# 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.

## 1 Answer

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.

• OK, I have $R_{y}^{-1}(\beta) L_{z} R_{y}(\beta) = e^{iL_y\beta}L_z e^{-iL_y \beta}$. When I apply BCH I was expecting the series to truncate, but it doesn't, as $[iL_y\beta , L_z] = i \beta Lx$. I'm still not coming up with the result you gave.
– user282846
Commented Feb 12, 2021 at 0:08
• The series does not truncate but there are terms in $L_z$ and $L_x$ that can each be resummed to trig functions. Commented Feb 12, 2021 at 0:10
• Thanks, I get $\cos{(\beta)}L_z - \sin{(\beta)}L_x$, so a small change in sign from what you've written.
– user282846
Commented Feb 12, 2021 at 0:50
• that could be right I may be had my signs wrongs in the exponentials. Commented Feb 12, 2021 at 2:30
• I was able to make it to the final result, but I can't make any sense of it. I've written things up in a new question. If you might have any insight, I'd be grateful.
– user282846
Commented Feb 12, 2021 at 4:58