Gleason's Theorem and the Measurement Problem

[vanhees71]

Why isn't this unitary? You just have the usual S-matrix of QED to describe this process, and that's unitary.

 

[A. Neumaier]

QT explains what happens in a single measurement, according to what's observed. The outcome of the measurement is in general random,

To say that in a single measurement, the outcome is random is not to have explained the outcome.

and that's what's observed (e.g., in the paradigmatic double-slit experiment with single particles or photons).

If you only observe a single measurement, you cannot observe whether the outcome is random. Instead you just observe a number.

 

[A. Neumaier]

But afterwards there's no photon. The usual collapse assumption means that after the measurement the photon's state is in an eigenstate of the measured observable. However, in this case you have the QED vacuum and no single-photon state.

The field always exists. Observed is a state of the electromagnetic field containing a superposition of 0 and 1 photon. After the observation this state collapsed to the 0-photon state.

Talking about the single photon is only an informal statement. For there might have been more photons and only one was observed. Then the electromagnetic field was in a superposition of 0,1,2,...,N photons, and its state collapsed to one with 0,1,2,...,N-1 photons.

 

[vanhees71]

Nowadays there are true one-photon preparations ("heralded photons"), and thus it's no longer an "informal statement". All I say is that the registration of a photon with a detector absorbing it is a drastic example that the collapse hypothesis is at best an oversimplification. The collapse hypothesis would lead to the conclusion that after the measurement of the photon, I'd have prepared the photon in a state with determined values of the measured observables (momentum and polarization, if you resolve these single-photon observables), but that's obviously not true, because there's no more photon after it's absorbed by the detector.

BTW, the same argument holds also if you have a coherent state (maybe of low intensity such as to have an average number of photons below 1): The measurement of the photon doesn't prepare a single-photon state as claimed by the collapse postulate, because this measured photon has been absorbed.

 

[A. Neumaier]

Nowadays there are true one-photon preparations ("heralded photons"), and thus it's no longer an "informal statement".

Even heralded photons have a significant nondetection rate, hence are superpositions of 0 and 1 particle states. In any case, photons are part of QED, hence photon states are actually states of the electromagnetic fields, and not states from a 2-dimensional helicity space.

The collapse hypothesis is at best an oversimplification.

Like all of quantum mechanics, since the unsimplified theory is an interacting relativistic quantum field theory.

if you have a coherent state (maybe of low intensity such as to have an average number of photons below 1): The measurement of the photon doesn't prepare a single-photon state as claimed by the collapse postulate, because this measured photon has been absorbed.

Whether a photon has been absorbed is immaterial to the properties of the state.
A coherent state remains unchanged under absorption of any number of photons.

 

[PeterDonis]

Why isn't this unitary? You just have the usual S-matrix of QED to describe this process, and that's unitary.

The S-matrix describes the probabilities of various possible final states given an initial state: in other words, the unitary process that the S-matrix describes starts from the initial state and gives you a probability-weighted superposition of final states.

But the evolution you described was not from an initial state to a probability-weighted superposition of final states. It was from an initial state to just one final state. That is not a unitary process.

 

[vanhees71]

It is of course such a process. Photoabsorption is the prozess photon+atom -> excited atom. The semicpassical treatment can be found in many QM textbooks as application of time-dependent 1st-order perturbation theory, e.g., in Sakurai or Landau-Lifshitz vol. 3.

 

[PeterDonis]

Photoabsorption is the prozess photon+atom -> excited atom.

Yes, but it does not happen with 100% probability. Nor is it guaranteed to happen with one particular atom. The unitary process involved has an amplitude for no absorption, and amplitudes for absorption by each of multiple atoms. The unitary process does not tell you which of those possibilities will actually happen. But the state transition you wrote down was to one particular outcome. If you are claiming that that one particular outcome is the result of a unitary process, I challenge you to write down the explicit unitary operator that realizes it.

The semicpassical treatment can be found in many QM textbooks as application of time-dependent 1st-order perturbation theory, e.g., in Sakurai or Landau-Lifshitz vol. 3.

None of these treatments answer the question I posed above. All of them produce predictions of probabilities using unitary operators. None of them predict the one particular outcome that actually happens using a unitary operator.

