Past exam of the quantum theory course of the University of Oxford / 2023 (A11, A13761W1)

Solutions to all three questions of the University of Oxford Part A quantum theory paper, sat on 22 June 2023. Source: original released PDF, bundle pages 153–156; paper pages 1–4. Question statements are omitted. The symbol $\hslash$ denotes the reduced Planck constant.

The free particle has Hamiltonian operator $H=P^2/(2m)$, with momentum operator $P=-i\hslash {\partial }_x$. Its time-dependent Schrödinger equation isHere $x$ measures arc length for the particle on a circle, so the wavefunction normalization uses $dx$, rather than $d(x/R)$.

A stationary state has $\Psi (t,x)=e^{-iEt/\hslash }\psi (x)$. Substitution into the Schrödinger equation gives the energy eigenvalue equationThe trial wavefunction $C{e}^{ikx}$ satisfies it with $E={\hslash }^2{k}^2/(2m)$. The periodic boundary condition requires $e^{2\pi ikR}=1$, hence $k=n/R$, $n\in \mathbb{Z}$. The wavefunction normalization is $1=2\pi R|C{|}^2$, so a convenient common global phase gives the orthonormal basisIndeed, ${\int }_0^{2\pi R}{\varphi }_n^{\ast }{\varphi }_{q}dx={\delta }_{nq}$. This is the free particle spectrum on a circle. The ground state ${\varphi }_0$ has energy zero and is nondegenerate. Each positive energy level ${\hslash }^2{n}^2/(2m{R}^2)$, $n=1,2,\dots$, has energy eigenspace $\mathrm{span}\{{\varphi }_n,{\varphi }_{-n}\}$ and quantum degeneracy two. Real sine and cosine combinations give another basis of the same energy eigenspace.

Using ${\sin }^{2}u=\frac{1}{2}-\frac{1}{4}{e}^{2iu}-\frac{1}{4}{e}^{-2iu}$, the energy eigenstate decomposition isOnly the energy eigenvalues zero and $E_{\star }=2{\hslash }^2/(m{R}^2)$ occur. The last two terms are distinct momentum eigenstates in the same energy eigenspace.

The orthonormal basis makes the squared norm equal to the sum of squared expansion coefficients:Choose a positive real wavefunction normalization constant, fixing an irrelevant global phase:As a direct check, ${\int }_0^{2\pi R}{\sin }^4(x/R)dx=3\pi R/4$.

Under unitary time evolution, each energy eigenstate acquires its own dynamical phase. Since $E_2=E_{-2}=E_{\star }$, the wavefunction isAt $t=0$, the identity $1-\cos (2x/R)=2{\sin }^2(x/R)$ recovers the normalized initial wavefunction.

By the Born rule, the probability for a measured energy eigenvalue is the squared norm of the orthogonal projection onto its whole energy eigenspace. ThusEvery other energy eigenvalue has probability zero. These probabilities are independent of $T$: unitary time evolution under the unchanged Hamiltonian operator changes the dynamical phases but not the squared expansion coefficients. The two equal-energy terms describe one measurement outcome.

Let ${\Pi }_{\star }=|{\varphi }_2⟩⟨{\varphi }_2|+|{\varphi }_{-2}⟩⟨{\varphi }_{-2}|$ be the orthogonal projection for $E_{\star }$. The Lüders rule gives the normalized post-measurement stateThe factor $-e^{-i{E}_{\star }T/\hslash }$ is a global phase, so $\cos (2x/R)/\sqrt{\pi R}$ is an equivalent representative. The projective measurement preserves the quantum coherence between the two momentum eigenstates; it does not select one of them. This matters for the subsequent position probabilities.

Put $\tau =t-T$ and retain the global phase from the preceding Lüders rule update. Since $P{\varphi }_n=(\hslash n/R){\varphi }_n$, the same orthonormal basis diagonalizes the new Hamiltonian operator, withIn particular, ${\tilde{E}}_{\pm 2}=E_{\star }+m{s}^2/2\mp 2\hslash s/R$. Applying unitary time evolution from $T$ therefore givesThis is rigid translation from a momentum term in the Hamiltonian: the position probability density moves with velocity $-s$. The plus sign in $x+s\tau$ follows directly from the negative sign of $-sP$. At $\tau =0$ the expression agrees exactly with the preceding post-measurement state.

For an arc represented by $0\le x\le a$, take $0\le a\le 2\pi R$. The Born rule usesIntegrate using ${\cos }^{2}u=(1+\cos 2u)/2$:The result is zero at $a=0$ and one at $a=2\pi R$, checking the wavefunction normalization. At $s=0$ it reduces to $a/(2\pi R)+\sin (4a/R)/(8\pi )$, the stationary position probability. Longer intervals on the periodically identified coordinate require specifying the corresponding subset of the circle; integrating multiple turns would count the same positions repeatedly.

