Afshar page

Afshar PDF

Questions welcome (Afshar's blog)

I learned about this experiment from Wikipedia where it was uncritically described at various pages about the interpretation of quantum mechanics.

This Gentleman and colleague of ours has argued that he was able to falsify the Copenhagen interpretation as well as the Many Worlds Interpretation - WOW - and confirm an obscure interpretation - the so-called Transactional interpretation by an interference experiment of the type Welcher Weg (which way). This transactional interpretation, proposed by a physicist John Cramer, is based on signals being sent back and forth in time. :-)

Incidentally, an older article about the interpretation of quantum mechanics on this blog is here:

Causality and entanglement

How could he have falsified the standard interpretation(s)? He argued that he has falsified Bohr's complementarity principle itself. As you know from the play Copenhagen, Bohr had the complementarity and Heisenberg had the uncertainty. But from the current perspective, they're very related if they're applied at the momentum (inverse wavelength) and the position. Actually, the current interpretation of "complementarity" was probably constructed as an argument by Heisenberg, not Bohr, in the 1950s. In some sense, I feel that they developed these notions together, and they divided the credits. But let's not analyze history - it's not the main point here. Complementarity says that a photon or light either has particle-like properties, or wave-like properties, but you can never measure both of them simultaneously.

For the interference experiments, it means that you either

- see an interference pattern - which means that you measure the wave-like properties of light
- or you're able to determine which slit the photon picked - which means that you measure the particle-like properties of light,

One may also consider the intermediate cases in which we see the interference pattern with contrast or visibility V which is a number between 0 and 1 - roughly speaking, it's the difference of the intensity at the interference maxima and minima, divided by the sum of the same two intensities. And we may consider the situation in which the "which way information" is extracted with reliability K which stands for Knowledge - again between 0 and 1. The principle of complementarity then implies something like "V^2 plus K^2 is never greater than one". More concretely, it is impossible for both V and K to be greater than 0.9, for example.

This statement is, of course, true as long as you define the variables V,K properly.

Afshar's strategy is simply to arrange an experiment in which he argues that he can see a sharp interference pattern with V=1 (or at least, greater than 0.99), but he can also say which slit the photon chose (which means K=1). That would mean that he gets 1+1=2, which is greater than one, and the principle of complementarity would be violated maximally.

Of course, it's a silliness - the same silliness like saying that "x" commutes with "the derivative with respect to x". Nevertheless, this silliness was described as a "quantum bombshell"

- in the New Scientist, July 24th, 2004
- on NPR Science, July 30th, 2004
- on EI Cultural, September 9th, 2004
- in the Philosophers' Magazine, October 2nd, 2004
- in The Independent, October 6th, 2004
- at various seminars, including one at Harvard University (I did not see it)

Afshar's FAQ

Afshar's reply says, roughly speaking, that Bill Unruh is like an engineer who designs planes without wings which cannot fly and who argues that it means that no planes can ever fly.

Unfortunately for Afshar, Unruh's description of the situation is rather transparent and it reveals some common errors in the interpretation of these experiments:

Rebel via UBC

See also the discussion on the blog of the daughter of that Cramer who invented the time machine interpretation of quantum mechanics - and she is happy about Afshar's statements because she admires her father:

Cramer about Dad

Unruh uses a more transparent experimental setup where some issues can be explained very easily, without various technical complications of Afshar's setup. The only respect in which Unruh's setup seems to be oversimplified is that his framework only contains "completely dark" and "completely bright" area, and therefore the subtlety "thickness of the wires is small" does not arise in Unruh's specific approach. (Thanks to William Unruh for his patient explanation of the isomorphism.)

Which photons we consider?

However, Unruh's explanation addresses Afshar's error assuming that Afshar only wants to consider the events with both pinholes open.

In this context there are many ways how to define the problematic notion of "contrast" of the interference picture (that you never really see directly). If you decide to define it using the "both pinholes open" situations, then you must still measure the influence of the wire grid located at the interference minima, as well as the wire grid located at the maxima. If you put the wires at the maxima, however, the direct link between the detectors and the pinholes starts to disappear. This is a point due to Unruh.

In the conventional and straightforward interpretation of the variables V,K, Afshar's error would be even more obvious.

In the inequality "V^2+K^2 is smaller than one" V is normally meant to measure the contrast of the actual picture that we see at the place where the photons under consideration are absorbed: in Afshar's case, it means the contrast of the pictures in the detectors themselves. Of course, these pictures don't show any sharp interference minima at all - the minima are the dark blue regions on the figure 1 screen below, and they're never inside the disk. That would mean that V=0, too.

One again the important point of the previous paragraph:

- In the complementarity principle, we first determine which set of photons we consider, and then we calculate both V as well as K from this set of photons. The contrast is computed from the pictures created by all these photons, and we also want to determine the "which way" information of all of them if we want to claim that K=1
- If you read the section 3 of his PDF file, Afshar seems to compute the contrast of his interference picture from a very small subset of his photons that he uses for the calculation of K - only from the photons that interact with the wire grid.

- If true, that's of course silly. We can always arrange an experiment with 2 million photons - the first million will be used to create a perfectly sharp interference picture, and for the second million we will be exactly able to determine the pinhole. But this does not mean that we have K=1, V=1. We must consider the same set of photons if we want to determine K, V.

Because of the differences between Afshar's and Unruh's setups, it is perhaps useful to talk directly about Afshar's experiment. The text below could very well be the first public critique of Afshar's claims in his own setup. Below, I will assume that Afshar wants to measure the wave-like character of the laser light by comparing the influence of the wire grid on the outcome of experiments with 1 or 2 pinholes open - this seems to be the last case in which we want to prove that the contrast is also going to zero as the thickness of the wires approaches zero.

First we must describe his bombshell. See the picture. Laser light goes through two pinholes and through the lens. The geometry of the lens is such that if you close the upper pinhole 1, the photons coming through the lower pinhole 2 will always be detected in the upper detector on the right, and vice versa. You only need the geometric optics of the light rays to see why. Each detector records a full image. This allows us to determine which pinhole the photon chose from the detector where we see the photon.

Fine: we can measure the position i.e. the pinhole through which the photon traveled - by looking which detector "beeped". But we don't see any wave properties. The interference pattern at the lens is not recorded.

