I Making the wave observable in the double-slit experiment

[PeterDonis]

There is an interaction I presume between each possible individual MWI branch when an observed particle interacts with the observer environment.

No. There is no interaction between branches. The branches are the outcome of the interaction.

For example, consider a single electron passing through a Stern-Gerlach device. Say the electron starts in the z-spin up state, and the Stern-Gerlach device is oriented in the x direction. Then the state before the interaction is

$$
\ket{z+} \ket{\text{input beam}}
$$

and the state after the interaction is

$$
\frac{1}{\sqrt{2}} \left( \ket{x+} \ket{\text{x-spin up beam}} + \ket{x-} \ket{\text{x-spin down beam}} \right)
$$

The former state is a product state, but the latter state is an entangled state; the entanglement is produced by the interaction between the electron and the Stern-Gerlach apparatus.

This interaction by itself does not necessarily produce decoherence, since the beams can in principle be recombined; to complete a spin measurement one needs to put detectors in each output beam and observe which one fires. But the entangled state produced by that further interaction looks simliar to the above, just with an additional ket for the detector system.

I don't know what you would say is the entangled attribute/observable though.

In the Stern-Gerlach case, the entanglement is between the spin and the momentum of the electron. The specific degrees of freedom that are entangled will depend on the particular interaction.

Or what the conserved quantity is.

I'm not sure why a conserved quantity would be relevant.

 

[vanhees71]

But that's what the equations say. It's not just pulled out of thin air, it is literally what the math (without collapse) tells you happens during any interaction. Measurement is just a type of interaction.No, it wouldn't. The entanglement created by an interaction does not have to be maximal. Non-maximal entanglements can be spread out among an arbitrary number of degrees of freedom. That is the basis of decoherence theory.

That's of course right: In an ideal von Neumann filter measurement the measurement appartus's "pointer state" gets maximally entangled with the system's state. That's the definition of a von Neumann filter measurement.

Take the ideal Stern Gerlach experiment: You measure a spin component by letting the Ag atoms through an appropriate magnetic field (large constant part in direction of the spin component to be measured and some inhomogeneous part) to entangle the spin component with the position of the particle. Then measurement of the position is 100% correlated with measuring the spin component.

 

[vanhees71]

This interaction by itself does not necessarily produce decoherence, since the beams can in principle be recombined; to complete a spin measurement one needs to put detectors in each output beam and observe which one fires. But the entangled state produced by that further interaction looks simliar to the above, just with an additional ket for the detector system.

Indeed, the time evolution of a closed system is always unitary and (in principle) reversible. Decoherence occurs when looking at a sub system of a larger system. The effective description of the state evolution of the subsystem, i.e., its reduced statistical operator is described in terms of some non-unitary master equation, which involves decoherence through the interaction of the "system" with "the environment".

By chance, there's just a nice Nature paper about the issue, how the 2nd law of thermodynamics (H-theorem) for subsystems is compatible with the unitary evolution of the closed system:

https://doi.org/10.1038/s41467-023-38413-9 (open access!)

 

[kered rettop]

TL;DR Summary: Physicist becomes convinced Schrodinger's equation describes real waves, because they can be made visible in the double-slit experiment using a laser and smoke machine.

I don't see how the smoke makes the wave visible. Each photon is scattered by the smoke at a different point, so it's equivalent to having the screen at that point. You are not watching the evolution of a single photon's wave function, you're just seeing its final state (where it's scattered). You can assume that they all behave the same way, but that just begs the question.

 

[vanhees71]

Obviously Youtube has not only very good explanations of QT. Very likely a good textbook is the better choice for learning it!

 

[kered rettop]

Going back to the first video, I have a naive question about measurement. Both the screen and the smoke act as measurements, but they don't destroy interference, even if one photon were emitted at a time. So why does using a detector cause decoherence? Is it because it's detecting photons in one specific location, or is it because the detector is a complex macroscopic object?

Debates around measurement, collapse and decoherence seem to confuse the issue as to exactly when and why a measurement destroys the interference pattern. If the environment is causing decoherence, then why does the smoke and screen still allow for an interference pattern?

Yes they do destroy interference. If the photon is scattered by the smoke or reflected off the screen, you can't coax interference out of it. The unscattered photons carry on interfering.

The environment only causes decoherence if the particle interacts with it. It can't do anything just by being.