Kim's delayed choice quantum eraser -- Is it really delayed?

[andrewkirk]

I have been mulling over various aspects of delayed quantum erasure and came upon the following puzzle. It is about the famous 'delayed choice quantum eraser' of Kim, Kulik, Shih and Scully.

The paper says (1st para of 2nd column on p2) that the path length from the BBO crystal that generates the entangled pair to the detectors D1, D2, D3 of the 'idler' photon is 2.5m longer than the path from BBO to D0, the detector of the 'signal' photon (from which interference fringes may/may not be observed). This equates to a delay of 8ns, which is substantially longer than the 1ns response time of the detectors.

It occurred to me though, on looking at the diagram (bottom left of last page), that the 'choice' to erase or not erase is not made at the detectors D1-D3 but at the beam splitter BSA. That is what determines whether a photon is deflected downwards to D3, thereby preserving which-path info, or transmitted on towards mirror M_A. After M_A. As soon as a photon is transmitted, it becomes inevitable that it will be registered by either detector D2 or D3, neither of which can discern which-path info, so the erasure has happened as soon as the photon clears BSA.

So the 'choice' to erase or not erase has been made as soon as the photon has traversed BSA. That choice is only 'delayed' if the path length from BBO to BSA is greater than from BBO to D0. The authors do not report that path length. I know the diagram is not to scale but on the diagram it looks shorter.

If that analysis is right then I can see no good reason to regard the erasure as delayed.

Agree / disagree?

If the analysis is right then the natural response would be to perform an experiment in which the path BBO to BSA is longer than from BBO to D0 by the same sort of margin (about 2.5m). Have such experiments been done? If not, is that because there is a practical difficulty in doing them? Looking at the diagram I can't see any reason why there should be a practical difficulty.

Here's a link to a diagram of the experiment on wikipedia. The setup in that diagram has one extra detector D4 of photons carrying which-path info, compared to Kim's diagram, but D0-D3 all denote the same detectors in both diagrams, as do BBO and BSA (BS_a).

 

[jerromyjon]

If that analysis is right then I can see no good reason to regard the erasure as delayed.

You are missing the point that the slits are at the beginning, which is the "source" of the interference patterns. First D0 says "we have a hit" then 8ns later you get either erasure which means we know what path it took, or you get interference which can only occur if both pathways were possible, prior to detection at D0 which indicates it is already beyond the slits so how did it know it was going as a particle or as waves?

 

[Nugatory]

It occurred to me though, on looking at the diagram (bottom left of last page), that the 'choice' to erase or not erase is not made at the detectors D1-D3 but at the beam splitter BSA.

The beamsplitter isn't "making the choice" because a beamsplitter doesn't collapse the wavefunction. We know this because we can produce an interference pattern by illuminating a screen with the two outputs of the splitter.

 

[andrewkirk]

The beamsplitter isn't "making the choice" because a beamsplitter doesn't collapse the wavefunction. We know this because we can produce an interference pattern by illuminating a screen with the two outputs of the splitter.

I don't see that production of an interference pattern in that way clarifies the issue. The experiment already produces both interference and non-interference patterns. The interference pattern comes from hits at D0 that synchronise with hits at D1 and D2. The non-interference pattern comes from hits at D0 that synchronise with hits at D3 and D4.

If the transmitted photons from BSA and BSB were directed towards a common screen, rather than being directed along Kim's paths that destroy which-path info, they would indeed produce an interference pattern (as too would the reflected photons). But that's a different experiment and we can't use that to say what is happening in Kim's without assuming counterfactual definiteness - which we don't want to do since CFD is one of the things being investigated by these Bell-related experiments.

Maybe what I'm asking for is a mathematical analysis that details the state at each step of the process. Is there a good presentation of that somewhere online? I haven't read the Maths in Kim's paper yet. Maybe that does it, but often I find in these papers that state descriptions at intermediate stages are omitted.

 

[Nugatory]

I don't see that production of an interference pattern in that way clarifies the issue. The experiment already produces both interference and non-interference patterns. The interference pattern comes from hits at D0 that synchronise with hits at D1 and D2. The non-interference pattern comes from hits at D0 that synchronise with hits at D3 and D4.

I'm sorry, I wasn't clear. If you take a beam splitter of of the type used in Kim's experiment and instead of sending the two output beams into the complex arrangement of mirrors, detectors, and coincidence counters you project the two output beams onto a screen... You'll find interference effects between the two beams. That tells us that the beam splitter interaction is not collapsing the wave function.

 

[andrewkirk]

Thank you Nugatory. I'm afraid I find it hard to convince myself with a prose explanation because all sorts of uncertainties and ambiguities arise. For instance, 'collapse of the wavefunction' seems to mean different things in different contexts, and can raise questions of what wavefunction we are talking about. It also seems strange that absorption of the idler photon by one of D1-D4 would collapse the wf when the earlier absorption of the signal photon by D0 did not. We need a mathematical analysis to elucidate the differences between those two absorptions. Only a mathematical analysis can tell us whether the 'collapse' is a consequence of the second absorption or of something subsequent to that, like the signal-idler coincidence being identified by the counter, or a physicist looking at the experimental results.

