Say I have [tex]{S_{x}=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}(adsbygoogle = window.adsbygoogle || []).push({});

0 & 1 & 0\\

1 & 0 & 1\\

0 & 1 & 0\\

\end{array}\right)}

[/tex]

Right now, this spin operator is in the Cartesian basis. I want to transform it into the spherical basis. Since, [tex]{\vec{S}}[/tex] acts like a vector I think that I only need to transform [tex]S_{x}[/tex] like a vector:

[tex]{

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)}

[/tex]

So my transformation looks like:

[tex]S_{spherical}=RS_{x}[/tex]

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 [tex]{\theta=\frac{\pi}{2} \hspace{5mm} \phi=0}[/tex] 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?

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

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