Steven Weinberg on the interpretation of quantum mechanics

[vanhees71]

This is one of many things you have read about Weinberg that you simply dismiss. You are of course entitled to your opinions, no issue about that. You should take care to label your opinions as your own to provide suitable notice to the reader.

I.e. do that instead of labeling standard views of quantum theory as questionable (which you did here), and quoting yourself as an authority in the counter case. I would say, for example, that Weinberg's statements - as I quoted - are fairly innocuous... and they fall closely in line with what most physicists believe. The burden is on you, my friend, not on Weinberg. Show us where Weinberg is wrong by quoting SOMEONE ELSE.

Which standard views of QT did I label as questionable? I've not said that Weinberg is wrong. I said that I don't understand his argument. That's a big difference!

Again, I do not understand, how Weinberg comes to the conclusion that one needs a cut. This may have looked as a conclusion from the formalism, where thanks to Bohr and Heisenberg the collapse played an (in my opinion unjustified) important role. Today we are about 80 years further and have a much better understanding about open quantum systems, coarse-graining, decoherence and all that.

 

[A. Neumaier]

I don't know, what an "assembly" is.

Peres defined it in his book.

What you measure related to "vacuum" QFT are squared S-matrix elements (transition probabilities), and that's what's measured in terms of cross sections. Of course, some mathematical tools like nn-point Green's functions are not directly observable but a calculational tool to get the observable transition probabilities. Where is here a problem?

The only problem here is that you had erroneously claimed that the state of a quantum field is observable:

States of quantum fields describe many (if not all) phenomena. Why are you claiming they are unobservable?

Measuring transition probabilities is not the same as measuring a state.

By quantum tomography, one can measure to some limited accuracy the state of a single monochromatic photon modeled in the paraxial approximation, and perhaps in the future that of two entangled such photons (to even lower accuracy).

But the work grows exponentially with the number of degrees of freedom, and is already impossible for the state of an unrestricted photon, which has infinitely many degrees of freedom, so that quantum tomography would take more resources that the universe offers. Similarly, quantum tomography of the quantum states regularly used in many-body theory is impossible.

Therefore the state of a quantum field is not observable.

Of course, some mathematical tools like nn-point Green's functions are not directly observable but a calculational tool to get the observable transition probabilities. Where is here a problem?

In the same way, the state of the universe is not directly observable but a calculational tool to get observable coarse-grained approximations, such as Newtonian physics. Where is here a problem?

 

[Demystifier]

I don't understand this one quote by Weinberg: "Of course, according to present ideas a measurement in one subsystem does change the state vector for a distant isolated subsystem - it just doesn't change the density matrix."

By density matrix of the subsystem he means the object obtained by partial trace of the full unitary evolving state, while by state of the subsystem he means an object the evolution of which involves also something akin to the collapse. Does it make sense now?

 

[Demystifier]

Today we are about 80 years further and have a much better understanding about open quantum systems, coarse-graining, decoherence and all that.

Today there is a consensus among most (though not all) experts in open quantum systems, coarse-graining and decoherence that those things alone do not solve the measurement problem.

 

[Demystifier]

States of quantum fields describe many (if not all) phenomena. Why are you claiming they are unobservable?

The state (of quantum field) is a vector (or ray) in a Hilbert space. It is not observable in the sense that it is not a hermitian operator or an eigenvalue of a hermitian operator.

 

[DrChinese]

Which standard views of QT did I label as questionable? I've not said that Weinberg is wrong. I said that I don't understand his argument. That's a big difference!

How about what you picked on: " Of course, according to present ideas a measurement in one subsystem does change the state vector for a distant isolated subsystem ..."

I'm really not sure why you keep pushing a contrary position to this simple statement, when I have challenged you repeatedly to provide authoritative quotes. Stop quoting yourself! Your question is really not a question, it is a statement of your opinion. You want to attack the answer. Well, here you go:

ANY (of many) measurement on a component of an entangled system throws the distant subsystem into state in which the outcome of another measurement can be predicted with certainty. This is not a "coincidence", as Weinberg is saying. One measurement "causes" the other outcome, although causal direction is not certain (the reason we have interpretations).

 

[vanhees71]

This contradicts the microcausality condition fulfilled by QFT. It's also in Weinberg's QFT books as it is in any other book on QFT: There is no causal influence over space-like distances by construction! That's why I'm very surprised that Weinberg seems to claim the contrary in his newer QM book.

 

[martinbn]

This contradicts the microcausality condition fulfilled by QFT. It's also in Weinberg's QFT books as it is in any other book on QFT: There is no causal influence over space-like distances by construction! That's why I'm very surprised that Weinberg seems to claim the contrary in his newer QM book.

That is not what they are saying. Consider the usual example, two spin one half particles in a state |10>-|01>. The statement is that if you measure the left particle along an axis and you get the result 1, then the right particle will be in the state |0> for that axis. Are you saying that this is not correct?

