I Entanglement swapping, monogamy, and realism

[Nullstein]

They most certainly are. Basing the signal on what happens at the BSM is very different from basing it on something from the initial preparation, which is what you said before.

Fine, if you want to study the more complicated situation, it just means that you are even less allowed to conclude causal relationships from the collected data. It means you need to to perform even more checks of the causal structure and take even more possible causal relationships into account, before any conclusion can be made. If I don't want to discuss these complications, it just means I'm being generous to you.

Not quite. The point of the event ready signal in these experiments is to select those events in which photons 2 & 3 are projected into the particular Bell state that the BSM is set up to uniquely detect. That doesn't mean the other events aren't successful measurements; they're just measurements whose results can't be distinguished by the humans reading the output of the apparatus. But they still involve projecting photons 2 & 3 into a Bell state and everything associated with that (for example, that photons 1 & 4 are also projected into a Bell state).

Well, I agree, but this just means that you need to perform an even more complicated analysis of the experimental setting, before you can perform proper causal inference. So it just got more complicated for you to argue for your position. Me agreeing to discuss the idealized experiment rather than the real world experiment is a generosity, not an excuse.

What you need to do to infer a causal relationship now is to draw a causal diagram and take all the effects into account that I explained in post #150. You then calculate the corrected correlations that account for these effects and you'd have to show that there is still correlation after the effects have bene taken into account.

 

[PeterDonis]

The paper https://arxiv.org/abs/1606.04523 shows that this is also well understood in the case of entanglement swapping.

Where is entanglement swapping discussed in the paper? I see a "swap" operator defined, but I don't see how it relates to what the BSM does in the entanglement swapping experiment we have been discussing.

 

[PeterDonis]

if you want to study the more complicated situation

What more complicated situation? I wasn't comparing two situations, I was correcting your misstatement about the "event ready" signal in the one situation we have been discussing all along.

 

[Nullstein]

Where is entanglement swapping discussed in the paper? I see a "swap" operator defined, but I don't see how it relates to what the BSM does in the entanglement swapping experiment we have been discussing.

The swap gate is a mathematical abstraction of the physical situation in entanglement swapping. It abstractly models the transition that we previously denoted by $\rho_{1234,\text{before}}\rightarrow\rho_{1234,\text{after}}$ in terms of a 4 qubit system. The paper applies to all such swapping experiments, independent of details such as e.g. the temporal and spatial ordering. (As discussed earlier, the temporal and spatial ordering has no effect on the measured statistics, so it can be abstracted away.)

What more complicated situation? I wasn't comparing two situations, I was correcting your misstatement about the "event ready" signal in the one situation we have been discussing all along.

The more complicated situation is what you get if you want to include the events, that did not constitute a proper BSM projection, into the description. In that case, you would need to introduce another variables X = "Has a proper BSM projection occured?" and study the causal graph that includes this variable, taking into account the issues mentioned in #150.

 

[PeterDonis]

It abstractly models the transition that we previously denoted by $\rho_{1234,\text{before}}\rightarrow\rho_{1234,\text{after}}$ in terms of a 4 qubit system.

I only see two qubit systems discussed in the paper.

 

[vanhees71]

I think so too. The only possible difference is that Fuchs and Peres are also willing to talk about the likelihood of single events as well as frequencies of events in ensembles, which I am not sure vanhees is willing to do.

"When we are told that the probability of precipitation tomorrow is 35%, there is only one tomorrow." -- Fuchs and Peres

That's the problem I have with interpretations of probability, no matter whether they are predicted by QT or other means of calculating statistical predictions (e.g., by classical statistical physics), which abandon the "frequentist interpretation", like "QBists".

What does it mean that the probability for rain tomorrow is 35%? Of course there's only one tomorrow, and I can't prepare the wheather situation today many times and check the frequency that it rains or not tomorrow. For me the prediction of some weather model means that at weather condistions like today, experience and/or projections via model simulations, tells me that in 35% of all such cases it will rain. A good statistical analysis also gives a level of significance and an uncertainty of this number, which also is based on experience with such weather models, and then I have some objective estimate for the probability of rain or good weather to plan my barbecue party tomorrow. Of course also this interpretation is after all based on the "frequentist interptation" of the statistical meaning of probabilities. From a single event, you of course can not decide whether your prediction that there's 35% chance for rain tomorrow is a good estimate or not.

 

[mattt]

That's the problem I have with interpretations of probability, no matter whether they are predicted by QT or other means of calculating statistical predictions (e.g., by classical statistical physics), which abandon the "frequentist interpretation", like "QBists".

What does it mean that the probability for rain tomorrow is 35%? Of course there's only one tomorrow, and I can't prepare the wheather situation today many times and check the frequency that it rains or not tomorrow. For me the prediction of some weather model means that at weather condistions like today, experience and/or projections via model simulations, tells me that in 35% of all such cases it will rain. A good statistical analysis also gives a level of significance and an uncertainty of this number, which also is based on experience with such weather models, and then I have some objective estimate for the probability of rain or good weather to plan my barbecue party tomorrow. Of course also this interpretation is after all based on the "frequentist interptation" of the statistical meaning of probabilities. From a single event, you of course can not decide whether your prediction that there's 35% chance for rain tomorrow is a good estimate or not.

Yes, that's what it usually means. The criteria of what exactly "conditions like today" means (among several other things) is what defines the probabilistic method used in each case.

 

[vanhees71]

1&2 are in a Bell state as well as 3&4, i.e., each of these pairs is maximally entangled. There are no correlations between 2&3 or 1&4 in the initial state. I haven't claimed anything like that.

Now if you project to a Bell state of 2&3, in the so prepared subensemble 2&3 are fully entangled, because they are prepared to be in a Bell state. Due to the entanglement of 1&2 and 3&4 this implies that for this subensemble also 1&4 are in a Bell state.

It's a easy, although some elaborate calculation, as described by Jennewein et al in

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

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.

Quantum theory is inherently nonlocal, although signal locality is respected. You - as many - have formulated requirements for some definition of nonlocality that cannot be met. Specifically, you basically reject experimental nonlocality proofs which some central observer does not experience until all of the information arrives - which is limited by signal. Circular reasoning - you have proved what you assume.

I don't know, how you can insist on "nonlocality" without clearly telling what you mean by it. It's a mathematical fact that the standard relativistic QFTs (used to formulate the standard model of HEP, which includes also QED, which is the theory describing all quantum-optics experiments) is a local theory, i.e., due to the enforced microcausality conditions on local observables, there are no causal influences between space-like separated events possible, and that's what's usually understood as "nonlocality". This discussion goes again in circles :-(((.

What is clearly nonlocal, as we have discussed here: An experimentalist here can create - or not - a distant biphoton from components that have never existed in a common light cone. There really is no disputing this experimental fact as I have characterized it. If you choose to reject this because it does not demonstrate signal nonlocality, that's... circular.

