Relativistic description of a fixed number of particles implies that they can't interact. The reason, essentially, is that, in the presence of such an interaction, particles accelerate and, inevitably, the energy in the center of mass frame of any two such particles will exceed the rest mass of any finite number of particles, which will lead to particle production, thus the number of particles can't remain fixed. So the only way for the number to remain fixed is for there to be no interaction-which includes the case of an integrable, many body, system, since that's a system of free particles in disguise.

Not all quantum field theories are invariant under global Lorentz transformations and space-time translations, nor are all Poincaré invariant classical theories the limits of corresponding quantum theories.

Hi, Stam,

Can you concentrate on the case in which the particles don't interact? I am asking the present question in connection with entanglements. After two particles are entangled, they are usually let fly far away from one another, s.t. of course they don't interact anymore. Whatever additional particles are generated together with these two, are removed by all sort of techniques. So, we remain to experiment with exactly two particles.

By the way, I watched the 1st Coleman's lecture, and I saw that he explains that in interactions, QFT has to take in considerations multiple pair-production. But, as said above, I am interested in the situation after interaction and filtration out of the unwanted particles. 

To be specific, can you tell me how do transform the operators â† and b̂† under a Lorentz transformation? Assume that â† and b̂† describe two spinless bosons, e.g. two 4He particles. Assume also that in the lab frame the particles differ only by that after being produced, one of them was let fly to the East and the other to the West.

With thanks in advance,

Sofia

The creation and annihilation operators of the free scalar field are defined as the Fourier transforms of the field operator that correspond to positive, respectively negative, frequencies. Therefore, since the scalar field transforms as a scalar, under Lorentz transformations, so do these operators. (Their arguments, of course, transform as 4-vectors.) The reason is that the Fourier transform is a unitary transformation, that commutes with the Lorentz transformations, so doesn't change the Lorentz transformations of the fields. That's all there is to it. 

However, while ak+ can be defined as the creation operator of a He4 nucleus, of momentum k, since all such particles are identical, there doesn't exist another operator for a second such particle. The two-particle state of two identical particles  is given by an expression of the form |k1,k2>=ak1+ak2+ |0>, where |0> is the vacuum state (up to symmetrization, of course). In the center of mass frame, the energy-momentum 4-vectors  have the same value for the energy and opposite values for the momenta. The point is that ak+ is the Fourier transform of φ(x),  a scalar field, so ak+ has the same Lorentz properties as φ(x), where -k refers to the 4-vector, where only the space-like components are flipped.

If the problem refers to two different particles, excitations of two different scalar fields, φ1(x) and φ2(x), same considerations apply, only there, indeed, the creation operators are different, e.g. ak+ for φ1 and b-k+ for φ2, in the center of mass frame, so the 2-particle state is ak+b-k+  |0>, for example. These creation (and corresponding annihilation) operators are Lorentz scalars. 

No, a theory of non-interacting particles doesn't have any issues with unitarity. Example: any integrable quantum field theory. The interaction is a fiction of the parametrization used and the description of any  fixed number of particles is consistent, since the number operator commutes with the Hamiltonian, which is hermitian, so the evolution operator is unitary. 

Perturbation theory isn't relevant in this discussion. Nor is the number of spacetime dimensions. 

It's not useful to make appeals to authority-especially when they can be easily checked to be wrong. Constructive quantum field theory is definitely not limited to perturbation theory, nor to finite volume, where it's known for more than a century  that no phase transitions occur (since tunneling occurs in finite volume); the study of phase transitions is  one of the main topics of the book quoted-that  deserves being studied, also. Nor are all expansions described therein identical to perturbative expansions in the coupling constant. And there's been considerable work since the last edition. However quantum field theory isn't limited to perturbation theory or to constructive techniques for certain classes of theories-for more than forty years it's known how to study quantum field theories using computer simulations in a way that can probe effects that are beyond strong or weak coupling expansions, or non-analytic in the coupling constant. The cutoff independence is realized in the scaling limit, where the lattice spacing is taken to zero, the volume is taken to infinity and coupling constant(s) are tuned to critical values. Finite volume isn't enough-and a test that the algorithms are correct is that tunneling can be measured in finite volume.  Of course the limiting situation is described by an extrapolation from finite values. However the final result is a description of fields. If they are interacting, then particle number isn't conserved; if they are free fields, then it is.

