From: utzoo!decvax!harpo!eagle!mhtsa!ihnss!ihuxv!lew
Newsgroups: net.misc
Title: derivation of curve in Sci Am QM article
Article-I.D.: ihuxv.123
Posted: Wed May 12 22:52:01 1982
Received: Thu May 13 03:49:47 1982

Here is a derivation of the curve on pg 174 of "The Quantum Theory and
Reality" by Bernard d'Espagnat in Nov '79 Sci Am.

Say observers A and B observe particles a and b. Suppose the particles
are emitted along the y-axis and the spins are measured in the x-z plane.
Let the states |a+>, |a->, |b+>, |b-> be the states in which particle
a(b) is always observed with spin up(down) ALONG THE Z-AXIS.
The expectation value of a spin measurement is given by the bra-ket of the
spin operators Sza, Sxa, Szb, Sxb. Since the one particle states are
eigen states of the Sz operators:

Sza|a+> = |a+> ; Sza|a-> = -|a-> ; Szb|b+> = |b+> ; Szb|b-> = -|b->

(We are really using the spin operators times 2 to get rid of factors of 1/2)

>From the Pauli matrix for Sx we find:

Sxa|a+> = |a-> ; Sxa|a-> = |a+> ; Sxb|b+> = |b-> ; Sxb|b-> = |b+>

The particles are emitted in the singlet state:

        |emit> = ( |a+>|b-> - |a->|b+> ) / sqrt(2)

so the correlation of spin measurements by A along the z-axis, with
those by B along an axis making an angle u with the z-axis is given by:

 corr(u) = 

But  == 0 since  ==  ==  == .

This gives  corr(u) = cos(u) *  =

cos(u)( - +  +  -  )/2

or simply -cos(u) .  Compare this with the graph. Why isn't the QM curve
on the graph equal to -1 at the origin?

					Lew Mammel Jr. - BTL Indian Hill