Where does a quantum experiment *begin*?

[vanhees71]

Yes, it's pretty easy to make it a "von Neumann filter measurement". Supposed you have a setup such that the partial beams of definite $\sigma_z$ are well separated you can just block all partial beams except the one with the $\sigma_z$ you want. Then you have a beam of pure $\sigma_z$ states. For a full quantum treatment of the SG experiment, see

Potel et al, PHYSICAL REVIEW A 71, 052106 (2005)
http://arxiv.org/abs/quant-ph/0409206

 

[ShayanJ]

Yes, it's pretty easy to make it a "von Neumann filter measurement". Supposed you have a setup such that the partial beams of definite $\sigma_z$ are well separated you can just block all partial beams except the one with the $\sigma_z$ you want. Then you have a beam of pure $\sigma_z$ states. For a full quantum treatment of the SG experiment, see

Potel et al, PHYSICAL REVIEW A 71, 052106 (2005)
http://arxiv.org/abs/quant-ph/0409206

So, when does the update happen in a SG setup?

 

[vanhees71]

It happens as soon as the observer is finding a particle at the position referring to a certain $\sigma_z$.

 

[Demystifier]

At least there seems to be even a difference in Bohr's and Heisenberg's view, and that's what made Einstein&Co. rebel (in my opinion righteoulsy) against "the Copenhagen doctrine", because it's not a sharp scientific definition but a quite fuzzy set of philosophical thoughts.

Would you also put Ballentine in that company of Einstein&Co?

 

[vanhees71]

Good question. The problem is that Einstein&Co. didn't give a specific interpretation but simply considered QT incomplete based on their classical prejudices. I think Ballentine's book is great in discussing the various aspects of the different interpretations and why the minimal interpretation is the least problematic one (and, in my opinion, the only necessary one too).

 

[ShayanJ]

It happens as soon as the observer is finding a particle at the position referring to a certain $\sigma_z$.

I know it seems clear, but its actually not!
You block one of the beams and the other creates a spot on the screen and so you will update the wave-function to the corresponding eigenstate. But what is the system? The whole beam? Only the unblocked beam? If its only the unblocked beam, when did it become a separate system from the other beam? How do you explain that separation?
Also, what if you don't block any of the beams? You get two spots. So what is the updated wave-function now?

 

[Demystifier]

But also this has to be taken with a grain of salt. There are only a very few very special experimental situations where you do a von Neumann filter measurement, and only then you can use the postulate that you can update your knowledge on the system to be represented by an eigenvector of the measured quantity to the eigenvalue found in the measurement.

In other words, the most general measurements are POVM measurements, while projective measurements are only a small subclass.

But even that should be taken with a grain of salt. By Neumark's theorem, POVM measurement in a Hilbert space ${\cal H}$ can always be represented by a projective measurement in a larger Hilbert space ${\cal H}' \supset{\cal H}$.

For instance, consider a photon detection. Since it involves a photon destruction, it is not a projective measurement in the space of 1-photon states. Nevertheless it is a projective measurement in the space of all QED states.

In this sense, all measurements are projective at a fundamental level. But for practical laboratory purposes it is more convenient to work with a smaller Hilbert space which describes only those degrees which are of direct experimental interest. That's why, for laboratory purposes, it makes sense to say that most measurements are not projective.

 

[Demystifier]

I think Ballentine's book is great in discussing the various aspects of the different interpretations and why the minimal interpretation is the least problematic one (and, in my opinion, the only necessary one too).

I certainly agree that Ballentine's book is great and that minimal interpretation has many merits.

Nevertheless, would you agree with me that Ballentine's conclusions about quantum Zeno effect (watched pot paradox) are wrong? I mean page 343 where he says:
" ... we have been led to the conclusion that a continuously observed system never changes its state! This conclusion is, of course, false."