 

[PeterDonis]

It is of course such a process.

Since we are in the interpretations forum, I will point out that there is an interpretation of QM that does say that the dynamics is always unitary: the MWI. But in the MWI, a process like the one you describe does not have a single outcome. All possible outcomes occur, each with their appropriate amplitudes. The final wave function is an entangled superposition of all of them. But that isn't what you wrote down.

 

[GarberMoisha]

Of course they have well-defined quantum states, i.e., the reduced statistical operators, who follow some kind of master equation rather than a unitary time evolution.

The holy grail of science.
You assume what others are saying is missing.

 

[Demystifier]

It is of course such a process. Photoabsorption is the prozess photon+atom -> excited atom. The semicpassical treatment can be found in many QM textbooks as application of time-dependent 1st-order perturbation theory, e.g., in Sakurai or Landau-Lifshitz vol. 3.

If we denote the unitary S-matrix operator with $S$, the process is not unitary in the sense that $|{\rm excited\, atom} \rangle \neq S |{\rm process\; photon+atom \rangle}$.

 

[vanhees71]

Of course it is! The S-matrix describes all kinds of creation and destruction processes, including absorption and emission of photons by an atom. It's just the inner photoeffect, where an electron is going from one atomic state to another one at higher energies. For a photodetector you also use the outer photoeffect, i.e.,
$$\text{atom} + \gamma \rightarrow \text{atom} + \text{e}^-.$$
Then the electron is going to a scattering state instead of to a bound state.

All this is included in the unitary S-matrix, which describes the corresponding transition-probability rates as its matrix elements squared.

 

[PeterDonis]

Of course it is!

You keep making this assertion without ever actually addressing the repeated responses about it. For example, see post #46 by @Demystifier. If you are claiming that the equation he wrote down in that post is wrong, i.e., that $\ket{\text{excited atom}} = S \ket{\text{process photon plus atom}}$, then please show your work. Everyone else except you appears to recognize that what you actually get from $S \ket{\text{process photon plus atom}}$ is a superposition of things like $\ket{\text{atom 1 excited}}$, $\ket{\text{atom 2 excited}}$, etc., as well as $\ket{\text{no atom excited, photon not absorbed}}$. That superposition is not the same as $\ket{\text{excited atom}}$ for any specific atom. You cannot simply keep ignoring this and make the assertion you are making.

 

[PeterDonis]

All this is included in the unitary S-matrix, which describes the corresponding transition-probability rates as its matrix elements squared.

Yes, but that probability-weighted sum of alternatives is not the same as any particular alternative. But we only observe one particular alternative as the actual result. You have repeatedly claimed that we can somehow get just one particular alternative as the actual result from a unitary process; yet you contradict yourself by admitting that the S-matrix gives only a probability-weighted sum of alternatives. You can't have it both ways.

 

[vanhees71]

Of course, I didn't claim that, I only claim that these probability (rates) is all that is relevant to describe all these experiments, given the confirmation of the quantum theoretical predictions concerning Bell's inequality. In other words, there's not the slightest hint at hidden variables.

 

[PeterDonis]

Of course, I didn't claim that,

Yes, you did. That is what the transition you wrote down in post #11 is claiming, whether you want to admit it or not.

Once again, trying to discuss these issues with you is a waste of time because you refuse to be consistent.

I only claim that these probability (rates) is all that is relevant to describe all these experiments

What you wrote down in post #11 wasn't a probability. It was a definite transition between definite states.

Once again, you are shifting your ground whenever you are challenged, which makes having these discussions with you a waste of time.

 

[PeterDonis]

I try not to participate anymore in these discussions.

Apparently you couldn't keep this up for very long. You responded to @A. Neumaier only 24 minutes after you posted what I quoted above, and that subthread has now gone on for three days, at which point we are where I described in my last post.

I am closing this thread for moderation.

 

[bhobba]

As the person who asked for this thread, if someone was interested in the role Gleason plays in the formalism of QM, reviewing recent posts, it has deviated from the thread's purpose. I see no purpose in it continuing, and will remain closed.

Thanks
Bill