(i) Consider two quantum systems with angular momentum states $|j m\rangle$ and $|1 q\rangle$. The eigenstates corresponding to their combined angular momentum can be written as

$$|J M\rangle=\sum_{q m} C_{q m}^{J M}|1 q\rangle|j m\rangle,$$

where $C_{q m}^{J M}$ are Clebsch-Gordan coefficients for addition of angular momenta one and $j$. What are the possible values of $J$ and how must $q, m$ and $M$ be related for $C_{q m}^{J} M \neq 0$ ?

Construct all states $|J M\rangle$ in terms of product states in the case $j=\frac{1}{2}$.

(ii) A general stationary state for an electron in a hydrogen atom $|n \ell m\rangle$ is specified by the principal quantum number $n$ in addition to the labels $\ell$ and $m$ corresponding to the total orbital angular momentum and its component in the 3-direction (electron spin is ignored). An oscillating electromagnetic field can induce a transition to a new state $\left|n^{\prime} \ell^{\prime} m^{\prime}\right\rangle$ and, in a suitable approximation, the amplitude for this to occur is proportional to

$$\left\langle n^{\prime} \ell^{\prime} m^{\prime}\left|\hat{x}_{i}\right| n \ell m\right\rangle,$$

where $\hat{x}_{i}(i=1,2,3)$ are components of the electron's position. Give clear but concise arguments based on angular momentum which lead to conditions on $\ell, m, \ell^{\prime}, m^{\prime}$ and $i$ for the amplitude to be non-zero.

Explain briefly how parity can be used to obtain an additional selection rule.

[Standard angular momentum states $|j m\rangle$ are joint eigenstates of $\mathbf{J}^{2}$ and $J_{3}$, obeying

$$J_{\pm}|j m\rangle=\sqrt{(j \mp m)(j \pm m+1)}|j m \pm 1\rangle, \quad J_{3}|j m\rangle=m|j m\rangle .$$

You may also assume that $X_{\pm 1}=\frac{1}{\sqrt{2}}\left(\mp \hat{x}_{1}-i \hat{x}_{2}\right)$ and $X_{0}=\hat{x}_{3}$ have commutation relations with orbital angular momentum $\mathbf{L}$ given by

$$\left[L_{3}, X_{q}\right]=q X_{q}, \quad\left[L_{\pm}, X_{q}\right]=\sqrt{(1 \mp q)(2 \pm q)} X_{q \pm 1}$$

Units in which $\hbar=1$ are to be used throughout. ]