Or must the spin state being measured be in the same direction as $\hat S_x $?

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    $\begingroup$ What you're asking doesn't really make sense. $\hat{S}_x|\uparrow_z\rangle$ is not an operator, it's a vector, so you can't compute its eigenvalue, because it doesn't have one, because it's not an operator. You can compute the eigenvalues and eigenvectors of $\hat{S}_x$, though. They are $\pm\hbar/2$ and $|\uparrow_z\rangle \pm |\downarrow_z\rangle$. Can you clarify your question? Also, "must the spin state being measured be in the same direction as the operator" doesn't make sense either... $\endgroup$
    – march
    Mar 15 at 16:03
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    $\begingroup$ ...because the quantity measured corresponds to an operator, so we choose the operator based on what measurement we're making. Can you clarify? $\endgroup$
    – march
    Mar 15 at 16:03
  • $\begingroup$ My apologies, I've corrected the terminology now. $\endgroup$
    – Tarius
    Mar 15 at 16:10
  • $\begingroup$ The question is still incoherent. What do you mean by “the vector $S_x$?” $S_x$ is not a vector. $\endgroup$
    – Sandejo
    Mar 15 at 16:19
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    $\begingroup$ $\hat{S}_x$ is an operator, $|\uparrow_z\rangle$ is a state(-vector) and so is $\hat{S}_x|\uparrow_z\rangle$. It only makes sense to talk about eigenvalues and eigenvectors of an operator, i.e. $\hat{S}_x$. Also: $\hat{S}_x$ does not indicate a direction, since it is an operator, and not a vector in the 3-dimensional coordinate space. $\endgroup$
    – Samuel
    Mar 15 at 16:31

1 Answer 1


Is it possible to calculate the eigenvalue for the spin vector $ S_x |↑_z⟩ $?

If you are asking whether $S_x|\uparrow_z\rangle$ is an eigenvector of some operator, the answer is yes. $$ S_x|\uparrow_z\rangle = \frac{1}{2}|\downarrow_z\rangle\;, $$ which is an eigenvector of $S_z$ with eigenvalue $-1$.

If you are asking whether or not it is possible to find the eigenvectors of $S_x$, the answer is yes.

In terms of eigenvectors of $S_z$ you can write the eigenvectors of $S_x$ as: $$ |+\rangle =\frac{1}{\sqrt{2}}\left(|\uparrow_z\rangle + |\downarrow_z\rangle\right) $$ and $$ |-\rangle =\frac{1}{\sqrt{2}}\left(|\uparrow_z\rangle - |\downarrow_z\rangle\right) $$

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    $\begingroup$ I think it should be $S_x|\uparrow\rangle = \frac{1}{2}(S_++S_-)|\uparrow\rangle=|\downarrow\rangle/2$. $\endgroup$
    – Samuel
    Mar 15 at 16:54
  • $\begingroup$ Yes. You are right. $S_x = \sigma_x/2$. I will update. $\endgroup$
    – hft
    Mar 15 at 17:03

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