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Consider an EPR style experiment where two electrons in a singlet state with total angular momentum zero head out in opposite directions. Then they are subjected to the measurements in this order,

  1. First vertical spin of each is measured, which yields one of them spin up and one spin down.
  2. Then horizontal spin is measured for each which makes the vertical spin uncertain again.
  3. Then we measure the vertical spin again.

Is it possible that the two electrons after measurement 3 are found to be both spin up or both spin down? If so, would this not violate conservation of angular momentum? Where did this net angular momentum come from?

If they are not ever found to be both up or both down after measurement 3, then how are they still entangled even after repeated measurements?

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A Stern-Gerlach apparatus is an inhomogeneous $B_z$ along some $z$-direction, with approximate $y$-translation and $x$-translation symmetry.

Most of the question is kind of a red herring. You don't need to fuss with entanglement really, or have more than one apparatus.

Your entire question is, assume we prepare a spin-$+y$ state for all electrons in a beam that goes into this Stern-Gerlach apparatus measuring spin-$x$, those electrons will split into two beams, one that tilts up and one that tilts down, and the expected $L_y$ of both beams, is known to be zero. So we can prepare a nonzero $L_y$ beam that becomes two on-average-zero $L_y$ beams after passing through this magnetic field, how does that work?

Well, the classical force on a magnetic dipole in an inhomogeneous magnetic field is $$\mathbf F = \nabla (\mathbf m \cdot \mathbf B)$$and we have to be very careful with how we apply this to a spin-state of definite $L_y$.

The key thing to understand is that the electron has total spin angular momentum $\hbar \sqrt{3/4}$ and only $\hbar/2$ of that is pointed in the $+y$ direction. The definite-$L_y$ state therefore does not have $L_z = 0$ (and indeed it is limited from attaining that ideal by Heisenberg uncertainty).

So if you imagine that you happened to have an ensemble of rings pointed mostly-$+y$ and a little $+z$ or $-z$, the $+z$ rings feel a force upwards a d the $-z$ rings feel a force downwards. Speaking purely classically, on entering the magnetic field the forces would be inhomogeneous across the rotating ring of charge and the inhomogeneous force would create a torque in the $x\times z=-y$ direction which I think would tilt rings spinning slightly upwards from horizontal to tilt more upwards, and rings spinning slightly downwards to tilt more downwards, reducing $L_y$ in each case?

But the point would be that the reaction force felt by the magnet, while still in the $z$-direction, varies along the $x$ axis and this variation is a torque along the $L_y$ direction: the angular momentum was conserved because it was absorbed by the Stern Gerlach apparatus.

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As already stated the total angular momentum is conserved but I'll add something that may help to understand. Let's first consider only spin and neglect the orbital angular momentum.

After the first measurement you don't have anymore a singlet state and I think that this is clear. What you have is just a product state of the two electrons. That state is no more an eigenstate of the total angular momentum!! You are conserving the Z component since after the application of a $B_z$ field you have a term of interaction in the Hamilton that commutes with the $S_z$ but the product state $| \uparrow \rangle | \downarrow \rangle $ is already no more an eigenstate of the total S, it is indeed the linear combination of $S_z=0$ states of the singlet and triplet. Does it mean that therefore the total angular momentum is not conserved? Considering only the system of two electrons yes since it is no more an integral of motion. Wait... But the interaction term in the Hamiltonian is proportional to Sz therefore it should commute with total S and conserve it, right? Unfortunately here we are speaking of measurement theory that is a much subtle subject with a big field of study by it's own, the only thing we can say for sure (I know almost nothing about measurement theory) is that even without knowing what happens during the measurement, we come out with a state that is not an eigenstate of S total therefore it is not conserved after the measurement for the system of two electrons. If you were able to measure the total angular momentum of system+electrons then it would be conserved and you should not use what you know about measurement theory because it would simply be quantum mechanics.

I hope that helped you to clarify everything 🙂

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  • $\begingroup$ I guess my question is then how the measuring device takes in the angular momentum? $\endgroup$
    – David Santo Pietro
    Commented Sep 21, 2021 at 16:25
  • $\begingroup$ Exactly. And the effect of a measure is something that is still on debate. I would pose it even in a better way as "What is the mechanism that leads to the collapse of the wavefunction?" and this is not clear. Maybe searching for "Objective-collapse theory" can give you some hints on theories you may want to look into. $\endgroup$
    – SmauG
    Commented Sep 21, 2021 at 19:01
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The total angular momentum is conserved (electron+instrument).

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  • $\begingroup$ OK, well how does the instrument absorb the L? $\endgroup$
    – David Santo Pietro
    Commented Sep 21, 2021 at 14:32

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