I How indistinguishable are photons?

[Killtech]

Assume we have a photon pair prepared just so they are suited for a two photon interference. Let's call the corresponding beams A and B. Then we run each beam individually through a 50:50 beam splitter such that we get beams A1, A2, B1, B2 such that we can symbolically write the state of the system as $\frac 1 2 (|A1\rangle + |A2\rangle)(|B1\rangle + (|B2\rangle)$.

Now we can do two different experiments by either bringing beams A1 and A2 into an interference (classic single photon interference) or beams A1 and B1. In either case the amplitudes of the incoming beams should be identical. My question is if and how nature distinguishes between those two scenarios and how does QT/QFT notation handle it.

Normally if A1 and B1 had each an amplitude of 1 then we would expect a two photon interference to happen. But what happens if the amplitudes are reduced to the same level as for a single photon inference? Will we still observe the Hong-Ou-Mandel effect in a quarter of events (two photon detection) with another quarter being no detection at all, or can two halfs of distinct photons create a new whole one (single photon detection all the time).

My question therefore is if the quantum field contains any additional information to track some particle data, or whether it is purely wavelike for photons. If it is the prior, then where would i find this information in the formalism?

 

[Cthugha]

Yes, you will still find the HOM effect in a certain percentage of events.

Assume, that, e.g., at the input port of beam splitter A, you have the ports $A_3$ and $A_4$.
You can then define the beam splitter operation in terms of how the photon operators for the output port modes look in terms of the photon modes at the input ports. Assuming reflection and transmission coefficients r and t (and potentially t' and r' for asymmetric beam splitters), respectively, you get:

\hat{a}_1=r \hat{a}_3 +t' \hat{a}_4
and
\hat{a}_2=t \hat{a}_3 +r' \hat{a}_4.

Considering a 50-50 lossless beam splitter and the phase shift inflicted upon reflection (which reverses one sign), you get:
\hat{a}_1=\frac{1}{\sqrt{2}} (\hat{a}_3 +\hat{a}_4)
and
\hat{a}_2=\frac{1}{\sqrt{2}} (\hat{a}_3 - \hat{a}_4).

Now you can reverse these expressions and express the input modes in terms of the output modes:

\hat{a}_3=\frac{1}{\sqrt{2}} (\hat{a}_1 +\hat{a}_2)
and
\hat{a}_4=\frac{1}{\sqrt{2}} (\hat{a}_1 - \hat{a}_2),

so that when considering the light field states at the output ports in the vacuum state $|0_{A1}0_{A2}\rangle$, you get, e.g., the following:
\hat{a}_3^\dagger |0_{A1}0_{A2}\rangle =\frac{1}{\sqrt{2}} ( |1_{A1}0_{A2} + |0_{A1}1_{A2})
for the output state if you input a single photon in port A3.

The same line of reasoning goes for beam splitter B, of course.
Now you can just go ahead and repeat this for the next beamsplitter and get a state consisting of the photon numbers in these output modes depending on whatever you consider as input modes. You will find that there is a possibility to have one photon at each input port of the second beam splitter and HOM interference if you input beams from A and B, but that there is no possibility to have two photons at the input ports of the second beam splitter if you take both outputs from beam splitter A. Just go ahead and do the math. It is not too tough.

 

[Killtech]

Yes, you will still find the HOM effect in a certain percentage of events.

Assume, that, e.g., at the input port of beam splitter A, you have the ports A3 and A4.
You can then define the beam splitter operation in terms of how the photon operators for the output port modes look in terms of the photon modes at the input ports.

Ah, thanks. I thought as much that HOM effect would still appear but i got uncertain because the formalism didn't naturally distinguish between the cases. Okay, so the trick is to rewrite everything in terms of the modes emitted by the original sources.

The problem i initially started with however was about how the HOM effect handles entangled states and whether there is something non-local to it. But in order to formulate the question i had to be sure about some basics/fundamentals first. Maybe you can help me out with this question, too.

The thing with HOM is that in the calculus the canceling out of the $|1,1\rangle$ states only works when the input modes are indistinguishable and if now two photons interfere that are both in a Bell-like entanglement, it gets less trivial: each input mode is also multiplied by the remote mode of its entangled photon. So for the interference terms to still cancel out it would seem it requires also the remote modes of both photons to be indistinguishable. This is what i am interested in understanding if there is some non-locality in there.

So if we ran a HOM interference on a entangled Bell photon with a normal one, then we would have to make the calculation with modes like $a^\dagger|0_{A1+} 0_{A1-} 0_{A2+} 0_{R1+} 0_{R1-}\rangle$ where R denotes the mode of the remote entangled photon and the +/- their spin configuration. So even though R is no where near the beam splitter it still seems to take a significant role in the interaction and in this case disables the effect in any case.

However, if we had two Bell photon pairs (4 photons in total) then it would seem one can get the interference to work again. While on the one side of the experiment we can interfere one photon of each pair first - but as their remote modes will be distinguishable the effect shouldn't appear. Doing the same for the other side of the experiment but just a little later changes things since for those two their remote modes cannot be distinguished anymore after they were interfered which should allow the middle terms in the HOM calculation to cancel out.

My problem is that i am not sure how the calculation really handles the remote modes and my first try at it lead me to this.

 

[Cthugha]

That is quite a long story that depends on the actual states you look at and what you do behind the beam splitter. Maybe the following overview article helps:
https://arxiv.org/pdf/2006.09335.pdf

It might be instructive to first look at the easier case described in section I.C.: Here, the text discusses what happens if two photons from an entangled photon pair meet at a beam splitter for maximally entangled states. The result actually depends on which of the maximally entangled states (or Bell states) are used. For 3 of them you will get the HOM dip. For the fourth state you will not.

The case of two photons from different pairs is a bit more difficult. It is treated briefly in section IV.A and references therein. In a nutshell, this is the basic setup for performing entanglement swapping and what happens depends on what states and measurements you use.

 

[Killtech]

That is quite a long story that depends on the actual states you look at and what you do behind the beam splitter. Maybe the following overview article helps:
https://arxiv.org/pdf/2006.09335.pdf

It might be instructive to first look at the easier case described in section I.C.: Here, the text discusses what happens if two photons from an entangled photon pair meet at a beam splitter for maximally entangled states. The result actually depends on which of the maximally entangled states (or Bell states) are used. For 3 of them you will get the HOM dip. For the fourth state you will not.

The case of two photons from different pairs is a bit more difficult. It is treated briefly in section IV.A and references therein. In a nutshell, this is the basic setup for performing entanglement swapping and what happens depends on what states and measurements you use.

This is exactly what i was looking for, thanks. I haven't read it all yet but the sections you mention are just what i was looking for... except that they stop when it gets interesting.

IV.A a) is indeed exactly what i had in mind and good to know the proper terminology for it: entanglement swapping. It mentions that after doing HOM with AB, A'B' end up becoming entangled just as i figured. Now section I.C discusses that certain type of entanglements give a characteristic pattern when run through a HOM interference and looking at the polarizations statistics we obtain from them we can uniquely identify. Furthermore with a few repeats it allows to distinguish them from not entangled photons to high enough confidence. This does only require the information gathered locally and nothing else.

And that is where my question arises: It does seem like A'B' now allows to check if entanglement swapping was done with AB or whether someone decided to simply destroy the Bell entanglements - without any need to look at the observations made with AB.

My line of thought is the following: The projection operator onto a Bell state is an observable that carries the information about correlations that make the violation of Bell inequalities happen. But normally this observable cannot be measured locally and requires to combine the measurements data from both locations. So i was wondering how QT prevents there to be a way to rotate it first (in the state space) onto an observable that can be measured locally (rotations are unitary operations). The above mentioned playing around with the HOM effect seemed like a candidate to explain on an example why it doesn't work in principle... except that i am now even less sure.

HOM is an unitary interaction and not a measurement operator, so it neatly circumvents the no-communication theorem prerequisites to begin with (this generalizes to all interactions done before measurement). On the other hand the microscopic causality postulate applies to interactions, but is a postulate, so without a rigorous proof there is no guarantee to be consistent with the other postulates (situation is same as with AC in mathematical set theory). Lastly all description of beam splitter interaction has a kind of empirical character to it, which make me question a bit how this all relates to QFT postulates... this part of the theory doesn't exactly give a first impression of a waterproof boat.

 

[Cthugha]

And that is where my question arises: It does seem like A'B' now allows to check if entanglement swapping was done with AB or whether someone decided to simply destroy the Bell entanglements - without any need to look at the observations made with AB.

My line of thought is the following: The projection operator onto a Bell state is an observable that carries the information about correlations that make the violation of Bell inequalities happen. But normally this observable cannot be measured locally and requires to combine the measurements data from both locations. So i was wondering how QT prevents there to be a way to rotate it first (in the state space) onto an observable that can be measured locally (rotations are unitary operations). The above mentioned playing around with the HOM effect seemed like a candidate to explain on an example why it doesn't work in principle... except that i am now even less sure.

It is difficult to answer without making lengthy references to the literature or being too simplistic. Quantum teleportation does not only work for pure states, but also for mixed states and entanglement swapping is essentially a way of performing quantum teleportation on mixed states - as one of the entangled subsystems viewed on its own is in a mixed state. Teleporting a qubit using two-qubit entanglement requires an unambiguous two-qubit Bell state measurement which works, however, only probabilistically as long as linear optics is used. There is no way to find out what happened to the entangled partners using only local information. It helps to analyze a concrete experimental setting, e.g.: https://arxiv.org/ftp/arxiv/papers/1203/1203.4834.pdf

 

[Killtech]

There is no way to find out what happened to the entangled partners using only local information. It helps to analyze a concrete experimental setting, e.g.: https://arxiv.org/ftp/arxiv/papers/1203/1203.4834.pdf

Thanks, and now we are getting somewhat closer. But in the example in the quoted paper is still a simpler case and not yet there. I have to say though, reading how the delayed choice of Victor is interpreted there sounds kind of spooky. I mean looking at the raw result and how the calculation works the outcome isn't actually that surprising... until people start with interpretations.

