I Confused by nonlocal models and relativity

Mentz114

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How about this quote: "Of course there's no change of photons 1&4 due to manipulations on photons 2&3." See anything like that? :smile:

I have represented my position (which is standard, generally accepted physics) as best I can at this point. In fact, I have probably repeated myself too much already. I will leave it to the readers to make their own judgments and assessments of these fabulous and groundbreaking experiments. Consequently, I will bow out of this thread and thank everyone for their time and comments - especially for helping me sharpen my LaTeX from terrible to poor. :biggrin:
Well, thank you. You have stated your position coherently it is just superluminal influences that are hard to digest.

For me it is very instructive to see the QED approach.
There the solution is to expand the state that exists after the two pairs 1,2 and 3,4 are created (creation operations on the vacuum) using Bell state bases. The result shows that the 'final' state is already in there by preparation. When the entangled pairs are created, the result is inevitable and does not require any further interaction. The symmetries ensure that.

I used to find that unsatisfactory but now it seems harmonious.

But I am easily led by equations.:smile:
 

vanhees71

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How about this quote: "Of course there's no change of photons 1&4 due to manipulations on photons 2&3." See anything like that? :smile:

I have represented my position (which is standard, generally accepted physics) as best I can at this point. In fact, I have probably repeated myself too much already. I will leave it to the readers to make their own judgments and assessments of these fabulous and groundbreaking experiments. Consequently, I will bow out of this thread and thank everyone for their time and comments - especially for helping me sharpen my LaTeX from terrible to poor. :biggrin:
Indeed, that's what I'm saying. There's no change of photons 1&4 due to manipulations on photons 2&3. All you do is to select a partial ensemble based on measurements on photons 2&3, and due to the preparation in the initial state, describing the full ensemble the so selected partial ensemble is described by the state ket $$|\psi_{14}^- \rangle \otimes |\psi_{23}^- \rangle.$$
That's indeed standard, generally accepted physics.
 

vanhees71

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Well, thank you. You have stated your position coherently it is just superluminal influences that are hard to digest.

For me it is very instructive to see the QED approach.
There the solution is to expand the state that exists after the two pairs 1,2 and 3,4 are created (creation operations on the vacuum) using Bell state bases. The result shows that the 'final' state is already in there by preparation. When the entangled pairs are created, the result is inevitable and does not require any further interaction. The symmetries ensure that.

I used to find that unsatisfactory but now it seems harmonious.

But I am easily led by equations.:smile:
Well, there's no other chance to understand what's going on in the physical world than math and equations (most importantly group theory, which is the guiding principle; all the rest is just calculational techique).
 

Lord Jestocost

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The only thing she does is to select a subensemble....
One doesn't select a subensemble, a measurement "creates" - so to speak - a subensemble. Thinking about the post-measurement situation cannot be conveyed to the pre-measurement situation.
 
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The states @vanhees71 has been writing down also include all four photons.
Yes, and he has also been arguing about the selected subsystem of photons 1&4 that DrChinese keeps saying is affected in a "spooky" way by what happens to 2&3, so they are both singling out 1&4 but the relevant thing is doing it with the right math.
Now the only way to mathematically speak coherently about the subsystem 1&4 other than in the Hilbert space of the 4 photons that everyone agrees about is using a reduced density matrix for 1&4, and this reduced density matrix is not affected by 2&3, as vanhees says and DrChinese rejects.
My hunch is that DrChinese is clinging to the 4 photon entangled pure state when he says the subsystem 1&4 is affected by 2&3 but by definition as mentioned before the entangled 4 photon state is inseparable from preparation to measurement and it doesn't matter that the particular photons 1&4 never interacted before, they are prepared as entangled in the 4 photon entangled state.

DrChinese himself said in a previous post:

"Weinberg goes on to say as follows: "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." Which is what I assert: A measurement on Alice's particle changes the physical state of Bob's remote entangled particle (what is observed)."
apparently mixing the mathematical sense of state vector that refers to the entangled total system(that being in superposition makes not well defined mathematically what a subsystem would be, Weinberg clearly takes an ordinary language license there) with some "physical state" that can't be atributed to 1&4 alone excep using the right math of the density matrix that Weinberg , of course correctly, says doesn't change in the same quote.
 
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One doesn't select a subensemble, a measurement "creates" - so to speak - a subensemble. Thinking about the post-measurement situation cannot be conveyed to the pre-measurement situation.
See my last post. It can be conveyed as long as we are talking about the prepaired entangled state in the system of 4 photons that is inseparable by definition. Knowledge of the measurement effectively subselects information applying quantum probabilities.
 
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vanhees71

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Of course, one is selecting a subensemble. That's the whole purpose of the experiment. As the authors (Jennewein, Zeilinger et al) have demonstrated, it doesn't matter who measures their photon pair first. They can even measure it at space-like separation of the measurement events. It also doesn't matter, when Victor chooses the subensemble from the measurement protocol.

