The discussion centers on the application of Feynman path integrals to entangled photons, specifically addressing whether the amplitude of one photon should include the amplitude of its entangled partner. Participants clarify that for two entangled photons, one must consider them as a single entity in a 6-dimensional configuration space, where each photon contributes to the overall amplitude. The conversation emphasizes that traditional interference patterns do not apply to entangled photons, and the concept of wavefunction collapse is not straightforward in this context. The Hong-Ou-Mandel effect is cited as an example of interference involving indistinguishable entangled photons.
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Amplitude for what? In order to even do a path integral, you need to define the endpoints.Quant said:Should we add to the usual amplitude (sum over all extremal paths of this photon) also the amplitude of the other photon of the pair?
Of course from a to b.PeterDonis said:Amplitude for what? In order to even do a path integral, you need to define the endpoints.
Surely. a and b.PeterDonis said:Amplitude for what? In order to even do a path integral, you need to define the endpoints.
So you think the amplitude ofar the other particle must be added?Demystifier said:I guess we are talking about nonrelativistic QM, not about QFT. If you have two particles, each moving in 3 dimensions, you can think of it as one particle in 6 dimensions. Write the path integral by pretending it's one particle in 6 dimensions, and you will get the right result.
In position space? Usually this formulation is used in momentum space.Quant said:Of course from a to b.
No, both photons together are "one particle" in 6 dimensions.Quant said:photon 1 is one particle in 6 dimensions
Quant said:Of course from a to b.
What are a and b? You're assuming it's obvious. It's not. Once you answer what a and b actually are, you will know the answer to the question you posed in your OP.Quant said:Surely. a and b.
That's not what @Demystifier said. What he said is that you need to first decide what you are computing the path integral for. Are you computing it for just one photon? Or for the system consisting of both of them?Quant said:So you think the amplitude ofar the other particle must be added?
I don't know Feynman reformulation of QM about entangled pair. For single particle it is 100 % sure that we consider all paths between any two point (usually emitter at a and detector at b) and as the path is surely function of xyz one is in (configuration) space. I don't see anywhere momenta here. They doesn't matter at all. I've not done it explicitly but I've seen how is done.PeterDonis said:That's not what @Demystifier said. What he said is that you need to first decide what you are computing the path integral for. Are you computing it for just one photon? Or for the system consisting of both of them?
Different answers to that question will result in different path integrals--and different answers to my question, what are a and b?
Is the Feynman integral in configuration space or in momentum space?Demystifier said:No, both photons together are "one particle" in 6 dimensions.
I understand this as one particle going two distant paths at the same time together.(e.g photon1 goes also the paths of photon2 which would be otherwise inaccessible for it?Demystifier said:No, both photons together are "one particle" in 6 dimensions.
@Demystifier described how this works in configuration space in post #4. See further comments below.Quant said:I don't know Feynman reformulation of QM about entangled pair.
You can do it either way. Doing it in momentum space is much more common because incoming and outgoing momenta and energies are what actual experiments usually measure.Quant said:Is the Feynman integral in configuration space or in momentum space?
No. The configuration space for two particles, at least in the approximation @Demystifier described, is 6-dimensional: you have 3 position coordinates for particle 1 and three position coordinates for particle 2. So every point in the 6-dimensional configuration space represents particle 1 being at some position ##x_1##, ##y_1##, ##z_1##, and particle 2 being at some position ##x_2##, ##y_2##, ##z_2##.Quant said:I understand this as one particle going two distant paths at the same time together.
Entangled photons do not produce traditional interference patterns in the manner you imagine. It’s not possible to explain it without a sufficiently detailed example, but the short story is that such photons do not self interfere and they don’t interfere with each other. However, 2 different entangled photons can interfere if indistinguishable. The Hong-Ou-Mandel effect is an example of that.Quant said:Suppose photon 1 of the twin pair goes to a beam splitter BS1 and photon 2 can go in the other port of the BS1 but it doesn't (it goes tru a BS2 which outputs are on BS1 and on a point c and is registered after a while in point c. Will photon 1 interfere on the BS1 with the "empty" wavefunction of the photon 2? If we add the amplitude pretending photon 1 is one particle in 6 dimensions it will have to interfere?
