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The figure above is taking from the Feynman lectures.Here $S$ ,$T$,$P_1$ and $U$ are Stern-Gerlach apparatuses.

$S$ and $T$ measures the spin in the $z$ direction where $z$ is pointing out of the page. The eigenstate associated with these operator are $z$ and $-z$.

$P_1$ is an apparatus that produces a state $+x$ and $U$ measures the spin in the $x$ direction so, its eigenstate is $+x$ and $-x$

As it is shown in the figure, Feynman is claiming that $U$ will measure always the state $+x$. I am not seeing why the state leaving $T$ is $+x$, since $T$ measures the spin in the $z$ direction ,so, the state leaving $z$ should be $-z$ or $+z$.

What I am misunderstanding here?

amilton moreira
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  • I haven't clicked on the link to read the lecture so this isn't a definitive answer, but just based on the figure it looks like there is a black dot right before $U$, and from what you wrote I understand the black dot represents a device to prepare a $+x$ state. – Andrew Apr 24 '21 at 13:19
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    @ Andrew Feynman says that the $T$ apparatus puts out just what it takes in – amilton moreira Apr 24 '21 at 13:23
  • Sure but there is a black dot between the $T$ and $U$ boxes. – Andrew Apr 24 '21 at 13:24
  • Feynman is considering some idealised "improved" Stern-Gerlach experiment which is able to use a magnetic field to spatially separate out a linear combination of spin states then recombine them back to the original combination. No measurement has actually taken place, no decoherence, so no wavefunction collapse. I've always wondered whether such a device can actually be built. – isometry Apr 25 '21 at 05:36
  • @duality could you expand your comment in an answer? – amilton moreira Apr 25 '21 at 05:44
  • I have written up what I was thinking. I need to check that is what Feynman was thinking. – isometry Apr 25 '21 at 07:44

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A standard Stern-Gerlach experiment, as described in most QM textbooks, functions as follows:

Consider a particle system described by a Gaussian wavepacket moving in the x direction, which is in a superposition of spin.

\begin{equation} \lvert \Psi\rangle = \psi(x) \otimes (\lvert + \rangle + \lvert - \rangle ) \end{equation}

The system interacts with an external magnetic field aligned to deflect the particle either to the left or the right depending on the value of spin. This results in an entanglement between the spin part and the spacial part of the particles Hilbert space:

\begin{equation} \lvert \Psi\rangle = \psi_{left}(x) \otimes \lvert + \rangle + \psi_{right} (x) \otimes \lvert - \rangle \end{equation}

Detectors (LEFT and RIGHT) are placed to detect the presence of a particle in the left or right beam. So schematically the complete experiment can be described as

\begin{equation} \lvert \Psi\rangle \otimes \lvert {\rm LEFT~OFF} \rangle \otimes \lvert {\rm RIGHT~OFF} \rangle \to \lvert + \rangle \otimes \lvert{\rm LEFT~ON} \rangle + \lvert - \rangle \otimes \lvert {\rm RIGHT ~ON} \rangle \end{equation}

At this point you apply your favorite interpretation of quantum mechanics to justify that only one part of the sum on the rhs actually exists.

Consider instead a magnetic field tuned so that the left and right spacial wavefunctions end up back in a common central location before reaching a central detector. Then the entangled state could in principle be converted back into a disentangled state

$$\lvert \Psi\rangle = \psi_{center}(x) \otimes (\lvert + \rangle + \lvert - \rangle ) $$

So as far as future measurements are concerned, the magnetic field has not altered the spin state in any way. No measurement projection has occurred.

For analysis of the decoherence aspects of this setup please check this answer.

isometry
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