0
$\begingroup$

This is a follow-up to a previous question. I am trying to compute $\langle (\Delta L_z)^2\rangle$, for a general coherent state in the coherent state system for $SU(2)$, where I get all the coherent states via rotations of the state $|j,j\rangle$. I have already computed $\langle L_z\rangle = \langle j,j | L_z\cos{\beta} - L_x\sin{\beta}|j,j\rangle$. I am struggling to make sense of the computation of $\langle L_z^2\rangle$ \begin{align} \langle L_z^2\rangle&=\left\langle j, j\left|R_{y}^{-1}(\beta) L_{z}^2 R_{y}(\beta)\right| j, j\right\rangle\\ &=\left\langle j, j\left|e^{iL_y\beta} L_{z}^2 e^{-iLy\beta}\right| j, j\right\rangle \end{align} I use the following identity to compute the operator product within the expectation value here, \begin{align} e^{i G \lambda} A e^{-i G \lambda}=A+i \lambda[G, A]+\frac{(i \lambda)^{2}}{2 !}[G,[G, A]]+\ldots+\frac{(i \lambda)^{n}}{n !} \underbrace{[G,[G,[G, \ldots[G}_{n \text { times }}, A]]] \ldots]+\ldots \end{align} I have computed the following commutators, in hopes of seeing a pattern that I can resum into trigonometric functions, \begin{align} [L_y,L_z^2] &= 2i(L_zL_x) \\ [L_y,[L_y,L_z^2]] &= 2(L_z^2-L_x^2) \\ [L_y,[L_y,[L_y,L_z^2]]] &= 4i(L_zL_x + L_xL_z) \\ [L_y,[L_y,[L_y,[L_y,L_z^2]]]] &= 8(L_z^2-L_x^2) \\ [L_y,[L_y,[L_y,[L_y,[L_y,L_z^2]]]]] &= 16i(L_zL_x + L_xL_z) \\ \end{align} and so on. Now I think I can ignore all of the $L_zL_x + L_xL_z$ terms, because, \begin{align} \langle j,j|L_zL_x|j,j\rangle = \langle j,j|L_z\left(\frac{L_+ + L_-}{2}\right)|j,j\rangle = 0 \end{align} This leaves me with the expression, \begin{align} e^{iL_y\beta} L_{z}^2 e^{-iLy\beta} &= (L_z^2 - L_x^2)\left(\frac{(i\beta)^2}{2!}2 + \frac{(i\beta)^4}{4!}8+ \frac{(i\beta)^6}{6!}32 +\dotsm \right)\\ &= (L_z^2 - L_x^2)\left(\sum_{n=1}^\infty \frac{(-1)^n(\beta)^{2n}}{(2n)!}\cdot 2^{2n-1}\right)\\ &= \frac{1}{2}(L_z^2 - L_x^2)\left(\sum_{n=1}^\infty \frac{(-1)^n(2\beta)^{2n}}{(2n)!}\right)\\ &= \frac{1}{2}(L_z^2 - L_x^2)\cos{2\beta}\\ \end{align} Putting it all together, \begin{align} \left\langle\left(\Delta L_{z}\right)^{2}\right\rangle&=\left\langle L_{z}^{2}\right\rangle-\left\langle L_{z}\right\rangle^{2}\\ &= \langle j,j|\frac{1}{2}(L_z^2 - L_x^2)\cos{2\beta}|j,j\rangle - \left(\left\langle j, j\left|L_{z} \cos \beta-L_{x} \sin \beta\right| j, j\right\rangle\right)^2\\ &= \cos{2\beta}\left(\frac{j^2}{2} - \frac{1}{2}\langle j,j|L_x^2|j,j\rangle\right) - j^2\cos^2{\beta}\\ &= \cos{2\beta}\left(\frac{j^2}{2} - \frac{1}{2}\langle j,j|\left(\frac{L_+ + L_-}{2}\right)^2|j,j\rangle\right) - j^2\cos^2{\beta}\\ &= \cos{2\beta}\left(\frac{j^2}{2} - \frac{1}{2}\langle j,j|\left(\frac{L_+ + L_-}{2}\right)^2|j,j\rangle\right) - j^2\cos^2{\beta}\\ &= \cos{2\beta}\left(\frac{j^2}{2} - \frac{1}{2}\langle j,j|\left(\frac{1}{4}L_+L_-\right)|j,j\rangle\right) - j^2\cos^2{\beta}\\ &= \cos{2\beta}\left(\frac{j^2}{2} - \frac{j}{4}\right) - j^2\cos^2{\beta}\\ &= \left(2\cos^2{\beta} - 1\right)\left(\frac{j^2}{2} - \frac{j}{4}\right) - j^2\cos^2{\beta}\\ &=\frac{j}{2}\left(\frac{1}{2} - j - \cos^2{\beta}\right) \end{align}

And I don't have any intuitive sense of this result, which makes me think there's a mistake. I thought the outcome here would show a simple, $\propto \sin^2{\beta}$ relationship, where the coherent states have maximum dispersion when they're positioned at the equator.

$\endgroup$
0
$\begingroup$

Start from $e^A B e^{-A}=C$ then $e^AB^2e^{-A}=(e^{A}Be^{-A})(e^A B e^{-A})=C^2$.

Note that your current calculation is likely technically incorrect somewhere since for large enough $j$ you would have $\langle L_z^2\rangle<0$ and the average value of a non-negative operator is surely positive.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy