Correlation coefficient of two-particle half-spin entangled state

[Skiggles]

Hi, I can't get my head around this question.

Homework Statement

The Bell state:
$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|\uparrow_1\rangle|\downarrow_2\rangle +|\downarrow_1\rangle |\uparrow_2\rangle \right)$$

Find the correlation coefficient of the measurement of the spins in the directions $$z,\phi$$.

Homework Equations

$$C(\phi) = \langle S_{z1} S_{\phi 2} \rangle$$

S takes the values $$\pm 1$$

The Attempt at a Solution

I know that I need to calculate the probability of the states being parallel and subtract the probability that the states are anti-parallel. But how do you calculate those?

$$Pr(\uparrow_{z1}\uparrow_{\phi 2}) = \left|\langle\uparrow_{z1}\uparrow_{\phi 2} \left |\frac{1}{\sqrt{2}}\left(|\uparrow_1\rangle|\downarrow_2\rangle +|\downarrow_1\rangle |\uparrow_2\rangle \right)\right| \uparrow_{z1}\uparrow_{\phi 2} \rangle \right|^2$$

Where do I go from here?

Thanks

 

[Skiggles]

I think I might be on to something with this, but am not quite there yet:

$$\hat{S}_z =\frac{\hbar}{2} \begin{pmatrix}1 & 0 \\ 0 & -1 \\ \end{pmatrix}$$

Rotate by angles $$\theta, \phi$$ to get

$$\hat{S}_\phi = \frac{\hbar}{2} \begin{pmatrix} \cos \theta & \sin \theta e^{-i\phi} \\<br /> \sin \theta e^{i\phi} & -\cos \theta \end{pmatrix}$$

Can I use this somehow to get the correlation coefficient?

Unfortunately the notation is getting a bit mixed up because in this post $$\phi = 0$$ and in the post above $$\phi$$ is used in place of $$\theta$$.