# Matrix representation of $xp_y - yp_x$ in two-dimensional harmonic oscillator

I have a pretty simply problem that is causing me some confusion. I am asked to consider a two-dimensional harmonic oscillator and compute the matrix representation of the operator $Q = xp_y - yp_x$ in the subspace spanned by the set $\{\Psi_0(x,y), \Psi_1(x,y) \} = \{\psi_0(x) \psi_1(y), \psi_1(x) \psi_0(y) \}$ of product states, where $\psi_0$ and $\psi_1$ are the usual ground state and first excited state.

I think I understand the diagonal terms. For instance,

$$Q_{11} = \langle \Psi_1 \rvert Q \lvert \Psi_1 \rangle = \langle \psi_0 \rvert \langle \psi_1 \rvert \big(x \lvert \psi_0 \rangle p_y \lvert \psi_1 \rangle - p_x \lvert \psi_0 \rangle y \lvert \psi_1 \rangle \big) = \langle \psi_0 \rvert x \lvert \psi_0 \rangle \langle \psi_1 \rvert p_y \lvert \psi_1 \rangle - \langle \psi_0 \rvert p_x \lvert \psi_0 \rangle \langle \psi_1 \rvert y \lvert \psi_1 \rangle.$$

But $\langle x \rangle = 0$ for stationary states of the harmonic oscillator by symmetry, and by Ehrenfest's theorem $\langle p \rangle = 0$, so the above evaluates to zero. (Which is the correct answer.)

However, I was hoping that someone could walk me through computing the off-diagonal elements. The answers are $Q_{12} = i\hbar$ and $Q_{21} = -i\hbar$, so I assume I have to find the canonical commutator somewhere in the above, but I can't see where.

Also, if the above argument for $Q_{11} = 0$ is incorrect, I would also appreciate corrections. I want to treat the product states as tensor products, but I don't know if that's okay.

Your calculation of $Q_{11}$ is correct. The off-diagonal elements would be of the form $$\langle \psi_0\vert x\vert \psi_1\rangle \langle \psi_1\vert p_y\vert\psi_0\rangle$$ and are $\ne 0$ by the very kind of argument you made to show that diagonal ones are $0$.

The observable $x p_y-y p_x$ is nothing but $L_z$. If you go to cylindrical coordinates, you will find that $$L_z\mapsto -i\hbar\frac{\partial}{\partial \phi}$$ while $x\pm i y=r\sin\theta e^{\pm i\phi}$ so you can see how the combinations $x\pm iy$ will occur naturally as they are eigenstates of $L_z$: indeed if you write $$Q=i\hbar \left(\begin{array}{cc} 0&1\\ -1&0\end{array}\right)$$ you will find the eigenstates are... $\left(\begin{array}{c}1 \\ \pm i\end{array}\right)$.

Note that the states you have $\psi_0(x)\psi_1(y)$ and $\psi_1(x)\psi_0(y)$ have the same energy, so linear combinations of them like $\psi_0(x)\psi_1(y)\pm i \psi_1(x)\psi_0(y)$ also have the same energy. As a matter of fact $L_z$ commutes with the Hamiltonian so the new basis $\psi_0(x)\psi_1(y)\pm i \psi_1(x)\psi_0(y)$ is simply one where the states are eigenstates of $L_z$ and $H$, and indeed you can verify that the angular part of $\psi_0(x)\psi_1(y)\pm i \psi_1(x)\psi_0(y)$ is $\sin\theta e^{\pm i\phi}$ and proportional to the spherical harmonics $Y_{1,\pm 1}(\theta,\phi)= \mp \sqrt{3/(8\pi)} \sin\theta e^{\pm i\phi}$, which are explicitly eigenstates of $L_z$.

I figured it out myself. Obviously, it's a good idea to express $Q$ in terms of the creation and annihilation operators:

$$Q = xp_y - yp_x = i\hbar (a_x a_y^\dagger - a_y a_x^\dagger)$$

Everything follows straightforwardly from here.