Intensity of sequential spin measurments in QM

The following is a question from Sakurai's book Modern Quantum Mechanics, 2nd edition, page 60:

A beam of spin $\frac12$ atoms goes through a series of Stern-Gerlach-type measurements as follows:

1. The first measurement accepts $s_z=\hbar/2$ atoms and rejects $s_z=-\hbar/2$ atoms.
2. The second measurement accepts $s_n=\hbar/2$ atoms and rejects $s_n=-\hbar/2$ atoms, where $s_n$ is the eigenvalue of the operator $\mathbf S\cdot\hat{\mathbf n}$, with $\hat{\mathbf n}$ making an angle $\beta$ in the $xz$-plane with respect to the $z$-axis.
3. The third measurement accepts $-s_z=\hbar/2$ atoms and rejects $s_z=\hbar/2$ atoms.

What is the intensity of the final $s_z=-\hbar/2$ beam when the $s_z=\hbar/2$ beam surviving the first measurement is normalized to unity? How must we orient the second measuring apparatus if we are to maximize the intensity of the final $s_z=-\hbar/2$ beam?

Proposed Solution: I assume the intensity is defined as the probability amplitude squared of getting $s_z = - \frac{\hbar}{2}$ in the final measurement. So you start with the state $| \psi \rangle = \frac{1}{\sqrt{2}} | + \rangle + \frac{1}{\sqrt{2}} | - \rangle$ before any measurements are made, you then project onto the eigenstates which correspond to the measurements you accept, hence you first project onto $| + \rangle \langle + |$ and then $| +; \hat{n} \rangle \langle +; \hat{n}|$ (taking $\alpha = 0$) and finally $|-\rangle \langle - |$, this process hence yields the intensity given by $$\bigg(\frac{1}{\sqrt{2}} \langle + | + \frac{1}{\sqrt{2}} \langle - |\bigg)\bigg(|- \rangle \langle - |+; \hat{n} \rangle \langle +; \hat{n} | + \rangle \langle + |\bigg) \bigg(\frac{1}{\sqrt{2}} | + \rangle + \frac{1}{\sqrt{2}} | - \rangle\bigg).$$

Apparently this is not correct, does anyone know where I went wrong?

• Why are you looking for $\langle\psi|(|+\rangle\langle-|+;\hat n\rangle\langle +;\hat n|+\rangle\langle +|)|\psi\rangle$? Your beam is normalised after the first projection. – dalum Mar 29 '17 at 13:14
• @Dalum Could you elaborate on what you mean? What should I be looking for? – user100411 Mar 29 '17 at 13:16
• As I see it, the problem says you should in fact ignore the first projection, and just assume that your state starts in $|+\rangle$, properly normalised. – dalum Mar 29 '17 at 13:17
• @Dalum What is the basic protocol for finding the probability density of this type of sequential measurement. I only know the case where we want to find the probability of say some measurement eigenvalue $a$ corresponding to some operator $\hat{A}$, we then project onto the eigenstate to get $\langle \psi|a\rangle \langle a | \psi \rangle$ which is the probability amplitude. But in this case we start with state $| + \rangle$ and we end with $| - \rangle$. – user100411 Mar 29 '17 at 13:49
• @Dalum What then is the protocol for the sequential measurements starting and ending with different states, and does it follow from the postulates of QM as in the simple case I outlined above? – user100411 Mar 29 '17 at 13:51

Using the notation you posted in the comments, $s_z |\pm\rangle = \pm\frac \hbar2 |\pm\rangle$ and $s_n |\pm,n\rangle = \pm \frac\hbar2|\pm,n\rangle$, then you have correctly summarized the effect of the experiment as the right-to-left application of the projectors $$|- \rangle \langle - | \cdot |+; \hat{n} \rangle \langle +; \hat{n} | \cdot | + \rangle \langle + |,$$ where you want to maximize the norm of the wavefunction after it's been acted on by that operator.

On the other hand, I've got no idea you got the superposition $| \psi \rangle = \frac{1}{\sqrt{2}} | + \rangle + \frac{1}{\sqrt{2}} | - \rangle$ from, but it's wrong; the text does not make any reference to the state of the system before it reaches the first magnet. If your thinking there was "take a state that's equally likely to be in $|+\rangle$ as it is in $|-\rangle$, then that's not how you describe it, because the state you wrote down is coherent and it doesn't appear up by magic. Why not write down, say, $| \psi \rangle = \frac{1}{\sqrt{2}} |+\rangle - \frac{1}{\sqrt{2}} |-\rangle$ instead?

Thus, you don't know what the initial state is, but you do know that you want the norm after the first projector comes up to be one, or in other words, what you really want to maximize is $$p=\frac{\bigg\||- \rangle \langle - |\cdot|+; \hat{n} \rangle \langle +; \hat{n} |\cdot| + \rangle \langle + | \cdot |\psi\rangle\bigg\|^2}{\bigg\|| + \rangle \langle + | \cdot |\psi\rangle\bigg\|^2}.$$ Now, this seems like it depends on what the initial state is, but it doesn't really (since the experiment doesn't, either) unless you happened to start with on $|\psi\rangle = |-\rangle$, in which case the experiment doesn't make sense. As such, you're free to take whatever arbitrary state you want, for convenience, so you can just start with $|\psi\rangle = |-+\rangle$ and be done with $|\psi\rangle$, in which case the calculation reduces to maximizing $$p=\bigg\||- \rangle \langle - |+; \hat{n} \rangle \langle +; \hat{n} | + \rangle \bigg\|^2 =\bigg| \langle - |+; \hat{n} \rangle \langle +; \hat{n} | + \rangle \bigg|^2 .$$

"The s_z=|+> beam surviving the first measurement is normalized to unity". "What is the intensity of the final s_z=|-> beam"? You seem to start and end with superpositions from which I don't see where you get them from. But those won't matter anyway, since there is only one Stern-Gerlach device that really does something. The first one creates the particles and the last one annihilates them. What remains is an amplitudo

$\langle-|+;n\rangle\langle+;n|+\rangle$.

As for the intensity, you should take the squared norm of this amplitudo, as usual in QM

• Thanks for your answer. Do you mind giving a basic explanation of what we are doing? So after the first measurement we are in the state $| \psi \rangle = | + \rangle$. And after the last we are in the state $\langle - |$. So are you saying that for this reason in order to find the probability amplitude we consider $$\langle - |(|-\rangle\langle-|+;n\rangle\langle+;n|+\rangle \langle+|)|+ \rangle?$$ – user100411 Mar 29 '17 at 13:45
• @John that's what I meant indeed – Wouter Mar 29 '17 at 14:36