Afshar was probably thinking like this:

- What will you do if you want to see which pinhole the photon chose, and also simultaneously see the wave-like properties of the photon? Well, you put a lot of wires just before the lens (the red dots in the second and the third picture above). You put the wires in the interference minima obtained from the interference between both pinholes.
- If the wires are sufficiently thin, it guarantees that we will see the same images created on both detectors (see the third picture) like the images from the first picture without the wires. It's because no photons are absorbed by the wires - the wave function is essentially zero near the wires.
- The fact that the situations "1" and "3" give you identical images proves that we are observing some wave-like properties of the photon. The situations "1" and "3" only lead to the same outcome if the spacing between the wires agrees with the shape of the wave function near the lens, for example.
- Obviously, the wires will have a more significant effect if you close one of the pinholes - this is visualized on the second picture in the middle. Because there is no interference from two pinholes in this case, the wave function is never zero, not even near the wires. Therefore the image is visibly weaker and more measurably disrupted, if compared to the single-pinhole no-wire situation.
- OK, return to the third picture with both pinholes open and with the wires arranged before the lens. Recall that we observe the interference from both pinholes - this is displayed by the fact that the presence of the wires does not make any measurable difference.
- That proves that we see the interference pattern with the maximum constrast or visibility, V=1 (sic !!!)

- But at the same moment, we are also able to say which pinhole the photon chose - if the photon appears in the upper detector, it chose the lower pinhole, and vice versa.

- Therefore we have showed that Bohr's complementarity principle is violated (sic !!!).

It is also important that 3 is not taken out of the context - 3 is only correct if you also say 4.

There is still complementarity in action, but one must express it a bit more quantitatively. Imagine that the relative thickness of the wires - the fraction of the space that they block - is T (effectively something below 0.1 in the actual experiment). At T=0, the wires don't really exist, and you can't say that you observed the wave phenomena because the image is unchanged not only in the two-pinhole interference case, but also in the single-pinhole case.

If T grows, then you can actually see that the two-pinhole interference situation is affected much less by the wires than the single-pinhole setup. This proves that you "start to see" the wave phenomena. But at the same moment, as T grows, in the second, middle picture, the photons also start to produce a weak image of the photon in the "wrong" detector. This starts to invalidate the connection between the detectors and the pinholes.

The more clearly you see some interference pattern, the less certainly you can claim that you know which pinhole was chosen. It is completely analogous to the situation in which the two pinholes are being closed randomly and independently: some photons will interfere, and some won't. Those that interfere will create a "weak" interference pattern, and those that don't interfere (coming through one pinhole only) will allow you to guess correctly which pinhole they went through.

But the main error of Afshar is that

- he thinks that the contrast of his interference pattern is V=1 or very close to one, i.e. he observes the wave-like properties "sharply", just because his thin wires don't spoil the signal in the case in which the photons interfere near the lens.

- The definition of V is a bit cumbersome because he really does not measure any interference pattern directly, but we can talk about the indirect V anyway. Actually, V is very close to zero. You only prove the existence of the interference pattern if you compare the effect of the wires in the case of two open pinholes with the case of one open pinhole.
- But it's important that for thin wires, even their effect on the situation of the second, middle picture is very small. You see that the colorful picture in the middle is disrupted "just a little bit". Even here, most photons don't care about the wires. It's exactly this disruption that defines how much we observe the wave-like properties - i.e. how much we observe the difference between putting the wires at the interference minima, and putting them into places where no interference occurs.
- Because this signal (disruption) from the second, middle picture is small (equivalently, it only affects a very small portion of the photons), the contrast V is also very small, and goes to zero for infinitely thin wires.

The only way how we observe the interference patterns with the thin, but visible wires is that while they don't affect the outcome of the experiment with interference (both pinholes open), they do influence the situation with one closed pinhole. We must consider an ensemble of two different kinds of experiment. But even in the latter case, the influence is small, and it is this influence that measures the contrast or visibility of the wavelike properties. As the thickness of the wires goes to zero, the visibility goes to zero, too.

I can try to be a bit quantative. If the relative thickness for the wires is T (effectively below 0.1 in the actual experiment), and the wires are put to every minimum, then the probability of the wires absorbing/reflecting the photon goes like T for the single-pinhole case, and T^3 in the case of interference. However, the probability that the wires will send the photon (with one pinhole closed) into the "wrong" detector goes like T^4 or something like that (about 10^-6 in the actual experiment) - this probability will be irrelevant. The wave phenomena are observed by seeing that the single-pinhole picture is damaged much more than the double-pinhole picture. This means that T must be rather large, but then the error in the determination of the pinhole is rather large, too.

In fact, the main error in determining the pinhole does not come from the photons that appeared in the "wrong detector". The main contribution comes from the photons that were absorbed or reflected by the wires - we are definitely unable to determine the "which way" information for these photons!

T^4 may look like a large power, but it is important to see that T itself must be comparable to one if we want to claim that we've seen the wave phenomena "with visibility near one", and then T^4 is also of order one, which means that the error of our identification of the pinhole is of order one, too.

There is absolutely nothing mysterious about Afshar's experiment. In fact, quantum mechanics allows him to predict much more than just qualitative statements whether he can guess. It predicts the full probability distribution for both detectors and for every setup he chooses. If he did science quantitatively, he could have just tested that these numbers exactly agree. And if he treated quantum mechanics seriously, he would not waste time with these configurations because physics behind such experiments has perfectly well understood since the experiments of Thomas Young in the 19th century. And of course, the conventional quantum mechanics is compatible with the principle of complementarity.

The variables V,K once again:

V=0+epsilon, not 1-epsilon

Let me use Afshar's variables V,K for a while - see the PDF file linked at the beginning of this article. V is the "visibility" or "contrast" of the interference picture (0 for no interference, 1 for perfect minima with zero probability), while K is the reliability how you can determine "which way" the photon went (0 if you cannot, 1 if you can say it perfectly). The inequality from complementarity is something like "V^2+K^2 is not greater than one". Afshar more or less argues that "V^2+K^2" in his case is "1+1=2", which is not realistic. While K is pretty close to one, V can by no means be set to one because he does not really see any interference picture. If the thickness T goes to zero, it's totally obvious that the information about the wave character of the photon goes away - the superthin wires affect neither the interfering photon nor the single-pinhole photon. The interference is not seen, no photons near the lens are absorbed or affected.

Because he does not directly see any interference pattern, it is not quite clear how to assign a nonzero value of V, but morally it is true that V^2 behaves like something of the form T^4 while K^4 behaves like 1-T^4, as argued above. I will have to re-check and fix the powers, but the main conclusion is not affected:

If the wires are thin, the visibility V of the wave character of the light is close to zero, not close to one as Afshar argues!

**Is it a quantum experiment?**

What I find most unjustified about this experiment is that it has actually nothing to do with quantum mechanics. You can describe everything about this type of experiment using classical electrodynamics simply because the number of photons is large - he has not really measured individual photons. If he measured the individual photons, the probability density would simply be proportional to the energy inflow determined from the classical theory.

The classical electromagnetic waves propagate according to Maxwell's equation. Of course, the standard quantum mechanics - quantum electrodynamics - in such limits reduces to classical electrodynamics. Afshar therefore

- either argues that he has also falsified Maxwell, which I consider enough to call someone a crank; my feeling is that he does not argue that

- or he admits that his experiment is just another trivial, unsurprising double-slit experiment that agrees with Maxwell's equations and quantum mechanics - but then I really don't understand how can he say all these big statements about "quantum bombshells" and "violation of the principle of complementarity"...

