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?

  • $\begingroup$ 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. $\endgroup$ Nov 11, 2021 at 23:00
  • $\begingroup$ I answered this for spin-3/2 here: physics.stackexchange.com/q/607218 $\endgroup$
    – TEH
    Nov 11, 2021 at 23:01

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).

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

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

  • $\begingroup$ and then the eigenstates of Sx and Sy in the Sz basis are just the eigenvectors of those matrices? $\endgroup$
    – jboy
    Nov 11, 2021 at 23:29
  • $\begingroup$ @jboy Yes, correct. $\endgroup$ Nov 11, 2021 at 23:30
  • $\begingroup$ 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) $\endgroup$
    – jboy
    Nov 11, 2021 at 23:33
  • $\begingroup$ @jboy Yes, all 3 matrices would be different. $S_x$ would be diagonal., and $S_y$ and $S_z$ would be non-diagonal. $\endgroup$ Nov 11, 2021 at 23:36
  • $\begingroup$ 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? $\endgroup$
    – jboy
    Nov 12, 2021 at 0:32

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