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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Pricklebush Tickletush Dec 24 '12 at 17:32

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 $\hat{S}_x=\frac{\hbar}{2}\sigma_x$ .

  • $\begingroup$ Thanks, it took me awhile to realize everything is based on the z direction. $\endgroup$ – Luka Milic Dec 26 '12 at 19:37

$$ \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ 1 \end{pmatrix} $$ 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 +.

  • $\begingroup$ But I thought the way in which you arrive at the "vector" is by computing the probabilities for each spin state? $\endgroup$ – Luka Milic Dec 26 '12 at 13:19

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