I Confused by nonlocal models and relativity

[PeterDonis]

Not being a specialist in the field of quantum physics, it is a matter of understanding.

Yes, and your understanding appears to be mistaken. Here is a simple example to illustrate the correct understanding:

Suppose Alice and Bob each have one of a pair of entangled spin-1/2 particles. They are about to measure them at spacelike separated events.

The operators (for example) "Measure z-spin on Alice's particle" and "Measure x-spin on Alice's particle" do not commute. Similarly, the operators "Measure z-spin on Bob's particle" and "Measure x-spin on Bob's particle" do not commute.

But the operators, for example, "Measure z-spin on Alice's particle" and "Measure x-spin on Bob's particle" do commute.

 

[PeterDonis]

i) You measure p on A and get value P. You now know remote p on B, and q is completely indeterminate.
ii) You measure q on A and get value Q. You now know remote q on B, and p is completely indeterminate.

These descriptions assume that there is an invariant ordering to the measurements. But if the measurements are spacelike separated, there isn't. And if there isn't, you simply can't help yourself to a description that assumes that there is. So I do not accept either of these descriptions. I want to see a description that does not assume that the measurements occur in a particular order; only such a description can be consistent with relativity, since in relativity the ordering of spacelike separated events is not invariant.

For example:

i-a) You measure p on A and get value P. You measure p on B and get value not-P. This will always happen when those two measurements are combined.
i-b) You measure p on A and get value P. You measure q on B and get value Q. Then you repeat that pair of measurements many times, and the measurements on A and B show zero correlation.

ii-a) You measure q on A and get value Q. You measure q on B and get value not-Q. This will always happen when those two measurements are combined.
ii-b) You measure q on A and get value Q. You measure p on B and get value P. Then you repeat that pair of measurements many times, and the measurements on A and B show zero correlation.

the above outcomes are completely different as quantum descriptions of A and B. And that is entirely because p and q don't commute, and the observer's choice steers the results

Both observers' choices. Not just one. It's not either observer's choice in isolation, but the combination of the two, that makes the difference--in one case, they both choose to measure in the same direction, in the other, they choose to measure in two orthogonal directions. Neither choice by itself "steers" the results; only the combination of the two does.

Not everyone will entirely agree that this demonstrates a quantum nonlocal effect. Different interpretations account for this differently.

I guess that depends on what "quantum nonlocal" is supposed to mean. That's why I asked earlier if by that term you meant "violates the Bell inequalities". Any interpretation has to agree that the results violate the Bell inequalities, because that's an experimental fact.

 

[microsansfil]

Yes, and your understanding appears to be mistaken. Here is a simple example to illustrate the correct understanding:

Suppose Alice and Bob each have one of a pair of entangled spin-1/2 particles. They are about to measure them at spacelike separated events.

The operators (for example) "Measure z-spin on Alice's particle" and "Measure x-spin on Alice's particle" do not commute. Similarly, the operators "Measure z-spin on Bob's particle" and "Measure x-spin on Bob's particle" do not commute.

But the operators, for example, "Measure z-spin on Alice's particle" and "Measure x-spin on Bob's particle" do commute.

Ok Thank

The operators commute even if it doesn't apply in the same Hilbert space?

if you formulate the problem in the tensor product space of the two particles ε_A ⊗ ε_B because the two particles are entangled.

Does the tensor product extension "Measure z-spin on Alice's particle" Z_A ⊗ I_B and the tensor product extension "Measure x-spin on Bob's particle" I_A ⊗ X_B also commute ?

/Patrick

 

[PeterDonis]

The operators commute even if it doesn't apply in the same Hilbert's space?

They are operators on the same Hilbert space. See below.

if you formulate the problem in the tensor product space of the two particles εA ⊗ εB because the two particles are entangled.

Yes, that's the appropriate Hilbert space. There is no "if"; only one Hilbert space is valid, and it's that one.

Does the tensor product extension "Measure z-spin on Alice's particle" ZA ⊗ IB and the tensor product extension "Measure x-spin on Bob's particle" IA ⊗ XB also commute ?

