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To CSTAR

Dear CSTAR

Please do not delete entries that conform to the encyclopedic style! My entry is almost 1:1 taken from a lecture by Bob I. Eisenstein, Professor of Physics at Department of Physics, University of Illinois at Urbana-Champaign [1] so it is real science, not like Drezet's pseudo-scientific speculations. So I try to make the article more encyclopedic and scientific, and you delete my entry. Before you delete the entry, please be informed IF it is true or not. If you don't have knowledge in physics, ask someone who is professional physicist.

If you need to read the perfect lecture presented by prof. Eisenstein see [2]

Best,

Danko

(a) Said "encyclopedic" entries are in powerpoint. 'nuff said.
(b) Please do not make personal attacks.
--CSTAR 20:46, 22 Jun 2005 (UTC)

Danko, Physics is obviously written in the mathematical language (I love math and even better I know them). However a physicist when he understand something can explain it without math. but only with words and images : There is two small and negligeable physicists who agree with me on that point: Einstein (for whom things should be so simple that even children can understand them) and Dirac ( for whom the art of physics is to solve a problem without writing equation). Bohr agree with me too since complementarity can be expressed only with words. This is a little bit subtile but I believe that a great physicist like you are will understand this simply.


--Drezet

To Drezet

Dear Drezet,

I have read your commentary about Bohm trajectories [in the section on crossed beams] and your comment seems to say exactly what I am saying - the assumtion that the trajectory of the photon goes to its optical image as is in the single pinhole case is not verifiable and exactly this is disproved by Afshar's experiment. Because THERE IS CONTRADICTION WITH THIS ASSUMPTION, NOT CONTRADICTION WITH COMPLEMENTARITY.

Now, I would like to present again my recent work that is very insightful and shows that the popular interpretation of complementarity should be replaced with the most accurate interpretation in the form of logical gates.

Investigate the case (i) when you close in a random fashion one of two slits. Then you will observe two "clumping" peaks. What you know about the slits is that either one is open but not both! You know S1 XOR S2.

Now you investigate the case (ii) with both slits open. Then you observe interference image. You know that this is not S1 XOR S2, so you have S1 XNOR S2. But it is impossible for both slits to be closed, so from XNOR you obtain S1 AND S2.

Now you can interprete this S1 AND S2 in the framework of "superposition is real" and assume that photon is at 1/2 going thorugh both slits. I personally like this.

The other possibility is to accept that the "open second slit" perturbs in some way at a distance the going photon to deviate its trajectory - this is Bohm's picture. But here in order to eliminate the superposition you accept even more ridiculous claim - an empty hole [slit] affects misteriously at a distance another object [photon].

Now back on the case of the lens. Well, as I said the actual image is not what Afshar presents, because he shows just the two "peaks" and that's all. Actually calculating superposition of waves psi1+psi2 and then squaring them will give us also some higher order maxima - but they CANNOT be explained in the classical ray optics. That is why Afshar has presented only the central part of the image plane in order to hide the higher order maxima [that may have not so great intensity]. This is really "fraud" and an example for pseudo-scientific behavior.

Danko


Additional note: I am very puzzled where the good communication was interrupted. I have seen that you have a lot of serious publications, so I will not delete your name from my preprint. Actually I quote you where you suggest that the photons at the image plane and those out-of-focus have fundamentally different wavefunctions, where my thesis is that the wave function psi(L) is the same, the only difference is the length from the slits L so the probability distribution at the new plane out-of-focus will be different.

Actually in your comment on Bohm trajectories you say what I am saying, so did you withdraw your words?

Danko


Dear Georgiev, I did not withdraw my words. I was clear since the biginning: Bohr's complememtarity is not a statement on path and wave like you can believe but a statistical result. The concept of path is not included in the description of QM since this can not be defined univocally. I think that it was the ideaof Unruh in his discussion with you. In reality it is not because I accept this ambiguity that I agree with you. It is even the opposite. Bohr's principle is a statement concerning observable not metaphysical and hidden parameter. The quantity observable are click in one or other plane (image of interference). The statistical pattern can not be build up with the same particles and that's it. The rest is not the subject of Bohr's principle. Your analyzis is wrong because you believe that there is not distinguishability in the image plane. This is a basic error : You dont understand that depending on the position of your photon detector you can have or not which path information (which path I emphasize again means: distinguishability) and you can see or not fringes. This is just a basic properties of the wave . A general mistake is that you consider the collapse of the wave like a physical attribute of the photon labeling the particle since the beginnig of his life to the end of his journey . This is a non sense since QM is the physics of information not the science of what is a photon an electron or even a cat.

