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

[UChr]

Walborn = ‘Double-slit quantum eraser’ with double slit and eventually quarter wave plates.
http://grad.physics.sunysb.edu/~amarch/Walborn.pdf


p. 5 section V: ”Figure 2 shows the standard Young interference pattern obtained with the double slit placed in the path of photon s, without quarter-wave plates QWP1 and QWP2, and with POL1 absent from detector Dp .”

So he just detects p and obtains anyway interfere with the 's'-beam.

We should then just have a different type of measurement on p – which do not give s interference - and we have a 'communication system'.

One possibility would be a PBS (0) - which forces p photons to choose between becoming horizontally or vertically polarized.

p. 2 section I: “Because of their momentum, time, and polarization correlation properties, photon pairs generated by spontaneous parametric down-conversion play an important role in the experimental demonstrations of quantum erasure”
“Because of their […] polarization correlation properties” the s-photons then should be vertically and horizontally polarized.

The vertically p-photons has been reflected - which gives a half wave difference.
and “Because of their […] time […] correlation properties” this gives the horizontally s-photons also a half wave difference .

Is this the argument? - So we need only a half silvered mirror at p to ensure that s consists of two types with half-wave of difference?

‘FTL’ because of coincidence counter - but with a greater distance to double slit and ideal conditions it maybe could work without.

 

[xts]

In my view this experiment brings nothing new or important neither to physics nor to philosophy. It just may be decomposed to:
1. "mystery of entanglement" - two photons are correlated. In this case we don't even have "Bell's mystery" - we may explain the results in terms of random series of H/V polarized pairs;
2. "double slit mystery" - still mysterious, despite over 200 years since Young noticed it - photons coming through double slit form fringe pattern;
3. "phase change" - not so mysterious - we may make an interferometer - as we change phase on one slit then fringe pattern shifts.

There is no need for sound words like 'which path information', 'erasure', 'FTL', etc. We just made a filter able to filter out one fringe pattern from its shifted counterpart, which mixed together form a bulky spot. Pattern and anti-pattern are created by two sub-series of events, which differ by the phase shift between slits. Other arm (p) allows us to tell what is the polarisation (H or V) of incoming photon, thus to which of those sub-series particular event belongs.

 

[xts]

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.

 

[DrChinese]

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 ...

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

No, it won't. You only get that result with entangled photons.

 

[DrChinese]

Walborn = ‘Double-slit quantum eraser’ with double slit and eventually quarter wave plates.
http://grad.physics.sunysb.edu/~amarch/Walborn.pdf


p. 5 section V: ”Figure 2 shows the standard Young interference pattern obtained with the double slit placed in the path of photon s, without quarter-wave plates QWP1 and QWP2, and with POL1 absent from detector Dp .”

So he just detects p and obtains anyway interfere with the 's'-beam.

We should then just have a different type of measurement on p – which do not give s interference - and we have a 'communication system'.
...

You do NOT get a communication system from this setup. A reasonable person would assume that top scientists might have noticed such a property of the system and would already be collecting there prizes if there were such.

But that doesn't happen. Instead, you can successfully transmit a bunch of random information faster than light. For whatever good that does anyone.

Haven't we had this conversation previously? Seems so familiar.

 

[xts]

No, it won't. You only get that result with entangled photons.

Would you accept a challenge, DrChinese?
I bet a bottle of (it is A.Z.'s favourite, btw) Beerenauslese Rotgipfler 2005 from Herr Kamper's winery, versus bottle of young Californian Chardonnay ;)

Challenge: Would you demonstrate calculations showing the results differ?

The only they (Walborn et co.) used of entanglement is a perfect correlation in predefined, fixed base. So it doesn't differ from 'classical' correlation based on common history.

 

[Cthugha]

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

No, it won't. You only get that result with entangled photons.

Yes and no. Thermal light has the tendency to bunch which indeeds mimics some of the properties of entangled light and you can indeed get increased coincidence count rates in coincidence count experiments (which by the way even made it to PRL, see "Two-Photon Imaging with Thermal Light" by A. Valencia et al., PRL 94, 063601 (2005)). While being not exactly the same experiment, DCQE and thermal ghost imaging are very closely connected.
Scarcelli and Shih did a myriad of experiments on that stuff.

