NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/sci/physics/18933

Internet Message Format | 1992-11-15 | 3.6 KB

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From: [e-mail] (Keith Ramsay)
Newsgroups: sci.physics
Subject: Re: Ramsay's questions to Sarfatti
Date: 16 Nov 1992 03:11:06 GMT
Organization: University of British Columbia, Vancouver, B.C., Canada
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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 <[e-mail]> [e-mail]
(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.

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?

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).

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,+>}.

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.

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.

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. 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.

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.

Keith Ramsay
[e-mail]