Angular momentum operator for 2-D harmonic oscillator

[Rabindranath]

1. The problem statement

I want to write the angular momentum operator $L$ for a 2-dimensional harmonic oscillator, in terms of its ladder operators, $a_x$, $a_y$, $a_x^\dagger$ & $a_y^\dagger$, and then prove that this commutes with its Hamiltonian.

The Attempt at a Solution

I get

$L = (x + y) × (p_x + p_y) = -i \hbar ((a_x^\dagger +a_y^\dagger) × (a_x + a_y) + a_x × a_x + a_y × a_y)$.

Does this seem to make any sense?

As for the commutator, with Hamiltonian $H = \hbar \omega (a_x^\dagger a_x + a_y^\dagger a_y + 1)$, I consequently get all sorts of exotic product terms, e.g.

$\left ( (a_x^\dagger +a_y^\dagger) × (a_x + a_y) \right ) a_x^\dagger a_x$

and the like, and I'm not comfortable with handling these. I want all of them to go away, for said operators to commute, but am not sure how to do/see that.

Any comments would be appreciated.

 

[hilbert2]

You should rather show that the operator $L^2$, which is the same as $L_{z}^{2}$ for motion on the xy-plane, commutes with the Hamiltonian. This actually happens only when the "spring constants" are the same in x and y directions and the potential energy $V(r,\theta )$ depends only on the radial coordinate $r$.

 

[Chandra Prayaga]

You can first work out the commutator of L_z with the Hamiltonian. In your attempt, the way you wrote L is wrong. You should work out the cross product of the position vector and the momentum vector.