Quantum Entanglement Explained by Jacobson Radical + PI

From: Osher Doctorow (mdoctorow_at_comcast.net)
Date: 07/09/04 

Date: Fri, 9 Jul 2004 01:17:21 +0000 (UTC)

 From Osher Doctorow mdoctorow@comcast.net

To see how PI and the Jacobson radical explain Quantum Entanglement,
let's start with three definitions of which the first two are based
on equations that apply under actual particular conditions.

1) P(A U B)_I = P(A) + P(B) - P(A)P(B)

2) P(A U B) = P(A) + P(B) - P(AB)

3) P(A U B)_M = P(A) + P(B)

Equation (2) always applies in probability-statistics provided that
P(AB) is not 0 or undefined. Equation (3) holds always when A, B
are mutually exclusive events (they can't both occur simultaneously
or together) and applies to "conjugate" quantum events which are not
even simultaneously defined supposedly. The symbol _ indicates
a subscript, and _I is the subscript "independent" which stands for
statistically independent - when A, B are statistically independent
in the sense of not affecting each other as in tosses of widely
separate coins, then (1) holds. The symbol _M refers to mutually
exclusive.

Now it is rather easy to prove the Lemma:

LEMMA. P(A U B) < = P(A U B)_I < = P(A U B)_Q

if A, B are positively quadrant dependent, that is to say P(AB) > =
P(A)P(B) in the sense of Lehmann (1967) - at least in the case of
P(A U B) on the far left hand side of the inequality. (Here Q
replaces M since we refer to quantum probability.)

PROOF. P(A U B)_Q = P(A) + P(B) > = P(A) + P(B) - P(A)P(B) = P(AUB)_I
> = P(A) + P(B) - P(AB) since all the probabilities are nonnegative
and since -P(A)P(B) > = -P(AB) iff P(AB) > = P(A)P(B) which is assum-
ed with positive quadrant dependence. Q.E.D.

REMARK 1. P(A U B)_I is just x o y, the Jacobson radical, with x =
P(A), y = P(B). So the probability of A and/or B for two statistical-
ly independent events A, B is the Jacobson radical. Notice this
further lends support to the "pathological" description of the
Jacobson radical since A, B in P(A U B)_I don't influence each other
in the statistically "dependent" sense.

REMARK 2. Since according to the Lemma, quantum probability is on
the opposite side of the Jacobson radical from Probable Influence
(PI), and since PI is non-pathological while the Jacobson radical
is pathological, it follows that quantum probability is patho-
logical. However, from the laser bombardment that produces twin
particles supposedly imitating each other in quantum entanglement,
it cannot be supposed any longer that the intersection of the two
wave-particles is either undefined or empty, and since they are
also not statistically independent since they imitate each other,
 from the Lemma cases P(A U B)_Q and P(A U B)_I are excluded and we
are left with P(A U B) which is P(A' --> B) from P(A' --> B) =
P{(A')') + P(AB) = P(A) + P(AB) = P(A --> B). Quantum entanglement
is a case of ordinary Probable Influence (PI).

Osher Doctorow

 

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