have introduced the Hamiltonian of an ordinary 1-dimensional harmonic oscillator $$H_0 = \omega \left(b^\dagger b + \frac{1}{2}\right),$$ and the angular momentum operator has been identified as $$L = a^\dagger a - b^\dagger b.$$ Interestingly, the creation and annihilation operators $b^\dagger$ and $b$ commute with the total energy $H$ because they raise (lower) $H_0$ by $\omega$, while they lower (raise) $L$ by 1, such that indeed $$[H,b] = [H_0,b] + \omega [L,b] = 0, \
[H,b^\dagger] = [H_0,b^\dagger] + \omega [L,b^\dagger] = 0.$$

Energy Spectrum and Energy Eigenstates
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Since the algebraic structure of the problem (but not the exact form of the Hamiltonian) is the same as for the 2-dimensional harmonic oscillator, we can construct the physical states accordingly. First of all, we construct a state $|0 0\rangle$ that is annihilated by both $a$ and $b$, i.e. $$a|0 0\rangle = b|0 0\rangle = 0.$$ Then we define states $$|n n'\rangle = \frac{\left(a^\dagger\right)^{n}}{\sqrt{n!}} 
\frac{\left(b^\dagger\right)^{n'}}{\sqrt{n'!}} |0 0\rangle,$$ which are eigenstates of the total energy $$H |n n'\rangle = \omega \left(n + \frac{1}{2}\right) |n n'\rangle,$$ as well as of the angular momentum $$L |n n'\rangle = (n - n') |n n'\rangle = m |n n'\rangle.$$ It should be noted that the quantum number 