 

[vanhees71]

I think it's wrong, because for a "continuously observed" quantum system you have to take the entire Hamiltonian including the apparatus into account, and then it can well be that the decay of an unstable quantum state is prevented due to this interaction. It has been confirmed (in full consistency with QT) by various experiments, as one can read in the references of the Wikipedia article

https://en.wikipedia.org/wiki/Quantum_Zeno_effect

 

[kith]

You block one of the beams and the other creates a spot on the screen and so you will update the wave-function to the corresponding eigenstate. But what is the system? The whole beam? Only the unblocked beam? If its only the unblocked beam, when did it become a separate system from the other beam? How do you explain that separation?

What the system is, should be decided by the experimenter. We could use only the unblocked beam or we could use both beams. For subsequent measurements on the unblocked beam, the spatial overlap between the measurement apparatuses and the blocked beam is zero. So if our state is |unblocked> + |blocked>, all probability amplitudes involving |blocked> will be zero. Thus if we redefine the system to include only the unblocked beam, we get the same probabilities as if we continue to use the whole system. In the spirit of this, the "updating of the state" is more a redefinition of the system.

(I am not sure if this reasoning can be extended to Bell tests. It's been a while since I've been thinking about this.)

 

[Mentz114]

If "collapse" is just a bad name for update, then it's obvious how can it be consistent with relativistic local QFT.

As Asher Peres briliantly said:
"Quantum phenomena do not occur in a Hilbert space, they occur in a laboratory."

The same idea is expressed by my signature.

No, the idea expressed in your signature is that someone who disagrees with you is a fool.

It is not brilliant like Peres' statement and could be offensive to some.

 

[Demystifier]

No, the idea expressed in your signature is that someone who disagrees with you is a fool.

It is not brilliant like Peres' statement and could be offensive to some.

That's why I put a smile at the end. :)

 

[vanhees71]

If you like to discuss with physicist you shouldn't be to sensitive. Discussions can get pretty tough, but it's usually not meant personally ;-)).

 

[vanhees71]

What the system is, should be decided by the experimenter. We could use only the unblocked beam or we could use both beams. For subsequent measurements on the unblocked beam, the spatial overlap between the measurement apparatuses and the blocked beam is zero. So if our state is |unblocked> + |blocked>, all probability amplitudes involving |blocked> will be zero. Thus if we redefine the system to include only the unblocked beam, we get the same probabilities as if we continue to use the whole system. In the spirit of this, the "updating of the state" is more a redefinition of the system.

(I am not sure if this reasoning can be extended to Bell tests. It's been a while since I've been thinking about this.)

This also underlines that quantum states refer to ensembles. Whether of not you block partial beams decides about the preparation of the ensemble. If you don't block anything, the ensemble in your example is represented by $|\psi_1 \rangle=(|\text{blocked} \rangle + |\text{unblocked} \rangle)/\sqrt{2}$, otherwise in $|\psi_2 \rangle=|\text{unblocked} \rangle$.

 

[Demystifier]

If you like to discuss with physicist you shouldn't be to sensitive. Discussions can get pretty tough, but it's usually not meant personally ;-)).

Exactly!

 

[atyy]

What means "physical"?

In theory you describe a point particle in a 6-dimensional phase space, i.e., three position and three momentum variables. These variables are real numbers and part of the mathematical description, but that's trivial, isn't it?

How can collapse be consistent with local interactions in the usual sense of relativistic local QFT? The collapse claims that you affect an extended system (like the polarization-entangled biphotons in Aspect-like experiments) instantaneously all over space via a local (!) interaction of part of it. That's obviously self-contradictory!

"Physical" is your term. You are the who introduced the term to discuss "collapse" and "point particle". You should explain it to me, not me to you.

 

[atyy]

I think he does not reject the update. He just refuses to call it collapse, because that expression has misleading connotations.

I couldn't agree more!

Then it is just semantics - but I find it odd that vanhees71, who is supposedly promoting the minimal interpretation always brings up "physicality" and "interpretation" and "interaction" ... when I use the term collapse, in a perfectly standard way, without any meaning of physicality.

