Combining paths in a Stern-Gerlach experiment

Question

I am trying to understand what exactly happens when paths are combined in Stern-Gerlach experiments. For the following setup:

My textbook argues that the state that is input into the 3rd analyser is simply $|{+z}\rangle$ since combining the channels is essentially equivalent to not making a measurement on the input $|+z\rangle$ state at all.

However, I am unable to apply the same reasoning as before in the following case:

Clearly, a measurement is being made at the second analyser; yet it seems that state that is input into the 3rd analyser is still not one of the eigenstates of the $X$ operator but a linear combination of the first two.

Would it be accurate to say that combining two out of the 3 channels constitutes a "partial" measurement of sorts, since we are not able to measure the first two eigenvalues, only the 3rd. If so, how does one explain this via the postulate:

If the result of measuring a physical quantity $Q$ on a system in state $\ket{\psi}$ is $\omega _{n}$, then the state of the system immediately after measurement is the normalised projection of $\ket{\psi }$ onto the eigenspace associated with eigenvalue $\omega_{n}$ of the corresponding observable $\Omega$

Does it need to be modified?

Answer 1

The key point in your first example is that we do not perform a measurement on the $X$-direction: Although the particles are split into two $X$ components, these components are never measured. Hence, they are not projected -- they keep their relative phase relation. This can be describes as "each particle took both paths". Therefore, measuring the $Z$-component again, we obtain the original state $|Z,+\rangle$.

In contrast, in your second example the $|X,-\rangle$ component is removed. Thus, it is no longer true that each particle took all paths. However, in the $X$-measurement the particles, which proceed to the $Z$ measurement, are not projected onto the two vectors $|X,+\rangle$ and $|X,0\rangle$, separately. Instead, they are projected onto the two dimensional plane generated by these two vectors. Thus, a part of the relative phase is kept. Just think of it in terms of a matrix $$ \begin{pmatrix} ii & ij & ik \\ ji & jj & jk \\ ki & kj & kk \end{pmatrix} \to \begin{pmatrix} ii & 0 & 0 \\ 0 & jj & jk \\ 0 & kj & kk \end{pmatrix} $$ where the non-diagonal elements describe the phase relationship (coherence) between different components.

Answer 2

You do not have to change the postulates (at least not for this reason), but you need to change the position of the Heisenberg cut, i.e., the place where you put in the measurement. For all practical purpose it does not matter whether you treat the detection screen as the observer, the computer storing the signals from the detection screen, or your self observing the computer. The only important thing is that you do not apply the projection postulate too early. In the many worlds jargon, "too early means" that the different branches $|\Psi_1\rangle$, $|\Psi_2\rangle$ of the wave function $|\Psi_1\rangle + |\Psi_2\rangle$ still interfere. Therefore, applying the projection postulate yields wrong predictions.

In your case, treating the second layer as observer is too early. So the assumption.

Clearly, a measurement is being made at the second analyser

is wrong. E.g., assume you measure the spin by some magnetic field, then the gradient of the magnetic field will spacially separate the spin up part of the wave function from the spin down part. But you could still build an interferometer that makes them interfere again. Another example are so called quantum eraser experiments.

It is not always clear how far away the Heisenberg cut has to be moved, and where the "classical world" begins. Therefore, John Bell used to call it "the shifty split". A nice treatment by him can be found here (unfortunately there is no free access version of the article.)