From: sarfatti@well.sf.ca.us (Jack Sarfatti)
Subject: re:Texas A&M inquiry on FTL controversy.
Organization: Whole Earth 'Lectronic Link
Date: Tue, 17 Nov 1992 20:17:31 GMT


From WMA6617@VENUS.TAMU.EDU Tue Nov 17 10:20:58 1992
Date:    Tue, 17 Nov 1992 12:21:15 -0600 (CST)
From: WMA6617@ZEUS.TAMU.EDU
Subject: Re: Ramsay's claim of Sarfatti contradiction 1=0.
To: sarfatti@well.sf.ca.us
X-Vmsmail-To: SMTP%"sarfatti@well.sf.ca.us"

In article <Bxp4I5.Bxy@well.sf.ca.us>, you write...
Dear sir,
I have jumped into this conversation in the middle.  Could you post or
E-mail a summary or a FAQ list which whould bring us late comers up to
date on this discussion.
Many Thanks
Bill Ashe
Stored Ion Lab
Texas A&M University

Sarfatti replies:

Basic idea, pair source,transmitter,receiver.
transmitter is calcite with half-wave plate in one path (o) and phase plate
(phi) in other path (e). Both paths brought together to one counter to make
an interferometer. Receiver is calcite ant angle theta to transmitter
calcite with two counters detecting o' and e' paths.

Pair source creates 2-photon entangled state

|a,b> = [|a,e,+>|b,e,+> + |a,o,->|b,o,->}/sqrt2

where, for example, |a,o,-> means photon a in o path with linear
polarization normally corresponding to that path. However, a half-wave
plate will cause |a,o,-> -> |a,o,+> which is photon a still on o path but
with polarization rotated 90 degrees from its normal value. Since the e and
o paths are going to be recombined erasing the space distinction, and since
the half-wave plate erases the polarization distinction, I say that |a,e,+>
and |a,o,+> are "parallel" differing only by a phase factor. That is,

<a,e,+|a,o,+> = |<a,e,+|a,o,+>|e^iarg<a,e,+|a,o,+>

|<a,e,+|a,o,+>| = 1

In contrast,
<a,e,+|a,o,-> = 0. However, Ramsay claims <a,e,+|a,o,+> = 0 as well.

Therefore, the distorted "disentangled" pair state is:

|a,b>' = [e^iphi|a,e,+>|b,e,+> + |a,o,+>|b,o,->}/sqrt2

       = [e^iphi|a,e,+>|b,e,+> + |a,e,+><a,e,+|a,o,+>|b,o,->}/sqrt2

       = [e^iphi|a,e,+>|b,e,+> + |a,e,+>e^iarg<a,e,+|a,o,+>|b,o,->}/sqrt2

       = |a,e,+>[e^iphi|b,e,+> + e^iarg<a,e,+|a,o,+>|b,o,->}/sqrt2


note that <a,b,|a,b> = 1 and '<a,b|a,b>' = 1 also.

Photons a and b in |a,b>' each have their own local pure states of zero
entropy which is not true for initial state |a,b>.

That is,

|a,b>' = |a>'|b>'

|a>' = |a,e,+>

|b'> = [e^iphi|b,e,+> + e^iarg<a,e,+|a,o,+>|b,o,->}/sqrt2

The local receiver detection probabilities are

p(b,e') = |<b,e',+'|b>'|^2 = [1 + sin(2theta)cos(phi - arg<a,e,+|a,o,+>]/2


p(b,o') = |<b,o',-'|b>'|^2 = [1 - sin(2theta)cos(phi - arg<a,e,+|a,o,+>]/2

Therefore, the quantum connection signal is

p(b,e') - p(b,o') = sin(2theta)cos(phi - arg<a,e,+|a,o,+>)

The transmitter detection probability is

p(a) = |<a,e,+|a,e,+>|^2 = 1

some red herrings:
1) the transformation |a,b> -> |a,b>' is not unitary - it obviously is.
2) local unitarity for arbitrary one-photon superpositions require that
cos(phi - arg<a,e,+|a,o,+> = 0. That's comparing apples to oranges - one
cannot take a reult in one photon Fock space with zero connection and
assume blindly that it is true in two-photon Fock space with non-zero
connection because the two situations have different total experimental
arrangements and must be analyzed separately (i.e., Bohr's principle).