How do I write the matrix for the operator $\hat{S_y}$ in the basis $\{|{↑}_x⟩ , |{↓}_x⟩\}$? I know how to write the matrix for the operator $S_y$ in the $\{|{↑}_z\rangle , |{↓}_z\rangle\}$ basis, but I don't understand how to write it in the $\{|{↑}_x\rangle , |{↓}_x\rangle\}$ basis.
Wouldn't the calculation become very complex? Any help will be appreciated.
 A: The four matrix elements you are looking for are
$$
\langle {\uparrow}_{x} |\hat{S}_{y}| {\uparrow}_{x} \rangle, \ 
\langle {\uparrow}_{x} |\hat{S}_{y}| {\downarrow}_{x} \rangle, \ 
\langle {\downarrow}_{x} |\hat{S}_{y}| {\uparrow}_{x} \rangle, \ 
\langle {\downarrow}_{x} |\hat{S}_{y}| {\downarrow}_{x} \rangle.
$$
As you probably know,
\begin{align}
| {\uparrow}_{x} \rangle=\frac{1}{\sqrt{2}}(| {\uparrow}_{z} \rangle+| {\downarrow}_{z} \rangle), \\
| {\downarrow}_{x} \rangle=\frac{1}{\sqrt{2}}(| {\uparrow}_{z} \rangle-| {\downarrow}_{z} \rangle)
\end{align}
and
\begin{align}
| {\uparrow}_{y} \rangle=\frac{1}{\sqrt{2}}(| {\uparrow}_{z} \rangle+i| {\downarrow}_{z} \rangle), \\
| {\downarrow}_{y} \rangle=\frac{1}{\sqrt{2}}(| {\uparrow}_{z} \rangle-i| {\downarrow}_{z} \rangle).
\end{align}
Play around with the equations above and knowing that $\hat{S}_{y}| {\uparrow}_{y} \rangle=| {\uparrow}_{y} \rangle$ and $\hat{S}_{y}| {\downarrow}_{y} \rangle=-| {\downarrow}_{y} \rangle$, you should get to your result.
A: You know what $S_y$ looks like in the $z$ basis, and you should know what $\alpha$ and $\beta$ are in
$$
|{\uparrow}_x\rangle =\alpha |{\uparrow}_z\rangle+\beta|{\downarrow}_z\rangle
$$
so just plug into
$$
\langle {\uparrow}_x|S_y| {\uparrow}_x\rangle= [\alpha^*,\beta^*] \left[\matrix{0&-i\cr i&0}\right]\left[\matrix{\alpha\cr \beta}\right].
$$
and similarly for the other three entries.
