hslash }\psi (x)$. The periodic boundary condition identifies $x$ and $x+2\pi$, giving a particle on a circle with wavefunction normalization over one period.

For a momentum eigenstate with eigenvalue $p$, the equation $-i\hslash {\psi }^{\prime }=p\psi$ has solution $\psi (x)=C{e}^{ipx/\hslash }$. The periodic boundary condition imposes $e^{2\pi ip/\hslash }=1$, so $p/\hslash \in \mathbb{Z}$. The wavefunction normalization gives $1=2\pi |C{|}^2$. Choosing a convenient global phase yieldsThese momentum eigenstates form an orthonormal basis, the Fourier basis for the particle on a circle:Here ${\delta }_{nk}$ is the Kronecker delta.

Since $P{\psi }_n=n\hslash {\psi }_n$, applying the Hamiltonian operator givesThus the energy eigenvalues and corresponding full stationary states areThe ground state is ${\psi }_0$ with energy zero and quantum degeneracy one. For each $n=1,2,\dots$, the positive energy level $n^2{\hslash }^2/(2m)$ has quantum degeneracy two: its energy eigenspace is $\mathrm{span}\{{\psi }_n,{\psi }_{-n}\}$. This is the free particle spectrum on a circle.

The identity ${\cos }^{2}x=\frac{1}{2}+\frac{1}{4}{e}^{2ix}+\frac{1}{4}{e}^{-2ix}$ gives the momentum eigenstate expansionUsing the orthonormal basis from part (a), the wavefunction normalization isHence $|A|=2/\sqrt{3\pi }$. The wavefunction normalization leaves a global phase free; the positive real choice used in the requested expansion givesThe two nonzero eigenvalues of the momentum operator have the same energy eigenvalue $E_{\star }=2{\hslash }^2/m$. Under unitary time evolution, their coefficients acquire the same dynamical phase. With $\Omega =E_{\star }/\hslash =2\hslash /m$,At $t=0$, $1+\cos (2x)=2{\cos }^{2}x$ recovers the initial wavefunction. Its probability density, useful below, is

The ground state is the single energy eigenstate ${\psi }_0$. Its coefficient in the normalized wavefunction is $\sqrt{2/3}$, so the Born rule givesThis probability is independent of $t$ because unitary time evolution changes only the dynamical phases of the energy eigenstates.

The coefficient of the momentum eigenstate ${\psi }_2$ is $e^{-i\Omega t}/\sqrt{6}$. The Born rule therefore gives $\mathrm{Pr}(P=2\hslash )=1/6$. Conditional on this outcome, the projective measurement prepares the post-measurement state $e^{-i\Omega t}{\psi }_2$, equivalent to ${\psi }_2$ up to global phase.

Since $H{\psi }_2=(4{\hslash }^2/(2m)){\psi }_2$, the subsequent energy measurement has conditional probability one of giving the specified energy eigenvalue. The probability for the entire sequence is consequentlyAny intervening unitary time evolution under the same Hamiltonian operator preserves this energy eigenstate, up to global phase.

For a position measurement, the Born rule integrates the probability density over the chosen interval. The three terms from part (b) have integralsandThusThe oscillating term comes from quantum interference between the zero and positive energy components. The probability lies between $1/8-1/(3\pi )$ and $1/8+1/(3\pi )$, both within $[0,1]$; integrating the same probability density over the whole circle gives one.

The parity operator exchanges ${\psi }_n$ and ${\psi }_{-n}$. Their coefficients in the wavefunction are equal, soThus the wavefunction is a parity operator eigenstate with eigenvalue $+1$ at every time. The orthogonal projection onto the $-1$ eigenspace is $(I-R)/2$, which annihilates it. By the Born rule,

The position operators and momentum operators are self-adjoint operators. Using conjugate linearity of the adjoint operator, $(cA+dB{)}^{\ast }=\overline{c}{A}^{\ast }+\overline{d}{B}^{\ast }$, givesWrite $D=\sqrt{2m\omega \hslash }$. Bilinearity and antisymmetry of the commutator, together with the canonical commutation relations, give $[P_j,X_k]=-i\hslash {\delta }_{jk}I$. HenceThe terms involving two position operators or two momentum operators vanish by the canonical commutation relations. Finally, reversing products under the adjoint operator, $(AB{)}^{\ast }=B^{\ast }{A}^{\ast }$, gives $[a_j,a_k{]}^{\ast }=-[a_j^{\ast },a_k^{\ast }]$. ThereforeHere $I$ is the identity operator. These are the relations for two independent pairs of creation and annihilation operators. The operator algebra in this question is understood on a common invariant operator domain, such as finite linear combinations of the number states constructed in part (c).

First apply the commutator product rule in the form $[AB,C]=A[B,C]+[A,C]B$:Repeated use of the commutator product rule gives, for $n\ge 1$,This is the commutator with powers of an eigenoperator. For $n=0$, the commutator with the identity operator is zero, agreeing with the formula.

Each $a_k^{\ast }{a}_k$ is unchanged by taking the adjoint operator, so $N^{\ast }=N$ in this algebra. Taking the adjoint of a commutator yields $[N,(a_j^{\ast }{)}^n{]}^{\ast }=[a_j^n,N]=n{a}_j^n$. By antisymmetry,In particular, each creation operator raises the number operator eigenvalue by one, and each annihilation operator lowers it by one.

Inverting the definitions of the creation and annihilation operators givesSubstituting in the orbital angular momentum $L$ givesOperators belonging to distinct modes commute. Therefore the terms with two annihilation operators and those with two creation operators cancel, and $a_1^{\ast }{a}_2=a_2{a}_1^{\ast }$, $a_2^{\ast }{a}_1=a_1{a}_2^{\ast }$. The surviving terms give the orbital angular momentum in oscillator ladder operators:

From part (a), $[N,a_j]=-a_j$ and $[N,a_j^{\ast }]=a_j^{\ast }$. The commutator product rule givesConsequentlyEach term in $L$ transfers one excitation between the two modes, preserving the total number operator eigenvalue. This is an instance of the oscillator bilinear commutator; every number operator eigenspace is an invariant subspace of $L$.

The common Fock vacuum obeys $N\psi ={\sum }_j{a}_j^{\ast }(a_j\psi )=0$. Put $B=(a_1^{\ast }{)}^{n_1}(a_2^{\ast }{)}^{n_2}$. By the commutator product rule and part (a),Moving $N$ through $B$ therefore givesDividing by $\sqrt{n_1!n_2!}$ provesThe constructed number state is nonzero: the relation $a_j(a_j^{\ast }{)}^r\psi =r(a_j^{\ast }{)}^{r-1}\psi$ follows from $[a_j,a_j^{\ast }]=I$ and $a_j\psi =0$. Commuting different modes and successively moving each creation operator across the inner product yieldsThus the given factorial denominator also gives unit norm, as expected for a number state.

Write $u={\psi }_{1,0}=a_1^{\ast }\psi$ and $v={\psi }_{0,1}=a_2^{\ast }\psi$. The canonical commutation relations for the creation and annihilation operators giveFor example, $a_1{a}_1^{\ast }\psi =(I+a_1^{\ast }{a}_1)\psi =\psi$. The expression for $L$ from part (b), with distinct modes reordered, then yieldsThus $\mathrm{span}\{u,v\}$ is an invariant subspace, and the matrix of $L$ in the ordered orthonormal basis $(u,v)$ isIts characteristic polynomial is ${\ell }^2-{\hslash }^2$, giving the eigenvalues and normalized eigenstatesIndeed, $L(u\pm iv)=\pm \hslash (u\pm iv)$ and $\Vert u\pm iv{\Vert }^2=2$ by the assumed orthonormality. Both eigenstates remain in the $N=1$ eigenspace.