An experimentalist can create such the biphoton (14) only by projection due to a local measurement on (23), and you have to do coincidence measurements on all four photons. I.e., you need a measurement protocol which enables you to select the subensemble for which (14) is entangled. This is only possible when you bring together the information contained in all measurement protocols in their future light cone. The temporal order of measurements to obtain these measurement protocols is irrelevant, which shows that there's no faster-than-light influence of the projection measurement on (23) on photons 1 and 4 necessary to explain the entanglement of the biphotons in the corresponding subensemble. As @PeterDonis has said above, you can even do the measurement on photons 1 and 4 before you perform the projection of (23) to a Bell state, i.e., by "postselection".

Logically, if there is something called "quantum nonlocality" (as I say there is) and it can be demonstrated (as it can be), but it does not feature FTL signaling: then it cannot be judged by a "macroscopic" experiment where component results must be brought together before you accept the nonlocality. Macroscopic Alice can write her results down and compare them with the measurement results of Macroscopic Bob and Victor anytime. Obviously, their combined story indicates that there is quantum nonlocality. You can't wave your hands and ignore the obvious.

That's a contradiction in it self. If there's no FTL "signalling", i.e., no causal connection between space-like separated events in your model, then your model is not "nonlocal", and standard relativistic QFT is indeed not "nonlocal", as we've discussed zillions of times now.

You can interpret any theory in any way you like. The only restriction is that your interpretation must not contradict the mathematical properties of the model you want to interpret (here it's relativistic QFT).

 

[vanhees71]

I believe their approach to probabilities is Bayesian, which treats probabilities as states of epistemic belief and so has no problem assigning probabilities to single events.

Indeed, but what's the meaning of probabilities for single events? See #156 for further details concerning my question.

 

[Fra]

That's the problem I have with interpretations of probability, no matter whether they are predicted by QT or other means of calculating statistical predictions (e.g., by classical statistical physics), which abandon the "frequentist interpretation", like "QBists".

What does it mean that the probability for rain tomorrow is 35%? Of course there's only one tomorrow

For me the important question is:
What difference does it make what the we(or an agent) THINK is going to happend tomorrow? This question is motivated by the fact that noone can ever know until its too late!

I would say the difference is that, that a rational agents actions now is guided by it's expectations of the future, it encodes now. Decisions are typically based on incomplete information, so it's essentially a game of expectations. That in a nutshell is my view. And I think essentiall the qbist view as well.
Here one could even argue that wether the expectations are right or wrong, are in fact completely irrelevant. A badly inform agent, will indeed be expected to make "irrational decisions" judged by someone that has different expectations - which should be in principle observable even! This is not a "problem", it's part of the game and learning process.

In qbism the probability is guiding, rather than descriptive. Obviously we can not describe the future, only the past. And there is no deductive inference of the future from the past (or present). And the best inference seem unavoidably relative.

/Fredrik

 

[Fra]

Indeed, but what's the meaning of probabilities for single events? See #156 for further details concerning my question.

I think to elaborate on this in detail. Ie. to "explain" how one can infer a guiding probability from "emprics" without huge amounts of repeats and preparatations, may require a new formalism that unavoidably becomes speculative and involving hypothesis that doesn't belong here.

This is one of the things I expect from a future theory. And while you usually don't associate to it, for me it naturally relates to QG, as the cosmological perspective vs subatomic perspective exactly illustrates the difference between statistics that we can understnad as ensembles, and generalized probability that we can understand without ensembles/prepartion paradigm, and that makes sense even when the ensemble production cant be realized.

/Fredrik

 

[Simple question]

That's a contradiction in it self.

No, that was sound reasoning.
A contradiction is to name "local" a principle like "space-like observable should commute".
The contradiction is in bold.

If there's no FTL "signalling", i.e., no causal connection between space-like separated events in your model,

That's not what FLT "signaling" means. "FLT signaling" means "FLT signaling": communication with no delay.

standard relativistic QFT is indeed not "nonlocal", as we've discussed zillions of times now.

Aside being based on your philosophical misconception of its premises, and AFAIK, standard relativistic QFT did not get rid of entanglement and inseparability. So spatially-extended (non-local) two-particle properties is the bread and butter of QM.

You can interpret any theory in any way you like.

No you cannot. You have to use logic and reason. Unless your prefer bad philosophy and running in circle.

The only restriction is that your interpretation must not contradict the mathematical properties of the model you want to interpret (here it's relativistic QFT).

Mathematical facts don't exist in Nature. Mathematical truth exist in heads. Facts are collected in the laboratory, where your claims of the purported "mathematical facts" have been proved wrong.

 

[DrChinese]

1. These events taken together, i.e. prior to post-selection, constitute the full ensemble.

The photons 1&4 will be uncorrelated in the full ensemble. The statement is true for the post-selected subensembles only.

If you don't look at a subensemble, there will not be entanglement in the data. The correlation will be zero. This follows strictly from the math.

1. No, the full ensemble is maximally entangled. You just don't know how. If you run an experiment, and can identify 1/2 the cases as to Bell state (which is feasible), that portion will violate the CHSH bound. The other half will not show any correlation, as you say. That does not mean they aren't entangled, QM predicts they are.

2. You are being imprecise on purpose here. Quantum non-locality means nothing more than the fact that Bell's inequality is violated.

3. What you want to imply is that there are non-local cause and effect relationships. The broad consensus among physicists is that no non-local cause and effect relationship can be inferred from the data. QM can be interpreted either way.

2. Quantum nonlocality is a lot more than violating a Bell Inequality! The predictions of QM embed the nonlocality. Specifically: cos^2(Alice-Bob) prediction is inherently nonlocal, as Alice and Bob and their measurement settings need not be local to each other. Note that there are no other variables in this equation.

And as we have demonstrated repeatedly in our extended discussions: the particles that Alice and Bob are observing need never have interacted or even overlapped in a common light cone. That's a lot of nonlocality! And how do quantum repeaters work in an extended quantum network? They are built on this nonlocality.3. I never said there was a cause-and-effect relationship in the Einsteinian sense, because there isn't. The action at a distance: a) does not change when the ordering changes (effect precedes cause); and b) and the effect cannot be said to occur at any specific time. You can make a statement about the final context, and say the swap created the entangled 1 & 4 pair. But that is mostly a convenience in describing the situation.

There are other ways to describe it, but since you always have the option to execute a swap or not: this is the most useful. The decision to execute the swap gives you something to point to as the "cause" even though it may be after the "effect". If you feel more comfortable saying the "spontaneous" creation of the (1 & 4) biphoton causes the BSM to succeed, then that is your choice and is just as correct in the final analysis. But again, I would not say this description is useful because the BSM is where the "on/off switch" is located.

 

[PeterDonis]

what's the meaning of probabilities for single events?

Read the Bayesian literature for a detailed answer to this question. The short answer is, it's a measure of a degree of belief. If you want to tie it to something concrete, think of betting odds: a 35% probability of rain tomorrow means that a person would pay 35 cents for a bet that pays 1 dollar if it rains tomorrow.