**Note added later:** let me mention that Kastner has submitted another paper criticizing Afshar's conclusions. In my opinion both Unruh as well as Kastner replace Afshar's experiment by a completely different experiment that does not capture the main flaw of Afshar's reasoning. The main flaw is that Afshar does not realize that for a tiny grid, only a very tiny percentage of photons is used to observe the wave-like properties of light; these are essentially the photons for which the which-way information is completely lost. Because most photons go through the lens without any interactions and interference, Afshar is not allowed to say that he observes the wave-like phenomena with visibility close to one. In fact, it is close to zero if a consistent set of photons is used to define both V and K.

Dear Prof. Afshar,

ReplyDeleteI studied your experiments a bit. Of course, I don't think that your

conclusions are correct. You can find an article about it at

http://motls.blogspot.com/2004/11/violation-of-complementarity.html

A short summary would be the following one:

You don't ever see any interference picture directly, so any nonzero

assignment to the visibility V is problematic. If the thickness of the

wires goes to zero, then the wires will not change the pictures with both

pinholes open - but they will not the pictures with 1 pinhole open either!

For a supertiny value of the thickness, you therefore don't observe any

wave phenomena of the light at all, so it is completely ridiculous to set

V=1. I have not rechecked the powers, but morally - for a small relative

thickness T of the wires - the visibility goes like V = T^2, while the

reliability K of the "which way" data goes like 1-T^4 or something like

that (the difference from one measures how many photons are scattered, by

the wires, to the wrong detector).

If you sum T^4 and 1-2T^4, you will get less than one. There should be

numerical coefficients everywhere, and the powers should be rechecked.

Of course that the bound is always satisfied. What I find most strange

about your statements is that your experiment with laser light can

actually be explained directly by classical electrodynamics - you don't

need any quantum mechanics. Of course that the orthodox quantum mechanics

(Copenhagen, and anyone who wants to compete) reduces to the classical

theory if the number of photons is large. This means that if you found

problems in this limit, it would also be a problem of Maxwell's equations,

and it's just silly.

All the best

Lubos

Hey Lubos, Travis again (I just commented down below on the anthropic thread). So are we then to conclude that you are a Many-Worlds'er since, after all, the brain is just ~10^26 atoms? ;^) I'm probably obligated to also mention unitary time time evolution (even weird non-hermitian theories like Carl Bender's PT symmetry have it - e.g. hep-th/0303005) and decoherence (which is why it's hard to get quantum computers working - although josephson junction models seem promising). It does seem like the only reasonable interpretation...

ReplyDeleteHey travis, good to see you here. But could you please separate your statements and questions more carefully? ;-) I am not quite following what you say: it seems to mix too many things.

ReplyDeleteFrom: Shahriar S. Afshar

ReplyDeleteDear Lubos,

"Therefore we have humiliated Bohr, Heisenberg, Dirac, the Copenhagen interpretation, complementarity, the uncertainty principle, quantum mechanics as well as the rest of physics."

From the content of your response, I can only conclude that you have not fully read my preprint:

www.irims.org/quant-ph/030503/

The only claim I have made (and still make) is that Bohr's Complementarity Principle (PC) fails in my experiment. The very idea of trying to "humiliate" a luminary like Bohr and others (for whom I have the greatest respect), is beyond any scientific code of conduct. My only aim has always been to search for the truth, (the Crimson “Veritas”!)

In my paper, (as well as my talk at Harvard, which you probably did not attend) I have explained that my work is ABSOLUTELY in line with QM and no violations of QM is suggested by the results. My experiment shows that PC does not follow from QM when non-perturbative measurements are made (see my paper for more details). The uncertainty principle (UP) is certainly NOT violated by my experiment, and it is wrong to assume that Compelementarity is the same as Heisenberg's UP. Please read my manuscript for ref.s on the distinction between UP and PC. Your above quotation--presumably refereing to my motivations--is truly irresponsible, and does not belong in scholarly discussions.

As for the argument that my experiment conducted in the high-flux regime is not a QM experiment, let me inform you (as I have mentioned in the NPR interview back in July, and other places) that the single-photon version of my experiment conducted at Rowan University has produced the exact same results, therefore you assertions to the contrary are factually incorrect.

I suggest that you visit my Blog here: http://irims.org/blog/index.php/2004/09/25/questions_welcome

Kindly read through the comments made there, where I have attempted to answer all questions related to my experiment including all the issues raised in your critique. Your considered/and well-researched comments are most welcome there.

As to the Unruh critique, I have not been at Rowan to upgrade my FAQ page there, but I will post a more detailed response in my Blog above soon.

Finally, I would appreciate if you find the facts about exactly what I have said, before going out there and making inaccurate quotations on my behalf.

Looking forward to hearing from you.

Best regards.

Shahriar S. Afshar

E-mail: afshar@rowan.edu

P.S. I am currently in Boston, and would be glad to meet with you to talk about my experiment. I gave a copy of my preprint to Prof. Arkani-Hamed (as well as others) before my talk back in March.

Dear All,

ReplyDeleteIn Topological GeometroDynamics (TGD) framework the result of Afshar's experiment can be seen as a support for what I call quantum classical correspondence. Quantum classical correspondence states that not only quantum states but also quantum jump sequences have space-time correlates. In particular, the states of the system before and after quantum measurement should correspond to space-time regions.

This is possible by the classical non-determinism of the basic variational principle determining space-time surfaces as absolute minima of what I refer to as Kaehler action. Maxwell action for a Maxwell field defined by the Kaehler form of CP_2 projected to the space-time surface in M^4xCP_2 (replace the points of the Minkowski M^4 space by complex projective space CP_2 of 4 real dimensions). Instead of gauge potentials the primary dynamical variables are imbedding space coordinates and only the second "topological" half of empty space Maxwell's equations (the absence of magnetic charges and Faraday law) is satisfied as such. The character of field equations is hydrodynamical: they state the conservation of classical 4-momentum and color currents (CP_2 has color group SU(3) as isometries).

The non-determinism is basically due to the U(1) gauge

degeneracy implying that the induced Kaehler form vanishes when the CP_2 projection of space-time surface belongs to a generically 2-dimensional Lagrange sub-manifold of CP_2. This gigantic vacuum degeneracy does correspond to a genuine gauge degeneracy since it it is absent for non-vacuum extremals and is broken by classical gravitation assignable to the induced metric. The degeneracy is more like spin glass degeneracy, and space-time surface as a 4-D spin glass is good metaphor for what happens.

The degeneracy implies that also absolute minima resulting as small deformations of the vacuum extremals have huge degeneracy and decompose into maximal deterministic regions representing (unfaithfully of course) quantum states at space-time level. The collections of these maximal deterministic regions represent quantum jump sequences at space-time level.

