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- Newsgroups: sci.physics
- Path: sparky!uunet!well!sarfatti
- From: sarfatti@well.sf.ca.us (Jack Sarfatti)
- Subject: re: Ramsay's objections to Sarfatti's FTL expt.
- Message-ID: <Bxtu21.GEL@well.sf.ca.us>
- Sender: news@well.sf.ca.us
- Organization: Whole Earth 'Lectronic Link
- Date: Mon, 16 Nov 1992 20:54:00 GMT
- Lines: 165
-
-
- *Sarfatti comments*
- Subject: Re: Ramsay's questions to Sarfatti
- References: <BxqM80.KDw@well.sf.ca.us>
-
- 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
-