Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 2 / Section A(ii) / 15

A physical bound-state wave function must be single-valued, finite, square-integrable and normalizable. Across a finite step in a regular potential, both $\psi$ and its first derivative are continuous when the particle mass is the same on either side; integrate the Schrödinger equation across an infinitesimal interval to obtain the derivative condition. The second derivative need not be continuous.

An infinite hard wall is an exception to derivative matching: the interior wave function vanishes at the boundary, but its derivative need not equal the exterior zero derivative. A delta-function potential also permits a prescribed derivative jump, and an effective-mass interface requires the appropriate mass-weighted matching condition. Thus ordinary finite-step conditions should not be asserted without qualification for every singular potential.

The wave function vanishes in regions I and III. In region II it is sinusoidal and satisfies $\psi (0)=\psi (L)=0$. The time-independent equation there isA nonzero stationary state therefore has standing-wave nodes at the two walls; it cannot have a nonzero constant ground-state wave function.

For $E>0$, write $\psi =A\sin kx+B\cos kx$ with $k^2=2mE/{\hslash }^2$. The condition at zero gives $B=0$; the condition at $L$ gives $kL=n\pi$, where $n=1,2,\dots$. The $n=0$ solution is identically zero and is not a state. Normalization gives $A^2{\int }_0^L{\sin }^2(n\pi x/L)dx=A^{2}L/2=1$. HenceChoosing positive $A$ fixes an irrelevant overall phase.

Only $n=1,2,3$ have $E_n<10c$. The printed squared coefficients sum to $S=0.1^2+0.4^2+0.3^2+0.86^2=0.9996$, rather than exactly one. Normalizing those numbers givesTreating the coefficients as rounded values of an already normalized state gives the intended approximately $0.26$ answer. Probabilities are sums of squared amplitudes because the energy eigenstates are orthonormal.

Let $(a_1,a_2,a_3,a_4)=(0.1,0.4,0.3,0.86)$ and $S=0.9996$. For the time-independent well,It remains zero outside the well. Each expansion coefficient acquires a phase of unit magnitude, so every energy probability, including the preceding $26.0\%$, is time-independent. The position probability density generally changes because cross terms between different energies oscillate.

The measured energy selects $n=3$. Immediately after an ideal projective measurement, the normalized state is ${\psi }_3(x)$ up to an overall phase, with no surviving $n=1,2,4$ components. At later times before any further disturbance it is $e^{-i9c(t-t_0)/\hslash }{\psi }_3(x)$, where $t_0$ is the measurement time. A repeated energy measurement returns $9c$ with probability one.