How Does the Double-Slit Quantum Eraser Work with Walborn's Setup?

[xts]

Actually, we may try something even more funny - quantum eraser and FTL at home:

1. point-like source of more or less monochromatic non-polarised light
2. double slit with QWP's
3. polarisation preserving screen (such as for 3D movie projector)
4. (polarisator glasses for viewing 3D movies)

Now we should see bulky spot.
But as we put 3D-glasses on, we should see fringe with left eye and anti-fringe with right eye.

I must do that with kids

Of course, the story to be told then is that 3D-glasses erase which-path info, acting backward in time. Now the FTL communication is even more apparent than at Walborn's setup...

Please note that the experiment in such setup is equivalent to Walborn's one.

 

[Cthugha]

It is not stated clearly if the crystal was located far away from slits or maybe the exciting laser beam had been focused as very narrow, but efectively he used point-like source. What was shown by the fact, that when he removed QWP from s arm, and polariser from p arm, he observed fringe. He still counted coincidences, but in this case detector p acted only as a trigger, reducing experimental noise, clicking (ideally) for every pair produced.

The distance was 42 cm. Usually you need to place the double slit in a distance of at least 100 cm from the BBO to get coherence and single-photon interference for commonly used spot diameters at the BBO. Also you cannot reduce the spot diameter to arbitrary small diameters because achieving phase matching becomes a real nightmare if you have a small depth of focus. In my opinion the sentence "Before the quantum eraser experiment was performed, Bell’s inequality tests were performed to verify that entangled
states were being detected" means that there should not be coincidences in single photon count rates. The section around equation 7 sounds as if the opposite was the case. However, the total phase difference mentioned there is not known which makes the case somewhat unclear.

However, I agree that the paper would be much more transparent if it was clearly shown in a figure whether interference is seen in single photon count rates or not. If the light was coherent and you could see interference in single-photon count rates, the experiment indeed would be pretty pointless.

A more sensible and complete discussion of this and similar experiments is by the way given by mostly the same authors in "Spatial correlations in parametric down-conversion" which can be found on the ArXiv: http://arxiv.org/abs/1010.1236" .

This is more like a review article and less ambiguous. However, even in this article the exact dependence on things like aperture diameters is discussed only for the twin photon double slit, but not for the quantum eraser. However, it discusses/cites also experiments where the case "no interference in single-photon detection rates, but interference in coincidence counts" is demonstrated in a clear manner.

 

[xts]

The light was spatially coherent to degree consistent with contrast they got.

Walborn (bottom of left column p.5):
The pump beam is focused onto the crystal plane using a 1 m focal length lens to increase the transverse coherence length at the double slit. The width of the pump beam at the focus is approximately 0.5 mm
You may also find the rest of geometry: the slits were 42 cm from the crystal, and detector was 56 cm from the slits.
So at the detector plane the image of the source is 0.66 mm in diameter, while pattern has strips separated by 2 mm.

Corrected:
slit-detector distance was 83 cm, not 56, so the image of the source should be 1 mm in diameter. But it is still consistent with contrast they got.

 

[Cthugha]

Ok, that gives us some numbers to work with. Zeilinger once performed similar calculations for his experiments and gave the following equation for the maximum possible slit distance R under which an interference pattern is seen in single photon count rates (citing B.E.A. Saleh, M.C. Teich, Fundamentals of Photonics, J. Wiley&Sons Inc. (1991) as a reference):

R=1.22 \frac{\lambda}{\theta_Q}

where \theta_Q=\frac{S}{D}, where S is the source diameter and D is the distance between the source and the double slit.

Taking 700 nm for the wavelength, 0.5 mm for the source diameter and 42 cm for the distance between source and slit, I get roughly R=717 nm. The slits, however, are separated by 200 micrometers. I suppose you should not see any interference pattern and have rather incoherent light.

 

[xts]

Great! You got 717 μm, not nm. This result, being an upper limit is 3.5 times bigger than actual distance. Thus the pattern may be perfectly seen!

Geometrical explanation is pretty simple, it works the same for continuous light, single photon count, whatever you like. Let's assume the ideal pattern is created with point source. Now we have spatially spanning source, of some angular size as seen by our double-slit ensamble. The image on the screen is now created as a convolution of ideal image and a circle of the same angular diameter as our source.
In our case blurring circle is 1 mm in diameter, while strips are separated by 2 mm - we lose some contrast, but the image is still pretty good.

