From: sarfatti@well.sf.ca.us (Jack Sarfatti)
Subject: re: Ramsay's objections to Sarfatti's FTL expt.
Organization: Whole Earth 'Lectronic Link
Date: Mon, 16 Nov 1992 20:54:00 GMT


*Sarfatti comments*
Subject: Re: Ramsay's questions to Sarfatti

Organization: University of British Columbia, Vancouver, B.C., Canada


I asked:
|What exactly is it that you are saying is invalid about the notation
|you've been using?  Why was it a mistake for you to represent the
|action of the phase plate upon the "A" photon as a unitary operator
|U(1/2) (which you now replace with two operators)?

In article <BxqM80.KDw@well.sf.ca.us> sarfatti@well.sf.ca.us
(Jack Sarfatti) writes:
|Because physically there are different pieces of equipment in the two
|paths. Each piece is represented by a local unitary operator.

This is not very much of an explanation.

*Of course it is.*

What constitutes a separate "piece of equipment"? Suppose we weld all
of the mirrors and such which are in the "transmitter" end of the
apparatus to the lab bench, so as to make them one solid object. In
what sense are you *obliged* to treat them as "separate", and use
multiple unitary operators in the way you now plan to do? What was
wrong, in your mind, with regarding them as producing one unitary
evolution of the photon-- aside from its predicting that your "effect"
is not there?

*Then you tell me. How would you do it? How do you describe:
1. action of a variable phase plate on a one-photon ket.
2. action of a half-wave plate on a one-photon ket.*


Suppose we consider just the transmitter end of the apparatus, and
"feed" it a photon in prepared in a state z1|a,+>+z2|a,->, where
|z1|^2+|z2|^2=1 for normalization. You have the `+' polarized state
|a,+> evolve into |a,e,+> and the `-' state evolve into a state
|a,o,+> (by separating it off and changing its polarization).

*OK*

The bracket of this evolved state z1|a,e,+> + z2|a,o,+> with itself is

        |z1|^2 <a,e,+|a,e,+> + |z2|^2 <a,o,+|a,o,+>
        z1z2*  <a,e,+|a,o,+> + z2zq*  <a,o,+|a,e,+>

       =  1 + 2 Re{ z1 z2* <a,e,+|a,o,+>}.

*OK*

Now, this evolved state represents a state of the photon, so it has to
be normalized: the above has to be 1. The only way to do this for all
relevant choices of z1 and z2 is for <a,e,+|a,o,+> = 0. Any other
assumed value of that bracket yields real problems.

*Mathematically false, Ramsay! Yes,as I wrote several times already:

<a,e,+|a,o,+> = 0 is a sufficient condition but not necessary! Another
possible solution is:

Re{ z1 z2* <a,e,+|a,o,+>} = 0

that is,

argz1 - argz2 + arg<a,e,+|a,o,+> = pi/2

IF we are dealing with a one-photon coherent superposition. However, and
this is crucial - in the actual problem of quantum connection communication
we are not dealing with a one-photon coherent superposition but with an
initially entangled photon-pair - this makes an enormous difference in the
mathematics!  Specifically z1 and z2 are not c-numbers as they are in the
one-photon case that you describe above. Rather,

z1 -> z1|b,e,+>

z2 -> z2|b,o,->

and <b,o,-|b,e,+> = 0

Therefore,

Re{ z1 z2* <a,e,+|a,o,+>} is replaced by

Re{ <b,o,-|b,e,+>z1 z2* <a,e,+|a,o,+>}

which is automatically zero without any unitarity constraint on

argz1 - argz2 + arg<a,e,+|a,o,+>

which is entirely an artifact of the one-photon problem with zero
connection. The pair connection removes that particular unitarity
constraint.

The main reason your critique is wrong is that you have not sufficiently
distinguished the essential mathematical difference between the one-photon
problem with zero connection and the two-photon problem with non-zero
connection (entanglement).*


The reason that "probability is conserved" for your pair of photons,
with your calculations, is that you have two such errors compensating
each other. The magnitude of one component is decreased and the
magnitude of another is increased, by exactly this same spurious
quantity. If you fed the apparatus just one component or the other, as
shown here, you'd find that "probability is conserved" breaks down for
your calculations. If, however, you believe what the calculations
really were telling you-- and this was a mathematical consequence of
things which you wrote down-- that <a,e,+|a,o,+> remains 0, there is
no problem.

*What you say here is all wrong for reasons I just gave.  What do you mean
by "one component"? Do you mean the one-photon problem which is physically
irrelevant. Do you mean putting only e beam thorugh -blocking off o beam?
You seem to ignore the "wholistic" nature of quantum mechanics - changing
the conditions of an experiment changes the experiment as Bohr stressed. I
think, Ramsay, you are arguing aganst my model by comparing apples to
oranges.*

The only justification you've given for asserting that the states
describing photons in the two beams, immediately before being detected
in the detector, have a non-zero bracket with each other is that they
are detected in the same "spot" roughly.

*No that's not the only justification. 1) I gave above a demonstration that
the math does not require <a,e,+|a,o,+> = 0 in the pair case, not even in
the one-photon case. Furthermore, on physical grounds, the kets are simply
another way to describe Feynman histories, so on physical grounds since the
polarization distinction is erased by the half-waveplate, and since both
beams will , either be sent through a recombiner, or detected by one
counter - clamped so that individual recoils are not detected, the
amplitudes for the two "indisitinguishable alternatives" add coherently!
To avoid confusion, lets restrict ourselves to the relevant photon pair
case. I am then talking about adding two pair amplitudes coherently. Each
amplitude has the receiver photon b arrive at the same counter, say e'+
while its twin transmitter photon a either takes e or o path to the same
fixed transmitter counter.*


This is still a very handwavy, and incorrect, treatment of the optics. If
you squint your eyes, the two states arriving in the detector look a lot
alike. But they are distinguishable by their momenta-- so they couldn't
possibly be parallel or differ only by a phase.

*False! Read Bohr's account of his discussion with Einstein on this very
question. If the counter is clamped there is no recoil measurement that can
distinguish different momenta! Also |a,e(o),+> are primarily spin states.
When the counter is clamped - recoil taken up by entire earth so to speak-
the space-part of the photon state is a coherent superposition of momentum
eigenstates like a converging spherical wave if it is focused by a lens to
the counter - so there is no collapse into linear momentum recoil states.
L.S. Bartell (Dept. of Chem, U Mich Ann Arbor published papers in Physical
Review on all this in late 70's.*

You are now having to make this rather arbitrary "adjustment" in your
discussion of how the "transmitter" photon is affected by passing
through the optical equipment in order to cover for it.

*No, I think your physical picture is not right.*

Keith Ramsay
ramsay@unixg.ubc.ca