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

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

• 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... Mar 15 at 16:03
• ...because the quantity measured corresponds to an operator, so we choose the operator based on what measurement we're making. Can you clarify? Mar 15 at 16:03
• My apologies, I've corrected the terminology now. Mar 15 at 16:10
• The question is still incoherent. What do you mean by “the vector $S_x$?” $S_x$ is not a vector. Mar 15 at 16:19
• $\hat{S}_x$ 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. Mar 15 at 16:31

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 $$|-\rangle =\frac{1}{\sqrt{2}}\left(|\uparrow_z\rangle - |\downarrow_z\rangle\right)$$

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