It's a bit like Ballentine - the nominal claim to support the minimal interpretation, but not the actual support of it. It is people like me who just want to shut up and calculate who are the true believers in the minimal interpretation, not vanhees71, and not Ballentine.

Also, if you look at his statement about "collapse contradicting foundations of relativistic QFT", you will find that the statement cannot be right whether collapse is physical or not. So it is just plain wrong for any interpretation of collapse.

Edit: Furthermore, see David Lewis's post below - what vanhees71 thinks about collapse is wrong. There is no accepted proof that collapse is the same as the classical Bayesian update (without "nontrivial" additional assumptions). Perhaps there is one, and many have pointed out the similarity, but the analogy to the die is not justified at this stage.

 

[David Lewis]

E.g., take a dice (thought of as a classical system) and you just say that the indeterminism of the outcome of throwing it comes from the unknown initial conditions, there will always be a clear and not a somehow "smeared" outcome. Nobody would come to the idea of a "collapse... Where is the difference to QT?

While the die is spinning through the air, each face still has some definite, fixed number of spots (1 through 6) even when nobody is looking at it.

 

[atyy]

The point of controversy in the "interpretation question" is also not so much the dynamics but rather the interpretation of the states themselves, which in the standard theory is just Born's Law (no collapse necessary), i.e., the usual probabilistic content of the state. There is, however, nothing which makes a collapse postulate necessary. You just state that QT predicts probabilities for measurement given the preparation of the measured system (and preparation can be very crude, e.g., you describe a gas by the usual thermodynamical quantities like temperature, volume of the container, and density of conserved charges), which allows the association of a statistical operator to the system.

From this point of view, if you introduce a collapse into QT, you have to introduce it for the probabilities in classical statistical physics too. E.g., take a dice (thought of as a classical system) and you just say that the indeterminism of the outcome of throughing it comes from the unknown initial conditions, there will always be a clear and not a somehow "smeared" outcome. Nobody would come to the idea of a "collapse", i.e., it just turns up with a specific result, and throughing many times leads to an experimental test of the prediction of any theoretical probability. You can also envoke some theory behind how to postulate such probabilities like information theory a la Shannon and say that as long as you don't know anything about the dice you say each outcome will have a probability of 1/6 (maximum-entropy probability). Then you can do the experiment and confirm or refute the estimate of the probabilities with some confidence level given the experimental outcomes of your measured ensemble.

Where is the difference to QT? The only difference is that, according to the minimal interpretation, the observables that are not determined by the preparation, are "really random", i.e., they have indeed no determined value and not only because we don't know them. Then a lot of philosophical mumbling is done about, how it can be that one has a clear outcome of any proper measurement of such observables. My point is that this is due to the construction of the measurement device, which works with very good precision as a classical system, and classicality can be explained satisfactorily by quantum statistics and coarse graining. It's just the usual quantum theoretical dynamics ("unitary evolution") of this interaction, and this interaction is (according to the best QT we have, which is relativistic local QFT) local and thus there cannot be an instantaneous influence of a measurement at a position A to another far-distant measurement at position B, but that's what's postulated when "envoking" a collapse.

To add to David Lewis's post above, the whole of this quote from vanhees71 is wrong.

In short, I object to attacks on Many-Worlds or Bohmian Mechanics (or other approaches to the measurement problem) based on incorrect Minimal Interpretations.

 

[vanhees71]

It's a bit like Ballentine - the nominal claim to support the minimal interpretation, but not the actual support of it. It is people like me who just want to shut up and calculate who are the true believers in the minimal interpretation, not vanhees71, and not Ballentine.

We should clarify this point, because I think it's wrong what you claim. The collapse in the usual meaning of textbooks claims that the interaction of the measured system with the measurement device (a local interaction according to local relativistic QFT) acts instantaneously over the entire space, which is contradicting itself, because that would mean the interaction is not local. In the mathematical foundations, however it is built in that the interaction is local. So collapse cannot be part of relativistic local QFT.

