How Many Hidden Variables Would It Take to Model Photon Polarization?

In summary: EPR concept were true, we would need to encode a hidden variable for each degree angle in order to get results that looked like perfect correlations between particles. And we would need to encode a hidden variable for every 2 degrees.
  • #1
DrChinese
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Suppose the EPR* concept were true as it would be applied to entangled photon polarization. (Please note: I am not saying it is.) They thought QM was incomplete, because there must exist "elements of reality" (hidden variables) that supplied the highly correlated results on entangled particle pairs (label them A and B). By inference, we deduce that those results must in fact be predetermined - even if they are random. Of course Bell showed us that such an approach would require an appeal to non-local action, but that is not the issue I am asking about here. My questions are trying to get an idea of how much "hidden" information (if it existed) would be needed to actually provide results consistent with EPR (perfect correlations) and, to some "fair" degree, with the QM statistical predictions (Bell correlations). For our purposes, we are only modeling polarization and are ignoring momentum and any other observables. So the EPR idea is that A and B contain the same set of hidden variables, and they are to be measured by Alice and Bob respectively and independently.

If there were only a single (1) hidden variable - as Bell considered in his paper (Section III - "a unit vector") - we wouldn't get close enough to the quantum expectation value at many Alice/Bob angle pairs (although a single hidden variable does give exactly the right stats for perfect correlations). And polarized unentangled photons would not follow Malus' Law, which would be a major problem. We need something approximating the cos^2(theta) rule too. So we need more than 1 hidden variable to support this hypothetical approach, so I am wondering... how many? I don't believe I have ever seen a calculation on this.

So I am looking for the "back of a napkin" answer for these 2 numbers:

A.
How many (hidden) binary bits would be required to model entangled photon polarization, given that a typical BBo crystal produces entangled photon pairs with perfect correlations at any angle setting. We also need to be able to come "close" to the QM expectation value at other angles, but obviously there are no local hidden variable combinations that can exactly reproduce those (as Bell proved). For simplicity, let's only consider angles from 0 to 90 degrees. Would we need 90 hidden variables, one for each degree? (Obviously, there is nothing special about a degree itself.) Do you need a hidden variable for every 2 degrees, requiring 45? Or?

B. How many permutations of those are possible, assuming A bits are random. I might guess 2^A, assuming binary bits.

-------------------------

When we produce entangled photon pairs, we know that they will be perfectly correlated (or anti-correlated depending on PDC Type) at every angle when that same angle is used for both photons of a pair. Suppose Alice and Bob both set their polarizing beam splitters (PBS) to 0 degrees. We expect them to see both H or both V when looking at A and B. We set the PBS=14 degrees, we expect them to see both H or both V. We set the PBS=37 degrees, we expect them to see both H or both V. Ditto for all angles. Presumably there is randomness added to the process, as we never know whether the outcomes would be both H or both V.

Concerning A:
Suppose we ask: what is the smallest angle difference that can be discriminated in a Bell test? When we set Alice to 0 degrees and shift Bob from a setting of 0 degrees to 1 degree: the predicted match percentage drops from 100.00% to 99.97% - not much difference. Moving Bob's setting to 6 degrees (holding Alice at 0 degrees) yields an expectation of 98.91%, a difference of about 1%. That might be enough to discriminate in a Bell test.

Of course, there are settings in which even a single degree difference yields a much higher difference. Alice=0 degrees, Bob=45 degrees, you get a match of 50.00%. Change Bob by a single degree, to 46 degrees, and the match rate drops to 48.26% - a difference of nearly 2 degrees. I would say that would easily be detectible in a Bell test.

So do we need a new hidden variable every 6 degrees? Or a new hidden variable every 1 degree? Or a new hidden variable every 1/2 degree?

Concerning B:
Obviously, there is a relationship between 2 very close angle settings. I don't know if that makes the amount of total information required less or not. But clearly there must be a point at which a completely independent bit of information must be encoded, so that the randomness is apparent.

-------------------------

So I might start as follows: I guess you'd need at least A=30 hidden variables containing polarization outcomes (spaced 3 degrees apart) in order to make a hidden variable approach viable (close enough to the QM predictions). B=2^30 permutations is about 1 billion. That is a bit less than the same number of permutations in the order of a single suit in a deck of cards (13! = 6 billion). What are the A and B values on your napkin?

Hopefully you follow my example, let me know if the idea is not clear to you. Is any of this a reasonable approach to getting the answers for A and B? If so, I might conclude: Any entangled pair sporting "hidden variables" would be about as likely to match another randomly selected entangled pair as to 1 in B, whatever your B value comes to.