Luiz,

This is not a way to address people. I suggest you to correct your post to Stam and withdraw the insulting words. Insults are not a scientific arguments, they show lack of arguments. I trust that you are able to find scientific arguments. Keep in mind that this is just a scientific debate, nobody here is AT WAR with nobody, we are all friends. No need for nerves!

Besides, both you and Stam argue about interacting particles, while I am intersted in NON-interacting particles, e.g. two particles that fly far away from one another. This is the case typical to an entanglement. As to the additional particles that may be created when the pair of two particles is prepared, they are not interesting in entanglements s.t. the experimenters carefully filter them out. Thus, the experiment refers in continuation to the two particles that separate spacely.

So, would you kindly refer to the case in which I am interested? That is what would help me.

Best regards,

Sofia

If the particles are ``non-interacting'', then it's possible to show that the generators of the Poincaré group, expressed in terms of the fields-or their Fourier transforms, the creation and annihilation operators-do  commute with the number operator, as mentioned above. So, in that case, it is possible to have a quantum field theory, that is Lorentz invariant,  with a fixed number of particles. What this means is that the Hilbert space breaks up into sectors labeled by particle number and there aren't any transitions from one to another.  This is consistent with Leutwyler's theorem.

The quotation marks are to stress that non-interacting means no potential term; but that one must impose commutation or anticommutation relations on the fields, and, thus, on the creation and annihilation operators, in three spatial dimensions, depending on their spin, and the particles do, in fact, have an effective interaction, this way. This interaction is not inconsistent with Leutwyler's theorem, because in three spatial dimensions, Poincaré invariance can be shown to imply the spin-statistics theorem. 

Luiz, 

Allow me this. Entanglement does not have anything to do with relativistic speed in any case. Relativistic considerations are based on the limited speed of energy. Entanglement does not involve energy exchanges.

In terms of fields one is computing a 2->2 scattering process, where the boundary conditions define the particular problem, so one is computing a 4-point function in a free field theory, that can be expressed in terms of 2-point functions (propagators) from Wick's theorem. The (anti)commutation relations of the fields automatically take into account the corresponding properties of the particles. In fact, in this case, since the number operator commutes with the evolution operator, one is assured to stay within the subspace defined by the initial value of the number operator. It's the (anti)commutation relations that describe the entanglement of the particles, since they imply long range correlations. But this can be seen, in this case,  not to lead to violations of Poincaré invariance, nonetheless. 

Luiz,

NOTHING, really NOTHING, justifies insults. We are all FRIENDS here. Besides, you are the only one who sees arrogance in Stam. He is a kind fellow, and he tells you what is known to him, as you are telling what is known to you.

Please see, I am not a specialist in QFT - to my deep regret - I currently investigate in the non-relativistic QM. But I am a real-time designer, programmer, debugger. When a program goes wrongly, I check statement by statement until I find the wrong one.

So, here is what I see in the clash between you and Stam. Stam refers to non-interacting particles, as I asked him to do. He says, about non-interacting particles:

==> . . . the description of any fixed number of particles is consistent, since the number operator commutes with the Hamiltonian, which is hermitian, so the evolution operator is unitary. 

On the other hand, you speak of particles that do interact:

==> There is not a relativistic INTERACTING PERTURBATIVE Quantum Field Theory with a fixed number of particles.

So, up to that point you spoke of different things. In such a situation, even if each one of you is right, you'd still clash. But you began to insult him.

For the moment, I stop here with the "debugging" of your controversy. I repeat, in debugging one goes step by step, s.t. further parts of the controversy I'll discuss later.