Also if photon 2 is registered prior to photon 1 in c and one expects collapse of its part of the wavefunction the possible path to BS for photon 1 does not seem to stop existing and photon1 must interfere on BS1.
That's wrong. Have you ever seen any equation in quantum mechanics for a system with two or more particles?Quant said:I understand this as one particle going two distant paths at the same time together.(e.g photon1 goes also the paths of photon2 which would be otherwise inaccessible for it?
Maybe. But in experiment like DSE and MZI one measure the positions by positioned photon detectors. I am interested in only these experiments.PeterDonis said:@Demystifier described how this works in configuration space in post #4. See further comments below.
You can do it either way. Doing it in momentum space is much more common because incoming and outgoing momenta and energies are what actual experiments usually measure.
The problem here is strictly in Feynman paths not any other formulations. So I want to know just the amplitude of photon 1 from 'a' (birth of twins) to 'b' (detector). Photon 2 starts also from 'a' but can go to either 'b' or 'c' tru point 'd' (BS 50/50). If it goes to c (is registered there) does the amplitude of photon 1 also include the path where photon 2 goes 'adc'? I hope of an explicit answer.Demystifier said:That's wrong. Have you ever seen any equation in quantum mechanics for a system with two or more particles?
When you have two entangled photons, there is no such thing as amplitude of photon 1. There is only an amplitude for both photons together.Quant said:The problem here is strictly in Feynman paths not any other formulations. So I want to know just the amplitude of photon 1 from 'a' (birth of twins) to 'b' (detector). Photon 2 starts also from 'a' but can go to either 'b' or 'c' tru point 'd' (BS 50/50). If it goes to c (is registered there) does the amplitude of photon 1 also include the path where photon 2 goes 'adc'? I hope of an explicit answer.
Thanks Dr. Chinese.DrChinese said:Entangled photons do not produce traditional interference patterns in the manner you imagine. It’s not possible to explain it without a sufficiently detailed example, but the short story is that such photons do not self interfere and they don’t interfere with each other. However, 2 different entangled photons can interfere if indistinguishable. The Hong-Ou-Mandel effect is an example of that.
There is also no sense in talking about collapse of one vs collapse of the other. There is no scenario in which you can precisely say “one is collapsed, the other is not” and back that up with experimental differences. Time sequence is generally not a factor, and you are free to attempt to interpret it as you like.
Yes, but in that case you are no longer considering two photons. And by the way, the expression "Feynman rules" means something entirely different, it's related to perturbative Feynman diagrams, not to Feynman path integrals.Quant said:And when the one photon (2) is registered in the detector (c) there would be just one photon (1). Does it has no amplitude? Can Feynman rules not say where we would chance to detect it?
An entangled photon will behave as what you call “normal” and will produce the usual interference pattern if you place a single slit in front of the double slits. That creates a coherent state and stops the entanglement for that photon. But… don’t think that means the other photon will evidence any change. It won’t. If it did, you could perform FTL signaling. Right?Quant said:Thanks Dr. Chinese.
Are you saying we can not destroy photon 1 of a entangled pair and leave photon 2 on an infinite journey in all inertial reference frames?
Where it should be an unentangled 'normal' photon which will begin to interfere with itself?
This is why I asked him Saturday if he had ever done such a calculation and if he really wanted a graduate-level answer. Based on his writings, I believe he has not done such a calculation, and that he is ill-served by insisting on a graduate-level answer.Demystifier said:the expression "Feynman rules" means something entirely different,
Why should we put a single slit before DS? If the (say idler) photons of entangled pairs do not make DS interference pattern one will disentangle the signal photons and the idler must start to do DS interference. One could send FTL easy then. Consider 10 groups of 1000 idler photons. One can disentangle number 1, 5, 9 of these groups and this is a message.DrChinese said:An entangled photon will behave as what you call “normal” and will produce the usual interference pattern if you place a single slit in front of the double slits. That creates a coherent state and stops the entanglement for that photon. But… don’t think that means the other photon will evidence any change. It won’t. If it did, you could perform FTL signaling. Right?