Well anyway, in the example above the photons are send to Alice and Bob each and the only thing they do is to measure them. If measurement is the only interaction allowed, then indeed no communication can be possible and this is perfectly covered by the no-communication theorem. That part i perfectly understand.

The situation however changes if Alice-Bob decide to run the photons through some other interactions first before measuring them. Funny thing is that even Wikipedia mentions this in in the first comment here about the no-com theorem: if the preceding interaction is non-local the theorem won't necessarily hold. Now the interaction of a beam splitter cannot be projected onto the local parts of the state, i.e. it isn't exactly local. So if we modify the example above and make Alice-Bob one person that now receives both photons, then they can now run then through a beam splitter to make them interact (HOM), then we therefore leave the scope of the no-com theorem. The outcome of this specific case is what i am asking about, as even the theory itself does not seem to rule out the possibility for Alice-Bob to find out what happened at Victors in this special case. No theorem seems to cover this particular case, hence why i ask how the calculation works for the second HOM interference.

Also, i am not sure how this is exactly related to quantum teleportation. While it uses a lot of the same techniques, the underlying information sent by Victor is no qubit. It is instead rather classically specified instead by his delayed choice of entanglement swapping or not i.e. the choice which quantum interaction he runs his photons through.

 

[Cthugha]

The situation however changes if Alice-Bob decide to run the photons through some other interactions first before measuring them. Funny thing is that even Wikipedia mentions this in in the first comment here about the no-com theorem: if the preceding interaction is non-local the theorem won't necessarily hold. Now the interaction of a beam splitter cannot be projected onto the local parts of the state, i.e. it isn't exactly local. So if we modify the example above and make Alice-Bob one person that now receives both photons, then they can now run then through a beam splitter to make them interact (HOM), then we therefore leave the scope of the no-com theorem.

Sorry, but the more you explain, the less I can follow. A beam splitter is just a passive element that mixes the two states. I have no idea what "the example above" is in this case, but there is nothing non-local in beam splitters. If you want to discuss a certain setup, then present it in detail, but throwing in few random parts is not sufficient for any reasonable discussion.

 

[Killtech]

Sorry, but the more you explain, the less I can follow. A beam splitter is just a passive element that mixes the two states. I have no idea what "the example above" is in this case, but there is nothing non-local in beam splitters. If you want to discuss a certain setup, then present it in detail, but throwing in few random parts is not sufficient for any reasonable discussion.

Oh, sorry I was referring to the example you provided in your previous post you linked (the "Experimental delayed-choice entanglement swapping" paper). I was also using their Alice, Bob, Victor terminology and their setup and then tried to explain the difference to the setup i have in mind and which my question is based on (i.e. more or less merging Alice-Bob into one and using a similar setup to that of Victor on their end as well).

 

[Killtech]

A beam splitter is just a passive element that mixes the two states. I have no idea what "the example above" is in this case, but there is nothing non-local in beam splitters.

Yeah, of course that is everyone's intuition, but this isn't exactly how the calculus in QT works. Whether two photons are in one way or another indistinguishable cannot be determined entirely locally since not all states are separable. However in the calculus this makes a difference in potentially every interaction, for example to determine if two terms cancel each other out or not - and that is also relevant for the HOM effect. So if any operation/interaction cannot ignore such non local information when dealing with non separable states then the operation isn't fully local itself.

 

[Cthugha]

Yeah, of course that is everyone's intuition, but this isn't exactly how the calculus in QT works. Whether two photons are in one way or another indistinguishable cannot be determined entirely locally since not all states are separable. However in the calculus this makes a difference in potentially every interaction, for example to determine if two terms cancel each other out or not - and that is also relevant for the HOM effect. So if any operation/interaction cannot ignore such non local information when dealing with non separable states then the operation isn't fully local itself.

But this is pretty exactly how it works. One of the subsystems of an entangled pair taken on its own is a mixed state. When doing the HOM effect with two entangled photons from separate pairs, you get the same results as if you use two independent mixed states. Indistinguishability is evaluated on the ensemble level as an ensemble average anyway.

Of course you will get cross-correlations when comparing the result to whatever you do to the entangled partners, but just by observing only the results on the HOM side of the setup you do not get any information about what is going on on the other side. All the measurements reproduce the mixed state distribution in the ensemble average.

Things would be different if you could force the entangled partner into some certain state. However, if that was possible, ftl communication would be possible anyway in simpler scenarios - which is obviously not the case.

 

[DrChinese]

And that is where my question arises: It does seem like A'B' now allows to check if entanglement swapping was done with AB or whether someone decided to simply destroy the Bell entanglements - without any need to look at the observations made with AB.

I don't see your reasoning. The swap to entangle A'B' is done using the AB pair. If that pair is ignored - regardless of whether you ran them through the BSM or not - all you see with A'B' is a bunch of random bit pairs containing no useful information from AB to tell you anything more.

Are you thinking that the A'B' pair would be subjected to HOM interference analysis (i.e. look for the dip)? And then the presence or absence of the dip would say something about what happened to AB? Or?

 

[Killtech]

But this is pretty exactly how it works. One of the subsystems of an entangled pair taken on its own is a mixed state. When doing the HOM effect with two entangled photons from separate pairs, you get the same results as if you use two independent mixed states. Indistinguishability is evaluated on the ensemble level as an ensemble average anyway.

Of course you will get cross-correlations when comparing the result to whatever you do to the entangled partners, but just by observing only the results on the HOM side of the setup you do not get any information about what is going on on the other side. All the measurements reproduce the mixed state distribution in the ensemble average.

It's quite likely i get quite a few things wrong about how it works, but there is a reason i ask. What i want for now is to see how to calculate the HOM interference for two independent Bell states. For as long as i don't understand that, i won't be able understand what is possible or not - and since work keeps me busy this week i had no time to go through the papers in details to check if that is maybe specifically calculated there.

In QT the distinction between ensembles and superpositions of corresponding states is a tricky business and measurement alone usually cannot do that. For example in the single photon interference the distinction is made pretty clear, yet no statistics obtained from input beams alone is capable to achieve it. So it is indeed only the specific quantum interaction (interference) preceding measurement that makes this information accessible. In general inference seems to be the most reliable way to do this kind of distinction but it get's tricky to find the right interference for more complicated states. Two particle interference is even more interesting as it allows to extract information on one particle from two. This is why i am looking at HOM to understand what information it can make accessible and in what case.

Are you thinking that the A'B' pair would be subjected to HOM interference analysis (i.e. look for the dip)? And then the presence or absence of the dip would say something about what happened to AB? Or?

Yes, i hoped that HOM could be able to show if two particles are in an identical state (like it does for other pure states), specifically even if they are in a Bell state i.e. i hope there is any dip at all for two such states. But truly, i don't yet know for how to calculate this case for two independent Bell states... so i don't know and ask.

I am unsure, if A'B' would show a dip, because in order for the $|1,1\rangle$ terms to cancel out they need to be the same, yet they don't seem to be because each Bell photon has a factor for it's sister photon and those are not the same between A' and B' (may differ by location, but everything else should match).

So i really want to understand what condition exactly is needed for the $|1,1\rangle$ terms to cancel out. I was thinking that the sister photon factors could be made more indistinguishable if they were both run through the same beam splitter. So practically if a HOM ist done with AB after A'B' HOM. I want to understand the argument that explains me why they still don't cancel out in the calculation - at least for the second/last HOM interference.

But i think i have found one error, already. If both A' and B' are in a superposition made out 2 terms each, then the resulting HOM will have 4 terms for each possible combination of the 2x2 terms (and same for AB). 2 out of those 4 terms should look like interferences of photons with opposing polarization contributing a perfectly uniform probability distribution 50% of time (so no dip, but at least just an uniform background) and i have forgotten about them entirely to be honest. The other two however could still contribute to a dip (if the $|1,1\rangle$ terms cancel out). So in total at best the dip would be halved.

Finally, underlying the condition of when $|1,1\rangle$ cancel out a situational non local part of the beam splitter interaction may be hidden. Or probably it isn't. But i need to know how it works.

 

[Cthugha]

In QT the distinction between ensembles and superpositions of corresponding states is a tricky business and measurement alone usually cannot do that. For example in the single photon interference the distinction is made pretty clear, yet no statistics obtained from input beams alone is capable to achieve it. So it is indeed only the specific quantum interaction (interference) preceding measurement that makes this information accessible. In general inference seems to be the most reliable way to do this kind of distinction but it get's tricky to find the right interference for more complicated states. Two particle interference is even more interesting as it allows to extract information on one particle from two. This is why i am looking at HOM to understand what information it can make accessible and in what case.

Maybe we should go back one step. It might be the case that you indeed got some of the basics wrong. Single photon interference is as non-quantum as it gets. There is really nothing in there which requires any quantum treatment. Yes, it gets interesting for quantum states, but simple single-photon interference does not show any quantum features. Higher-order interference does as you need commutation relations to sort them out.

Also, I would be hesitant to call interference an interaction unless one defines these terms very well beforehand. Usually, people tend to interpret interactions as irreversible processes, which simple HOM interference is not. For example, if you realize a geometry where you have a 50% chance that two photons undergo HOM and separate the photon pair again afterwards and direct them into two fibers and a 50 % chance that the photons directly go into the same fibers, you have no chance to find out whether HOM interference has actually taken place.

