Why the Quantum | A Response to Wheeler's 1986 Paper - Comments

[vanhees71]

There you have made the subjective classical-quantum cut.

Not again this wrong statement. You cannot admit at the same time that the classical behavior is derivable from QT and then claim that there is a cut. That's a contradictio in adjecto!

 

[vanhees71]

Whatever your version of QM is, individual particles have no speed or momentum or whatnot. And yet, nature only "hidden ontology", can only be approached and probed, (and this is even more true in QM), with unique individual event in the lab (and that is a that individual level, that all conservation law work).
Your view that only "ensemble of identically prepared thing" exist, and this is a complete physical phenomenology of nature, is highly incoherent with the fact that quanta do exist, and only them, and their individual interaction, are ever observed in a laboratory, or elsewhere.

Quantum objects have all the observables which can be defined on them. For massive particles these are particularly their energy, momentum, angular momentum, and their position. The point in QT is that not all the observables can take determined values at once (some cannot take determined values at all, which is the case for all observables having a continuous spectrum only like energy, momentum, and position).

Your view that only "ensemble of identically prepared thing" exist, and this is a complete physical phenomenology of nature, is highly incoherent with the fact that quanta do exist, and only them, and their individual interaction, are ever observed in a laboratory, or elsewhere.

That's not what I claim. Of course single electrons exist, and we can prepare them in many quantum states quite accurately. All I say is that within the ensemble interpretation quantum theory only describes the probabilities, and these probabilities can be empirically measured only on ensembles of equally prepared systems.

 

[A. Neumaier]

The observer constructs the measurement device to measure the observable he likes to measure. If you now start to discuss the ability of this free choice of the observer

I don't object to the choice. But after the observer has chosen the device (by whatever rule), there remains the pure quantum problem to show that the device actually produces on each reading the numbers that qualify as a measurement, in the sense that they satisfy Born's rule.

This is the measurement problem! It has nothing to do with the observer but is a purely quantum mechanical problem.

 

[zonde]

Of course single electrons exist, and we can prepare them in many quantum states quite accurately. All I say is that within the ensemble interpretation quantum theory only describes the probabilities, and these probabilities can be empirically measured only on ensembles of equally prepared systems.

But each individual system produces definite result (or appears to produce definite result). And either you have something to say about that, and then you participate in discussions about "collapse" and alternatives, or you keep agnostic position and do not say anything like "collapse is superfluous"/"collapse is required".

 

[Boing3000]

That's not what I claim. Of course single electrons exist,

So far so good...

and we can prepare them in many quantum states quite accurately.

No you cannot. There is no phenomenon as "state preparation" in nature. You (the observer) only do it in a lab, because you need that to match the ensemble with the esoteric Hilbert space, by using an ad-hoc Born rule. This ensemble exist only in your head. CM does not need any ad-hoc projection for observation, nor does nature (as per QM Schrodinger equation).

All I say is that within the ensemble interpretation quantum theory only describes the probabilities, and these probabilities can be empirically measured only on ensembles of equally prepared systems.

You say much more then that. You made make hidden assumptions (often circular) and quite astonishing claim (like QM completeness). You say noticeably that there is no epistemological difference with CM.
But in CM you don't need to make measurement to "create" the value of any observable (out of probability or whatnot).

 

[martinbn]

Of course classical mechanics is a limit of quantum mechanics. One can see this in the saddle point approximation to the path integral.

However, what you are not understanding and which Landau and Lifshitz state clearly, is that quantum mechanics cannot be formulated without "classical concepts" also in its assumptions. It is not possible to derive classical physics from "purely quantum" assumptions.

One can use different language to state this assumption, but they are all essentially equivalent - measurement has a different status than the interactions described in the Hamiltonian.

There is a difference between this and the statement that QT doesn't make any predictions without a cut.

 

[vanhees71]

I don't object to the choice. But after the observer has chosen the device (by whatever rule), there remains the pure quantum problem to show that the device actually produces on each reading the numbers that qualify as a measurement, in the sense that they satisfy Born's rule.