 

[DrChinese]

This just in! From an experiment described in a paper deposited 15 Jan 2023:

https://arxiv.org/abs/2301.06091

"Photon-pair sources based on spontaneous parametric down-conversion (SPDC) serve as such a resource of entanglement, as they can be matched in frequency and bandwidth to the ionic transitions [4–7]. When entanglement between distant nodes has been established, quantum teleportation [8] provides a suitable protocol, alternative to direct transmission, for quantum state transfer [9]. Application of teleportation has been shown between photons over large distances [10–15], between quantum nodes [16–20], and from a photon to a quantum node [21–24]. In this paper we show quantum teleportation of an atomic qubit, encoded in the Zeeman sub-levels of the 40Ca+ ion, to the polarisation qubit of a single 854 nm photon. The teleportation protocol employs heralded absorption of a single photon by the ion [4, 9, 25, 26], which functions as a Bell measurement. In contrast to photonic teleportation protocols, this Bell measurement can identify all four Bell states."

All 4! Where's the subensemble "out" now?

They ran 500 million swaps (teleportations) over a 50 hour period. They identified all 4 Bell states with approximately 81% fidelity.

Pure optical setups may be limited to discerning a maximum of 2 Bell state due to limitations inherent in Avalanche Photon Detectors (APD). That's because a single APD cannot distinguish 2 photons that arrive in close proximity from a single photon, which would be necessary to identify 2 of the 4 Bell states. They can't reset fast enough - at least not at this time.

But technology advances, and red herrings fall. There is nothing about the BSM (pre- or post-) selection process that is can be considered "cherry picking".

As I was saying in the entanglement swapping scenarios: All of the cases (in principle) where we have (1 & 4) matched clicks indicate entanglement swapping. This can be seen in this experiment, although it is expressed differently than in the swaps we have been discussing. Hopefully everyone can see clearly: there are no (1 & 2) pairs where there is "coincidental" or "accidental" entanglement between 1 & 4 waiting to be revealed. If there is no swap (teleportation), there is no entanglement to be seen. If and only if...

 

[akvadrako]

All 4! Where's the subensemble "out" now?

There are 4 subensembles. You still need to know which runs belongs to which to get entanglement.

This doesn't change anything about the discussion; as several people have pointed out, it would have been clearer to just consider the ideal experiment, where there is no need for an event ready signal.

 

[DrChinese]

1. There are 4 subensembles. You still need to know which runs belongs to which to get entanglement.

2. This doesn't change anything about the discussion; as several people have pointed out, it would have been clearer to just consider the ideal experiment, where there is no need for an event ready signal.

1. And they did. Case closed.

2. The original post-selection "out" was that only *some* of the pairs were being considered. Most of the optical versions could only identify 1 or 2 of the Bell states, and some folks saw that as explaining that only the "cherry-picked" sample would show entanglement - but the full data set would not.

Obviously: If you can identify the entanglement of the full universe, then the cherry-picking argument fails. It never made sense anyway, as *all* of the 1 & 4 pairs are entangled according to the predictions of QM (perfect case of course). So unless you have alternate and different predictions to make, I don't see the point of that line of reasoning.

I hope the debate about "subensembles" will shift to something more appropriate. Like the obviously nonlocal nature of entanglement swapping in which a CHSH inequality is violated by systems that have never been in a common light cone (and need not have ever even communicated with systems in a common light cone).

 

[mattt]

For each of the four (1,4) subensembles, (1,4) will violate a Bell inequality (after all, for each of those (1,4) subensembles, the state that describes its statistics is, respectively, a maximally entangled state).

But for the four subensembles taken together, its state (which describes its statistics) tells you that it will not violate a Bell inequality (the whole (1,4) ensemble).

 

[Morbert]

1. No, the full ensemble is maximally entangled. You just don't know how. If you run an experiment, and can identify 1/2 the cases as to Bell state (which is feasible), that portion will violate the CHSH bound. The other half will not show any correlation, as you say. That does not mean they aren't entangled, QM predicts they are.

This distinction you are implying (unentangled pairs as distinct from hidden entangled pairs) isn't inherent in the formalism. It's also a double-edged sword. The consistent histories interpretation lets us split the full ensemble into subensembles of entangled (14) pairs even if the BSM on (23) is never performed.

 

[akvadrako]

2. The original post-selection "out" was that only *some* of the pairs were being considered. Most of the optical versions could only identify 1 or 2 of the Bell states, and some folks saw that as explaining that only the "cherry-picked" sample would show entanglement - but the full data set would not.

You are missing the point. It was always an aspect of the idealized experiment that you select 4 sub-ensembles, each entangled. And nobody was arguing that it was only the non-ideal experiment that allowed an "out" due to low fidelity.

One still needs to "cherry-pick" those 4 sub-ensembles to show entanglement. If you perform the entanglement swapping but then throw away the information of which runs are in which sub-ensemble, you will never be able to show any entanglement.

Which I would say is the same as saying there is no entanglement, because entanglement is not an objective property of pairs of systems, but a statistical correlation. It's some subjective information. The monogomy of entanglement means you can never have information like pairs A&B are fully entangled while B&C are fully entangled.

 

[DrChinese]

For each of the four (1,4) subensembles, (1,4) will violate a Bell inequality (after all, for each of those (1,4) subensembles, the state that describes its statistics is, respectively, a maximally entangled state).

But for the four subensembles taken together, its state (which describes its statistics) tells you that it will not violate a Bell inequality (the whole (1,4) ensemble).

This is one of the strangest comments I have ever seen, completely meaningless and essentially contradictory.

We're scientists. You may as well be saying the results from looking at only the 1 stream are random. True, but of no value whatsoever. I guess that means Bell tests are evidence of nothing.

 

[DrChinese]

This distinction you are implying (unentangled pairs as distinct from hidden entangled pairs) isn't inherent in the formalism. It's also a double-edged sword. The consistent histories interpretation lets us split the full ensemble into subensembles of entangled (14) pairs even if the BSM on (23) is never performed.

When the BSM is turned on, all of the 1 & 4 pairs are entangled. When the BSM is not performed, none of the 1 & 4 pairs are entangled. How many ways can I say the same thing?

 

[PeterDonis]

The consistent histories interpretation lets us split the full ensemble into subensembles of entangled (14) pairs even if the BSM on (23) is never performed.

Please give a reference for this very surprising statement.

 

[PeterDonis]

One still needs to "cherry-pick" those 4 sub-ensembles to show entanglement.

No, you don't. The subensembles for photons 1 & 4 are defined by the results of the BSM on photons 2 & 3. "Cherry picking" would be using the results on photons 1 & 4 to pick "subensembles" for photons 1 & 4.

 

[DrChinese]

One still needs to "cherry-pick" those 4 sub-ensembles to show entanglement. If you perform the entanglement swapping but then throw away the information of which runs are in which sub-ensemble, you will never be able to show any entanglement.

Which I would say is the same as saying there is no entanglement, because entanglement is not an objective property of pairs of systems, but a statistical correlation. It's some subjective information. The monogomy of entanglement means you can never have information like pairs A&B are fully entangled while B&C are fully entangled.

Again, a strange comment.

In science, we look for patterns and pattern exceptions. Yes, if you throw half of your experimental data away, then probably you won't see much of a pattern. Who does that? And why would any scientist suggest that?

