Talk:Afshar experiment/Archive 20

message: This is getting ridiculous! This talk page already has more entries than Einstein's talk page. Please refrain from posting general discussions here. This page is not a forum or a weblog. Any discussion not directly related to the article and its content/structure should be conducted elsewhere. For now, I will move any such discussion to the user page of the individuals involved, no exception.-- Prof. Afshar 07:45, 7 December 2006 (UTC)
replies: text: Thanks, it is better! Danko Georgiev MD 07:50, 7 December 2006 (UTC)

text: Thanks from me too :) --Carl A Looper 08:06, 7 December 2006 (UTC)
thread_title:  Enough already! 

message: For those of you who are interested, here's a link to a major conference on the nature of light where most of the giants of the field including Scully (A&M), Schleich (Ulm), Boyd (Rochester), Khrennikov (Vaxjo), Eggleton (Sydney U.), Myers (Harvard), Walmsley (Oxford), and others are attending (the papers are screened by the members of the Program Committee):  THE NATURE OF LIGHT: WHAT ARE PHOTONS? (OP201) -- Prof. Afshar 03:16, 8 December 2006 (UTC)
replies: text: Thanks Prof Afshar. --Carl A Looper 05:29, 8 December 2006 (UTC)
thread_title:  2007 Conference on "What is a Photon" 

message: I redistributed the 12 archives of this page (and also some content of the current talk page) into 19 archive pages because the size of some archives was very large (not suitable for slow internet connections). I fixed the links in the archive box as can be seen above. --Meno25 01:48, 13 December 2006 (UTC)
replies: text: Shokran Meno25. -- Prof. Afshar 02:41, 13 December 2006 (UTC)
thread_title:  Archives 

message: Dear Afshar, one really good US physicist producing papers on quantum entanglement, delayed choice quantum erasure, etc., has evaluated my paper in the following way (name of the physicist will not be announced yet): 
*"Thank you for sending me your paper.  I am reading it with great interest.  I would caution you against relying on anything from Wikipedia, but otherwise I am finding your arguments and explanations very appealing.  I hope to give you specific comments soon.  Best regards, xxx" 
I suggest you to restore again the link to the deleted paper of mine, and please do not characterize in the future the work of others, before you read and understand what is the whole point about: 
* Georgiev, Danko (2006) Correct exposition of complementarity in Unruh's and Afshar's experiments. PhilSci 3048
Yet, this is not a big deal, but I think that the process triggered by my paper will be irreversible. Mathematical proof can not be simply busted away, so it's time to re-evaluate your position and behaviour. Regards, Danko Georgiev MD 10:33, 29 December 2006 (UTC)
replies: text: *Dear Danko, I'm sure that if that "Physicist" sees your record on the internet, he/she would quickly distance him/herself from you. Some people may be so critical of my work that anything resembling an argument against my position would be good enough for them initially. However, rest assured that I will respond to that "Physicist" with the same rigor I have responded to Bohr's views once he/she makes his support of your "paper" public. As for restoring the link to the "paper" in the article, until you have published it in a peer-reviewed physics journal I will remove it as crackpottery because it contains a clear violation of the conservation laws, pointed out by myself, Drezet and prominent physicists such as Prof. Unruh. Regarding the "math", they are the standard formalism of QM discussed by innumerable physicists for decades, thus not novel. Again your statements such as "the process triggered by my paper will be irreversible" is a standard crank stance. You should let others say that for you.-- Prof. Afshar 16:44, 29 December 2006 (UTC)

text: Afshar, don't consider yourself the "center of the Universe", the mentioned physicist has nothing to do with your setup, so he is interested mainly in the discussion of complementarity that I have provided. There has been some claims already in the literature that complementarity is result from uncertainty relations, or entanglement. I show that complementarity indeed depends on the very nature of quantum superposition, and thus is tighly connected with the density matrix of the qubit. Thus entanglements that will lead to diagonalization of the reduced density matrix of the qubit are important, but complementarity is much deeper than that. Complementarity is intimately connected to the density matrix of the qubit. If one claims which way information in the case of your setup where the photon has pure state density matrix, then he\she is inconsistent, and this was done by the claimed by you "famous" Unruh, Drezet, etc. Danko Georgiev MD 17:39, 29 December 2006 (UTC)
thread_title:  Inconsistency of Afshar's interpretation 

