Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 3 / 6D (Quantum Mechanics)

The spatial wavefunction $\chi (x)$ is the position amplitude of a stationary quantum state, so $|\chi (x){|}^2$ is its probability density after normalization. The real function $U(x)$ is the potential energy, $E$ is the energy eigenvalue, $m$ is the particle mass, and $\hslash$ is the reduced Planck constant.

As $|x|\to \infty$, the equation approachesIf $E>U_0$, the asymptotic solutions are oscillatory and do not decay, so there are no bound states. If $0<E<U_0$, the asymptotic solutions are exponential; one decaying branch can be selected at each end. Bound states are therefore possible, although only energies satisfying the global matching conditions actually occur.

Suppose ${\chi }_1$ and ${\chi }_2$ are two bound-state solutions with the same energy. Their Wronskiansatisfies $W^{\prime }=0$ because the Schrodinger equation contains no first-derivative term. Hence $W$ is constant. Both wavefunctions and their derivatives decay at infinity, so $W\to 0$ there and therefore $W\equiv 0$. The two solutions are linearly dependent. Thus every one-dimensional bound-state energy is nondegenerate.