If two photons meet at a beam splitter, the correct way of treating the problem is to have a look at the mode composition of the photons. Assume that they are both in a mixed state and assume thatthe modes considered are orthogonal. Now you need to consider all possible combinations of modes that may arrive at the beam splitter at the two sides. For all combinations where the same mode enters at both sides, you will get photon pairs at the exit port with a 50/50 chance for each individual exit port. For all combinations where different modes enter, you will get a 25% chance for photon pairs for each exit port and a 50% chance for one photon leaving from each port. Now the total distribution is given by the weighted average of all of this cases.
This is not a measurement! If you consider, e.g., a photon in mode A entering the beam splitter from port 1 and a photon in a 50/50 superposition of mode A and mode B entering from the other side, the result will also be a superposition of the following cases:
1) two photons in mode A leaving from port 1
2) two photons in mode A leaving from port 2
3) 1 photon in mode A and 1 photon in mode B leaving from port 1
4) 1 photon in mode A leaving from port 1 and 1 photon in mode B leaving from port 2
5) 1 photon in mode A leaving from port 2 and 1 photon in mode B leaving from port 1
6) 1 photon in mode A and 1 photon in mode B leaving from port 2.

All of this is still in a superposition. The superposition is only broken as soon as you perform a projective measurement. The interference alone gives you no information at all without actually performing a measurement.

 

[Killtech]

Maybe we should go back one step. It might be the case that you indeed got some of the basics wrong. Single photon interference is as non-quantum as it gets. There is really nothing in there which requires any quantum treatment. Yes, it gets interesting for quantum states, but simple single-photon interference does not show any quantum features. Higher-order interference does as you need commutation relations to sort them out.

Communication isn't what is interesting about single photon interference. It's a good example to study some basic properties of quantum states and their measurement in QT - particularly what information they store and what can be measured and what cannot.

Look at just the situation i have described for the single photon interference - in case you don't know whether the incoming two beams come from a single beam 50:50 split in two or whether they are just probabilistically redirected photons (so 100% transmission 50% of the times, 100% reflection elsewise). Measuring the individual beams will give you identical statistics (including correlations) in both cases, so it is impossible to give a prediction of what happens if instead of measuring the individual beams they are brought to interference. Classically we could just measure the amplitudes exactly to distinguish between the ensembles or deterministic states, but that's not an option in QT. But after the interference the information can be easily derived from the statistics: in the former case all photons leave the splitter at one channel, in the latter they are uniformly distributed amongst the two outputs.

Also, I would be hesitant to call interference an interaction unless one defines these terms very well beforehand. Usually, people tend to interpret interactions as irreversible processes, which simple HOM interference is not. For example, if you realize a geometry where you have a 50% chance that two photons undergo HOM and separate the photon pair again afterwards and direct them into two fibers and a 50 % chance that the photons directly go into the same fibers, you have no chance to find out whether HOM interference has actually taken place.

Fair, enough the term "interaction" isn't normally attributed to interference and beam splitters. What i mean is the effective unitary time evolution as described by an effective operator that translate states from before they entered the beam splitter to the states afterwards. The time evo usually includes all interactions, hence why i would use the term, even if that's just the self-interaction of the EM-field. Then again, in theory the passing of a photon through a half mirror is actually a very complex quantum interaction between the whole atomic lattice of the mirror and the EM field - so a lot of complex physics behind the oversimplified effective unitary trafo we use for beam splitters.

But QFT time evo including interaction terms between fields stays unitary, so it remains time reversible. It does so in theory even in case of decoherence. So by your conditions measurement would end up being the very only interaction in existence :P... so your meaning of the word "interaction" doesn't sound useful to me either.

If two photons meet at a beam splitter, the correct way of treating the problem is to have a look at the mode composition of the photons. Assume that they are both in a mixed state and assume thatthe modes considered are orthogonal

Okay, here i see a significant misunderstanding. Bell states are pure states and as long as no measurement has happened and as long as the system is viewed as a whole. In fact, if we consider that some unitary transformations we work with might have a non-local aspect to them it would be wrong to do it differently. Either way, calculating this globally has to work.

The calculation of the mixed states is significantly easier and just as you nicely showed it's just most basic probability theory. Doing it for the pure entangled state is the part i find really tricky as there is no easy probability calculation but instead i have to write the input states in terms of quite complicated combinations of various creation operators and pass them through the unitary transformation of the beam splitter... only here in this calculation it becomes relevant to correctly distinguish if two creation operators are the same or different in order to know if they cancel each other out or not.

And even here, i can get around how most of the calculation goes for the first HOM. But that yields even more complex state for the total system which then needs to undergo another unitary transformation of the second HOM beam splitter. that's where i fail so far. Specifically, because the trick you showed me in your first answer in this thread how to distinguish photons in this formalism in a slightly tricky situation just won't work anymore.

My whole question centers around the comparison between the calculation of the mixed state vs pure state HOM. Apparently it's therefore not enough to focus just on the mixed state, so let's get into the details of HOM calculating for pure Bell states - where all my questions so far are focused on.

 

[Cthugha]

Communication isn't what is interesting about single photon interference. It's a good example to study some basic properties of quantum states and their measurement in QT - particularly what information they store and what can be measured and what cannot.

Where did you get communication from? I cannot follow.

Look at just the situation i have described for the single photon interference - in case you don't know whether the incoming two beams come from a single beam 50:50 split in two or whether they are just probabilistically redirected photons (so 100% transmission 50% of the times, 100% reflection elsewise). Measuring the individual beams will give you identical statistics (including correlations) in both cases, so it is impossible to give a prediction of what happens if instead of measuring the individual beams they are brought to interference. Classically we could just measure the amplitudes exactly to distinguish between the ensembles or deterministic states, but that's not an option in QT. But after the interference the information can be easily derived from the statistics: in the former case all photons leave the splitter at one channel, in the latter they are uniformly distributed amongst the two outputs.

I have no idea what you are talking about. I assume that you talk about single photon interference? What are "probabistically redirected photons"? Which of the 5 scenarios you have already discussed is "the situation you have described"? What are "deterministic states"? I assume that you are talking about a standard interferometer-type setup. Even then, "all photons leave the beam splitter at one channel" is only the case for a very special relative phase between the two beams. Even if they are derived from the same beam. If you manage to phase-stabilize two otherwise unrelated beams, you will get the same result. All of this is unrelated to photons.

Fair, enough the term "interaction" isn't normally attributed to interference and beam splitters. What i mean is the effective unitary time evolution as described by an effective operator that translate states from before they entered the beam splitter to the states afterwards. The time evo usually includes all interactions, hence why i would use the term, even if that's just the self-interaction of the EM-field. Then again, in theory the passing of a photon through a half mirror is actually a very complex quantum interaction between the whole atomic lattice of the mirror and the EM field - so a lot of complex physics behind the oversimplified effective unitary trafo we use for beam splitters.

Self-interaction means something entirely different. Indeed the processes taking place at a beam splitter are complex and may include surface plasmons and the like. However, the only consequence one really needs to take into account is the phase shift that occurs during reflection - and losses of course for a bad beam splitter.

But QFT time evo including interaction terms between fields stays unitary, so it remains time reversible. It does so in theory even in case of decoherence. So by your conditions measurement would end up being the very only interaction in existence :P... so your meaning of the word "interaction" doesn't sound useful to me either.

You can subdivide into reversible and irreversible interactions if you like.

Okay, here i see a significant misunderstanding. Bell states are pure states and as long as no measurement has happened and as long as the system is viewed as a whole. In fact, if we consider that some unitary transformations we work with might have a non-local aspect to them it would be wrong to do it differently. Either way, calculating this globally has to work.

Ok, I am afraid that you are indeed missing some very basics. Bell states involve two photons. That automatically puts any individual photon of the pair into a mixed state. And every single experimental result you get by looking at this individual photon alone is completely described by what happens to the mixed state. Entanglement then allows you to draw conclusions on the other particle and to calculate correlations. What hits a single entrance port in a HOM experiment is a single photon out of the photon pair, not the total Bell state.

The calculation of the mixed states is significantly easier and just as you nicely showed it's just most basic probability theory. Doing it for the pure entangled state is the part i find really tricky as there is no easy probability calculation but instead i have to write the input states in terms of quite complicated combinations of various creation operators and pass them through the unitary transformation of the beam splitter... only here in this calculation it becomes relevant to correctly distinguish if two creation operators are the same or different in order to know if they cancel each other out or not.

My whole question centers around the comparison between the calculation of the mixed state vs pure state HOM. Apparently it's therefore not enough to focus just on the mixed state, so let's get into the details of HOM calculating for pure Bell states - where all my questions so far are focused on.

No, you do not have to write the input states in some complicated combinations to find out what happens behind the beam splitter (unless of course you interfere both photons of an entangled pair on the same beam splitter). That is the simple part about it. You need to do that if you want to establish some correlations to what is happening to the partners. But the results you will get at the HOM side ALONE are completely given by what happens to the mixed states. The mixedness is even a good measure of how strong the entanglement in the state is. Sorry, but you are simply assuming something that is not there. There is simply nothing additional that happens here. The additional stuff are the correlations between photon pairs. These may become slightly more complicated. However, if you do not make any use of an entangled partner photon in the experiment, knowledge of the reduced density matrix of the photon subsystem you actually use is fully sufficient.

Summarizing it in a nutshell: In order to change the outcome of an experiment that depends on the properties of the state of a single photon (such as the HOM interference), an experiment performed on the entangled partner would need to change the reduced density matrix of the entangled partner. This is not possible (and would imply FTL information transfer). Accordingly, it is sufficient to consider the reduced density matrix where the entangled partner is traced out.

That is really something that is taught at the very beginning of course on quantum information or similar topics, see e.g.: https://www.physics.umd.edu/grt/taj/623f/QInfo.pdf

 

[Killtech]

Where did you get communication from? I cannot follow.

last few words of your quote i was answering to?

I have no idea what you are talking about. I assume that you talk about single photon interference? What are "probabistically redirected photons"? Which of the 5 scenarios you have already discussed is "the situation you have described"? What are "deterministic states"? I assume that you are talking about a standard interferometer-type setup. Even then, "all photons leave the beam splitter at one channel" is only the case for a very special relative phase between the two beams. Even if they are derived from the same beam. If you manage to phase-stabilize two otherwise unrelated beams, you will get the same result. All of this is unrelated to photons.

you quote a part of my post which start with me explicitly saying that in that paragraph i talk about the single photon interference, why do you have to ask? But i admit i had trouble with some terminology since the other ensemble case is rarely discussed.

let's maybe schedule the way more complicated stuff for later and get this very basic case straighten out first.