message: ::*To Danko, and any other OR pusher: Please refrain from posting ad.s on your OR in Wikipedia. Publish your papers in a peer-reviewed physics journal, and if relevant, it may be included in the article. All forms of OR will be removed and violators reported to admin.s Best regards. --Prof. Afshar 18:03, 29 December 2006 (UTC)
thread_title:  Original Research (OR) alert 

message: I have added direct links to Afshar's papers available at ArXiv. Now one can have a better idea what exactly Afshar says. Some of the details were not available in his 2003 paper. Particularly in his SPIE paper he says that detection at the image plane leads to prediction P = |\psi_{1}|^2 + |\psi_{2}|^2 with the prediction of no interference pattern at the focal plane. Yes, if you interprete the setup in which way you must accept that claim, plus accepting that the photon's density matrix is of mixed type (Afshar says "decoherent" in his paper). So yes, all of this is true, and is clearly stated in my PhilSci 3048 paper. I call this "Unruh's which way interpretation" which is consistent, but is experimentally disprovable. Then I show that at the image plane the P = |\psi_{1} + \psi_{2}|^2 distribution is producing the same picture as the P = |\psi_{1}|^2 + |\psi_{2}|^2 distribution. So there is possibility to obtain the same image at the image plane without "which way" information. Basicly if the photons are coherent as in Afshar's setup without grid there is no which way information at first place. If one puts different polarization filters on the pinholes he will receive mixed density matrix and will have really which way information, but then putting the grid before the lens will absorb photons. Finally, if one interprets the results as which way plus pure state density matrix of the photon he will be mathematically inconsistent. How one mathematically arrives at the same intensity distributions for pure state and for mixed state setups? Well, I have shown that for mixed which way setup one has e.g. at image A the amplitudes \psi_{1}=1/\sqrt{2}, \psi_{2}=0, while for pure state setup one has e.g. at image A amplitudes \psi_{1}=1/\sqrt{8}, \psi_{2}=1/\sqrt{8}. I think the link to my paper should be restored again, and better be done by Afshar himself as he vandalized the main article. Danko Georgiev MD 02:57, 13 January 2007 (UTC)
thread_title:  Afshar's papers available at ArXiv 

message: I am not happy about the detailed analysis presented in the recent addition "Fundamental flaw in Afshar's argument".  It is not appropriate to have such a claim without this having been vetted by other experts.  As a non expert it does not seem to  make sense because a photon is detected at either detector on a photon by photon basis, and accouting for all photons.  The experiement works one photon at a time.  Since each photon is detected on either detector, according to the principles of optics and lenses, this appears to be which way information.  It can't just be ignored, unless the truth is that the photons always pass through both, and detection shows something else.  The current paragraph is presenting itself as a difinitive proof and I can't just take the word of one person that it is valid.  There needs to be debate on this first.  So I have reverted again. Especially take note of the claim "Same way, in Afshar's experiments, after interference, each wave packet gives contribution to both the detectors.".  This is appears to contradict the experimental evidence that after interference each photon is detected only on one detector, not both.  Please do not revert again until other voices have expressed themselves here.  I am reverting on the basis of NPOV until others have had a chance to comment.  Thank you. Dndn1011 14:39, 17 January 2007 (UTC)
replies: text: Incidentally I reviewed the paper, and along with many other claimed refutations of violation of complementarity, immediately simplifies the experient in order to develop the argument.  It appears (although I am not an expert as I said) to then apply an argument related to Unruh's, using a half silvered mirror.  The first thing this papaer does is remove the incovenient lenses.  The paper might be making the error of taking the context of Afhsar's experiment operating with a stream of light and then converting that into a single photon equivalent.  This probably is in error as Afhsar's experiment on a photon by photon basis cannot be so easily correlated with an experiement in which complementarity cannot be violated.  In essence I think it is like saying "This experiement is really just the same as a whole class of experiements that do not violate complementarity".  In any case, I believe it possible that Complementarity is not falsifiable because in order to prove it false it is nescessary to know the path of a photon without destroying it.  Since this is impossible, it may be impossible to prove Complementarity wrong.  If a theory is not falsifiable then it is not scientifically valid, apparently Dndn1011 15:16, 17 January 2007 (UTC)
replies: text: OK I did it, it was not hard.  I believe I have proved that complementarity is not falsifiable.  Please feel free to take my argument apart.  But it seems pretty water tight to me. Dndn1011 15:47, 17 January 2007 (UTC)
thread_title:  Recent addition 

