Exact time evolution of Stern-Gerlach (SG) apparatus

Background: I was always under the impression that when considering the Stern-Gerlach (SG) Experiment, the interpretation of the split of the beams is that the spin $$1/2$$ particle get measured the first time when it moves into the $$B\neq 0$$ area, and is subsequently diverted upwards or downwards, depending on the outcome $$\pm 1/2$$ of the spin measurement. Finally the particle will get measured a second time at the screen, to be registered by us.

Confusion: Now I came across several sources, e.g. Coherence and Entanglement in SG experiment or Modern analysis of SG experiment that model the interaction part of the magnetic field exactly with the Hamiltonian

$$$$H = \frac{p^2}{2m} + \mu \vec{B} \cdot \vec{s}$$$$

and hence model the time evolution of the beam not via "measurement + unitary time evolution" but only through "unitary time evolution" according to the above Hamiltonian. They assume the initial state to be

$$$$|\phi \rangle = \left( \int |z\rangle \phi_0(t) \right) \otimes (\alpha |+ \rangle + \beta |- \rangle$$$$

$$$$\phi_0(z,t) := \frac{1}{\sqrt{\sqrt{2\pi} \sigma} } e^{-\frac{z^2}{4\sigma^2}}$$$$

The exact time evolution is then

$$$$|\phi (t) \rangle = \int |z \rangle \otimes \left( \alpha \phi_+(z,t) |+ \rangle + \beta \phi_-(z,t) |- \rangle \right)$$$$

$$$$\phi_{\pm}(z,t) := Ne^{-i\theta(z,t)} e^{-\frac{1}{4\sigma(t)^2}(z \mp \overline{\Delta z}(t))^2}$$$$

where $$N$$ is a normalisation factor, $$\theta(z,t)$$ is represents a phase shift and

$$$$\overline{\Delta z}(t) = \mu \frac{t^2}{2m} \frac{\partial B}{\partial z}$$$$

So, interpreting the final state, when the particle hits the screen / gets measured at the screen, due to having two gaussian packets whose Centers are separated, one can expect to have two peaks, separated vertically.

Question: This replacement of the 1. measurement by a unitary time evolution seems to be contradictory as I thought measurement and unitary evolution describe inherently different time evolutions - so might it be that the spin is not measured when it flies into the $$B\neq 0$$ area ? But if it is measured, it seems like the measurement description is an approximation... and then in similar vein one could (in principle) replace the particle hitting the screen thereafter for the 2. measurement with another unitary time evolution? Or what else seems to be my misconception here?

• Commented Jan 15 at 9:10

Measurement is a macroscopic action which creates some reliable records, it does not describe time evolution. In case of the Stern-Gerlach experiment, the measurement happens when the atom impacts the screen and leaves a permanent mark there. By checking the position of the mark, we can retrodict what state the spin was in when the particle impacted the screen.

Measurement does not happen just by magnetic field interacting with the spin. Spin just like magnetic moment in classical theory interacts with magnetic field in a continuous way, the spin/magnetic moment rotates its direction in space (precession) and in non-uniform field, it also may change its axis of precession.

Application of the projection postulate(which changes the state vector "by hand", based on the new knowledge of some measurement result) produces a new state vector that is valid only at the time just before the measurement result is created (when the atom impacts the screen), and also some time before that, if the atom is in region of space where spin interaction with magnetic field(and its rotation in space) is negligible.

But applying the projection to the quantum state at any time during the flight in the magnet, when the atom leaves no record of its state, is unnecessary and wrong. The Schroedinger-Pauli equation is applicable there, so there is no reason to project the state. It would be futile to apply the projection postulate at all times the spin interacts with the magnet.

We apply the projection postulate when we want the state vector to reflect some newly obtained fact, like the position of the mark on the screen.

However, since experiments show the SG magnet splits the beam into two beams which, in the region of weak magnetic field far from the magnet, have opposite spin projections, people sometimes simplify the language and say the interaction of the spin with the SG magnet is the measurement, and causes the spin of any single atom to change to a state that is one of two possible projections of the original state. But this is a language short-hand, no definite single atom is measured here, and the split into two beams does not actually happen at some definite time - change of spin state in the magnet is a continuous process described by the Schroedinger-Pauli equation.

• thanks, would installation of a polarisation filter right after the magnet then be considered a measurement, because it is some interaction with something macroscopic? Commented Jan 14 at 19:17
• I'm not a fan of inflating the meaning of "measurement" to other actions that are not meant to measure anything. A filter filters, it prepares a beam of atoms in a known spin state. I do not see why one has to call such preparation a measurement. It just confuses people. Commented Jan 14 at 20:18
• If you redefine the meaning of measurement as "irreversible change of quantum state" (I wouldn't, because measurement is about results, not about quantum states) then even such filter doing the filtering does measurement in this new sense, because randomly some atoms are captured and some go through, so those that go through get their spin state changed irreversibly from many possible initial states to the single final state. Commented Jan 14 at 20:27

Separating the paths of of the $$|+\rangle$$ and $$|-\rangle$$ parts of the wavefunction is not a measurement, because it can be reversed. It is only the preparation for the subsequent position measurement, which then effectively measures the spin on the particle (see "In quantum mechanics, can we measure anything else than position?").

You can undo the entanglement of spin and position by sending the spin particle through another reversed Stern-Gerlach apparatus, as in my answer to "How can anything be unentangled?". The trick is to erase all information about the $$|+\rangle$$ and $$|-\rangle$$ parts of the wavefunction taking different paths.

Ján Lalinský and A. P. gave great answers, but your question also raises good philosophical objections to Copenhagen interpretation. I do not like Many Worlds interpretation, but they have a good point, that there is no reason why the interaction at the wall ought to disobey Schrödinger equation; i.e. it should also be a unitary time evolution.

But it really has very little to do with Many Worlds. Any interpretation that has decoherence will give a similarly satisfactory resolution, even if they do not completely resolve the measurement problem. But it does show that your question is interpretation dependent.

As Naturally Consistent notes, decoherence in the standard approach to quantum theory does not really solve the measurement problem. This issue is discussed in detail here, and a resolution is proposed: https://ijqf.org/archives/5951. (I am the author of this paper.)