$S_x$ and $S_y$ states for spin 1

Question

I asked a question earlier but it looks like I misunderstood something Convert eigenvectors to different basis. I'm considering the case of a spin 1 object, where the eigenvalues of $S_z$ are 1,0,-1 so the $S_z$ diagonal basis is just {|1⟩,|0⟩,|−1⟩} and from this we can just write the $S_z$ operator as

$$S_z = \hbar \begin{bmatrix}1&0&0\\0&0&0\\0&0&-1\end{bmatrix}.$$

In most text, the discussion for the expression of the $S_x$ and $S_y$ states in terms of the $S_z$ basis is discussed only for a spin-$1/2$ system. How do I do this for spin-1?

Answer 1

Like all angular momentum operators the spin-$1$ operators ($S_x,S_y,S_z$) need to satisfy the commutator relations:

$$\begin{align} [S_x,S_y]&=i\hbar S_z \\ [S_y,S_z]&=i\hbar S_x \\ [S_z,S_x]&=i\hbar S_y \end{align}$$

Given the matrix for $S_z$ you can find matrices for $S_x$ and $S_y$, so that all these commutator relations are satisfied (see for example Spin operators and matrices).

$$S_x = \frac{\hbar}{\sqrt{2}} \begin{bmatrix}0&1&0\\1&0&1\\0&1&0\end{bmatrix}$$

$$S_y = \frac{\hbar}{\sqrt{2}} \begin{bmatrix}0&-i&0\\i&0&-i\\0&i&0\end{bmatrix}$$