However, the difference is: You get similar results with a lightbulb, but the visibilty of the interference pattern is of course not as good as using entangled photons. It is not a surprise that it is reduced to the maximum amount allowed by the inequalities that distinguish classical and non-classical light so that everything stays classical from the DCQE point of view.
From the imaging point of view it is the same. You get non-classical superresolution when using entangled photons and the commonly allowed resolution when using thermal light.

 

[xts]

You get similar results with a lightbulb, but the visibilty of the interference pattern is of course not as good as using entangled photons

I believe it would be even better, as lots of experimental noise would be reduced.
I may repeat the challenge thrown to Dr.C (I still have several bottles in my cellar) - show calculations!

I am not speaking about any entanglement-like coincidences. Just, perfectly correlated in a fixed base (eigenstate) of H/V, stream of independent single photons. They produce perfectly matching pattern/anti-pattern.

EDITed>>>
I am not claiming that entanglement may be explained by 'common-history-correlations' or other hidden local variables. No. Just contrary.
But this particular experiment does not violate Bell's inequality and is equally explainable in terms of common history.
Walborn does not use entanglement in a "Bell-sophisticated" way - he uses it just to obtain one-to-one correlation between orthogonal polarisations in a fixed base. The same as a lightbulb and Nicol's prism...

 

[Cthugha]

That depends on the circumstances. For the same mean count rate, quantum light performs better than thermal light. However, the mean intensity is of course higher for thermal light. However, you have a large noise background as thermal light is not perfectly correlated which provides an intrinsic background noise source even on the low-light level limit.

You can find a whole page of boring calculations in the appendix of O’Sullivan et al, "Comparison of the signal-to-noise characteristics of quantum versus thermal ghost imaging", Phys. Rev. A 82, 053803 (2010).

edit: I thought I understood what you mean, but now I do not see what you are getting at. When you are indeed assuming independent photons you will usually get no interference pattern at all because such experiments are usually carried out in a manner that the light at the double slit does not produce an interference pattern on its own anyway. If the photon in the other arm is completely independent of the other photon, no pattern will arise.

 

[xts]

OK. I won't bet good wine in challenge on supremacy of candle light over BBO crystals regarding detector noise
Anyway, as I played a bit with quantum cryptography devices, their single-photon sources are solid lasers attenuated to a bit less than one photon per clock tick. Even N.Gisin uses BBO crystals only on Sundays

EDITed>>>
N.Gisin is a co-owner of Id-Quantique - the company manufacturing QKD devices. He is also (at first!) recognised physicist in quantum optics, it was he, who performed lots of experiments extending Aspect's one.
Sundays in a Quantum Engineer’s Life: http://arxiv.org/abs/quant-ph/0104140v1

 

[xts]

I thought I understood what you mean, but now I do not see what you are getting at.

I mean that photons in p and s arms are totally independent by means of entanglement, phase, frequency, time, or whatever else you like. They just have the opposite linear polarisation - both went through the same Nicol's prism, and one got reflected in one direction, the second in other direction.

 

[Cthugha]

Yeah, I know. Nevertheless, I am the kind of guy for which every day is sunday. I second what the physicist in Gisin's paper says ”weak
pulses are not good, because they may contain two photons”. Using faint light is ok if you know what that means and where the fundamental limitations are, but very often people new to quantum optics have the misconception that decreasing the light intensity really gives you single photons. That is not the case. No antibunching, no quantum light - even if that Clavis2 thing from idquantique uses attenuated laser light.

My main point, however, was: It is the fact that the photons in a thermal light beam are not statistically independent which gives you a coincidence count spectrum looking similar to the one in Walborn's paper.

edit: Reply to second post. I suppose that would only work in situations where you would also see an interference pattern in one arm without coincidence counting, but not in situations where the light is too incoherent to produce an interference pattern in one arm alone. I mean - this is one of the major surprises of using entangled light. The pattern is only in the coincidence counts, but not in the single count rates. Imho it is not possible to reproduce that using independent photons.

 

[xts]

My main point, however, was: It is the fact that the photons in a thermal light beam are not statistically independent which gives you a coincidence count spectrum looking similar to the one in Walborn's paper.

Really? So explain really independent experiment like this.