In Afshar's experiment the space-time region between the screen and lens would represent the situation before the state function reduction and the region between lens and detectors the situation after the state function reduction. The addition of the rings does not affect appreciably the situation since the interference indeed occurs in this region.

There is a little article titled "Double Slit Experiment and Classical Non-Determinism" at

http://www.physics.helsinki.fi//articles/doubleslit.pdf

With Best Regards,

Matti Pitkanen

I've sent another reply on Shahriar's blog. It has some scripting problems, however. Here it is:

ReplyDeleteDear Shahriar,

after half a day of research and reading, I would agree with you that Prof. Bill Unruh's setup is not equivalent, and there is a different problem with your interpretation.

The problem with your interpretation is the point where you say that your V (contrast/visibility) is close to one, and the you see the wave character "very sharply".

The reality is that you don't directly observe almost any influence of the thin wires - because they are so thin. This may be a very confusing point. So let me add a couple more sentences.

You claim that you observe the interference pattern near the lens "sharply". But the only sense in which you "observe" the interference pattern is that the wires don't modify the outcome of the experiment, almost at all.

But you're wrong if you think that it is because you see the wave-like properties sharply. The real reason is that the wires are simply thin, and they have a very small effect on the outcome of the experiment.

The only real evidence that you observe the interference near the lens is if you compare the two experiments with wires - one of them with both pinholes open, the other with one pinhole closed. Then, indeed, the signal is very little changed in the former case, but visibly more changed in the latter case.

But it's not changed much even in the latter case. Your real contrast is only obtained if you compare the influence on the signals caused by the wires in these two situations (1 hole open, 2 holes open). And this difference is very small for thin wires, scaling like a power law for small t, which means that your contrast V scales like some power of t and is very tiny in your actual setup.

Yes, K is close to one, but it is slightly smaller than one - because some photons ARE sent, by the wires, to the "wrong" detector. At any rate, K is slightly below one, and V is slightly above zero, and the inequality V^2+K^2 < 1 holds.

Sincerely,

Lubos Motl, ass. professor at Harvard Univ.

See this URL for more details:

http://motls.blogspot.com/2004/11/violation-of-complementarity.html

I have been enjoying your posts as they are quite informative for me. As soon as you mentioned Shahriar S. Afshar I immediately recogize it from another time.

ReplyDeleteThe point about quantum mechanics might have been raised from Kathryn Cramer as I think you suggested?

If you put,< a href=url target=_blank >it will open in another page.

Lubos,

ReplyDeleteI think you are pretty much on the money with this one. To me a destruction of the complementarity principle means that I can get correct results computing with probabilities rather than probability amplitudes.

Trying to spell it out in one more way:

With no wire grid, a photon that starts out with a wave function localized around the lower pinhole arrives at the upper detector with an amplitude that corresponds to probability 1, and thus does not arrive at the lower detector at all.

With the wire grid in position, a photon that starts out localized from the lower pinhole now has some non-zero probability amplitude of ending up at the lower detector (where without the wire grid this amplitude was zero), and the amplitude at the upper detector has reduced. This is what is observed. Then when both pinholes are open and wire grid is present, if the amplitude of the photon wave function at the two detectors is unchanged as compared to when the grid was absent, this has to be because of at the upper detector, in the presence of the wire grid, some amplitude is contributed by the upper pinhole. In the absence of the grid, zero amplitude was contributed by the upper pinhole at the upper detector.

With the wire grid in place we do not know any more with certainty that measurement at the upper detector means origin of the photon at the lower pinhole.

Let me say few words. As far as I remember QM says there is interference before that lens (both slits opened). So when Afshar says there is interference he cannot be wrong. He can be wrong when he says his V(isibility) = 1. That can be wrong. But the question is: can we improve that V(isibility)? Of course if we improve V(isibility), the K(nowledge of path) becomes <<1, because photons are quanta, and they are destroyed in the interference pattern. So the choice would be to take a portion of that virtual interference pattern, existing before the lens, and to test, using that portion, the real V(isibility) of fringes. Now the real mystery, imo, is that the lens-detectors system follows geometric optics (straight ray) while what is between the slits and the lens follows wave optics. How can we know, then, which was the _entire_ path of each photon from a specific slit to a specific detector? Notice that complementarity specifically forbids 100% knowledge of the _entire_ path (from a slit to a detector) and 100% Visibility of fringes at same time (same experiment). s.

ReplyDeleteHi Lubos,

ReplyDeleteIt seems we've both come to some of the same conclusions (and done the same thought experiments!)

One difference seems to be that you are not a fan of a sort of counterfactual knowledge: "if the interference pattern was not there, something else would have happened, ergo, it is there." I'm not sure what my philosopher friends would say about that.

Here is how I would say it (succinctly.)

Afshar measures angular momentum and angle (two conjugate variables.) The angular momentum here is that 'about' the center of the source, i.e., the actual AM is some crazy superposition of lots of l's.

Anyway, he measures AM by putting in wires, angle by putting in lenses. What he gets wrong is the idea that by putting in wires, he has a precise measurement of the angular momentum of the photon. (If he did, that would violate complementarity.)

Instead, Afshar gets a very imprecise measurement of the AM (he has a rough idea where some of the nodes are.) And as his knowledge of the AM goes up (he puts in more wires), his certainty about which slit he has goes down (diffraction around the wires.)

Unfortunately for Afshar, I believe, his new method of measurement, using these wires, is fundamentally limited. He can only get a measurement of the number of nodes of the wavefunction; in the limit where the wires go to zero (and he integrates for an infinite amount of time), he knows the value of the wavefunction on a set of measure zero -- not much help at all.

Anyway, I am not enough of an expert to say much more, but I believe this is roughly consistent with what you write, seen from a different angle.

Yours,

Simon DeDeo

Perhaps this is a stupid question. Has the effect of wires been tested by putting them on different positions than the minima of interference pattern. Putting the wires at the maxima could prove right or wrong the claim that the the wires are too thin to have any effect.

ReplyDeleteMatti Pitkanen

This comment has been removed by a blog administrator.

ReplyDeleteI've deleted an anonymous comment that had absolutely nothing to do with the technical questions discussed in this article and whose only effect could be to lower the quality of the discussions on my blog.

ReplyDeleteHi Matti!

ReplyDeleteOf course that the wires had some effect, but a small one. They had a bigger effect if you put them in the maxima than if you put them to the minima or generic places. You must have misunderstood the text completely if you think that I said that in the actual experiment the wires have *no* effect whatsoever.

If you look at the colorful picture in the article, you see that the second, middle picture on the right side is somewhat distorted - this is the effect of the wire grid. The second picture corresponds to having no interference on the lens at all - so there are no minima (and maxima), and the wires do matter.