 

[vanhees71]

Of course it's correct, but this correlation is not caused by the measurement but by the preparation of the entangled state, you have written down in the beginning, i.e. before (!) the measurement is done. There is no instantaneous change of the state due to the measurement.

 

[martinbn]

Of course it's correct, but this correlation is not caused by the measurement but by the preparation of the entangled state, you have written down in the beginning, i.e. before (!) the measurement is done. There is no instantaneous change of the state due to the measurement.

The point is that prior to the measurement you cannot say that the right particle was in the state |0>. Because it has to be true for any axis and there are no states like that.

 

[vanhees71]

Of course, I can't say this before the measurement, because the single-particle states are given by the partial traces and thus $\hat{\rho}_1=\hat{\rho}_2=\frac{1}{2} \hat{1}$.

Of course, it's true for any axis, because you have prepared a spherically symmetric state (total spin $S=0$). Now doing a local measurement at particle 1 Alice knows that Bob will find (or has already found) with certainty the opposite result when both are measuring in the same spatial direction, but that she knows because this 100% correlation of measurement results is due to the preparation of the two particles in this state. Alice's measurement has no instantaneous influence on Bob's spin and vice versa.

At least that's what's commonly agreed upon when interpreting all the Bell tests of recent experiments, i.e., when the measurement events at A's and B's place the common understanding is that there cannot be a causal effect of one of the measurements at the other (or are there refereed physics papers claiming otherwise?).

 

[Lord Jestocost]

....but by the preparation of the entangled state,...

Does it mean, as your are focusing on "preparation", that each particle emerges from the singlet state with, in effect, a set of pre-programmed instructions for what spin to exhibit at each possible angle of measurement?

 

[vanhees71]

No, there's nothing "pre-programmed". The two-spin state is just the singlet state you prepare it in. You can get such a state by, e.g., decaying a spin-0 particle to two spin-1/2 particles.

 

[Lord Jestocost]

Of course it's correct, but this correlation is not caused by the measurement but by the preparation of the entangled state, you have written down in the beginning, i.e. before (!) the measurement is done. There is no instantaneous change of the state due to the measurement.

J. Bell in "Bertlmann's Socks and the Nature of Reality"

“Let us summarize once again the logic that leads to the impasse. The EPRB correlations are such that the result of the experiment on one side immediately foretells that on the other, whenever the analyzers happen to be parallel. If we do not accept the intervention on one side as a causal influence on the other, we seem obliged to admit that the results on both sides are determined in advance anyway, independently of the intervention on the other side, by signals from the source and by the local magnet setting. But this has implications for non-parallel settings which conflict with those of quantum mechanics. So we cannot dismiss intervention on one side as a causal influence on the other.”

 

[martinbn]

Of course, I can't say this before the measurement, because the single-particle states are given by the partial traces and thus $\hat{\rho}_1=\hat{\rho}_2=\frac{1}{2} \hat{1}$.

Of course, it's true for any axis, because you have prepared a spherically symmetric state (total spin $S=0$). Now doing a local measurement at particle 1 Alice knows that Bob will find (or has already found) with certainty the opposite result when both are measuring in the same spatial direction, but that she knows because this 100% correlation of measurement results is due to the preparation of the two particles in this state. Alice's measurement has no instantaneous influence on Bob's spin and vice versa.

At least that's what's commonly agreed upon when interpreting all the Bell tests of recent experiments, i.e., when the measurement events at A's and B's place the common understanding is that there cannot be a causal effect of one of the measurements at the other (or are there refereed physics papers claiming otherwise?).

So, you agree that before and after Alice's measurement the state of Bob's particle is different, but you disagree with saying that the measurement changed it.

 

[EPR]

Lots of physicists think in terms of broken realism akin to 'everything is a field'(the 'particles' are approximate, momentary notions).

Then nonseparability comes out 'naturally' - hence your misunderstanding.

 

[DrChinese]

Of course it's correct, but this correlation is not caused by the measurement but by the preparation of the entangled state, you have written down in the beginning, i.e. before (!) the measurement is done. There is no instantaneous change of the state due to the measurement.

Quote someone to back up your position. Other than yourself. As requested too many times to count.

 

[EPR]

We can say that because we have evidence that realism and determinism are broken in the quantum world, everything there is fields.

This notion once accepted, there is no issue with entanglement. There certainly are no causal inflences between Alice and Bob.
Only when one thinks in terms billiard balls, do classical issues arise.
There is nothing wrong with vanheesh71's statements - some people need to adjust their worldview. The billiard balls do not exist.