Thanks in advance.-DrC

*EPR: https://journals.aps.org/pr/pdf/10.1103/PhysRev.47.777
 
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  • #2
DrChinese said:
If there were only a single (1) hidden variable - as Bell considered in his paper
I don't think this is correct. Bell used one symbol, ##\lambda##, but I don't think he intended that to mean there was just one hidden variable. He intended it to stand for however many hidden variables there were.
 
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  • #3
PeterDonis said:
I don't think this is correct. Bell used one symbol, ##\lambda##, but I don't think he intended that to mean there was just one hidden variable. He intended it to stand for however many hidden variables there were.
Sorry, I wasn't clear. Yes, you are correct that his ##\lambda## meant a set of HVs, or functions, or whatever.

But I was referring (poorly) instead to his (section III) Illustration, formula (4). In that, he has a single hidden variable that is a unit vector. He is able to rule out that. I edited the post to reference this.

My point is that clearly, 1 hidden variable can't come close enough to the quantum statistics. If fact, Bell shows that no set of HVs would work either. I am wondering how many HVs it might take to come pretty close.

Or phrased in a different way: how unique is any entangled superposition of polarizations of 2 photons? Do they come in 4 flavors, 90 flavors, 1 billion flavors? Or perhaps every one is completely unique.
 
  • #4
DrChinese said:
how unique is any entangled superposition of polarizations of 2 photons?
What is your definition of "uniqueness"? There is a continuum of entangled 2-photon states.
 
  • #5
Forget entangled photon states. How many hidden variables are needed to specify the spin of a single electron about any possible axis?
 
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  • #6
PeterDonis said:
What is your definition of "uniqueness"? There is a continuum of entangled 2-photon states.
I would agree with that, probably no two are alike (I think you are saying the continuum would be infinite). But I don't think that can be proven by direct experimental means.

We have been discussing the concept that in entanglement swapping scenarios, where there are independent sources for 2 initial entangled pairs (1 & 2) and (3 & 4): That in some of them the 1 & 4 pairing was also entangled to start with. (Of course I reject that entirely.)

So I wondered: If there were only a small number of possible entangled pairs, would that support the hypothesis that there could be a subset where the entanglement between (1 & 2) and (3 & 4) was in fact identical to begin with. But I don't believe that concept is tenable, because there must be a large number of permutations - which would then have too low of a probability of getting matched together.

PeroK said:
Forget entangled photon states. How many hidden variables are needed to specify the spin of a single electron about any possible axis?

I would say a large number (hypothetically assuming spin outcomes are predetermined). Suppose you start with an electron oriented Up in the X direction. You need it to follow the cos^2(theta/2) rule across a range of angles when you perform a second measurement. But it must follow a outcome map with randomness thrown in. And if there are a pair of entangled electrons, each must possess an identical map.
 
  • #7
DrChinese said:
probably no two are alike (I think you are saying the continuum would be infinite).
AFAIK, yes. Basically, you can form a continuous infinity of linear combinations of Bell states, all of which (except possibly for a set of measure zero) are entangled. (I don't think they're all maximally entangled, but they're all entangled to some degree).

DrChinese said:
I don't think that can be proven by direct experimental means.
In the sense that in practice state tomography only has finite accuracy, yes. But there is no theoretical limit to how accurate it can be.
 
  • #8
PeterDonis said:
AFAIK, yes. Basically, you can form a continuous infinity of linear combinations of Bell states, all of which (except possibly for a set of measure zero) are entangled. (I don't think they're all maximally entangled, but they're all entangled to some degree).In the sense that in practice state tomography only has finite accuracy, yes. But there is no theoretical limit to how accurate it can be.
Don't think we are referring to the same scenario.

If I have a pair of maximally entangled photons (1 & 2) in a specific Bell state from Source A (here and now), and another pair of maximally entangled photons (3 & 4) in the same specific Bell state from Source B (located anywhere else at some other time): what is the likelihood they are in an identical superposition prior to measurement (again assuming hypothetically there are hidden variables) ? I.e. that they would yield identical outcomes at any possible angle setting for all 4. I say there are too many possible permutations (if such existed) for there to be much of a chance. (Of course, there's probably no chance because I would say no 2 superpositions would ever quite be the same.)

So I am trying to work through the logic of those who assert that there is common entanglement in some pairs from different sources. I don't think their argument stands up to scrutiny, and leads to contradictory assertions. So I am hoping someone who thinks that different entangled pairs (from different sources) can be mutually entangled (which they could appear to be, if there were a finite number of "flavors" of entanglement) will step up and defend that position. For their position to make sense, the number of flavors of 2 photon entanglement in a specific Bell state must be very small - perhaps 4. Anything over perhaps 20 is probably untenable.
 