1) the classic single photon interference with the usual incoming pure state is rather straight forward and we don't have to discuss much here. Except that for the curiosity we decide to measure the beams before the interference happens (therefore absorbing the photons and preventing it), just so we can answer the question what information and statistics we can get from there for one and to compare it to anther case for the other.

2) The other case is the same, we have beams A and B, but here the photon comes in a mixed state instead - an ensemble of two cases: we can think of it as two independent beams A and B but one is always randomly turned off so effectively we have only a single beam but we don't know which it is. What i had in mind was that one could construct it by using a mirror which randomly rotates and reflecting an incoming beam to go either in direction A or B purely probabilistically.

What i now meant is that if one were only to measure A and B and gather all possible statistics, it won't be possible to distinguish the pure state 1) from the mixed one 2) at all. All statistics will look exactly the same in either case, right?

But if we apply a unitary transformation (the beam splitter) onto the states, things change and we can clearly detect a difference since there will be an interference in 1), none in 2). What this is meant to show, is that some information about the state cannot be obtained from measurement alone. But the information has to be there in order to decide if interference will happen and unitary transformation applied to the states can make a big difference in terms of if we can extract it or not.

In particular the distinction between the pure and mixed states calculation may only become visible once we find the right unitary trafo to apply to make it measurable.

Ok, I am afraid that you are indeed missing some very basics. Bell states involve two photons. That automatically puts any individual photon of the pair into a mixed state. And every single experimental result you get by looking at this individual photon alone is completely described by what happens to the mixed state. Entanglement then allows you to draw conclusions on the other particle and to calculate correlations. What hits a single entrance port in a HOM experiment is a single photon out of the photon pair, not the total Bell state.

The Bell state is a pure two photon state, and as long as the state was not subject to a projective measurement i would argue there is no safe way to sperate the state into individual photons at all, specifically when it comes to applying unitary transformations to it. If you do that anyway you practically assume a projective measurement of the second photon of the pair has happened. But that is precisely what i do not want. This is why HOM gets complicated for me: even in the case of interfering two beams i have the creation operators for all 4 photons to deal with, even if only two should be really active in the beam splitter while the other two should be passive. Think of that i want the time evo of the whole system, not just the 2-photons that really meat. And all question that arise are exactly about the passive photons that are not actually taking part, yet strangely they still seem to have some say in the outcome... and they have to, or otherwise Bell inequalities wouldn't work.

 

[Cthugha]

you quote a part of my post which start with me explicitly saying that in that paragraph i talk about the single photon interference, why do you have to ask? But i admit i had trouble with some terminology since the other ensemble case is rarely discussed.

You are frequently using words and terms which mean something entirely different than what you seem to think they mean. This makes this discussion pretty hard.

For example The single photon interference" alone can refer to several dozens of different experiments, even when just considering two beams meeting at a beam splitter. Most importantly, "single-photon interference" just means that interference is happening at the single photon level. It does not mean that single photons arrive. We could be talking about a coherent beam arriving, about single photons arriving, about sunlight arriving and so on. You need to specify the light ource you consider. Do you consider Fock states? These have no well-defined phase anyway, so single-photon interference becomes pointless.

1) the classic single photon interference with the usual incoming pure state is rather straight forward and we don't have to discuss much here. Except that for the curiosity we decide to measure the beams before the interference happens (therefore absorbing the photons and preventing it), just so we can answer the question what information and statistics we can get from there for one and to compare it to anther case for the other.

There is no "the usual pure state". This is what I was aiming at. Single photon interference depends on the relative phase of two fields and not on whether the states are pure. Please define clearly: At which entrance ports of the beam splitter do light fields arrive? What is the photon statistics of the state you consider (single photon, coherent, thermal, ...). If several beams arrive, is there some fixed relative phase between these fields? What is the polarization state of these fields? This is the bare minimum one needs to discuss these cases. Please also note that single photons do not have a preferred phase and you need two fields for interference to occur.

2) The other case is the same, we have beams A and B, but here the photon comes in a mixed state instead - an ensemble of two cases: we can think of it as two independent beams A and B but one is always randomly turned off so effectively we have only a single beam but we don't know which it is. What i had in mind was that one could construct it by using a mirror which randomly rotates and reflecting an incoming beam to go either in direction A or B purely probabilistically.

A mixed state of what? Polarization? Photon number? Are A and B beams arriving at the same entrance ports? They should be, but then you intend to talk about sending the beams in different directions. Your experiment is ill-defined.

What i now meant is that if one were only to measure A and B and gather all possible statistics, it won't be possible to distinguish the pure state 1) from the mixed one 2) at all. All statistics will look exactly the same in either case, right?

See above. I do not know. Your experimental setting is completely unclear.

But if we apply a unitary transformation (the beam splitter) onto the states, things change and we can clearly detect a difference since there will be an interference in 1), none in 2).

But you already had a beam splitter present above - otherwise there would not have been any interference.

In particular the distinction between the pure and mixed states calculation may only become visible once we find the right unitary trafo to apply to make it measurable.

This sounds very wrong or maybe uses technical terms in an incorrect way. Maybe it makes more sense if you define what pure and mixed states you exactly have in mind.

The Bell state is a pure two photon state, and as long as the state was not subject to a projective measurement i would argue there is no safe way to sperate the state into individual photons at all, specifically when it comes to applying unitary transformations to it. If you do that anyway you practically assume a projective measurement of the second photon of the pair has happened. But that is precisely what i do not want. This is why HOM gets complicated for me: even in the case of interfering two beams i have the creation operators for all 4 photons to deal with, even if only two should be really active in the beam splitter while the other two should be passive. Think of that i want the time evo of the whole system, not just the 2-photons that really meat. And all question that arise are exactly about the passive photons that are not actually taking part, yet strangely they still seem to have some say in the outcome... and they have to, or otherwise Bell inequalities wouldn't work.

You have not had a single look at the reference I posted, did you?
Again: Of course you can do the full math, consider all possible combinations and see what comes out. This is physically absolutely trivial and has been done thousands of times. You simply find that the "passive" photons have absolutely no effect on the ENSEMBLE AVERAGE of what happens at the HOM side which acts as a mixed state. Yes, you will find that there is some correlation between the results you get on the HOM side and what the entangled partners will show. That is of course trivial as well. But what happens on the HOM side ALONE does not in any way depend on whatever has been done to the entangled partner. So your statement about "passive photons" that have "some say in the outcome" depends on what "the outcome" is. - The bare photon count statistics behind the HOM beam splitter? No effect at all.
- The correlations between the bare photon count statistics behind the HOM beam splitters and some measurement you do on the entangled partner? Yes, of course there will be some influence of the entangled partner. However, in that case you of course need to define what kind of measurement you actually do on the entangled partner.

So if you really want to calculate the full set of what may happen and all of the two-photon states, just go ahead and do so. The math is pretty simple.

 

[Killtech]

Let me first start that i am sorry for my unclear and ambiguous writing. the discussion is frustrating for us both when i don't make it clear enough what i am talking about. I try my best to get better.

You are frequently using words and terms which mean something entirely different than what you seem to think they mean. This makes this discussion pretty hard.

For example The single photon interference" alone can refer to several dozens of different experiments

Oh... actually now that you say it out loud, i just noticed i was being an idiot not to notice how ambiguous "single photon interference" is. I am truly sorry for that, i mean i wasted quite a bit both of our time with this. Thanks for still trying answer me.

1) What i meant was an interference like in a Mach Zehnder interferometer at the second beam splitter where the source beam for the interferometer producing a beam with suited coherence length with a fixed polarization and wave length. Beam should be tuned to low intensities so that we only observe individual photons. Admittedly i am too far away from experimental physics to specify more about the light source usually used for these kind of experiments and i hope you tell me in if i am missing something. Second beam splitter needs to be setup such that the phases of the arriving beams are suited for interference, but i guess that's covered by the MZI setup.

My argument additionally needs to obtain statistics for just before the beams enter the second beam splitter, i.e. by placing photon detectors directly in front of it (this will of course absorb the photons, but i need all statistics of the beams before the interference).

2) The second experiment would use also a MZI setup, except that the first beam splitter would be exchanged for a mechanical rotated mirror with two configurations: one letting the beam pass unhindered, the other reflecting it by 90 degrees to go the other route. The mirror should randomly switch between those configurations. The idea is that this should produce an ensemble/mixed state for the second beam splitter.

But when i think of it, i guess there are probably much easier ways to achieve the same, i.e. putting the whole MZI in a cloud chamber should do the trick for the beams to lose coherence producing the same mixed state and results. And that should work for photons, too.

You have not had a single look at the reference I posted, did you?

Oh, sorry, i somehow indeed missed your last link and i honest i don't remember seeing the last part of your post. I fear work had made me quite tired and i must have lost attention, sorry for that. Or perhaps by any chance did you edit that part of your post and my browser showed me a cached outdated version?

No, you do not have to write the input states in some complicated combinations to find out what happens behind the beam splitter (unless of course you interfere both photons of an entangled pair on the same beam splitter). That is the simple part about it. You need to do that if you want to establish some correlations to what is happening to the partners. But the results you will get at the HOM side ALONE are completely given by what happens to the mixed states. The mixedness is even a good measure of how strong the entanglement in the state is. Sorry, but you are simply assuming something that is not there.

Okay, i know that this is the general consensus. My problem is that i don't see what would guarantee that to be true in general. The link you posted shows that a pure 2 qubit state reduces effectively to a mixed state when looking only at a single qubit. It demonstrates this in a "proof by example" way. However i know when considering only observables (like in the link) that you are right, this is still adequate in every possible case and that can be rigorously proven. But at the instance where we involve a time evolution, i.e. an unitary transformation, things become not as clear anymore.