This is state of the art experimentation, and if you choose to ignore the evidence of entanglement: I really have nothing more to discuss with you. Not sure why you would participate if you think looking at ALL of the data is "cherry-picking". And why would you say there is NO entanglement when all 4 Bell states shows entanglement?

 

[PeterDonis]

entanglement is not an objective property of pairs of systems, but a statistical correlation.

This is not correct as a statement about the actual math. The actual math contains an objective definition of an entangled state, and that definition has nothing to do with statistical correlation: it's an easily testable objective property of the mathematical state. Whether you think that mathematical state represents particular two-qubit systems, or only ensembles of such systems, is a matter of interpretation.

 

[PeterDonis]

This distinction you are implying (unentangled pairs as distinct from hidden entangled pairs) isn't inherent in the formalism.

Yes, it is. See my post #176 just now.

 

[Morbert]

This distinction you are implying (unentangled pairs as distinct from hidden entangled pairs) isn't inherent in the formalism. It's also a double-edged sword. The consistent histories interpretation lets us split the full ensemble into subensembles of entangled (14) pairs even if the BSM on (23) is never performed.

When the BSM is turned on, all of the 1 & 4 pairs are entangled. When the BSM is not performed, none of the 1 & 4 pairs are entangled. How many ways can I say the same thing?

In the original paper, the four bell states are $\psi^+_{ij},\psi^-_{ij},\phi^+_{ij},\phi^-_{ij}$. I will relabel them as $\psi^1_{ij},\psi^2_{ij},\psi^3_{ij},\psi^4_{ij}$ respectively.

Including the collective degrees of freedom of the BSM measurement apparatus, the experiment is characterised by the preparation $$|\Psi\rangle= |\psi^2_{12},\psi^2_{34},M^0\rangle =\sum_{i=1}^4c_i|\psi^i_{14},\psi^i_{23},M^0\rangle$$ where $M^0$ denotes the ready state of the detector (The detector registers the BSM result $\psi^i_{23}$ with the state $M^i$) and $c_i$ is +1/2 or -1/2. The measurement begins at time $0$ and concludes at time $t$. The time evolution is $$U(t,0)|\Psi\rangle =\sum_{i=1}^4c_i|\psi^i_{14},M^i\rangle$$We construct four history operators of the form $$C_i = P_{M^i}(t)P_{\psi^i_{14}}(0)$$ where $P$ are projectors. These operators are complete $$\sum_i C_i|\Psi\rangle = |\Psi\rangle$$ and consistent $$\mathrm{tr}\left[C^\dagger_iC_j|\Psi\rangle\langle\Psi|\right] = 0, i\neq j$$ and so they partition the ensemble of experimental runs into into four subensembles, resolved by the measurement outcome, each with probability $$p_i = \mathrm{tr}\left[C^\dagger_iC_i|\Psi\rangle\langle\Psi|\right]$$Each subensemble has the (14) pair in a bell state. No major controversy yet, except perhaps the history operators projecting to the bell states for (14) at time $0$, which is not an issue for CH.

Now consider an alternative "experiment", where the measurement apparatus never interacts with the (23) pair. The new time evolution is $$U(t,0)|\Psi\rangle = \sum_{i=1}^4c_i|\psi^i_{14},\psi^i_{23},M^0\rangle$$We can similarly construct the complete and consistent history operators $$C_i = P_{M^0}(t)P_{\psi^i_{14}}(0)$$that partition the experimental runs into four subensembles, each with the (14) pair in an entangled bell state. But these subensembles aren't useful to the experimenter, because they aren't resolved by the singular outcome $M^0$.

Please give a reference for this very surprising statement.

I don't know of a paper on the consistent histories interpretation that deals explicitly with entanglement swapping. But I have applied the consistent histories account of measurement presented in Roland Omnes's "Understanding Quantum Mechanics" (chapter 21) and RB Griffiths's paper here (pdf)

 

[akvadrako]

This is not correct as a statement about the actual math. The actual math contains an objective definition of an entangled state, and that definition has nothing to do with statistical correlation: it's an easily testable objective property of the mathematical state. Whether you think that mathematical state represents particular two-qubit systems, or only ensembles of such systems, is a matter of interpretation.

I think your view is contradicted by this post-selection swapping experiment, most clearly if I assume there are no FTL or backwards in time influences.

Are you trying to say that I must accept FTL influences? Because if not, the idea that entanglement is statistical correlation instead of an objective property is a self-consistent position to hold.

Also, how can entanglement be objective when it's frame dependent? From https://www.nature.com/articles/s41467-018-08155-0:

From the examples considered it is clear that the notions of superposition and entanglement are reference-frame dependent.
...
From A’s point of view B and C are entangled
...
A and B are entangled in an EPR state from C’s point of view

 

[akvadrako]

No, you don't. The subensembles for photons 1 & 4 are defined by the results of the BSM on photons 2 & 3. "Cherry picking" would be using the results on photons 1 & 4 to pick "subensembles" for photons 1 & 4.

I was just trying to use Dr. Chinese's terminology. My point was you still need to pick sub-ensembles; that's what the BSM results let you do.

Without knowing the sub-ensembles there is no entanglement, at least no possibility to show entanglement.

Edit: I think I get it now – Dr. Chinese thought the "cherry-picking" method only worked when picking 1 or 2 of the subsets of runs (I won't use the word sub-ensemble for this). This is incorrect; it works for all 4 subsets.

 

[vanhees71]

Read the Bayesian literature for a detailed answer to this question. The short answer is, it's a measure of a degree of belief. If you want to tie it to something concrete, think of betting odds: a 35% probability of rain tomorrow means that a person would pay 35 cents for a bet that pays 1 dollar if it rains tomorrow.

Yes, and this makes also sense, but it's not convincing in the application of probability theory to physics. In physics we aim at theories/models that describe/predict the outcome of experiments and observations, i.e., if I have a theory like QT that predicts probabilities for the outcome of measurements, I need to repeat that experiment under the same conditions ("preparation") many times and use statistical analysis (which itself is of course also based on probability theory) to decide, whether the prediction compares well or does not with experiment. In this sense quantum states describe ensembles rather than single realizations of measurements.

 

[vanhees71]

This is not correct as a statement about the actual math. The actual math contains an objective definition of an entangled state, and that definition has nothing to do with statistical correlation: it's an easily testable objective property of the mathematical state. Whether you think that mathematical state represents particular two-qubit systems, or only ensembles of such systems, is a matter of interpretation.

But entanglement describes correlations, i.e., to test the predictions of entanglement you need to measure the predicted correlations, e.g., by testing for the validity of Bell's inequalities vs. the predictions of QT, and indeed the outcome is objectively in favor of QT in all experiments done yet.

 

[gentzen]

Dreischner calls the account of Peres + Fuchs a minimal instrumentalist interpretation and cites Friebe (though I do not have access to the cited text). By minimal statistical interpretation, do you mean a minimal ensemble interpretation? I don't know if they are as different as you imply, but the minimal interpretation as described in this thread and others has an instrumentalist character (for example, the association of a quantum state with a preparation procedure). If there is some important distinction or misrepresentation then please be explicit.

