Delayed Choice Quantum Eraser - double-photon explanation

[jwalker]

It is claimed that DCQE is equivalent to a single-photon double-slit setup. However, In this experiment, the 351.1nm Argon ion pump laser beam is divided by a double-slit, which means the actual photon rate is 10¹³ times higher than in a single-photon setup.

Consequently, it makes it possible for both A and B regions to generate photon pairs simultaneously enough to allow corresponding photons from different pairs to interfere further in the set.

I believe that the interpair interference explains the experimental results. If this hasn’t been discussed yet, I’d like to develop this idea step by step, so please comment.

 

[San K]

It is claimed that DCQE is equivalent to a single-photon double-slit setup. However, In this experiment, the 351.1nm Argon ion pump laser beam is divided by a double-slit, which means the actual photon rate is 10¹³ times higher than in a single-photon setup.

Consequently, it makes it possible for both A and B regions to generate photon pairs simultaneously enough to allow corresponding photons from different pairs to interfere further in the set.

I believe that the interpair interference explains the experimental results. If this hasn’t been discussed yet, I’d like to develop this idea step by step, so please comment.

it's just a "filtering" trick.

the interference pattern is "hidden/embedded" in the non-interference pattern

 

[jwalker]

it's just a "filtering" trick.

the interference pattern is "hidden/embedded" in the non-interference pattern

Not sure what you mean. Do you support or oppose quantum eraser explanation?

 

[San K]

Not sure what you mean. Do you support or oppose quantum eraser explanation?

there are many explanations for the quantum eraser..:), i support the one which says superimposed/embedded...as explained in the paper by kim yoon.

 

[jwalker]

i support the one which says superimposed/embedded...as explained in the paper by kim yoon.

Let's not go that far for now. What do you think of photons generated at A and B at the same time?

 

[questionpost]

Let's not go that far for now. What do you think of photons generated at A and B at the same time?

Photons aren't "generated" into two spots, it's a wave, waves span out over distance, what's the big deal? When you measuring something all your doing is making it's probability equal to a specific point, so there's nothing that happens in-between, the photon's position is just equal to being in a specific point once it interacts with the screen. This is why quantum mechanics is where math meets reality. Things work because they mathematically act that way or they "appear" in a given place without going through the intervening space from a previous measurement is because of the fact that their probability is equal to a non-zero number in that place in that distance.

 

[jwalker]

Photons aren't "generated" into two spots

I don't think you're referring to the experiment.

 

[questionpost]

I don't think you're referring to the experiment.

Are you talking about that double slit experiment? Because that's not two photons, that's the same photon interfering with itself.

 

[Cthugha]

At first, you should quote which DCQE experiment you are talking about. Your quote makes me assume you mean the following one: http://arxiv.org/abs/quant-ph/9903047.

Now let us start at the very beginning.

It is claimed that DCQE is equivalent to a single-photon double-slit setup.

This is not claimed anywhere. You see an equivalent pattern. That does not mean the experiments are equivalent.

 

[jwalker]

This is not claimed anywhere. You see an equivalent pattern. That does not mean the experiments are equivalent.

Well, not in original paper, but not anywhere, too. There are many comments emphasizing simultaneous emissions from A and B are not possible as only one photon at a time passes through the double slit. Thus my original point, I believe simultaneous emissions do occur. Do you agree?

 

[Cthugha]

Well, not in original paper, but not anywhere, too.

Ehm, where is what claimed, exactly? It does not make any sense to discuss someone said sometime without knowing what is exactly meant.

There are many comments emphasizing simultaneous emissions from A and B are not possible as only one photon at a time passes through the double slit. Thus my original point, I believe simultaneous emissions do occur.

Again: Who claims that where? By the way, that does not have anything to do with DCQE anyway. The physics of single-photon interference patterns is well understood and the physics of two-photon interference patterns is important for DCQE, but very different from single photon interference.

Do you agree?

Not really.

In this experiment, the 351.1nm Argon ion pump laser beam is divided by a double-slit, which means the actual photon rate is 10^13 times higher than in a single-photon setup.

Where do you get this number from? The original paper does not even give the intensity or coherence properties of the pump beam, so one actually cannot judge the actual photon number rate.

 

[jwalker]

Ehm, where is what claimed, exactly? It does not make any sense to discuss someone said sometime without knowing what is exactly meant.

It was in the articles and presentations I looked through.

the physics of two-photon interference patterns is important for DCQE

I believe simultaneous emissions do occur. Do you agree?

