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

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1. Jun 22, 2016

### chi_rho

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?

2. Jun 22, 2016

### Truecrimson

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 $S_x$.

3. Jun 22, 2016

### Jilang

4. Jun 22, 2016

### Truecrimson

5. Jun 23, 2016

### blue_leaf77

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 $S_{-1}$.