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

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.