We are also still stuck with the problem of counterfactual definiteness. Without assuming CFD we cannot assume that measurements made in the different experiment you describe in post 5 tell us anything about measurements that 'could have been' but were not made in the Kim experiment. Sometimes we can't avoid assuming CFD but it seems to me that if we assume it in an experiment like this, most of the interest disappears from the findings.

Do you know of a source that mathematically analyses the evolution of the state in an experiment like this? I spent a little time looking but wasn't able to quickly find something that addressed it in sufficient detail.

Thanks

Andrew

 

[Derek Potter]

It depends what you think the experiment sets out to demonstrate. As Nugatory points out, beam splitters do not collapse the wavefunction any more than passage through air does. Wavefunction collapse occurs when D1 to D4 register. However if the experiment were intended to confirm this, then you would certainly want to know the distance to the beamsplitters, not to the detectors. In other words, the experiment does demonstrate delayed choice but it leaves a massive loophole if you are prepared to re-write QM.

 

[Derek Potter]

It also seems strange that absorption of the idler photon by one of D1-D4 would collapse the wf when the earlier absorption of the signal photon by D0 did not.

It seems strange because it would be strange. The detection at D0 does collapse the wavefunction! How is this possible? Simple - the wavefunction is the wavefunction of both photons. After detection at D0, the wavefunction would be collapsed with respect to the signal photon but not with respect to the idler.

I say "would be" because, in fact, the signal photon is absorbed. That's just an additional complication - if a similar experiment were performed with electrons, they could be detected and not disappear. You can ignore the ultimate fate of the signal photon, the important part is the measurement at D0.

Thus, after detection at D0, the wavefunction collapses into a probability distribution of "photon detected at D0" and "no photon detected at D0". Since we ignore the no-photon cases, we simply have "photon detected at D0". But remember, D0 is movable so the measurement at D0 is actually a measurement of position, x. But wait! The x-outcomes are correlated with "which-way information kept" and "which-way information erased". This has not been decided yet, so the wavefunction is not collapsed with respect to "the choice".

So the purported mystery, the reverse causality etc are right there, introduced because we assumed that a partial collapse occurred at D0. If the entire collapse is assumed not to occur until next day :) then the system remains in superposition until then. There is then no question of apparent reverse causality because the final collapse makes the "choice" and determines the outcome at D0 all at once.

Of course Kim et al will have looked at the data before then. Exactly how their lives are put on hold overnight is another matter.

 

[Nugatory]

We are also still stuck with the problem of counterfactual definiteness. Without assuming CFD we cannot assume that measurements made in the different experiment you describe in post 5 tell us anything about measurements that 'could have been' but were not made in the Kim experiment.

Counterfactual definiteness is a red herring here because the beam splitter is a macroscopic device and decoherence will cheerfully produce unobserved but definite outcomes in a macroscopic system. The spin of an electron is undefined if I haven't measured it, but a tossed coin is heads or tails even if no one ever looks.
(This is a good thing, as otherwise we would be unable to trust any experimental apparatus ever).

 

[Derek Potter]

Counterfactual definiteness is a red herring here because the beam splitter is a macroscopic device and decoherence will cheerfully produce unobserved but definite outcomes in a macroscopic system. The spin of an electron is undefined if I haven't measured it, but a tossed coin is heads or tails even if no one ever looks.
(This is a good thing, as otherwise we would be unable to trust any experimental apparatus ever).

I don't see why not. An improper mixed state is as good as a proper one FAPP.

 

[andrewkirk]

Ahah, or Eureka, or whatever.

It looks like the authors became troubled by the same question that I was. If we compare the arXiv version I linked above to the version finally published in Phys Rev Letters, we see that the latter has changed both the diagram and the description of the experimental setup so that now the paths from BBO to the beam splitters BSA and BSB are longer than from BBO to D0. See the following sentence in column 2 of page 2:

"The experiment is designed in such a way that L0, the optical distance between atoms A, B and detector D0, is much shorter than LA (LB), the optical distance between atoms A, B and the beam splitter BSA (BSB) where the which-path or both-path “choice” is made randomly by photon 2."

 

[Derek Potter]

Ahah, or Eureka, or whatever.

It looks like the authors became troubled by the same question that I was. If we compare the arXiv version I linked above to the version finally published in Phys Rev Letters, we see that the latter has changed both the diagram and the description of the experimental setup so that now the paths from BBO to the beam splitters BSA and BSB are longer than from BBO to D0. See the following sentence in column 2 of page 2:

"The experiment is designed in such a way that L0, the optical distance between atoms A, B and detector D0, is much shorter than LA (LB), the optical distance between atoms A, B and the beam splitter BSA (BSB) where the which-path or both-path “choice” is made randomly by photon 2."

