Past exam of the quantum theory course of the University of Oxford / 2023 / 1

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.