QFT & Entanglement: NRQT vs. PCT Transform

[stevendaryl]

Yes of course, you can do that. But then the other two photons won't be entangled.

No, that's not true. I went through it already. The original 4-particle state is this:

$\frac{1}{2} (|H_A H_{EC} H_{ED} H_{B}\rangle + |H_A H_{EC} V_{ED} V_{B}\rangle + |V_A V_{EC} H_{ED} H_{B}\rangle + |V_A V_{EC} V_{ED} V_{B}\rangle$

The first two photons are correlated (they either are both horizontally polarized, or both vertically correlated) and the last two are correlated, but there is no correlation between the first two and the last two. Act on this state with the projection operator $\Pi_A$. If you get a 1, then the "collapsed" state of the other two is:

$\frac{1}{\sqrt{2}} (|H_A H_{B}\rangle + |V_A V_{B})\rangle$

You get the same result, whether or not you use the projection operator $\Pi_A$ or $\Pi_1$.

This is using the quantum "recipe" that if you measure an observable, then the state after the measurement is the projection of the original state onto the subspace consisting of eigenstates of the observable with the appropriate eigenvalue.

 

[zonde]

What @akvadrako was suggesting was not that A&D had pre-existing measurement results (which would violate Bell's inequality), but that they had pre-existing entanglement. The way that I described things, Eric makes a measurement and depending on the result of his measurement, afterward, either A&D are correlated (if Eric gets one result) or anti-correlated (if Eric gets the other result). @akvadrako was suggesting that A&D were already either correlated or anti-correlated, and we just didn't know which until Eric performs his measurement.

I'm thinking that that assumption also should be ruled out by statistics along the lines of Bell's theorem, but it's not immediately obvious how to show that it's impossible.

Okay, I would like to make a distinction between pre-existing entanglement and pre-existing correlation.
Yes it seems fine to say that A&D had preexisting entanglement and B&C just sorts A&D pairs into four possible pre-existing entangled states.
So if we for some reason would think that there is a problem with Bell inequalities and that there actually is a way how to explain correlations between measurements using LHVs then entanglement swapping would be no problem.
So in a sense entanglement swapping does not add any new "mystery" in addition to violation of Bell inequalities.

 

[stevendaryl]

Okay, I would like to make a distinction between pre-existing entanglement and pre-existing correlation.
Yes it seems fine to say that A&D had preexisting entanglement and B&C just sorts A&D pairs into four possible pre-existing entangled states.
So if we for some reason would think that there is a problem with Bell inequalities and that there actually is a way how to explain correlations between measurements using LHVs then entanglement swapping would be no problem.
So in a sense entanglement swapping does not add any new "mystery" in addition to violation of Bell inequalities.

It just shows that the intuition that particles become entangled by interaction is not correct. Particles don't need to have ever interacted in order to be entangled.

 

[zonde]

No, that's not true. I went through it already. The original 4-particle state is this:

$\frac{1}{2} (|H_A H_{EC} H_{ED} H_{B}\rangle + |H_A H_{EC} V_{ED} V_{B}\rangle + |V_A V_{EC} H_{ED} H_{B}\rangle + |V_A V_{EC} V_{ED} V_{B}\rangle$

The first two photons are correlated (they either are both horizontally polarized, or both vertically correlated) and the last two are correlated, but there is no correlation between the first two and the last two. Act on this state with the projection operator $\Pi_A$. If you get a 1, then the "collapsed" state of the other two is:

$\frac{1}{\sqrt{2}} (|H_A H_{B}\rangle + |V_A V_{B})\rangle$

You get the same result, whether or not you use the projection operator $\Pi_A$ or $\Pi_1$.

This is using the quantum "recipe" that if you measure an observable, then the state after the measurement is the projection of the original state onto the subspace consisting of eigenstates of the observable with the appropriate eigenvalue.

