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The angular momentum operator $L_z$ can be expressed in terms of $L_z = xp_y - yp_x$ where $ p = \hat{p} = -i\sqrt{hwm/2}(a^{\dagger} - a)$ and $x = \hat{x}\sqrt{\hbar/2mw}(a^{\dagger} + a)$.

We want to express $L_z$ in terms of the ladder operators $a$ and $a^{\dagger}$. By direct insertion we then have

$L_z = i \frac{\hbar}{2}((a_x^{\dagger} + a_x)(a_y^{\dagger} - a_y) - (a_y^{\dagger} + a_y)(a_x^{\dagger} - a_x))$

Here's where my problems begin. I've multiplied it out, and tried to invoke both some commutation relations and using the fact that $[a, a^{\dagger}] = 1$ to rewrite the expression, but I cant seem to get the correct answer. Im not that secure on the commutations and relations that can be used the rewrite these operators, obviously. Specifically, I'm not sure what commutations are valid when we're concerned with two directions.

The correct answer is $L_z = i\hbar(a_xa_y^{\dagger} - a_x^{\dagger}a_y)$. Would someone be kind enough to give me some hints or tricks to better manipulate these?

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Position and momentum operators related to different spatial directions commute between themselves. As a simple proof, notice that $\hat{p}_x = -i \hbar \frac{\partial}{\partial x}$ and $\hat{p}_y = -i \hbar \frac{\partial}{\partial y}$ commute due to the fact that partial derivatives commute, $\hat{x} = x$ and $\hat{y} = y$ commute because they act by means of multiplication by a real number and $\hat{p}_x$ and $\hat{y} = y$ commute because they concern different variables.

As a consequence of this fact, it can be show (I'll let you fill in the details) that $[\hat{a}_x, \hat{a}_y] = [\hat{a}^{\dagger}_x, \hat{a}_y] = [\hat{a}^{\dagger}_x, \hat{a}^{\dagger}_y] = 0$ and so on. This can be done by rewriting the ladder operators in terms of positions and momenta and expanding the expressions in the commutators.

With this at hand, your expression will undergo a couple of cancellations, leading to the answer you provided. Phys.SE doesn't recommend very detailed computations on homework-like problems, so I'm leaving a few blanks for you to fill in, but if you're still having a hard time just mention what's cloudy in the comments and I'll elaborate further.

By the way, notice that the argument I gave on the first paragraph holds for position and linear momenta. Angular momentum operators on different directions do not commute.

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