Past exam of the quantum theory course of the University of Oxford / 2021 / 1

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,