Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 15B (Quantum Mechanics)

Use the normalized eigenfunctions of the infinite square well,The sine identity expands the initial wavefunction asThe normalization factor $4/\sqrt{5a}$ yields precisely these coefficients. By the Born rule, the energy measurement probabilities are $9/10$ for $E_1$ and $1/10$ for $E_3=9E_1$. Thus the energy expectation isThe two probabilities sum to one, also checking the wavefunction normalization.

Each energy eigenstate acquires the phase $e^{-i{E}_{n}t/\hslash }$ under unitary time evolution. Consequently at the requested time $t=1$,and the wavefunction is zero outside the well. Equivalently, in elementary functions,The distinct phases retain their relative sign from the initial expansion; an overall global phase would not change measurement probabilities.

For a bound state, $0<E<U_0$: the Hamiltonian operator is nonnegative, and an energy at or above the exterior potential does not give a nonzero normalizable wavefunction. The Time-independent Schrodinger equation and the hard wall at zero therefore giveContinuity of the wavefunction and its derivative at the finite step gives $k\cot (ka)=-\kappa$. Put $z=ka$ and $z_0=a\sqrt{2m{U}_0}/\hslash$. The matching equation becomesIt requires $\cot z<0$, so solutions lie in the intervals $((j+1/2)\pi ,(j+1)\pi )$ for $j=0,1,\dots$. Squaring, with that sign restriction retained, gives $z_0=z/|\sin z|$. On each such interval this function increases strictly from $(j+1/2)\pi$ to infinity: its derivative is $(|\sin z|-z\mathrm{sgn}(\sin z)\cos z)/{\sin }^{2}z>0$, because the second term in the numerator is positive there. Each interval therefore contributes exactly one bound state if and only if $z_0>(j+1/2)\pi$. These are the bound-state thresholds for a square well with one hard wall.

Exactly one normalizable bound state exists if and only ifThe printed strict upper inequality misses the threshold case. At the upper endpoint, the prospective second state has $E=U_0$ and $\kappa =0$. Its exterior solution is constant or linear, hence is not square-integrable unless zero; the zero exterior solution and matching would force the entire solution to vanish. Thus there is still exactly one bound state at that endpoint. The strict upper inequality is valid if threshold values are additionally excluded, but that exclusion is not part of the stated assumptions.