The expectation value of the spin of a proton which is entangled

Question

Suppose we have the Bell state $$|\text{ }\psi\rangle=\frac{1}{\sqrt{2}}\left(|\uparrow\rangle_1|\downarrow\rangle_2+|\downarrow\rangle_1|\uparrow\rangle_2\right)$$ and denote the spin operator of the first and second particle by $s_1$ and $s_2$ respectively. I want to calculate the expectation values $\langle s_i\rangle$. If the particles would not be entangled, it would be really easy to find it, because then $\langle s_1\rangle=|\phi\rangle_1^T S_z|\phi\rangle_1$ where $$S_z=\frac{\hbar}{2}\left(\begin{array}{cc} 1&0\\ 0&-1 \end{array}\right)$$ The problem is however that the states are entangled, so now I have no idea if this approach also works. Could you point me in the right direction? Thanks.

Answer 1

A quantum state is a quantum state is a quantum state. Regardless of what kind of state $\lvert \psi \rangle$ is, the expectation value of any operator $\mathcal{O}$ is $\langle \psi \rvert \mathcal{O} \lvert \psi \rangle$.

Now, in your example, you must represent the operators $s_i$ from the single particle Hilbert spaces $\mathcal{H}_i$ properly on the two-particle space $\mathcal{H}_1 \otimes \mathcal{H}_2$, which is by letting them act as unity on the respective other particle, i.e. $s_1 \otimes \boldsymbol{1}_2$ and $\boldsymbol{1}_1 \otimes s_2$.