Expectation value of two spin 1/2 particles where particle 1 along $z$ axis and particle 2 along another axis? We are supposed to show that for a two spin 1/2 particles, the expectation value of $\langle S_{z1} S_{n2} \rangle$ is $-\frac{\hbar^2}{4}\cos \theta$ when the system is prepared to be in the singlet state $$|00\rangle=\frac{1}{\sqrt{2}}\left(|\uparrow\rangle|\downarrow\rangle-|\downarrow\rangle|\uparrow\rangle\right).$$
It is given that the matrix $S_{z1}$ returns particle 1's z component. Where
$$S_{z1}=\frac{\hbar}{2}\begin{pmatrix}1& 0 \\0&-1\end{pmatrix}
$$
Now for particle 2 the matrix $S_{n2}$ gives the component of spin angular momentum along an axis denoted by the unit vector $\hat{n}$. where $\theta$ is the angle between the z-axis and $\hat{n}$.
My guess is there must be some relation between $S_n$ and $S_z$ to get the required relation? Any hints would be greatly appreciated!
 A: I'm thinking that you know
$\hat n = \cos\theta\ \hat z + \sin \theta\ \hat x\,$,
perhaps with a suitable choice of $\hat x$ orientation, and so from that you infer $S_n = \cos\theta\ S_z + \sin\theta\ S_x\,$. (Perhaps it's not obvious that this step should be allowed?)
Then you want to find
$⟨(S_z\otimes I)(I\otimes S_n)⟩=
⟨\cos\theta\ \ S_z\otimes S_z\ +\ \sin\theta\ \ S_z\otimes S_x⟩$.
Since your state is expressed in terms of $S_z$ eigenstates, it would be preferable to write $S_x$ in the same basis. So you're finding
$\cos\theta\ ⟨S_z\otimes S_z⟩ + \frac\hbar2\sin\theta\ ⟨S_z\otimes 
(|{↑}⟩⟨{↓}|+|{↓}⟩⟨{↑}|)\ ⟩$.
You can also simplify it further noting that
$⟨00|S_z\otimes A|00⟩=\frac\hbar4(⟨{↓}|A|{↓}⟩-⟨{↑}|A|{↑}⟩)$.
A: 
My guess is there must be some relation between $S_n$ and $S_z$ to get the required relation?

Spot on! You are "measuring" particle 2 states antialigned to those of particle 1 in the z-axis basis! So you have to express $S_{n2}$ in that basis through the Wigner d rotation matrices for 2-spinors, namely  $d^{1/2}_{1/2~1/2}=cos(\theta/2)$ and $d^{1/2}_{1/2~-1/2}=-\sin(\theta/2)$; in this basis,  $S_{n2}$ is off-diagonal,
$$
S_{n}= {\hbar\over 2} \begin{pmatrix}\cos {\theta\over 2}& -\sin {\theta\over 2} \\ \sin{\theta\over 2}&\cos {\theta\over 2}\end{pmatrix} \begin{pmatrix}1& 0 \\0&-1\end{pmatrix}\begin{pmatrix}\cos {\theta\over 2}& \sin {\theta\over 2} \\-\sin {\theta\over 2} &\cos {\theta\over 2}\end{pmatrix}\\
 = {\hbar\over 2} \begin{pmatrix}\cos \theta& \sin \theta \\ \sin\theta&-\cos \theta\end{pmatrix},
$$
traceless hermitian, alright, with the proper vanishing angle limit!
Now apply this matrix to the second particle,
$$
|\!\downarrow\rangle \mapsto {\hbar\over 2} (- \cos  \theta ~|\downarrow\rangle + \sin  \theta ~|\uparrow\rangle ) \\
|\! \uparrow \rangle \mapsto {\hbar\over 2} ( \sin \theta ~|\downarrow\rangle + \cos \theta ~|\uparrow\rangle ),
$$
with the correct vanishing angle limit.
Apply to your operator tensor product, and you are done:  $\langle S_{z1} S_{n2} \rangle =-\frac{\hbar^2}{4}\cos \theta$, with the correct vanishing angle limit.
