that $a^\dagger$ and $a$ also raise and lower the angular momentum. Interestingly, we have seen before that $$[L,R_\pm] = \pm R_\pm, \ [R_+,R_-] = \frac{2}{e B} = \frac{2}{M \omega}.$$ Hence, $R_+$ and $R_-$ also act as raising and lowering operators of the angular momentum. Indeed, we can identify another set of creation and annihilation operators $$b = \sqrt{\frac{M \omega}{2}} R_+, \ b^\dagger = \sqrt{\frac{M \omega}{2}} R_-,$$ which obey $$[L,b] = b, \ [L,b^\dagger] = - b^\dagger.$$ As a result, $b$ raises and $b^\dagger$ lowers the angular momentum by one unit. It is straightforward to derive the commutation relations $$[a,b] = [a^\dagger,b] = [a,b^\dagger] = [a^\dagger,b^\dagger] = 0, \ 
[b,b^\dagger] = 1.$$ Interestingly, just like a 2-dimensional harmonic oscillator, the particle in a magnetic field is described by two sets of commuting creation and annihilation operators. However, in contrast to the 2-dimensional harmonic oscillator, the Hamiltonian of the particle in a magnetic field contains only $a^\dagger a$, but not $b^\dagger b$.

Alternative Representation of the Hamiltonian
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Interestingly, the Hamiltonian can also be expressed as $$H = \frac{1}{2} M \omega^2 \left(R_x^2 + R_y^2\right) + \omega L =
\omega \left(b^\dagger b + \frac{1}{2} + L\right) = H_0 + \omega L.$$ Here we 