What do you mean "also"? These are exactly the operators that I already said commute.

The operators that don't commute are pairs like $Z_A \otimes I_B$ and $X_A \otimes I_B$, or $I_A \otimes Z_B$ and $I_A \otimes X_B$.

 

[DrChinese]

But the operators, for example, "Measure z-spin on Alice's particle" and "Measure x-spin on Bob's particle" do commute.

I say they don't commute (assuming they are entangled). Yours is actually a variation of the EPR argument. So here is a specific point of departure, and I believe we are speaking the same terminology. A Bell inequality is violated in this case, indicating that Alice and Bob are not independent (i.e. separable). If they commute, then there are Product State statistics. If they don't, you see Entangled State statistics.EDIT: I'm sure you know that conjugate observables of entangled particles do not commute, and in fact I think you have written on that previously.

 

[microsansfil]

They are operators on the same Hilbert space. See below.

My understand is that Alice particle "live" in the Hilbert Space ε_A and Bob particle "live" in the Hilbert Space ε_B . However, Alice and Bob particles "live" in the Hilbert Space ε_A ⊗ ε_B which is different from ε _A or ε_B.

This is not a question of physics, but of mathematics.

/Patrick

 

[DrChinese]

1. Both observers' choices. Not just one. It's not either observer's choice in isolation, but the combination of the two, that makes the difference--in one case, they both choose to measure in the same direction, in the other, they choose to measure in two orthogonal directions. Neither choice by itself "steers" the results; only the combination of the two does.

2. I guess that depends on what "quantum nonlocal" is supposed to mean. That's why I asked earlier if by that term you meant "violates the Bell inequalities". Any interpretation has to agree that the results violate the Bell inequalities, because that's an experimental fact.

1. Concur completely.

2. I said yes earlier, which is why I label anything that violates a Bell Inequality as quantum non-local. That's pretty standard too. And of course I agree that any viable interpretation must follow experiment. I don't really know what quantum non-locality is, but I know how to identify it when I see it.

 

[PeterDonis]

I say they don't commute (assuming they are entangled).

Then I disagree with you and I have no idea why you would say that, since it appears to contradict basic QM. The two operators are $Z_A \otimes I_B$ and $I_A \otimes X_B$, which obviously commute since $Z_A$ and $I_A$ commute and $I_B$ and $X_B$ commute.

A Bell inequality is violated in this case, indicating that Alice and Bob are not independent (i.e. separable). If they commute, then there are Product State statistics. If they don't, you see Entangled State statistics.

I don't think this is correct. The Bell inequality is violated because the joint probability function does not factorize, i.e., it violates Bell's locality assumption. But that is not the same as the two operators not commuting.

 

[PeterDonis]

My understand is that Alice particle "live" in the Hilbert Space εA and Bob particle "live" in the Hilbert Space εB . However, Alice and Bob particles "live" in the Hilbert Space εA ⊗ εB which is different from ε A or εB.

This makes no sense. First you say each particle lives in its own Hilbert space. Then you say the two particles live in a single Hilbert space. Those two statements contradict each other.

The correct statement is that, if you have two particles, the Hilbert space of the two-particle system is the tensor product one. If the two particles are not entangled, then the state of the two particles is separable--it can be expressed as a single product of an "A" state and a "B" state. But such a state is still a state on the tensor product Hilbert space.

 

[kurt101]

But the operators, for example, "Measure z-spin on Alice's particle" and "Measure x-spin on Bob's particle" do commute.

The order you make spacelike separated measurements on entangled particles does not affect the resulting probability. However for a given instance of entangle particles, measuring one particle first may have given you a different result than if you had measured it second. In other words the order of measurements may commute for the probability, but not for a single pair of measurements. I think we should be clear on this point, because otherwise you might incorrectly be mislead into thinking a non-local causal interpretation is not allowed because of this.

 

[PeterDonis]

for a given instance of entangle particles, measuring one particle first may have given you a different result than if you had measured it second.

First, this is not what the math of QM says; I already gave that in a previous post. The math is clear that the two operators commute.

