QFT & Entanglement: NRQT vs. PCT Transform

[Nusc]

Why is it that nonrelativistic quantum theory (NRQT) does not have PCT, whereas QFT does? How would you characterize the collapse of the wave function in terms of PCT? What is its PCT transform?

How does QFT address EPR?

Many physicists universally accept the Copenhagen interpretation and move on as with renormalization.

Those who were discontent with renormalization, acknowledge it, and those like Dirac and Feynman made attempts to tackle the issue. Dirac disregard the notion of the point charge and attribute structure to the electron in handling radiation reaction. He was not successful but made some interesting remarks; he tried to resurrect the notion of the aether arguing how it was not incompatible with relativity when one considers quantum indeterminancy, the other involved monopoles and emphasized the use of strings.

 

[Demystifier]

PCT theorem refers to a property of the Lagrangian or Hamiltonian of the system. The collapse, on the other hand, is not governed by the Lagrangian or Hamiltonian. Hence PCT theorem is not relevant for the collapse.

More generally, collapse and EPR are just two aspects of the measurement problem in quantum theory. The core of this problem does not depend much on the Lagrangian or Hamiltonian of the system. Hence, as far as the measurement problem is concerned, it is not so much important whether one works with NRQM or RQFT, the problem is essentially the same.

 

[vanhees71]

I'd say it's even simpler. While P and T (space reflection and time reversal) are transformations (for many aspects of the world even symmetry transformations) which can be formulated in both special relativity and Newtonian physics and thus in both relativistic and non-relativistic QT. However C (charge conjugation) cannot be even defined in non-relativistic physics, because there's no way to define what an anti-particle is, while this is a necessary concept for any relativistic local and microcausal QFT, including the corresponding symmetry for all physics except if the weak interaction is involved. BTW, it is now independently measured that the weak interaction violates all the discrete symmetries built with C, P, and T, i.e., C, P, T themselves and also CP. What it doesn't violate, however, is the combined "grand reflection" CPT, and that's in some sense good news since any relativistic local and microcausal QFT, the fundament of the entire Standard Model of elementary particle physics, necessarily implies symmetry ander the "grand reflection", CPT.

There's no need to bring EPR up here at all. 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's only the nonsensical idea of collapse introduced by some early Copenhagen physicists. AFAIK Bohr himself never insisted on the collapse; in this sense he was a wise guy. Of course, Einstein was on the right track in criticizing this Copenhagen flavor of QT vehemently, although the EPR paper is infamous in the sense as Einstein himself rightfully didn't like it, because he had a much better understanding of the putative problem and tension between locality, causality and non-classical long-range correlations described by entanglement. The problem for Einstein was not so much this tension but the long-ranged correlations themselves, but that he correctly described as the "inseparability" of long-distant parts of quantum systems which can be independently observed (like two entangled photons can be registered in principle light years away and still have the strong correlations in entangled properties, usually their polarization state in the usual Bell experiments with polarization-entangled photons from downconversion sources).

 

[Nusc]

Thanks for your input!

BTW, it is now independently measured that the weak interaction violates all the discrete symmetries built with C, P, and T, i.e., C, P, T themselves and also CP.

Do you have the reference article for this?

Also can you respond to this thread?
https://www.physicsforums.com/threads/what-is-mass.959594/#post-6085102
Thanks again!

 

[vanhees71]

One example is the following experiment from the BABAR collaboration

https://arxiv.org/ct?url=https://dx.doi.org/10.1103%2FPhysRevLett.109.211801&v=8b33df54
https://arxiv.org/abs/1207.5832

 

[stevendaryl]

There's no need to bring EPR up here at all. 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".

I would say that the word "clearly" is not appropriate here, since it's possible to come to the opposite conclusion.

 

[DarMM]

There's no need to bring EPR up here at all. 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.

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.

 

[stevendaryl]

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.

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).

 

[vanhees71]

I would say that the word "clearly" is not appropriate here, since it's possible to come to the opposite conclusion.

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

 

[vanhees71]

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, e.g., for polarization entanglement of photon pairs you can use parametric down conversion at an appropriate BBO.