This is the measurement problem! It has nothing to do with the observer but is a purely quantum mechanical problem.

Ok, that's true. Of course, it's only possible for very simple cases in a strict way (like the famous analysis of tracks of charged particles in vapour chambers by Mott or the measurement of spin components in the Stern Geralach experiment).

 

[vanhees71]

So far so good...

No you cannot. There is no phenomenon as "state preparation" in nature. You (the observer) only do it in a lab, because you need that to match the ensemble with the esoteric Hilbert space, by using an ad-hoc Born rule. This ensemble exist only in your head. CM does not need any ad-hoc projection for observation, nor does nature (as per QM Schrodinger equation).You say much more then that. You made make hidden assumptions (often circular) and quite astonishing claim (like QM completeness). You say noticeably that there is no epistemological difference with CM.
But in CM you don't need to make measurement to "create" the value of any observable (out of probability or whatnot).

Of course we can handle electrons pretty well in accelerators and thus prepare, e.g., electrons with a pretty well determined energy and momentum to make all kinds of scattering experiments with them for decades. The ensemble doesn't exist only in my head, but it's realized with accelerators. That's why they aim at ever higher luminosities to "collect statistics as quickly as possible".

I don't claim the completeness of any physical theory we have so far. QM is incomplete because there is no satisfactory quantum description of the gravitational field yet. Indeed, I don't see any epistemological difference with CM. There's an ontological difference though.

 

[Boing3000]

Of course we can handle electrons pretty well in accelerators and thus prepare, e.g., electrons with a pretty well determined energy and momentum to make all kinds of scattering experiments with them for decades. The ensemble doesn't exist only in my head, but it's realized with accelerators. That's why they aim at ever higher luminosities to "collect statistics as quickly as possible".

Circular reasoning. You cannot prepare an electron in a pretty well defined state without measuring it first
The ensemble preparation is an laboratory artifact. You cannot, in the real world (or even based on QM phenomenology), propose an experiment to "ask/probe" an electron to find its companions in "an ensemble". This ensemble is not real. And the prediction are only more and more accurate with respect to the ensemble size.

Indeed, I don't see any epistemological difference with CM. There's an ontological difference though.

CM does not need an ad-hoc rule to connect the evolution formalism to the lab event. It does not need ensemble either. And CM don't treat measurement and interaction differently.

 

[zonde]

Because "ensemble" can not be defined in terms of QM, minimal QM is not a selfcontained model. It requires CM as a starting platform.

 

[stevendaryl]

But also #2 doesn't distinguish measurements from other interactions.

I think it definitely does.You measure a property and you get an eigenvalue ##\lambda## of the operator corresponding to the observable being measured. That means that the measuring device is in a specific state---the state of "having measured ##\lambda##". But treating the device as a physical system and treating the measurement as a physical interaction leads to a different state--where the measuring device is not in a specific state, but is entangled with the system being measured. Those are two different situations in QM, and are described by different quantum-mechanical states and those states have theoretically different statistical properties, leading to different predictions for future states. The two possible quantum-mechanical states are different, with different (in theory) observable consequences. They can't both be correct.

Now, I stuck the phrase "in theory" in there, because I think that the difference between an entangled macroscopic system and one that has a specific macroscopic properties may be undetectable in practice, but they are different states in QM. So you get different answers depending on whether you're treating the macroscopic system as a physical system following Schrodinger's equation or as a measuring device obeying the Born rule.

 

[stevendaryl]

But each individual system produces definite result (or appears to produce definite result). And either you have something to say about that, and then you participate in discussions about "collapse" and alternatives, or you keep agnostic position and do not say anything like "collapse is superfluous"/"collapse is required".

That's my feeling. A true minimalist interpretation, in the sense of making minimal assumptions, is not a denial of the collapse interpretation or the Many-Worlds Interpretation or the Bohmian interpretation, but should open to any of those possibilities. It should be silent on the question of what happens during a measurement.

 

[atyy]

Not again this wrong statement. You cannot admit at the same time that the classical behavior is derivable from QT and then claim that there is a cut. That's a contradictio in adjecto!