Second, there is no invariant sense in which either particle is measured first if the measurements are spacelike separated. So I don't see how to even formulate what you are claiming consistently.

And third, we have no way of measuring counterfactuals, so even if there were an invariant sense in which one measurement was before the other, we have no way of measuring what would have happened if the order had been opposite. Our only basis for talking about counterfactuals at all is theory, and the theory--the math of QM--says, as noted above, that the operators commute, which means that the results must be independent of the order in which the measurements are done.

you might incorrectly be mislead into thinking a non-local causal interpretation is not allowed

What do you mean by a "non-causal interpretation"?

 

[microsansfil]

This makes no sense. First you say each particle lives in its own Hilbert space. Then you say the two particles live in a single Hilbert space. Those two statements contradict each other.

I agree it's misspoken. I just want to point out that ε_A ⊗ ε_B is different to ε_A and also different to ε_B. in ε_A ⊗ ε_B you can have entangled particles. i. e. which is not factorized as the tensor product, and thus can't be expressed as a single product of an "A" state and a "B" state.

For the Operators in ε_A ⊗ ε_B which are tensor product of an operator of ε_A and an operator of ε_B, then this operator commute :

However, just as with vectors, there exist operators in ε_A ⊗ ε_B which are not tensor products of an operator of ε_A and an operator of ε_B.

Thus what about the commutation of this kind of operator?

/Patrick

 

[Tendex]

The order you make spacelike separated measurements on entangled particles does not affect the resulting probability. However for a given instance of entangle particles, measuring one particle first may have given you a different result than if you had measured it second. In other words the order of measurements may commute for the probability, but not for a single pair of measurements. I think we should be clear on this point, because otherwise you might incorrectly be mislead into thinking a non-local causal interpretation is not allowed because of this.

If by "non-local causal" interpretation you mean for instance the Bohmian Interpretation this indeed is allowed as long as the interpretation can not be made relativistic in which case you have the spacelike versus non-spacelike interval distinction that doesn't exist in a purely non-relativistic interpretation. Without this distinction the QFT microcausality prescription is moot and you can have "spooky actions at a distance" in the way DrChinese is using them to contradict vanhees71. To do it one just has to ignore that relativistic QFT is considered more fundamental and general than non-relativistic quantum mechanics.

One can only avoid confusing debates on Bell's theorem when this point is not ignored.

 

[vanhees71]

Where do they exclude that entanglement can be created via post-selection?

To the contrary, the here discussed experiment confirms the possibility to create entanglement by post-selection (as also other experimental realizations of delayed-choice protocols clearly demonstrate).

 

[Tendex]

If you accept that an entangled pair of photons is indeed one physical system, demonstrating quantum non-locality: then you must accept that the nature of a measurement of one member of the pair physically changes the state of the other.

I say they don't commute (assuming they are entangled). Yours is actually a variation of the EPR argument. So here is a specific point of departure, and I believe we are speaking the same terminology. A Bell inequality is violated in this case, indicating that Alice and Bob are not independent (i.e. separable). If they commute, then there are Product State statistics. If they don't, you see Entangled State statistics.EDIT: I'm sure you know that conjugate observables of entangled particles do not commute, and in fact I think you have written on that previously.

As explained by PeterDonis it is only if applied to spacelike-separated operators that they commute. If you treat the particles nonrelativistically as one system there is no spacelike separation consideration possible and the operators don't commute, but then mathematically there is no "spooky action at a distance" to be puzzled about. If one insists in factoring in that distance there is no other accepted way than going to the wider scope relativistic theory that acknowledges a spacelike separation for which the operators must commute.

 

[kurt101]

First, this is not what the math of QM says; I already gave that in a previous post. The math is clear that the two operators commute.

The QM math does not describe what happens when you measure a single instance of entangled particles. The QM math speaks to the probability you will get when you measure many samples of entangled particles.

Second, there is no invariant sense in which either particle is measured first if the measurements are spacelike separated. So I don't see how to even formulate what you are claiming consistently.

If the universe can be simulated (in a theoretical sense) and has hidden variables that describe its state and you can pause the simulation then you can say what measurement happened first in the simulation.