Another very simple way to prepare a highly entangled state between a proton and an electron is to put them in a box together and wait unti hydrogen atoms have formed ;-))).

 

[vanhees71]

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).

2. is just defining the physical meaning of the formalism. There's nothing nonlocal going on. As any part of the theory its justification is simply that it works with real-world experiments/observations all the time!

Of course, this minimal interpretation leaves many philosophers unsatisfied since they want for some reason more from the natural sciences than these are supposed to deliver.

 

[DarMM]

But as for any state the cause of entanglement is the preparation procedure to bring the system into this state, e.g., for polarization entanglement of photon pairs you can use parametric down conversion at an appropriate BBO.

Another very simple way to prepare a highly entangled state between a proton and an electron is to put them in a box together and wait unti hydrogen atoms have formed ;-))).

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.

Secondly, although the preparation might be a cause, it's not a decorrelating explanation, a sub condition of Reichenbach's principle. See: https://arxiv.org/abs/1402.0351
Hence the preparation is necessary, but does not fully account for the correlation.

Of course, this minimal interpretation leaves many philosophers unsatisfied since they want for some reason more from the natural sciences than these are supposed to deliver.

Many people don't find this satisfying, but I don't think they're all philosophers, many physicists think the same.

I mean what you are saying is equivalent to "put water in the ground and the plant grows, anything further is philosophy". A perfectly consistent position and a valid one for QM (where AntiRealist interpretations have strong arguments in their favour). However I don't think the equivalent of "I see, but how exactly does the plant grow?" is silly nonsense that only a philosopher would worry about.

 

[DrChinese]

Firstly can't you have entanglement between systems that have never been in physical contact ...

Entanglement swapping is such an example. In fact, the entangled particles need not have ever co-existed.

 

[kurt101]

Entanglement swapping is such an example. In fact, the entangled particles need not have ever co-existed.

But with entanglement swapping there is still a mutual friend particle that each of the particles that never met each other had to physically interact with in order for the entanglement swapping to occur. So even though there is no direct physical contact, there is indirect physical contact (i.e. physical contact through the mutual friend or chain of friends). At least that is my understanding of it, please correct me if I am wrong!

 

[stevendaryl]

But with entanglement swapping there is still a mutual friend particle that each of the particles that never met each other had to physically interact with in order for the entanglement swapping to occur. So even though there is no direct physical contact, there is indirect physical contact (i.e. physical contact through the mutual friend or chain of friends). At least that is my understanding of it, please correct me if I am wrong!

I don't think so. 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). In terms of "mutual friends", $B$ and $C$ never met until after $A$ and $D$ were far away.

 

[DarMM]

then a measurement involving $B$ and $C$ can force $A$ and $D$ to become entangled (I think)

...and I thought QM was odd before. I think my head just short-circuited!

 

[Hendrik Boom]

I don't think so. 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). In terms of "mutual friends", $B$ and $C$ never met until after $A$ and $D$ were far away.

Wouldn't the person doing both measurements then be the particle connecting both systems?

 

[stevendaryl]

Wouldn't the person doing both measurements then be the particle connecting both systems?

Let's make up a cast of characters: Alice, Bob, Charlie, Daniel, Eric.

1. Charlie creates a pair of particles and sends one to Alice and the other to Eric.
2. Daniel creates a pair of particles and sends one to Bob and the other to Eric.
3. Eric performs a measurement on the pair of particles he received.
4. Afterwards, Alice's particle and Bob's particle are entangled.

Alice and Bob have no mutual friends. Alice has only interacted with Charlie. Bob has only interacted with Daniel. Eric has never interacted with either Alice or Bob. There is an interaction chain connecting Alice to Bob, but it goes both forward and backward in time: Bob backward to Daniel, forward to Eric, backward to Charlie, forward to Alice. So Eric is sort of the mutual friend connecting Alice and Bob, but he's not in their past---they never met Eric or received anything from him.

 

[Mentz114]

Let's make up a cast of characters: Alice, Bob, Charlie, Daniel, Eric.