Best regards, and, please, I insist on FRIENDSHIP,

Sofia

The answer regarding Hardy's paradox is that it isn't relevant, because free fields can be described in a subspace of fixed number of particles. 

Lucien Hardy isn't the same person as Godfrey Harold Hardy. The first is a contemporary physicist, working on quantum computing,  the second a mathematician of the twentieth century, known for work in number theory. There doesn't exist any Ramanujan-Hardy-Titchmarsh-Littlewood paradox. 

The free field Hamiltonian can be expressed in terms of the number operator, that's why evolution is always in a subspace of fixed particle number. However since the particle excitations of a field are indistinguishable, one can only state unambiguously that one is working in a two-particle subspace; the labels of each particle don't have any invariant meaning, under evolution of the state, whether in non-relativistic, or relativistic, quantum mechanics (the latter's known, also, as relativistic quantum field theory).

Dear Luiz,

the quantum entanglement is also a property of bosons which do not obey to the Fermi Dirac statistics and cannot be represented by Spinors. Anton Zelinger proved the possibility to enable transitions of phase in entangled photons at different locations much faster than the speed of  signals (light) connecting the two locations.

Any relativistic equation is meaningless in this case, since 

a) no energy is involved in such process

b) no fermions Dirac or Majorana are involved 

c) the speed of "signals" linking the two photons would not even have a finite value, so they are not signals...

i know you are a very good mathematician but not all well written math can match a correct description of Physical reality.

The only difference between non-relativistic and relativistic quantum mechanics (the latter is known, also, under the name of (relativistic) quantum field theory), for free particles is that, in the non-relativistic approximation the spin-statistics relation is a postulate, whose consequences have to be imposed on the multiparticle states, while in the relativistic case it's a theorem, whose consequences can be taken into account automatically, since they imply that the vacuum state exists, since the energy is bounded from below only if bosons commute and fermions anticommute at space-like separations. Therefore, once one states the spin of the particle excitations, the statistics implies that the 2-particle states will be, by construction of the theory, properly defined-and entanglement, indeed, is an automatic property since it's inevitable that the amplitudes don't factorize. 

For both cases it turns out that the 2->2 evolution process can be described by computing the 4-point function , where F=field, whether boson or fermion.  Since the fields are free, this is expressed in terms of the 2-point functions, by Wick's theorem:

= + +

where 1,2,3,4 label the spacetime points, sources and detectors.  The 2-point functions are known from the action of free fields, that's given-and Lorentz invariance and the stability of the ground state imply the (anti)commutation properties. This expression is then inserted in the expression for the transition amplitude, cf. the first pages of http://physics.indiana.edu/~sg/p622/qft_S5-11.pdf , for the details, and one can immediately remark that the final expression describes everything correctly, entanglement included, which describes the non-factorizability of the quantities involved. One way to understand this is by noticing that the expression for the 4-point function consists of more than one term and, thus, when computing the amplitude and then its modulus squared, the result won't be one product but a sum of products. That's why entanglement requires a separate discussion in non-relativistic quantum mechanics, but not in relativistic quantum mchanics-it's, already, taken into account in the latter case.

if the bosons are photons, where a relativistic description is required by definition, then some care is required, since it's necessary to impose a gauge-fixing condition to define the 2-point function, but, while the amplitude depends on this condition, it is a good exercise to show that the cross section, the transition probability, essentially, that's the measurable quantity,   does not. Another subtle point is that, since photons are massless particles, particular care is required to deal with infrared divergences, that arise and are absent when considering massive particles, such as electrons.

Manifest Lorentz invariance implies, in particular, that no physical signal propagates superluminally. If a non-covariant gauge-fixing condition is imposed, Lorentz invariance is recovered for the final result, that is gauge invariant, thus, independent of the condition.

Dear Stefano,

What is the "phase" of which you speak? Phase, and phase transition, have many meanings in physics. What is the phase transition enabled by Zeilinger?