Which version of Landau and Lifshitz are you reading? Perhaps the German translation is different from the English one. There is a possibility the English version is biased towarda my views, since John Bell apparently had a role in it.

 

[martinbn]

Which version of Landau and Lifshitz are you reading? Perhaps the German translation is different from the English one. There is a possibility the English version is biased towarda my views, since John Bell apparently had a role in it.

The english translation of that section is faithfull to the original russian text.

 

[vanhees71]

Circular reasoning. You cannot prepare an electron in a pretty well defined state without measuring it first
The ensemble preparation is an laboratory artifact. You cannot, in the real world (or even based on QM phenomenology), propose an experiment to "ask/probe" an electron to find its companions in "an ensemble". This ensemble is not real. And the prediction are only more and more accurate with respect to the ensemble size.CM does not need an ad-hoc rule to connect the evolution formalism to the lab event. It does not need ensemble either. And CM don't treat measurement and interaction differently.

How do you come to these conclusions? We can prepare single electrons, even single photons, very well nowadays. And an ensemble can (among other ways to prepare them) consist of many repetitions of such single-quanta states. If this was not the case, we couldn't have ever checked that QT is describing things right in terms of the predicted probabilities.

Quantum mechanics doesn't treat measurement and interaction differently (I won't again repeat the obvious arguments I've stated several times in this thread again).

 

[Boing3000]

How do you come to these conclusions? We can prepare single electrons, even single photons, very well nowadays

The circularity of that claim is obvious.
But maybe that "preparation" is yet another kind of physical process I am not aware off, and described in your version of QM that is neither interaction nor measurement.

And an ensemble can (among other ways to prepare them) consist of many repetitions of such single-quanta states. If this was not the case, we couldn't have ever checked that QT is describing things right in terms of the predicted probabilities.

OK then how do you prepare an entangled pair of electron or photon that have probability 1 to be polarized at such angle along such axes...

 

[stevendaryl]

The circularity of that claim is obvious.
But maybe that "preparation" is yet another kind of physical process I am not aware off, and described in your version of QM that is neither interaction nor measurement.OK then how do you prepare an entangled pair of electron or photon that have probability 1 to be polarized at such angle along such axes...

I'm sort of in agreement with you that in QM, measurement and preparation seem very similar, but there are some circumstances where it is possible to get particles in a particular state without measuring them. For example, if you send electrons through a Stern-Gerlach device, the ones that are spin-up will go in one direction and the ones that are spin-down will go in another direction. Then if you perform an experiment on just one of the two streams, you can be assured that the electrons are in a specific spin state even though you didn't measure the spin.

 

[Boing3000]

I'm sort of in agreement with you that in QM, measurement and preparation seem very similar, but there are some circumstances where it is possible to get particles in a particular state without measuring them. For example, if you send electrons through a Stern-Gerlach device, the ones that are spin-up will go in one direction and the ones that are spin-down will go in another direction. Then if you perform an experiment on just one of the two streams, you can be assured that the electrons are in a specific spin state even though you didn't measure the spin.

But didn't you just describe a measurement ? How can you say you didn't measure their spin ? Or are you saying you are no more interested by spin, but want to measure some other property (maybe loosely coupled with spin) ?

 

[Fra]

The ensemble doesn't exist only in my head, but it's realized with accelerators. That's why they aim at ever higher luminosities to "collect statistics as quickly as possible".

I agree this process is important, and this is where the probabilistic abstractions are attached to physics.

This requires two things to actually make sense:

- The timescale of the processes we observer must be "small" so that we can prepare, decode data, and repeat enough statistis fast on a relative timescale
- The experimental control requires the system of study to be small relative to the lab so that we can control its boundary.

This is certainly true for HEP where we can observe scattering on the boundary, but fails for cosmology (here a new paradigm for inference is needed! which one?)

If we can do this we have good foundation for the probabilistic predictions, as well as extracting timeless patterns that stay constant over trials (symmetries). This how the standard model of particle physics is designed. But if these premises fail, not only do "probability" loose its original meaning, we also loose the ability in inferring symmetries, either because its too much data and limiting processing power or because of insufficient data to with any reasonable accuracy make statistical statements.