 

[Cthugha]

Argh, I never should perform calculations by hand if PhD students asking questions rush in and out. The mm just vanished. Yes, right. However, then their claim that "Before the quantum eraser experiment was performed, Bell’s inequality tests were performed to verify that entangled states were being detected" becomes pretty pointless as the states are just polarization-entangled, but not at all momentum-entangled. So they do not need entangled photons besides for figures 7-9 where they have some boiled-down kind of delayed choice. At least unless you use 3d glasses.

Then I do not understand why they did not perform the experiment under incoherent circumstances. That experiment is much cooler. Maybe it is just a "publish or perish" issue, so they could publish two papers...

 

[xts]

Sure, they could use spatial/momentum entanglement, and use additional lens in p arm (such that 1/f = 1/lens-crystal + 1/lens-detector). Maybe that was a subject for next grant.
But such experiment is still equivalent to my "quantum eraser at home" - it does not really utilize entanglement Bell's mystery: it just uses it to select cases of some state in well defined fixed base.
Currently they use entanglement to select pairs (fixed-H,V). If utilising spatial entanglement they would just put additional condition to the trigger: (k=fixed value, k'=thus also fixed), while now k' is getting fixed by narrow focused pumping beam. Such experiment would be still essentially equivalent to the one with point-like source of polarised light.

 

[Cthugha]

I tend to disagree. Using spatial entanglement under incoherent conditions means that you really need the information from the other arm to get the desired subset of photons in the double-slit arm.

In the setup used in the paper they could have placed the linear polarizer also in the double slit-arm and erase the which-way info there without having any need to use the twin photon. For spatial entanglement, there is no means to get the pattern without coincidence counting (unless you choose conditions where the light in that arm alone is already coherent, which, however, automatically breaks entanglement - that would be bogus).

 

[xts]

I tend to disagree. Using spatial entanglement under incoherent conditions means that you really need the information from the other arm to get the desired subset of photons in the double-slit arm.

That's wahat I said - as you set the trigger in p arm to some transverse momentum (direction) you reject all cases with photons of direction differing from some fixed value in arm s. So such coincidence trigger is essentially equivalent to a pinhole in s arm installed just by the crystal.

In the setup used in the paper they could have placed the linear polarizer also in the double slit-arm and erase the which-way info there without having any need to use the twin photon. For spatial entanglement, there is no means to get the pattern without coincidence counting (unless you choose conditions where the light in that arm alone is already coherent, which, however, automatically breaks entanglement - that would be bogus).

I see those two cases quite similar:
- coincidence trigger on polarisation H in p is equivalent to V-polariser in s.
- coincidence trigger on direction in p is equivalent to pinhole (of the same angular size, as trigger detector) in s

The only difference is that it is impossible to modify spatial-entanglement/pinhole experiment further to "really delayed choice" - something like 3D-glasses do for polarisation.

 

[Cthugha]

I see those two cases quite similar:
- coincidence trigger on polarisation H in p is equivalent to V-polariser in s.
- coincidence trigger on direction in p is equivalent to pinhole (of the same angular size, as trigger detector) in s

Ok, but this is a limited similarity. You would place the polarizer behind the double slit, but you would have to place the pinhole before the double slit which is effectively nothing else as breaking entanglement. There is still no means to get the pattern once the incoherent photon passed the double slit. That might sound like nitpicking, but I think the distinction is important (as you acknowledge in terms of delayed choice).

 

[xts]

Of course, you can't put a pinhole nor anything else behind slits to restore the pattern for spatially non-coherent light.
But that is not a delayed choice experiment! It is fully explainable in terms of common history.
Replace your spatially large crystal with a matrix of point-like pixels (display of your phone).
Now generate a random sequence of flashing pixels, only one pixel at a time. Don't note the sequence - it'll be recovered by measurements in p.
You have no pattern visible at s. But as you turn on coincidence trigger (p focused on just one pixel), you'll recover the pattern at s.

 

[Cthugha]

Of course this is not yet a delayed choice experiment, but you can build one by expanding that setup.