Not really.

So, it's both yes and no at the same time, got it .
Joking aside, what are the two interfering photons you are referring to?

Where do you get this number from? The original paper does not even give the intensity or coherence properties of the pump beam, so one actually cannot judge the actual photon number rate.

My bad, that was my quick estimate, and I just considered a 1 W laser, which is probably too much, and it appears I even made a mistake by one order. 1 W creates 10¹⁸ photons, and SPDC (that is used for one-photon source) rate is 10⁶, so it is 10¹² difference.

 

[Cthugha]

It was in the articles and presentations I looked through.

Do you by chance have a link or something? There is everything from valid peer-reviewed publication to obscure crackpottery about DCQE.

So, it's both yes and no at the same time, got it .
Joking aside, what are the two interfering photons you are referring to?

The entangled photons after the down-conversion process are in a state that lacks first-order coherence, but has phase sensitive correlations that can end up yielding so-called two-photon interference patterns. Such patterns show up ONLY in coincidence counting, but never on one detector alone.

My bad, that was my quick estimate, and I just considered a 1 W laser, which is probably too much, and it appears I even made a mistake by one order. 1 W creates 10¹⁸ photons, and SPDC (that is used for one-photon source) rate is 10⁶, so it is 10¹² difference.

That is hard to judge. I do not know whether the laser power is given in the manuscript and whether the laser is pulsed or cw. Also, the efficiency of the SPDC crystal will depend on its thickness and how well it is adjusted. 10^12 photons (if equidistant) would correspond to a mean difference of 300 microns between consecutive photons which is already quite much. That should be long compared to the coherence length of the SPDC photons.

 

[jwalker]

Do you by chance have a link or something? There is everything from valid peer-reviewed publication to obscure crackpottery about DCQE.

I could try to find the links again, but they're really irrelevant here.

The entangled photons after the down-conversion process are in a state that lacks first-order coherence, but has phase sensitive correlations that can end up yielding so-called two-photon interference patterns. Such patterns show up ONLY in coincidence counting, but never on one detector alone.

So, is it correct that it is not a real interference but something that just produces an interference-like pattern?

That is hard to judge. I do not know whether the laser power is given in the manuscript and whether the laser is pulsed or cw. Also, the efficiency of the SPDC crystal will depend on its thickness and how well it is adjusted. 10^12 photons (if equidistant) would correspond to a mean difference of 300 microns between consecutive photons which is already quite much. That should be long compared to the coherence length of the SPDC photons.

I know, but the key word is "mean", and the rate is high.
Now that I'm thinking about it, since the distance between A and B is quite big, the emissions must not be simultaneous for interference to occur somewhere down the set. However, my original question still stands -- is it the interpair interference what causes the interference patterns? That is something for me to think about further, but do you agree that technically this interference is possible?

Update: since Glan-Thompson prism was used to split signals and idlers, the distance between A and B gets compensated, so that if A and B are emitted at the same time, they can interfere later. Thoughts?

 

[Cthugha]

So, is it correct that it is not a real interference but something that just produces an interference-like pattern?

Two-photon interference patterns are just as real as single-photon interference patterns. It is just a different mechanism. There is no singular mechanism which is "the real" interference mechanism.

I know, but the key word is "mean", and the rate is high.
Now that I'm thinking about it, since the distance between A and B is quite big, the emissions must not be simultaneous for interference to occur somewhere down the set. However, my original question still stands -- is it the interpair interference what causes the interference patterns? That is something for me to think about further, but do you agree that technically this interference is possible?

I still do not get your point. The photons are either inside the same coherence volume and therefore indistinguishable per se, so it is meaningless to talk about "interpair" interference or they are not from one coherence volume. In the latter case they are distinguishable and do not interfere at all.

 

[jwalker]

Two-photon interference patterns are just as real as single-photon interference patterns. It is just a different mechanism. There is no singular mechanism which is "the real" interference mechanism.

To me there is an interference when photons (either single or double) pass through a double slit and interact with each other or themselves. On the contrary, in the experiment, the entangled photons never interact, the only connection between them is the coincidence circuit. Can you still call this an interference even though the result chart looks like interference? I hope we'll not get at interpretations or philosophy here.

I still do not get your point. The photons are either inside the same coherence volume and therefore indistinguishable per se, so it is meaningless to talk about "interpair" interference or they are not from one coherence volume. In the latter case they are distinguishable and do not interfere at all.