QM is much more subtile taht you think and I guess that you should study more the subject. Qm is a study for a life and even more. Anyway since even Eisntein was concious of that we can forget your obstinacy.

Drezet Aurelien

Dear Aurelien,

thanks for your reply, but you are still resisting my arguments and deviate the topic about my knowledge in physics. I will focus again on the topic and I will re-state my argument in very clear and formal mathematical way. So once formulated you have to reply with mathematics, not with meta-pseudo-stuff.

You know that a quantum system in state |K> can pass a test as if it were in a state |L> with probability P = |<L|K>|^2. If |K> and |L> are orthogonal states then the probability is zero!

Now define state |D1> of a photon being localized in space at D1. Define also state |D2> of a photon being localized in space at D2. These states are orthogonal because being in say state |D1> the photon cannot pass a test for being at D2. Formally |<D2|D1>|^2 = 0.

Now Afshar states that the photons at the image plane are in a mixture of states D1 and D2. You claim the same thing, because you say that there is "which way" info. The density matrix is mixed one and can be written as rho = 1/2|D1><D1| + 1/2|D2><D2|.

Now my claim is that the photons ARE IN SUPERPOSITION IN SPACE AT THE IMAGE PLANE. SO THEY ARE IN STATE D3 = 1/root_of_2 (|D1>+|D2>). Now if you make a measurement with the detectors, you make projection on the basis vectors |D1> and |D2>. You get probabilities of 1/2 for photon to be detected at D1 or D2.

So do you get my point? You cannot distinguishthe superposition from mixture if you measure the position of the photon at the image plane. Because even if it is in superposition you will "collapse it" or "project it" on one of the two detectors. But if you let those photons go out of the image plane they will follow different distributions. If they are in a mixed state they will form two "clumping peaks" out-of-focus, and if they are in superposition they will produce typical "interference fringes".

I hope you finally get my point. I know perfectly that you cannot measure a photon twice, but IF the experimental setup is the same, then the photon's wavefunction psi is not changing between experiments. Also the probability is obtained directly by squaring psi, so your remarks on what I am doing are "shots in the empty space". First do understand what I am saying before accusing me in ignorance in QM.

Please prove that the photons are not in superposition of states at the image plane, so that each photon is at half at both detectors at once. So what i am saying is that the detector "projects" the photon in basis "being at D1" ("..or D2") and does not confirm actually that the photon has BEEN in pure state |D1> [defined as above, orthogonal with |D2>].

Best,

Danko

Dear Danko you are really really dificult to understand. Are you claiming that in a which path experiment the photon is not in a pure state? personally I never say that. It is obvious for me that in the image plane we have before the measurment the state |D1>+|D2> (without normalization) which collapse in either |D1>or |D2> after the measurement (if the measurement is done). If the measurment is not done the supperposition occurs still and in particular the fringes will be observed if the photon is recorded much later. I dont see your problem?

In fact I understand you perfectly well : I understand that there is nothing to understand in your empty non-argument (this justify what it is effectively a non-argument CQFD). You are saying that because the fringes can be recovereded if we observe the photon after the image plane there is no which path at all in this experiment. This is true if you record your photon as you propose after the image plane . However if you record a photon in the image plane you have two narrow peaks and no fringes . This is the key point which prove me that you dont understand a picobit of QM: the so called which path information or the interference information are available in different point of space but not with the same photons....

I was perfectly clear with my explanation of the Heisenberg-von Neuman collapse : In the image plane we have before the measurment the state |D1>+|D2> (without normalization) which collapse in either |D1>or |D2> after the measurement (if the measurement is done). If the measurment is not done the supperposition occurs still and in particular the fringes will be observed if the photon is recorded much later.