Well spotted Andrew. Glad they didn't leave that loophole open.

 

[DrChinese]

If you are concerned about the delayed choice quantum eraser because of path length, why not look at the following delayed choice setup in which it should be quite clear:

http://arxiv.org/abs/quant-ph/0201134

"Such a delayed-choice experiment was performed by including two 10 m optical fiber delays for both outputs of the BSA. In this case photons 1 and 2 hit the detectors delayed by about 50 ns. As shown in Fig. 3, the observed fidelity of the entanglement of photon 0 and photon 3 matches the fidelity in the non-delayed case within experimental errors. Therefore, this result indicate that the time ordering of the detection events has no influence on the results and strengthens the argument of A. Peres [4]: this paradox does not arise if the correctness of quantum mechanics is firmly believed."

In all scenarios such as these, the ordering of detections does not affect the outcome.

 

[Derek Potter]

You are missing the point that the slits are at the beginning, which is the "source" of the interference patterns. First D0 says "we have a hit" then 8ns later you get either erasure which means we know what path it took, or you get interference which can only occur if both pathways were possible, prior to detection at D0 which indicates it is already beyond the slits so how did it know it was going as a particle or as waves?

Actually the slits do not create the interference pattern. Each entangled pair is created in one of two small regions within the BBO. The paper is very clear about that. There is therefore no interference due to the slits. The interference is due to the fact that D1 and D2 "see" the source effectively through a twin aperture after the BBO. The actual interference pattern is "wasted" as D1 to D4 are fixed, but coincidence between D1 or D2 and D0 create "ghost interference". But because the interference is due to a twin aperture rather than two slits, the pattern is far less well-defined than in a good Young's Slits experiment. As is clearly visible.

 

[Derek Potter]

You are missing the point that the slits are at the beginning, which is the "source" of the interference patterns. First D0 says "we have a hit" then 8ns later you get either erasure which means we know what path it took, or you get interference which can only occur if both pathways were possible, prior to detection at D0 which indicates it is already beyond the slits so how did it know it was going as a particle or as waves?

There are a number of misconceptions here. Firstly, as above, interference does not and cannot occur at the slits, interference occurs when two parts of a wave meet at the same point, hence the term - one part interferes with another. Secondly, erasure means we do not know what path the photon took - it is the "which-path" information that is erased. Thirdly, even Kim et al perpetuate the nonsense about "behaving as a particle" but nonsense it is, the photon presumably behaves as a particle precisely to the extent that it is a particle (said to pacify those who insist it's "only" an excitation of a photon field!) and does so at all times. The distinction is between a two-slit interference pattern and a single-slit diffraction pattern. Which sounds a lot less dramatic than wave/particle dualism but has the merit of being a lot more correct.

Hence the important distance is from D0 to BSA and BSB. Assuming, that is, that a photon chooses whether to erase its own "which-path" information.

 

[jerromyjon]

I have been trying to reread the entire paper from the start again but haven't had time yet. The words as usual fail to express my understanding of this "interference". I'm not certain but it appeared to me that you have 2 sets of data, from each of 2 detectors where interference occurs between anonymous "pathways" but does not occur on the 2 detectors which have path dependence. The slits or paths only function to create a constraint on the path to "isolate" an observable interference "zone" so it can be observed.

 

[jerromyjon]

In all scenarios such as these, the ordering of detections does not affect the outcome.

"Compared to the 1 ns response time of the detectors, a 2.3 m delay is thus sufficient for the “delayed erasure.” Although there is an arbitrariness in the time when a photon is detected, it is safe to say that the choice of photon 2 is delayed with respect to the detection of photon 1 at D0 since the entangled photon pair is created simultaneously."
Totally agree, it's not an issue.

First D0 says "we have a hit" then 8ns later you get either erasure which means we know what path it took, or you get interference which can only occur if both pathways were possible

Bad use of erasure, don't remember what I was thinking. You get either diffusion? or interference. I'm still trying to get to the heart of the phase difference, though.

 

[andrewkirk]

I am trying to work through the maths in the paper to understand how it predicts what was observed. I am finding it difficult because it uses different terminology from that with which I am familiar from papers like Bell's and those on some other double slit experiments (Aspect, Walborn).

In particular, on page 2, 2nd column, I am not familiar with the 'Glauber formula' on which it relies and I don't know what it means by "positive and negative-frequency components" (how can a frequency be negative?). Nor do I know the meaning of the 'creation operators' for photons, denoted by ##a_i^\dagger ##. The notation looks like that of raising and lowering operators used in solving harmonic oscillator problems, but I don't think that's what they mean.

