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

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1. Jul 18, 2015

### 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 MA. After MA. 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 (BSa).

2. Jul 18, 2015

### jerromyjon

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?

3. Jul 18, 2015

### Staff: Mentor

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.

4. Jul 18, 2015

### andrewkirk

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.

5. Jul 18, 2015

### Staff: Mentor

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.

6. Jul 19, 2015

### 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

7. Jul 20, 2015

### 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.

8. Jul 20, 2015

### Derek Potter

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.

9. Jul 20, 2015

### Staff: Mentor

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).

10. Jul 21, 2015

### Derek Potter

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

11. Jul 21, 2015

### 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."

12. Jul 21, 2015

### Derek Potter

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

13. Jul 21, 2015

### 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.

14. Jul 21, 2015

### Derek Potter

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.

15. Jul 22, 2015

### Derek Potter

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.

16. Jul 22, 2015

### 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.

17. Jul 22, 2015

### jerromyjon

"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.

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.

Last edited: Jul 22, 2015
18. Jul 22, 2015

### 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.

19. Jul 23, 2015

### Derek Potter

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.

20. Jul 23, 2015

### andrewkirk

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.