And third, we have no way of measuring counterfactuals, so even if there were an invariant sense in which one measurement was before the other, we have no way of measuring what would have happened if the order had been opposite.

Agreed and that is the point I am making.

Our only basis for talking about counterfactuals at all is theory, and the theory--the math of QM--says, as noted above, that the operators commute, which means that the results must be independent of the order in which the measurements are done.

What we observe in the experiment is also a basis for talking about counterfactuals.

What do you mean by a "non-causal interpretation"?

In the general sense like John Bell said in his Bertlmann's socks paper "Thirdly, it may be that we have to admit that causal influences - do go faster than light.".

In a more specific way as I previously described in a previous post on this thread about the entanglement swapping paper that DrChinese mentioned. For example I said "1 is measured and non-locally gives the result of this measurement to 2".

 

[kurt101]

If by "non-local causal" interpretation you mean for instance the Bohmian Interpretation this indeed is allowed as long as the interpretation can not be made relativistic in which case you have the spacelike versus non-spacelike interval distinction that doesn't exist in a purely non-relativistic interpretation. Without this distinction the QFT microcausality prescription is moot and you can have "spooky actions at a distance" in the way DrChinese is using them to contradict vanhees71. To do it one just has to ignore that relativistic QFT is considered more fundamental and general than non-relativistic quantum mechanics.

I am not referring to any formal interpretation, but to how we interpret what is observed in the entanglement experiments. And my point is that nothing in this informal interpretation of these entanglement experiments contradicts relativistic QFT or special relativity.

 

[Tendex]

I am not referring to any formal interpretation, but to how we interpret what is observed in the entanglement experiments. And my point is that nothing in this informal interpretation of these entanglement experiments contradicts relativistic QFT or special relativity.

Sure, but it is important to realize that this informal interpretation doesn't contradict relativistic QFT just as long as it ignores its relativistic content and the current consensus about non-relativistic quantum mechanics being less fundamental than relativistic QFT.

 

[Lord Jestocost]

... where "Bell's local causality" is defined?

Caslav Brukner in “Elegance and Enigma: The Quantum Interviews” (ed. by Maximilian Schlosshauer), p. 166:

"Bell’s theorem is a no-go theorem that states that no “local causal” or “local realistic” theories can ever reproduce all of the predictions of quantum mechanics. The desire for a local causal theory is based on the following three assumptions:

(1) There exist “causes” that determine measurement outcomes, or probabilities of outcomes, for all possible experiments that could be performed on an individual system, no matter whether any experiment — and which experiment — is actually performed (and so, in this sense, would be “real”).

(2) The actually measured outcome (or the probability for the outcome), and equally those outcomes that could be potentially measured, can only be influenced by local causes (that is, other events in the backward light cone) and not by any event in spacelike separated regions (“locality”).

(3) The experimenter’s choice of the measurement setting is independent of the causes that determine the actually measured outcome (“freedom of choice”)."

 

[DrChinese]

Then I disagree with you and I have no idea why you would say that, since it appears to contradict basic QM. The two operators are $Z_A \otimes I_B$ and $I_A \otimes X_B$, which obviously commute since $Z_A$ and $I_A$ commute and $I_B$ and $X_B$ commute.

The question is whether entangled Alice(p) commutes (or not) with Bob(q). Or substitute any conjugate pair for p and q (such as spin at various angles). To be specific, let's discuss entangled spin-1 particle spins x and z (singlet). These are not separable, do NOT commute, and the uncertainty principle should be applied.

a. There is no question that if Alice(x) and Bob(x) are measured, they will be opposite. Similarly, if Alice(z) and Bob(z) are measured, they too will be opposite. The uncertainty principle is not a constraint.

b. In the EPR-B case, as summarized by Weinberg, Lectures on Quantum Mechanics, 12.1 Paradoxes of Entanglement: "... the observer could have measured the x-component of the spin of particle 1 instead of its z-component, and by the same reasoning, if a value h/2 or −h/2 were found for the x-component of the spin of particle 1 then also the x-component of the spin of particle 2 must have been −h/2 or h/2 all along. Likewise for the y-components. So according to this reasoning, all three components of the spin of particle 2 have definite values, which is impossible since these spin components do not commute..."