Now, since you speak of faster-than-light "signals" linking photons, NO! Nothing travels between entangled particles. What connects particles distant from one another is NOT something that travels between them. It is the JOINT AMPLITUDES OF PROBABILITY !!!!!!!!!!! 

Let me try a DETAILED EXAMPLE. Look at this wave-function:

(1) |ψ> = α|x>A |x>B + β|y>A |y>B,      |α|2 + |β|2 = 1,

where A and B are photons, and x, y, polarization. Assume that each photon may come to the polarization beam-splitter (PBSA respectively PBSB) on two ways, a, b, respectively u, v, in a way that if photon A takes the path a (b), the photon B takes the path u (v). The calculus of the wave-function would look as follows

(2) |ψ'> = γa|x>A(ζu |x>B + ηu |y>B) + γb|x>A(ζv |x>B + ηv |y>B)

+ δa|y>A (λu |x>B + ξu |y>B) + δb(λv |x>B + ξv |y>B).

But what the equation (1) tells us is that the joint amplitudes add up, as complex numbers, with the following result:

(3) γaζu + γaζv = α;     δaξu + δbξv = β;     γaηu + γbηv = δaλu + δbλv = 0.

Note that the joint amplitudes are products of local amplitudes. However, dispite the distance between the measurement sites, these products add up as if the particles were at the same place.

In short, if the photon A produces a result "x", the photon B won't be able to produce a result "y" because the cancellation of the joint amplitudes won't let a wave-packet "y" be accessible at the output of the PBSB.

I hope that my example is clear.

Zeilinger measured the phase coherence of photons that were separated by a distance of more than 100 kilometers; nothing to do with any phase transition. So he had a 2-photon state that far apart, that he could show wasn't a product of 1-photon states, but in a superposition of the product.http://www.nature.com/nphys/journal/v3/n7/full/nphys629.html Very nice. 

It isn't correct that the products add as if the particles are in the same place. (First of all the expressions aren't those of the wavefunctions-but of the states. These are distinct quantities.)  That's why Zeilinger's measurement isn't a triviality. The relative phase is the relevant quantity and how to calculate it has been presented more than once. 

It has a part that depends on the statistics of the particles, 0 or π and a part that depends on the evolution. In one dimension one would have 

Ψ(x,y,t)=(ψ1(x,t)ψ2(y,t)+eiθψ1(y,t)ψ2(x,t))/sqrt(2)

wth θ=0 or π, depending on whether the particles are bosons or fermions. If the initial conditions are wavepackets, 

ψ1(x,0)= f(x), ψ2(x,0)=g(x)

with f(x) peaked at x=X and g(x) peaked at x=Y, the initial positions of the particles.

Then 

ψa(x,t)=exp(-iHt)ψa(x,0)

where a=1,2 labels the particles and with H = p2/(2m) for a massive particle; for a photon one really needs to work with electromagnetic fields, they're free, else one gets into all sorts of contorsions.

So it's obvious one uses Fourier transforms, and finds that the two terms have an, additional, relative phase because the wavefunctions in position space aren't eigenfunctions of the evolution operator.

One obtains in this way the expression for Ψ(x,y,t). The joint probability density is 

ρ(x,y,t)=|Ψ(x,y,t)|2

from which one can compute everything. In particular one finds that 

|Ψ(x,y,t)|2≠|ψ1(x,t)|2|ψ2(y,t)|2

which is what entanglement means. And it's due to the relative phase θ, that implies that the 2-particle wavefunction doesn't factorize, in general. It might do so, if the relative phase, due to the evolution could cancel θ; but this can't happen for all times and all positions. 

This example can be readily generalized to more space dimensions.

For fermions, i.e. when θ=π, one does find that ρ(x,x,t)=0, for all x and t.  But the average value of the relative separation is

(t) = integral( (x-y)ρ(x,y,t) dxdy)/integral( ρ(x,y,t) dxdy)

It has a width: 

(t) 

that doesn't vanish identically. So there's no problem.

Stam,

You write in hurry s.t. it's not clear what you talk about.

"Zeilinger measured the phase coherence of photons . . ."