message: This poster keeps reverting the article to their own agenda without discussion here.  I cannot revert again without risking  a three revert infringement.  As for you 220.224.21.140, you are disregarding the accepted protocols for dispute reslution and ask you to remove your edits pending a discussion here.  Wikipedia is not meant to be a platform for your opinions.  I have questioned your addition and you have not responded this is unacceptable behavior. Dndn1011 16:37, 17 January 2007 (UTC)
thread_title:  220.224.21.140 is not working to concensus 

message: Sorry about the additions, without discussion. Was unaware of Wikipedia rules/ethics.
The argument is quite simple. Afshar (and most others) assumes that the two wave-packets coming from two slits just pass through unaffected to the detectors. The basis for this is just the observation that when one slit is closed, only one detector clicks. This doesn't imply that when both slits are open, the same will hold. Because now the two wave-packets meet and interfere. What happens in the interference region is that part of the two wave-packets cancel each other (dark fringes), and part of the wave-packets reinforce each other.
The following simple example will make it clear. Let there be a state which
is a superposition of two opposite momentum states. This can signify a superposition of two
waves travelling in opposite direction, and crossing at some point. Say the part coming from slit A is \exp(\imath kx) and that coming from B is exp(- \imath kx). Clearly, the two parts are moving in different directions, so that one will reach one detector and the other will reach a different detector. (Ideally we should be dealing with wave-packets, but the arguments easily extends to that) So, when the two parts meet in the interference region,
the state at the point of crossing can be written as:
* \psi(x)  = \exp(\imath kx) + \exp(- \imath kx)
Now this state can also be written as
* \psi(x)  = [\cos(kx) + \imath \sin(kx)] + [\cos(kx) - \imath \sin(kx)]
The two square brackets represent parts coming from different slits.
The two \sin(kx) terms cancel out to give the dark fringes. What is left is:
* \psi(x)  = [exp(\imath kx) + exp(-\imath kx)]/2 + [\exp(\imath kx) + \exp(- \imath kx)]/2
Thus even the part which had momentum k, gives a superposition of momentum
k and -k because of interference. So, the part which was moving in one direction with momentum k, now gives a SUPERPOSITION of two parts, one moving in one direction ad the other moving in the other direction. Same holds for the part which was moving with momentum -k. Clearly, if you are in place of any detector, you will see identical parts coming from both the slits towards you. Thus, each detector gets contributions from both the slits. But as the parts reaching the two detectors belong to one single particle, only one detector will register it (randomly)  at a time (wave-function collapse or whatever...).
So, there is no which-path information in the detector clicks. —The preceding unsigned comment was added by 220.224.21.140 (talk) 18:50, 17 January 2007 (UTC)
replies: text: Note: dear anonymous, please use the inbuilt laTex function in Wikipedia. I have edited your math equations to make them more clear. Danko Georgiev MD 09:34, 18 January 2007 (UTC)

text: Comment why the proof by anonymous (220.224.21.140) is invalid. Dear anonymous, it is very easy to speak about the focal plane and explain how the interference occurs, but let me say WHY there is a small problem when you discuss the lens action at the image plane. Please note that when the only pinhole A is open at image B reaches NO intensity. So the overall quantum amplitude is ZERO but not as unstructured mathematical zero (perceived as lack, complete emptiness, etc.), but "structured zero" that is result from destructive self-interference. So if some quantum aplitude comes from A to image B, then squaring this complex number will always give as result summation of the complex and the imaginary components |a + \imath b|^2 = a^2 + b^2 , so by this mathematical argument the zero intensity at image B must be obtained only via squaring of zero real part and zero imaginary part coming from slit A. Yet, here is the deep question of this "self-destructive interference zero". But fortunately the loophole that your argument has can be filled in, by much deeper argument on distinguishability/indistinguishability of quantum states and showing that the principle of complementarity (PC) is superior to the which way claim. Indeed PC is tightly connected with the density matrix of the qubit, and all which way claims must follow from the density matrix. For details, see my PhilSci entry. I recommend that the argument by anonymous in its current form NOT be included in the main text since it is with loophole. Danko Georgiev MD 09:57, 18 January 2007 (UTC)
thread_title:  Sorry about that