# Commutation Relations, 2D Harmonic Oscillator

1. May 17, 2017

### BOAS

1. The problem statement, all variables and given/known data
Consider a two-dimensional harmonic oscillator, described by the Hamiltonian

$\hat H_0 = \hbar \omega (\hat a_x \hat a_x ^{\dagger} + \hat a_y \hat a_y^{\dagger} + 1)$

Calculate $\hat H_0 \hat L | n_1, n_2 \rangle$ and $\hat L \hat H_0 |n_1, n_2 \rangle$. What does this imply for $[\hat H_0 , \hat L]$

2. Relevant equations

3. The attempt at a solution

I can solve this problem by "brute force", having shown that $\hat L = \hat x \hat P_y - \hat y \hat P_x = i \hbar (\hat a_y^{\dagger} \hat a_x - \hat a_x ^{\dagger} \hat a_y)$. I can apply one operator and then the other, but the results are messy making me think this can be done in a smarter manner.

Trying to combine these operators into a single simplified operator gives me

$H_0 \hat L = i \hbar^2 \omega (\hat a_x ^{\dagger} \hat a_x (\hat a_y ^{\dagger} \hat a_x - \hat a_x ^{\dagger} \hat a_y) + \hat a_y^{\dagger} \hat a_y (\hat a_y^{\dagger} \hat a_x - \hat a_x^{\dagger} \hat a_y) + (\hat a_y^{\dagger} \hat a_x - \hat a_x^{\dagger} \hat a_y))$

I need to deal with this $(\hat a_y ^{\dagger} \hat a_x - \hat a_x ^{\dagger} \hat a_y)$ term but I am unsure of how to do it. I know that $[\hat a_i, \hat a_i^{\dagger}] = 1$ and that the x and y operators are mutually commuting.

If I set $\hat N = \hat a_y^{\dagger} \hat a_x$, then $(\hat a_y ^{\dagger} \hat a_x - \hat a_x ^{\dagger} \hat a_y) = [\hat N, \hat N^{\dagger}]$ but the numbered states or not eigenstates of my invented operator...

So, is there anything I can realise to make this problem less of an accounting exercise?

I hope it's clear what i'm asking; I can solve this problem, but it does not feel efficient.

Thank you for your help!

2. May 17, 2017

### MisterX

Do you realize that the Hamiltonian on a number state $\mid n_1,n_2\rangle$ has eigenvalue related to the total $n_1+n_2$?
\begin{align*} H& = \hbar \omega (a_x ^{\dagger}a_x + a_y^{\dagger}a_y \ + 1) \\
H &= \hbar\omega(N_x + N_y + 1)\\
H|n_1, n_2> &= \hbar\omega(n_1 + n_2+1)|n_1, n_2>\end{align*}
So a question you can ask is "does $L$ change the total $n_1 + n_2$ ?"
You can examine the operator $L$ or try it out.

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