Are you trying to talk about coherent beams?

"It isn't correct that the products add as if the particles are in the same place."

Products of what?

If you are bringing an argument against what I said, the only connection that I see between your post and mine, is your phrase "It isn't correct that the products add as if the particles are in the same place." And I have the very strong feeling that, as it happens many times, people disagree because they don't pay attention that they speak of different things.

Now, not everybody knows that experiment of Zeilinger, and not everybody has time to read an article just for seeing what you talk about. If you mean the dialogue to be between you and Stefano, it's O.K. If you want more people to participate, then maybe you'd say what is that experiment, what it proves. Just recall that people are busy.

Best regards,

Sofia

Just do the calculations and compare them. I spelled out practically all the steps, so it should be straightforward to adapt to any particular example. The only implicit steps involve the Fourier transforms: write ψa(x,t)=Σexp(i E t)φa(x) and so on.

Of course  there are some Gaussian integrals to do.

While I don't have a particular example,it's clear that a calculation that can't capture a relative phase is suspect. So the conclusion in boldface can't be as stated and isn't helpful, since the complete calculation is possible, as indicated.

The reason is that the wavfeunction of particle 2 has a phase difference with respect to that of particle 1, initially, by the fact thst it's peaked in another position. So, in my example g(x)=f(x+a), with a=Y-X. Now f(x+a) = exp(iPa)f(x), where P = -id/dx, the translation operator. This is the relative phase factor, that appears by virtue of the particles not being at the same place initially. And it appears in the solution explicitly. Of course if the distance is macroscopic, its contribution is correspondingly small and does require skill to probe.(hbar=1 has been used).That's why, in the classical limit, its effects are very hard to spot-and it suppresses the effects of the ststistical phase, θ, too,  so the statistics of the particles becomes Boltzmann, as expected. 

Dear Sofia,

"Now, since you speak of faster-than-light "signals" linking photons, NO! Nothing travels between entangled particles. What connects particles distant from one another is NOT something that travels between them. It is the JOINT AMPLITUDES OF PROBABILITY !!!!!!!!!!! "

I know, if you read well I say that it is impossible to have faster than light signals.

The phase transitions you are right I expressed not properly. It is any characteristics of the entity which if changed in a region it is immediately changed in the region where the entangled particle is.

http://www.nature.com/ncomms/2015/150324/ncomms7665/full/ncomms7665.html

Dear Stam,

In my last post I explained how the entanglements work. You seem to have criticised me. But I did not understand what's the criticism.

Let me repeat in general what I said there, for clarity. So, let me name the two particles A and B, and use the subscripts A and B for what pertains to the respective particle.

1) Local results are obtained first of all in base of local amplitudes - the amplitudes of the wave-packets impinging on the local detectors.

2) However, we want to get probabilities of JOINT RESULTS.

3) The nature works with AMPLITUDES, not with probabilities as thought John Bell. The joint amplitude of a measurement result RA and a measurement result RB, is given by the formula

(1) CR_A,R_B = C1,A C1,B + C2,A C2,B + ... Cn,A Cn,B ,

where Ci,A and Ci,B are local amplitudes generated if a particle reaches the respective detector by more than one path. However, the product Ci,A Ci,B is non-local and is a complex number, s.t. the sum of these products may eventually vanish.

4) If at LabA is obtained the result RA, however, the joint amplitude CR_A,R_B = 0, no wave-packet is accessible at the detector meant to record the result RB.

Now, what I ask you is, if you disagree with me, please tell me on which point you disagree and why. You say for instance "It isn't correct that the products add as if the particles are in the same place." Yes, so it is! The joint amplitude doesn't care of the distance between the particles. What's your problem with this?

Also, you mention a certain 2-particle Zeilinger experiment. Stam, my friend, I don't have access to the journal and I leave far from Technion, in another town. I won't buy that article, so, please describe the experiment in a few words or lines, and also which counter-argument you found there.