1. Charlie creates a pair of particles and sends one to Alice and the other to Eric.
2. Daniel creates a pair of particles and sends one to Bob and the other to Eric.
3. Eric performs a measurement on the pair of particles he received.
4. Afterwards, Alice's particle and Bob's particle are entangled.

Alice and Bob have no mutual friends. Alice has only interacted with Charlie. Bob has only interacted with Daniel. Eric has never interacted with either Alice or Bob. There is an interaction chain connecting Alice to Bob, but it goes both forward and backward in time: Bob backward to Daniel, forward to Eric, backward to Charlie, forward to Alice. So Eric is sort of the mutual friend connecting Alice and Bob, but he's not in their past---they never met Eric or received anything from him.

Fascinating. I think that Eric has to perform something that leads to 'unitary collapse' - like total interference or state recombination. If Eric uses an interference device then we have to pick those cases where no photon emerges. This is based on my recollection of the 2016 experiment* in Delft where two diamond nitrogen vacancies are entangled by swapping with emited photons.

*arXiv:1508.05949v1

 

[DrChinese]

I don't think so. 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). In terms of "mutual friends", $B$ and $C$ never met until after $A$ and $D$ were far away.

That's correct. The "meeting" of B & C can occur at any time, either before or after measurement of entangled A & D.

 

[DrChinese]

...and I thought QM was odd before. I think my head just short-circuited!

Yes, entanglement swapping is one of the strangest. The final entangled particles, A & D, can even be from fully independent laser sources.

 

[stevendaryl]

Fascinating. I think that Eric has to perform something that leads to 'unitary collapse' - like total interference or state recombination. If Eric uses an interference device then we have to pick those cases where no photon emerges. This is based on my recollection of the 2016 experiment* in Delft where two diamond nitrogen vacancies are entangled by swapping with emited photons.

*arXiv:1508.05949v1

I don't know how it would work with a feasible experiment, but in principle, the 4-particle state would initially look like this:

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

where $H_A$ means "Alice's photon is horizontally polarized", $H_B$ means "Bob's photon ...", $V_A$ means "Alice's photon is vertically polarized", $V_B$ means "Bob's photon", $H_{EC}$ means "Eric's photon from Charlie is Horizontally polarized", $H_{ED}$ means "Eric's photon from Daniel is horizontally polarized, etc.

Now, what I don't know how to do, practically, is to measure a property of a pair of photons. But in principle, if there is some operator $A$ such that:

• $A$ has eigenvalues 0 or 1
• $|H_{EC}, H_{ED}\rangle$ and $|V_{EC}, V_{ED}\rangle$ have eigenvalue 1
• $|H_{EC}, V_{ED}\rangle$ and $|V_{EC}, H_{ED}\rangle$ have eigenvalue 0

then if Eric measures 1 for his photon pair, then that will put Alice's and Bob's photons into the entangled state

$\frac{1}{\sqrt{2}} (|H_A, H_B\rangle + |V_A, V_B\rangle)$

If Eric measures 0 for his photon pair, then that will put Alice's and Bob's photons into the entangled state

$\frac{1}{\sqrt{2}} (|H_A, V_B\rangle + |V_A, H_B\rangle)$

 

[kurt101]

I don't think so. 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). In terms of "mutual friends", $B$ and $C$ never met until after $A$ and $D$ were far away.

I understand what you are saying: $A$ or $B$ could not actually pass something to $D$ and vice versa without the entanglement behavior of instant propagation of state over any distance.

 

[akvadrako]

This experiment is always very fascinating. I've tried to understand it better a few times and one question I still have is: how do we know that Eric's measurement doesn't just bin A & D in two buckets, revealing pre-existing correlations? Maybe A & D are always entangled along one of two bases, even if B & C never meet. I have no problem with a more exotic explanation, but it's not clear why it's needed.

 

[stevendaryl]

This experiment is always very fascinating. I've tried to understand it better a few times and one question I still have is: how do we know that Eric's measurement doesn't just bin A & D in two buckets, revealing pre-existing correlations? Maybe A & D are always entangled along one of two bases, even if B & C never meet. I have no problem with a more exotic explanation, but it's not clear why it's needed.

