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

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,