Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2B / Section A / 26

Inside the infinite square well, the Hamiltonian is purely the kinetic energy operator,Its admissible wavefunctions obey the boundary conditions $\psi (0)=\psi (L)=0$ and vanish outside the interval. Equivalently write $\hat{H}={\hat{p}}^2/(2m)+V(x)$ with $V=0$ inside and an infinite confining wall outside. The boundary conditions are part of the specification of this Hamiltonian; the differential expression alone does not determine the spectrum.

Twice differentiating gives $d^2\sin (kx)/d{x}^2=-k^2\sin (kx)$, so inside the infinite square wellThe boundary condition at zero is automatic. The other requires $\sin (kL)=0$, hence $kL=n\pi$. A nonzero eigenfunction can therefore be labelled byIf desired, its normalized form is $\sqrt{2/L}\sin (n\pi x/L)$. The positive lowest energy follows from confinement and the required spatial variation of a nonzero wavefunction.

For $n=0$ the proposed wavefunction vanishes identically and cannot be normalized. A negative integer gives $\sin (-|n|\pi x/L)=-\sin (|n|\pi x/L)$, which differs from the corresponding positive-integer wavefunction only by a physically irrelevant global phase. Both have the same positive energy, proportional to $n^2$. Thus zero is inadmissible and negative labels are redundant. A negative integer is not forbidden by a separate negative-energy argument.