Particles A and D were created by two difference processes. The only connection between them is something that Eric does. So saying that they were already entangled would seem to be saying that every pair of particle is entangled. Or more precisely, every entangled particle is entangled with every other entangled particle.

That sounds like a hidden-variable theory, but instead of the hidden variable describing measurement outcomes, the hidden variables describe entanglement relationships. I have to think about whether Bell's theorem could be used to disprove this idea. I'm not sure.

 

[zonde]

Now, what I don't know how to do, practically, is to measure a property of a pair of photons. But in principle, if there is some operator $A$ such that:

• $A$ has eigenvalues 0 or 1
• $|H_{EC}, H_{ED}\rangle$ and $|V_{EC}, V_{ED}\rangle$ have eigenvalue 1
• $|H_{EC}, V_{ED}\rangle$ and $|V_{EC}, H_{ED}\rangle$ have eigenvalue 0

then if Eric measures 1 for his photon pair, then that will put Alice's and Bob's photons into the entangled state

$\frac{1}{\sqrt{2}} (|H_A, H_B\rangle + |V_A, V_B\rangle)$

If Eric measures 0 for his photon pair, then that will put Alice's and Bob's photons into the entangled state

$\frac{1}{\sqrt{2}} (|H_A, V_B\rangle + |V_A, H_B\rangle)$

Something is not quite right here. You have to sort out the quartets into four bins not two.
You have to have measurement that tells apart at least one of these four states (usually it's two states that can be singled out and the other two go together):
$\frac{1}{\sqrt{2}} (|H_{EC}, H_{ED}\rangle + |V_{EC}, V_{ED}\rangle)$
$\frac{1}{\sqrt{2}} (|H_{EC}, H_{ED}\rangle - |V_{EC}, V_{ED}\rangle)$
$\frac{1}{\sqrt{2}} (|H_{EC}, V_{ED}\rangle + |V_{EC}, H_{ED}\rangle)$
$\frac{1}{\sqrt{2}} (|H_{EC}, V_{ED}\rangle - |V_{EC}, H_{ED}\rangle)$

 

[zonde]

This experiment is always very fascinating. I've tried to understand it better a few times and one question I still have is: how do we know that Eric's measurement doesn't just bin A & D in two buckets, revealing pre-existing correlations? Maybe A & D are always entangled along one of two bases, even if B & C never meet. I have no problem with a more exotic explanation, but it's not clear why it's needed.

It's because you can violate Bell inequality using measurements of A & D, so A & D can't have pre-existing correlations.

 

[stevendaryl]

Something is not quite right here. You have to sort out the quartets into four bins not two.
You have to have measurement that tells apart at least one of these four states (usually it's two states that can be singled out and the other two go together):
$\frac{1}{\sqrt{2}} (|H_{EC}, H_{ED}\rangle + |V_{EC}, V_{ED}\rangle)$
$\frac{1}{\sqrt{2}} (|H_{EC}, H_{ED}\rangle - |V_{EC}, V_{ED}\rangle)$
$\frac{1}{\sqrt{2}} (|H_{EC}, V_{ED}\rangle + |V_{EC}, H_{ED}\rangle)$
$\frac{1}{\sqrt{2}} (|H_{EC}, V_{ED}\rangle - |V_{EC}, H_{ED}\rangle)$

Yes, that's a complete set of two-photon states. But you if you let $\Pi_1$ be the projection operator for state 1, and $\Pi_2$ be the projection operator for state 2, etc., then you can define a new projection operator $\Pi_A = \Pi_1 + \Pi_2$. This operator has 2 eigenstates with eigenvalue +1 (the first two) and two eigenstates with eigenvalue 0 (the last two).

 

[stevendaryl]

It's because you can violate Bell inequality using measurements of A & D, so A & D can't have pre-existing correlations.

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.

 

[zonde]

Yes, that's a complete set of two-photon states. But you if you let $\Pi_1$ be the projection operator for state 1, and $\Pi_2$ be the projection operator for state 2, etc., then you can define a new projection operator $\Pi_A = \Pi_1 + \Pi_2$. This operator has 2 eigenstates with eigenvalue +1 (the first two) and two eigenstates with eigenvalue 0 (the last two).

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

 

[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:

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