{"id":4459,"date":"2010-01-23T10:01:16","date_gmt":"2010-01-23T10:01:16","guid":{"rendered":"http:\/\/scienceblogs.com\/principles\/2010\/01\/23\/neil-bates-owes-me-160\/"},"modified":"2010-01-23T10:01:16","modified_gmt":"2010-01-23T10:01:16","slug":"neil-bates-owes-me-160","status":"publish","type":"post","link":"http:\/\/chadorzel.com\/principles\/2010\/01\/23\/neil-bates-owes-me-160\/","title":{"rendered":"Neil Bates Owes Me $160"},"content":{"rendered":"

That’s the bill for the time that I spent on deciphering his supposed falsification of decoherence<\/a>. I don’t want anyone to fall for his false argument, so here’s the correct explanation of the scenario, to save other people the trouble.<\/p>\n

The center of his so-called “proof” is this modified Mach-Zehnder Inteferometer:<\/p>\n

\"i-9d2485ef8503c637f7fe07ace2dd873e-neil_apparatus.png\"<\/p>\n

Light enters at the lower left, is split by a beamsplitter (which I’m representing as a beamsplitter cube, because that’s what I usually use, but it could be anything), redirected by two mirrors to a second beamsplitter where the beams A and B are recombined, then the recombined outputs C and D are themselves recombined at a third beamsplitter, after which they are detected at points E and F. The greenish box represents a phase shift caused by the environment, which people (and dogs) who understand quantum mechanics say causes decoherence, which ultimately makes the outcome look classical, with a 50\/50 chance of showing up at either C or D rather than showing an interference pattern.<\/p>\n

If you follow the link above, you’ll find a bunch of inordinately confusing notation attempting to describe this scenario mathematically, leading to the claim that the third beamsplitter somehow “undoes” the effects of decoherence. This is because Neil has done the math wrong, and leaped to interpret his wrong result as support of his predetermined conclusion.<\/p>\n

I’ll go through the correct math below the fold, for those who care. If you don’t ordinarily like seeing equations full of complex exponential functions, skip the rest of this post. There’ll be something more enjoyable later.<\/p>\n

<\/p>\n

The first step is to establish what the wavefunctions are at points A and B, immediately before the second beamsplitter:<\/p>\n

\"i-17ad476f7826e8275f689715700e5783-neil_AB.png\"<\/p>\n

(Apologies for the big clunky graphics, but I don’t have a better way of getting these in quickly.)<\/p>\n

The wavefunction at A involves three exponential factors, one for the phase shift of π\/2 that the beam gets on reflection from the first beamsplitter, one for the phase shift φ due to interaction with the environment, and one factor eikL<\/i><\/sup> due to the propagation of the wave. The wavefunction at B has not been reflected or shifted, so it just gets the eikL<\/i><\/sup> for the propagation. To keep things simple, we’ll assume all of the path lengths are equal.<\/p>\n

The second beamsplitter combines these two beams to form the outputs at C and D that are usually the end of a Mach-Zehnder experiment. Taking these one at a time, we have:<\/p>\n

\"i-0e0f3b429fd2393968968e64464e7474-neil_C.png\"<\/p>\n

The wavefunction at C is the combination of the wavefunction from A with an additional phase shift of π\/2 due to the reflection, and the wavefunction from B with no additional phase shift. When you add these together, the two phase shifts of the A beam give a total phase shift of π, which is a factor of -1. Taking the norm of the wavefunction, to get the probability of detecting the light at point C, we end up with a sinusoidal function of the phase shift φ, exactly as we expect for an interference experiment.<\/p>\n

The same process at D gives:<\/p>\n

\"i-08172a23d3654457da962c24d1d258a3-neil_D.png\"<\/p>\n

Here, the extra reflection phase shift goes on the B wavefunction, and as a result, we get a cosine rather than a sine, giving us the required phase difference between the two outputs. No matter what the phase shift φ is, the two sum to one, and when one is at a maximum, the other is at a minimum. That’s interference, which is unmistakable wave behavior.<\/p>\n

If there were no shift, that is, φ=0, the wavefunction would be exactly 1 at D and exactly 0 at C. If you think about this process in a particle picture of light, rather than a wave picture, this means that each individual photon entering the interferometer would always end up at D. This is very different from the classical particle picture, which would have the particle end up at C 50% of the time and D 50% of the time.<\/p>\n

“Decoherence” is the name given to the process by which random interactions with the environment destroy the coherent interference that leads to the quantum result, and push the system back toward the classical 50\/50 result. In the particle picture, we can think of this as repeating the single-photon experiment many times, each time with a different value of φ. The final probabilities are obtained by adding together the results of many repeated measurements with individual photons.<\/p>\n

To account for the effects of decoherence mathematically, we would average over some range of φ depending on the strength of the environmental interactions. If the range of φ is large enough (2π or more), both sin2<\/sup> and cos2<\/sup> average to 1\/2, giving us the 50% probability of ending up at either detector that we expect for a classical situation. Thus, the random phase introduced by environmental interactions makes the quantum waves look like classical particles.<\/p>\n

(Strictly speaking, if you really wanted to investigate the effects of decoherence in a meaningful way, you would do this all with density matrices rather than wavefunctions. There are limits to the amount of mathematical effort I’m willing to put into this, though.)<\/p>\n