 

[bruce2g]

Walborn et al have a more approachable article entitled "Quantum Erasure" available at http://www.fsc.ufsc.br/~lucio/2003-07WalbornF.pdf" . It first appeared in American Scientist, and is targeted toward a technically literate audience, but leaves out the Dirac brackets.

They first describe the experiment without SPDC, i.e., with just a simple laser beam and a double slit. They get interference. Then, they insert the QWP's. Interference goes away. Then, they insert a polarizer between the qwp's and the screen, and interference comes back if it is at 0 or 90 degrees, but goes away if it's at + or - 45 degrees.

I think this is similar to one of XTS' experiments.

It's a good article, because they then show that if you then use entangled beams, and put the polarizer in the P beam and coincidence count, you get the same result.

One other note on XTS' experiment with the randomly oriented Nicol's prism:

To deprive the experiment from mystery, try something similar, but (almost) classical:
Throw out the laser and BBO crystal, and put a light bulb in place of laser (very dim lightbulb, it should emit so little light that single photons should be distinguishable by our detectors) and Nicol's prism in place of crystal. Then made series of experiments: you turn prism to randomly chosen position of two such, that H polarised light goes to p and V to s or vice versa (don't note its position in the lab-book, that secret must be revealed by measurement in p!), then you turn on the lightbulb for a while (not too long, just to score a at least one click in both of p and s detectors, and note it as a coincidence event), then start again: choose randomly position...

The outcome of the experiment will be exatly the same, as of Walborn's.

Would you still say that 'measurement in arm p erases the which-path information in arm s' ?

But it is too trivial, so it won't probably be awarded by publication in Phys.Rev. nor by grant for next similar experiment.

Basically, he is saying that you randomly orient the prism, wait to get a count in both the p and s detectors, then repeat. What this means is that the P photon basically records the polarization of the S photon (one came out of the ordinary beam, the other from the extraordinary beam, since Nicol's prism is a polarizing beam splitter). So, if you put a horizontal polarizer in the P beam and detect a photon there, then XTS assumes the corresponding photon in the S beam would be vertically polarized, and so you would get interference. I believe that this assumption is flawed, because the S and P photons were polarized in specific orthogonal angles by the pbs, and will not have perfectly correlated results if tested with other polarization angles.

PS - sorry to come into this so late, but I've been away from the forum for a while, and I had to reread all of the articles to sort of catch up again. Thanks for your discussions in this thread, it really stimulated me to think this stuff through again from a fresh point of view.

 

[xts]

...if you put a horizontal polarizer in the P beam and detect a photon there, then XTS assumes the corresponding photon in the S beam would be vertically polarized, and so you would get interference. I believe that this assumption is flawed, because the S and P photons were polarized in specific orthogonal angles by the pbs, and will not have perfectly correlated results if tested with other polarization angles.

It is not flawed in context of Walborn's experiment setup - he never measures their polarisation in a rotated base - other than H/V. Only what he takes from "entanglement mystery" is a source of mixed (H/V) and (V/H) pairs, of which those in eigenstate (H/V) are selected by coincidence trigger.

I never claimed that Bell was wrong... (just contrary) But this experiment has nothing in common with it.

My Occamian point is that people like Walborn tend to use high pitch rethorics of 'quantum erasure' where simple 19th-century wave optics is sufficient to explain the experiment.

PS.
I just received my QWP (ordered few days ago), and my 18 years old nephew is coming back from his holidays, so we are ready to build our Quantum Eraser at Home...

 

[bruce2g]

It is not flawed in context of Walborn's experiment setup - he never measures their polarisation in a rotated base - other than H/V. Only what he takes from "entanglement mystery" is a source of mixed (H/V) and (V/H) pairs, of which those in eigenstate (H/V) are selected by coincidence trigger.

Actually, after additional reflection, I think the xts experiment will work. If the Nicol's prism PBS is oriented at 90 degrees, then the P beam will be polarized horizontally, so it will be detected through a horizontal polarizer, and the S photon will be 90 degrees and the QWP's will not mess up its interference pattern. As the PBS angle moves away from 90 degrees, the probability P will be detected declines. If the PBS angle is 45 degrees, then the S photons will be non-interfering, but only 50% of the P photons will register, so the fringes should still be visible above the noise. And, when the angle goes to 90 degrees, none of the P photons will register, so none of the anti-fringe S photons will register, so the anti-fringes will not mess up the fringes. So, the interference pattern will be less dramatic, but I believe it will appear.