You should learn more of basic QM. PS(1): I am Bohred with your misleading and circular argumentation and I will not answer anymore to your aberation. It is not my job to teach you QM and if you reply to me you should now that I will not react anymore : please dont consider that I will refuse to listen your truth I amk sufficiently clear on what is truth in your case. PS(2): please remove my name from your paper. In 1000 year when some archeologist will find this paper they could think wrong things about me. Drezet

Dear Drezet,

I will not remove your name from my paper. And I finally got you in the trap!!! You just agreed that the photon at the image plane is in a state 1/root_of_2 (|D1>+D2>), so you supported my main claim.

Now - you make the same fallacy as Afshar - you unconsciously link the "two peaks" with "which way". THIS IS WRONG! The two peaks are result from the lens action and do not tell you that there is which way info.

If I follow your wrong thesis THEN I CAN PROVE THAT THERE IS WHICH WAY INFO AT EVERY PLANE, EVEN IF THERE IS INTERFERENCE PICTURE.

The argumentation following your wrong thesis goes: the photon at the detection plane is in superposition of spatial possibilities x1, x2, x3, ... x_n. But the "projection" measures every photon only at given point x_i, so that after the measurement the photon is now in a state |x_i>. Actually if we perform a measurement upon a bunch of coherent photons [of a laser beam] we will convert the status of the coherent beam into mixture of states |x1>, |x2>, ... |x_n>, with probability coefficients equal to psi_i^2. Since each state x_i is not superposed, therefore you could argue that this was a "which way" experiment so that each photon had to follow a classical straight ray trajectory to that point. Alas, this is WRONG!

This is certainly what your words suggest and show that you don't understand that an experiment is "which way" ONLY IF THE STATE BEFORE THE MEASUREMENT was in a mixture, not converted into a mixture AFTER the measurement.

You have shown completely ignorant in this topic, so now I have your words "black on white" [i.e. metaphorically "written on paper"]. So there is no way to withdraw your claim without bearing the consequences. You should publically confess your error, and appologize to my personality for many personal insults that you have raised.

Kind regards,

Danko Georgiev

p.s. Summary:

1. if the state of photons BEFORE the measurement is linear superposition of spatial locations [pure state], the experiment is NOT which way.

2. if the state of the photons BEFORE the measurement is mixed state then the experiment IS which way.

3. Following the projection postulate, AFTER any measurement done in basis x [position], you get mixture of localizations in space. So, the conclusion is that AFTER the measurement you get mixture in both cases: (i) "which way" and (ii) "no which way" experiment. THEREFORE, what is the state AFTER the measurement is not determining the type of the experiment, and in ALL cases you get mixture!

So, dear Drezet learn your homework: 1 & 2 are important for discussing Afshar's experiment, not your wrong thesis based on misapplication of 3.

Danko

On the Distinguishability criterion!

I have just seen the formula presented in the main body of the article concerning the Afshar's experiment. I have repeatedly stated that Afshar's experiment IS NOT WHICH WAY, and this is clealry evident from the given formula. The distinguishability D is defined as

D = ||psi_A|^2 - |psi_B|^2|

and in the double slit D = 0, as calculated

D = ||1/root_2|^2 - |1/root_2|^2| = 1/2 - 1/2 = 0

I hope everything is clear now.

Danko


Oh danko danko every thing is effectively clear completely clear....... Alea jacta est

Drezet

Dear Drezet, you probably have added this text in the main article:

"The experiment of afshar contains no detecting device using entanglement. In the formalism presented previously this means that D=0 and V=1. However in this experiment the photon are actually detected in the image plane of the lens and not in the Fourier plane (e.g in the focal plane)."

I fully AGREE that D=0 and V=1. But this says that Afshar's setup IS NOT WHICH WAY. There is no difference where you will detect the photon - in the image plane or in the Fourier plane, or any other plane. Possibly you cannot swallow the fact that I am the first one who has really shown where Afshar's error is, and what possibly annoys you most is that I am MD and not professional physicist. Since I am the first one who has pointed the error in Unruh, Motl, Kastner, Cramer et al. I think that I am the person who wins [at least morally] the 1000$ award annouced by Afshar for disproving him.

Danko

reading your message I can only agree on one point : you are effectively a MD the rest is abberation and non sequitur Drezet PS I can swallow very well but not when i am reading your physics: I stop to swallow and to breath then imediately in order to laugh in peace.