Your recent comment (I know, this is cheating!) on the above quote was: "Weinberg is not talking about successive measurements. He's talking about alternative possibilities for single measurements (one on each particle). He's simply pointing out that, since measuring the same spin component of both particles will always give opposite results, no matter which component is measured (x or y or z), any hidden variable model, i.e., any model that attributes the correlations between these measurements to pre-existing properties of the particles, would have to attribute definite spin components (+1 or -1) in all three directions (x and y and z) to each particle, i.e., the pre-measurement state of each particle would have to have definite values for spin-x, spin-y, and spin-z. But that is not possible because no quantum state can have definite values for multiple non-commuting operators. "

I think you agree with Weinberg that the Uncertainty Principle applies to non-commuting components of entangled pairs. And again, any analysis of EPR-B is going to say virtually the same as Weinberg. Weinberg further comments: "There is a troubling weirdness about quantum mechanics. Perhaps its weirdest feature is entanglement, the need to describe even systems that extend over macroscopic distances in ways that are inconsistent with classical ideas." . One of the classical ideas being local causality.

c. And in fact, Weinberg goes on to say as follows: "Of course, according to present ideas a measurement in one subsystem does change the state vector for a distant isolated subsystem - it just doesn't change the density matrix." Which is what I assert: A measurement on Alice's particle changes the physical state of Bob's remote entangled particle (what is observed). Although if you and vanhees71 are instead referring to the density matrix, you would be right about that. (Weinberg personally believes in the reality of the reduced density matrix rather than the state vector, I believe, but that does not seem a common interpretation.)

d. This from Wikipedia, which is not cited as a source but rather to indicate the generally accepted viewpoint and related derivation:

EPR Paradox
"...how does Bob's [entangled] positron know which way to point if Alice decides (based on information unavailable to Bob) to measure x (i.e., to be the opposite of Alice's electron's spin about the x-axis) and also how to point if Alice measures z, since it is only supposed to know one thing at a time? The Copenhagen interpretation* rules that the wave function "collapses" at the time of measurement, so there must be action at a distance..." Of course, the mathematical presentation shows that Alice's x and Bob's z are constrained by the Uncertainty Principle and clearly do not commute. That portion is identical to nearly any presentation of this problem.*Copenhagen simply being one viable interpretation, there are many others too; such would not change the sense of this passage.

 

[kurt101]

Sure, but it is important to realize that this informal interpretation doesn't contradict relativistic QFT just as long as it ignores its relativistic content and the current consensus about non-relativistic quantum mechanics being less fundamental than relativistic QFT.

How does this informal interpretation contradict QFT by ignoring its relativistic content?

What does non-relativistic QM vs relativistic QFT have to do with this interpretation?

And the interpretation that I am discussing is what I said to PeterDonis:
In the general sense like John Bell said in his Bertlmann's socks paper "Thirdly, it may be that we have to admit that causal influences - do go faster than light.".

In a more specific way as I previously described in a previous post on this thread about the entanglement swapping paper that DrChinese mentioned. For example I said "1 is measured and non-locally gives the result of this measurement to 2".

 

[DrChinese]

As explained by PeterDonis it is only if applied to spacelike-separated operators that they commute. If you treat the particles nonrelativistically as one system there is no spacelike separation consideration possible and the operators don't commute, but then mathematically there is no "spooky action at a distance" to be puzzled about. If one insists in factoring in that distance there is no other accepted way than going to the wider scope relativistic theory that acknowledges a spacelike separation for which the operators must commute.

I don't believe relativity has anything to do with it. There are no predictive differences for a local setup versus a non-local setup for anything related to entanglement (at least AFAIK).

Conjugate operators of distant entangled particles do not commute. If they did, the EPR Paradox would not have been solved. EPR's argument was precisely that you could learn non-commuting values on a particle by reference to the entangled partner. If the partner's conjugate property DID commute, you could do that. But you can't.

