How do I write the matrix for the operator $hat{S_y}$ in the basis ${|{↑}_x⟩ , |{↓}_x⟩}$?

Question

I know how to write the matrix for the operator $S_y$ in the ${|{↑}_zrangle , |{↓}_zrangle}$ basis, but I don't understand how to write it in the ${|{↑}_xrangle , |{↓}_xrangle}$ basis.
Wouldn't the calculation become very complex? Any help will be appreciated.

mike stone · Accepted Answer

You know what $S_y$ looks like in the $z$ basis, and you should know what $alpha$ and $beta$ are in
$$
|{uparrow}_xrangle =alpha |{uparrow}_zrangle+beta|{downarrow}_zrangle
$$
so just plug into
$$
langle {uparrow}_x|S_y| {uparrow}_xrangle= [alpha^*,beta^*] left[matrix{0&-icr i&0}right]left[matrix{alphacr beta}right].
$$
and similarly for the other three entries.

Milarepa · Answer

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.

How do I write the matrix for the operator $hat{S_y}$ in the basis ${|{↑}_x⟩ , |{↓}_x⟩}$?

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