Ladder operators to prove eigenstates of total angular momen

[Logan Rudd]

Homework Statement

Consider the following state constructed out of products of eigenstates of two individual angular momenta with $j_1 = \frac{3}{2}$ and $j_2 = 1$:
$$
\begin{equation*}
\sqrt{\frac{3}{5}}|{\tiny\frac{3}{2}, -\frac{1}{2}}\rangle |{\tiny 1,-1}\rangle + \sqrt{\frac{2}{5}}|{\tiny\frac{3}{2},-\frac{3}{2}}\rangle|{\tiny1,0}\rangle
\end{equation*}
$$

(a) Show that this is an eigenstate of the total angular momentum. What are the values of $j$ and $m_j$ for this state?

Homework Equations

$$
\begin{equation}
J^2|\Psi\rangle=\hbar ^2j(J+1)|\Psi\rangle
\end{equation}
$$

$$
\begin{equation}
J_\pm=J_x\pm iJ_y
\end{equation}
$$

$$
\begin{equation}
J^2 = (J_1)^2 + (J_2)^2 + 2(J_1\cdot J_2)
\end{equation}
$$$$
\begin{equation}
J_1 \cdot J_2 = J_{1x}J_{2x} + J_{1y}J_{2y} + J_{1z}J_{2z}
\end{equation}
$$

The Attempt at a Solution

I want to use (2) to write (4) in terms of the ladder operators so I can prove the left hand side of (8) does work out to be the right hand side. Re-writing (4) gives me:$$
\begin{equation}
J_1 \cdot J_2 = J_{1\pm}J_{2\pm} \mp i J_{2y}J_{1\pm} \mp i J_{1y}J_{2\pm} + J_{1z}J_{2z}
\end{equation}
$$

I know how $J_{\pm}$ and $J_z$ act on $|\Psi\rangle$ however I don't know how $J_y$ acts on $|\Psi\rangle$ Can someone tell what I can do to the $J_y$ terms?

 

[fzero]

You can solve (2) for both $J_x$ and $J_y$ by computing $J_+ \pm J_-$.

 

[Logan Rudd]

Ah great. Thanks!