 Solution

The angular momentum commutation relations are $[L_x,L_y]=i\hslash L_z$ and cyclic permutations. The product rule yields $[L_z,L_x^2]=i\hslash (L_y{L}_x+L_x{L}_y)$ and $[L_z,L_y^2]=-i\hslash (L_x{L}_y+L_y{L}_x)$, so $[L_z,L^2]=0$. Simultaneous eigenstates have eigenvalues ${\hslash }^{2}l(l+1)$ and $\hslash m$; orbital angular momentum has integer $l\ge 0$ and $m=-l,\dots ,l$.

By cyclic symmetry, $L^2$ commutes with every component and hence every squared component. Therefore $[H,L^2]=0$: an energy eigenbasis can be chosen with definite $l$. For unequal transverse coefficients, $L_z$ need not commute with $H$, so definite $m$ is not generally available.

Expanding the ladder operators gives $L_{+}^2+L_{-}^2=2(L_x^2-L_y^2)$ and $L^2-L_z^2=L_x^2+L_y^2$. Comparing coefficients givesFor $l=1,m=0$, both squared ladder operators vanish, since two raising or lowering steps would leave the multiplet. Thus $E_0=2\beta {\hslash }^2={\hslash }^2(a_x+a_y)$. In the remaining basis $\{|1,1⟩,|1,-1⟩\}$ the Hamiltonian isIts symmetric and antisymmetric combinations have energies ${\hslash }^2(a_x+a_z)$ and ${\hslash }^2(a_y+a_z)$, respectively. ConsequentlyThe energies are related by permutation of the principal axes, not generally equal. Two coincide when the corresponding coefficients coincide; all three become $2a{\hslash }^2$ in the isotropic limit.