reilly said:
You are right to point out the inadequacy of my density matrix example.
After a good bit of thought, I still conclude that your convincing argument about conditional probabilities is, in fact incorrect.
It all boils down to the requirement that if
P(X) = P(X|A) P(A) + P(X|B) P(B)
then events A and B must be disjoint and independent. That is, they cannot occur simultaneously, so there is no interference involved in the above probability scheme.
I think you misunderstand me. To address Fra's justified remark, I think that if one is going to say that something has "a probability interpretation", it has to be in one and the same probability space (one and the same event universe with Kolmogorov measure over it). It simply doesn't make sense, IMO, to talk about probabilities in different spaces when talking about something physical (eg. an actual, objective, physical event), because probability just means a description of our ignorance, and grossly justified by a frequentist repetition of the situation.
As such, when I say that the probability that you went through the left door of the building yesterday is 30%, then that means two things: it 1) describes my state of knowledge about that event and 2) it must correspond to a kind of "generic situation" compatible with all I know, that in 30% of the cases you go to the left door (and let's assume, 70% of the time you take the right door).
Of course, after asking you through which door you went, this probability changes, say, from 30% to 100% when you say "I went through the left door" (assuming you're not lying). But that is because my state of knowledge changes (namely, the additional phrase "I went through the left door" is added), so the set of generic events must now be that set where you also utter that phrase, and in THAT set, you ALWAYS go through the left door.
But all these events are part of one and the same probability space. The only thing that changes is the "generic subspace of events compatible with my knowledge", which is expressed in probability language with conditional probabilities.
If you have different probability spaces, you can always combine them in one bigger probability space, by adding a generic tag to each one. If Omega1 is a probability space, with probability measure p1 over it, and Omega2 is another probability space with probabiliity measure p2 over it, then it is possible to construct a new probability space,
Omega = Omega1 x {"1"} union Omega2x{"2"}
(I added the tags 1 and 2).
The probability measure p over Omega can be of different kinds, but we can, for instance, define p(x) = a p1 if x in omega1 and b p2 if x in omega2 ; when a + b = 1, then p will be again a probability measure. P(omega1) = a, and P(omega2) = b.
As such, we can recover the old measures as: p1(x) = P(x | x in omega1) and p2(x) = P(x | x in omega2).
It is in this sense that I understand that we work in ONE SINGLE probability space, and that it is in THIS unique global space that we cannot interpret the wavefunction, when it is at the height of the slits, as giving a probability density.
Indeed, let us assume omega1 as the universe of events of measured impacts on the screen. p1 is then nothing else but the wavefunction squared of the interference pattern. Right. But now we want to interpret the wavefunction at the height of the slits ALSO as a probability generating function. This would be then our Omega2 which consists of 2 events, namely "through slit 1" and "through slit 2", and p2 would then be 50 - 50.
It is the building of an overall probability space as I showed above, that doesn't work in this case. THIS is what I mean that you cannot see the wavefunction with a probability interpretation when it is "in between measurements".
In the quantum situation, the situation is very different. For all practical purposes there are three independent( well almost) distinct events for the two slit expt., and they are the initial experimental conditions, slit A open, or slit B open, or slits A and B open. Properly normalized and orthogonalized, the QM measurement pattern will be reproduced as the sum of conditional QM probs based on the three slit states.
This only works when considering only as events the "preparation" and the "measurement" events, and NOT when considering the "in between" states as EVENTS (necessary to give them a probability interpretation).
I agree of course with you that P(measured at x | slit situation 1) will give you the right answers. But we only consider "omega" here as consisting of "preparation" events and "measurement" events. This is fully in line with the Copenhagen view that you *cannot talk about what happens in between". And that "cannot talk" is not only about IGNORANCE, but simply about non-talkability! If it were *ignorance* then it could be seen as events in a probability space and that's what is impossible.
If the wavefunction at the height of the slit is to have an *ignorance* or probability interpretation - instead of NOT having an interpretation such as in Copenhagen - then it is to be seen as giving us a probability for "going through slit 1" and "going through slit 2".
And once we do that, we have completely "forgotten" the conditional probabilities of the experiment setup, simply because the wavefunction at the slits contains all "information" for going further.
Let us consider the following events: A = slit 1 open, B = slit 2 open, C = slit 1 & 2 open.
X = particle through slit 1, Y = particle through slit 2
U(x) impact at position x on screen
P(U(x) | A ) gives us a blob behind 1, P(U(x) | B) gives us a blob behind 2, and P(U(x) | C) gives us an interference pattern. That's what you get out of quantum mechanics, and out of Copenhagen. X and Y don't exist in the event space of Copenhagen.
But if we insist on having X and Y in there (probability interpretation of the wavefunction all the time), then the situation is the following:
P(U(x) |A and X) = P(U(x) | X). Indeed, if we KNOW that slit A was open, AND we know that the particle went through slit 1, then we can limit ourselves to just knowing that the particle went through slit 1.
(A and Y) = 0 so no conditional prob can be defined here
(B and X) = 0 idem
P(U(x) | B and Y) = P(U(x) | Y)
We know this also from QM: if we evolve the state "at slit 1" onto the screen, we will find the blob behind 1.
And now comes the crux:
P(U(x) |(C and X)) = P(U(x) |X). Indeed, IF WE KNOW that the particle went through slit 1, even if we opened slits 1 and 2, (for instance by measuring it at the slit), then we find simply the blob behind 1. In other words, knowing that the particle went through slit 1 is sufficient, and the "extra information" that both the slits are open doesn't change anything on the conditional probability.
And now we're home:
P(U(x) | (C and X) ) = P(U(x) | X ) = P(U(x) | (A and X))
In other words, when we count X and Y as events, and we take them conditionally, we don't care what were the preparations.
If that's true, thenP(X|C) * P(U(x) | (C and X)) + P(Y|C) P(U(x) | (C and Y) ) should be P(U(x) | C) and it isn't, because P(U(x) | (C and X) ) = P(U(x) | X ) = P(U(x) | A) and in the same way P(U(x) | (C and Y) ) = P(U(x) | B). P(X|C) = P(Y|C) = 0.5 (under the probability interpretation of the wavefunction)
so we should have: 0.5 P(U(x) | A ) + 0.5 P(U(x) | B) = P(U(x) |C), which is not true.