'short time' = such that probability of finding more than two clicks is less than 1%
Perform billion times:
1. select randomly (note the position) polarizer setting as H or V
2. light up the candle
3. wait for a short time and note if the click happened in arm 'p'

Wait a fortnight, and move with your equipement to Australia

Perform billion times:
1. select the polarizer setting as H or V the same as in previous series
2. light up the Australian candle
3. wait for a short time and note if the click happened in arm 's'

Perform correlation analysis.
Does it differ from what I've shown, or what Walborn got?

 

[xts]

Reply to second post. I suppose that would only work in situations where you would also see an interference pattern in one arm without coincidence counting,

No. The pattern is not seen. We have bulky blob.
But you may use results from one (p) arm as a filter splitting the results from (s) arm into two classes. One class produce fringe, the second one produce anti-fringe. Mixing them - you have that blob.

 

[DrChinese]

Would you accept a challenge, DrChinese?
I bet a bottle of (it is A.Z.'s favourite, btw) Beerenauslese Rotgipfler 2005 from Herr Kamper's winery, versus bottle of young Californian Chardonnay ;)

Challenge: Would you demonstrate calculations showing the results differ?

The only they (Walborn et co.) used of entanglement is a perfect correlation in predefined, fixed base. So it doesn't differ from 'classical' correlation based on common history.

Mmmm, you are definitely talking my language with the wine. Certainly sounds like you have some nice stuff.

Hey, I don't think we are talking quite the same thing... although maybe we are. The reason we see the main results in DCQE setup is because the photons are part of a single (entangled) system. You cannot expect to learn more about any component than the HUP allows - which explains why the outcome at either arm will be consistent. Now if your setup mimics that result, then the source photons must likewise be entangled in some basis.

If you have a reference for a setup using thermal light, sure, I'd love to learn some more.

 

[xts]

my language with the wine. Certainly sounds like you have some nice stuff.

You must visit Hr.Kamper's winery - it is a fair chance to meet A.Z there ;)

the main results in DCQE setup is because the photons are part of a single (entangled) system.

I don't agree. The whole experiment relies on one-to-one classical (anti)correlation: horizontal-in-p-arm <=> vertical-in-s-arm

 

[DrChinese]

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.

OK, let's decompose this. The prism is like a polarizing beam splitter (PBS). Fine. The light source is one which emits 2 photons around the same time, right? Well, you already have an issue because you are trying to say they take an identical path to the PBS, were created at indistinguishable times and are of *opposite* polarization. I don't see how that happens but I stand open for correction.

Is this setup similar? Take a SINGLE Type I BBo crystal which outputs 2 photons of identical known polarization (and are therefore not polarization entangled). Shift Alice via wave plate so she is opposite to Bob. Bring them both back together again in such a way that they are going through the PBS at very nearly the same time. I say these are not polarization entangled although they might be entangled in another base. I say you cannot use polarization orientation (via the PBS) or any other polarization technique to get Walborn-like results.

 

[DrChinese]

You must visit Hr.Kamper's winery - it is a fair chance to meet A.Z there ;)

It is in Austria, I take it? I may need to plan a trip! Although I am partial to reds, I hope they might be found there too. Or maybe not...

 

[xts]

I say you cannot use polarization orientation (via the PBS) or any other polarization technique to get Walborn-like results.

Ouch?!
So point a flaw in my explanation!
Or present calculations showing that (classically probabilistic) mixture of H/V gives different results than you expect.

it is in Austria, I take it?

Gumpoldskirchen - far southern suburb of Wien. In a range of metropolitan transport. One of the nicest places for niche wines...

Austrian red wines are, well, how to say..., they cultivate some red grapes, esp. Pinot Noir (Schwarzburgunder), but for those I'd rather recommend Spain or New World, esp. South Africa...

 

[DrChinese]

Gumpoldskirchen - far southern suburb of Wien. In a range of metropolitan transport. One of the nicest places for niche wines...

Austrian red wines are, well, how to say..., they cultivate some red grapes, esp. Pinot Noir (Schwarzburgunder), but for those I'd rather recommend Spain or New World, esp. South Africa...

I was afraid that might be the case. At any rate, would love to join you for a glass of the white, perhaps after dinner, and enjoy the countryside!

Next time I am that way, because my travel budget is a little soft right now.

 

[Cthugha]

No. The pattern is not seen. We have bulky blob.
But you may use results from one (p) arm as a filter splitting the results from (s) arm into two classes. One class produce fringe, the second one produce anti-fringe. Mixing them - you have that blob.

