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

  • #1
10
0
Say I have [tex]{S_{x}=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}
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?
 

Answers and Replies

  • #2
263
86
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##.
 
  • #5
blue_leaf77
Science Advisor
Homework Helper
2,629
784
First off, the matrix ##S_x## you have there is written in basis formed by the eigenvectors of ##S_z##, it has nothing to do with the Cartesian basis vectors ##\mathbf i##, ##\mathbf j##, and ##\mathbf k##. Secondly, I don't feel familiar with matrix ##R## as you wrote above. If it's a rotation matrix then it should contain three independent parameters, but your ##R## matrix only has two.

As for your main question, I got the impression that you confused the way vector operator ##\mathbf S = S_x\mathbf i+ S_y\mathbf j+S_z\mathbf k## is expanded with the matrix element of each of the spin components. Although it's true that ##\mathbf S## transforms like an ordinary vector under rotation, but it doesn't connect to the matrix elements. There is the so-called Spherical Tensor, though, which transforms like spherical harmonics under rotation. Especially for a vector operator like ##\mathbf S##, you can simply rearrange the basis vectors ##\mathbf i##, ##\mathbf j##, and ##\mathbf k## such that ##\mathbf S## is expressed in rank-1 spherical tensors ##S_{+1}##, ##S_{0}##, and ##S_{-1}##.
 

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