 

[vanhees71]

To add to David Lewis's post above, the whole of this quote from vanhees71 is wrong.

In short, I object to attacks on Many-Worlds or Bohmian Mechanics (or other approaches to the measurement problem) based on incorrect Minimal Interpretations.

I don't have an opinion on many worlds. I never understood its point. Just to claim the universe splits at each measurement or observation into many unobservable new universes, is just empty. It's nothing physically observable. Otherwise it doesn't provide anything new concerning the observable predictions of QT. Bohmian mechanics is similar. It introduces trajectories of particles in non-relativistic quantum theory that are not observable either. There's no merit to calculate them. Also there is no convincing Bohmian interpretation for relativistic QFT.

 

[Demystifier]

There is no accepted proof that collapse is the same as the classical Bayesian update (without "nontrivial" additional assumptions).

In fact, there is a proof of something opposite. The PBR theorem and its variations shows that wave function is something more than (the square root of) epistemic probability.

 

[Demystifier]

The collapse in the usual meaning of textbooks claims that the interaction of the measured system with the measurement device (a local interaction according to local relativistic QFT) acts instantaneously over the entire space,

It would be nice to have an exact quote from some standard textbook. Could you give some?

 

[Demystifier]

I don't have an opinion on many worlds. I never understood its point. Just to claim the universe splits at each measurement or observation into many unobservable new universes, is just empty. It's nothing physically observable. Otherwise it doesn't provide anything new concerning the observable predictions of QT. Bohmian mechanics is similar. It introduces trajectories of particles in non-relativistic quantum theory that are not observable either. There's no merit to calculate them. Also there is no convincing Bohmian interpretation for relativistic QFT.

Classical general relativity does something similar. It claims that behind horizon (either black-hole horizon or cosmological horizon) there is a lot of space (perhaps even infinite space in the cosmological case) which is completely unobservable to us. Does it mean that spaces behind horizons of classical general relativity are physically meaningless?

 

[Demystifier]

It's a bit like Ballentine - the nominal claim to support the minimal interpretation, but not the actual support of it.

Promoting minimal interpretation is like promoting non-existence of free will. You can promote it in abstract discussions of the deepest principles of nature, but as a human being you cannot think that way in all situations of practical interest. Whether one promotes it or not may depend on the level of discussion (fundamental vs practical).

Think of the sentence
"I decided to write a new argument why free will does not exist."
and note that it makes sense because it involves thinking at two different levels.

Similarly, the statement
"I have written a paper where I explain why minimal interpretation is good. The paper is in the tray."
makes sense for a similar reason.

 

[ddd123]

In fact, there is a proof of something opposite. The PBR theorem and its variations shows that wave function is something more than (the square root of) epistemic probability.

Could the Bayesian inference be intended as referring to measurement outcomes instead of the physical underlying states (whether psi is ontic or epistemic)? In that case the PBR theorem doesn't apply.

 

[Demystifier]

Could the Bayesian inference be intended as referring to measurement outcomes instead of the physical underlying states (whether psi is ontic or epistemic)? In that case the PBR theorem doesn't apply.

Nobody said that psi cannot be used as a Bayesian tool. It can. But the PBR theorem shows that psi contains also something more than that. This is like showing that swiss knife is something more than a knife, which does not stop you from using it only as a knife.

 

[vanhees71]

It would be nice to have an exact quote from some standard textbook. Could you give some?

In nearly any introductory textbook you find the collapse postulate, i.e., it says that if you make a measurement of an observable $A$ and find a value $a$ which is necessarily in the spectrum of the representing self-adjoint operator $\hat{A}$ and if $|a,\beta \rangle$ is a complete orthonormal basis of the eigenspace to $a$ and the system as been prepared in the state represented by $|\psi \rangle$, then after the measurement the system is immediately in the state
$$|\psi' \rangle = \sum_{\beta} |a,\beta \rangle \langle a,\beta|\psi \rangle.$$

 

[vanhees71]

Classical general relativity does something similar. It claims that behind horizon (either black-hole horizon or cosmological horizon) there is a lot of space (perhaps even infinite space in the cosmological case) which is completely unobservable to us. Does it mean that spaces behind horizons of classical general relativity are physically meaningless?