Does anybody know of a less cryptic presentation of the mathematics, or can explain the above terms?

Thank you.

 

[Derek Potter]

I am trying to work through the maths in the paper to understand how it predicts what was observed. I am finding it difficult because it uses different terminology from that with which I am familiar from papers like Bell's and those on some other double slit experiments (Aspect, Walborn).

In particular, on page 2, 2nd column, I am not familiar with the 'Glauber formula' on which it relies and I don't know what it means by "positive and negative-frequency components" (how can a frequency be negative?). Nor do I know the meaning of the 'creation operators' for photons, denoted by ##a_i^\dagger ##. The notation looks like that of raising and lowering operators used in solving harmonic oscillator problems, but I don't think that's what they mean.

Does anybody know of a less cryptic presentation of the mathematics, or can explain the above terms?

Thank you.

I can't answer your question but I'd just like to ask you something. I'm not sure whether you are specifically interested in unravelling the maths or whether you just want to understand the DCQE experiment itself. For the latter, you need to understand the nature of the entanglement that is being exploited and also the production of the interference patterns. Can you answer these questions?

What entanglement property is being exploited?
Where is the interference taking place?
How could the system be modified to produce interference all the time?
In the Wikipedia picture (which has coloured ray paths), does the optical match of the blue and red paths matter?
Likewise, why are the alternative paths drawn as straight lines and not allowed to fan out by diffraction?
Is the which-path information erased from the quantum state or just ignored?
Does a beam splitter enable a photon to decide its own fate a) at all b) retrospectively?
Could the system be set up with random optical switches instead of beam splitters?
Is there any interference between photons or photon pairs from the two regions of the BBO?
Where is Schrodinger's Cat in all this?
What would MWI say happens?

OK, that last one was a bit tongue-in-cheek but if you allow D0 to remain entangled with the *other* photon until D1-D4 detect it, then all the mystery disappears. The entanglement of the photon pair is transferred to the detectors. The "reverse causality" is an artifact of assuming wavefuntion collapse, i.e. disentanglement, at D0.

 

[andrewkirk]

I can't answer your question but I'd just like to ask you something. I'm not sure whether you are specifically interested in unravelling the maths or whether you just want to understand the DCQE experiment itself.

I see the two as inextricably tied up together. It's not possible to do one without the other.

What entanglement property is being exploited?

Immediately after generation of the pair at the BBO crystal, the entanglement is of the type where the state is ##\frac{1}{\sqrt{2}}(|0\rangle_{p1}|1\rangle_{p2}+|1\rangle_{p1}|0\rangle_{p2})##. The signal and idler photons are entangled with orthogonal polarisations. This is outlined on page 2 of the Phys Rev Letters version of the paper, just above the diagram.

It is what happens to that entanglement as the photons go through the various change points (absorption of signal at D0, splitting at BSB or BSA, absorption at one of D1-D4) that I do not understand, and why I am trying to understand the mathematics.

 

[Derek Potter]

I see the two as inextricably tied up together. It's not possible to do one without the other.

Immediately after generation of the pair at the BBO crystal, the entanglement is of the type where the state is ##\frac{1}{\sqrt{2}}(|0\rangle_{p1}|1\rangle_{p2}+|1\rangle_{p1}|0\rangle_{p2})##. The signal and idler photons are entangled with orthogonal polarisations. This is outlined on page 2 of the Phys Rev Letters version of the paper, just above the diagram.

It is what happens to that entanglement as the photons go through the various change points (absorption of signal at D0, splitting at BSB or BSA, absorption at one of D1-D4) that I do not understand, and why I am trying to understand the mathematics.

Well that's the sort of thing I'm getting at: the entanglement by polarization is not utilized in the experiment! If you feel that understanding the experiment will emerge through understanding the mathematics, don't let me stop you going that route. But for me, getting a grip on what the experiment does comes first. The maths then becomes a matter of describing the step-by-step development in ket notation, so there's no calculation or nasty formulae to grapple with, but the logic is set out clearly. If you want to calculate the actual diffraction pattern you'll need some calculation of course, but it's only a Fourier transform once you've identified the source aperture correctly.

 

[jerromyjon]

Is there any interference between photons or photon pairs from the two regions of the BBO?

Of course. The A and B are the paths, slits or whatever which gives you two distinguishable angles where interference occurs.

Hence the important distance is from D0 to BSA and BSB. Assuming, that is, that a photon chooses whether to erase its own "which-path" information.

I really can't understand where you are coming from.

 

[Derek Potter]

Of course. The A and B are the paths, slits or whatever which gives you two distinguishable angles where interference occurs.

You certainly need two paths to get interference but I did not ask about "whatever gives you two angles", I asked specifically about the slits. I've already said what I think about the subject in post #14.

I really can't understand where you are coming from.