The facts are the following:

The total ensemble (which Victor of course also can consider from the measurement protocol) is described by the initial state and is unaffected by the measurements of A on photons 2&3 and B on photons 1&4. Due to the entanglement between the photons 1&2 as well as 3&4 (i.e., the maximal Bell-state entanglement beyond the minimal entanglement bosons that are not in the same single-particle state) the subensemble chosen based on the measurement of A leads to the entanglement between B's photons 1&4 in the subensemble. There's nowhere any causal effect of A's measurement on B's photons. This becomes particularly clear by the fact shown in the experiment in the Jennewein at al paper, where A's measurement is done clearly after B's measuremet. All one does is to choose a subensemble.

Another very nice paper I stumbled over is

https://link.springer.com/article/10.1007/s10701-019-00278-8

However one has to read this paper such that it is about two spin-entangled distinguishable particles (say an electron and positron from ##\pi^0 \rightarrow \mathrm{e}^+ + \mathrm{e}^-## decay).
 

Mentz114

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this reduced density matrix is not affected by 2&3, as vanhees says and DrChinese rejects.
I don't think @DrChinese rejects that the reduced density matrix of 1&4 is not changed by the measurement that is made on 2&3. He agreed with me that the measurement on 2&3 doesn't affect the probabilities for measurements on 1&4, which is the same thing.

I think what @DrChinese is rejecting is the claim that the measurement on 2&3 does not change photons 1&4, i.e., he does not agree that "no change to the reduced density matrix" means "no change at all". The measurement on 2&3 certainly changes the wave function of the system as a whole, and 1&4 are part of the system as a whole. It also changes the entanglement relationships--before the measurement, 1&2 are entangled and 3&4 are entangled; after the measurement, 1&4 are entangled and 2&3 are entangled. That change shows up in the correlations between the appropriate pairs of photons violating the Bell inequalities.

At least some of the disagreement might be just interpretation. When @vanhees71 talks about post-selection of a particular sub-ensemble, he is taking an ensemble interpretation viewpoint, in which QM does not describe individual photons or individual experimental runs, but only ensembles of them. When @DrChinese talks about the entanglement relations changing in an individual run of the experiment, he is taking the opposite viewpoint, that QM describes individual quantum systems and measurements.

I'll also repeat once more my suggestion to not use vague ordinary language but instead look at the math. The math is unambiguous, but there are many different ways of describing in ordinary language what the math is telling us, and those ways often seem like they contradict each other, even though they're all describing the same math and the same predictions for experimental results. To me that just means we should stop arguing about the ordinary language since it's superfluous anyway. But not everyone takes that view.
 
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I don't think @DrChinese rejects that the reduced density matrix of 1&4 is not changed by the measurement that is made on 2&3. He agreed with me that the measurement on 2&3 doesn't affect the probabilities for measurements on 1&4, which is the same thing.
Bell's inequalities and their violations are all about probabilities and ensembles so I don't know what else he might be arguing mathematically by the "spooky" change to 1&4.
I think what @DrChinese is rejecting is the claim that the measurement on 2&3 does not change photons 1&4, i.e., he does not agree that "no change to the reduced density matrix" means "no change at all".
I haven't seen anyone saying " no change at all". There is consensus about change in the 1&2&3&4 entangled state.

The measurement on 2&3 certainly changes the wave function of the system as a whole, and 1&4 are part of the system as a whole. It also changes the entanglement relationships--before the measurement, 1&2 are entangled and 3&4 are entangled; after the measurement, 1&4 are entangled and 2&3 are entangled. That change shows up in the correlations between the appropriate pairs of photons violating the Bell inequalities.
As you stress this is all change in the system as a whole, the product Hilbert space ##\mathcal H = \mathcal H_1 \otimes \mathcal H_2 \otimes \mathcal H_3 \otimes \mathcal H_4##, the system where we can have 4 entangled photons and different subsystems selected depending on the measurement setup.

At least some of the disagreement might be just interpretation. When @vanhees71 talks about post-selection of a particular sub-ensemble, he is taking an ensemble interpretation viewpoint, in which QM does not describe individual photons or individual experimental runs, but only ensembles of them. When @DrChinese talks about the entanglement relations changing in an individual run of the experiment, he is taking the opposite viewpoint, that QM describes individual quantum systems and measurements.
Again, this is a fine viewpoint but Bell's violations are about ensembles, an individual run is not relevant here except for selling pop-sci books about QM mysteries.
 
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Bell's inequalities and their violations are all about probabilities and ensembles
Bell's particular inequalities are, yes. But a lot of work in this area has been done since Bell, including finding cases where QM predicts results that are impossible according to local hidden variable models, so that the latter models can be ruled out with 100% certainty by observing such an "impossible" result, with no probabilities or statistics or ensembles required. For example, the GHZ experiment:


I haven't seen anyone saying " no change at all".
More precisely, this:

There's no change of photons 1&4 due to manipulations on photons 2&3.
Which, mathematically, refers to the fact that the reduced density matrix of 1&4 does not change. But the wave function does.
 
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Bell's particular inequalities are, yes. But a lot of work in this area has been done since Bell, including finding cases where QM predicts results that are impossible according to local hidden variable models, so that the latter models can be ruled out with 100% certainty by observing such an "impossible" result, with no probabilities or statistics or ensembles required. For example, the GHZ experiment:

Right.