My point is that if two coherent photons are generated at A and B, they can interfere (actually, not just by means of a coincidence circuit) further in the set. And to me it looks like this real interference (namely, on BS) is what causes interference patterns when D₁ or D₂ is used, but not with D₃. Also it simply and clearly explains the phase shift between D₁ and D₂ charts.

 

[Cthugha]

To me there is an interference when photons (either single or double) pass through a double slit and interact with each other or themselves. On the contrary, in the experiment, the entangled photons never interact, the only connection between them is the coincidence circuit. Can you still call this an interference even though the result chart looks like interference? I hope we'll not get at interpretations or philosophy here.

Interference occurs whenever you have two or more different, but completely indistinguishable ways to get from some initial to some final state. Photons do not have to meet somewhere to interfere. That has been tested by Pittman, Shih and others.

My point is that if two coherent photons are generated at A and B, they can interfere (actually, not just by means of a coincidence circuit) further in the set. And to me it looks like this real interference (namely, on BS) is what causes interference patterns when D₁ or D₂ is used, but not with D₃. Also it simply and clearly explains the phase shift between D₁ and D₂ charts.

If what you see was a single photon interference pattern created by coherent photons from A and B, you should be able to see the patterns without coincidence counting by just moving D1 and D2 around. This is not the case.

 

[jwalker]

Interference occurs whenever you have two or more different, but completely indistinguishable ways to get from some initial to some final state. Photons do not have to meet somewhere to interfere. That has been tested by Pittman, Shih and others.

Are you referring to "Optical imaging by means of two-photon quantum entanglement"? This is so good that you mentioned it, as it makes it easier for me to explain my idea--see below.

If what you see was a single photon interference pattern created by coherent photons from A and B, you should be able to see the patterns without coincidence counting by just moving D1 and D2 around. This is not the case.

If they moved them, which I believe they did not in the original experiment, they wouldn't get the pattern, but not because interference isn't happening there. The reason is that in reality A and B photons each time have different directions, and the resulting image is a sum of multiple interference patterns randomly shifted along the detector scanning axis, which makes it a simple hump.
Now if you are able to only select photons having specific direction, you'll get a normal interference pattern out of this mess. And you are--you have the movable D₀ that catches entangled photons, and entangled means their directions are correlated with the photons interfering on BS.
It works almost like in the experiment mentioned above, only instead of letters aperture you have an interference pattern in front of the fixed detector.

 

[jwalker]

By the way, I'm not sure if passing the double slit and hitting the BBO with signal-idler generation collapses photon's wave function. If it doesn't, things are even simpler because simultaneity doesn't apply and interference on BS is obvious.

 

[San K]

If what you see was a single photon interference pattern created by coherent photons from A and B, you should be able to see the patterns without coincidence counting by just moving D1 and D2 around. This is not the case.

well said!

Photons do not have to meet somewhere to interfere. That has been tested by Pittman, Shih and others.

can you elaborate (on the experiment) or post the paper?

aps won't allow...

 

[Cthugha]

Are you referring to "Optical imaging by means of two-photon quantum entanglement"? This is so good that you mentioned it, as it makes it easier for me to explain my idea--see below.

No, I was not even talking about entangled photons. Probably the clearest demonstration of what I mean is "Can Two-Photon Interference be Considered the Interference of Two Photons?" by Pittman et al., Phys. Rev. Lett. 77, 1917–1920 (1996). The NIST hosts a version of that paper here: http://physics.nist.gov/Divisions/Div844/publications/migdall/psm96_twophoton_interference.pdf. They start from the Hong-Ou-Mandel effect and test whether it is really necessary that the photons actually meet at the beam splitter.

If they moved them, which I believe they did not in the original experiment, they wouldn't get the pattern, but not because interference isn't happening there. The reason is that in reality A and B photons each time have different directions, and the resulting image is a sum of multiple interference patterns randomly shifted along the detector scanning axis, which makes it a simple hump.

This is somewhat puzzling. See my next response for my explanation why.

Now if you are able to only select photons having specific direction, you'll get a normal interference pattern out of this mess. And you are--you have the movable D₀ that catches entangled photons, and entangled means their directions are correlated with the photons interfering on BS.
It works almost like in the experiment mentioned above, only instead of letters aperture you have an interference pattern in front of the fixed detector.

Indeed you are using D0 to filter out a specific emission angle - it acts as a kind of spatial coherence filter. But you do not need to have "different" photons interfere to achieve that. Exactly that is happening in the low photon rate regime, too. Entangled photons are spatially incoherent and do not produce an interference pattern. You place D0 somewhere, get a small subset of entangled photons and the photon subset found in coincidence counting with D0 will be spatially coherent. That is a two-photon interference effect.

By the way, there are also DCQE-experiments performed in the single photon count regime like "Experimental realization of Wheeler’s delayed-choiceGedanken experiment" by Jacques et al. (Science 315, 966 (2007).

 

[zonde]

Indeed you are using D0 to filter out a specific emission angle - it acts as a kind of spatial coherence filter. But you do not need to have "different" photons interfere to achieve that. Exactly that is happening in the low photon rate regime, too. Entangled photons are spatially incoherent and do not produce an interference pattern. You place D0 somewhere, get a small subset of entangled photons and the photon subset found in coincidence counting with D0 will be spatially coherent. That is a two-photon interference effect.

I do not say that I understand what jwalker is talking about but it does not seem like you are providing explanation either.

In two outputs of last BS (detector D1 and D2) relative phase between path A and path B differs by $\pi$ independently from any considerations about spatial direction. The same is true about detector D0 compared when it is located at crest and trough points (for D0/D1 or D0/D2 coincidence graphs).
And nowhere spatial coherence comes into consideration.

... or am I missing something?

 

[Cthugha]

Ok, maybe my explanation was a bit short. The Kim et al. version is also not really the most pedagogical example for DCQE experiments out there, so let me try again.

In two outputs of last BS (detector D1 and D2) relative phase between path A and path B differs by $\pi$ independently from any considerations about spatial direction.

The phase shift introduced by reflection or transmission at beam splitters and miroors is always $\pi$, completely correct.

And nowhere spatial coherence comes into consideration.

Hmm, it is indeed not easily seen. The arm of the experiment which goes to detectors D1 and D2 is somewhat similar to a modified Mach-Zehnder interferometer. You have two indistinguishable ways to get to the final beam splitter and whether reflection or transmission occurs depends on the relative phase acquired along these to ways. In a Mach-Zehnder one would now move a small piezo-based mirror to increase the length of one of the paths to vary the phase difference. In this DCQE experiment that phase difference exists because the emitted light is spatially incoherent which means that there is a large spread of emission angles. The effective difference in path lengths from emission spot A to the beam splitter and emission spot B to the beam splitter will differ for each possible emission angle. The sum over all possible emission angles will loo like no interference at all.

However, placing D0 at some position in the Fourier plane behind the lens in the other arm changes things. As D0 is located in the Fourier plane, each detector position directly corresponds to photons emitted under some angle. That means that the emission angle of the corresponding photons on the other side is also well defined. Therefore, in coincidence counting one kind of filters out one single emission angle for each detector position at D0. That means you have a fixed phase difference on the Mach-Zehnder like setup on the other side and therefore photons will either be more likely to go to D1 or D2. When you now move D0 around, you continuously change the phase difference at the Mach-Zehnder like part and also get a similar interference pattern.

As spatial coherence basically is a measure for how big the spread in the emission angles from some light source as seen at some position is, I tend to think of this kind of experiments as sophisticated spatial coherence filters. However, that is way more appropriate and easier to identify in the Walborn version of the DCQE experiment.

Did this explanation make more sense?

 

[San K]

Interference occurs whenever you have two or more different, but completely indistinguishable ways to get from some initial to some final state. Photons do not have to meet somewhere to interfere.

ref: http://physics.nist.gov/Divisions/Div844/publications/migdall/psm96_twophoton_interference.pdfThey do not have to meet at the beam splitter, got that.

However do they not have to meet, somewhere along the path, after the interferometer, either?can we not tell the path the photon took based on the delay (order of arrival) at the detector?

if so, then...

how are the amplitudes made indistinguishable in the above paper/experiment?

 

[zonde]

However, placing D0 at some position in the Fourier plane behind the lens in the other arm changes things. As D0 is located in the Fourier plane, each detector position directly corresponds to photons emitted under some angle.

It sounds like we should get image of spot A and spot B in that plane where D0 is moved. But then we would observe two peaks corresponding to A and B not interference between them.

Btw are you assuming frequency filters before detectors or no? Paper does not say this but it's hard to imagine that there where no filters before detectors.

 

[Cthugha]

However do they not have to meet, somewhere along the path, after the interferometer, either?

can we not tell the path the photon took based on the delay (order of arrival) at the detector?

They do not meet somewhere, but compensating the delay is of major importance. In that paper the delay is compensated by having light beams of perpendicular polarization and by adding a delay line which cause light of a given polarization to take a longer path. At the end polarization information is of course erased. In that way you can adjust the time the photons take to the detectors. In this experiment, the delay is not even exactly cancelled, but D2 will always fire before D1. However, the setup is fine-adjusted as to keep this delay constant for any possible path the photons could take, so that D1 always fires at a fixed delay to D2, thus keeping the pathways indistinguishable.

It sounds like we should get image of spot A and spot B in that plane where D0 is moved. But then we would observe two peaks corresponding to A and B not interference between them.

On page 2 the authors claim to have placed D0 in the Fourier transform plane. If that is correct, you cannot get images of spots A and B because the emitted light does not carry any information about its position of emission in that plane, but emission angle position is completely present.

Btw are you assuming frequency filters before detectors or no? Paper does not say this but it's hard to imagine that there where no filters before detectors.

Hmm, I suppose one should use filters to reduce influence of background noise. In the paper they just assume some filter having some spectral transmission function in the calculations, so I suppose they used filters, too.

 

[jwalker]

The effective difference in path lengths from emission spot A to the beam splitter and emission spot B to the beam splitter will differ for each possible emission angle. The sum over all possible emission angles will loo like no interference at all.

However, placing D0 at some position in the Fourier plane behind the lens in the other arm changes things. As D0 is located in the Fourier plane, each detector position directly corresponds to photons emitted under some angle. That means that the emission angle of the corresponding photons on the other side is also well defined. Therefore, in coincidence counting one kind of filters out one single emission angle for each detector position at D0. That means you have a fixed phase difference on the Mach-Zehnder like setup on the other side and therefore photons will either be more likely to go to D1 or D2. When you now move D0 around, you continuously change the phase difference at the Mach-Zehnder like part and also get a similar interference pattern.

That's what I said, no?

So, can this still be seen as an eraser since there is a more logical explanation?

 

[zonde]

On page 2 the authors claim to have placed D0 in the Fourier transform plane. If that is correct, you cannot get images of spots A and B because the emitted light does not carry any information about its position of emission in that plane, but emission angle position is completely present.

Hmm, I suppose one should use filters to reduce influence of background noise. In the paper they just assume some filter having some spectral transmission function in the calculations, so I suppose they used filters, too.

Ok, I think these two things are clear to me now and would like to return to your explanation.

In this DCQE experiment that phase difference exists because the emitted light is spatially incoherent which means that there is a large spread of emission angles. The effective difference in path lengths from emission spot A to the beam splitter and emission spot B to the beam splitter will differ for each possible emission angle. The sum over all possible emission angles will loo like no interference at all.

After light from path A and path B is mixed in BS it has the same intensity in either output (so it does not have certain phase difference). As I understand you are claiming that this is because light in path A and path B is spatially incoherent.
Then you say that in detector D0 we get spatially coherent light. And yet we see no interference in detector D0 (so it does not have certain phase difference between A and B either).

So I see no grounds for your "because".

As I see, by moving detector D0 we introduce change in relative phase difference between path A and path B. In modified Mach-Zehnder interferometer we again introduce change in relative phase difference between path A and path B. And what we see is that for entangled photons phase difference between path A and path B is not certain but it is correlated for two photons in entangled pair.

It's just like with polarization - polarization entangled photons do not have certain polarization but two photons in entangled pair have correlated polarization i.e. polarization varies but it varies from pair to pair.

 

[Cthugha]

That's what I said, no?

Well, you claimed the effect arises due to interference arising from the pump beam photons. Although it may sound like semantics, this is not the case and imho it is an important difference that you really need two-photon interference.

So, can this still be seen as an eraser since there is a more logical explanation?

Well, it is still an eraser, no? It should be noted that the pseudoscientific explanations on crackpot sites out there having a very special opinion on what the eraser experiment means should not be taken as the standard reference what the DCQE experiment means.

After light from path A and path B is mixed in BS it has the same intensity in either output (so it does not have certain phase difference). As I understand you are claiming that this is because light in path A and path B is spatially incoherent.

Yes. However that means that the intensity behind both outputs is equal only when averaging over all realizations of possible phase differences.

As I see, by moving detector D0 we introduce change in relative phase difference between path A and path B. In modified Mach-Zehnder interferometer we again introduce change in relative phase difference between path A and path B. And what we see is that for entangled photons phase difference between path A and path B is not certain but it is correlated for two photons in entangled pair.