Hmm, what is then the possible composite state if the two pairs have respective states like that?
$\frac{1}{\sqrt{2}} (|H_A H_{B}\rangle - |V_A V_{B}\rangle)$
$\frac{1}{\sqrt{2}} (|H_{EC} H_{ED}\rangle - |V_{EC} V_{ED}\rangle)$

 

[stevendaryl]

Hmm, what is then the possible composite state if the two pairs have respective states like that?
$\frac{1}{\sqrt{2}} (|H_A H_{B}\rangle - |V_A V_{B})\rangle$
$\frac{1}{\sqrt{2}} (|H_{EC} H_{ED}\rangle - |V_{EC} V_{ED})\rangle$

I'm not sure I understand the question. My original state was a product state in which the first two particles had the state:

$\frac{1}{\sqrt{2}}(|H_A H_{EC}\rangle + |V_A V_{EC}\rangle)$
and the last two particles had the state
$\frac{1}{\sqrt{2}}(|H_{ED} H_{B}\rangle + |V_{ED} V_{B}\rangle)$

The product has 4 terms:
$\frac{1}{2}(|H_A H_{EC} H_{ED} H_B \rangle + |H_A H_{EC} V_{ED} V_B \rangle + |V_A V_{EC} H_{ED} H_B \rangle + |V_A V_{EC} V_{ED} V_B \rangle)$

If instead you say that particles $A$ and $B$ are in the state
$\frac{1}{\sqrt{2}} (|H_A H_{B}\rangle - |V_A V_{B})\rangle$
and particles $EC$ and $ED$ are in the state
$\frac{1}{\sqrt{2}} (|H_{EC} H_{ED}\rangle - |V_{EC} V_{ED})\rangle$

then the 4-particle composite state would be:

$\frac{1}{2}(|H_A H_{EC} H_{ED} H_B \rangle - |H_A V_{EC} V_{ED} H_B \rangle - |V_A H_{EC} H_{ED} V_B \rangle + |V_A V_{EC} V_{ED} V_B \rangle)$

 

[akvadrako]

Or more precisely, every entangled particle is entangled with every other entangled particle.

Something along these lines is what I'm thinking, though because of the monogamy of entanglement, it can't be that simple.

The reason it looks strange is you can assume Erik will measure 1 and do a Bell test on A&D, then throw away the results where Erik doesn't actually measure 1. Viola, A&D were entangled . If Erik decides not to measure them, it'll still be possible to bin them into buckets that violate a Bell test, either by knowing about the relative phases or the Bell test outcomes.

So maybe Erik's measurement is just as way to discover the relative phases of A&D, allowing the bucketing.

 

[Mentz114]

Something along these lines is what I'm thinking, though because of the monogamy of entanglement, it can't be that simple.

The reason it looks strange is you can assume Erik will measure 1 and do a Bell test on A&D, then throw away the results where Erik doesn't actually measure 1. Viola, A&D were entangled . If Erik decides not to measure them, it'll still be possible to bin them into buckets that violate a Bell test, either by knowing about the relative phases or the Bell test outcomes.

So maybe Erik's measurement is just as way to discover the relative phases of A&D, allowing the bucketing.

Eric cannot measure anything but relative phase because that is unitary. Yes, the data has to be selected (based on Erics outcome) before the Bell violations can be seen.

 

[zonde]

I'm not sure I understand the question. My original state was a product state in which the first two particles had the state:

$\frac{1}{\sqrt{2}}(|H_A H_{EC}\rangle + |V_A V_{EC}\rangle)$
and the last two particles had the state
$\frac{1}{\sqrt{2}}(|H_{ED} H_{B}\rangle + |V_{ED} V_{B}\rangle)$

The product has 4 terms:
$\frac{1}{2}(|H_A H_{EC} H_{ED} H_B \rangle + |H_A H_{EC} V_{ED} V_B \rangle + |V_A V_{EC} H_{ED} H_B \rangle + |V_A V_{EC} V_{ED} V_B \rangle)$

Ok, I will take a step back.

Following the way how it is explained in https://arxiv.org/abs/1203.4834:

