Measuring spin with Stern-Gerlach apparatus

Question

I have a beam of electrons prepared in the state $|\Psi\rangle = cos(\theta/2)|+\rangle_z + sin(\theta/2)|-\rangle_z$ passing through a Stern-Gerlach apparatus which is free to rotate along the y-axis, and set with an angle $\alpha$ so as to measure the spin projection along the axis $cos(\alpha)u_z + sin(\alpha)u_x$

I'm supposed to find out wether or not I can determine $\theta$ in three different situations:

• Sending all electrons through the SG apparatus with $\alpha = 0$
• Sending half of the electrons with $\alpha = 0$ and the other half with $\alpha = \pi$
• Sending half of the electrons with $\alpha = 0$ and the other half with $\alpha = \pi/2$

So, how can I do that ?

EDIT:

Using the matrix for the spin projection in an arbitrary direction $$S_u = \left(\begin{array}{ccc} cos(\alpha) & sin(\alpha)e^{-i\phi} \\ sin(\alpha)e^{i\phi} & -cos(\alpha) \\ \end{array}\right)$$

(in the basis $\left\{|+\rangle_z, |-\rangle_z\right\}$, with $\phi = 0$)

I find that

$$S_u |\psi\rangle = \left(\begin{array}{ccc} cos(\alpha - \theta/2) \\ sin(\alpha - \theta/2) \\ \end{array}\right) = cos(\alpha -\theta/2)|+\rangle_z + sin(\alpha -\theta/2)|-\rangle_z$$

So $P_r(\hbar/2) = cos^2(\alpha - \theta/2)$ and $P_r(-\hbar/2) = sin^2(\alpha - \theta/2)$

$\theta = 2\left( \alpha - arcos(\sqrt{P_r(\hbar/2)})\right) = 2\left( \alpha - arcsin(\sqrt{P_r(-\hbar/2)})\right)$

But I don't think it's correct because, if it were the case, I could determine $\theta$ regardless of $\alpha$'s value...

Answer 1

Your spin projection matrix $S_u$ actually looks like a rotation, not a projection: A projection has only degenerate $+1$ eigenvalues, but your $S_u$ has eigenvalues $\pm 1$. Also, I'm not sure the problem is meant to require a general solution for arbitrary $\alpha$.

In any case, it's easier, even though less elegant, to go case by case:

• For $\alpha = 0$ the apparatus measures along $u_z$ and gives $$ \tan^2\frac{\theta}{2} = \frac{|\langle -|\Psi\rangle|^2}{|\langle +|\Psi\rangle|^2} $$ This leaves two solutions for $\tan \frac{\theta}{2}$ (or multiple pairs $\cos \frac{\theta}{2}$, $\sin \frac{\theta}{2}$) and no information to select among them.

• If half the electrons are sent through $\alpha =0$ and half through $\alpha = \pi$, the projections are along $u_z$ and $-u_z$, and no additional information is gained.

• Sending half the electrons through $\alpha = 0$ and half through $\alpha = \frac{\pi}{2}$ retrieves the info above from the projection along $u_z$ and adds a projection along $u_x$. For the latter the probabilities read $$ P(\pm u_x) = \Big|\frac{1}{\sqrt{2}}\langle +|\Psi\rangle \pm \frac{1}{\sqrt{2}}\langle -|\Psi\rangle\Big|^2 = \frac{1}{2}\left(\cos \frac{\theta}{2}\pm \sin \frac{\theta}{2} \right)^2 = \frac{1}{2} \pm \sin \theta $$ and resolve the correct value of $\theta$ from the ones compatible with the $u_z$ results.