$S_x$ and $S_y$ states for spin 1

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?

• Just write the matrix expression for $S_y$ or $S_x$ and diagonalize…. you will get the eigenstates as combo of the $S_z$ eigenstates. Commented Nov 11, 2021 at 23:00
• I answered this for spin-3/2 here: physics.stackexchange.com/q/607218
– TEH
Commented Nov 11, 2021 at 23:01

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).

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

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

• and then the eigenstates of Sx and Sy in the Sz basis are just the eigenvectors of those matrices?
– jboy
Commented Nov 11, 2021 at 23:29
• @jboy Yes, correct. Commented Nov 11, 2021 at 23:30
• thanks! just as clarification, the result I would get would be different for Sx in the Sx diagonal basis right? (which is also something I still cant figure out)
– jboy
Commented Nov 11, 2021 at 23:33
• @jboy Yes, all 3 matrices would be different. $S_x$ would be diagonal., and $S_y$ and $S_z$ would be non-diagonal. Commented Nov 11, 2021 at 23:36
• How would I find $S_x$ in the $S_x$ diagonal basis? Would it be as if it were the $S_z$ that is the eigenstates are 1, 0, -1?
– jboy
Commented Nov 12, 2021 at 0:32