The paper of Peres + Fuchs simply does not live up to its title. It is so often quoted only for two reasons: first of course its title, and second that it gives you a convenient straw man to attack, if you associate some interpretation you don't like (for example QBism) with that paper. (See for example the "discussion" of QBism in "Making Sense of Quantum Mechanics" by Jean Bricmont. He extensively quotes from Peres + Fuchs. But it should be obvious that Peres is not a QBist.)

I am an instrumentalist, and I want to distance myself explicitly here from that paper by Peres + Fuchs. If you want, attack my instrumentalist position, but not by claiming that this paper would have anything to do with it. I am fine if you associate me with positions explained by Roland Omnès.

 

[vanhees71]

No, that was sound reasoning.
A contradiction is to name "local" a principle like "space-like observable should commute".
The contradiction is in bold.

In theoretical physics we have clear mathematical definitions, and a QFT is called local if the microcausality constraint on local observables hold and that the Hamilton density of the theory is a local observable. Microcausality means that two local observable-operators commute if their arguments are space-like separated. For a thorough discussion of all this, see Weinberg, Quantum Theory of Fields, Vol. 1.

That's not what FLT "signaling" means. "FLT signaling" means "FLT signaling": communication with no delay.Aside being based on your philosophical misconception of its premises, and AFAIK, standard relativistic QFT did not get rid of entanglement and inseparability. So spatially-extended (non-local) two-particle properties is the bread and butter of QM.

Of course not. Any QT has entanglement and inseparability. If QFT would not enable entanglement and inseparability it would have been proven wrong by all the Bell tests we discuss here. That's not the case. Obviously you misunderstand the meaning of locality and microcausality.

No you cannot. You have to use logic and reason. Unless your prefer bad philosophy and running in circle.Mathematical facts don't exist in Nature. Mathematical truth exist in heads. Facts are collected in the laboratory, where your claims of the purported "mathematical facts" have been proved wrong.

No, but there are mathematical facts about the theories we use to describe nature, and local QFTs are local, and any interpretation must not contradict the mathematical properties of the theory that it claims to interpret. Philosophy, of course, often is vague, because it doesn't consider the mathematical properties of the theories carefully enough, and that's why it rather causes confusion than clarification of the natural sciences.

 

[Morbert]

The paper of Peres + Fuchs simply does not live up to its title. It is so often quoted only for two reasons: first of course its title, and second that it gives you a convenient straw man to attack, if you associate some interpretation you don't like (for example QBism) with that paper. (See for example the "discussion" of QBism in "Making Sense of Quantum Mechanics" by Jean Bricmont. He extensively quotes from Peres + Fuchs. But it should be obvious that Peres is not a QBist.)

I am an instrumentalist, and I want to distance myself explicitly here from that paper by Peres + Fuchs. If you want, attack my instrumentalist position, but not by claiming that this paper would have anything to do with it. I am fine if you associate me with positions explained by Roland Omnès.

What does the article get wrong re/instrumentalism beyond perhaps the conceit of the title?

 

[Fra]

This is a side note in the thread, but i couldn't resist commenting.

No, but there are mathematical facts about the theories we use to describe nature,

Yes, but they have a limited domain of corroboration. Which means that the "mathematical model" that is corroborated for subatomic systems can not be universally applied in a deductive sense. One can certainly TRY it, and see where it leads us, this is FINE and in many cases the method is extremely powerful, but it's technically a speculation or hypothesis only; it's not "mathematical truth" in any way.

and local QFTs are local, and any interpretation must not contradict the mathematical properties of the theory that it claims to interpret. Philosophy, of course, often is vague, because it doesn't consider the mathematical properties of the theories carefully enough, and that's why it rather causes confusion than clarification of the natural sciences.

There flip side of the coin is that extrapolating "mathematical models" beyond it's domain of corroboration, may cause just as much confusion!

IMO a potential example of this is extrapolating mathematical symmetry principles from the smallest things we can distinguish, to the larger scale. For example applying the paradigm of QFT to arbitrary macro scales and cosmology. It may very well be that the symmetries for small subsystems seem solid, simply because they change too slow to be nocticable.

Smolin frequently has in his books, quoted Roberto Unger who calles this the "poisoned gift of mathematics to physics". https://www.newscientist.com/article/mg19125701-100-do-the-laws-of-nature-last-forever/
Which is also in a nutsheel the core topic of this book as well; we may have reached the time in theory development (when seeking unification of all forces) where this method is reaching its limits.
https://www.amazon.com/dp/0544245598/?tag=pfamazon01-20

/Fredrik

 

[vanhees71]

This is a side note in the thread, but i couldn't resist commenting.

Yes, but they have a limited domain of corroboration. Which means that the "mathematical model" that is corroborated for subatomic systems can not be universally applied in a deductive sense. One can certainly TRY it, and see where it leads us, this is FINE and in many cases the method is extremely powerful, but it's technically a speculation or hypothesis only; it's not "mathematical truth" in any way.

I'm not sure what you mean. Of course, current QT has the limitation of not being able to consistently describe the gravitational interaction. Here we talk about QED, and we apply it to quantum-optical phenomena. There are no known limits of validity of QED in this domain yet.

There flip side of the coin is that extrapolating "mathematical models" beyond it's domain of corroboration, may cause just as much confusion!

What we discuss here is a paradigmatic example for the application of QED. Nowhere have we applied it to anything that's not well established. To the contrary, the very experiments we discuss here confirm its validity at an amazing level of accuracy and significance.

IMO a potential example of this is extrapolating mathematical symmetry principles from the smallest things we can distinguish, to the larger scale. For example applying the paradigm of QFT to arbitrary macro scales and cosmology. It may very well be that the symmetries for small subsystems seem solid, simply because they change too slow to be nocticable.

The application of QFT to macroscopic physics is also amazingly successful. Ask any solid-state physicist, how well he can describe all kinds of macroscopic matter with QFT. In cosmology it's successfully applied to the description of the evolution of the universe. Of course, the further you go towards the big bang the more uncertain it gets, but at least after a few microseconds after the big bang we know at least about the known Standard Model particles. What we don't know is the complete mechanism behind CP violation (why are we here and not being annihilated already at the level of particles with an equal amount of antiparticles) and also what Dark Matter is and how to explain the value of the cosmological constant/Dark Energy content of the Universe. That are all open questions, but nothing indicates yet that QFT fails to describe it.

Smolin frequently has in his books, quoted Roberto Unger who calles this the "poisoned gift of mathematics to physics". https://www.newscientist.com/article/mg19125701-100-do-the-laws-of-nature-last-forever/
Which is also in a nutsheel the core topic of this book as well; we may have reached the time in theory development (when seeking unification of all forces) where this method is reaching its limits.
https://www.amazon.com/dp/0544245598/?tag=pfamazon01-20

/Fredrik

Well, Smolin...

 

[gentzen]

What does the article get wrong re/instrumentalism beyond perhaps the conceit of the title?