The Hamiltonian operator separates into two commuting one-dimensional quantum harmonic oscillators. WriteProducts of harmonic oscillator eigenfunctions giveHere $H_n$ is the physicists' Hermite polynomial, so $H_0=1$, $H_1(u)=2u$ and $H_2(u)=4u^2-2$. The first six energy eigenstates, ordered with multiplicity, areFor this two-dimensional oscillator level ordering, the excitation energies obeyEvery state with $n_1+n_2\ge 3$ costs at least $3\hslash {\omega }_1$, so none has been missed. Equalities within the list occur when ${\omega }_2={\omega }_1=\omega$. Then the two-dimensional isotropic harmonic oscillator hasThe quantum degeneracy counts $n_1=0,\dots ,N$ with $n_2=N-n_1$. The six listed states occupy three distinct energy levels, of quantum degeneracies one, two and three.

Since $X$ and $P$ are self-adjoint operators, the adjoint operator relation requires $c_{+}=c_{-}=c$ and $d_{+}=-d_{-}=-d$. Thus $a_{-}=cX+idP$ and $a_{+}=cX-idP$. The canonical commutation relation $[X,P]=i\hslash I$ givesMatching the $X^2$, $P^2$ and identity operator coefficients in $H=\hslash \omega (a_{+}{a}_{-}+\lambda I)$ yieldsConsequently all real solutions areThe conventional creation and annihilation operators use $\epsilon =1$. The other choice multiplies both by minus one and satisfies all the given equations. This harmless freedom is the oscillator ladder-operator sign convention; the subsequent formulas use the positive choice in both coordinates.

Use the canonical commutation relationsWriting $c_j=\sqrt{m{\omega }_j/(2\hslash )}$, $d_j=1/\sqrt{2m\hslash {\omega }_j}$, direct expansion of the commutator givesThe Fock vacuum ${\psi }_{0,0}$ is annihilated by both $a_{j,-}$. Successive creation operators generateHence, in the order of part (a), the six energy eigenstates are proportional toIf the Fock vacuum has unit norm, dividing the general expression by $\sqrt{n_1!n_2!}$ gives normalized number states. Normalization is not required here, but using this convention makes the later change of basis unambiguous.

The orbital angular momentum $L_3=X_1{P}_2-X_2{P}_1$ generates rotations in the plane. The canonical commutation relations giveUse the commutator product rule $[A,B^2]=[A,B]B+B[A,B]$. Since different coordinates commute,Thus, for arbitrary positive angular frequencies,At ${\omega }_1={\omega }_2$, this becomes $\boxed{[L_3,H]=0}$. The rotational symmetry of the two-dimensional isotropic harmonic oscillator allows a basis of simultaneous energy eigenstates and orbital angular momentum eigenstates.

Set ${\omega }_1={\omega }_2=\omega$. Inverting the creation and annihilation operators givesSubstitute these expressions in $L_3=X_1{P}_2-X_2{P}_1$. Terms with two creation operators or two annihilation operators cancel because different modes commute. The remaining orbital angular momentum in oscillator ladder operators isEquivalently it is $-i\hslash (a_{1,+}{a}_{2,-}-a_{2,+}{a}_{1,-})$, consistent with the circular ladder operators of a two-dimensional oscillator.

Choose the first three energy eigenstates consistently as ${\psi }_{0,0}$, ${\psi }_{1,0}=a_{1,+}{\psi }_{0,0}$ and ${\psi }_{0,1}=a_{2,+}{\psi }_{0,0}$. The Fock vacuum is annihilated by both lowering operators, soFor example $a_{1,-}{a}_{1,+}{\psi }_{0,0}={\psi }_{0,0}$, while $a_{2,-}{\psi }_{1,0}=0$, giving the middle formula. Thus the simultaneous eigenstates can be chosen asTheir pairs of eigenvalues $(H,L_3)$ are respectively $(\hslash \omega ,0)$, $(2\hslash \omega ,+\hslash )$ and $(2\hslash \omega ,-\hslash )$. The factor $1/\sqrt{2}$ normalizes the combinations when the original three states are normalized. Mixing only inside the degenerate first-excited energy eigenspace preserves the energy eigenvalue while diagonalizing $L_3$.

Write $\alpha =m\omega /\hslash$. The first Hermite polynomials give ${\psi }_{0,0}\propto e^{-\alpha (x_1^2+x_2^2)/2}$ and ${\psi }_{1,0},{\psi }_{0,1}$ proportional, with the same constant, to $x_1{e}^{-\alpha (x_1^2+x_2^2)/2}$ and $x_2{e}^{-\alpha (x_1^2+x_2^2)/2}$. Thus in Cartesian coordinates the new wavefunctions areUsing polar coordinates $x_1=r\cos \phi$, $x_2=r\sin \phi$ givesThe orbital angular momentum is $L_3=-i\hslash {\partial }_{\phi }$, which immediately verifies its eigenvalues $0,\pm \hslash$. Multiplying these spatial wavefunctions by $e^{-i\omega t}$ for ${\chi }_0$ and $e^{-2i\omega t}$ for ${\chi }_{\pm }$ gives the corresponding full stationary states.