What I am explaining in this article is that because the effect of the wires is always rather small (I don't say exactly zero), you're not allowed to say that you see the interference pattern with visibility/contrast one.

Roughly speaking, the photons that don't interact with the wires don't give you any interference pattern at all, and therefore they don't contribute to the contrast of the picture. The photons that *do* interact with the wires contribute to the contrast, but the relation between the pinholes and the detectors is violated for them.

All the best

Lubos

Those little wires intercept no light.

ReplyDeleteThis has been tested by Afshar.

So how can they diffract?

Or have any other effect?

s.

Hi S.! If you look at the colorful picture in the article, focus on the middle picture (of the three situations). Do you see the circles on the right? They're distorted, right? This is the influence of the thin wires. In the middle case, the thin wires are at generic places because no interference minima exist in this situation - see that one of the pinholes is closed.

ReplyDeleteThe effect of the wires is much smaller in the 3rd picture because the wires are located at the interference minima (both pinholes open).

The influence of the wires is very small, because the wires are thin, but it's nonzero. However, the contrast with which we can measure the wave-like behavior, is very small too - exactly because the wires probing the interference are almost invisible.

Thanks for Lubos for explaining the details of the pictures. The figure did not tell much to my eyes since I was unable zoom it.

ReplyDeleteJust a few comments about complementarity principle, uncertainty principle, and somewhat unexpectedly, about the generalization of string diagrams to 4-D case.

1. Macroscopic quantum entanglement instead of complementarity?

I think that the uncertainty principle is much more clearer mathematically and also all that is needed. Instead of using complementarity to explain what happens in the double slit experiment, one can consider quantum entanglement between macroscopic detector states and photon states.

When a photon is detected in the detector A, a quantum jump leading from the state

|detector A fires>|photon goes path a>+ |detector B

fires>|photon goes path b>

is reduced to its first component. The minimal assumption in accordance with my first posting is that path a/b correspond to a path leading from the lens to the detector A/B. If the identification of the space-time surface as a correlate for a quantum jump sequence is accepted, the situation between the screen and lens need not change in the tick since it represents the situation before the state function reduction.

2. Generalized stringy diagrams and photons travelling

simultaneously via two paths

Also the space-time correlate for the notion that a photon travels simultaneously along paths a and b should exist. If photons correspond to tiny 3-surfaces, the counterpart for a photon travelling along paths a and b simultaneously would be a photonic 3-surface, which splits into two parts at lens and goes through the detectors. Similar splitting could occur also for the photons entering the screen and the branches of the photonic 3-surfaces could lead through the pinholes and join back together behind the screen. Quite generally, photonic 4-surfaces would be kind of multi-head dragons and put their heads through all possible holes. Second quantized free induced spinor fields at these branches would be the fundamental fields and indeed propagate along the branches and interfere again.

3. What happens when photon is detected?

What is then the space-time counterpart for the tick in the detector? Lens would cause a branching of a photonic 3-surface to paths a and b. A tick in the detector A could mean that the branch of the photonic 4-surface leading to the detector B degenerates in the quantum jump to a vacuum extremal for which the fermionic Fock space would reduce to a mere vacuum state. Vacuum lines are indeed the basic new element of the generalized Feynman diagrams, and are very natural in view of the enormous vacuum degeneracy of the Kaehler action. They seem to be also unavoidable: if you have an absolute minimum of Kaehler action you get a new one by fusing to it vacuum lines.

Vacuum lines are also forced also by the braid theory inspired construction of generalized Feynman diagrams based on the idea that generalized Feynman diagrams with loops are equivalent to tree diagrams. This notion can be formulated axiomatically by generalizing the axioms of topological quantum field theories using generalization of braid category axioms. See

"Equivalence of Loop Diagrams with Tree Diagrams and

Cancellation of Infinities in Quantum TGD" at

http://www.physics.helsinki.fi/~matpitka/tgd.html#bialgebra .

4. What are the space-time counterparts of particle decay vertices?

The 4-D counterparts of stringy diagrams would not describe the decay of a particle to two particles but a simultaneous travel of a particle via two different routes. The counterpart of the particle decay vertex should be something different. In ordinary Feynman diagram particle decay corresponds to a branching of 1-manifold so that it becomes singular. So why not also in higher-dimensional case? In n-vertex the n four- surfaces would be glued together along their ends. The outcome would be also the possibility of n>3-vertices: the counterparts of stringy n-particle vertices decompose always to 3-vertices by a small deformation.

The 4-D "Feynman diagram" would correspond to a singular 4-manifold whereas the vertices itself would be smooth 3-manifolds. For the stringy option the 4-D Feynman diagram would be a smooth 4-manifold whereas the vertex would be singular 3-manifold. The construction of vertices would be dramatically simplified. The effect is amplified by the effective 2-dimensionality reducing the calculations to 3-D light-like causal determinants so that vertex becomes a 2-surface.

Matti Pitkanen

Lubos said:

ReplyDeleteIf you look at the colorful picture in the article, focus on the middle picture (of the three situations). Do you see the circles on the right? They're distorted, right? This is the influence of the thin wires. In the middle case, the thin wires are at generic places because no interference minima exist in this situation - see that one of the pinholes is closed.

The effect of the wires is much smaller in the 3rd picture because the wires are located at the interference minima (both pinholes open).

The influence of the wires is very small, because the wires are thin, but it's nonzero. However, the contrast with which we can measure the wave-like behavior, is very small too - exactly because the wires probing the interference are almost invisible."

lubos also wrote:

"The only real evidence that you observe the interference near the lens is if you compare the two experiments with wires - one of them with both pinholes open, the other with one pinhole closed. Then, indeed, the signal is very little changed in the former case, but visibly more changed in the latter case.

But it's not changed much even in the latter case. Your real contrast is only obtained if you compare the influence on the signals caused by the wires in these two situations (1 hole open, 2 holes open). And this difference is very small for thin wires, scaling like a power law for small t, which means that your contrast V scales like some power of t and is very tiny in your actual setup."

Fortunately science evolves by abstract papers that are well reviewed. If you read the paper you see that the The normalized radiant flux blocked by the wires is found to be R ~ = (6.6± 0.2)% when only one hole is open and with both open the data show that the attenuation of the transmitted light in this case is

negligible, R = (-0.1± 0.2)%. That is a huge difference.

Dear anonymous fan of well-reviewed (but otherwise wrong) papers,

ReplyDeleteno one is questioning the fact that there is a difference between putting a wire to the interference minimum, and putting the wire to a generic place - or the interference maximum.

In the case of the minimum, the effect of the wire is tiny and almost unmeasurable; in the other case it is measurable but very small.

But exactly because it is very small, even in the second case, the visibility of the wave-like properties is very small, too. The contrast is near zero, and my guess is that most readers have understood it.