$\frac{1}{2}(|H_A H_{EC} H_{ED} H_B \rangle + |H_A H_{EC} V_{ED} V_B \rangle + |V_A V_{EC} H_{ED} H_B \rangle + |V_A V_{EC} V_{ED} V_B \rangle)$
can be rewritten as:
$\frac{1}{4}( (|H_A H_{EC} H_{ED} H_B \rangle + |V_A H_{EC} H_{ED} V_B \rangle + |H_A V_{EC} V_{ED} H_B \rangle + |V_A V_{EC} V_{ED} V_B \rangle) +$
$+ (|H_A H_{EC} H_{ED} H_B \rangle - |V_A H_{EC} H_{ED} V_B \rangle - |H_A V_{EC} V_{ED} H_B \rangle + |V_A V_{EC} V_{ED} V_B \rangle) +$
$+ (|H_A H_{EC} V_{ED} V_B \rangle + |V_A H_{EC} V_{ED} H_B \rangle + |H_A V_{EC} H_{ED} V_B \rangle + |V_A V_{EC} H_{ED} H_B \rangle) +$
$+ (|H_A H_{EC} V_{ED} V_B \rangle - |V_A H_{EC} V_{ED} H_B \rangle - |H_A V_{EC} H_{ED} V_B \rangle + |V_A V_{EC} H_{ED} H_B \rangle) )$
and it is equal to:
$\frac{1}{4}( (|H_A H_B\rangle + |V_A V_B\rangle) \otimes (|H_{EC} H_{ED}\rangle + |V_{EC} V_{ED}\rangle) +$
$+ (|H_A H_B\rangle - |V_A V_B\rangle) \otimes (|H_{EC} H_{ED}\rangle - |V_{EC} V_{ED}\rangle) +$
$+ (|H_A V_B\rangle + |V_A H_B\rangle) \otimes (|H_{EC} V_{ED}\rangle + |V_{EC} H_{ED}\rangle) +$
$+ (|H_A V_B\rangle - |V_A H_B\rangle) \otimes (|H_{EC} V_{ED}\rangle - |V_{EC} H_{ED}\rangle) )$

Does it seem right so far?

 

[DrChinese]

If Erik decides not to measure them, it'll still be possible to bin them into buckets that violate a Bell test, either by knowing about the relative phases or the Bell test outcomes.

If Erik makes no measurement: All you would ever see at A & D would be completely random combinations. No pattern, no Bell test violations.

 

[stevendaryl]

Ok, I will take a step back.

Following the way how it is explained in https://arxiv.org/abs/1203.4834:

$\frac{1}{2}(|H_A H_{EC} H_{ED} H_B \rangle + |H_A H_{EC} V_{ED} V_B \rangle + |V_A V_{EC} H_{ED} H_B \rangle + |V_A V_{EC} V_{ED} V_B \rangle)$
can be rewritten as:
$\frac{1}{4}( (|H_A H_{EC} H_{ED} H_B \rangle + |V_A H_{EC} H_{ED} V_B \rangle + |H_A V_{EC} V_{ED} H_B \rangle + |V_A V_{EC} V_{ED} V_B \rangle) +$
$+ (|H_A H_{EC} H_{ED} H_B \rangle - |V_A H_{EC} H_{ED} V_B \rangle - |H_A V_{EC} V_{ED} H_B \rangle + |V_A V_{EC} V_{ED} V_B \rangle) +$
$+ (|H_A H_{EC} V_{ED} V_B \rangle + |V_A H_{EC} V_{ED} H_B \rangle + |H_A V_{EC} H_{ED} V_B \rangle + |V_A V_{EC} H_{ED} H_B \rangle) +$
$+ (|H_A H_{EC} V_{ED} V_B \rangle - |V_A H_{EC} V_{ED} H_B \rangle - |H_A V_{EC} H_{ED} V_B \rangle + |V_A V_{EC} H_{ED} H_B \rangle) )$
and it is equal to:
$\frac{1}{4}( (|H_A H_B\rangle + |V_A V_B\rangle) \otimes (|H_{EC} H_{ED}\rangle + |V_{EC} V_{ED}\rangle) +$
$+ (|H_A H_B\rangle - |V_A V_B\rangle) \otimes (|H_{EC} H_{ED}\rangle - |V_{EC} V_{ED}\rangle) +$
$+ (|H_A V_B\rangle + |V_A H_B\rangle) \otimes (|H_{EC} V_{ED}\rangle + |V_{EC} H_{ED}\rangle) +$
$+ (|H_A V_B\rangle - |V_A H_B\rangle) \otimes (|H_{EC} V_{ED}\rangle - |V_{EC} H_{ED}\rangle) )$

Does it seem right so far?

It looks right.

 

[zonde]

It looks right.