  • #9
DrChinese said:
I would agree with that, probably no two are alike (I think you are saying the continuum would be infinite). But I don't think that can be proven by direct experimental means.
It's not a question of proof. It's a question of the idea of hidden variables being dubious from the outset.
DrChinese said:
We have been discussing the concept that in entanglement swapping scenarios, where there are independent sources for 2 initial entangled pairs (1 & 2) and (3 & 4): That in some of them the 1 & 4 pairing was also entangled to start with. (Of course I reject that entirely.)

I would say a large number (hypothetically assuming spin outcomes are predetermined). Suppose you start with an electron oriented Up in the X direction. You need it to follow the cos^2(theta/2) rule across a range of angles when you perform a second measurement. But it must follow a outcome map with randomness thrown in.
The ##cos^2## is the expected value. That itself cannot be a hidden variable, but the distribution of hidden variables in each orientation.

You either have a very large number or an infinity of individual variables. Or, some sort of cut-off point.

Even if it doesn't succumb to an analytic proof like Bell's inequality, it's very odd that the distribution agrees with QM and that ##\cos^2## term.

IMO, if you try to take hidden variables seriously for a single electron you have an inexplicable distribution of expected values.
 
  • #10
DrChinese said:
If I have a pair of maximally entangled photons (1 & 2) in a specific Bell state from Source A (here and now), and another pair of maximally entangled photons (3 & 4) in the same specific Bell state from Source B (located anywhere else at some other time): what is the likelihood they are in an identical superposition prior to measurement (again assuming hypothetically there are hidden variables) ?
Um, 100 percent, because you specified the states that way?
 
  • #11
PeroK said:
It's not a question of proof. It's a question of the idea of hidden variables being dubious from the outset.

...

You either have a very large number or an infinity of individual variables. Or, some sort of cut-off point.
I was hoping to get your idea of a number. If it is very large, as you say, then the permutations would be extremely large. My point is that IF there were any possibility that the superpositions from 2 different entangled sources were identical (i.e. the same permutation, each yielding the exact same outcomes at all possible angles, but coincidentally so), that likelihood would be so low as to effectively be zero.

It would be like independently shuffling 2 decks of 52 cards and discovering they were in the same order (that would be about the same as 225 hidden bits, i.e. 1 chance in 2^225, on the order of 1 in 10^67).

PeterDonis said:
Um, 100 percent, because you specified the states that way?
Not much chance (of course we are using the term "identical" in different senses)!

I can have 2 independent sources of entangled photons, A (for entangled photons 1 & 2) and B (for 3 & 4), created by Alice and Bob in an identical Bell state from Type I BBo. Although the pairs are in identical Bell states (as we agree), they are no more identical than a deck of cards in terms of the expected outcomes for various polarization tests. If all 4 are tested at any identical angles, say 0 degrees: 1 & 2 are always the same outcome, 3 & 4 are always the same outcome, but 1 & 4 only have a 50-50 chance of having the same outcome at that single angle. And that is true for any angle we choose. If these pairs were clones of each other, that would not be true. Their quantum states are described the same, yes, but their observables will not all have the same values upon measurement. And for our purposes, we need them to generate the same outcomes (if they are pre-existing).

And keep in mind that we can teleport an unknown quantum state, even an entangled one. 1 & 2 are in an unknown quantum state that is NOT the same exact unknown state as 3 & 4. But we can teleport the 2 state to the 4 state - we wouldn't need to do this if they were already the same up and down the line.

So again: how many bits would it take (if there were hidden variables) to fully describe the prepared/pre-existing state of an entangled photon pair? What does your napkin say?
 
  • #12
DrChinese said:
Although the pairs are in identical Bell states (as we agree), they are no more identical than a deck of cards in terms of the expected outcomes for various polarization tests. If all 4 are tested at any identical angles, say 0 degrees: 1 & 2 are always the same outcome, 3 & 4 are always the same outcome, but 1 & 4 only have a 50-50 chance of having the same outcome at that single angle. And that is true for any angle we choose. If these pairs were clones of each other, that would not be true.
Ah, I see. I'm not sure the underlying question here is answerable, since AFAIK, while, as you say, we have had posters in previous threads here try to account for Bell inequality violations and entanglement swapping by arguing that it's somehow just due to statistics (claims with which I don't agree any more than you do), none of them have any actual model that backs up those claims. (As I argued in one of those threads, cherry picking results after the fact is not a model. If I flip a coin 1000 times and pick out all the heads after the fact, that doesn't mean there was some kind of "pre-existing correlation" at work.)