/Fredrik

 

[stevendaryl]

But didn't you just describe a measurement ?

No. After sending an electron through a Stern-Gerlach device, I know that:

• If the electron went left, then it must have been spin-up
• If the electron went right, then it must have been spin-down

But I don't know which is the case, so I haven't actually measured the spin.

 

[Boing3000]

But I don't know which is the case, so I haven't actually measured the spin.

Do you mean someone else have chosen which stream (left or right, or apparatus angle) and that you just don't know on which one you are working on ?

 

[stevendaryl]

Do you mean someone else have chosen which stream (left or right, or apparatus angle) and that you just don't know on which one you are working on ?

Well, it depends on exactly what is done with the two streams. If I perform a measurement of the electrons that go through one of the streams and get some result, then I'm indirectly measuring which stream the electron went in (since only one of the streams is measured), and so that indirectly counts as a spin measurement. But the measurement occurs at the moment I measure something about the electron. The separation into streams did not constitute a measurement.

To see that the separation by itself is not a measurement, I could redirect both streams back together into a single stream, and then no measurement of spin would ever be performed.

So a preparation does not necessarily count as a measurement (although it can be a preliminary step in a measurement).

 

[Mentz114]

Well, it depends on exactly what is done with the two streams. If I perform a measurement of the electrons that go through one of the streams and get some result, then I'm indirectly measuring which stream the electron went in (since only one of the streams is measured), and so that indirectly counts as a spin measurement. But the measurement occurs at the moment I measure something about the electron. The separation into streams did not constitute a measurement.

To see that the separation by itself is not a measurement, I could redirect both streams back together into a single stream, and then no measurement of spin would ever be performed.

So a preparation does not necessarily count as a measurement (although it can be a preliminary step in a measurement).

If you recombine the beams you do not get a thermal state, but you may have had one before the projections (depending on your preparation !) .
All projective 'measurements' are preparations. Nothing has been measured and all information about the previous state is lost.

This is elementary stuff which most people choose to ignore.

 

[stevendaryl]

If you recombine the beams you do not get a thermal state, but you may have had one before the projections (depending on your preparation !) .
All projective 'measurements' are preparations. Nothing has been measured and all information about the previous state is lost.

I'm not sure what you mean. Suppose I do the following:

1. Start with a stream of electrons that are spin-up in the x-direction
2. Separate it into two streams by sending electrons that are spin-up in the z-direction to the left, and the ones that are spin-down in the z-direction to the right.
3. Now, I recombine the two beams into a single beam
4. Finally, I measure the spin of the combined beam in the x-direction

If step 2 were a measurement, then step 4 would yield spin-up or spin-down, with equal probability. If step 2 is not a measurement, then step 4 will only produce the result spin-up.

 

[Mentz114]

I'm not sure what you mean. Suppose I do the following:

1. Start with a stream of electrons that are spin-up in the x-direction
2. Separate it into two streams by sending electrons that are spin-up in the z-direction to the left, and the ones that are spin-down in the z-direction to the right.
3. Now, I recombine the two beams into a single beam
4. Finally, I measure the spin of the combined beam in the x-direction

If step 2 were a measurement, then step 4 would yield spin-up or spin-down, with equal probability. If step 2 is not a measurement, then step 4 will only produce the result spin-up.

If a coherent state is prepared before the splitting/recombination and coherence is maintained then there will be state reconstruction. In those circumstances the splitting is a 'reversible measurement' because it tells us nothing about the previous state - i.e. like having no 'which-path' information.

 

[stevendaryl]

If a coherent state is prepared before the splitting/recombination and coherence is maintained then there will be state reconstruction. In those circumstances the splitting is a 'reversible measurement' because it tells us nothing about the previous state.

I would call it "not a measurement" rather than "a reversible measurement".

 

[Mentz114]

I would call it "not a measurement" rather than "a reversible measurement".