If you could use arbitrary unitary trafos it's not difficult to construct a case where reducing the system onto single qubit subspace and applying the trafo afterwards yields inconsistent results compared with doing the same with the global pure state. This can be used to identify a trafo as non local. More precisely this article debates the issue probably much more comprehensibly then i do: https://arxiv.org/pdf/quant-ph/9906036.pdf.

Note that the link you gave me contains no mention whatsoever about time evolutions or other unitary transformation preceding the measurements, yet that is what undermines the assumption that it has no effect if you calculate it locally or globally.

Note also, that some people here suggest that QFTs should be capable to explain measurement in principle as regular physical interaction like any other without the need of an additional on top schema of measurement. That however would imply that its time evo operator in general requires to include some non locality to be able to achieve that. So if, QFTs were to come with some non-locality to boot, it's fair to ask if we can bring it to bear in any other instance other then classic measurement and i am wondering if the beam splitter trafo already might do it: given the complex structure of correlations it can produce for Bell states and alike it's very doubtful that it factorizes into a product of operators acting on the qubit sub-spaces in general (it may work for some states, but it's hard to achieve it for all) and if it doesn't it would imply it is non-local in general (but if one could sperate it, it would make calculations for entangled states significantly simpler, so one would do it?). Think of it this way: a photon detection looks just as much local, as beams going through a beam splitter but in QT that alone doesn't count for much. We have to query the math for such questions instead.

Again, i am really sorry my worked up posting made you use caps lock. i guess i deserved that. Even more thanks therefore for replying despite that.

 

[Cthugha]

1) What i meant was an interference like in a Mach Zehnder interferometer at the second beam splitter where the source beam for the interferometer producing a beam with suited coherence length with a fixed polarization and wave length. Beam should be tuned to low intensities so that we only observe individual photons. Admittedly i am too far away from experimental physics to specify more about the light source usually used for these kind of experiments and i hope you tell me in if i am missing something. Second beam splitter needs to be setup such that the phases of the arriving beams are suited for interference, but i guess that's covered by the MZI setup.

Thanks a lot for clarifying. Yes, I guess I understand what you intend to discuss.

My argument additionally needs to obtain statistics for just before the beams enter the second beam splitter, i.e. by placing photon detectors directly in front of it (this will of course absorb the photons, but i need all statistics of the beams before the interference).

Ok, so I understand that you intend to get the data about how many photons travel along the left half of the MZI and how many travel along the right side on average? Okay. That can be done.

2) The second experiment would use also a MZI setup, except that the first beam splitter would be exchanged for a mechanical rotated mirror with two configurations: one letting the beam pass unhindered, the other reflecting it by 90 degrees to go the other route. The mirror should randomly switch between those configurations. The idea is that this should produce an ensemble/mixed state for the second beam splitter.

Ok, so when describing the state inside the interferometer in a left/right-basis, you intend to have a state that is fully diagonal in the second experiment, while you have coherences (off-diagonal elements) in the first kind of setup, right? Yes, I can understand that point.

But when i think of it, i guess there are probably much easier ways to achieve the same, i.e. putting the whole MZI in a cloud chamber should do the trick for the beams to lose coherence producing the same mixed state and results. And that should work for photons, too.

There are a lot of things one could do. A cloud chamber would usually result in photons being absorbed or scattered away. However, you could foe example introduce photon detectors that then trigger a single photon source to "reemit" the photon that has just been annihilated. You could also rotate the polarization (although that change is in principle reversible). You could also replace the mirrors inside the MZI by extremely light mirrors mounted on a spring or cantilever. If the mirror is so light that it shows recoil upon reflection of the photon, it serves as a non-destructive path detector.

Oh, sorry, i somehow indeed missed your last link and i honest i don't remember seeing the last part of your post. I fear work had made me quite tired and i must have lost attention, sorry for that. Or perhaps by any chance did you edit that part of your post and my browser showed me a cached outdated version?

I am not sure. I think I did one edit to remove a typo, but the link should have been in there. Well, it does not matter, I guess.

Okay, i know that this is the general consensus. My problem is that i don't see what would guarantee that to be true in general. The link you posted shows that a pure 2 qubit state reduces effectively to a mixed state when looking only at a single qubit. It demonstrates this in a "proof by example" way. However i know when considering only observables (like in the link) that you are right, this is still adequate in every possible case and that can be rigorously proven. But at the instance where we involve a time evolution, i.e. an unitary transformation, things become not as clear anymore.

If you could use arbitrary unitary trafos it's not difficult to construct a case where reducing the system onto single qubit subspace and applying the trafo afterwards yields inconsistent results compared with doing the same with the global pure state. This can be used to identify a trafo as non local. More precisely this article debates the issue probably much more comprehensibly then i do: https://arxiv.org/pdf/quant-ph/9906036.pdf.

Oh, well. I do not see the primary reason to bring in time evolution here. It serves to illustrate the point, but it is not fundamental. Essentially, the argument given in the manuscript can be reduced to the point that - if you have an entangled system consisting of part A and B - one would need to invoke a change in the reduced density matrix of part A when B is traced out to see the kind of effect you are looking for and the authors suggest that such things may be achieved by what they call nonlocal Hamiltonians. The time evolution operator contains the Hamiltonian and it more or less acts as a trojan horse here: Assuming non-local Hamiltonians, time evolution just becomes non-trivial (and non-local) because it involves a non-local operator. I see nothing that can be gained from discussing time evolution in addition to what can be gained from discussing the Hamiltonian.

Note that the link you gave me contains no mention whatsoever about time evolutions or other unitary transformation preceding the measurements, yet that is what undermines the assumption that it has no effect if you calculate it locally or globally.

See above. I do not see that this point is relevant. It seems like the author wants to motivate non-local Hamiltonians.

Note also, that some people here suggest that QFTs should be capable to explain measurement in principle as regular physical interaction like any other without the need of an additional on top schema of measurement. That however would imply that its time evo operator in general requires to include some non locality to be able to achieve that. So if, QFTs were to come with some non-locality to boot, it's fair to ask if we can bring it to bear in any other instance other then classic measurement and i am wondering if the beam splitter trafo already might do it: given the complex structure of correlations it can produce for Bell states and alike it's very doubtful that it factorizes into a product of operators acting on the qubit sub-spaces in general (it may work for some states, but it's hard to achieve it for all) and if it doesn't it would imply it is non-local in general (but if one could sperate it, it would make calculations for entangled states significantly simpler, so one would do it?). Think of it this way: a photon detection looks just as much local, as beams going through a beam splitter but in QT that alone doesn't count for much. We have to query the math for such questions instead.

Well, I still do not see the point. The discussion seems like a proxy to me. The beam splitter indeed introduces a lot of correlations that depend on the statistics of the incoming light. I am not sure what you mean by "it's very doubtful that it factorizes into a product of operators acting on the qubit sub-spaces in general". The beam splitter operation actually is a very simple and linear operation that simply depends on the density matrix of the incoming beams (in an appropriate basis of course). You would need to show that some measurement on an entangled partner B can change the reduced density matrix of partner A to see a difference in what the beam splitter does. However, that line of discussion does not really need the detour via the beam splitter operator.

In my opinion you need to look at a different place: QFTs usually involve a thing called microcausality, which essentially means that spacelike separated field variables commute. This is the algebraic field theory version of formulating locality. One finds that such a formulation is quite beneficial and reasonable, but indeed it is an axiom in algebraic field theory, as far as I know. So essentially, my guess is that what you really want to look at are different versions of QFTs and you may want to understand how and why they incorporate locality and what kind of non-local QFT might be reasonable. However, this may be tough and I do not know your level of expertise. If I remember correctly, the first book on QFT by Weinberg discusses some of these points. However, I might be mistaken and certainly there will be more suitable sources. However, there are several other people around in these forums which are much more knowledgeable with respect to this topic than I am. I am about as useful as a hedgehog asked about spacecraft design here.

But you will certainly not find what you are looking for by studying the beam splitter operation - which is, however, still very fascinating.

 

[Killtech]

Oh, well. I do not see the primary reason to bring in time evolution here. It serves to illustrate the point, but it is not fundamental. Essentially, the argument given in the manuscript can be reduced to the point that - if you have an entangled system consisting of part A and B - one would need to invoke a change in the reduced density matrix of part A when B is traced out to see the kind of effect you are looking for and the authors suggest that such things may be achieved by what they call nonlocal Hamiltonians

Yes, and my problem is that i see absolutely nothing in QT that would prevent the reduced density matrix of part A being affected by operations done outside that part, albeit it obviously won't happen for simple states and i couldn't put my finger on it yet, what i was exactly looking for.

However the discussion with you and some contemplation helped me a lot refine this question. I think it may be easier to change the situation and look at more specialized states that challenge the concept better. Those states are however exotic as they need at least 3 deeply entangled beams and at least 2 particles, so more complicated then usual Bell states.

Let's however still keep the notation as in one of the papers of beams A, B going to one side and A' and B' going towards the other with a single source producing all 4 entangled beams for the experiment. Let's consider the two following example hypothetical states:

1. a 2-photon 4 beam entanglement with same polarizations:
   $$|\phi\rangle = |H_A, H_{A'}\rangle + |H_B, H_{B'}\rangle$$
   So this case is a mirror experimental setup with a mirror photon state - whichever path one photon chose (A or B), his mirror photon had to do the same but mirrored.
2. a 4-photon 4 beam Bell state
   $$|\psi\rangle = |H_A, H_{A'}, V_B, V_{B'}\rangle + |V_A, V_{A'}, H_B, H_{B'}\rangle$$
   Here the distinction is that the HOM interferences will differ when also taking into account polarization depending on the loss of coherence.

both scenarios are constructed in such a way that any reduction density matrix to 2 of the 4 beams still needs to be a pure state. On the other hand measurement of any of the 4 beams gives full information about the remaining ones. If measurement causes any collapse to a mixed state, it would require the remaining beams to lose coherence to each other. While that alone has no immediately measurable effect since coherence alone does not affect the expectation values of observables, passing two such beams through a beam splitter will actually uncover it. This kind of setup can only work with at least 2 particle systems since we definitively need more then 1 measurement.