All the best

Lubos

Dear Lubos,

ReplyDeleteI have just read your comments on Afshar's blog and found them interesting. However, I was wondering if you had considered the following line of argument.

Assume that Afshar's wires do measure the interference pattern perfectly (I see that you disagree with this point, but please bear with me). Measurement of perfect interference implies the photon came from both pinholes and that the wavefunction at the wires is an equal sum of the wavefunctions from the two pinholes. If at every point in the plane of the wires the wavefunction is an equally weighted sum of contributions from both pinholes then subsequent evolution of this wavefunction is also an equally weighted sum of these contributions. Therefore a photon measured at either detector came from both pinholes and there is no which way information.

I.e. if there is wave interference of the wavefunctions from the two pinholes, we cannot subsequently measure which pinhole the photon came from.

This is the opposite of the traditional experiment showing complementarity where you first try to measure which slit the photon came from, disallowing a subsequent observation of an interference pattern.

Morgan Harvey

Dear Lubos

ReplyDeleteQuote "no one is questioning the fact that there is a difference between putting a wire to the interference minimum, and putting the wire to a generic place - or the interference maximum."

Huh? The figures are with one hole open 6.2% With both holes open 0.1%. The wires are never moved when the figures are taken. I never mentioned minimum or maximum. This is how interference is demonstrated in the experiment.

"In the case of the minimum, the effect of the wire is tiny and almost unmeasurable; in the other case it is measurable but very small."

6.2 and 0.1 Both measures are in the calculated minimum. one figure is 60 times greater than the other!

"But exactly because it is very small, even in the second case, the visibility of the wave-like properties is very small, too. The contrast is near zero, and my guess is that most readers have understood it."

I am not so sure that most have understood it.

I am a fan of science. You can win me over to your side of thinking with a logical, well researched argument!

6.2% is much greater than 0.1%, but it is much less than 100%. The contrast V with which you observe the wave phenomena is just proportional to 6.2% (or some positive power of it), certainly not 100% which is just an illegitimate rescaling.

ReplyDeleteIf you wanted to observe the interference with a bigger contrast, you would have to put thicker wires, but then the correlation between pinholes and detectors would no longer hold.

I am sure that others have already understood it, and I don't care terribly much whether exactly you will be able to join those who have understood it, especially because you're anonymous. It's your decision whether you want to understand interference or not.

Perhaps the following, if correct, will help unconfuse folks.

ReplyDeleteSuppose we draw two planes 1 and 2, very close to each other just before the plane of the lens (and on either side of where we are planning to put the wire grating).

We think in terms of a propagator that propagates the wavefunction on plane 1 to plane 2.

A photon at the Upper Pinhole |UP> propagates to plane 1 as |U1>, subsequently to plane 2 as |U2> and finally to the Lower Detector as |LD>. Likewise, a photon at the Lower Pinhole |LP> propagates to plane 1 as |L1>, thenceforth to plane 2 as |L2> and finally to the Upper Detector as |UD>.

Now, when both pinholes are open, sans normalization, we have at the plane 1 |U1> + |L1>. Without the wire grid, this propagates to plane 2 as |U2> + |L2>.

What Afshar does is put some new interaction, namely his wires, at the zeroes of |U1> + |L1>. We all expect , even with the wires present, for |U1> + |L1> to continue to propagate to |U2> + |L2>.

What underlies Afshar's claim, and what is wrong, is that, because this one state is propagated unchanged in the presence of the wires, one knows which pinhole from which photon originated.

But just because the propagator has not changed for one state, it does not mean the propagator has not changed.

We know that |U1> in the presence of the wires, no longer propagates to |U2>, but to something else, say |UW2>, because now the upper detector also fires, and a sole |U2>would end up in the lower detector |LD>.

Likewise, |L1> now propagates to |LW2>.

Now, while physics ensures that |UW2> + |LW2> = |U2> + |L2>, it also ensures that |<U2|UW2>| != 1 and |<U2|LW2>| != 0.

To repeat what has been said all along - with the wires in place, there is a non-zero amplitude for the upper photon to reach the upper detector.

Or to put it yet another way, given some physical system described by a Hamiltonian, and a particular physical situation described by a particular wavefunction. Suppose I perturb the system by putting interactions that are non-zero only where this particular wavefunction is zero. Then it appears that this particular wavefunction is not affected at all. However, the interpretation of this wavefunction is certainly affected, as seen by taking a set of basis states, some of which are non-zero at the sites of the perturbation.

Dear Lubos,

ReplyDeleteAs you know, in Quantum Optics, the total radiant flux (or total photon count for a given time interval) Phi_t = Phi_Coherent + Phi_Decoherent. That is all sub-ensembles are either coherent, decoherent or a superposition of both, so that if Phi_Decoherent=0, you have perfect interference and if Phi_Coherent=0, then you have a perfectly decoherent distribution, and of course you may have intermediate distributions (the same applies to the wavefunctions where you have full or partial dephasing).

Actually contrary to your belief, V=0 is not the majority view from the hundreds of e-mails I have received and a few that have sent letters to editors of different press publications. In fact you are the ONLY one so far that has claimed V=0, and there is a good reason for it: If V=0, a certain amount of light must have been blocked by the wires, and we simply don't see that kind of reduction in the flux. As soon as you have decoherent light (which is technically different from incoherent light, but has the same spatial distribution), the photons are RANDOMLY distributed. There is NO way (using QM formalism) in which you can produce a decoherent distribution WITHIN which you have regions where the photons persistently avoid. The only possible explanation for such minima (as trivially discussed in the theorems in my preprint) is interference. If you can show me a decoherent distribution with zeros in it, then I'll accept your argument.

Quantum Optics is a very tricky field, and there is a huge literature on welcher-weg experiments which I am sure you have not had the time to read. My experiment does look ridiculously simple, but don't let the appearances deceive you. Like a good geometry question, the problem looks simple, but it is devilishly difficult to tackle!

See you at my Blog where I'll give a more detailed response your arguments, but it would really help to use numbers instead of adjectives like “small” and “big” to make your point precisely clear.

Regards.

Shahriar S. Afshar

P.S. Thank you for toning down your earlier remarks, I completely understand how one initially reacts to my results. I myself felt absolutely sick for a whole month, and even now have not fully recovered!!!

This comment has been removed by a blog administrator.

ReplyDeleteDear Shahriar,

ReplyDeletethanks for your comments!

Well, if I am the only one who thinks that with infinitely thin wires, you can only see the interference with infinitely small contrast, and therefore V goes to zero for infinitely tiny wires, then you should perhaps say to all these experts you mentioned that an irrelevant Lubos Motl from Harvard thinks that they should think about it twice because they have not gotten the point yet.

With no wires at all, you don't see *any* interference at all. These are the pictures you see on the detectors - NO INTERFERENCE, vanishing contrast. No effect for wires at minima, and no effect for wires at other places. With very thin wires, you start to see *some* interference, but V does not jump to one just because you added some superthin wires. Of course, if you define V incorrectly so that it's scaled to 100%, then you get 100%. But you have not ruled out the complementarity principle.

