I "Filming" a quantum measurement

[EPR]

I can, no it isn't.

No, you can't. It is your personal opinion.

In the MWI every interaction is a measurement.

 

[PeterDonis]

In the MWI every interaction is a measurement.

First, any claims about particular interpretations belong in a separate thread in the interpretations forum, not here. (And before you try to post your claim about the MWI in any thread in the interpretations forum, I strongly suggest that you check it first.)

Second, the 7 Basic Rules of QM (already linked once in this thread, I believe, and they are one of the sticky links at the top of this forum) make clear what a "measurement" is as far as basic QM is concerned, and "every interaction" is not what they say a measurement is.

 

[Demystifier]

No, you can't.

I just did.

 

[vanhees71]

Measuring a voltage is a classical process so it is a bad example.

If I am doing a double slit experiment with a photon, I don't call the effect of the slits on the photon a "measurement". The measurement is when the photon hits the detector screen and makes a dot. Nobody has a specific mathematical description of this either, the mathematical description is all about the effect of the slits, but that doesn't mean the effect of the slits on the photon is the measurement.

Ok, at least I now know, what you mean by measurement: You call only a process a measurement, where the quantum system interacts with a macroscopic measurement device which stores the result irreversably. That I can agree with. Then in the here discussed paper, what they call a "measurement" is indeed a "preparation procedure" and only the detection of the fluorescence photons is the measurement. So what's demonstrated here is the time-dependence of a preparation procedure leading in the limit of high intensity of the 422nm laser field to a preparation procedure as in what's commonly called a "von Neumann filter measurement". All this nitpicking on words is not that important, because it's clear from the description of the experiment what's meant by the words.

NB: It's of course not true that nobody has a mathematical description of photon detection. Usually it's just the photoelectric effect which FAPP can be described in 1st-order (time-dependent) perturbation theory in the dipole approximation. This also theoretically shows that the photon-detection probability is proportional to the energy density of the photons, i.e., the quantum version of the classical "intensity" $\propto \vec{E}^2(t,\vec{x})$ for a plane-wave mode. That's important, because that's indeed the only physical, i.e. gauge invariant, quantity you can define for free photons with the physical meaning of an intensity. Of course this is discussed in textbooks on quantum optics at length and nobody cares anymore in papers, because this is settled standard knowledge. You find a very nice description of all that in, e.g.,

J. Garrison and R. Chiao, Quantum optics, Oxford University
Press, New York (2008),
https://doi.org/10.1093/acprof:oso/9780198508861.001.0001

 

[PeterDonis]

You call only a process a measurement, where the quantum system interacts with a macroscopic measurement device which stores the result irreversably.

Yes.

All this nitpicking on words is not that important, because it's clear from the description of the experiment what's meant by the words.

It's clear to you and me (after some discussion and clarification), yes, but note that what is actually happening in the experiment, given the definition of "measurement" we have just agreed on, is not accurately described by the phrase "filming a quantum measurement". What is being "filmed" is a preparation process, not a measurement. So I think calling out the misleading nature of those words, which is what I have been doing, is important. Particularly since the misleading phrase is the very title of this thread.

 

[vanhees71]

Yes, in the quantum-foundation community you have the tendency to make some "headlines" and "advertisement" of research, which sometimes is not as exciting as this advertisement promises. At least that's an impression I have as an interested non-expert in this field.