Wave tank experiments in water

I don't get it. It seems to me that the entire Afshar experiment could be reproduced in a wave tank, using suitably arrayed pylons instead of wires, and changes of depth instead of lenses. Exactly the same single-slit focusing phenomena, and dual-slit interference would be observed. There is absolutely no quantum mechanics involved in a ripple-tank experiment; yet exactly the same experimental results are accorded. Am I missing something? What would that be? (And don't say I'm missing the concept of wave function collapse; collapse is known to be insane; I wrote up a short article on an old, traditional paradox in spherical decay experiment, for which, unfortunately, I do not have any handy references.) linas 1 July 2005 19:21 (UTC) Postscript: I was conflating two different ideas: the Mott problem, and the Renninger negative-result experiment. linas 21:01, 7 December 2005 (UTC)

"Am I missing something?"
No.--CSTAR 1 July 2005 21:30 (UTC)
Well I admit that's somewhat of snarky answer, since "single photon" makes no sense for water waves. --CSTAR 1 July 2005 22:14 (UTC)
Yes, well, all quantum measurement/wave-function collapse effects are always single-particle measurements. (Well, all but the superconductor / superfluid / condensate type experiments). And the neutron interferometer already provides a pretty compelling single-particle experiment. So I still don't understand how the Afshar experiment provides any new insight onto anything that isn't already manifest in the traditional quantum paradoxes. What is new or different (if anything) needs to be more precisely explained. linas 2 July 2005 04:33 (UTC)
BTW the new stuff on complimentary is getting kind of wordy. There is an article by Stephen Durr in Phys. Rev. A Quantitative wave-particle duality in multibeam interferometers which gives a lucid and general exposition of this.--CSTAR 1 July 2005 22:14 (UTC)

Dear linas,

your example is straight at the target. I have mathematically proved that there is NO WHICH WAY experiment in Afshar's setup regardless of the fact is there grid or not! I have said EXACTLY what your classical water model says - at the image plane we can have two peaks produced by interference. Now your water tank scenario is perfect illustration of my main argument. Please contact me by e-mail at dankomed@gmail.com - I would like to quote you in my paper for providing illustration of my words.

Danko Georgiev

I can't write today, linas at linas dot org is inundated by spam; I don't take email there any more :(. However, if I don't respond by 5 July, write me at linas at us dot ibm dot com. linas 2 July 2005 15:33 (UTC)

I agree, linas, I don't see how this experiment adds anything new at all. How do the wires change anything? You still have the photons acting like waves when they go through the pinholes, and then being detected as particles. In fact, you could take the ordinary double slit experiment with single photons and make the exact same argument: each photon acts like a wave when it goes through the slits, and then like a particle when it is detected. Also, I don't think there is any "which way information" to begin with. The photons go through both slits, then through the lens, and then to both detectors, and are detected by one or the other, randomly. Pfalstad 19:34, 19 December 2005 (UTC)

Dear Pfalstad, actually there were a link to a paper where I clearly have shown the error in Afshar's argument, yet I claimed that John G. Cramer, Lubos Motl, Drezet, Unruh, [including Zelinger!] make a fundamental error in claiming that the image at the image plane carries 'which way' info. I have shown that mathematically this is not so, and if the photon wavefunctions passing through both pinholes are coherent, then the image is 'hologram' - it is pure state and photon's wavefunction is superposition of states at both detectors. The situation is IDENTICAL with the ordinary two-slit experiment where on the screen the photon's wavefunction is superposition of many states, but it is the "projection/measurement postulate" that destroys this quantum coherence, and the projection postulate says that if the wavefunction was in superposition x1+x2+...xn, after the measurement the state is now xi, so the coherence is irreversibly lost at the time of measurement - i.e. the photon is detected only as a particle somewhere on the screen. Occasionally I was accused as non-specialist [despite that my current research in on optical solitons in neurons] and felt useless to continue the conversation in Wikipedia. Now I am glad to find that there is somebody who thinks clearly and sees the truth. Afshar is fraud and profits from this. Danko_Georgiev_MD 31 Dec 2005

The things get complicated

Dear Drezet,

I have found very interesting experiment performed by prof. Zeillinger using SPDC, lens and a double-slit. He used the conventional interpretation to explain his results - namely measuring at the focal plane, and measuring at the image plane. I think his interpretation rather inconvincing, and I think I can explain Zeillinger's experiment in another way without invoking this duality of the image vs. focal plane.

The fact is that this duality of the lens planes is possibly commonly used in textbooks, that is why nobody wants to study my argument more carefully.

From very basic principle for continuity of the wavefunction, IF out-of-focus there is interference double-slit pattern, then this pattern cannot "jump there" without passing through the "image plane". Therefore the function |psi_1+psi_2|^2 must be continuous and produce similar image at the "image plane" as if it is produced by classical ray optics [with some residual effects of the interference].

Danko Georgiev Dear Danko, you can send a email to Anton directly to precize your new non-thought. I think that it will be a pleasure for him to give you a non-answer to your non-idea. (In fact he will probably no find the time for this strange purpose so probably I should give you a non-answer for him:)

Without jokes Are you serioulsy trying to rewrite quantum mechanics?

Aurelien Drezet

Please do NOT remove the POV tag

Please do NOT remove the POV tag. This article contains pseudo-scientific information, so the reader must know about the RISK of reading it.Danko_Georgiev_MD

Fellas, please sign your posts! --Reuben 19:40, 5 December 2005 (UTC)

Dear Reuben, thanks for your suggestion, the posts will be signed Danko_Georgiev_MD

I reverted, Danko is right, the article is rather slanted in favour of Afshar, and fails to emphasize that Afshar is basically just plain wrong. It needs to have a POV sticker on it until it gets cleaned up. linas 01:31, 6 December 2005 (UTC)

Problems with the theory section

There are problems with the theory section. I tried to clean it up, but promptly got tangled. I'll try to fix it in time, but any discussion here would help clarify what the intent of this section was. linas 02:06, 7 December 2005 (UTC)

The theory section was appearently added by anonymous user User:143.50.61.66 around 28 June 2005, but no matter how I try to fix it, it just seems broken. I'm punting for now. linas 04:03, 7 December 2005 (UTC)
So... anyone care to help fix, or at least discuss, the theory section of this article? linas 15:25, 16 January 2006 (UTC)

Hunter Monroe's deletion bid

I moved the following statement by User:HMonroe to the discussion page, because (1) it does not belong in the main page. (2) it is entirely unsubstantiated as far as the sockpuppet allegation is concerned.

Prof. Afshar plays an active role in editing this article, under his own name and under pseudonyms (sock puppets), which undermines its objectivity.

Here's my suggestion to Hunter Monroe if he is indeed not a sockpuppet himself: Reveal your true identity or send me an e-mail and explain what you find to be wrong with my experiment or my interpretation. My e-mail address is: afshar@rowan.edu Alternatively, you can write up a critique like other academics and post it at my blog, the arxiv, or your own web-page. I expect you to behave like a real scholar, rather than a prankster who abuses the free access to Wikipedia to push his/her agenda. What you are currently doing is entirely against scientific process of debate and exploration. You cannot stop truth from marching on. At best, you can slow its progress for a while, that is, to your own historic demise.--Afshar 18:42, 20 December 2005 (UTC)

I replaced the substantiated part of HMonroe's remark, namely "Prof. Afshar plays an active role in editing this article." Surely we can all agree on that, and it is certainly appropriate for people to know that Afshar frequently edits a page that is all about his own work. Note that I do not support HMonroe's deletion bid, irrespective of whether Afshar's interpretation of his experiment is correct. Dave Kielpinski 22:52, 20 December 2005 (UTC)

Dear Dave, I think people would see that by checking the page history, but that's fine. I'm still waiting for your response to my last e-mail, and mostly your critique. Regards. -- Afshar 23:11, 20 December 2005 (UTC)

I've already thought through the critique, but I'm in the midst of moving to Australia right now and writing it up is not my maximum priority. Dave Kielpinski 01:15, 21 December 2005 (UTC)

If we have doubts about the neutraility of the article, rather than making our own notice a good old-fashioned {{npov}} marker would do the trick. DJ Clayworth 15:01, 21 December 2005 (UTC)

