Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S3 / 2

For a stationary nonrelativistic state, $\boxed{[-{\hslash }^2{\nabla }^2/(2m)+V(\mathbf{r})]\psi =E\psi }$. The wavefunction $\psi$ is a probability amplitude, $m$ the particle mass and $E$ its energy eigenvalue; the potential energy $V$ is measured in the same energy units. In one dimension the Laplacian becomes $d^2/d{x}^2$.

Inside the well, ${\psi }^{\prime \prime }+k^2\psi =0$. Vanishing at zero removes the cosine term; vanishing at $L$ requires $k_{n}L=n\pi$. The nonzero states have positive integer $n$ because $n=0$ gives the zero function and negative $n$ only repeats a state up to sign. Normalization follows from ${\int }_0^L{\sin }^2(n\pi x/L)dx=L/2$. ThusThe wavefunction vanishes outside the infinite barriers, so there is no additional contribution to its norm.

A sudden wall displacement leaves the initial wavefunction unchanged: it is $\sqrt{2/L}\sin (\pi x/L)$ on $[0,L]$ and zero on $(L,2L]$. It is not an eigenfunction of the new Hamiltonian, so there is no single certain new energy. Its energy expectation immediately afterward remains $E_1^{\mathrm{old}}$, as seen from ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)$; the measurement outcomes are $E_n^{\mathrm{new}}=n^2{E}_1^{\mathrm{old}}/4$.

The new ground state is $L^{-1/2}\sin (\pi x/(2L))$. Its overlap is $4\sqrt{2}/(3\pi )$, soThe discontinuity of the derivative at the former wall does not invalidate the finite quadratic-form expectation, although it matters when applying the differential Hamiltonian pointwise. Adiabatic expansion would instead keep the state in the ground level and reduce its energy by four.