But isn't it now obvious that detecting together
$(|H_{EC} H_{ED}\rangle + |V_{EC} V_{ED}\rangle)$ with $(|H_{EC} H_{ED}\rangle - |V_{EC} V_{ED}\rangle)$
would put together
$(|H_A H_B\rangle + |V_A V_B\rangle)$ with $(|H_A H_B\rangle - |V_A V_B\rangle)$
so it won't be entangled state?

 

[stevendaryl]

But isn't it now obvious that detecting together
$(|H_{EC} H_{ED}\rangle + |V_{EC} V_{ED}\rangle)$ with $(|H_{EC} H_{ED}\rangle - |V_{EC} V_{ED}\rangle)$
would put together
$(|H_A H_B\rangle + |V_A V_B\rangle)$ with $(|H_A H_B\rangle - |V_A V_B\rangle)$
so it won't be entangled state?

Why not? After projection with $\Pi_A$, you'll be in the state:

$1/2 (|H_A H_{EC} H_{ED} H_B\rangle + |V_A V_{EC} V_{ED} V_B\rangle)$

Alice's and Bob's photons seem entangled to me: they're either both H or both V, but it's not predetermined which.

 

[zonde]

Why not? After projection with $\Pi_A$, you'll be in the state:

$1/2 (|H_A H_{EC} H_{ED} H_B\rangle + |V_A V_{EC} V_{ED} V_B\rangle)$

Alice's and Bob's photons seem entangled to me: they're either both H or both V, but it's not predetermined which.

But they have to be correlated in diagonal basis too. But they won't be.

 

[kurt101]

It just shows that the intuition that particles become entangled by interaction is not correct. Particles don't need to have ever interacted in order to be entangled.

What do you mean by interacted? When $B$ interacts with $C$ you could hypothesis that $B$ collapses the result of its interaction to $A$ and $C$ collapses the result of it interaction to $D$ so that when you measure $A$ and $D$ they are correlated. If it happens in this way then interaction is a requirement for the entanglement.

 

[stevendaryl]

What do you mean by interacted? When $B$ interacts with $C$ you could hypothesis that $B$ collapses the result of its interaction to $A$ and $C$ collapses the result of it interaction to $D$ so that when you measure $A$ and $D$ they are correlated. If it happens in this way then interaction is a requirement for the entanglement.

But the entanglement between A and B happens afterward when Eric performs his measurement---long after A and C have gone their separate ways, and long after D and B have gone their separate ways.

 

[stevendaryl]

But they have to be correlated in diagonal basis too. But they won't be.

What definition of "entangled" are you using?

 

[zonde]

What definition of "entangled" are you using?

I don't understand why it matters. We are talking about best known type of entanglement - two polarization entangled photons. You can test your definition of entanglement with this type of entanglement and not really the other way around.

 

[kurt101]

You originally said $A$ and $B$ were one of the entangled pairs before the measurement:

If you produce one pair of entangled particles, $A$ and $B$, and another pair, $C$ and $D$, then a measurement involving $B$ and $C$ can force $A$ and $D$ to become entangled (I think)

Now you are saying $A$ and $B$ are entangled after the measurement:

But the entanglement between A and B happens afterward when Eric performs his measurement---long after A and C have gone their separate ways, and long after D and B have gone their separate ways.

Did you change the experiment?

 

[stevendaryl]

You originally said $A$ and $B$ were one of the entangled pairs before the measurement:

I may have gotten my own naming convention screwed up, but no. Originally, we have 4 particles:

1. $A$ and $EC$, which are entangled. $A$ is sent to Alice and $EC$ is sent to Eric.
2. $ED$ and $B$, which are entangled. $B$ is sent to Bob and $ED$ is sent to Eric.
3. At this point, $A$ and $B$ are NOT entangled.
4. But Eric performs a measurement on $EC$ and $ED$ which causes $A$ and $B$ to become entangled.

If I ever said that A and B were an entangled pair to start with, I screwed up my names. That was never the intention.

 

[stevendaryl]

I don't understand why it matters.

Because they certainly are entangled, by the definition that I know of: Two particles are entangled if their measurement results are correlated, but neither measurement result is predetermined.

 

[stevendaryl]

Because they certainly are entangled, by the definition that I know of: Two particles are entangled if their measurement results are correlated, but neither measurement result is predetermined.