More precisely, this:



Which, mathematically, refers to the fact that the reduced density matrix of 1&4 does not change. But the wave function does.
I thought they agreed on the wave function.
 
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I thought they agreed on the wave function.
They do. But @DrChinese thinks that the change in the wave function (and the entanglement relations) is sufficient to say that something changes about photons 1&4 when photons 2&3 get measured, while @vanhees71 does not. At least, that's my understanding of the viewpoints they have been expressing.
 

vanhees71

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As you stress this is all change in the system as a whole, the product Hilbert space ##\mathcal H = \mathcal H_1 \otimes \mathcal H_2 \otimes \mathcal H_3 \otimes \mathcal H_4##, the system where we can have 4 entangled photons and different subsystems selected depending on the measurement setup.
The 4-photon space is NOT this product space but it is the subspace spanned by all totally symmetrized (bosonic) product-basis states.
 

vanhees71

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They do. But @DrChinese thinks that the change in the wave function (and the entanglement relations) is sufficient to say that something changes about photons 1&4 when photons 2&3 get measured, while @vanhees71 does not. At least, that's my understanding of the viewpoints they have been expressing.
Despite the fact that there are no wave functions for photons, of course nothing done locally on photons 2&3 can change photons 1&4, at least not instantly in a "spooky action at a distance".

Concerning my interpretation, it's indeed the minimal statistical interpretation, including the frequentist interpretation of probabilities: All that's done is to select (or even post-select) a sub-ensemble. The observed correlations (entanglement) between photons 1&4 for this subensemble is implied by the initially prepared four-photon state.

In my opinion there's no other way to interpret QT without violating the one or the other fundamental property of physical theories, particularly Einstein causality in special-relativistic spacetime.
 

vanhees71

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Bell's particular inequalities are, yes. But a lot of work in this area has been done since Bell, including finding cases where QM predicts results that are impossible according to local hidden variable models, so that the latter models can be ruled out with 100% certainty by observing such an "impossible" result, with no probabilities or statistics or ensembles required. For example, the GHZ experiment:




More precisely, this:



Which, mathematically, refers to the fact that the reduced density matrix of 1&4 does not change. But the wave function does.
There's no wave function for photons!

What changes is the description of the sub-ensemble based on the selection depending on Alice's specific measurement of the pair 2&3.

Also in classical applications of statistics and probability theory a subensemble usually has some different probability distribution for some of its properties than the full ensemble. There's nothing mysterious in this.
 
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The 4-photon space is NOT this product space but it is the subspace spanned by all totally symmetrized (bosonic) product-basis states.
I don't see this, why can't the tensor product be used for the 4-photon system?
 

vanhees71

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Because bosons are bosons. It's a well-established fact of nature.
 
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Because bosons are bosons. It's a well-established fact of nature.
I fail to see how the boson-fermion distinction is relevant to this discussion, we could be talking about electrons and spin instead of photons and polarization and the situation would be essentially the same.
 

vanhees71

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Sure, but it's important to treat the photons as bosons (or electrons as fermions) particularly when it comes to entanglement.
 
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Sure, but it's important to treat the photons as bosons (or electrons as fermions) particularly when it comes to entanglement.
Ok, I agree it is important to always be precise in math descriptions, but in this particular experiment, does the Fock space instead of the tensor product description of the system add anything to the conceptual issue of swapping and entanglement we've been discussing with DrChinese? I just want to make sure I'm not missing something important.
 

vanhees71

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In this case it doesn't make much of a difference.
 

Lord Jestocost

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.......he is taking an ensemble interpretation viewpoint, in which QM does not describe individual photons or individual experimental runs, but only ensembles of them.
...........I'll also repeat once more my suggestion to not use vague ordinary language but instead look at the math. The math is unambiguous, but there are many different ways of describing in ordinary language what the math is telling us......
Regarding the "Ensemble Interpretation", one can clearly speak out how "murky" this interpretation is. See comment https://www.physicsforums.com/threads/confused-by-nonlocal-models-and-relativity.973876/post-6202479 or listen to Maximilian Schlosshauer in “Decoherence, the measurement problem, and interpretations of quantum Mechanics”, Section B. 1. Superpositions and ensembles (https://arxiv.org/abs/quant-ph/0312059):

“Put differently, if an ensemble interpretation could be attached to a superposition, the latter would simply represent an ensemble of more fundamentally determined states, and based on the additional knowledge brought about by the results of measurements, we could simply choose a subensemble consisting of the definite pointer state obtained in the measurement. But then, since the time evolution has been strictly deterministic according to the Schrödinger equation, we could backtrack this subensemble in time and thus also specify the initial state more completely (“postselection”), and therefore this state necessarily could not be physically identical to the initially prepared state on the left-hand side of Eq. (2.1).“
 

vanhees71

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I don't understand this criticism of the ensemble interpretation. Of course QED is T symmetric and thus all interactions of photons with charged matter are in principle reversible, but what has this to do with any specific interpretation?
 

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