But Neil thinks he can prove something by adding an additional beamsplitter, so let’s plunge ahead to point E:<\/p>\n

\"i-c67b1b8f86ce851c3d4937ec443a2526-neil_E.png\"<\/p>\n

The wavefunction at E is a combination of the wavefunction from C with an additional reflection phase shift, plus the wavefunction from D with no extra shift. When you add these together, something odd happens– the terms depending on φ cancel out, leaving you with only a single term in the wavefunction. Without the second term that was present at C and D, there is no interference– when you take the norm of the wavefunction, all the complex phases cancel out, and you’re left with just a constant probability of 1\/2.<\/p>\n

So, what happens at F? Well, the phase shift due to reflection is on the other path, but the net effect is similar:<\/p>\n

\"i-e5e2938d3128c8fccdb7d5bd9fbc80d8-neil_F.png\"<\/p>\n

Here, we find that the constant terms cancel out, leaving only the term that depends on φ. But again, without a two-component wavefunction, there is no interference. When we take the norm of the wavefunction, the complex phases cancel, and we end up with 1\/2 again.<\/p>\n

What’s going on, here? Loosely speaking, the amplitudes of the wavefunctions at C and D are out of phase with one another by π\/2. If you do a little bit of algebra on the equations above, you can easily make ΨC<\/sub> look like a sine function, and ΨD<\/sub> look like a cosine. When you recombine those two, the additional phase shift due to reflection puts them a full π out of phase with each other– the sine becomes a -cosine, or the cosine becomes a -sine. When you add those together, they destructively interfere, wiping out the earlier interference effect. It’s true that, as Neil claims, the phase shift introduced by decoherence doesn’t matter for the final result, but that’s because there is no interference<\/em>.<\/p>\n

(This depends on the path lengths all being equal. If there’s an overall path length difference, you should still get interference, but the math gets nasty. Update:<\/strong> It does not depend on the reflectivity of any of the beamsplitters– all you get from unequal intensity is a lowering of the contrast in the interference signal at C and D.)<\/p>\n

Where’s Neil’s mistake? He fundamental mistake is that he’s too attached to his notion that decoherence is wrong, and too enamoured of his own cleverness. When his incorrect math seemed to align with his pet notion, he grabbed onto that immediately, without checking it over.<\/p>\n

Mathematically, though, I believe he screwed up the reflection phase shift, but I can’t be bothered to sort through his garbled notation to figure it out for sure. When you actually work out the details, his conclusions are wrong, and the normal quantum understanding of the world is correct.<\/p>\n

And here’s an important note: I am not good at this sort of thing. I’m an experimentalist, not a theorist, and I do not do research in quantum foundations. The people who do this sort of thing professionally are vastly smarter than I am, and the chances that they somehow missed an effect as simple as adding a third beamsplitter to a standard Mach-Zehnder interferometer are totally negligible. If you see some outsider like Neil claiming to have blown away the last few decades of quantum foundations research, remember that.<\/p>\n

This post serves two purposes: first, it’s a public record of the error of Neil’s “disproof” of decoherence. Not that I expect this to shut him up– he’s never bothered to read and understand anything I’ve written on the subject of quantum mechanics before, so I don’t expect him to start now. The next time he starts touting this “falsification of decoherence,” though, you can point to this to show that he’s wrong.<\/p>\n

It is also a public statement that I am done dealing with Neil Bates. The errors in math and reasoning in Neil’s post are the sort of thing I expect from a sophomore physics major, and when I grade sophomore level work, I get paid for it. (Old academic joke: “I teach for free. They pay me to grade.”) So, until and unless Neil sends me $160, I will not be reading anything else he writes, in comments here or elsewhere.<\/p>\n

Goodbye, Neil. I’d say it’s been fun, but really, it hasn’t.<\/p>\n","protected":false},"excerpt":{"rendered":"

That’s the bill for the time that I spent on deciphering his supposed falsification of decoherence. I don’t want anyone to fall for his false argument, so here’s the correct explanation of the scenario, to save other people the trouble. The center of his so-called “proof” is this modified Mach-Zehnder Inteferometer: Light enters at the… Continue reading Neil Bates Owes Me $160<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"1","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7,23,11],"tags":[],"class_list":["post-4459","post","type-post","status-publish","format-standard","hentry","category-physics","category-quantum_optics","category-science","entry"],"_links":{"self":[{"href":"http:\/\/chadorzel.com\/principles\/wp-json\/wp\/v2\/posts\/4459","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/chadorzel.com\/principles\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/chadorzel.com\/principles\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/chadorzel.com\/principles\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"http:\/\/chadorzel.com\/principles\/wp-json\/wp\/v2\/comments?post=4459"}],"version-history":[{"count":0,"href":"http:\/\/chadorzel.com\/principles\/wp-json\/wp\/v2\/posts\/4459\/revisions"}],"wp:attachment":[{"href":"http:\/\/chadorzel.com\/principles\/wp-json\/wp\/v2\/media?parent=4459"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/chadorzel.com\/principles\/wp-json\/wp\/v2\/categories?post=4459"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/chadorzel.com\/principles\/wp-json\/wp\/v2\/tags?post=4459"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}