The conceit of the title is part of the problem, because Peres intentionally played with the expectations of its readers:

Asher Peres was a master of creating controversy for the sake of making a point. For instance, in 1982 he was asked to make a nomination for the Nobel prize in physics. He nominated Israeli prime minister Menachem Begin! Asher reasoned that Begin’s decision to invade Lebanon proved him as qualified for a Nobel physics prize as he was for his earlier peace prize.
It certainly wasn’t of the same magnitude, but Asher intended to make trouble when we wrote our 2000 “opinion piece” for Physics Today. Previous to our writing, the magazine had published a series of articles whose essential point was that quantum mechanics was inconsistent—it tolerated the unacceptable “measurement problem,” and what else could that mean but inconsistency? Quantum theory would need a patch to stay afloat, the wisdom ran—be it decoherence, consistent histories, Bohmian trajectories, or a paste of Everettian worlds.
To take a stand against the milieu, Asher had the idea that we should title our article, “Quantum Theory Needs No ‘Interpretation’.” The point we wanted to make was that the structure of quantum theory pretty much carries its interpretation on its shirtsleeve—there is no choice really, at least not in broad outline. The title was a bit of a play on something Rudolf Peierls once said, and which Asher liked very much: “The Copenhagen interpretation is quantum mechanics!” Did that article create some controversy! Asher, in his mischievousness, certainly understood that few would read past the title, yet most would become incensed with what we said nonetheless. And I, in my naiveté, was surprised at how many times I had to explain, “Of course, the whole article is about an interpretation! Our interpretation!

I certainly didn't knew this, when I wrote:

The conversation started, when I quoted Heisenberg’s own explanation:

… what one may call metaphysical realism. The world, i.e., the extended things, ‘exist’. This is to be distinguished from practical realism, and the different forms of realism may be described as follows: We ‘objectivate’ a statement if we claim that its content does not depend on the conditions under which it can be verified. Practical realism assumes that there are statements that can be objectivated and that in fact the largest part of our experience in daily life consists of such statements. Dogmatic realism claims that there are no statements concerning the material world that cannot be objectivated. Practical realism has always been and will always be an essential part of natural science. Dogmatic realism, however, …

and BT replied

That’s all a bit technical for me! I can’t tell whether he is in favour of “explanations” in the simple(-minded?) Deutsch sense or not.

After setting the passage in context, I got very concrete:

But I disagree that the quoted passage is technical. If he adheres to this passage, then Heisenberg cannot claim that Schrödinger’s cat would be both alive and dead, or that the moon would not be there if nobody watches.
Others, like Christopher A. Fuchs and Asher Peres in Quantum Theory Needs No Interpretation”, are apparently less sure whether (neo-Copenhagen) quantum theory is so clear about that fact. Hence they try to weasel out by claiming: “If Erwin has performed no observation, then there is no reason he cannot reverse Cathy’s digestion and memories. Of course, for that he would need complete control of all the microscopic degrees of freedom of Cathy and her laboratory, but that is a practical problem, not a fundamental one.”
This is non-sense, because the description of the experiment given previously was complete enough to rule out any possibility for Erwin to reverse the situation. Note the relevance of “… a consistent interpretation of QM as applied to what we do in a physical laboratory and how practitioners experience QM in that context.” If Erwin had access to a time machine enabling him to realistically reverse the situation, then it might turn out that Cathy and Erwin indeed lived multiple times through both situations (and experienced real macroscopic superpositions), as depicted in movies like “Back to the Future”.

According to Jan Faye in the SEP article on the Copenhagen Interpretation, even Bohr does not disagree with that conclusion: “Thus, Schrödinger’s Cat did not pose any riddle to Bohr. The cat would be dead or alive long before we open the box to find out.” However, I doubt that Bohr really said this, or replied to the content of Schrödinger’s article in any other direct way. (I read that he complained to Schrödinger about that article for indirectly supporting Einstein in his crusade against QM.)

You see, back then I assumed that Christopher A. Fuchs and Asher Peres were really unsure whether everything could be reversed as long as Erwin did not perform an observation. How should I have known that they used irony here? Or maybe they didn't, and really meant what they wrote? If you want, I can try to lookup the points that Jean Brickmont fell into to take serious. (On the other hand, maybe he wanted to fall for that trap. I certainly didn't want. I wanted to defend Heisenberg. But I am fully aware that vanhees71 won't join me in defending Heisenberg...)

 

[Fra]

I'm not sure what you mean.

My comment was mainly an principal comment, and a note that just like philosophy can be confusing, the apparent decuctive precision of mathematics can misguide. I prefer a balance. My personal view is that even in the hierarchy of the the interacitons besides gravity, some subtle symptoms of this appears. After all, there is not even a GUT. Because even within the GUT, there are large energy gaps between QCD and Electroweak. But it's harder to argue how this relates to the issue.

/Fredrik

 

[vanhees71]

Why should there be a GUT?

 

[Nullstein]

I only see two qubit systems discussed in the paper.

There are 3 systems, B, C and D in the paper. It discusses a general setting. System B corresponds to the 2&3 system in our case and systems C and D correspond to systems 1 and 4 respectively.

 

[Nullstein]

1. No, the full ensemble is maximally entangled. You just don't know how. If you run an experiment, and can identify 1/2 the cases as to Bell state (which is feasible), that portion will violate the CHSH bound. The other half will not show any correlation, as you say. That does not mean they aren't entangled, QM predicts they are.

No, the full system is not entangled. I have provided a mathematical proof several times and you had the opportunity to point at the equality sign that you do not comply with. You didn't bother to do that. On the other hand, you never provided any calculation at all. System 1&4 is in the product state $\frac 1 4\mathbb 1\otimes\mathbb 1$ and this is a mathematical fact. It takes at most 15 minutes of linear algebra, even for a beginner, to verify this.

2. Quantum nonlocality is a lot more than violating a Bell Inequality! The predictions of QM embed the nonlocality. Specifically: cos^2(Alice-Bob) prediction is inherently nonlocal, as Alice and Bob and their measurement settings need not be local to each other. Note that there are no other variables in this equation.

A correlation depends on spacelike separated variables. So what? Correlation does not imply causal relationships. In order to infer causal relationships, you must take the issues in post #150 into account. But if you do this, i.e. if you pass to the full ensemble, the correlation vanishes.

3. I never said there was a cause-and-effect relationship in the Einsteinian sense, because there isn't.

No cause and effect relationship can be inferred at all. That's the point. The statistical analysis is very clear on that.

 

[DrChinese]

No, the full system is not entangled. I have provided a mathematical proof several times and you had the opportunity to point at the equality sign that you do not comply with. You didn't bother to do that. On the other hand, you never provided any calculation at all. System 1&4 is in the product state $\frac 1 4\mathbb 1\otimes\mathbb 1$ ...

I don't follow your notation or what you what you are trying to communicate; but initially 1 and 4 are uncorrelated and therefore can be considered in a Product State as you say. I might represent that as: $\hat{\rho}_{14}=\hat{\rho}_{1} \otimes \hat{\rho}_{4}.$ But 1 is monogamously entangled with 2, and 4 is monogamously entangled with 3, so I prefer the notation: $\hat{\rho}_{1234}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}$ which is more descriptive. Typically each pair is going to be in either the $\psi+ or \phi+$ Bell state, depending on the Source (i.e. Type I or Type II PDC, or whatever).

