# Spin coherent states

I'm trying to prove that $$\langle {S_{z}} \rangle =S \cos \theta$$ using the notion of spin coherent states, so $$|\theta,\phi \rangle = e ^{ -i \phi S_{z} } e ^{ -i \theta S_{y} } |S \rangle$$, and i know that $$\langle S|S_{y}|S \rangle$$ = $$\langle S|S_{x}|S \rangle$$= 0 and $$\langle S|S_{z}|S \rangle=0$$.

But I'm stuck in the very beginning

$$\langle \theta,\phi| e ^{ i \phi S_{y} } e ^{ i \theta S_{z} } \theta S_{z} e ^{ -i \phi S_{z} } e ^{ -i \theta S_{y} } | \theta,\phi \rangle = \langle S| e ^{ i \phi S_{y} } e ^{ i \theta S_{z} } \theta S_{z} e ^{ -i \phi S_{z} } e ^{ -i \theta S_{y} } |S \rangle$$

I have the BCH formula $$e ^{ -X } Ye ^{ X } = Y + [X,Y]....$$ but i don't see how to use it with this $$i \theta$$

1. You forgot to interchange the order of rotations in your bra vector, i.e. if $$\vert\theta,\phi\rangle=R_z(\phi)R_y(\theta)\vert SS\rangle$$ then $$\langle \theta,\phi\vert=\langle SS\vert R_y^{-1}(\theta) R_z^{-1}(\phi)$$.
2. You have a typo and there is no $$\theta S_z$$, just $$S_z$$, i.e you need to compute $$\langle SS\vert R_y^{-1}(\theta)R_z^{-1}(\phi)S_zR_z(\phi)R_y(\theta)\vert SS\rangle.$$