The angular momentum commutation relations arewhere ${\epsilon }_{ijk}$ is the Levi-Civita symbol. The angular momentum eigenstates havewith angular momentum quantum number $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and magnetic quantum number $\mu =-j,-j+1,\dots ,j$. There are $2j+1$ such states in one irreducible spin representation, so the $J^2$ eigenspace has dimension $2j+1$ within that multiplet. The use of $\mu$ avoids confusing the magnetic quantum number with the particle mass in the earlier questions.

For completeness, the ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfy $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$ andThe upper and lower ends are $\mu =\pm j$; integer steps between them require $2j$ to be a nonnegative integer. This explains both the spectrum and the multiplet dimension. Pure orbital angular momentum permits only integer $j$, whereas general spin allows half-integers. If a larger Hilbert space contains several copies of the same irreducible spin representation, the full $J^2$ eigenspace includes all those copies: its dimension is $(2j+1)$ times their multiplicity, and can be infinite. The paper's dimension question is understood per irreducible multiplet.

In the ordered orthonormal basis $(|\frac{1}{2},\frac{1}{2}⟩,|\frac{1}{2},-\frac{1}{2}⟩)$, the spin one-half angular momentum operators are $J_i=(\hslash /2){\sigma }_i$, with Pauli matricesThe ladder operators areExplicit multiplication gives $J_{+}{J}_{-}={\hslash }^2\mathrm{diag}(1,0)$ and $J_{-}{J}_{+}={\hslash }^2\mathrm{diag}(0,1)$. Their commutator is therefore

Choose the ordered angular momentum eigenstates $(|1,1⟩,|1,0⟩,|1,-1⟩)$, so $J_3=\hslash \mathrm{diag}(1,0,-1)$. Since $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$, a nonzero matrix element of $J_{+}$ must increase the magnetic quantum number by one, and one of $J_{-}$ must decrease it by one. Thus only entries $(1,2),(2,3)$ of $J_{+}$ and $(2,1),(3,2)$ of $J_{-}$ can be nonzero. The adjoint operator identity $J_{-}=J_{+}^{†}$ givesFor the $j=1$ irreducible spin representation, both adjacent transitions are present; the following calculation determines their sizes.

The phases of the three basis vectors can be chosen so that $A$ and $B$ are nonnegative real numbers. Matrix multiplication then givesThe angular momentum commutation relations require this to equal $2\hslash J_3=2{\hslash }^2\mathrm{diag}(1,0,-1)$. Equating diagonal entries yieldsHenceThese are also the $j=1$ special case of the normalized spin ladder operator coefficients $\hslash \sqrt{j(j+1)-\mu (\mu \pm 1)}$.

Recover the angular momentum operators from $J_1=(J_{+}+J_{-})/2$ and $J_2=(J_{+}-J_{-})/(2i)$. The spin-one angular momentum matrices in the chosen descending-$J_3$ orthonormal basis areThey are self-adjoint operators, satisfy $[J_i,J_j]=i\hslash {\epsilon }_{ijk}{J}_k$, and give $J^2=2{\hslash }^{2}I$, as required for $j=1$.

The Hamiltonian operator $H=\omega J_3$ is already diagonal in the angular momentum eigenstate basis of part (b). Its three distinct energy eigenvalues and stationary states areThus the three column eigenvectors are $(1,0,0{)}^{\mathsf{T}}$, $(0,1,0{)}^{\mathsf{T}}$ and $(0,0,1{)}^{\mathsf{T}}$, with energies $+\hslash \omega$, zero and $-\hslash \omega$, respectively. Each is nondegenerate since $\omega >0$. Any nonzero scalar multiple gives the same stationary state after normalization; a quantum superposition of different energies is generally not stationary.

Diagonalizing the $J_1$ matrix gives the normalized eigenvectorswith eigenvalues $+\hslash ,0,-\hslash$, respectively. The first projective measurement prepares $v_{+}$ up to global phase, regardless of the earlier state, conditional on obtaining $+\hslash$.

The unitary time evolution under $H=\omega J_3$ gives, with $\theta =\omega t$,The probability amplitudes for the second projective measurement areThe Born rule therefore gives the general spin-one transverse measurement after axial precession:At the specified times $t_k=\pi k/\omega$, $\theta =\pi k$, soFor even $k$, the result is certainly $+\hslash$; for odd $k$, it is certainly $-\hslash$. Although zero belongs to the spectrum of the observable, it never occurs at these times. Indeed, for even $k$ the evolved vector is $v_{+}$, while for odd $k$ it is $-v_{-}$, where the minus sign is a global phase.