Computing dispersion of $L_3$ in spin coherent states I am trying to compute $\langle \left(\Delta L_3\right)^2\rangle$ for coherent states of $SU(2)$. I understand that a set of coherent states can be be formed from rotations of the the state $|j,j\rangle$,
$$\left\{T^{(j)}(g)|j,j\rangle, g \in S U(2)\right\}$$
A potentially helpful property is
$$\hat{n}\cdot \vec{L}|\hat{n}\rangle = j|\hat{n}\rangle$$
where $\hat{n}$ is a point on on the sphere, i.e. $\hat{n} = (\sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta)$. I want to understand how $\langle \left(\Delta L_3\right)^2\rangle$ changes as a function of $\varphi$ and $\theta$ for different coherent states in this coherent state system.
\begin{align}
\langle \left(\Delta L_3\right)^2\rangle &= \langle L_3^2 \rangle - \langle L_3\rangle^2\\
&= \langle \hat{n}|L_3^2 |\hat{n}\rangle - \langle \hat{n}| L_3|\hat{n}\rangle^2
\end{align}
I can define $|\hat{n}\rangle$ as,
\begin{align}
|\hat{n}\rangle = \exp (i \theta \hat{m} \cdot \hat{L})|j,j\rangle
\end{align}
where $\hat{m} = (\sin \varphi,-\cos \varphi, 0)$. Now I try to compute $\langle \hat{n}| L_3|\hat{n}\rangle$,
\begin{align}
\langle \hat{n}| L_3|\hat{n}\rangle = \langle j,j |\left(\exp (i \theta \hat{m} \cdot \hat{L})\right)^\dagger L_3\exp (i \theta \hat{m} \cdot \hat{L})|j,j\rangle
\end{align}
I haven't been able to progress past this point, not that I'm even sure I'm on the right track.
 A: You’re not using a good form for the rotation operator.  If instead you use the Euler factorization
\begin{align}
R(\alpha,\beta,\gamma)=R_z(\alpha)R_y(\beta)R_z(\gamma)
\end{align}
and take
\begin{align}
R(\alpha,\beta,\gamma)\vert j,j\rangle \sim 
R(\alpha,\beta,0)\vert j,j\rangle
\end{align}
since the two rotated states differ by an overall phase,
then
\begin{align}
\langle L_z\rangle 
&= \langle j,j\vert R^{-1}_y(\beta)R^{-1}_z(\alpha) L_z 
R_z(\alpha)R_y(\beta)\vert j,j\rangle \, ,\\
&= \langle j,j\vert R^{-1}_y(\beta) L_z R_y(\beta)\vert j,j\rangle
\end{align}
since $R_z$ commutes with $L_z$.  Then it’s a matter of computing
\begin{align}
R_y^{-1}(\beta)L_zR_y(\beta)
\end{align}
for which you can use the general expression
$$
e^{A}B e^{-A}= B+[A,B]+\frac{1}{2}[A,[A,B]]+\ldots
$$
or you can “guess” on physical grounds that
\begin{align}
R_y^{-1}(\beta)L_zR_y(\beta)\sim \cos(\beta)L_z +\sin(\beta) L_x\, .
\end{align}
The computation of $\langle L_z^2\rangle$ follows the same approach.
