Quantum Gravity 168.6: Portuguese Relate Quantum Phase Transitions to Probability



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Topic: Science > Physics
User: "OsherD"
Date: 01 Aug 2007 12:47:01 AM
Object: Quantum Gravity 168.6: Portuguese Relate Quantum Phase Transitions to Probability

From Osher Doctorow

Nikola Paunkovic and Vitor Rocha Vieira of Instituto de
Telecommunicacones (ISC) Technical U. Lisbon Portugal have
respectively 7 and 23 papers in arXiv, and in their latest paper
"Macroscopic distinguishability between quantum states defining
different phases of matter: fidelity and the Uhlmann geometry phase,"
arXiv: 0707.4667 v1 [quant-ph] 31 Jul 2007, 18 pages, they show that
phase transitions are marked by a sudden drop of fidelity near regions
of criticality in two thermal phase transitions, Stoner-Hubbard model
for magnetism and BCS theory of superconductivity. They point out
that this should not be surprising in effect since to name only a few
examples, macroscopic quantities of magnetism, superconductivity, and
superfluidity can only be explained by using Quantum Mechanics.
Where probability comes in, and from our viewpoint Probable Causation/
Influence (PI), is the fact that four pure states fidelity reduces to
sqrt(P(system in state |w2> passes test of being in state |w1>), or in
simple notation sqrt(P(w2|w1)).
But for constant P(A) not equal to 0, it turns out that P(A-->B) (PI,
in other words) is linearly proportional to P(B|A), so the same sudden
drop of fidelity arguably is signalled by a sudden drop in P(A-->B).
To prove this, look at the definitions:
1) P(A-->B) = 1 + P(AB) - P(A)
2) P(B|A) = P(AB)/P(A) if P(A) is not 0
Solve (1) and (2) for P(A) and equate them, yielding:
3) P(A-->B) + P(A) - 1 = P(B|A)P(A)
and solving (3) for P(A-->B) results in:
4) P(A-->B) = 1 - P(A) + P(A)P(B|A) = k1 + k2P(B|A) when P(A) is
constant, defining k1 = 1 - P(A), k2 = P(A).
Osher Doctorow
.


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