With kind regards,

Sofia

Dear Sofia, 

The experiment Stam is refering to is the Canary island teleportation experiment at 144 km distance which is alredy another thing. The quanglement was demonstrated in the experiment under the Danubio river in Vienna. 

http://www.nature.com/nature/journal/v489/n7415/fig_tab/nature11472_F1.html

One must check that the conditions claimed can actually be satisfied and the way to do that is to do the calculation. The statement ``no wavepacket is accessible'' can't be given any meaning; the detector measures a wavepacket, this means that it measures Ψ(x,y,t), which is the only wavepacket available, convolved with a function that contains the information of what particular property is measured (cf. below for an example).  That's the important point. As explained in a previous message, it isn't true that it's possible to measure, i.e. isolate  ψ1(x,t), once one has specified that there is a ψ2(x,t), also, around; one can only measure the  superposition

Ψ(x,y,t)=(ψ1(x,t)ψ2(y,t) + eiθψ!(y,t)ψ2(x,t))/sqrt(2). 

When one wishes to measure, at point x, at time t, some property of particle 1, this means, one must sum over all states of particle 2. 

Thank you Stefano!

But, please tell me a few more details. What is the "active feed-forward"? In simple words, what they do in their experiment?

I know in general that in teleportation experiments one entangles one member of a photon polarization singlet with a particle in some polarization state, and sends to the target location the other member of the plarization singlet. Then, by a classical channel, one sends to the target location two logical bits, e.g. 01, or 00, etc.

Is this what they do the experiment? I don't work currently with the teleportation, s.t. I am not aware of the new achievements.

Many thanks for your kindness! 

Stam,

Again I have no idea what you talk about.

==> " . . . it isn't true that it's possible to measure, i.e. isolate ψ1(x,t), once one has specified that there is a ψ2(x,t), also, around;"

Who are ψ1(x,t) and ψ2(x,t)? And what you believe that I isolate? You see, I don't want to make guesses and reply in base of guessing what you mean. Maybe you think that your notations are clear. No, they are not.

If you want your comment to be understood, you should use the notations in my comment; and if they are not clear, please tell me. If you introduce new notations, please define them.

==> " . . . when one  measures the density for the properties of particle 1 one is doing this, in fact,  by integrating over particle 2, |ψ1(x,t)|2=integral(|Ψ(x,y,t)|2 dy)/integral(|Ψ(x,y,t)|2 dxdy)."

Are you serious? I am speaking of joint measurements and you suggest integration over one of the particles? What you talk about?

Stam, it may be again that we have in mind different things, but I can't "see" more because of your independent notations.

My kindest regards!

Dear Luiz,

"I am not well qualified to talk about "entanglement" .But this kind of phenomena does not appear to make sense ("Mr Spock teleportation in Star War.....)."

These were the things which were refuted by Einstien for more than 20 years (1930 -1955) which he brough with himslef to the grave, including the Heisenberg indetermination prinicple. For this reason and many other reasons I think Einstein is overrated both in his discoveries and insigths. Newton and Hamilton were completely another level, they were able to create the math they needed and they applied it to make a formalism and principles  still surviving after centuries. Hamiltonians are the daily meal of any Physicist.

Einstein barely knew the math he needed to apply (tensors) and made  mistakes which I'm not here to talk about, including the  a priori exclusion of "non locality".

So Luiz if you don't accept Quantum entanglement you are in good company it seems. At that time it was the genious of Shrodinger and Heisenberg to bring it forward against many and there was no experimental evidence. Now though there are experiments which prove such concept, thanks God!!!

.As you can  see ,Schrodinger equation is really to be used on phenomenological grounds when applied to Atoms and Molecules and its interaction with relativistic Photons  !

Non relativistic quantum mechanics is embodied in the Shrodinger equation and this is more than enough to handle 98% of the chemistry. Relativistic quantum mechanics includes the spinorial description. Entaglement comes out of the non relativistic quantum mechanics  between photons ( spinors are not needed).