More generally, while the best that current QM can do is predict probabilities, and it can't tell us why results come out the way they do in a particular case (for example, why the 1&2 and 3&4 results only match 50 percent of the time in the scenario you describe), nobody knows why that is. Is it because there is some underlying model that we just haven't found yet? (Bohmians would disagree--they would say we do have such a model, it just grossly violates our intuitions about what such a model "should" be like. But even Bohmians have to fall back on things like the initial distribution of Bohmian particle positions just happening to turn out a certain way so that the quantum probabilities are satisfied; there is no deeper explanation for that.) Or is it because that uncertainty is built into nature at the most fundamental level and will never have any deeper explanation than "that's just the way it is"?

In short, to even try to come up with numbers to answer your question, we would need to have some kind of underlying model that could account for the results. And nobody does.
 
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  • #13
PeterDonis said:
Ah, I see. I'm not sure the underlying question here is answerable, since AFAIK, while, as you say, we have had posters in previous threads here try to account for Bell inequality violations and entanglement swapping by arguing that it's somehow just due to statistics (claims with which I don't agree any more than you do), none of them have any actual model that backs up those claims. (As I argued in one of those threads, cherry picking results after the fact is not a model. If I flip a coin 1000 times and pick out all the heads after the fact, that doesn't mean there was some kind of "pre-existing correlation" at work.)

...

In short, to even try to come up with numbers to answer your question, we would need to have some kind of underlying model that could account for the results. And nobody does.

:smile:

Well, I'm hoping all those folks who were quick to respond that "entanglement swapping is merely post-selection" would come forth with something to address my point. They all flopped when I pointed out that Monogamy of Entanglement prevents exactly that argument. Not a single poster could dispute that point, as you know. And despite every attempt I make to try to apply their approach to the case of independent/distant swapping examples, I cannot find a single such approach that fits with the experimental facts. I have read Griffiths' (and others) attempts to explain Bell entanglement swapping vis a vis Consistent Histories, for example, and to say it is hand-waving is kind. Frameworks (or agents, or updating of knowledge, or whatever), really? Experimental efforts have moved far past these weak attempts to dismiss quantum nonlocality. We need to accept that something IS occurring that violates Einsteinian causality, and move forward from that point. Where and how, I don't know, but these outdated interpretations need to be left behind. So...

Where's the beef? If we have (1 & 2) entangled and (3 & 4) entangled, then how do we get (1 & 4) entangled if nothing remote ever happened? Because if there are many different entangled permutations as described in the OP (1 billion or more) then those won't come along frequently enough to match existing experiments. They see a successful Bell swap far more often, perhaps near 100%* (not all of the 4 equally likely Bell states can be fully discriminated though). If the number of entanglement "permutations" is more than a very small number, then the swap would practically never occur.*100% being the cases where there are 2 clicks within the specified coincidence window, so that the source is indistinguishable. That ignores the cases where there is only 1 click within the coincidence window.
 
  • #14
DrChinese said:
I was hoping to get your idea of a number. If it is very large, as you say, then the permutations would be extremely large.
If we prepare an electron the "z-up" state, then we have a QM model that explains the ##\cos^2(\dfrac \theta 2)## probability distribution for spin measurements about other angles. There's no clear HV model that would produce this. Einstein never had to justify the hidden variable model, or submit an alternative theory to scrutiny. Eventually Bohmian mechanics achieved something like this - but with some elements that went beyond the simple HVs.

You could propose a HV model for the electron spin, where (for each electron) there is a cut-off ##\theta##: less than that and we get spin-up; greater than that and we get spin-down. And, ##\theta## needs to be chosen with the correct probability to align with experiment. This entails something equivalent to that part of the QM mathematics.

A secondary problem is addition of spin angular momentum, where adding pre-defined probabilities won't give the correct distribution. You need to add the amplitudes and then calculate the probabilities.

This is why, IMO, so many physicists abandoned the classical HV ideas and followed orthodox QM - because the experimental results didn't conform to simply adding probabilities classically.

This is the point I was making. Eventually the tests of Bell's Theorem were the death-knell for HV's. But, the writing was on the wall in terms of the theory of electron spin and addition of angular momentum.
 
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  • #15
DrChinese said:
How many (hidden) binary bits would be required to model entangled photon polarization, given that a typical BBo crystal produces entangled photon pairs with perfect correlations at any angle setting.
Infinitely many.

After all, one qubit contains, in principle, infinitely many bits.
 