 

[PeterDonis]

this correlation is not caused by the measurement but by the preparation of the entangled state, you have written down in the beginning, i.e. before (!) the measurement is done.

the common understanding is that there cannot be a causal effect of one of the measurements at the other

You have been focusing on the easy case, where Alice and Bob both measure spin in the same direction, so there is perfect anti-correlation and a "Bertlmann's socks" type of argument, which is the argument you are making, is workable.

However, such an argument is not workable for the hard cases, where Alice and Bob measure spin in different directions, at angles for which the correlations violate the Bell inequalities. For those cases, I don't think you can claim that "it's all just due to the previously prepared state", since the whole point of the Bell inequalities being violated is that there is no possible "previously prepared state" (no set of local hidden variables) that can account for the correlations. The QM "state" can do it, but only by being nonlocal, i.e., giving correlation probabilities that do not factorize as described in Bell's paper.

 

[PeterDonis]

There certainly are no causal inflences between Alice and Bob.

That's not what is "certain". The only thing that is "certain" is that Alice's and Bob's measurements must commute; but that does not require that there are no causal influences between them. It only requires that any such influences must not have a preferred direction, i.e., that the "Alice to Bob" and "Bob to Alice" causal directions must both be consistent with the results.

 

[A. Neumaier]

the whole point of the Bell inequalities being violated is that there is no possible "previously prepared state" (no set of local hidden variables) that can account for the correlations.

This is not quite cogent. Bell showed that there is no possible "previously prepared state" with classical hidden variables - but a quantum state is of course not covered by the argument. (Bell's argument are purely classical and assume nothing at all about quantum mechanics.)

 

[DrChinese]

We can say that because we have evidence that realism and determinism are broken in the quantum world, everything there is fields.

This notion once accepted, there is no issue with entanglement. There certainly are no causal inflences between Alice and Bob.

Not what Weinberg said. Again, if you want to contradict him, perhaps you’d care to provide a suitable reference that says different.

And your first statement does not follow anyway. Realism and determinism may indeed be broken. And we might throw in locality as well. That says absolutely nothing about QFT. And as everyone should know, QFT must respect Bell, regardless of one’s spin.

 

[PeterDonis]

Bell showed that there is no possible "previously prepared state" with classical hidden variables - but a quantum state is of course not covered by the argument.

Yes, that's why I said later in the same post that a quantum "state" can account for the correlations (but only by violating the Bell locality assumption).

 

[DrChinese]

That's not what is "certain". The only thing that is "certain" is that Alice's and Bob's measurements must commute; but that does not require that there are no causal influences between them. It only requires that any such influences must not have a preferred direction, i.e., that the "Alice to Bob" and "Bob to Alice" causal directions must both be consistent with the results.

Exactly, there is no clear causal direction. And there is nothing that explains the random outcome prior to measurement. That these confusing elements are present but not solved by any theory is why Weinberg would express dissatisfaction (which most of us share in some respect).

 

[EPR]

We can say that because we have evidence that realism and determinism are broken in the quantum world, everything there is fields.

This notion once accepted, there is no issue with entanglement. There certainly are no causal inflences between Alice and Bob.
Only when one thinks in terms billiard balls, do issues arise.

Exactly, there is no clear causal direction. And there is nothing that explains the random outcome prior to measurement. That these confusing elements are present but not solved by any theory is why Weinberg would express dissatisfaction (which most of us share in some respect).

Without realism/determinism, how can there be causal influences?

This is essentially the epr debate started anew.

A new, bigger theory superceded qm of single objects(qft) as a more comprehensive explanation of reality.
You will likely struggle forever with the ingrained notion of particles and billiard balls like thousands other physicists do.

 

[EPR]

That's not what is "certain". The only thing that is "certain" is that Alice's and Bob's measurements must commute; but that does not require that there are no causal influences between them. It only requires that any such influences must not have a preferred direction, i.e., that the "Alice to Bob" and "Bob to Alice" causal directions must both be consistent with the results.

They are certain in as much as the theory of relativity is certain. There are no ftl influences.

 

[PeterDonis]

Without realism/determinism, how can there be causal influences?

Without realism, how can there be anything? If you're going to say realism has been "broken", you need to explain how you exist before asking how there can be anything else.

As you can see from the above, we can go round and round forever making assertions about "realism" by simply using different definitions of what that word means. That will get us nowhere.

A new, bigger theory superceded qm of single objecta(qft) as a more comprehensive explanation of reality.

QFT still predicts violations of the Bell inequalities (as it must, since it agrees with experiment and such violations have been observed experimentally), so it does nothing at all to solve whatever problems people have with explaining the fact of such violations.

Gerrymandering definitions of words like "realism" does not help with those problems at all.

They are certain in as much as the theory of relativity is certain. There are no ftl influences.

The theory of relativity does not say "no ftl influences". It only says that any such "influences" must be independent of the time order of the events involved.

 

[EPR]

The theory of relativity does not say "no ftl influences". It only says that any such "influences" must be independent of the time order of the events involved.