I'm just trying to help Andrew get the right balance between grappling with mathematics and understanding how the experiment works.

 

[andrewkirk]

I am trying to work through the maths in the paper to understand how it predicts what was observed. I am finding it difficult because it uses different terminology from that with which I am familiar from papers like Bell's and those on some other double slit experiments (Aspect, Walborn).

In particular, on page 2, 2nd column, I am not familiar with the 'Glauber formula' on which it relies and I don't know what it means by "positive and negative-frequency components" (how can a frequency be negative?). Nor do I know the meaning of the 'creation operators' for photons, denoted by ##a_i^\dagger ##. The notation looks like that of raising and lowering operators used in solving harmonic oscillator problems, but I don't think that's what they mean.

Does anybody know of a less cryptic presentation of the mathematics, or can explain the above terms?

Thank you.

Surely somebody reading this thread understands the Kim paper well enough to explain these aspects of it? It is, after all, a very influential, oft-cited paper.

Anyone?

 

[Cthugha]

Surely somebody reading this thread understands the Kim paper well enough to explain these aspects of it? It is, after all, a very influential, oft-cited paper.

Some comments:

The Glauber formula basically means that you start from the initial state A, choose some final state B (photons being absorbed at the two detectors) and just integrate over all probability amplitudesto get from A to B and get the square of the absolute to get the coincidence count rate.

Positive and negative frequency components might not make much sense for a sine wave, but remember that a Fourier transform does not disassemble a signal into sinusoids, but into complex exponentials. These frequency components can be positive or negative, which determines whether the component spiral clockwise or counterclockwise in complex space.

The creation and annihilation operators of light behave pretty much exactly like the raising and lowering operators of the harmonic oscillator.

However, most of the detailed math is not necessary to get the basics of the DCQE experiment. Here is a link to a (really ancient) strongly simplified description of the experiment I posted here several years ago. Maybe it helps a bit. The text and equation format of that post is horrible. I apologize for that.
https://www.physicsforums.com/threa...ed-choice-quantum-eraser.320334/#post-2241460

 

[Derek Potter]

Some comments:

The Glauber formula basically means that you start from the initial state A, choose some final state B (photons being absorbed at the two detectors) and just integrate over all probability amplitudesto get from A to B and get the square of the absolute to get the coincidence count rate.

Positive and negative frequency components might not make much sense for a sine wave, but remember that a Fourier transform does not disassemble a signal into sinusoids, but into complex exponentials. These frequency components can be positive or negative, which determines whether the component spiral clockwise or counterclockwise in complex space.

The creation and annihilation operators of light behave pretty much exactly like the raising and lowering operators of the harmonic oscillator.

However, most of the detailed math is not necessary to get the basics of the DCQE experiment. Here is a link to a (really ancient) strongly simplified description of the experiment I posted here several years ago. Maybe it helps a bit. The text and equation format of that post is horrible. I apologize for that.
https://www.physicsforums.com/threa...ed-choice-quantum-eraser.320334/#post-2241460

In that post you say:
"You can treat this experiment as if every photon here originates from either point A or B (following the pictures in the DCQE paper by Kim). Now the setting is as following:
"In entangled photon experiments each photon on its own behaves like incoherent light. This means, there is no or at least not much correlation between the phases of the fields originating from points A and B at the same moment. For the sake of simplicity I will now treat the phase of these fields as completely random and the amplitude as constant and equal."

Do you think that there is a wavefunction of a downconverted photon coming simultaneously from/through slits A and B (and/or the regions of the BBO directly in front of them)? i.e. is the stage set for two-slit interference a la Young's slits with slits A and B? I have claimed here several times that this is not the case. If I am wrong then I need to be told why, but I believe I am correct, that there is no interference of this kind going on. In fact I would challenge anyone who says there is to tell me how the wavefunction of the downconverted photon(s) propagates back to the pump slits! Or, if you prefer the idea that there is interference due to the two small regions of the BBO rather than the slits themselves, how a photon generated in a single region has a wavefunction that "emanates", as it were, from both? Rest assured that for a long while I thought exactly this. As far as I could tell this could only happen if the pump photon's wavefunction did indeed split over each slit and then excite both regions in superposition, retaining phase coherence of the solid state excitation, and then the pair emission in both regions would have to retain the phase coherence resulting in a superposition of photon pairs emanating from A and B. That quite obviously cannot be the case as the down-conversion is described as "spontaneous".