I beg your pardon but Photons are not "relativistic", they are entities travelling at the maximum possible speed. Everything in term of matter  which tries to cope with  c speed in vacuum is usually defined  "relativistic" . The term relativisitc is something I don't see properly used, others define also photons as "ultrarelativistic". Everything with the word "relativistic" starts from the Fitzgerald Lorentz factor gamma introduced in two somehow equivalent ways:

a) preserve  conservations during scattering processes: Newtonian mechanics was not sufficient to comply to conservations.

b) preserve the invariance of the Maxwell equations: Galilean transformations were not enough to preserve their invariance in presence of EM radiation. 

and all this regards basically matter.

People .like Wechsler and Nicolis need to go to 21 Century and leave behind the primeval paradoxes of Einstein , Podolsk ,Cat of

I usually refrain to express judgements on people especially in a negative way. To be able to have a good view of what people know or don't know you cannot base yourself on something written in threads of a forum. Somebody wastes his energy in personal attacks and this is sad in a communitiy of people who is fond of science and is supposed to be more "intelligent" than others. Intelligence and self confidence should overcome our own pride or the scare of being ignorant about an argument.

Luiz I'm  basically ignorant with a incredible determination to learn though. I try to admit my limitations, I ask when I don't know, I refute  "ipse dixit" and I often challange them, I follow the principle of non contradiction and the basic principle of Physics. It is difficult for others to convince me of something since I've seen bad mistakes and misinterpretations made by people who are supposted to master the subject..

And as Sofia told me and I fully agree:

"Entanglements are a basic feature of quantum mechanics, one of the main features that distinguishes the quantum realm from the classical one. They are used in many applications, scientific and/or technological. To begin with the latter, quantum teleportation is one of the applications. About scientific, the proof of the famous violation of Bell's inequality is done in base of entanglements, the HBT effect is due to the entanglement of identical particles."

My dear Stam,

Maybe it would be a good idea to clarify some things that I said. Again, I believe that in fact there is no divergence of opinions between us, just the notations and terminology are the problem.

Therefore, please see the attached picture. Two photons exit the preparation region. For clarity I painted the photon A in orange, and the photon B in blue. The preparation produces the polarization singlet, |ψ> = ( |x> |x> + |y> |y>) and it's in the base { |x>, |y> } that each photon is ultimately measured.

However, for explaining the play of the amplitudes, I chose to pass each photon, first, through a preliminary polarization beam-splitter which discriminates in another base than { |x>, |y> }: PBS1 discriminates in the base { |a>, |b> }, and PBS2 in the base { |u> = |a>, |v> = |b> }. Thus, the polarization |a> of A is coupled with the polarization |u> of B, and |b> of A with |v> of B.

After merging for each photon the two polarizations (see PBS3, respectively PBS4), I do the final splitting in the base { |x>, |y> } to be measured (see PBS5 , PBS6). Thus, in the formula,

(2) |ψ'> = γa|x>A(ζu |x>B + ηu |y>B) + γb|x>A(ζv |x>B + ηv |y>B)

+ δa|y>A (λu |x>B + ξu |y>B) + δb|y>A(λv |x>B + ξv |y>B),

the coefficient preceding each vector, is a LOCAL amplitude. For instance: γa and γb are the projections of the vectors |a> and |b> on the vector |x>, while δa and δb are the projections of these vectors on the vector |y>. ζu is the product of two projections , while ηu = ; analogously, λv = , while ξv = . 

All the partial amplitudes for the photon A, as detailed above, are local. Similarly for B. However, the products as γaζu are NON-local. So is the product γbζv . The sum of these products gives the joint amplitude of obtaining the result |x>A |x>B. But each produce results from a different combination of paths.

Now, Stam, YOU KNOW all these things, they are trivial.

Best regards!

Dear Luiz,

."What QM says is that their interaction coalesce to zero far from scatter center as free particles !.By the way , this Free QM particle Behavior far from the scattering center does not holds for long range potentials like the Coulomb Potential ."

infact we are not talking about scattering, we are talking about generating pairs of photons from a single photon.

https://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion