Each of those four “complete input-and-output configurations” has as its set of “corresponding Machine 1 and Machine 2 congfigurations,” I take it, the very same set: {A, B}.
But how can the probability of each of those complete input-and-output configurations be the product of the probabilities of the corresponding Machine 1 configuration (in this case, “A”) and Machine 2 configuration (in this case, “B”)? Suppose the probability of Machine 1 being in configuration “A” is 1/3. Suppose the probability of Machine 2 bing in configuration “B” is 1/3. This should mean that each of the four complete input-and-output configurations listed above should have a probability of 1/9. But that causes the four probabilities to add up to 4/9. Shouldn’t they add up to 1/9? (Since they are four exclusive possibilities given the particular set of Machine 1 and Machine 2 configurations, shouldn’t their probabilities add up to the probability of that particular Machine 1 and Machine 2 configuration?
Now that I asked that, I see I didn’t have to go through all the rigmarole. What I didn’t understand was simply this: You said that there could be a probability distribution where the probability of (sorry for not knowing the notation here) “X and Y and Z” occuring is the same as the probability of “X and Y and W” occuring, both of which are the same as the probability of “X and Y” occuring. This could happen if Z and W both have a probability of 1. But otherwise, this doesn’t seem possible to me.
What did I misunderstand?
Am I wrong to think that if A and B are exclusive and exhaustive possibilities given C, then the probabilities of A and B add up to the probability of C?
ETA: Oh. I bet I’m wrong to think that the product of the probabilities of the two machine inputs is the same as the probability of both inputs occuring. Something about independent and dependent probabilities here. Curse you, my ignorance of probability.
No – the ground state of an atom, a molecule, a harmonic oscillator, or a particle in a box (even one with walls at infinite potential) is not at absolute zero – the particle wavefunction has a finite spread in space and a corresponding spread in momentum, neither of which is infinite or infintesimal in extent. The ground state is the lowest lying energy state, which isn’t the state with no momentum.
see a book on quantum mechanics, or look up subjects like “harmonic oscillator” and look at the quantum mechanical version.
Well, it’s certainly fine by me. (Incidentally, I have a partially typed up resurrection still intended in reply to your “Does physics support realism about transcendental numbers?” thread, which I keep putting off completing in the vain hopes that I can get the work I actually have to do done first. But when it comes, it will be glorious…)
By a “Machine 1 configuration”, I mean a specification of both the input and the output at Machine 1, and likewise for a “Machine 2 configuration”. So a machine 1 configuration might look like “Machine 1: A red”. What you’ve given is not a set of Machine 1 and Machine 2 configurations; it’s just a set of possible input values.
By a “complete input-and-output configuration”, I mean a pair consisting of a Machine 1 configuration and a Machine 2 configuration. So, for example, a complete input-and-output configuration might look like this: (Machine 1: A red, Machine 2: B blue). Its component Machine 1 configuration is “Machine 1: A red” and its component Machine 2 configuration is “Machine 2: B blue”. If the distribution fixes the common causes, then it needs to be the case that P(Machine 1: A red, Machine 2: B blue) = P(Machine 1: A red) * P(Machine 2: B blue), and so on for every other complete input-and-output configuration.
That I think was the most major confusion, but I’ll address some of the other questions as well:
Certainly, this would occur just in case, conditioned on X and Y, it is the case that W and Z has probability 1. But this needn’t mean either W or Z has an unconditioned probability of 1. For example, take X, Y, Z, and W to all be “The (sole) coin I flip comes up heads”.
No, you are not wrong about that; this is (basically) correct. [That is, it’s correct so long as A is entirely contained within C and likewise B is entirely contained within C]
Yes, P(X and Y) = P(X) * P(Y) if and only if X and Y are probabilistically independent.
(Thus, to say that a distribution fixes the common causes is to say that it makes machine 1 configurations probabilistically independent from machine 2 configurations.)
Hi Indistinquishable, crossing over from the other thread.
The math isn’t that hard to deal with conceptually. I need to go over this some more to get all the terminology straight, and try to work out some of the details.
But in general you seem to be saying a local variable would add a factor that would change the statistical results, I guess because a degree of randomness has been removed. Am I at least looking in the right direction?
Actually I’m figuring things out a little more. I guess probability is a better term than statitistics. So I’m hearing that the probabilities don’t match up with conditions that would result from a local variable being present. I’m going to look things over more so I can form better questions. But I’m trying to discern how the local variable changes those conditions. I’m sensing that the definition of ‘local variable’ is based on a determination of the spin at the time of entanglement, not a choice that can appear to be random later at the time of the measurement.
If that doesn’t make sense at all, I just need more time to look at this. Then I can ask ‘less dumb’ questions.
That last statement used words that don’t seem to correspond in context. A better way to say it may be that if there was a ‘hidden local variable’, it couldn’t have any affect on the probability of an outcome. The nature of a hidden variable would be a deterministic result in the spins. If the spin were determined at the time of the entanglement, the measured results wouldn’t match the expected probability.
I am getting the sense that comparison of results from both particles, and the introduction of other factors in the definition of the probability can highlight that. And that sounds like the area I have to study up more on.
Veering off, I think I’m distracted by the simplistic idea that the hidden variable could cause the measured spin to be just as apparently random as without it.
Other explanations have presented a case that the there would be an appearance of information sharing that doesn’t conform to the probability of the outcome if the resulting spin was truly random instead of predetermined. I understand that concept, but I’m not sure how the hidden variable would have that effect.
I’ll let this stuff gestate in my brain for a while. I’m probably still finding the slots to stick each term and definition into, and the connections and conclusions won’t come until that basic organization is established. Sort of like walking into a crowded room full of people you don’t know in mid-conversation. After you identify the people, and who’s talking to who, in a particular order, you can start to make sense of the conversations.