 

[Cthugha]

Do you think that there is a wavefunction of a downconverted photon coming simultaneously from/through slits A and B (and/or the regions of the BBO directly in front of them)? i.e. is the stage set for two-slit interference a la Young's slits with slits A and B? I have claimed here several times that this is not the case. If I am wrong then I need to be told why, but I believe I am correct, that there is no interference of this kind going on. In fact I would challenge anyone who says there is to tell me how the wavefunction of the downconverted photon(s) propagates back to the pump slits! Or, if you prefer the idea that there is interference due to the two small regions of the BBO rather than the slits themselves, how a photon generated in a single region has a wavefunction that "emanates", as it were, from both? Rest assured that for a long while I thought exactly this. As far as I could tell this could only happen if the pump photon's wavefunction did indeed split over each slit and then excite both regions in superposition, retaining phase coherence of the solid state excitation, and then the pair emission in both regions would have to retain the phase coherence resulting in a superposition of photon pairs emanating from A and B. That quite obviously cannot be the case as the down-conversion is described as "spontaneous".

Point 1: Of course there is no wavefunction of a downconverted photons. There are no wavefunctions for any photons. Just as for any other massless particle, you cannot apply the concept of a wavefunction to a photon. This is why you need to use probability amplitudes. And yes, these of course go back to the pump slits. But that is not a central point here.

Point 2: I do not understand at all how the photon coming simulataneously from A to B would set the stage for two-slit interference. Interference is defined in the ensemble average only. You need to average over many realizations to get the pattern. If the relative phase of two fields originating from the slits is the same for each realization, you have first-order coherence and will see an interference pattern. If the relative phase differs, you will average over many patterns which is the same as having no pattern. As this is first-order, this is not even sensitive to whether the light field is classical or not.

Point 3: I do not get your point about retaining phase coherence of the solid state excitation. Of course the phase of the pump beam matters as you need to get phase matching inside the crystal to get the downconversion to work. This translates to the (total or maybe better relative) phase of the entangled state created, which is the relative phase of the two light fields which you will create. The phase of each light field (signal/idler) on its own is random in each realization and will average out. The relative phase of signal and idler is of course not random, is determined by the pump light field and the geometry of the non-linear crystal and will "survive" ensemble averaging. Are you aware that such first-order interference effects and second-order interference effects are very different animals and even mutually exclusive? See, e.g. Phys. Rev. A 63, 063803 (2001) or simply follow this link:
http://arxiv.org/abs/quant-ph/0112065

 

[Derek Potter]

Sorry, I thought I was being perfectly clear. If it wasn't helpful forget it.

 

[jerromyjon]

Sorry, I thought I was being perfectly clear. If it wasn't helpful forget it.

It just appears more like a miniscule misconception but I can't quite isolate it. The OP is looking to understand math that isn't intrinsically related, so any elaborations could be helpful, as long as they legitimately convey accurate understanding. I thought this stuff was mind-blowing until I "got" it, now I say it really is quite simple.

 

[andrewkirk]

Thank you Cthuga for your post. I have copied that post in the other thread, to which you referred me, into a post of my own in that thread, and worked on the formatting to make it easier to read. I hope you don't mind. I did that purely for my own benefit, to improve my chances of understanding it.

 

[Derek Potter]

I am trying to work through the maths in the paper to understand how it predicts what was observed. I am finding it difficult because it uses different terminology from that with which I am familiar from papers like Bell's and those on some other double slit experiments (Aspect, Walborn).

In particular, on page 2, 2nd column, I am not familiar with the 'Glauber formula' on which it relies and I don't know what it means by "positive and negative-frequency components" (how can a frequency be negative?). Nor do I know the meaning of the 'creation operators' for photons, denoted by ##a_i^\dagger ##. The notation looks like that of raising and lowering operators used in solving harmonic oscillator problems, but I don't think that's what they mean.

Does anybody know of a less cryptic presentation of the mathematics, or can explain the above terms?

Thank you.

I can't follow the maths either, not being familiar with "standard quantum mechanical calculations" of any kind :) nor even creation operators, let alone the Glauber formula.

However I'll hazard a guess that the frequencies are spatial frequencies, ditto the phase shift of π, especially since the authors proceed to calculate the diffraction pattern using sinc functions: the far-field diffraction pattern is the Fourier transform of the diffractor aperture and the FT of a rectangular function is a sine(x)/x or "sinc" function). Negative frequencies are obvious for spatial sinusoids: you just reverse the sign of x e.g. sine(-x). This also makes sense since a standard FT produces components of both signs.

Yes, when the variable is time, a literal negative frequency is pretty meaningless, but you just move the sign around inside the term:
sin(ω.-t) = sin(-ωt) = -sin(ωt) i.e. sign reversal
cos(ω.-t) = cos(-ωt) = cos(ωt) i.e. no sign reversal

Notwithstanding my ignorance of second quantization, I'm pretty sure the dagger is a creation operator. I guess Kim et all want to introduce the random SPDC events properly and bundle them and the optics and everything else (except Schrodinger's cat for some reason) into one expression. It might be fun to see whether the creation simply propagates through as a "joint (potential) hit", meaning we could do as I do and ignore it, just assuming that photon pairs are created at random.