IFF the BSM occurs, each and every 1 & 4 are entangled (according to the predictions of QM). Then is new description is: $\hat{\rho'}_{1234}=\hat{\rho}_{14} \otimes \hat{\rho}_{23}.$ That can be simplified to $\hat{\rho}_{14}$ which is will be one of 4 possible Bell states. That system of 2 photons is a biphoton (Fock state = 2) and is nonseparable, therefore entangled. You can refer to the distant BSM results to learn which Bell state. Different BSMs can discern different of the Bell states, either 1 or 2 or probably soon to be all 4.

I don't think anyone is any longer disputing what I say here. The only thing left to discuss is whether the distant BSM should be considered "action at a distance". Is there any other open question at this point?

 

[DrChinese]

I was just trying to use Dr. Chinese's terminology. My point was you still need to pick sub-ensembles; that's what the BSM results let you do.

Without knowing the sub-ensembles there is no entanglement, at least no possibility to show entanglement.

Edit: I think I get it now – Dr. Chinese thought the "cherry-picking" method only worked when picking 1 or 2 of the subsets of runs (I won't use the word sub-ensemble for this). This is incorrect; it works for all 4 subsets.

There is entanglement whether or not you know the Bell state. That is a specific prediction of QM, there is no questioning this. The issue you have is whether it can be "proven" to be entangled in the cases when the Bell state cannot be determined. You are free to accept the predictions of QM, or say QM is wrong, but we all know what happens to those who deny the predictions of QM. You will need a towel to remove the egg on your face.

But you might want to know that recent (Jan 2023) experiments demonstrate teleportation with identification of all 4 Bell states. That will probably end up in a version someday in which the kind of swapping we are discussing can be identified similarly. Don't bet against improving technology either!

 

[gentzen]

IFF the BSM occurs, each and every 1 & 4 are entangled (according to the predictions of QM). Then is new description is: $\hat{\rho'}_{1234}=\hat{\rho}_{14} \otimes \hat{\rho}_{23}.$ ...

I don't think anyone is any longer disputing what I say here.

But your formula is simply wrong, especially if we take into account what you wrote in the next post:

But you might want to know that recent (Jan 2023) experiments demonstrate teleportation with identification of all 4 Bell states.

So you admit that there are 4 different Bell states, but still insist to write $\hat{\rho}_{14}$ for all of them? If they are 4 different states (even so all are maximally entangled), then you should also use 4 different symbols for them, for example $\hat{\rho}_{14}^1$, $\hat{\rho}_{14}^2$, $\hat{\rho}_{14}^3$, and $\hat{\rho}_{14}^4$. And then, wouldn't $\hat{\rho'}_{1234}= \frac{1}{4} \sum_{i=1}^4 \hat{\rho}_{14}^i \otimes \hat{\rho}_{23}^i$ be a more appropriate formula for the state, as long as no conditioning with respect to the 4 different states is done?

 

[Nullstein]

This just in! From an experiment described in a paper deposited 15 Jan 2023:

https://arxiv.org/abs/2301.06091

"Photon-pair sources based on spontaneous parametric down-conversion (SPDC) serve as such a resource of entanglement, as they can be matched in frequency and bandwidth to the ionic transitions [4–7]. When entanglement between distant nodes has been established, quantum teleportation [8] provides a suitable protocol, alternative to direct transmission, for quantum state transfer [9]. Application of teleportation has been shown between photons over large distances [10–15], between quantum nodes [16–20], and from a photon to a quantum node [21–24]. In this paper we show quantum teleportation of an atomic qubit, encoded in the Zeeman sub-levels of the 40Ca+ ion, to the polarisation qubit of a single 854 nm photon. The teleportation protocol employs heralded absorption of a single photon by the ion [4, 9, 25, 26], which functions as a Bell measurement. In contrast to photonic teleportation protocols, this Bell measurement can identify all four Bell states."

All 4! Where's the subensemble "out" now?

This is literally the exact thing we have been discussing all along. Each Bell states corresponds to one of the four subensembles we have been discussing the whole time. Nothing this paper changes anything about the discussing.

But technology advances, and red herrings fall. There is nothing about the BSM (pre- or post-) selection process that is can be considered "cherry picking".

Nobody said that post-selection was cherry picking. Post-selection generates subensembles by conditioning on a common effect. Thus the correlations in these subensembles is spurious as every statistician will tell you.

 

[Nullstein]

I don't follow your notation or what you what you are trying to communicate

This is the whole problem. You take a step back and try to understand what is being discussed and confirm the calculations on your own before replying.

but initially 1 and 4 are uncorrelated and therefore can be considered in a Product State as you say. I might represent that as: $\hat{\rho}_{14}=\hat{\rho}_{1} \otimes \hat{\rho}_{4}.$ But 1 is monogamously entangled with 2, and 4 is monogamously entangled with 3, so I prefer the notation: $\hat{\rho}_{1234}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}$ which is more descriptive. Typically each pair is going to be in either the $\psi+ or \phi+$ Bell state, depending on the Source (i.e. Type I or Type II PDC, or whatever).

I'm talking about the state of the 1&4 subsystem, which is given by $\rho_{14}=\mathrm{tr}_{23}\rho_{1234} = \frac 1 4 \mathbb 1\otimes\mathbb 1$. The state of the 1&4 system it not entangled, neither before nor after the BSM. This is the math and its simple linear algebra. It would be best if you tried that calculation on your own, since you don't seem to believe it. Nobody disputes it other than you. It's literally an undergrad calculation.

IFF the BSM occurs, each and every 1 & 4 are entangled (according to the predictions of QM). Then is new description is: $\hat{\rho'}_{1234}=\hat{\rho}_{14} \otimes \hat{\rho}_{23}.$

This is true for a subensemble. The full ensemble still is in a state such that the trace over 2&3 produces the state $\frac 1 4\mathbb 1\otimes\mathbb 1$.

I don't think anyone is any longer disputing what I say here.

Yes, we are. You deny that the full ensemble of 1&4 after the BSM is given by $\frac 1 4\mathbb 1\otimes\mathbb 1$. Everyone accepts that other than you.

 

[Fra]

Why should there be a GUT?

I guess that could or even should be questioned from a minimalist instrumentalist perspective - we could await building the GUT until explicitly driven by experimental deviations. This is a reasonable point. The same goes for QG. Who needs it anyway?

I am however driven by conceptual understanding of how physical interactions work and the nature of physical law from the perspective of inference. I do not work in a lab, I am not at the frontier of pushing subatomic interactions at higher energies. I see my personal quest as making sense of and find a coherent and efficient want to understand the massive body of data mankind already have. So far it's too much of a patchwork of effective theories, too many parameters that are experimentally determined that most likely should be predictable from a deeper theory. All the masses of the elementary particles for example. Predicting just a few of the manually tuned parameters or explaining some of the symmetries in the standard model would be an achievement worth asking the question even if no other experimental contradictions appear. It's how I see it. Is there a lack of ideas without experimental input? I don't think so?