Technical

I recently marked this article as too technical, referring specifically to the "Theoretical" section. One would need a couple of years of college physics to understand it; I'm not all that sure that it is suitable WP material. Stifle 22:10, 24 December 2005 (UTC)

WP has about 10,000 articles on math and physics that require college degrees in math or physics in order to be comprehended. Based on this, I'll remove the tag. Note, however, the theoretical section is broken, but that is a separate issue. (The total number of physics/math articles is displayed at Wikipedia:WikiProject Mathematics/Current activity and is 12,023 as of right now. Try a few links, and you'll see that the majority are quite advanced.) linas 17:30, 3 January 2006 (UTC)

NPOV tag

I've added an NPOV tag after noticing that the first paragraph states "Prof. Afshar plays an active role in editing this article.". --HappyCamper 22:52, 31 December 2005 (UTC)

  • I agree with the NPOV tag. I have offered to voice my views in the talk page and ask a neutral editor to edit the main article accordingly after we discuss it here. To start with, I take issue with the following subjective statement in the introduction: "the controversy has been mostly ignored within the mainstream academic physics community". This is in fact incorrect, and evidence to the contrary can be found by a simple search in the Google under "afshar experiment". Rarely has an experiment in quantum mechanics produced such tumult in academia as this experiment. I asked Linas to explain his reasoning to include the above statement (other than his own personal misgivings of the experiment) but have not received a response. There are other major problems with the article including the Overview, and the Theory section. But, I will hold on for now to see the kind of action Linas or other editors would take on the issue I have raised above. Looking forward to hearing from all interested Wikipedians. --Prof. Afshar 04:47, 1 January 2006 (UTC)
I'll change "the controversy has been mostly ignored within the mainstream academic physics community" to read "there has been very little published in professional physics journals on the experiment". (As far as I can tell, there has yet to be even one article published in a refereed physics journal). A google search reveals a variety of blogs. As far as I can tell, the "tumult" is more along the lines of irritation and disappointment that this is less of a "breakthrough" then was initally claimed. linas 17:48, 3 January 2006 (UTC)
There are serious problems in the theory section. It was added by some anon editor, and is fundamentally incorrect as it stands. If Prof. Afshar can discuss/outline an accetable set of equations, I can massage them into something appropriate for WP. linas 17:48, 3 January 2006 (UTC)

Dear prof. Afshar, the fact you resist any dialogue, and do not want to be engaged in discussion about the density matrices of the photons in your experiment, means that your entry does not deserve to be in this encyclopedia. I have clearly written down the density matrices of your experiment and have shown that your interpretation is wrong, yet I have not heard any reasonalbe reply. --Danko_Georgiev_MD 1 Jan 2006

Water or photons? - It does not matter - complementarity is about mathematics!

Dear CSTAR and dear linas,

I hope I can help your thinking with an experiment that I just 'constructed' in order to explain you more clearly MY DISPROOF of Afshar, where I used only math equations, but alas, nobody reads them. ALSO I show that linas should not worry about the photon being particle [unlike water] because Afshar VIOLATES MATHEMATICAL PRINCIPLES, and it is not relevant is the photon a wave or particle. I hope you will pay attention to this brief argument that will show that the wave-particle duality IS NOT TOUCHED by COMPLEMENTARITY which is mathematical rule for calculating probabilities. The fact that "complementarity" has some popular interpretation is irrelevant - the interpretation may be wrong, because it is the mathematica fundament that makes the complementarity principle valid, not what is released in the popular press.

Now a novel experiment to show that you may have WHICH WAY info, and at the same time INTERFERENCE - here "which way" info means that you have special way you calculate the probability distribution:

WHICH WAY experiment

[1] Suppose you have 4 slits. You may put different polarization filters on slits 1,2 compared to slits 3, 4 and see how you detect photons behind. In this case you will have "mixed state" which is to say that you will observe mixture of two interference double-slit scenarios each having maximum between slit 1,2 and between slit 3,4 respectively.