The complementarity principle says that the sharper you observe the particle-like properties (which way information), the less sharply you see the interference pattern. And this is what happens in your setup, too. Your interference pattern is very far from being sharp; it almost does not exist, and most of the time, you measure particle-like properties. Photons just directly go through one of the holes, and end up in one of the detectors. No interference. Trivial pictures on the detectors. No oscillating effect of the grid.

If you wanted to see the interference with better contrast (so that the middle picture would be significantly different from others), you would have to make the wires significantly thicker - as much as 50% of the area - but Bill Unruh's page then explains you that in this case, K would be going towards zero because the correlation pinhole-detector would no longer hold.

Best

Lubos

Shahiar, otherwise it's not so difficult to see that I've had enough time to learn all the necessary literature to understand interference experiments like that - in fact, I've learned such things at the high school.

ReplyDeleteOptics - and this one should not really be called "quantum optics" - is not *such* a tricky field.

Moreover, you tell me now that your blog contains a quantitative answer. I have not found any new numbers on your blog, just some insults.

Dear Lubos,

ReplyDeleteI would never insult anyone, nor would I engage in name-calling, it is not in my nature, and it certainly goes against my principles to be belligerent. Would you kindly show me where I have insulted anyone on my Blog, so I can apologize to that individual? My work is not about one-upmanship; it is about the pursuit of truth through scientific methodology! As for my "numbers", they are in the manuscript.

The interference pattern is NOT a single particle phenomenon and CAN be observed non-perturbatively. I never said you can measure V with infinitesimally thin wire (or detector). As you can find in my preprint, I have said a thin wire (not infinitesimally thin) would block a much larger amount of light for decoherent distribution as opposed to the coherent one. There is a world of difference between this and what you say.

Single-particle properties like the kinetic energy of a neutron can ONLY be measured by perturbaitve schemes, be it in a scintillator, or by means of QND.

My experiment is all about a better definition of “measurement”, the rest is details.

Regards.

Shahriar S. Afshar

P.S.- I'll have a new post on my Blog soon which might interest you.

Lubos:

ReplyDeleteA question for you.

Dirac (The Principles of Quantum Mechanics, Fourth Edition, Chapter 2, section 13) notes that "If they {two observables} do not commute a simultaneous eigenstate is not impossible, but is rather exceptional".

My question is - would we consider this eigenstate to be in violation of complementarity or some other principle? ( I think not, I want your thoughts).

How is this relevant to Afshar's experiment? Well, he has constructed such a state.

What I mean is, consider a Hamiltonian H, with eigenstates psi_n, so that

H psi_n = E_n psi_n

For a particular n, locate the zeroes of psi_n, say, y1,y2,y3,... and construct

the operator U(x) with the Dirac delta as

U(x) = delta(x-y1) + delta(x-y2) + ....

(We're constructing our wire grid to put at the interference minima )

Then,

[H,U(x)] != 0

However U(x) psi_n = 0 and so

psi_n is one of those exceptional simultaneous eigenstates of the

non-commutating observables H and U(x).

Afshar's experiment corresponds to measuring H and ( 1 - U(x)) simultaneously to full precision for the simultaneous eigenstate psi_n.

That looks like an excellent interpretation, Urs, especially if he were discussing the uncertainty principle. ;-)

ReplyDeleteYes, if [A,B]=C, then the error in A times error in B cannot be smaller than the expectation value of C.

If the eigenvalues of the operators are zero - like for J=0: Jx=0, Jy=0, Jz=0, then there is no error in their measurement - they effectively commute on this state.

Yes, it seems that it may be analogous, but i need more time for everything to make sense in this new language.

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ReplyDeleteSorry, "Urs" in my previous comment should have been "Arun".

ReplyDeleteI've deleted roughly 10 copies of Arun's message that were posted because of the lagged server blogger.com. All of us encounter these lags - it's sad, but we should survive.

Arun wrote 'My question is - would we consider this eigenstate to be in violation of complementarity or some other principle?'

ReplyDeleteWeak measurements, proposed by Aharonov, exploit quantum uncertainty. Aharonov's quantum detectors are so weakly linked to the experiment that any measurement moves the detector's pointer by less than the level

of uncertainty. There is a price to pay for these delicate readings, they are inaccurate. But while this might appear to make the whole process pointless, Aharonov has calculated that when repeated many times,

the average of these measurements approximates to the true value of the observable being measured. But I doubt the weak measurement is relevant here, since there is no repetition of measurement. It is more like a 'negative' measurement (Renninger, Dicke, that sort of measurements).

Complementarity principle. I think this is not defined. Which complementarity? Since there are many. 'Waves' vs. 'which slit'? 'Waves' vs. 'knowledge of *entire* paths'? V^2 + K^2 = 1 (or probabilistic complementarity)? Causality vs. space-time description (in Bohr's terms: 'any conclusion, based in an unambiguous manner upon the strict conservation

of energy and momentum, with regard to the dynamical

behaviour of the individual units obviously necessitates a complete renunciation of following their course in space and time')?

Regards,

s.

The idea of "where it was" is, I think, fairly easy to abstract. Visibility seems to be much harder.

ReplyDeleteE.g., suppose we have a physical system with "In" states, and a time evolution operator that carries "In" states to "Out" states. Suppose we can partition the "In" states into two disjoint sets (e.g. states localized around one or other pinhole). Now, given some particular "Out" state, we express it in terms a superposition of the time-evolved "In" states. In general, the "In" states involved will come from the two disjoint sets. In Afshar's set-up, in the initial configuration, all of one type of "Out" states (e.g., those that arrive at the lower detector) would have come from the upper pinhole.

In general, knowledge of "where was it" means resolving the amplitude of an "Out" state into contributions from the 2 disjoint sets of "In" states, and thus computing the probability of having originated from each of the sets.

Dear Prof. Afshar, I have a simple question:

ReplyDeleteI will refer and use the notation of figure 2, page 27 of your preprint:

http://www.irims.org/quant-ph/030503/

When BOTH PINHOLES ARE OPEN how do you know that EVERY photon detected by detector 1 (2) came from pinhole 1 (2)?

I think that you are reasoning this way: if you detected a photon at 1 (2) it must have followed one of the dashed (solid) paths, so it came from pinhole 1 (2).

But that reasoning is wrong! It is valid only when you have just one open pinhole and so there is no interference. WHEN BOTH PINHOLES ARE OPEN you cannot say that ANY, not even a single one, photon followed the dashed (solid) paths!

To prove it experimentally you can simply put wires ALONG the dashed (solid) paths in those points where we have destructive interference. And there is more than one wire position which satisfy this simple condition. Since no photon will be blocked in this case how can you say that ANY photon followed the dashed (solid) paths? You cannot. And so you cannot also determine the which-way of the photons detected at 1 (2).