The angular momentum eigenstates have $J^2$ eigenvalue ${\hslash }^{2}j(j+1)$, where the angular momentum quantum number is $j=0,\frac{1}{2},1,\frac{3}{2},\dots$. For each $j$, the magnetic quantum number runs in unit steps from $-j$ to $j$. Here that magnetic quantum number is $+1/2$, so only half-integer $j$ are compatible:Integer $j$ cannot contain a $J_3$ eigenvalue $\hslash /2$. Applying the angular momentum lowering operator and using $[J_3,J_{-}]=-\hslash J_{-}$ givesThe lowered vector is nonzero, since the adjoint operator identity $J_{-}^{\ast }=J_{+}$ givesFinally, $J_3$ is a self-adjoint operator, so its eigenstates for distinct eigenvalues are orthogonal. Explicitly,and hence $\boxed{⟨\phi \mid \psi ⟩=0}$.

For $\lambda =3/4$, the adjoint operator relation $J_{+}^{\ast }=J_{-}$ and the given product identity implyA vector of zero norm is zero, so $J_{+}\psi =0$. Part (i) givesThe commutator $[J^2,J_{-}]=0$ also gives $J^2\phi =(3/4){\hslash }^2\phi$, while $J_3\phi =-(\hslash /2)\phi$. Applying the other product identity to this unnormalized vector,ThereforeThese are the two angular momentum ladder endpoints of a spin one-half pair. In particular, ${\psi }_2=\phi /\hslash$ has unit norm, and $J_{+}{\psi }_2=\hslash {\psi }_1$ because $J_{+}{J}_{-}\psi ={\hslash }^2\psi$.

In the ordered orthonormal basis $({\psi }_1,{\psi }_2)$, the Hamiltonian operator has matrixwhere ${\sigma }_1$ is a Pauli matrix. Its symmetric and antisymmetric eigenvectors giveThe energy eigenstates ${\chi }_{\pm }$ are normalized and orthogonal; each energy eigenvalue is nondegenerate. Their full stationary states are $e^{-i\omega t}{\chi }_{+}$ and $e^{i\omega t}{\chi }_{-}$. Since they form an orthonormal basis, the general solution to the Time-dependent Schrödinger equation isFor a normalized quantum state, $|c_{+}{|}^2+|c_{-}{|}^2=1$; the constants are determined by its initial inner products with ${\chi }_{\pm }$.

The initial quantum state is ${\psi }_1$, up to global phase, and ${\psi }_1=({\chi }_{+}+{\chi }_{-})/\sqrt{2}$. Applying unitary time evolution givesSince $J_3{\psi }_1=(\hslash /2){\psi }_1$ and $J_3{\psi }_2=-(\hslash /2){\psi }_2$, the Born rule givesThe other outcome has probability ${\sin }^2(\omega t)$, verifying that the two probabilities sum to one. This is a two-level quantum return probability.

Continue with the initial quantum state ${\psi }_1$ from part (ii), and let $\Delta t=T/N$. Conditional on every preceding outcome being $+\hslash /2$, the post-measurement state after each ideal projective measurement is again ${\psi }_1$, up to global phase. Part (ii) therefore gives the same conditional probability ${\cos }^2(\omega \Delta t)$ for each successive outcome. Multiplying these conditional probabilities yields the repeated projective-measurement survival probabilityThis multiplication uses the quantum state reset at each successful projective measurement; the outcomes need not be assumed statistically independent.

For fixed $T$, put $z=\omega T$. The Taylor expansion ${\cos }^{2}u=1-u^2+O(u^4)$ gives the quadratic short-time quantum survival probability. For sufficiently large $N$, taking logarithms givesHere the remainder uses Big O notation. Exponentiating provesThis is the quantum Zeno effect: increasingly frequent ideal projective measurements inhibit the transition away from the initial quantum state. The departure probability in a short interval is quadratic in its duration, so $N$ intervals of duration $T/N$ produce a total leading departure probability proportional to $T^2/N$. Without intervening measurements, the return probability at time $T$ would be ${\cos }^2(\omega T)$. The result concerns obtaining $+\hslash /2$ at every measurement in the ideal limit of instantaneous measurements with unitary time evolution between them.