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For a spin 1 object, the eigenvalues of $S_z$ are 1,0,-1 so the $S_z$ diagonal basis is just $\{|1\rangle, |0\rangle, |-1 \rangle\}$ and the $S_z$ and $S_y$ operator are respectively

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

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

My question is, if I diagonalize the matrix $S_y$, does this then give me the eigenkets in the $S_y$ diagonal basis?

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  • $\begingroup$ No. It will give you the eigenkets if $S_y$ as combos of eigenkets of $S_z$. $\endgroup$ Nov 11, 2021 at 23:04

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No, if you diagonalize $S_y$ as written in the $S_z$-eigenvector-basis, you get the linear combinations of your $S_z$ basis vectors that are eigenvectors of $S_y$. But they will still be expressed in the $S_z$ basis.

The eigenvectors of $S_y$ in the $S_y$ basis should follow from exactly the same reasoning you opened your question with.

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  • $\begingroup$ Oh okay, I must have misunderstood. So how do I write the eigenvectors of Sy using the Sz basis? $\endgroup$
    – jboy
    Nov 11, 2021 at 21:22

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