Past exam of the physics A course of the University of Cambridge / 2022 / Paper 1 / Section B / B7

For a particle in a ring, the arc-length coordinate is $s=R\varphi$. The tangential kinetic energy isThe constraint removes radial motion; the constant potential energy along the ring can be chosen as zero. The tangential momentum operator is ${\hat{p}}_s=-i\hslash {\partial }_s$, or equivalently the angular momentum operator is ${\hat{L}}_z=-i\hslash {\partial }_{\varphi }$. ThusIts physical domain has periodic boundary conditions, $\psi (\varphi +2\pi )=\psi (\varphi )$ and ${\psi }^{\prime }(\varphi +2\pi )={\psi }^{\prime }(\varphi )$. These make the Hamiltonian self-adjoint and reflect that $\varphi$ and $\varphi +2\pi$ describe the same point. No magnetic flux or additional angle-dependent potential energy is present.

Solving the Time-independent Schrödinger equation with $\psi \propto e^{ik\varphi }$ gives $E={\hslash }^2{k}^2/(2m{R}^2)$. The periodic boundary condition requires $e^{2\pi ik}=1$, hence $k=n\in \mathbb{Z}$. With the angular integration measure $d\varphi$, wavefunction normalization givesso a convenient orthonormal eigenbasis isThe $n=0$ ground state is nondegenerate. For each $|n|>0$, $u_n$ and $u_{-n}$ span a two-dimensional energy eigenspace; real sine and cosine combinations are an equally valid basis. These opposite-circulation eigenstates have angular momentum $n\hslash$ and $-n\hslash$.

If instead the wavefunction is normalized with respect to physical arc length $ds=Rd\varphi$, its expression is $e^{in\varphi }/\sqrt{2\pi R}$. The two conventions describe the same probability, since $|u_n{|}^{2}d\varphi =|u_n/\sqrt{R}{|}^{2}ds$. The remaining parts use $d\varphi$.

Put $ϵ={\hslash }^2/(2m{R}^2)$. The two-particle product energy eigenstates areBecause the particles are distinguishable, $(n_1,n_2)$ and $(n_2,n_1)$ are distinct when $n_1\ne n_2$. Also, the signs of nonzero integers distinguish opposite angular momenta. The quantum degeneracy is therefore the number of ordered integer pairs at each sum of squares.

The first distinct energy levels are $0,ϵ,2ϵ,4ϵ,5ϵ,\dots$. This ordering matters: $4ϵ$ is the third excited level, while $5ϵ$ is the fourth excited level.

These counts concern energy eigenspaces; arbitrary quantum superpositions within each one do not create additional independent states.

Identical spin one-half particles are fermions. By fermionic exchange symmetry, the complete spatial-and-spin wavefunction must change sign when the two particle labels are exchanged. Define the normalized antisymmetric spin-one-half singlet stateand the three symmetric spin-one-half triplet statesThe Hamiltonian is independent of spin, so these spin functions carry no additional energy.

Both particles can occupy $u_0$ if their spin factor is antisymmetric. The unique ground state wavefunction isTwo equal-spin particles could not share this orbital, by the Pauli exclusion principle.

At the first excited energy, $E=ϵ={\hslash }^2/(2m{R}^2)$, one orbital is $u_0$ and the other is $u_s$ with $s=+1$ or $-1$. Define symmetric and antisymmetric normalized spatial functionsFor each $s$, all allowed wavefunctions are spanned byThere are $2(1+3)$ orthonormal states, hence first-excited-state degeneracy $g=8$. In each product, exactly one factor is antisymmetric, ensuring the required total fermionic exchange symmetry.