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audioloop said:travis, do you believe in CFD ?
Interesting question.
The first thing I'd say is: who cares? If the topic is Bell's theorem, then it simply doesn't matter. CFD *follows* from locality in the same way that "realism" / hidden variables do. That is: the only way to locally (and, here crucially, non-conspiratorially) explain even the perfect correlations is with a "realistic" hidden-variable theory with pre-determined values for *all* possible measurements, i.e., a model with the CFD property. So... to whatever extent somebody thinks CFD needs to be assumed to then derive a Bell inequality, it doesn't provide any kind of "out" since CFD follows from locality. That is, the overall logic is still: locality --> X, and then X --> inequality. So whether X is just "realism" or "realism + CFD" or whatever, it simply doesn't make any difference to what the correct answer to this thread's poll is.
So, having argued that it's irrelevant to the official subject of the thread, let me now actually answer the question. Do I believe in CFD? I'm actually not sure. Or: yes and no. Or: it depends on a really subtle point about what, exactly, CFD means. Let me try to explain. As I think everybody knows, my favorite extant quantum theory is the dBBB pilot-wave theory. So maybe we can just consider the question: does the pilot-wave theory exhibit the CFD property?
To answer that, we have to be very careful. One's first thought is undoubtedly that, as a *deterministic* hidden variable theory, of course the pilot wave theory exhibits CFD: whatever the outcome is going to be, is determined by the initial conditions, so ... it exhibits CFD. Clear, right?
On the other hand, I've already tried to make a point in this thread about how, although the pilot-wave theory assigns definite pre-existing values (that are then simply revealed in appropriate measurements) to particle positions, it does *not* do this in regard to spin. That is, the pilot-wave theory is in an important sense not "realistic" in regard to spin. And that starts to make it sound like, actually, at least in regard to the spin measurements that are the main subject of modern EPR-Bell discussions, perhaps the pilot-wave theory does *not*, after all, exhibit CFD.
So, which is it? Actually both are true! The key point here is that, according to the pilot-wave theory, there will be many physically different ways of "measuring the same property". Here is the classic example that goes back to David Albert's classic book, "QM and Experience." Imagine a spin-1/2 particle whose wave function is in the "spin up along x" spin eigenstate. Now let's measure its spin along z. The point is, there are various ways of doing that. First, we might use a set of SG magnets that produce a field like B_z ~ B_0 + bz (i.e., a field in the +z direction that increases in the +z direction). Then it happens that if the particle starts in the upper half of its wave packet (upper here meaning w.r.t. the z-direction) it will come out the upper output port and be counted as "spin up along z"; whereas if it happens instead to start in the lower half of the wave packet it will come out the lower port and be counted as "spin down along z". So far so good. But notice that we could also have "measured the z-spin" using a SG device with fields like B_z ~ B_0 - bz (i.e., a field in the z-direction that *decreases* in the +z direction). Now, if the particle starts in the upper half of the packet it'll still come out of the upper port... *but now we'll call this "spin down along z"*. Whereas if it instead starts in the lower half of the packet it'll still come out of the lower port, but we'll now call this *spin up along z*.
And if you follow that, you can see the point. Despite being fully deterministic, what the outcome of a "measurement of the z-spin" will be -- for the same exact initial state of the particle (including the "hidden variable"!) -- is not fixed. It depends on which *way* the measurement is carried out!
Stepping back for a second, this all relates to the (rather weird) idea from ordinary QM that there is this a correspondence between experiments (that are usually thought of as "measuring some property" of something) and *operators*. So the point here is that, for the pilot-wave theory, this correspondence is actually many-to-one. That is, at least in some cases (spin being one of them), many physically distinct experiments all correspond to the same one operator (here, S_z). But (unsurprisingly) distinct experiments can have distinct results, even for the same input state.
So... back finally to the original question... if what "CFD" means is that for each *operator*, there is some definite fact of the matter about what the outcome of an unperformed measurement would have been, then NO, the pilot-wave theory does *not* exhibit CFD. On the other hand, if "CFD" means that for each *specific experiment*, there is some definite fact of the matter about what the outcome would have been, then YES, of course -- the theory is deterministic, so of course there is a fact about how unperformed experiments would have come out had they been performed.
This may seem like splitting hairs for no reason, but the fact is that all kinds of confusion has been caused by people just assuming -- wrongly, at least in so far as this particular candidate theory is concerned -- that it makes perfect sense to *identify* "physical properties" (that are revealed or made definite or whatever by appropriate measurements) with the corresponding QM operators. This is precisely what went wrong with all of the so-called "no hidden variable" theorems (Kochen-Specker, etc.). And it is also just the point that needs to be sorted out to understand whether the pilot-wave theory exhibits CFD or not. The answer, I guess, is: "it's complicated".
That make any sense?