My Occamian point is that people like Walborn tend to use high pitch rethorics of 'quantum erasure' where simple 19th-century wave optics is sufficient to explain the experiment.

I basically agree. I'd say that Walborn's using the polarizer on the P beam to detect photons which will be (or have been) all phase shifted the same way by the qwp's, so that subset can show interference. In this sense it doesn't matter if the P photons are detected before the QWP's affect the beam, or afterwards.

One complication, however, is that in QM the polarization of the S photons is not defined when they go through the QWP's (Bell's theorem, no 'hidden variables'). The polarization is not defined until it's detected, which happens when the P photons go though the filter and get detected. So in this sense, it's hard to reconcile the classical optics version with the QM version.

PS.
I just received my QWP (ordered few days ago), and my 18 years old nephew is coming back from his holidays, so we are ready to build our Quantum Eraser at Home...

Looking forward to hearing about the results.

 

[xts]

One complication, however, is that in QM the polarization of the S photons is not defined when they go through the QWP's (Bell's theorem, no 'hidden variables'). The polarization is not defined until it's detected, which happens when the P photons go though the filter and get detected. So in this sense, it's hard to reconcile the classical optics version with the QM version.

In my QE@H you have the same - even more delayed choice, than in Walborn's!
Actually, mathematics is also the same as in QM - photon is a wavelet whos polarisation may be described as a complex superposition of H and V waves, which is unknown and gets defined not earlier than when I see it either with left (H) or right (V) eye, after reflecting from the screen
.
To remind the setup: QE@H consists of incoherent pointlike source, double slit with QWPs (oriented H behind one slit, V behind another), polarisation preserving screen and polarisation (linear, H/V) glasses to watch the result.

Walborn-like story is that photons come through slits with they polarisation marked by QWP, so no pattern is visible. But if you (very quickly!) put your glasses on, you are acting backward in time (as the marked photons already passed QWP and slits, and maybe even got reflected from the screen), erasing that information.

 

[bruce2g]

...
.
To remind the setup: QE@H consists of incoherent pointlike source, double slit with QWPs (oriented H behind one slit, V behind another), polarisation preserving screen and polarisation (linear, H/V) glasses to watch the result.
...

So QE@H is the same as the first setup in the Walborn paper at http://www.fsc.ufsc.br/~lucio/2003-07WalbornF.pdf :
laser->double slit->QWP->H polarizer-> screen
(except they use a laser, and you polarize after the screen).

I'm not sure you'll get interference if the source is not coherent, but if you play around with the source enough then you should get the same results they get: fringes viewed in one eye, and anti-fringes in the other eye. It sounds like you might be using a video projector setup, which is actually a nice repurposing of some home theater equipment.

I was thinking more about your other version, where a randomly oriented PBS creates S and P beams which are not entangled but are orthogonally polarized. In that one, you say that you will wait until you get a detection at S and P. I'm assuming there's a H polarizer in the P beam before the detector. In this case, there are orientations of the PBS which will result in nearly zero passage of P photons through the polarizer, so you could end up waiting a nearly infinite time for detection of a P photon. So in that version of the experiment you would obtain best results if you only leave the detectors on for a short period of time, and then re-randomize the PBS orientation if you do not detect a P photon within deltaT.

 

[UChr]

I am also curious to hear about xts succeed.


Meanwhile, a commentary on QWP in Walborns setup:


After the two QWP’s are passed, the two paths are circularly polarized Right = R or Left = L.

This gives 4 possible outcomes:Path1,Path2= R,L OR L,R OR R,R OR L,L.

If we put a mirror immediately after the QWP's and send the two ‘paths’ return we get:
R,L returns as (+/-) H, (+/-) H = Horizontal polarized.
L,R returns as (+/-) V, (+/-) V = Vertical polarized.
R,R and L,L returns as (+/-) H, (+/-) V OR (+/-) V, (+/-) H = polarized +45 (H + V) or -45 degrees (H – V).
This suggests that QWP start to polarize linearly in 4 combinations: H H or V V or V H or H V and then a little later circularly polarizes.
The linear polarization entanglement then stops and sets p-photon as linearly polarized perpendicular to one of the 4 polarizations.

Sounds fairly reasonable?