Unfortunately this kitchen sink approach obscures (to simple minds like mine) what is going on optically. Kim et al glibly take the FT of the two slits (which they do have the grace to put in quotation marks) but no-where do they show why this is appropriate. The photons are created one pair at a time in a probability distribution: either at A or at B as far as I can tell. They are not created behind a twin slit and there is no interference at D0 where the patterns are found.

I rather like the sound of that so I'll say it again. There is no interference at D0 where the patterns are found.

Instead, an interference pattern is created using a technique of Ghost Interference. I think the optical trick is to be seen by looking back at A and B from D1 and D2. As A and B only "flash" separately, there is no interference at D0 where the patterns are found. However, D1 and D2 see the two regions optically superimposed by the "half-silvered mirror" BSC. Hence the interference is A interfering with A, or B interfering with B, never A interfering with B. It occurs at D1 and D2 - not at D0 where the patterns are found*.

The question then is how the width of A, the pumped region, translates to a slit width. I have to assume that with SPDC, the photons are emitted across the entire excited region, all of it but no more - i.e. "as if" they were passing through a slit. That is a serious bit of physics to look into. Pity it gets glossed over.

Hence my concern as to whether the "slit" width is actually an artifact of the path aperture.

* Allegedly. In fact, spatial data is recorded there but the interference patten is encrypted by the D1-D4 data.

Does that help?

 

[Cthugha]

However I'll hazard a guess that the frequencies are spatial frequencies, ditto the phase shift of π, especially since the authors proceed to calculate the diffraction pattern using sinc functions: the far-field diffraction pattern is the Fourier transform of the diffractor aperture and the FT of a rectangular function is a sine(x)/x or "sinc" function). Negative frequencies are obvious for spatial sinusoids: you just reverse the sign of x e.g. sine(-x). This also makes sense since a standard FT produces components of both signs.

No,they are really talking about time. Take a simple cosine function and decompose it into complex exponentials:
[itex]\cos(\omega t)=\frac{1}{2}(e^{i \omega t}+ e^{-i \omega t})[\itex]
There you already have a positive and a negative frequency component. One simply finds that negative components are associated with photon annihilation operators and positive components are associated with photon creation operators. However, I am not sure doing all this math really helps for the problem at hand if one is not familiar with the math. The experiment is really not as complicated as it seems.

The photons are created one pair at a time in a probability distribution: either at A or at B as far as I can tell. They are not created behind a twin slit and there is no interference at D0 where the patterns are found.

One very important point: This is exactly NOT what is happening. If they are created et either slit, the experiment will not work. Only if it is completely indistinguishable from which slit the photons originated, you will be able to see interference.

I rather like the sound of that so I'll say it again. There is no interference at D0 where the patterns are found.

Of course not. And you are right, this is a very impoirtant point. Light created by SPDC is about as incoherent as it gets. It is more or less like using unfiltered sunlight, which will also not show interference in a usual size double slit experiment. Many people are not aware that in a common double slit experiment you can easily switch between seeing interference and not seeing interference by just placing the light source closer to the double slit and that you can change the positions of the maxima and minima of the interference pattern by moving the light source parallel to the double slit. In my opinion understanding the "simple" double slit is the most important prerequisite for understanding the DCQE.

Instead, an interference pattern is created using a technique of Ghost Interference. I think the optical trick is to be seen by looking back at A and B from D1 and D2. As A and B only "flash" separately, there is no interference at D0 where the patterns are found. However, D1 and D2 see the two regions optically superimposed by the "half-silvered mirror" BSC. Hence the interference is A interfering with A, or B interfering with B, never A interfering with B. It occurs at D1 and D2 - not at D0 where the patterns are found*.

If that was the case,one would get the same result by aiming at one slit 50% of the time and aiming at the other also 50% of the time. That does not work.

The question then is how the width of A, the pumped region, translates to a slit width. I have to assume that with SPDC, the photons are emitted across the entire excited region, all of it but no more - i.e. "as if" they were passing through a slit. That is a serious bit of physics to look into. Pity it gets glossed over.

Usually the crystal size more or less directly translates into a slit width. But yes, this is a nontrivial detail.

 

[Derek Potter]

No,they are really talking about time. Take a simple cosine function and decompose it into complex exponentials:
[itex]\cos(\omega t)=\frac{1}{2}(e^{i \omega t}+ e^{-i \omega t})[\itex]
There you already have a positive and a negative frequency component. One simply finds that negative components are associated with photon annihilation operators and positive components are associated with photon creation operators. However, I am not sure doing all this math really helps for the problem at hand if one is not familiar with the math. The experiment is really not as complicated as it seems.

I don't see any negative frequencies there, I see imaginary ones :)
But I'm not familiar with the math (of 2Q I mean) so if you are absolutely sure that the analysis is about creation and annihilation operators then I will duck out and leave Andrew to unravel it.