Ehm, no. This does not work. If performed under incoherent conditions, you can maybe get two different blobs, but you will not get an interference pattern for independent photons. If all the photons are truly independent (except for polarization of course) you do not get any additional information from coincidence counting which allows to get an interference pattern.

It is in Austria, I take it? I may need to plan a trip! Although I am partial to reds, I hope they might be found there too. Or maybe not...

While you are looking for wine in German speaking countries try to ask someone on the open street, where you might be able to get some "Nacktarsch". Have a look at the priceless face and afterwards translate the name of the (quite famous) Mosel wine "Kröver Nacktarsch" into English. You will be surprised. ;)

 

[xts]

This does not work. If performed under incoherent conditions, you can maybe get two different blobs, but you will not get an interference pattern for independent photons.

I am speaking about independent photons, sent one after another a short while after the previous one had been recorded. And those single, one in a time, photons, form fringe pattern. Strange and weird, but true...

If all the photons are truly independent (except for polarization of course) you do not get any additional information from coincidence counting which allows to get an interference pattern.

Really? So send billion of truly independent photons through double slit and then tell if what you see is a 'pattern' or not.

"Nacktarsch"

As a heir of loyal KuK subjects I can't agree. I am sharpening my bayonett and marching towards Sadowa to prove superiority of Austrian wines
Anyway, I admit, Nacktarsch is still uncompareably better than Californian grapes
Prost!

 

[DrChinese]

Anyway, I admit, Nacktarsh is still uncompareably better than Californian grapes
Prost!

I wouldn't expect any less... But they taste pretty good to me!

 

[xts]

I wouldn't expect any less...

Oh, you are Texan? Don't feel offended! Some of US wines are quite nice (Washington state). But Californian mass production?
You are not personally involved (do you plant grapes in Texas? ) as I am not (I live in Poland), so we are free to have external view.
Anyway, in my honest opinion, if you speak about white wines, Austria (sharpening bayonett to defend agains bloody Germans, espetially those from Mosel region...) is the best place in the world.
And A.Z. used to make his shopping at Kemper's winery, Gumpoldskirchen...

 

[DrChinese]

Oh, you are Texan? Don't feel offended! Some of US wines are quite nice (Washington state). But Californian mass production?
You are not personally involved (do you plant grapes in Texas? ) as I am not (I live in Poland), so we are free to have external view.
Anyway, in my honest opinion, if you speak about white wines, Austria (sharpening bayonett to defend agains bloody Germans, espetially those from Mosel region...) is the best place in the world.
And A.Z. used to make his shopping at Kemper's winery, Gumpoldskirchen...

Not offended at all, I like reds from west coast of US, Chile, Spain, Italy. Just don't much experience with French, Austrian bottles. Or from Poland.

 

[xts]

We have no wines growing in Poland. Well, just a few wineries planted 15 yrs ago, but no tradition.
Don't ever try Austrian reds! (except, maybe some fine Schwarzburgunder/Pinot Noir and Skt.Laurent, which are more like very dark rose than real reds)
But for whites - I am an advocate of Niederösterreich ...

 

[Cthugha]

I am speaking about independent photons, sent one after another a short while after the previous one had been recorded. And those single, one in a time, photons, form fringe pattern. Strange and weird, but true...
[...]
Really? So send billion of truly independent photons through double slit and then tell if what you see is a 'pattern' or not.

This is not necessarily correct. The interesting point about Walborn's experiment is that he uses incoherent light (meaning that the transverse coherence length is smaller than the slit distance). If you use incoherent light from a light bulb for a common double slit, you will of course not see an interference pattern. Now for entangled light you can use coincidence counting to "pick" some photon pairs which form a subset of higher coherence which offers the possibility to see interference patterns in the coincidence counts, but not in the single photon count rates.

For independent incoherent photons you will not see an interference pattern ever in single photon or coincidence count rates. For independent coherent photons you will always see an interference pattern in both single and coincidence count rates (unless of course you use polarizers to place which-way markers). But for independent photons from a light bulb there is no possibility to see the pattern in coincidence counting, but not in single-photon count rates. This is the main difference between entangled light and classical light in this setup.

As a heir of loyal KuK subjects I can't agree. I am sharpening my bayonett and marching towards Sadowa to prove superiority of Austrian wines
Anyway, I admit, Nacktarsch is still uncompareably better than Californian grapes
Prost!

