# When you apply the spin operator, what exactly is does it tell you?

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The example I'm trying to understand is:

$\hat{S}_{x} \begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{pmatrix} = 1/2 \begin{pmatrix} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} \end{pmatrix}$

My interpretation of this is that the vector shows you the probabilities of a particle being spin up or spin down if you square them.

And I've been told that $\hat{S}_{x}$ gives you the spin as an eigenvalue, but how? Since its 50:50 of getting -1/2 and 1/2. $\hat{S}_{x}$ has only given you one of them.

Is it that $\hat{S}_{x}$ only measures the magnitude of spin in the x direction?

This post imported from StackExchange Physics at 2014-03-22 16:56 (UCT), posted by SE-user 9k9
You operated on an eigenvector of the spin operator. Generally, wave functions will be linear combinations of these eigenvectors, and their coefficients represent how much of each eigenvector is in the total state.

This post imported from StackExchange Physics at 2014-03-22 16:56 (UCT), posted by SE-user santa claus

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$$\frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ 1 \end{pmatrix}$$ is the eigenvector of $\hat{S}_x=\frac{\hbar}{2}\sigma_x$ with eigenvalue $+\frac{\hbar}{2}$. $$\frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ -1 \end{pmatrix}$$ is the eigenvector of $\hat{S}_x=\frac{\hbar}{2}\sigma_x$ with eigenvalue $-\frac{\hbar}{2}$.

So for your vector it isn't 50/50 probability of getting + or -, it's 100% probability of getting +.

This post imported from StackExchange Physics at 2014-03-22 16:56 (UCT), posted by SE-user twistor59
answered Dec 24, 2012 by (2,500 points)
But I thought the way in which you arrive at the "vector" is by computing the probabilities for each spin state?

This post imported from StackExchange Physics at 2014-03-22 16:56 (UCT), posted by SE-user 9k9
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Your equation says that your "vector" is an an eigenvector of your operator, i.e., that the x-projection of the spin is certain an equal to 1/2. As well it says that probabilities to find certain z-projections are equal to 1/2.

This "vector" is not an eigenvector $\begin{pmatrix} 0\\ 1 \end{pmatrix}$ or $\begin{pmatrix} 1\\ 0 \end{pmatrix}$ of the spin z-projection $\hat{S}_z=\frac{\hbar}{2}\sigma_z$ , but is a superposition of them, that is why it is an eigenvector of a non-commuting with $\hat S_z$ operator $\hat{S}_x=\frac{\hbar}{2}\sigma_x$ .

This post imported from StackExchange Physics at 2014-03-22 16:56 (UCT), posted by SE-user Vladimir Kalitvianski
answered Dec 24, 2012 by anonymous
Thanks, it took me awhile to realize everything is based on the z direction.

This post imported from StackExchange Physics at 2014-03-22 16:56 (UCT), posted by SE-user 9k9

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