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  • #16
Demystifier said:
After all, one qubit contains, in principle, infinitely many bits.
Let me just state that I disagree with that statement. A finite volume of space at a finite energy can only contain a finite amount of information. A probability of ##1/\pi## doesn't contain infinitely many bits either.

(I have not followed this thread. So if somebody wants to discuss about that stuff, I guess it should be done in a separate thread, if at all.)
 
  • #17
gentzen said:
A finite volume of space at a finite energy can only contain a finite amount of information.
Do you have an argument for this statement?

Note that Bekenstein-Hawking bound (BHB) would not count as a good argument. The BHB essentially claims that the dimensionality of the Hilbert space associated with a system of finite size is finite. However, the qubit is associated with a 2-dimensional Hilbert space, so the infinite (or possibly finite but very large) amount of bits it contains has nothing to do with dimensionality of the Hilbert space.
 
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  • #18
Demystifier said:
Do you have an argument for this statement?
As I said, I don't want to discuss that stuff in this thread. Whether I want to discuss it at all is a different question. From my point of view, we still have this "monogamy of entanglement" argument in the room, where I am pretty convinced that I understand extremely well why it tells us something completely different than DrChinese has in mind. But even there, it would take me a huge amount of time and energy to try to explain it, and it would still likely fail. I write this with confidence, because I see others invest that time and energy, with very little progress so far.
 
  • #19
gentzen said:
A probability of ##1/\pi## doesn't contain infinitely many bits either.
That's because the number ##1/\pi## has finite Kolmogorov complexity. But almost all real numbers in the interval ##[0,1]## have infinite Kolmogorov complexity. Those they don't are a subset of measure zero.
 
  • #20
Demystifier said:
That's because the number ##1/\pi## has finite Kolmogorov complexity. But almost all real numbers in the interval ##[0,1]## have infinite Kolmogorov complexity. Those they don't are a subset of measure zero.
You nicely spotted one of my lines of defence. (I certainly didn't expect this.) But still, I don't want to discuss that stuff in this thread, and won't elaborate any further here.
 
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  • #21
gentzen said:
As I said, I don't want to discuss that stuff in this thread. Whether I want to discuss it at all is a different question. From my point of view, we still have this "monogamy of entanglement" argument in the room, where I am pretty convinced that I understand extremely well why it tells us something completely different than DrChinese has in mind. But even there, it would take me a huge amount of time and energy to try to explain it, and it would still likely fail. I write this with confidence, because I see others invest that time and energy, with very little progress so far.
So what precisely means "monogamy of entanglement". I'd prefer a clear mathematical statement without philosophical fog, if that's possible!
 
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  • #22
Demystifier said:
That's because the number ##1/\pi## has finite Kolmogorov complexity. But almost all real numbers in the interval ##[0,1]## have infinite Kolmogorov complexity. Those they don't are a subset of measure zero.
That's why there's an argument that real numbers are perhaps unphysical and natural quantities ultimately form only a finite or countable subset of the real numbers.
 
  • #23
vanhees71 said:
So what precisely means "monogamy of entanglement". I'd prefer a clear mathematical statement without philosophical fog, if that's possible!
What I have in mind is that each single one of two maximally entangled qubits will act like a totally independent unbiased random bit in interactions with other qubits. (As long as only one of the two qubits takes part in those interactions.) It is a property of the relation of those two maximally entangled qubits to other qubits, not an "objective property" of the two qubits themselves in isolation.
 
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  • #24
Sure, the state of a system which is part of a large system is simply given by tracing out the part which you are not interested in:
$$\hat{\rho}_A=\mathrm{Tr}_B \hat{\rho}_{AB}.$$
 
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  • #25
vanhees71 said:
Sure, the state of a system which is part of a large system is simply given by tracing out the part which you are not interested in:
$$\hat{\rho}_A=\mathrm{Tr}_B \hat{\rho}_{AB}.$$
That is indeed one good way to see why each single one of two maximally entangled qubits will act like a totally independent unbiased random qubit in interactions with other qubits.

So it seems like I basically "put the independence in by hand," because saying that "the remaining qubit does not take part in those interactions" is the same as saying that "the other qubit is totally independent of those other qubits". But after now reflecting about this, I think that this impression is wrong. The two maximally entangled qubits are really independent of all other qubits. That independence is already there, it is not "manually put in by hand".
 
  • #26
They are not independent in the sense that you can use measurements on ##B## to select subensembles, for which ##A## is in a different state, if ##A## and ##B## are entangled.

