If $A, B$, and $C$ are operators establish the identity

$$[A B, C]=A[B, C]+[A, C] B$$

A particle moves in a two-dimensional harmonic oscillator potential with Hamiltonian

$$H=\frac{1}{2}\left(p_{x}^{2}+p_{y}^{2}\right)+\frac{1}{2}\left(x^{2}+y^{2}\right) .$$

The angular momentum operator is defined by

$$L=x p_{y}-y p_{x}$$

Show that $L$ is hermitian and hence that its eigenvalues are real. Establish the commutation relation $[L, H]=0$. Why does this ensure that eigenstates of $H$ can also be chosen to be eigenstates of $L$ ?

Let $\phi_{0}(x, y)=e^{-\left(x^{2}+y^{2}\right) / 2 \hbar}$, and show that $\phi_{0}, \quad \phi_{x}=x \phi_{0}$ and $\phi_{y}=y \phi_{0}$ are all eigenstates of $H$, and find their respective eigenvalues. Show that

$$L \phi_{0}=0, \quad L \phi_{x}=i \hbar \phi_{y}, \quad L \phi_{y}=-i \hbar \phi_{x}$$

and hence, by taking suitable linear combinations of $\phi_{x}$ and $\phi_{y}$, find two states, $\psi_{1}$ and $\psi_{2}$, satisfying

$$L \psi_{j}=\lambda_{j} \psi_{j}, \quad H \psi_{j}=E_{j} \psi_{j} \quad j=1,2 .$$

Show that $\psi_{1}$ and $\psi_{2}$ are orthogonal, and find $\lambda_{1}, \lambda_{2}, E_{1}$ and $E_{2}$.

The particle has charge $e$, and an electric field of strength $\mathcal{E}$ is applied in the $x$ direction so that the Hamiltonian is now $H^{\prime}$, where

$$H^{\prime}=H-e \mathcal{E} x$$

Show that $\left[L, H^{\prime}\right]=-i \hbar e \mathcal{E} y$. Why does this mean that $L$ and $H^{\prime}$ cannot have simultaneous eigenstates?

By making the change of coordinates $x^{\prime}=x-e \mathcal{E}, y^{\prime}=y$, show that $\psi_{1}\left(x^{\prime}, y^{\prime}\right)$ and $\psi_{2}\left(x^{\prime}, y^{\prime}\right)$ are eigenstates of $H^{\prime}$ and write down the corresponding energy eigenvalues.

Find a modified angular momentum operator $L^{\prime}$ for which $\psi_{1}\left(x^{\prime}, y^{\prime}\right)$ and $\psi_{2}\left(x^{\prime}, y^{\prime}\right)$ are also eigenstates.