 Solution

For each mode in Cartesian coordinates, $[a_i,a_j^{†}]={\delta }_{ij}$ and $a_i^{†}{a}_i=H_i/(\hslash \omega )-1/2$. Consequently the two-dimensional isotropic harmonic oscillator hasThe normalized actions of the creation and annihilation operators areInvert their definitions as $x=\sqrt{\hslash /(2m\omega )}(a_x+a_x^{†})$ and $p_x=i\sqrt{m\hslash \omega /2}(a_x^{†}-a_x)$, and similarly for $y$. linear operators belonging to different coordinates commute, so substitution into $L_z=x{p}_y-y{p}_x$ givesEach term transfers one excitation between the two modes: $[N_x+N_y,a_x^{†}{a}_y]=(1-1)a_x^{†}{a}_y=0$, and likewise for its adjoint. Thus $[H,L_z]=0$ and, with no explicit time dependence, $d{L}_z/dt=(i/\hslash )[H,L_z]=0$. This is conservation of angular momentum under planar rotations.

Finally $L_z|1,0⟩=i\hslash |0,1⟩$ and $L_z|0,1⟩=-i\hslash |1,0⟩$. HenceThe angular momentum is $+\hslash$; division by $\sqrt{2}$ performs wavefunction normalization on this state without changing its eigenvalue. It is one excitation of the positive circular mode.