Take as the most simple example two particles with spin 1/2, being prepared in the Bell singlet state (I'm only looking at the spin)
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle-|-1/2,1/2 \rangle), \quad \hat{\rho}_{AB}=|\Psi \rangle \langle\Psi|.$$
Now indeed you can easily calculate
$$\hat{\rho}_A=\mathrm{Tr}_B \hat{\rho}_{AB}=\frac{1}{2} \hat{1},$$
i.e., you have a fully unpolarized particle when measuring particle ##A## (or letting it collide with other particles and measuring the cross sections etc.).

If you now measure particle ##B## and only select those where you get ##\sigma_z=+1/2## and then experiment further with the particles ##A## of the corresponding subensemble, in this subensemble, particle ##A## is prepared to have a definite spin component ##\sigma_z=-1/2##, i.e., when selecting (or even post-selecting) only this subensemble then particle A is fully polarized with ##\sigma_z=-1/2##.

There's nothing put in "by hand" in any way. It's just the possibility to prepare particles in entangled states.
 
  • #27
vanhees71 said:
So what precisely means "monogamy of entanglement". I'd prefer a clear mathematical statement without philosophical fog, if that's possible!
@vanhees71 I expect you have seen something like this, which is the "standard" starting point for MoE (eq. 12 in the reference):

https://arxiv.org/abs/quant-ph/9907047
$$τ{ab} + τ{ac} ≤ τ{a(bc)}.$$
It has been proved that the CHSH is monogamous: if three parties A, B and C share a quantum state and each chooses to measure one of two observables, then the trade-off between AB’s and AC’s violation of the CHSH inequality is given by:

|Tr(βABCHSHϱ)| + |Tr(βACCHSHϱ)| ≤ 4.

So if AB violate the CHSH inequality then AC cannot. In our examples of swapping: If (1 & 2) violates a Bell Inequality, then (1 & 4) cannot. And vice versa. That's the math.

When we see the CHSH violated by the (1 & 4) pairs in a known Bell state (after BSM events), that implies that there has been an objective change to original (1 & 2) state - because in the original state (1 & 4) cannot be entangled due to MoE. Unless of course you don't accept the reference as valid, or disagree with how MoE is applied here. But this is pretty much textbook, and is probably the most simplistic application you will ever see.
 
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  • #28
Thanks. Indeed here we look only at entangled pairs, which themselves are not entangled. Three-photon entanglement is another subject. Perhaps it's better to start another thread, if you like to discuss related issues like the GHZ experiment?
 
  • #29
vanhees71 said:
Thanks. Indeed here we look only at entangled pairs, which themselves are not entangled. Three-photon entanglement is another subject. Perhaps it's better to start another thread, if you like to discuss related issues like the GHZ experiment?
The Monogamy rules/measures/formulae get very complicated very quickly when you start looking at GHZ states and other higher dimension entanglement. None of the photons we are discussing are in the GHZ state.

My post #27 applies to our discussion topic, swapping, and 2 particle entanglement. Again, the basic rule is: If 1 & 2 are (maximally) entangled, there is no other particle in the universe either can also be entangled with. Two particle entanglement. Similarly, if 1 & 4 are maximally entangled, there is no other particle in the universe they can also be entangled with. Two particle entanglement. One or the other, there is only so much entanglement to go around. If 2 are entangled maximally (as CHSH shows), there is nothing left to share with others.

In conclusion: there is no "coincidental", "residual" or other correlation between 1 & 4 initially - as you have stated yourself. That should be obvious to everyone here, as there is absolutely no connection between pairs from distant independent sources that have never interacted.

And monogamy of entanglement means that something must interact to cause components of otherwise independent biphotons (1 & 2) and (3 & 4) to evolve into an entirely new biphoton (1 & 4).
 
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  • #30
DrChinese said:
The Monogamy rules/measures/formulae get very complicated very quickly when you start looking at GHZ states and other higher dimension entanglement. None of the photons we are discussing are in the GHZ state.
Exactly, and that's why I suggest to open another thread if you are interested in discussing it.
DrChinese said:
My post #27 applies to our discussion topic, swapping, and 2 particle entanglement. Again, the basic rule is: If 1 & 2 are (maximally) entangled, there is no other particle in the universe either can also be entangled with. Two particle entanglement. Similarly, if 1 & 4 are maximally entangled, there is no other particle in the universe they can also be entangled with. Two particle entanglement. One or the other, there is only so much entanglement to go around. If 2 are entangled maximally (as CHSH shows), there is nothing left to share with others.
Of course, that's the math of entanglement, but what's the point?

