Past exam of the physics A course of the University of Cambridge / 2012 / Paper 1 / Section B / B6 / Solution

The canonical commutation relations are $[x,p_x]=[y,p_y]=i\hslash$, with $[x,y]=[p_x,p_y]=[x,p_y]=[y,p_x]=0$. Operators from different Cartesian modes commute, so $(p_x{p}_y{)}^{†}=p_y{p}_x=p_x{p}_y$ and $(xy{)}^{†}=yx=xy$. Thus $J_1$ is Hermitian and represents an observable on the usual oscillator domain.

Introduce annihilation operators $a_x=\sqrt{m\omega /(2\hslash )}x+i{p}_x/\sqrt{2m\hslash \omega }$ and similarly $a_y$. Set $N_x=a_x^{†}{a}_x$, $N_y=a_y^{†}{a}_y$ and $N=N_x+N_y$. The Schwinger boson representation givesFor $A=a_x^{†}{a}_y$ and $B=a_y^{†}{a}_x$, $AB=N_x(N_y+1)$ and $BA=N_y(N_x+1)$. ThereforeThe two-mode oscillator Casimir identity isSince $[N,A]=[N,B]=0$, $[H,J_1]=0$, and $J_1$ is conserved. However, $[J_1,J_2]=i\hslash J_3$ is not the zero operator, so there is no general joint eigenbasis. The exception is important: a common eigenstate of two angular-momentum components must satisfy $J_3\psi =0$, then the other commutators force $J_1\psi =J_2\psi =0$. Thus only the $j=0$ sector admits a common state; here that is the oscillator ground state.

Using $J^2={\hslash }^{2}j(j+1)$, with $j=0,1/2,1,\dots$, and selecting the positive energy branch givesEquivalently $N=2j=0,1,\dots$, so $E_N=\hslash \omega (N+1)$ and the degeneracy is $N+1$, corresponding to $N_x=0,\dots ,N$.