So if such states were to exist, we would either be able to uncover "which-way" information (the name makes more sense in 1.) whilst still seeing an interference or be able to signal non-locally by checking if coherence was broken or not.

I see many things that could be potentially wrong with both of the two hypothetical states above, but i am not sure, so i think they may be worth a discussion.

The beam splitter indeed introduces a lot of correlations that depend on the statistics of the incoming light. I am not sure what you mean by "it's very doubtful that it factorizes into a product of operators acting on the qubit sub-spaces in general". The beam splitter operation actually is a very simple and linear operation that simply depends on the density matrix of the incoming beams (in an appropriate basis of course). You would need to show that some measurement on an entangled partner B can change the reduced density matrix of partner A to see a difference in what the beam splitter does. However, that line of discussion does not really need the detour via the beam splitter operator.

My issue with QT is hat one can easily write down complex entanglement states with such a non-trivial non local dependency structure that it becomes a mathematical impossibility to be able maintain their non-local dependency throughout any non-trivial time evolutions with those remaining entirely local. That is Hamiltonians will rarely be able to factorize like $H_{tot}=H_A \otimes 1_B + 1_B \otimes H_B$ whilst also being able to evolve any state.

But in general, the non-local dependency present in QT - aside from classical correlations of ensembles - can always be expressed in terms of coherence information. That information is however not accessible for measurement directly yet beams splitters have a magic way of uncovering it. My question therefore has refined to whether coherence information can be shared between different particles - because i think that is exactly what needs to be prohibited somehow.

But you will certainly not find what you are looking for by studying the beam splitter operation - which is, however, still very fascinating.

Most certainly, it is fascinating.

In my opinion you need to look at a different place: QFTs usually involve a thing called microcausality, which essentially means that spacelike separated field variables commute. This is the algebraic field theory version of formulating locality.

Yes, and i think QFT navigates around the issue differently. Afaik, QFT does not derive the transformation of the beam splitter, so there is for me no immediate guarantee it is entirely consistent with it. i am still not fully sure what i should be looking for in the theory, hence i want to go through the simple cases and make sure i understand the situation here sufficiently before i dig deeper into this aspect of QFT.

 

[PeterDonis]

this article debates the issue probably much more comprehensibly then i do: https://arxiv.org/pdf/quant-ph/9906036.pdf.

That paper, while interesting, only considers non-relativistic QM, which makes it very limited. Particularly in view of the title of this thread, since strictly speaking there is no such thing as non-relativistic QM for photons.

some people here suggest that QFTs should be capable to explain measurement in principle as regular physical interaction like any other without the need of an additional on top schema of measurement. That however would imply that its time evo operator in general requires to include some non locality to be able to achieve that.

There is no such thing as a "time evolution operator" in QFT. QFT is relativistic, which means it respects Lorentz invariance, which means it respects the fact that there is no such thing as absolute time, and therefore no such thing as absolute "time evolution". "Time" is not a parameter in QFT the way it is in non-relativistic QM.

Further, in QFT, there is no explicit restriction that spacelike separated measurements cannot "causally affect" each other. There is no explicit concept of "causally affect" at all. The only restriction QFT has on spacelike separated measurements is that they must commute--i.e., their results cannot depend on the order in which they are done (which makes sense since the time ordering of spacelike separated events is not invariant). I believe that restriction alone is sufficient to ground the QFT version of the no signaling theorem.

 

[Killtech]

That paper, while interesting, only considers non-relativistic QM, which makes it very limited. Particularly in view of the title of this thread, since strictly speaking there is no such thing as non-relativistic QM for photons.

That is in part for a reason. This kind of signaling entrenches on the very nature of relativity as "instantaneous" isn't exactly relative. The paper is to be understood as a response to the no-signaling theorems available which chose a similar framework. It's noteworthy that it is, while far from perfect, akin to the calculus used to describe the Bell-related experiments - which are the ultimate judge on these matters - and they mostly use photons i.e. the quantized EM-field locked into the speed of light and only in the laboratory frame. Even if it is very simplistic, it achieves its sole goal to make correct predictions about statistics produced by detectors in these experiments and is the formalism of experimental papers. So you cannot argue it is wrong in terms of this tiny part of relativistic physics it inspects.

Now, any theory must be accountable to experiments and not the other way around. In the normal case, it is based on the conclusions taken from experiments and generalizes their results into a set of axioms. The axiom of microcausality in QFT is be understood as assumption made specifically on the experimental evidence of no-signalig and theorems extrapolating these results. This situation makes it less useful to query a theory if one of its postulates is the very thing in question.

Not to forget the issue of the unitary transformation given by the beam splitter is hard to derive from QED. It is therefore unpractical to solely use its framework to contrast related experiments unless you are looking for inconsistencies/contradictions.

So in the end, the semi-empirical oversimplified purpose-specific framework build up by countless experience through a vast set of experiments is therefore closely related to the source. On top of that it is quite lightweight in that it does not presume more then it needs for those particular type of experiments. That together makes it better qualified to study this particular question.

 

[Cthugha]

Yes, and my problem is that i see absolutely nothing in QT that would prevent the reduced density matrix of part A being affected by operations done outside that part, albeit it obviously won't happen for simple states and i couldn't put my finger on it yet, what i was exactly looking for.

Well, there is also nothing fundamental in physics that would prevent conservation of energy being broken. Similar points can be raised with respect to the second law of thermodynamics. If you drop locality, you can of course get your theory to predict such stuff. If you drop conservation of energy, you can get your theory to predict processes that do not conserve energy.
However, this is not really a question of using "simple" or "sophisticated" states.

Let's however still keep the notation as in one of the papers of beams A, B going to one side and A' and B' going towards the other with a single source producing all 4 entangled beams for the experiment. Let's consider the two following example hypothetical states:

1. a 2-photon 4 beam entanglement with same polarizations:
   $$|\phi\rangle = |H_A, H_{A'}\rangle + |H_B, H_{B'}\rangle$$
   So this case is a mirror experimental setup with a mirror photon state - whichever path one photon chose (A or B), his mirror photon had to do the same but mirrored.

1. 
   
   And why would the mirror photon do that? What would be the entangled property that determines what a photon does inside a beam splitter? This is not a realistic example. One could think about using double slits, where the slit a photon passes through is correlated with the photon momenum (emission angle). But the scenario above is not physical.
   
   

   [*]a 4-photon 4 beam Bell state
   $$|\psi\rangle = |H_A, H_{A'}, V_B, V_{B'}\rangle + |V_A, V_{A'}, H_B, H_{B'}\rangle$$
   Here the distinction is that the HOM interferences will differ when also taking into account polarization depending on the loss of coherence.

both scenarios are constructed in such a way that any reduction density matrix to 2 of the 4 beams still needs to be a pure state. On the other hand measurement of any of the 4 beams gives full information about the remaining ones. If measurement causes any collapse to a mixed state, it would require the remaining beams to lose coherence to each other. While that alone has no immediately measurable effect since coherence alone does not affect the expectation values of observables, passing two such beams through a beam splitter will actually uncover it. This kind of setup can only work with at least 2 particle systems since we definitively need more then 1 measurement.

What does "to lose coherence to each other" mean? Are you talking about coherences in the 4 photon state? In two photon states? In 1-photon states? Having coherence in some of them excludes having coherences in the same basis in others. Also, coherence is a basis-dependent concept. I got one of my rather good PhD students to write a fun paper on that earlier this year. Essentially, quantum resource theory tells us that you can convert any kind of (quantum) coherence into some kind of entanglement. For example, coherent states of the light field are very classical states. However, when considering their density matrix in the Fock basis, there is a lot of coherence (off-diagonal matrix elements). When considering single photons, these are considered very non-classical, but they do not have any coherence in the Fock basis. However, of course they show a lot of coherence when considered in the coherent state basis. Both can be converted into entanglement. I do not see what this experiment is supposed to show.

So if such states were to exist, we would either be able to uncover "which-way" information (the name makes more sense in 1.) whilst still seeing an interference or be able to signal non-locally by checking if coherence was broken or not.

There are already like a gazillion different delayed choice quantum eraser experiments dealing with the issue of interference versus which-way information. This topic has been studied already extremely heavily.

My issue with QT is hat one can easily write down complex entanglement states with such a non-trivial non local dependency structure that it becomes a mathematical impossibility to be able maintain their non-local dependency throughout any non-trivial time evolutions with those remaining entirely local. That is Hamiltonians will rarely be able to factorize like $H_{tot}=H_A \otimes 1_B + 1_B \otimes H_B$ whilst also being able to evolve any state.

Sure, but why should that be a problem? The relevant part is the measurement.

But in general, the non-local dependency present in QT - aside from classical correlations of ensembles - can always be expressed in terms of coherence information. That information is however not accessible for measurement directly yet beams splitters have a magic way of uncovering it. My question therefore has refined to whether coherence information can be shared between different particles - because i think that is exactly what needs to be prohibited somehow.

Beam splitters are just phase-sensitive. And even here only to very few instances of coherence, namely classical first-order field coherence. While they are fun, you need other experiments to study higher-order coherence - typically correlation measurements (which may of course also utilize beam splitters, but do not have to). I just do not get your obsession with beam splitters. The beam splitter just mixes states. There is not even any non-linearity. If you want non-linearity you need to introduce it by measurement. This is, e.g., the very fundament of linear optical quantum computing. Even if there was some inherent non-locality existing, beam splitters would be pretty much the last place where I would go and look for it.

Yes, and i think QFT navigates around the issue differently. Afaik, QFT does not derive the transformation of the beam splitter, so there is for me no immediate guarantee it is entirely consistent with it. i am still not fully sure what i should be looking for in the theory, hence i want to go through the simple cases and make sure i understand the situation here sufficiently before i dig deeper into this aspect of QFT.

