Violation of complementarity?

It's kind of surprising that no one in the high-energy community has discussed this amusing experiment by Shahriar S. Afshar.

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,

but in the latter case, the interference pattern disappears. There is a complementarity principle that always restricts you - or the uncertainty principle for the photon, if you will. You can't see its position and the momentum (which is manifested by sharp interference patterns) simultaneously.

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)

On the main page, Afshar "responds" to Bill Unruh:

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.
  

Whatever reasonable way we choose to define V, we will always see that it equals zero as long as we're able to determine the "which way information", in agreement with the principle of complementarity.

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:

1. 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.
2. 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.
3. 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.
4. 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.
5. 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.
6. That proves that we see the interference pattern with the maximum constrast or visibility, V=1 (sic !!!)
   
7. 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.
   
8. Therefore we have showed that Bohr's complementarity principle is violated (sic !!!).
   

OK, where's the problem? If you think about it for a few minutes, you will see the problem. The statements 1,2,3,4,5 are correct, 7 is approximately correct, but 6 and probably 8 are incorrect - even though I can't guarantee that everything Bohr has ever said about complementarity was true; it's just my understanding of the complementarity principle that has not been violated by this experiment, and Shahriar may have found a statement due to Bohr in the libraries or archives that was proven incorrect by this experiment. (Thanks to Bert Halperin for discussions about this point.)

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.

If the explanation is not clear, we may try to be more extreme. Imagine that the wires are so thin - and we put just a couple of them - so that it is guaranteed that we will not be ever able to measure their effect. Would you then claim that we observe a sharp interference pattern? Of course not. The position of the superthin wires does not affect any experimental setup, and we don't observe any interference pattern whatsoever.

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.