Past exam of the quantum theory course of the University of Oxford / 2021 / 1 / a

For a free particle, the Hamiltonian operator is $H=P^2/(2m)$, where $P=-i\hslash d/dx$ is the momentum operator. The time-independent Schrödinger equation is thereforeThe full stationary state is $\Psi (x,t)=e^{-iEt/\hslash }\psi (x)$. The periodic boundary condition identifies $x$ and $x+2\pi$, giving a particle on a circle with wavefunction normalization over one period.

For a momentum eigenstate with eigenvalue $p$, the equation $-i\hslash {\psi }^{\prime }=p\psi$ has solution $\psi (x)=C{e}^{ipx/\hslash }$. The periodic boundary condition imposes $e^{2\pi ip/\hslash }=1$, so $p/\hslash \in \mathbb{Z}$. The wavefunction normalization gives $1=2\pi |C{|}^2$. Choosing a convenient global phase yieldsThese momentum eigenstates form an orthonormal basis, the Fourier basis for the particle on a circle:Here ${\delta }_{nk}$ is the Kronecker delta.

Since $P{\psi }_n=n\hslash {\psi }_n$, applying the Hamiltonian operator givesThus the energy eigenvalues and corresponding full stationary states areThe ground state is ${\psi }_0$ with energy zero and quantum degeneracy one. For each $n=1,2,\dots$, the positive energy level $n^2{\hslash }^2/(2m)$ has quantum degeneracy two: its energy eigenspace is $\mathrm{span}\{{\psi }_n,{\psi }_{-n}\}$. This is the free particle spectrum on a circle.