That's a good question. It's not observable in principle. So it's irrelevant for physics. Whether or not the prediction of GR that these regions of space time exist, cannot be checked by experience. That doesn't invalidate GR as long as anything predicted that's observable is not ruled out by observation.

 

[martinbn]

There is a difference though. In general relativity the space-time may contain causally disconnected regions, but the space-time itself is connected. At least those if physical interest. In the many world interpretation it seems that the worlds are completely disjoint.

 

[Demystifier]

In nearly any introductory textbook you find the collapse postulate, i.e., it says that if you make a measurement of an observable $A$ and find a value $a$ which is necessarily in the spectrum of the representing self-adjoint operator $\hat{A}$ and if $|a,\beta \rangle$ is a complete orthonormal basis of the eigenspace to $a$ and the system as been prepared in the state represented by $|\psi \rangle$, then after the measurement the system is immediately in the state
$$|\psi' \rangle = \sum_{\beta} |a,\beta \rangle \langle a,\beta|\psi \rangle.$$

Yes, but exact words are very important in this context. For instance, in the explanation above you don't mention that collapse has anything to do with interaction, while in another post you do. Similarly, in the explanation above you use the word "immediately", while some textbook may not use that word. It also matters whether one talks about "measurement" or about "observation". Depending on the exact words, the connotations may take different flavors.

 

[Demystifier]

In the many world interpretation it seems that the worlds are completely disjoint.

Not really. First, they are fully connected in the past. Second, even in the future the overlap of the two worlds exponentially decreases with time, so it is very small but not exactly zero. Third, even this exponential law is only an approximation, and after a very very long time (essentially the Poincare recurrence time for the many-body Schrodinger equation) the two worlds may join together again.

One of the most frequent misconceptions about many worlds is that this interpretation postulates that wave function splits at measurements. In reality, there is no such postulate. The only explicit postulate is the Schrodinger equation, while the split is derived from the Schrodinger equation.

 

[martinbn]

That confuses me. What exactly is a world in MWI? I assumed it is literally a space, so after the measurement there is a family of disjoint spaces. For example take a particle that has wave function that is the superposition of localized in region A and localized in region B. I make a measurement and detect it in region A. What is the description in MWI? I thought it is that now there are two separate particles in two separate spaces each in the corresponding region.

 

[Demystifier]

That's a good question. It's not observable in principle. So it's irrelevant for physics. Whether or not the prediction of GR that these regions of space time exist, cannot be checked by experience. That doesn't invalidate GR as long as anything predicted that's observable is not ruled out by observation.

Fair enough. But then also other worlds of MWI and unmeasurable trajectories of Bohmian mechanics (BM) do not invalidate MWI and BM, as long as anything predicted by MWI and BM that's observable is not ruled out by observation. You will ask: Yes, but what's the motivation for introducing other words or unobservable trajectories in the first place? And my answer is: What's the motivation for introducing spaces behind the horizon in the first place?

 

[Demystifier]

That confuses me. What exactly is a world in MWI? I assumed it is literally a space, so after the measurement there is a family of disjoint spaces. For example take a particle that has wave function that is the superposition of localized in region A and localized in region B. I make a measurement and detect it in region A. What is the description in MWI? I thought it is that now there are two separate particles in two separate spaces each in the corresponding region.

No. In MWI, particles do not have wave functions. In fact, particles do not exists at all in MWI. According to MWI, there is only a wave function and nothing but the wave function. Only one wave function, not many wave functions. However, evolution by Schrodinger equation is such that wave function often splits into branches, such that the overlap between the branches is very small. When the overlap is small, then each branch can approximately be thought of as an object by its own, not depending on the existence of other branches. In this case, each branch can approximately be thought of as a separate "world". That's what the world in MWI is.