# Transforming Spin Matrices (Sx, Sy, Sz) to a Spherical Basis

• A
Say I have $${S_{x}=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\\ \end{array}\right)}$$
Right now, this spin operator is in the Cartesian basis. I want to transform it into the spherical basis. Since, $${\vec{S}}$$ acts like a vector I think that I only need to transform $$S_{x}$$ like a vector:
$${ R=\left(\begin{array}{ccc} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos\theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\ -\sin \phi & \cos \phi & 0 \\ \end{array}\right)}$$
So my transformation looks like:
$$S_{spherical}=RS_{x}$$
I thought this would be fine, so I performed a cross check on the angles I obtained. My cross-check was determined from the fact that I think the spin operator "points" solely in the x-direction so after performing my rotation in order to make sure I'm right when $${\theta=\frac{\pi}{2} \hspace{5mm} \phi=0}$$ I should return with the original Sx in the Cartesian basis...This does not happen though. Is there something wrong with my transformation, or my reasoning behind the cross-check, or both?

The matrix of the component of the spin in the direction ##\hat{n} = (\sin \theta \cos \phi , \sin \theta \sin \phi, \cos \theta)## is ##\hat{n} \cdot \vec{S}## where $$\vec{S} = \left( \begin{array}{c} S_x \\ S_y \\ S_z \end{array} \right).$$ ##R## should act on the 3-vector, not just one component ##S_x##.