In this case complementarity says: The states |slits 1,2> and |slits 3, 4> are distinguishable so we have probability P = |psi1|^2 + |psi2|^2 here |psi1|^2 = <slits 1,2|slits 1,2> where the bra <| is the time inverse of the ket |> which means that the bra <| it is the complex conjugate of the ket |>. In the language of complex numbers this means that z=x + iy, then its complex conjugate z* =x - iy. Analogously |psi1|^2 = <slits 3,4|slits 3,4>

COMPLEMENTARITY mathematically says NOTHING about wave particle duality - because in this experiment you observe "mixture" of two interference pictures. The photons producing the probability distribution |psi1|^2 are distingushable from the photons that produce the probability distribution |psi1|^2. So in both cases the photons INTERFERE with themselves, and we HAVE WHICH WAY info because we can say whether the photon passed through slits 1,2 or it passed through slits 3,4.

Thus we see that WHICH WAY has mathematical formulation in which is said that IF two states are DISTINGUISHABLE, then the probability distribution is P = |psi1|^2 + |psi2|^2

NO WHICH WAY experiment

[2] Now let all four slits open without polarization filters, so the photon states are INDISTINGUISHABLE - therefore we observe one global four-slit INTERFERENCE picture.


The probability distribution is P = |psi1^2 + psi2|^2 Here we again see that COMPLEMENTARITY has nothing to do with wave-particle duality - COMPLEMENTARITY is RULE how to CALCULATE PROBABILITIES. Here and we DO NOT HAVE WHICH WAY info because we CANNOT say whether the photon passed through slits 1,2 or it passed through slits 3,4.

IF now (linas and CSTAR) you have realized what I am speaking about in clear mathematical language, you will immediately see that my entry must be considered immediately in the main portion of the Afshar article. You may use my math to update the main article [I cannot do it myself for understandable reasons] and please restore the original link to my pdf article where I have disproved Afshar's thesis [thus winning the 1000$ award at least morally].

WHY it is relevant to Afshar -

well, Afhsar like little child sees in textbook that it is written that the "image plane has the which way info" and without much thinking and not understanding mathematics goes into confusion.

Mathematically his error is like this - he observes some distribution P at the image plane and says "this is function P = |psi1|^2 + |psi2|^2, which is the function that calculates the probabilities in a WHICH WAY experiment".

Alas, my thesis is that in SPECIAL CASES you CANNOT MAKE DIFFERENCE between P_1 = |psi1|^2 + |psi2|^2 and P_2 = |psi1^2 + psi2|^2 by having the value of P only.

Mathematically I will show what does it mean.

Let's psi1 = x1 + i y1 and psi2 = x2 + i y2

P_1 = (x1 + i y1)(x1 - i y1) + (x2 + i y2)(x2 + i y2) = x1^2 + x2^2 -y1^2 - y2^2

P_2 = ((x1 + x2) + i (y1 + y2))((x1 + x2) - i (y1 + y2)) = x1^2 + x2^2 + 2x1x2 -y1^2 - y2^2 - 2y1y2

Mathematically it is seen that P_1 = P_2 when 2x1x2 - 2y1y2 = 0

For example use the following values: x1 = 2 x2 = 8 y1 = 4 y2 = 4

You will have P_1 = P_2 =36, and you will not be able to say by having the P value alone whether this is WHICH WAY distribution or NOT. In other cases with other values for x1,x2,y1,y2 you may be able just by having the P value to say what is the probability distribution.

Note: in QM you have normalization, but this does not change the essence of the argument, and it is important that normalization is used only for quantum coherent functions. The mixed state probability function does not describe a quantum coherent (pure) state, therefore each of the functions psi1 and psi2 should be normalized alone - in contrast in the four-slit experiment where the function psi1+psi2 must be normalized as a whole. (This is too advanced but relevant, and if you want you may skip this argument).

Application of all this theory to the image plane just says that both types of distributions may produce the same image at the image plane, but these two types of probability distributions DIFFER everywhere outside the image plane. So Drezet's nonsense objections that I have to write again the QM textbooks are ridiculous. What I provide is basic mathematics and it must be understandable for everyone that studies physics. IF NOT, my suggestion is: change the field of study. What about a hostile note by CSTAR that my quotation of prof. Bob Eisenstein is powerpoint, I can ask - SO WHAT??? Math is math - so IF you are not competent, then quit this discussion. --Danko_Georgiev_MD 5 Jan 2006