To me, there's two different phenomena: (1) Entanglement of states, and (2) Bell-violating correlations. The Bell-violating correlations imply that two systems have entangled states, but not vice-versa. There can be at most two systems that have Bell-violating correlations, but there can be any number of systems that are entangled.

I'm not sure that the terminology is standardized. But to me, entanglement is the underlying mechanism, and Bell-violations is the symptom. The mechanism doesn't always lead to Bell violations.

 

[akvadrako]

If Erik makes no measurement: All you would ever see at A & D would be completely random combinations. No pattern, no Bell test violations.

As far can be known, we would get the same random values if Erik makes a measurement. It's just that Erik picks out pairs that (on average) violate the Bell test along some basis. Seems like his actions don't have any effect on the individual pairs of A+D, like entangling them. A simple story of the experiment which doesn't contain any nonlocal aspect (except the Bell test) goes like this:

1. Two pairs of photons with entangled polarisations A+B (Alice) and C+D (Bob) are created independently.
2. We assume pair Alice is entangled with pair Bob along some random basis.
3. The relative phase between B+C is measured, revealing the relative phase between A+D.
4. Now we know along what basis A+D will violate a Bell Test.

I'm open to the possibility that this scheme may not work, but to understand what this experiment is telling us, it would be good to know why.

 

[stevendaryl]

As far can be known, we would get the same random values if Erik makes a measurement. It's just that Erik picks out pairs that (on average) violate the Bell test along some basis. Seems like his actions don't have any effect on the individual pairs of A+D, like entangling them. A simple story of the experiment which doesn't contain any nonlocal aspect (except the Bell test) goes like this:

1. Two pairs of photons with entangled polarisations A+B (Alice) and C+D (Bob) are created independently.
2. We assume pair Alice is entangled with pair Bob along some random basis.
3. The relative phase between B+C is measured, revealing the relative phase between A+D.
4. Now we know along what basis A+D will violate a Bell Test.

I'm open to the possibility that this scheme may not work, but to understand what this experiment is telling us, it would be good to know why.

I'm not sure what you mean by "we know along what basis A+D will violate a Bell Test". The Bell test involves measurements at multiple (at least two) angles.

I've already gotten in trouble with the names, but I specifically sent particle B to Bob.

Anyway, if Eric measures the middle two (one entangled with Alice's particle, one entangled with Bob's) and finds that the pair are in the state:

$|HH\rangle + |VV\rangle$

then that state is actually independent of the basis. If you take another orthogonal basis--$|X\rangle = cos(\theta) |H\rangle + sin(\theta) |V\rangle$ and $|Y\rangle = -sin(\theta) |X\rangle + cos(\theta) |Y\rangle$, then

$|HH\rangle + |VV\rangle = |XX\rangle + |YY\rangle$

 

[DrChinese]

...The relative phase between B+C is measured, revealing the relative phase between A+D...

Oops. This is experimentally incorrect. You can measure the relative phase between B & C with causing a swap. You simply measure B and C separately, not allowing them to be in a Bell state. Now you know their relative phase. A & D will not be entangled, and no Bell test violation.

 

[kurt101]

Of course, relativistic QFT has no problem with EPR, because it describes on the one hand the observed strong correlations described by entanglement without violating relativistic causality, and it does this by construction, fulfilling the socalled Linked Cluster Principle. This clearly shows that entanglement doesn't imply any "spooky action at a distance".

It describes that they exist, certainly, but not how they come about (i.e. what's causing the strong correlations) that's the contentious part.

But as for any state the cause of entanglement is the preparation procedure to bring the system into this state.

Firstly can't you have entanglement between systems that have never been in physical contact due to vacuum entanglement, it's not always our preparations that do it possibly? I might be wrong on this.

After reading through this thread many times, I think vanhees71 is correct to say "the cause of entanglement is the preparation procedure to bring the system into this state." I would like to see a counter to this.

In the example of entanglement swapping, a local interaction of the photons seems to be the cause of entanglement between the photons. Using the stevendaryl description, Eric performing the measurement is the local interaction that causes the non-local collapse of state that is propagated to Alice and Bob. In other words the cause of entanglement is the local interaction and the result is the non-local collapse.

That being said, I don't see how this relates to the discussion on how QFT clearly shows that entanglement doesn't imply any "spooky action at a distance".