Personally I must admit that I do just like the word "Nacktarsch". I am from the Ruhr area, so I have to prefer beer anyway. Everybody who lives here and likes wine better than beer is considered a snob.

 

[xts]

The interesting point about Walborn's experiment is that he uses incoherent light (meaning that the transverse coherence length is smaller than the slit distance).

(Bigger than?) Did he? As I understand his experiment, the photons are transverse coherent with accuracy much better than angular span of the strips on the interference pattern. How could it appear otherways (there is no angular momentum/position measurement in other branch, so no entanglement may help here)?

For independent coherent photons...

What are 'independent coherent' photons? Originating from the same point-like source - sure, that is our case.
Ruhr beer...? Have I ever tasted it...? I visited Dortmund once, but I don't remember local beer... I had apparently missed something...EDITed>>>

DAB??
I believe you must brew something more snobish too...

 

[Cthugha]

(Bigger than?) Did he? As I understand his experiment, the photons are transverse coherent with accuracy much better than angular span of the strips on the interference pattern. How could it appear otherways (there is no angular momentum/position measurement in other branch, so no entanglement may help here)?

Downconverted light is among the most incoherent stuff there is. In fact it is necessarily incoherent as entanglement and coherence are complementary. You can have either coherence or entanglement (or both to a lesser degree) but not both perfectly. See A.F. Abouraddy et al., "Demonstration of the complementarity of one- and two-photon interference", PRA 63, 063803 (2001). In fact this is not too surprising. A high degree of spatial coherence basically means basically that your light source appears like a point source and the spread in possible wavevectors as seen from the double slit is small. For entanglement however, you want to have a well defined total wavevector of the two emitted photons, but want a large spread in each of them so that the detection of one tells you something about the other. This becomes pretty pointless if you detect such a small range of wavevectors that it is always the same pair of momenta seen for the two "entangled" photons.

What are 'independent coherent' photons? Originating from the same point-like source - sure, that is our case.

That is rather difficult to define in short terms. The experimentalist version is: If it shows an interference pattern when subjected to a double slit, it is spatially coherent. The long version is that just as the coherence time is given by the Fourier transform of the spectral width, the coherence length is given by the Fourier transform of the spread in momenta. If the source seen from the double slit is point-like, it is very coherent. If the source looks large, it is incoherent. Basically this is just a question of distinguishability. If you have an extended source the difference in travel times from the different positions on the source to the two different slits will be larger than for an almost point-like source. Therefore in principle you could get information about which slit a photon went through by exactly analyzing the photon travel times. If you can do that, you will see no interference pattern.

Therefore roughly speaking: You place your source directly in front of the slits and you will have no interference pattern and incoherent photons. You place it far away (or insert a pinhole and mimic a point source like in Young's original double slit experiment) and you will have an interference pattern and coherent photons. For entanglement experiments for example, it is important to place the down-conversion crystal not too far from the double slit so that you still have an extended source. Otherwise you increase the coherence and entanglement gets lost.

Ruhr beer...? Have I ever tasted it...? I visited Dortmund once, but I don't remember local beer... I had apparently missed something...

Yes, definitely. Especially as there is not much else to see in Dortmund except that yelllow-black football stadium. Even the university is pretty ugly.

DAB?
I believe you make something more snobish ;)

Yuck, DAB and the infamous Hansa Pils are about the worst kind of beer you can get in Dortmund. Try Hövels. That is great.

 

[xts]

Great. You must however look at Walborn's setup. 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.

In this experiment only spin entanglement is claimed by authors to be used. Spatial/momentum entanglement is not used, and the source is treated as point-like.

As the trigger detector in p is preceded by H-polariser, they could add V-polariser to s arm, and no coincidence clicks would be affected. But now they don't need p arm at all!
Well - they need it to justify all that "philosophy".

If slits are equipped with QWP, they don't produce fringes when lit by nonpolarised light, but produce when lit by the light polarised in the plane parallel or perpendicular to QWP axes. Funny experiment to be performed in high-school, but not so mysterious to call that polariser "eraser of which-path info"

The pattern vanishes at the presence of QWP not because they mystically mark 'which-slit', but because they simply introduce phase shift depending on polarisation.
Hövels - I must remember that!