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!

  • $\begingroup$ Have you tried expressing $S_{n2}$ in the $S_z$ basis? You can easily figure out the representation geometrically, and then it's straightforward to operate on the state in the usual manner. $\endgroup$ – Kyle Arean-Raines Nov 23 '15 at 6:16
  • $\begingroup$ So I got to thinking that $S_{n2}=\frac{\hbar}{2}\cos \theta \begin{pmatrix}1& 0 \\0&-1\end{pmatrix}$ making the argument that the z-component of particle 2 is projected onto $\hat{n}$. It gave the right answer for the expectation value $\langle S_{z1}S_{\hat{n}2}\rangle$ but I am thinking it has to be wrong. That a measurement along $\hat{n}$ must also return the same eigenvalues as $S_{z1}$, namely $\pm\frac{\hbar}{2}$ thus $S_{z1}=S_{n2}$. Is this correct? $\endgroup$ – Wikkyd Nov 23 '15 at 6:29
  • $\begingroup$ You're correct in your intuition that $S_n$ and $S_z$ do not generally share the same eigenstates. Therefore, $S_n$ is not diagonal in the $S_z$ basis. You should be able to figure out how $S_n$ acts on $\mid + \rangle$ and $\mid - \rangle$ using basic geometry. What would you expect if $\theta=\frac{\pi}{2}$? Is this not equivalent to another spin operator? $\endgroup$ – Kyle Arean-Raines Nov 23 '15 at 6:39

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