Science > Physics > Probable Influence Type 2 in Quantum Theory Part 2
| Topic: |
Science > Physics |
| User: |
"OsherD" |
| Date: |
15 Jun 2005 09:53:18 AM |
| Object: |
Probable Influence Type 2 in Quantum Theory Part 2 |
From Osher Doctorow
So I've shown in Part 1 that the "deterministic" school of quantum
theory contradicts itself in its claim that the Schrodinger equation is
deterministic while "containing" the Born quantum probability ww*. In
the process, I defined a third type of PI, P*(A-->B), namely P(B) -
P(A) when P(B) > = P(A), which expresses a non-deterministic causation
or influence of A on B even when P(A-->B) = 1 + P(AB) - P(A) and
P'(A-->B) = 1 + P(B) - P(A) are both 1 because A is a subset of B.
Do we still retain one of the advantages of PI over conditional
probability, namely that PI is a probability on a set/event rather than
a ratio of probabilities? In other words, do we have:
1) P*(A-->B) = P(B) - P(A) = P(C)
for some set/event C? Of course, C can be defined "backwards" as one
of the undoubtedly many sets whose probability is P(B) - P(A), but what
we really have in mind is some way of relating C to intuition in the
way that A U B (A "and/or" B), AB (the intersection of A and B), and so
on are related, e.g., by Venn Diagrams.
Geometrically, the solution is simple. Since P(B) > = P(A), just move
the Venn diagram of A inside the Venn diagram of B, and P(BA') (the
probability of the part of B outside A) is equal in magnitude to P(B) -
P(A). Then separate the Venn diagrams in case A isn't a subset of B,
but retain the magnitude obtained by the above process. If this seems
strange, then you haven't looked at quantum erasure in computer
computer theory!
Osher Doctorow
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| User: "OsherD" |
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| Title: Re: Probable Influence Type 2 in Quantum Theory Part 2 |
15 Jun 2005 10:00:00 AM |
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From Osher Doctorow
I notice that P(BA' ), the probability of the part of B outside A,
didn't come out clearly in the type in the last paragraph of the
previous posting, because I put the parentheses directly after A' to
get P(BA'). So for those like me whose vision isn't perfect, it's
supposed to be P(BA' ).
Osher Doctorow
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