lew (05/13/82)
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) = <emit| Sza ( cos(u)*Szb + sin(u)*Sxb ) |emit> But <emit| Sza Sxb |emit> == 0 since <a+|a-> == <a-|a+> == <b+|b-> == <b-|b+>. This gives corr(u) = cos(u) * <emit| Sza Szb |emit> = cos(u)( -<a+|a+><b-|b-> + <a+|a-><b-|b+> + <a-|a+><b+|b-> - <a-|a-><b+|b+> )/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