Further, Bell experiments would have different outcomes. After all, the experiments have been done with strict Einsteinian non-locality a number of times. The predictions for Bob depend on the choice of measurement on distant Alice. The match statistics are dependent solely on the relative angle theta between, in a manner inconsistent with separability (Bell 1964). Bell showed there was a different relationship on theta if there were separability (i.e. and therefore they commuted). This difference does not show up in cases where the same measurement is performed on both Alice and Bob. Which makes it easy to gloss over.

 

[microsansfil]

Conjugate operators of distant entangled particles do not commute.

The commuting property of operators is independent of the states of the systems to which these operators may apply, isn't it? whether or not they're entangled.

Mathematically the two operators $Z_A \otimes I_B$ and $I_A \otimes X_B$ commute, regardless of whether Alice and Bob's particles are entangled, as shown here

/Patrick

 

[DrChinese]

The commuting property of operators is independent of the states of the systems to which these operators may apply, isn't it? whether or not they're entangled.

... as shown here

/Patrick

I am not sure where your citation is from (the image presented on the link). Could you pass that along for context? I admit that without any mention of entanglement, it is difficult to see how this relates.

On the other hand: I did mention the source for my (directly opposing) quotes in post #140, but did not provide a link. It is:

Weinberg, Lectures on Quantum Mechanics

 

[microsansfil]

I am not sure where your citation is from (the image presented on the link).

Quantum Mechanics by Claude Cohen-Tannoudji

/Patrick

 

[DrChinese]

Quantum Mechanics by Claude Cohen-Tannoudji

/Patrick

Thanks. The author is of course top notch. However, I believe this was originally written in 1973. It lacks a section on entanglement, even in the newer editions. I don't think this is a good fit for our discussion. The formulae presented do not describe entangled systems. (Or if it does, I couldn't locate it.)

 

[microsansfil]

It lacks a section on entanglement, even in the newer editions.

It is in volume 3.

/Patrick

 

[PeterDonis]

I just want to point out that εA ⊗ εB is different to εA and also different to εB.

Yes, and εA ⊗ εB is the only one of them that we need to be discussing here.

in εA ⊗ εB you can have entangled particles. i. e. which is not factorized as the tensor product, and thus can't be expressed as a single product of an "A" state and a "B" state.

Yes, this is the correct definition of an entangled state, as I think I've already said.

there exist operators in εA ⊗ εB which are not tensor products of an operator of εA and an operator of εB.

Yes, but none of them are relevant to the measurements we've been talking about.

what about the commutation of this kind of operator?

It would depend on the specific operators. But, as above, none of them are relevant to what we've been discussing. You're welcome to start a separate thread if you want to discuss operators like this, though it would be a good idea to first find a specific experiment that involves them.

 

[PeterDonis]

As explained by PeterDonis it is only if applied to spacelike-separated operators that they commute. If you treat the particles nonrelativistically as one system there is no spacelike separation consideration possible and the operators don't commute

You are misinterpreting what I said, and your statements are incorrect.

The operators I wrote down commute regardless of whether events A and B are spacelike separated or not.

Considering two entangled particles nonrelativistically as one system certainly does not rule out spacelike separation. It just means you don't make any use of it in your analysis. And the operators I wrote down commute regardless of whether you do a relativistic or a non-relativistic analysis.

 

[PeterDonis]

The QM math does not describe what happens when you measure a single instance of entangled particles.

That depends on which interpretation you adopt.

If the universe can be simulated (in a theoretical sense) and has hidden variables that describe its state and you can pause the simulation then you can say what measurement happened first in the simulation.

This is way off topic for this thread, and indeed this forum.

What we observe in the experiment is also a basis for talking about counterfactuals.

I don't see how, since we only observe what actually happened, not what might have happened counterfactually.

In the general sense like John Bell said in his Bertlmann's socks paper "Thirdly, it may be that we have to admit that causal influences - do go faster than light.".

Then I think you need to pick a specific interpretation that is based on this premise if you want to discuss it. Bell, IIRC, had Bohmian mechanics in mind as falling into this category.