I am inclined to agree with stevendaryl post in so much that #2 (the Born rule) is what describes the entanglement and collapse. And so clearly showing that there is no "spooky action at a distance" would need to involve it.

Right. Whether it's QFT or nonrelativistic quantum mechanics, the theory has two parts:
1. Probability amplitudes are computed using unitary evolution.
2. Upon a measurement, measurement results occur nondeterministically with probabilities given by the Born rule (the square of an appropriate amplitude)
The issue about nonlocality is about #2, not #1. So pointing out that QFT fulfills the Linked Cluster Principle is irrelevant. (Well, it's relevant to proving that part #1 satisfies locality, but it's not relevant to part #2).

 

[DarMM]

After reading through this thread many times, I think vanhees71 is correct to say "the cause of entanglement is the preparation procedure to bring the system into this state." I would like to see a counter to this.

It's a cause, but doesn't fully explain the correlations, that's more the point. In the sense that there is no event $E$ such that the probabilities of the entanglement measurements $A$ and $B$ obey:
$$p(A \wedge B | E) = p(A|E)p(B|E)$$
This is what is usually the minimum required for something to be an explanation statistically (see Reichenbach's principle). So there is no dispute that the preparation is a cause, I never disagreed with that, but it doesn't fully explain the entanglement.

See: https://arxiv.org/abs/1503.06413

Edit: In simpler language, we don't know how the correlations are achieved.

 

[akvadrako]

Oops. This is experimentally incorrect. You can measure the relative phase between B & C with causing a swap. You simply measure B and C separately, not allowing them to be in a Bell state. Now you know their relative phase. A & D will not be entangled, and no Bell test violation.

Maybe relative phase isn't the right word - I'll try to be more correct with my terminology. But you'll have to learn about a joint property of A&D that's incompatible with learning their values, which is what separate measurements of B&C will give you.

 

[akvadrako]

I'm not sure what you mean by "we know along what basis A+D will violate a Bell Test". The Bell test involves measurements at multiple (at least two) angles.

I've already gotten in trouble with the names, but I specifically sent particle B to Bob.

Sorry, I was using names from another paper; I'll stick to yours. What I meant by "basis" is that A&B start out entangled either along the same bases or cross bases.

If it's same, $AB = |HH\rangle + |VV\rangle$, Erik will measure 1, $E = |HH\rangle + |VV\rangle$.
If it's cross, $AB = |HV\rangle + |VH\rangle$, Erik will measure 0, $E = |HV\rangle + |VH\rangle$.

That's at least the basic idea and I think it has some nice properties. (1) There is no "unitary collapse" or retrocausality, just post-selection. (2) Without access to EA&EB, the A&B correlation is forever hidden. To be natural the assumed prior correlation should be fully general; could the prior correlation be any superposition of same and cross and would that ruin the statistics?

 

[Demystifier]

Sure, with philosophical rather than physical arguments ;-)) SCNR.

Said someone who uses philosophical arguments himself. ;-)) SCNR.

 

[stevendaryl]

Sorry, I was using names from another paper; I'll stick to yours. What I meant by "basis" is that A&B start out entangled either along the same bases or cross bases.

If it's same, $AB = |HH\rangle + |VV\rangle$, Erik will measure 1, $E = |HH\rangle + |VV\rangle$.
If it's cross, $AB = |HV\rangle + |VH\rangle$, Erik will measure 0, $E = |HV\rangle + |VH\rangle$.

That's at least the basic idea and I think it has some nice properties. (1) There is no "unitary collapse" or retrocausality, just post-selection. (2) Without access to EA&EB, the A&B correlation is forever hidden. To be natural the assumed prior correlation should be fully general; could the prior correlation be any superposition of same and cross and would that ruin the statistics?

Yeah, that's the possibility I referred to way back in https://www.physicsforums.com/threads/qft-and-entanglement.959593/page-2#post-6089411: Even if the measurement results may not be predetermined, maybe the entanglement relationships are. I don't think that that's consistent, but I don't have a quick argument why not. But note that Eric can measure different things about his pair of particles, leading to different conclusions about the correlation between Alice's particle and Bob's particle. So it doesn't seem to me to be consistent to assume that Alice's and Bob's particles have a pre-existing entanglement relationship.