Let me say it again: WITH BOTH PINHOLES OPEN block the dashed (solid) paths with wires localized in those points where we have destructive interference -- the so called dark zones. No single photon will be blocked, and so no single photon can be said to have followed the dashed (solid) paths, since those paths are now blocked by the wires! And so you have not determined the which-way information for the photons detected at 1 (2).

Best Regards!

Zweistein.

Dear Zweistein,

ReplyDeleteit's unlikely that Afshar reads it here - you should go to his blog. ;-)

However I think that you're not right. If you don't put any wires - or if you just put very thin wires - most of the photons won't care about anything near the lens, and they will just propagate "geometrically" and you can safely determine the "which way information" for these photons.

It does not matter that the photons' wavefunction - or simply the electromagnetic field - "interferes" with itself near the lens. The reason why it does not matter is that the light is not absorbed or measured over there.

The which way information only starts to disappear for the photons that kind of interact with the wires.

What I am saying is that in order to *see* the interference and the wave properties significantly - with a non-negligible contrast/visibility V, you must make the wires non-negligibly thick, and then, of course, the which way information is distorted by the wires.

At any rate, complementarity holds.

Best

Lubos

You don't understand my argument. Think about it again. I am not putting wires near the lens, as Afshar did to show that there is dark zones, and so interference, there.

ReplyDeleteI am putting new wires along the paths (see the preprint figure) where these same paths inteserct the various dar zones. When I do this nothing changes in the detectors, and so we have no right to infer the photon paths, since they are now blocked!

Dear Lubos,

ReplyDeleteI'm afraid I would disagree with your statement that the interference contrast V in Afshar's experiment is small because thin wires make a small effect. When you put the wires in, it will cause a reduction in the photon count at e.g. detector 2' equal to(1-V)*T*N where T is the ratio of the thickness of the wires to the period of the interference pattern and N is the photon count without wires (with higher order terms in T disregarded). Although this change will be small for thin wires, it can be measured accurately (for a large N) and, consequently, V can be reliably determined. Of course if T is zero then nothing could be measured but I don't think this is a good argument.

Let's say I've a detector with a low quantum effeciency. Would you argue that I cannot use it to measure a high interference contrast because it can only detect a small fraction of photons and because if the quantum efficiency were zero, I wouldn't be able to measure anything?

I don't think Afshar's experiment provides "which way" informartion but the interference is there.

cheers,

-Alex

alex.maznev@DELETETOEMAILphilips.com

Hi Alex,

ReplyDeleteI think that you got both answers in the wrong way.

Be sure that if the wires are almost invisibly thin, then you get the "which way" information almost always right. It's just geometric optics! Of course that this works. And if someone argues that the very existence of some irrelevant infinitesimal wires *completely* destroys the "which way" information that is clearly reconstructed for no wires, then he's just completely wrong.

Physics is continuous.

You are also using Afshar's incorrect definition of "V". You say that "because you can measure a decrease of the intensity accurately for large N, you accurately reconstruct V=1".

But that's just silly. The intensity (contrast) with which you observe the wave phenomena is EXACTLY large if you DON'T NEED many photons to see the wave phenomena. It's the whole point of the quantitative form of the complementarity principle. The correct visibility is your incorrect visibility multiplied by T, so to say.

Most photons in Afshar's actual experiment don't give you any interference whatsoever - they are absorbed in the detectors where they create a completely non-interfering picture.

You are making the very same elementary error as Afshar. You are only using the photons that reduce your intensity because of the wires to measure V - this is why you need "many photons" to measure the interference pattern near the lens exactly. But if you're defining V only from the ensemble of the photons that (potentially) interact with the wires, you must define K (the reliability of the which way information) using the same ensemble - and of course, for these particular photons that interact with the wires, the "which way" information will be completely destroyed.

But for most of the photons that just don't care about the wires, the which way information is completely preserved - but these photons are not helpful to tell you anything about the interference.

If you measure K with one ensemble of photons and V with another ensemble, of course that you can get V=1 and K=1. But this triviality has nothing to do with the complementarity principle which always talks about the *same* photons used to measure both V and K.

I assure you that if you will be using this wrong definition of the "sharp interference", you will also eventually understand that Afshar definitely DOES reconstruct the which way information with K>0.99, and therefore you will have to believe him that he is the next Bohr. ;-)

The true answer is, of course, that if you have thin wires and you want to use all photons that are emitted, you will have K>0.99, but you will get a tiny value of V because you should count all the photons that are absorbed in the detectors - without interference - as contributing the non-interfering "background noise" while the interference is only a small effect above it, caused by the photons that interact with the wires, roughly speaking.

Best

Lubos

Hi Lubos, in your reply to Afshar's experiment you seem to be violating your own complimentary views on other topics:http://www.physicsforums.com/showthread.php?t=88225

ReplyDeleteI think most criticisms have missed the essential flaw in Afshar's experiment. Firstly, Unruh's experiment is equivalent to Afshar's although not everybody is able to see this easily. Unruh's experiment is also more elegant because the dark fringe can be blocked without interfering with bright fringe.

ReplyDeleteThe catch in Afshar's experiment is that the photons reaching the detector do not contain any which-way information! As such there is not reason to believe that when both the slits are open, the photon detected at one detector has come from one particular slit.

However, there is an easy way to see that there is no which-way information.

Let us suppose that the two waves (or wave-packets) emerging from slits 1 and 2 are exp(ikx) and exp(-ikx) respectively. Let us suppose that the state in the region of interference is

psi(x) = [ exp(ikx) ] + [ exp(-ikx) ]

The square brackets denote part coming from different slits. Now this can also be written as

psi(x) = [ cos(kx) + i sin(kx) ] + [ cos(kx) - i sin(k) ]

The sin(kx) terms cancel out from the two parts (destructive interference!), to give what we call the dark fringes. Then what is left is the portion which we call the bright fringe. All light to the detectors now comes from the bright fringes. This is written as

psi(x) = [ cos(kx) ] + [ cos(kx) ]

So, the bright fringe has IDENTICAL contribution from the two slits. Hence NO WHICH WAY INFORMATION exists at this stage. This is also what one would expect from an interference. Bright fringe is formed by parts of the two waves that add up. Wave that add up have to be the same, they can't be distinguishable. The state that moves to the detectors is

psi(x) = 2 cos(kx) = exp(ikx) + exp(-ikx)

Now, this state looks very much like the original state, and exp(ikx) will move towards one detector whereas exp(-ikx) moves towards the other detector. BUT we can no longer say that exp(ikx) came from slit 1, because we KNOW that interference has mixed the two wave so that they can't be distinguished.

Hence there is not which-way information in Afshar's experiment, and complementarity is not violated!

- Tabish