I do not get your point. QFT does not derive the beam splitter? It just gives you the sum and the difference of the two input fields at the output ports. This is one of the simplest transformations there is. The only surprising thing is that one actually has to consider that one also needs to take vacuum fields into account if there is "nothing" present at some entrance port.

Besides that, I fully agree with PeterDonis.

 

[vanhees71]

In my opinion you need to look at a different place: QFTs usually involve a thing called microcausality, which essentially means that spacelike separated field variables commute. This is the algebraic field theory version of formulating locality. One finds that such a formulation is quite beneficial and reasonable, but indeed it is an axiom in algebraic field theory, as far as I know. So essentially, my guess is that what you really want to look at are different versions of QFTs and you may want to understand how and why they incorporate locality and what kind of non-local QFT might be reasonable. However, this may be tough and I do not know your level of expertise. If I remember correctly, the first book on QFT by Weinberg discusses some of these points. However, I might be mistaken and certainly there will be more suitable sources. However, there are several other people around in these forums which are much more knowledgeable with respect to this topic than I am. I am about as useful as a hedgehog asked about spacecraft design here.

But you will certainly not find what you are looking for by studying the beam splitter operation - which is, however, still very fascinating.

For QED, of course studying first a HEP book is a very good idea, because there all these issues are discussed in careful detail. If it comes to such fundamental questions Weinberg's QT of Fields (particularly Vol. 1) is the best source I know of. Another equally well organized and complementary source is Duncan, The Conceptual Framework of QFT, where issues like Haag's theorem etc. is covered, which is not in Weinberg's books. The key for a real understanding is the representation theory of the proper orthochronous Poincare group + some physical constraints, of which the most important one indeed is the microcausality condition for local observables. It is a way (afaik the only known way) to formulate a relativistic QT, which is causal and particularly for scattering theory leading to a Poincare-invariant S-matrix, fulfilling the cluster decomposition principle. In connection with QED also the specific case of (in this case Abelian) gauge symmetry is crucial too, and this is also explained via this group-theoretical approach developed in all generality and detail by Weinberg in his QFT books: A massless spin-1 field, as one possibility of a "physical, i.e., microcausal QFT" necessarily must be described as a gauge field.

Another application of QED is, of course, the now very wide field of quantum optics with all its possibilities to test the fundamental features of quantum theory, which are often thought of being counter-intuitive or even "weird". For this you rather need a pretty simple intuitive version of QED in the medium. The pragmatic way is to just start with the standard classical theory of em. fields in, e.g., dielectrica and quantize the corresponding em. field in the sense of an effective field theory. This gives you a nice description of all the "optical elements" needed to do all these amazing experiments (beam splitters, polarizers, wave plates, etc.). Roughly speaking, on the level of the usual linear-response approximation, it boils down to substitute field operators for the classical fields. For a working theory of photodetection you only need to understand the interaction of a bound electron with the em. field in the dipole approximation to describe the photoelectric effect. For understanding parametric down conversion also a simple form of "non-linear optics" is sufficient.

My favorite books on the subject (though I'm not a "quantum optician" myself, i.e., not an expert):

Scully, Zubairy, Quantum Optics
Garrison, Chiao, Quantum Optics

The bible seems to be

Mandel, Wolf, Optical coherence and quantum optics

 

[Killtech]

And why would the mirror photon do that? What would be the entangled property that determines what a photon does inside a beam splitter? This is not a realistic example. One could think about using double slits, where the slit a photon passes through is correlated with the photon momenum (emission angle). But the scenario above is not physical.

Because the math formalism dictates it. If we write a Hilbert spaces as $\mathcal {H}_{AB} \otimes \mathcal {H}_{A'B'}$ for the four beams 2 or 4 photon states, then this mathematical structure incorporates states like the ones i have written down. And the formalism allows to make consequent calculations for beam splitters and measurements acting on those example states. For those particular examples we see that it makes a difference whether a measurement of beam A is done before or after A'B' passes through their beam splitter and it is not mere change in correlations.

There is of course the likely possibility that these states are unphysical. But if so, there needs to be a solid reasoning - or in terms of a theory axioms or theorems - prohibiting such states. That's what i am interested in because so far i cannot find anything that would disallow them.

But if you were to declare them as unphysical/realistic then the assumption that one can write $\mathcal {H}_{tot} =\mathcal {H}_{AB} \otimes \mathcal {H}_{A'B'}$ is wrong, because that product explicitly allows these states, which is a serious dilemma.

What does "to lose coherence to each other" mean? Are you talking about coherences in the 4 photon state? In two photon states? In 1-photon states? Having coherence in some of them excludes having coherences in the same basis in others.

Math allows to write down such states (given the assumptions on the Hilbert space), and coherence can be calculated for these states according to it. What is of interest is the visible effect it has, so in this case coherence is to be measured by interfering two beams and measuring the statistics in the out channels of the beam splitter. We can calculate that for the states above and notice that if a measurement of another beam was conducted before the two beams met at the beam splitter, the calculation changes with a different result for the out channels - in particular some expectation values change.

What i find structurally interesting about it is that this is technically in no violation of the no-signaling theorem, even though these states allow to send signals. See the argument below.

Sure, but why should that be a problem? The relevant part is the measurement.

Because this is theoretically leaves a big gap to circumvent the no-signaling theorem. And of course measurement is key, but it just is one part of it and not all.

Let's stick to the 4 beam Hilbert spaces and notice that $[O_{AB}, O_{A'B'}]=0$ holds true for any local observables that could be reduce to either subspace of the total Hilbert space. Let $U_{A'B'}$ denote the unitary transformation of the beam splitter interfering beams A'B' (i.e. an element time evolution of whole the system). We find that this operator does not factorize like that and it cannot since it must operate on the whole space to be able to use it to calculate Bell correlations correctly. The least obvious consequence of this is there exists an observable $O$ in $\mathcal {H}_{AB}$ such that $[O, U_{A'B'}] \neq 0$, i.e. for this operator there are observables in either subspace that don't commute globally - which however by itself accounts to no measurable effect.

Let $S_X$ denotes the measurement operator of a photon (and its polarization) in a beam X, and let C',D' be the out channels of the A'B' beam splitter, then we find that while $S_{A}$ individually commutes with $S_{A'}$, $S_{B'}$, $S_{C'}$ and $S_{D'}$, it does not commute with something like $S_{C'}U_{A'B'}$. Now the latter is a linear operator and could be understood as an observable, but because $U_{A'B'}$ does not have a factorization into the subspaces, then $S_{C'}U_{A'B'}$ cannot be written purely in the subspace $\mathcal {H}_{A'B'}$ either - i.e. it represent a non local observable. As such it is excluded from consideration in all no-signalig theorems.

The states i have written above are constructed to make use of $[S_{A}, S_{C'}U_{A'B'}] \neq 0$ and translate it into something noticeable.

I just do not get your obsession with beam splitters. The beam splitter just mixes states. There is not even any non-linearity. If you want non-linearity you need to introduce it by measurement. This is, e.g., the very fundament of linear optical quantum computing. Even if there was some inherent non-locality existing, beam splitters would be pretty much the last place where I would go and look for it.

Exactly: $U_{A'B'}$ is linear (in the amplitudes) but does not factorize, i.e. it isn't mathematically local, and $S_{C'}$ isn't linear (in term of Born probabilities, not amplitudes) but it is local in the sense of factorizing. It is specifically the combination of both that is neither linear nor local - at least if the Hilbert space is not properly restricted. So beam splitter trafo is fascinating as it is the most simple operator that allows construction of observables no theorems account for. The non locality is however quite weak, so it requires to find special states which expectations depend most on the nonzero commutator. The ones i have posted may be related to the eigenstates of the commutators.

 

[PeterDonis]

This kind of signaling entrenches on the very nature of relativity

No, it doesn't, it just requires a correct understanding of what relativity actually requires. I stated that in the last paragraph of post #22.

 

[Cthugha]

Because the math formalism dictates it. If we write a Hilbert spaces as $\mathcal {H}_{AB} \otimes \mathcal {H}_{A'B'}$ for the four beams 2 or 4 photon states, then this mathematical structure incorporates states like the ones i have written down. And the formalism allows to make consequent calculations for beam splitters and measurements acting on those example states. For those particular examples we see that it makes a difference whether a measurement of beam A is done before or after A'B' passes through their beam splitter and it is not mere change in correlations.

Well, one can come up with a lot of math formalisms that do not have any resemblance to physical reality. What is that supposed to show? Entanglement is mostly about conserved quntities in non-linear proceesses such as SPDC. E.g., the total energy is conserved, so the energy of the two photons created in SPDC becomes entangled in the right scenario. Measurements that are sensitive to energy may then detect this entanglement. Similar reasoning works for momentum or polarization. There is no quantity that dictates "if a single photon is the only light field present at the input of a beam splitter, it needs to be transmitted or reflected". Accordingly, you will not find entanglement between which arm a photon takes. What you propose is not related to entanglement.

There is of course the likely possibility that these states are unphysical. But if so, there needs to be a solid reasoning - or in terms of a theory axioms or theorems - prohibiting such states. That's what i am interested in because so far i cannot find anything that would disallow them.

Well, the same reasoning goes for the physical theory that quantum theory is not really probabilistic, but Gurglphyx the friendly dragon will throw a dice in his castle at the center of the universe and physics just behaves according to this throw of a dice. There is nothing prohibiting that.
Maybe more importantly, we do not observe any effects of the kind you look for in any experiment.

But if you were to declare them as unphysical/realistic then the assumption that one can write $\mathcal {H}_{tot} =\mathcal {H}_{AB} \otimes \mathcal {H}_{A'B'}$ is wrong, because that product explicitly allows these states, which is a serious dilemma.

Sorry, but this is just plain ridiculous. Again, please explain which physical quantity it is which you would like to entangle that determines whether a photon is transmitted or reflected at a beam splitter.