One very important point: This is exactly NOT what is happening. If they are created et either slit, the experiment will not work.

According to the caption of fig 2 "A pair of signal-idler photons is then generated from either the A or the B region." Why would the authors say this if it not true?
If you insist that each photon is created across both psuedo-slits simultaneously then the big question is whether the part at A is created coherently with that at B or whether there is a random phase difference. If they were coherent then there would be interference at D0, regardless of coincidences. This does not happen. If there were a random phase difference then there could be no interference between the parts coming from A and from B, not even of the "coincidence" kind. So the premise must be false: the photon must be created in one or other of the psuedo-slits, just as Kim et al say.
An important point is that D1 and D2 must be able to "see" the doubled (both paths) A and the doubled B. But if that were not the case then there would be no interference however you cut it.

Only if it is completely indistinguishable from which slit the photons originated, you will be able to see interference.

I do not agree. If you have just one slit and divide the ray path into two before recombining them, so that it looks like a double slit from D1 or D2's viewpoint, then you will get interference. This is exactly what is done here.

If that was the case, one would get the same result by aiming at one slit 50% of the time and aiming at the other also 50% of the time. That does not work.

Why not? There would be no interference pattern at D0 - same result. There would be no interference at D0/D3 or D0/D4 - same result. You must mean there would be no interference at D0/D1 or D0/D2 which I am pretty sure is wrong for the reasons given.

 

[Cthugha]

I don't see any negative frequencies there, I see imaginary ones :)
But I'm not familiar with the math so if you are absolutely sure that the analysis is about creation and annihilation operators then I will duck out and leave Andrew to unravel it.

Yes. I am sure. I should retract some papers if the math was different. ;)
You can read up on that in pretty much every standard textbook on quantum optics in the chapter where quantization of the em field is introduced. Look it up in the Mandel/Wolf, Scully/Zubairy or in Schleich's book. It is really not as complicated as it sounds, though it is intimidating. :)

According to the caption of fig 2 "A pair of signal-idler photons is then generated from either the A or the B region." Why would the authors say this if it not true?

I do not see your point. Of course the caption is correct. The pair is created at the A or B region. No other regions are possible and the photons are emitted as a pair. Note that they do not say that it is necessarily defined from which region A or B the pair actually came.

If you insist that each photon is created across both psuedo-slits simultaneously then the big question is whether the part at A is created coherently with that at B or whether there is a random phase difference.

Well, if we have which-way information, we can of course say from which slit the pair was emitted. Otherwise, I have already explained the difference. Are you aware that two photon interference is a different concept than single photon interference? So, again: The two-photon state is coherent. The single photon states are of course not. They cannot be as explained in the paper I linked. Have you read it?

If they were coherent then there would be interference at D0, regardless of coincidences. This does not happen. If there were a random phase difference then there could be no interference between the parts coming from A and from B, not even of the "coincidence" kind. So the premise must be false: the photon must be created in one or other of the psuedo-slits, just as Kim et al say.

This is getting cumbersome. I explained already several times that coherence of the two-photon state does NOT imply coherence of the single photon states which make up the two-photon state. You are insisting that the mechanism of coherence for both is the same, which is wrong. I do not know, how to make this easier to understand besides linking to papers which describe the details as I do not know your level of expertise and how much of the concept is clear to you.
To boil it down: The two-photon probability amplitudes from A and B are of course coherent. The single photon probability amplitudes are not coherent. They vary from shot to shot. However, this description might be pointless, if you are not even aware of these concepts. It would be really helpful to know what level of expertise you have. That is why I asked, whether you are aware that even in a standard double slit not every light source will show an interference pattern.

I do not agree. If you have just one slit and divide the ray path into two before recombining them, so that it looks like a double slit from D1 or D2's viewpoint, then you will get interference. This is exactly what is done here.

You mean like in a Mach-Zehnder interferometer? That is somewhat similar, but usually rather used for measuring temporal coherence and gets more complicated as only spatial coherence matters in the DCQE. But it is remotely similar, yes.

Why not? There would be no interference pattern at D0 - same result. There would be no interference at D0/D3 or D0/D4 - same result. You must mean there would be no interference at D0/D1 or D0/D2 which I am pretty sure is wrong for the reasons given.

Indeed there would be no interference in the coincidence counts. This can be seen in the double slit version of the quantum eraser as discussed by Walborn (http://arxiv.org/abs/1010.1236). One can also see that in the quantum versions of DCQE (http://arxiv.org/abs/1206.4348), but I doubt that these are easy to understand without doing some math. The whole paper of Kim et al. is very clear about no interference being present as soon as there is which-way information, which you simply cannot erase if just one slit is pumped.

 

[Derek Potter]

Thanks Cthugha. We are somewhat at cross-purposes but the reason seems to be that I did not understand the actual optical set-up as I thought I did.
So, apologies to anyone whom I may have confused as a result.