One can discuss this only in terms of math, if one doesn't want to run into the problem of frequently misunderstanding each other. Let me repeat the arguments from the other thread. We discuss an experiment like the one described by Jennewein et al in

https://arxiv.org/abs/quant-ph/0201134

vanhees71 said:
It's more easily seen in the 2nd-quantization notation. The four Bell states of a photon pair with momentum labels ##j## and ##k## are created from the Vakuum by
$$\hat{\Psi}_{jk}^{\dagger \pm} \rangle=\frac{1}{\sqrt{2}}[\hat{a}^{\dagger}(\vec{p}_j,H) \hat{a}^{\dagger}(\vec{p}_k,V) \pm \hat{a}^{\dagger}(\vec{p}_j,V) \hat{a}^{\dagger}(\vec{p}_k,H)],$$
$$\hat{\Phi}_{jk}^{\dagger \pm} \rangle=\frac{1}{\sqrt{2}}[\hat{a}^{\dagger}(\vec{p}_j,H) \hat{a}^{\dagger}(\vec{p}_k,H) \pm \hat{a}^{\dagger}(\vec{p}_j,V) \hat{a}^{\dagger}(\vec{p}_k,V)].$$
Then the initial four-photon state can then be written in two forms,
$$|\Psi_{1234} \rangle = \hat{\Psi}_{12}^{\dagger -} \hat{\Psi}_{34}^{\dagger-} |\Omega \rangle,$$
but this can as well be written as
$$|\Psi_{1234} \rangle=\frac{1}{2} (\hat{\Psi}_{23}^{\dagger +} \hat{\Psi}_{14}^{\dagger +} - \hat{\Psi}_{23}^{\dagger -} \hat{\Psi}_{14}^{\dagger -}-\hat{\Phi}_{23}^{\dagger +} \hat{\Phi}_{14}^{\dagger +} + \hat{\Phi}_{23}^{\dagger -} \hat{\Phi}_{14}^{\dagger -})|\Omega \rangle.$$
The former notation shows that photon pairs (12) and (34) are each in the polarization-singlet Bell state but (14) and (23), i.e, are uncorrelated.

The latter notation shows that if you project pair (23) to either of the four Bell state the pair (14) must be found in the same Bell state. In Pan et al's work, which we discuss here, (23) has been projected to the polarization-singlet state, and it has been demonstrated that then also the pair (14) is then in the same polarization-singlet state. This happens with probability 1/4.
DrChinese said:
In conclusion: there is no "coincidental", "residual" or other correlation between 1 & 4 initially - as you have stated yourself. That should be obvious to everyone here, as there is absolutely no connection between pairs from distant independent sources that have never interacted.
In the subensemble due to the projection measurement on 2&3 you have also 1&4 in a Bell state. That's a fact from the above calculation as well as the experiment. That's after all what entanglement swapping means. It's a sophisticated version of "teleportation".

Of course also your socalled "monogamy theorem" is fulfilled since obviously in the subensemble 1&2 as well as 3&4 are now completely uncorrelated. That's why it's called "swapping".

The swapping is of course only possible, because the initial four-photon state, describing the "full ensemble" was such that 1&2 as well as 3&4 were in Bell states. The trick in all teleportation experiments is to use one photon of a maximally entangled pair and then entangle them with something else (here you take photon 2 and entangle it with photon 3) by a projection. In teleportation the original entanglement is then completely destroyed for the so prepared subensemble. I guess that's what you call "monogamy".
DrChinese said:
And monogamy of entanglement means that something must interact to cause components of otherwise independent biphotons (1 & 2) and (3 & 4) to evolve into an entirely new biphoton (1 & 4).
Of course photons 2&3 interact with the beam splitter and finally the detectors to enable you to select the subensemble by projection.
 
  • #31
vanhees71 said:
1. Exactly, and that's why I suggest to open another thread if you are interested in discussing it.

2. One can discuss this only in terms of math, if one doesn't want to run into the problem of frequently misunderstanding each other. Let me repeat the arguments from the other thread. We discuss an experiment like the one described by Jennewein et al in

https://arxiv.org/abs/quant-ph/0201134

In the subensemble due to the projection measurement on 2&3 you have also 1&4 in a Bell state. That's a fact from the above calculation as well as the experiment. That's after all what entanglement swapping means. It's a sophisticated version of "teleportation".

3. Of course also your so-called "monogamy theorem" is fulfilled since obviously in the subensemble 1&2 as well as 3&4 are now completely uncorrelated. That's why it's called "swapping".

... In teleportation the original entanglement is then completely destroyed for the so prepared subensemble. I guess that's what you call "monogamy".

Of course photons 2&3 interact with the beam splitter and finally the detectors to enable you to select the subensemble by projection.
1. I never brought up GHZ, you did.