The whole experiment amounts to prepaing the beam in +x state then measuiring in x and finding +x. The only measurement in this experiment is was the one where you prepared the initial beam. I stand by All projective 'measurements' are preparations. Nothing has been measured and all information about the previous state is lost.

 

[stevendaryl]

The whole experiment amounts to prepaing the beam in +x state then measuiring in x and finding +x. The only measurement in this experiment is was the one where you prepared the initial beam. I stand by All projective 'measurements' are preparations. Nothing has been measured and all information about the previous state is lost.

I don't know what you mean. I would have guessed that "information about the previous state" would cover "the electrons have spin-up in the x-direction". That information has not been lost.

Perhaps all measurements are preparations, but the issue is whether all preparations are measurements.

 

[Mentz114]

I don't know what you mean. I would have guessed that "information about the previous state" would cover "the electrons have spin-up in the x-direction". That information has not been lost.

In step 1 there is a measurement. You started with a thermal beam and separated out x+. That was a projection and the previous state is lost. Step 2 is not a measurement, nor is the final step a measurement because there was no projection, so nothing changed.

 

[stevendaryl]

In step 1 there is a measurement. You started with a thermal beam and separated out x+. That was a projection and the previous state is lost. Step 2 is not a measurement

Yes, I agree that it's not a measurement, but it is a preparation.

 

[Mentz114]

Yes, I agree that it's not a measurement, but it is a preparation.

Do you mean step 2 is a preparation but not a measurement ?

 

[stevendaryl]

Do you mean step 2 is a preparation but not a measurement ?

Yes, that's what I meant.

 

[Mentz114]

Yes, that's what I meant.

Whatever we call steps 2 and 3 we can ignore them and look at steps 1 and 4. The only actual projection ( which some people may call a collapse) happens in step 1. After that there is no further projection so no information is lost or gained. We prepared x+ and we've still got it.

It occurs that due to the idempotency of operators ##\hat{S}_x\hat{S}_x|\phi\rangle=\hat{S}_x|\phi\rangle##

 

[Boing3000]

The separation into streams did not constitute a measurement.

OK do you call it an interaction ? but one that nobody "observe" ? The problem is that by you own setup, you are going to work on one of the stream only...

To see that the separation by itself is not a measurement

I am trying hard to follow your argumentation. Here I am still wondering how any preparation is different with "knowing/measuring/projecting" some state.

, I could redirect both streams back together into a single stream, and then no measurement of spin would ever be performed.

But a measurement has been made nonetheless. There is no way for someone not knowing/measuring (that is taking note of which electron when by which path) to assert/prove/measure that a measurement had not been made. Sure he cannot detect it, but it doesn't mean nobody can.
I someone else (aware of the result) come an got a much more accurate result (let's say 100% correct), it does not mean then QM is wrong. It means something did happen to each individual electron, no mater ones ignorance of it.
That thing is a measurement, not an interaction, because the projection is done by a classical apparatus which is the only thing able to set a particle into some eigenvalue. If the apparatus wasn't classical in the first place, you simply could not even set it in some orientation in the first place.
However that process take place, the only formulation of it is the Born rule, which may or may not be deduced in some way (but isn't currently).

So a preparation does not necessarily count as a measurement (although it can be a preliminary step in a measurement).

Even with your second example in post #269, step 2 is a also a measurement (in another bases, but nonetheless). Why should it change step 4 ? But it does change the wavefunction (of this basis, and maybe in other, but then QM would predict it anyway).

Can you try to give another example where no classical apparatus is used to "prepare" a state ? I kind of think it is impossible given the very definition of quanta.

 

[Mentz114]

[..]
But a measurement has been made nonetheless. There is no way for someone not knowing/measuring (that is taking note of which electron when by which path) to assert/prove/measure that a measurement had not been made. Sure he cannot detect it, but it doesn't mean nobody can.
I someone else (aware of the result) come an got a much more accurate result (let's say 100% correct), it does not mean then QM is wrong. It means something did happen to each individual electron, no mater ones ignorance of it.
[..]

The experiment that @stevendaryl described has been analysed in terms of projection operators here in post#5
https://www.physicsforums.com/threads/spin-state-recombination.927182/