But all events pertaining to the Alice/Bob's measurements are causal. And time ordered. How can there be ftl influences without breaking SR?

 

[vanhees71]

You have been focusing on the easy case, where Alice and Bob both measure spin in the same direction, so there is perfect anti-correlation and a "Bertlmann's socks" type of argument, which is the argument you are making, is workable.

However, such an argument is not workable for the hard cases, where Alice and Bob measure spin in different directions, at angles for which the correlations violate the Bell inequalities. For those cases, I don't think you can claim that "it's all just due to the previously prepared state", since the whole point of the Bell inequalities being violated is that there is no possible "previously prepared state" (no set of local hidden variables) that can account for the correlations. The QM "state" can do it, but only by being nonlocal, i.e., giving correlation probabilities that do not factorize as described in Bell's paper.

Why is not workable? To the contrary as you say, the correlations violate the Bell inequalities precisely as predicted by QT, i.e., also in these cases the quantum formalism describes the outcomes correctly, and of course you have prepared just the entangled state you need to do this experiment. The preparation does not predetermine which measurements you do afterwards of course, but the correlations are described by these Bell states, including the violation of Bell's inequalities.

The state itself you may call nonlocal, but that's somewhat misleading, because it seems to imply non-local interactions. That's why I prefer Einstein's more precise word "inseparability", i.e., the entanglement describes precisely what's nowadays confirmed with these Bell tests, which at Einsteins lifetime was not yet known. So relativistic QFT is a theory which in a consistent way describes both the locality of interactions (implemented by the construction of local quantum fields obeying the microcausality condition, i.e., the commutation of local observables with the Hamiltonian density at space-like distances) as well as the inseparability of entangled subsystems leading to the long-ranged correlations between measurements, including the violation of Bell's inequalities for the appropriate measurements (in this example of course by choosing a certain set of measurements of spin components (or, for photons, polarizations) in non-collinear directions). Of course, and this is Bell's great achievement, it rules out any local deterministic hidden-variable models.

 

[vanhees71]

Quote someone to back up your position. Other than yourself. As requested too many times to count.

Read any experimental paper on Bell tests. There it's described, that the entangled state is prepared and then the measurements are done. It's demonstrated that the correlations are there even when measuring with space-like separated measurement events ("detector clicks"). I think this has been achieved first in the late 1990ies by Zeilinger's group with polarization entangled photons. Today we have several "loophole-free" measurements too.

 

[PeterDonis]

all events pertaining to the Alice/Bob's measurements are causal. And time ordered.

Not for both measurements taken together. For each measurement taken individually, yes. But the whole point is that we are looking at the correlations between the measurement results; that means we have to look at the complete set of events involved in both measurements. And that set includes events that are spacelike separated from each other and therefore do not have an invariant time ordering.

 

[PeterDonis]

Why is not workable?

Because, as Bell explained in his "Bertlmann's socks" paper, that kind of argument implicitly presupposes local hidden variables, and local hidden variables are ruled out if the Bell inequalities are violated.

The preparation does not predetermine which measurements you do afterwards of course

Exactly. So the preparation by itself is not sufficient to account for the actual measurement results. But you have been asserting the contrary.

 

[PeterDonis]

Read any experimental paper on Bell tests.

Referring to the experimental results does not support any particular claim about interpretation, since all interpretations make the same predictions for experimental results. So you are not responding to the request @DrChinese made by referring to the experimental results. We all know the experimental results violate the Bell inequalities, and we all know the math of QFT correctly predicts that. Stating those things as though they supported your particular position is not responsive.

 

[vanhees71]

Indeed, and it has been demonstrated that the correlations are as predicted even when the measurement events at Alice's and Bob's place are space-like separated. Any experimental paper I know takes this as an indication that thus the correlations can NOT be caused by an influence of A's measurement on B's subsystem when measured by him and vice versa. DrChinese knows all these papers very well. We have discussed about them at length for some time in this forum.

 

[DrChinese]

Without realism/determinism, how can there be causal influences?

Well, ask Weinberg. As I repeatedly request, a citation saying something different than what HE says would be helpful. By the lack of presentation of such, which I think we know is not around, one might conclude these are your personal opinions and not backed by scientific consensus.

But to address your question: as best I can: As I said earlier, there is no causal direction that is evidenced by experiment. Nonetheless, experiment shows that anyone of a large number of measurements that can be performed lead to an exact prediction of the result of a similar faraway measurement. So that is the type of influence Weinberg describes. It's not a coincidence, and you can't wave your hand and make it disappear. It can't be predetermined, as Bell showed us (excepting of course the Bohmian solution).

Again, waiting for your suitable quote* to counter Weinberg's well spoken words. (*Doesn't it make you wonder when such a quote cannot be located, and you must say it yourself?)