(a) Consider the angular momentum operators $\hat{L}_{x}, \hat{L}_{y}, \hat{L}_{z}$ and $\hat{\mathbf{L}}^{2}=\hat{L}_{x}^{2}+\hat{L}_{y}^{2}+\hat{L}_{z}^{2}$ where

$$\hat{L}_{z}=\hat{x} \hat{p}_{y}-\hat{y} \hat{p}_{x}, \quad \hat{L}_{x}=\hat{y} \hat{p}_{z}-\hat{z} \hat{p}_{y} \text { and } \hat{L}_{y}=\hat{z} \hat{p}_{x}-\hat{x} \hat{p}_{z} .$$

Use the standard commutation relations for these operators to show that

$$\hat{L}_{\pm}=\hat{L}_{x} \pm i \hat{L}_{y} \quad \text { obeys } \quad\left[\hat{L}_{z}, \hat{L}_{\pm}\right]=\pm \hbar \hat{L}_{\pm} \quad \text { and } \quad\left[\hat{\mathbf{L}}^{2}, \hat{L}_{\pm}\right]=0$$

Deduce that if $\varphi$ is a joint eigenstate of $\hat{L}_{z}$ and $\hat{\mathbf{L}}^{2}$ with angular momentum quantum numbers $m$ and $\ell$ respectively, then $\hat{L}_{\pm} \varphi$ are also joint eigenstates, provided they are non-zero, with quantum numbers $m \pm 1$ and $\ell$.

(b) A harmonic oscillator of mass $M$ in three dimensions has Hamiltonian

$$\hat{H}=\frac{1}{2 M}\left(\hat{p}_{x}^{2}+\hat{p}_{y}^{2}+\hat{p}_{z}^{2}\right)+\frac{1}{2} M \omega^{2}\left(\hat{x}^{2}+\hat{y}^{2}+\hat{z}^{2}\right) .$$

Find eigenstates of $\hat{H}$ in terms of eigenstates $\psi_{n}$ for an oscillator in one dimension with $n=0,1,2, \ldots$ and eigenvalues $\hbar \omega\left(n+\frac{1}{2}\right)$; hence determine the eigenvalues $E$ of $\hat{H}$.

Verify that the ground state for $\hat{H}$ is a joint eigenstate of $\hat{L}_{z}$ and $\hat{\mathbf{L}}^{2}$ with $\ell=m=0$. At the first excited energy level, find an eigenstate of $\hat{L}_{z}$ with $m=0$ and construct from this two eigenstates of $\hat{L}_{z}$ with $m=\pm 1$.

Why should you expect to find joint eigenstates of $\hat{L}_{z}, \hat{\mathbf{L}}^{2}$ and $\hat{H}$ ?

[ The first two eigenstates for an oscillator in one dimension are $\psi_{0}(x)=$ $C_{0} \exp \left(-M \omega x^{2} / 2 \hbar\right)$ and $\psi_{1}(x)=C_{1} x \exp \left(-M \omega x^{2} / 2 \hbar\right)$, where $C_{0}$ and $C_{1}$ are normalisation constants. ]