From my perspective, the quest for a GUT goes hand in hand with the quest for QG, it's one quest, not two. Many programs, except string theory, try to solve them separately. String theory seems to have gotten stuck though IMO. I think solving the two problems together will be easier, than solving them separately. The problem does get "bigger", but I think some of the keys to make it work lies in seeing the relation.

/Fredrik

 

[vanhees71]

I don't follow your notation or what you what you are trying to communicate; but initially 1 and 4 are uncorrelated and therefore can be considered in a Product State as you say. I might represent that as: $\hat{\rho}_{14}=\hat{\rho}_{1} \otimes \hat{\rho}_{4}.$ But 1 is monogamously entangled with 2, and 4 is monogamously entangled with 3, so I prefer the notation: $\hat{\rho}_{1234}=\hat{\rho}_{12} \otimes \hat{\rho}_{34}$ which is more descriptive. Typically each pair is going to be in either the $\psi+ or \phi+$ Bell state, depending on the Source (i.e. Type I or Type II PDC, or whatever).

That's all right. Just once more for clarity in the notation you seem to prefer to the QFT-notation with creation operators: The notation for the Bell states of photon pairs $(ij)$ (where (ij) labels, e.g., the momentum):
$$|\Psi_{ij}^{\pm} \rangle= \frac{1}{\sqrt{2}} (|\vec{p}_i,H;\vec{p}_j, V \rangle \pm (|\vec{p}_i,V;\vec{p}_j, H \rangle,$$
$$|\Phi_{ij}^{\pm} \rangle= \frac{1}{\sqrt{2}} (|\vec{p}_i,H;\vec{p}_j, H \rangle \pm (|\vec{p}_i,V;\vec{p}_j, V \rangle.$$
The initial state is $\hat{\rho}_{1234}=|\Psi \rangle \langle \Psi |$
$$|\Psi \rangle=|\Psi_{12}^{-} \rangle \otimes |\Psi_{34}^{-} \rangle.$$
It is a bit of work but easy to see that this is the same as
$$|\Psi \rangle=\frac{1}{2} (|\Psi_{23}^{+} \otimes |\Psi_{14}^{+} \rangle - |\Psi_{23}^{-} \otimes |\Psi_{14}^{-} \rangle -|\Phi_{23}^{+} \otimes |\Psi_{14}^{+} \rangle +|\Phi_{23}^{-} \otimes |\Psi_{14}^{-} \rangle).$$
That means that the subsystem consisting of photons 1 and 4 are in the reduced state
$$\hat{\rho}_{14} = \mathrm{Tr}_{23} \hat{\rho}_{1234}=\frac{1}{4} \hat{1}_{14},$$
i.e., the photons (14) are both unpolarized and uncorrelated.

IFF the BSM occurs, each and every 1 & 4 are entangled (according to the predictions of QM). Then is new description is: $\hat{\rho'}_{1234}=\hat{\rho}_{14} \otimes \hat{\rho}_{23}.$ That can be simplified to $\hat{\rho}_{14}$ which is will be one of 4 possible Bell states. That system of 2 photons is a biphoton (Fock state = 2) and is nonseparable, therefore entangled. You can refer to the distant BSM results to learn which Bell state. Different BSMs can discern different of the Bell states, either 1 or 2 or probably soon to be all 4.

Indeed if 2&3 are projected to the state $|\Psi_{23}^- \rangle$, which is what you call BSM (Bell-state measurment), then for the so selected subensemble 1&4 are also in the state $|\Psi_{14}^- \rangle$. In math it's completely clear that you do something to photons 2&3 but nothing (!!) to photons 1 and 4, i.e., when projecting 2&3 to $|\Psi_{23}^- \rangle$, then there's nothing interacting with photons 1 and 4:
$$\mathcal{N} |\Psi' \rangle = |\Psi_{23}^{-} \rangle \langle \Psi_{23}^{-} \rangle \otimes \hat{1}_{14} |\Psi \rangle=\frac{1}{2} |\Psi_{23}^{-} \rangle \otimes |\Psi_{14}^{-1} \rangle.$$
Thus the projection to this specific Bell state occurs with probability $(1/2)^2=1/4$, and the (renormalized) state of the subensemble is
$$|\Psi' \rangle=|\Psi_{23}^{-} \rangle \otimes |\Psi_{14}^{-} \rangle,$$
i.e., now (23) and (14) are in Bell states, but the pairs are completely uncorrelated.

The state $|\Psi_{23}^{-} \rangle$ is chosen by the experimenters, because with a polarizing beam splitter it's simply identified by the fact that only in this state both output channels register a photon (since photons are bosons and the two photons are in an odd polarization state, the momentum state must be also odd, i.e., the photons cannot exit the beam splitter in the same momentum state).

I think about these facts we agree, and I hope we also agree on the math.

I don't think anyone is any longer disputing what I say here. The only thing left to discuss is whether the distant BSM should be considered "action at a distance". Is there any other open question at this point?

If we agree on the math and the meaning of a projection measurement, then nothing has be done to photons 1 or 4 when projecting photons 2 and 3 to a Bell state, which is indeed a local measurement, because you have to bring these two photons together at the polarizing beam splitter and the detectors in the two exit channels of this splitter, i.e., this is something where the two photons interact with equipment in the pretty small space-time region.

It is also completely irrelevant, in which temporal order you do the BSM and the measurements on photons 1 and 4. You can also perform the measurements space-like separated, so that there cannot be any actions at a distance by the BSM measurement on photons 1 and 4, if you accept the microcausality principle.

Interestingly enough it has been explicitly tested that indeed Q(F)T prevails also when tested against a type of "multisimultaneity model" (using energy-time entangled photon pairs):

https://arxiv.org/abs/quant-ph/0110117
https://doi.org/10.1103/PhysRevLett.88.120404

 

[Nullstein]

Just to complete @vanhees71's correct calculation: In total we then get 4 such subensembles, depending on the measurement outcome:
$$\rho_1 = P_{|\Psi_{23}^{+} \rangle \otimes |\Psi_{14}^{+} \rangle}\quad\rho_2 = P_{|\Psi_{23}^{-} \rangle \otimes |\Psi_{14}^{-} \rangle}\quad\rho_3 = P_{|\Phi_{23}^{+} \rangle \otimes |\Phi_{14}^{+} \rangle}\quad\rho_4 = P_{|\Phi_{23}^{-} \rangle \otimes |\Phi_{14}^{-} \rangle}\quad$$
We can now calculate the full ensemble by summing over them with weights given by the probabilities $p_i = \frac 1 4$, according to which they are distributed. This gives us the state of the full ensemble after the BSM:
$$\rho_{1234,\text{after}}=\frac 1 4\left(\rho_1+\rho_2+\rho_3+\rho_4\right)$$
It's then easy to see that the state of the 1&4 subsystem after the BSM is given by:
$$\rho_{14,\text{after}}=\mathrm{tr}_{23}\rho_{1234,\text{after}}=\frac 1 4\mathbb 1\otimes\mathbb 1$$
This is just the completeness relation, because the partial trace gives us the 4 Bell states, which constitute an orthogonal basis.