2. Fully agreed. This is just another recap of the experiment. I will point out your statement: "...due to the projection measurement on 2&3 you have also 1&4 in a Bell state." I couldn't say it better myself. :smile:

3. The "so-called" MoE insures exactly as you say: 1 & 2 are no longer entangled, and 3 & 4 are no longer entangled, even though they previously were.

-------------------

We started with 2 biphotons. We end up with a single biphoton (1 & 4). It was created by the action of a projection measurement upon (2 & 3), exactly as you say.

However, both 1 and 4 are distant from each other when that happens (since they don't need to be in a common light cone). And the projection measurement is distant as well (far from 1, and far from 4). We are fully agreed to this point.

So the only thing you need to do is make the leap to the most obvious element here: accept the action at a distance which causes the creation of the (1 & 4) biphoton. How you are able to hand-wave this straightforward deduction away is a mystery. The BSM event creates a new and distant quantum system that did not previously exist. From the abstract of the reference: "Demonstrating quantum nonlocality for photons that never interacted... the nonlocality is confirmed by observing a violation of Bell's inequality by 4.5 standard deviations." That's the math we're talking about. In the words of the experimentalists we're talking about.

We walked up to the door together, we knocked on the door together, but only one of us is willing to step inside and say hello to quantum nonlocality. :smile: Come on in, this home is very comfortable inside.
 
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  • #32
DrChinese said:
1. I never brought up GHZ, you did.
You brought up a discussion about entanglement of more than 2 photons.
DrChinese said:
2. Fully agreed. This is just another recap of the experiment. I will point out your statement: "...due to the projection measurement on 2&3 you have also 1&4 in a Bell state." I couldn't say it better myself. :smile:

3. The "so-called" MoE insures exactly as you say: 1 & 2 are no longer entangled, and 3 & 4 are no longer entangled, even though they previously were.
So finally we agree about the mathematical facts in agreement with the experiments we discuss ;-).
DrChinese said:
-------------------

We started with 2 biphotons. We end up with a single biphoton (1 & 4). It was created by the action of a projection measurement upon (2 & 3), exactly as you say.

However, both 1 and 4 are distant from each other when that happens (since they don't need to be in a common light cone). And the projection measurement is distant as well (far from 1, and far from 4). We are fully agreed to this point.

So the only thing you need to do is make the leap to the most obvious element here: accept the action at a distance which causes the creation of the (1 & 4) biphoton. How you are able to hand-wave this straightforward deduction away is a mystery. The BSM event creates a new and distant quantum system that did not previously exist. From the abstract of the reference: "Demonstrating quantum nonlocality for photons that never interacted... the nonlocality is confirmed by observing a violation of Bell's inequality by 4.5 standard deviations." That's the math we're talking about. In the words of the experimentalists we're talking about.
There is no action at a distance. This claim simply contradicts the properaties of standard relativistic QFT, for which microcausality is enforced by construction for the very reason to avoid any action at a distance or faster-than-light signal propagation. The BSM doesn't create anything new, it just selects a subensemble, and it's a local measurement on photons 2&3. The correlations due to the entanglement of 1&4 in this subensemble are due to the fact that in the initial state 1&2 were entangled as well as 3&4 (but not (2&3) and (1&4)). The selection of one of the 4 Bell states for (2&3) implies that also (1&4) is in (in this case the same) Bell state for this subensemble. Which of the 4 Bell states in each measurement is selected, of course, is random with a probability 1/4 for either outcome.
DrChinese said:
We walked up to the door together, we knocked on the door together, but only one of us is willing to step inside and say hello to quantum nonlocality. :smile: Come on in, this home is very comfortable inside.
There are "nonlocal correlations", i.e., strong correlations between far distantly observed parts of a quantum system but no violations of Einstein causality, because in relativistic QFT this is excluded by construction, i.e., only consider such QFTs, for which the microcausality constraint for all local observables is fulfilled. You should adapt your view to the mathematical facts. Any interpretation of a theory must respect the mathematical properties of this theory!
 
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  • #33
How Many Hidden Variables Would It Take to Model Photon Polarization?

I claim that this number is larger than or equal to ##x##, where ##x## is the number of angels that can dance on the head of a pin. (SICR :oldbiggrin: )
 
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  • #34
Demystifier said:
How Many Hidden Variables Would It Take to Model Photon Polarization?

I claim that this number is larger than or equal to ##x##, where ##x## is the number of angels that can dance on the head of a pin. (SICR :oldbiggrin: )
Are angels bosons or fermions?
 
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  • #35
anyons...
 
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