That is roughly correct, I think.

So I see no grounds for your "because".

Hmm, I tend to think in German and then translate that into English. Sometimes the translations may get a bit sloppy. I apologize for that. I intended to say that spatially incoherent light is necessary in order have some fixed phase difference for the two-photon state - which is exactly the correlation you mention - but not in any of the single photon states alone. Or to put it differently, spatial incoherence is a prerequisite for momentum entanglement.

 

[jwalker]

Well, you claimed the effect arises due to interference arising from the pump beam photons.

I never considered the pump photons behind BBO. Like I said before, I noticed that there are claims that the experiment equals to double-slit. If it does, the wave function won't collapse on BBO emission. To me it looked questionable, so I proposed an option of two photons hitting both A and B to later cause an interference on BS. It absolutely has nothing to do with the original beam.

Well, it is still an eraser, no?

Technically it is, but since there is a quite simple logical explanation to it, "delayed choice quantum eraser" sounds very misleading, doesn't it?
And the reason you don't get the interference pattern on R₀₃ is not because the photons are distinguishable, but because there is simply no interference for those photons. Correct?

 

[Cthugha]

Technically it is, but since there is a quite simple logical explanation to it, "delayed choice quantum eraser" sounds very misleading, doesn't it?

Does it? The original experiment was designed to ask clever questions about complementarity and does that very well. However, there were also people "selling" their papers who make it sound more spectacular than it really is, sure. The term "choice" is somewhat ill-chosen as the only choice of the experimentalist lies in which subset to pick. He does not actually change the outcomes after the detections happened as is claimed by some crackpot sites. Peer-reviewed papers never claimed that. At least not the papers I know.

And the reason you don't get the interference pattern on R₀₃ is not because the photons are distinguishable, but because there is simply no interference for those photons. Correct?

Is there a difference? The photon paths are distinguishable - there is only one way going to D3 after all. As you need two indistinguishable pathways to get interference there is of course also no interference. Interference and indistinguishability are so closely entwined that I would not det a dividing line there. You can even formulate a duality relation between distinguishability of photon paths and visibility of the resulting interference pattern called Englert-Greenberger duality relation. Imho, one of the more important results of DCQE is that it stresses the role of complementarity.

 

[zonde]

I never considered the pump photons behind BBO. Like I said before, I noticed that there are claims that the experiment equals to double-slit. If it does, the wave function won't collapse on BBO emission. To me it looked questionable, so I proposed an option of two photons hitting both A and B to later cause an interference on BS. It absolutely has nothing to do with the original beam.

If you would say that pump photon is in superposition of hitting region A and B ("superposition" is hitting region A and B ) it would be common QM terminology.

The way you said it sounds like explaining interference as two photons arriving at screen at the same time. This can be discussed but then it would be reasonable to base this discussion on simple double slit instead of DCQE.

Technically it is, but since there is a quite simple logical explanation to it, "delayed choice quantum eraser" sounds very misleading, doesn't it?
And the reason you don't get the interference pattern on R₀₃ is not because the photons are distinguishable, but because there is simply no interference for those photons. Correct?

In context of Cthugha's comment I would like to add that distinguishability can be trivial or rather nontrivial, for example, arrival time at the same detector.
As I understand your objection is that distinguishability is trivial in this case - only path A leads to R₀₃ and there is no way how path B can end up in that detector.

 

[San K]

just to make sure I got this right:

Does

"indistinguishablity of the various two-photon amplitudes"

means that

the paths are indistinguishable in the sense that either of the entangled photons could have took it and we cannot tell

i.e.

we cannot tell if photon A took path A or (its entangled twin) photon B took path A

same argument for path B

where path A and path B could, for example, be the upper arm or lower arm of the mach-zehnder

 

[zonde]

just to make sure I got this right:

Does

"indistinguishablity of the various two-photon amplitudes"

means that

the paths are indistinguishable in the sense that either of the entangled photons could have took it and we cannot tell

i.e.

we cannot tell if photon A took path A or (its entangled twin) photon B took path A

same argument for path B

where path A and path B could, for example, be the upper arm or lower arm of the mach-zehnder

The paths are indistinguishable in the sense that single photon could have taken either path to detection event.
For two-photon case it's when we can not tell apart from coincidence event case:
photon 1/path 1A & photon 2/path 2A
from case:
photon 1/path 1B & photon 2/path 2B