Math allows to write down such states (given the assumptions on the Hilbert space), and coherence can be calculated for these states according to it. What is of interest is the visible effect it has, so in this case coherence is to be measured by interfering two beams and measuring the statistics in the out channels of the beam splitter. We can calculate that for the states above and notice that if a measurement of another beam was conducted before the two beams met at the beam splitter, the calculation changes with a different result for the out channels - in particular some expectation values change.

And exactly that does not happen for the standard beam splitter interaction. There is not a single example state where a measurement on another beam changes the statistics of the whole ensemble. Such changes may occur in subensembles or in postselection - which requires coincidence counting experiments.

Let's stick to the 4 beam Hilbert spaces and notice that $[O_{AB}, O_{A'B'}]=0$ holds true for any local observables that could be reduce to either subspace of the total Hilbert space. Let $U_{A'B'}$ denote the unitary transformation of the beam splitter interfering beams A'B' (i.e. an element time evolution of whole the system). We find that this operator does not factorize like that and it cannot since it must operate on the whole space to be able to use it to calculate Bell correlations correctly.

No, sorry. Again: this is the part where people will already disagree. You only need the beam splitter operation to be non-local if you want to see that the reduced density matrix changes - which, however, is your own very personal scenario which has never been observed by anyone. Otherwise you do not need a non-local beam splitter operation to reproduce Bell correlations. There is little else I can add to the discussion, I am afraid. I am not really interested or have the time to discuss sophisticated multiple beam entanglement scenarios based on flawed premises.

 

[Killtech]

Well, one can come up with a lot of math formalisms that do not have any resemblance to physical reality.

You could say the same for Bell inequalities which violation was derived solely on the basis on the existing math formalism of QT and only after it was tested it became a physical reality that surprised of quite a few people. Isn't it how theory works in general? you look in on what the know formalism implies specifically in terms of unexpected stuff and then check back how it fits with experiments?

Your statement about the math formalism sounds a little brush TBH, like you would throw Einstein, Bohr and quite a few other theoretical physicist out of the window with their weird ideas they based on some mathematical behavior of some equations in the abstract Gedankenexperiments they put them in. ;)

Entanglement is mostly about conserved quntities in non-linear proceesses such as SPDC. E.g., the total energy is conserved, so the energy of the two photons created in SPDC becomes entangled in the right scenario. Measurements that are sensitive to energy may then detect this entanglement. Similar reasoning works for momentum or polarization. There is no quantity that dictates "if a single photon is the only light field present at the input of a beam splitter, it needs to be transmitted or reflected". Accordingly, you will not find entanglement between which arm a photon takes. What you propose is not related to entanglement.

So the first 2 photon 4 beams scenario is about the photon number which in this kind of experiments must be a conserved quantity, the second one is about polarization i.e. angular momentum and is really constructed with the same mindset as Bell states but extrapolated to 4 photons. Why would the 2 photon Bell state be any more physical then the 4 photon entanglement i wrote down? So both cases are about conserved entangled quantum numbers, albeit i wrote down the first example maybe not in the most appropriate basis.

I personally thought initially the first scenario to be quite artificial so i made the second one that sounded just as realistic as a Bell state. The first one however was still very interesting as it perfectly isolated the weird mathematical structure the formalism allows for and which i have trouble dealing with.

Of course there is no way one could construct such a weird entanglement via a beam splitters, it would have to be emitted like this from a source that produces all 4 beams already in this state. But now that i had a little time to think about it, i even think it might be realistically constructed: One would need some source that can create photon pairs that are always emitted in opposed directions along a specific axis. Now if circumstances existed that the source could be brought into a superposition of emitting the pair equally likely along the x-Axis as for y-Axis, then this would indeed create such an entangled state. Looking at the wave pattern of a lateral quadrupole (somehow properly aligned by an EM-field) it would also fit if it could be additionally constrained to emit only a single photon pair for which the total momentum cancels out i.e. an emission that does nets out to produce no recoil for the source.

So i hope this clarifies your question about what quantum numbers and conservations the example states i have written are and i hope upon a second inspection you can confirm that.

Sorry, but this is just plain ridiculous. Again, please explain which physical quantity it is which you would like to entangle that determines whether a photon is transmitted or reflected at a beam splitter.

You might find it ridiculous, but that is an argument used in papers, particularly the ones about no-go theorems. It is indeed an important assumption how to translate locality and many proofs fall apart if that argument would not hold.

And if the factorization of Hilbert spaces was wrong indeed, it would be also theoretically very interesting since the math would have to be extended by a new exclusion principle like Pauli's. It also concerns which unitary transformations are allowed since, if there are unphysical states in the product space, then valid trafos cannot take an physical state and map it onto an unphysical one - which begs the question of what is the group of allowed/physical trafos.

It is generally quite important to root out all the unphysical stuff from the math formalism, if it allows for such things. Or at least properly understand it. But as far as i know the situations where something like this occurred were usually decided by making experiments to show that it was just some nonsense math rather then actually something real (e.g. tachyons)... or find out the math gave the right clue (e.g. Bell, Einstein).

So either way there is insight to be gained in contrasting what math allows with what experiments can reproduce... and testing how our understanding fits with either.

No, sorry. Again: this is the part where people will already disagree. You only need the beam splitter operation to be non-local if you want to see that the reduced density matrix changes - which, however, is your own very personal scenario which has never been observed by anyone. Otherwise you do not need a non-local beam splitter operation to reproduce Bell correlations. There is little else I can add to the discussion, I am afraid. I am not really interested or have the time to discuss sophisticated multiple beam entanglement scenarios based on flawed premises.

How would you observe something if no one ever would look at it? We would not observe Bell non-locality if someone didn't came up with an idea for an experiment to look for it. If you never heard of Bell's ideas and corresponding experiments, how would you judge the scenario some theoretical guy wrote down on paper based on some math only?

The argumentation i used there is indeed about the math formalism, but that is the same tool all the related theorems use. If there is something wrong with the math, then there is something wrong with the theorems concluded from it. And again, the normally unimportant contextual non-locality of the beam splitter allows to construct observables which are all by themselves mathematically not local - that is something that cannot be brushed away that easily and is important because i haven't seen anyone account for something like that but instead found a paper (the one i linked) that stresses that this is actually a loophole in the no-signaling theorems. So that non-locality of the beam splitter is of interest until one can show it's of no consequence in every possible case.

The way i am used to think, it would help me a great deal understand QT a lot better if i could understand what these states are. A part of my intuition of the quantum world i took from the theory was difficult to pinpoint but this analysis helped me understand that it is such states that i struggle with since the math seemed to allowed for some more interesting things that did not seem to align with a few no-go theorems. these states destill that part of math. Trying to understand proofs of no-go theorems isn't too difficult but what struck me was that their initial assumption were motivated too much by classical intuition not taking into account some weirdness that the math of QT allows in principle.

 

[DrChinese]

For those particular examples we see that it makes a difference whether a measurement of beam A is done before or after A'B' passes through their beam splitter and it is not mere change in correlations.

I've lost track of your various examples. But if you are talking about the AB + A'B' setup I commented on in #12: There is no difference in observable outcomes regardless of ordering of observations. That is true of all EPR entanglement setups, especially swapping setups.

Unless you have a reference that says otherwise, this statement should be retracted as being personal speculation.

 

[Killtech]

've lost track of your various examples. But if you are talking about the AB + A'B' setup I commented on in #12: There is no difference in observable outcomes regardless of ordering of observations. That is true of all EPR entanglement setups, especially swapping setups.

No, i have reduced the structure i am querying about to the following two states in a similar 4 beam setup but a different source emitting a different initial state instead of two independent Bell pairs.

1. a 2-photon 4 beam entanglement with same polarizations:
   |ϕ⟩=|HA,HA′⟩+|HB,HB′⟩
   So this case is a mirror experimental setup with a mirror photon state - whichever path one photon chose (A or B), his mirror photon had to do the same but mirrored.
2. a 4-photon 4 beam Bell state
   |ψ⟩=|HA,HA′,VB,VB′⟩+|VA,VA′,HB,HB′⟩
   Here the distinction is that the HOM interferences will differ when also taking into account polarization depending on the loss of coherence.

But you'll need to read a little more about the following discussion to follow the context, specifically to understand my statement about observable outcomes - since you need to know what what observables are being observed. The issue is that within the formalism the observable in question is technically non-local with the way no-signaling theorems use this word (and therefore failing to commute with observables in either part of the space), albeit it remains is a simple measurement.

The question moved on how physical such state would be since just because the formalism does allow to write them down doesn't make them necessarily realistic.

 

[DrChinese]

1. a 4-photon 4 beam Bell state
|ψ⟩=|HA,HA′,VB,VB′⟩+|VA,VA′,HB,HB′⟩2. "...For those particular examples we see that it makes a difference whether a measurement of beam A is done before or after A'B' passes through their beam splitter and it is not mere change in correlations."

1. A 4 photon Bell state (presumably photons A A' B B') does not produce polarization as you have tried to write. There are a lot more outcomes to consider, such as |VA,HA′,HB,VB′⟩. To be clear: if these 4 are entangled together, there is no requirement that any pair subset follow perfect anti-correlations. On the other hand, if any 2 follow perfect anti-correlations, they are prohibited from being entangled to other particles due to monogamy of entanglement.

2. And again, I'll ask you to retract your statement per my challenge in #30. Or produce an appropriate reference. If you don't, I will pass this to the moderators as being personal speculation that is not generally accepted.

 

[PeterDonis]

For those particular examples we see that it makes a difference whether a measurement of beam A is done before or after A'B' passes through their beam splitter and it is not mere change in correlations.

@DrChinese has already raised a valid challenge for this claim of yours:

I'll ask you to retract your statement per my challenge in #30. Or produce an appropriate reference. If you don't, I will pass this to the moderators as being personal speculation that is not generally accepted.

The moderators have already seen it.

@Killtech, I am closing this thread. If you are unable to supply a reference to back up your claim quoted above, the thread will remain closed. If you do have a reference, please PM me a link and the moderators will review it.