From: ramsay@unixg.ubc.ca (Keith Ramsay)
Newsgroups: sci.physics
Subject: Re: Ramsay's objections to Sarfatti's FTL expt.
Date: 18 Nov 1992 00:44:03 GMT
Organization: University of British Columbia, Vancouver, B.C., Canada
NNTP-Posting-Host: unixg.ubc.ca
Summary: Sarfatti objected to his own notation-- which was correct.

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.

I replied:
|This is not very much of an explanation.

In article <Bxtu21.GEL@well.sf.ca.us> sarfatti@well.sf.ca.us 
(Jack Sarfatti) writes:
|*Of course it is.*

And later:
|*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.*

This is silly, Sarfatti. There *is* no problem with the way you were
representing the action of the phase plate as a unitary operator. The
two pieces of equipment can be represented by two operators, each
acting on the state of the photon. That agrees with quantum mechanics.
The reason you changed the notation is that it was starting to become
too clear, and the result wasn't agreeing with your claims about it.

I wrote:
|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).

Sarfatti:
|*OK*

I:
|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,+>}.

Sarfatti:
|*OK*

I:
|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.

Sarfatti:
|*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!

In another article:
|I forgot to add on your remark "for all relevant choices" - that only
|applies to one-photon problem, not at all relavant to actual photon
|pair problem, as I show in detail above. 

If your method of calculation were correct, it would be possible to
apply it to any prepared state of the photon, for the given
experimental set-up. It would be possible to apply it to the simpler
cases, where the photon states are not "entangled". If you're going to
talk about states such as

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

Then you'd better be able to talk about such a state as

    |a,b> = |a,e,+> [ z1 |b,e,+> + z2 |b,o,-> ]
           +|a,o,-> [ z3 |b,e,+> + z4 |b,o,-> ],

where normalization requires |z1|^2+|z2|^2+|z3|^2+|z4|^2=1.

If z3=z4=0, then this reduces to a state of two unentangled photons.
If z1=z2=0, likewise. You *must* be able to deal with those cases too,
and have the operators involved apply unitarily. It's simple raw
nonsense to deny it.

|Also, even for one-photon
|problem - the formal "all possible choices" suppose a kind of gauge
|symmetry which may or may not be physically correct. It has to be tested.

Standard quantum theory says that states can be superposed freely--
deny it if you like, but quit pretending that the result conforms to
anyone's theory other than the one of your own invention.

|For example, in one-photon case, send |a,e,+> and |a,o,+> through a
|beam recombiner with two counters getting each of the two output
|interferograms - if "fringes" are seen in the count rates, then your
|remark <a,e,+|a,o,+> = 0 is disproved experimentally in the one-photon
|problem! I'll show this in detail later.

I'll be waiting to see your attempt.

|*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.*

One has to be able to talk about the action of the operators on the
*whole* state-space, which means talking about states which you're not
planning to use or see in the experiment. In your calculation, you
also refer to the components-- so it must be acceptable to do so, or
you're up a creek. The superposition principle, which is implicitly
invoked everywhere in the calculation, only can make sense if it is
possible to talk about the components separately.

I wrote:
|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.

Sarfatti:
|*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. 

This is not a justification. (1) The claim is wrong: <a,e,+|a,o,+>=0
*is* required. (2) The logic is wrong: showing that something *else*
is not "required" doesn't support the claim that |<a,e,+|a,o,+>|=1. It
has to be supported on its own.

|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!

This is just an elaborate way of saying that they are detected in
roughly the same spot. Just failing to make a measurement doesn't make
the alternatives indistinguishable. There is a continuum of distinct
alternatives, depending upon where the photon ends up when it finally
is detected.

Recall your quotes from Isham's paper. 

|... and P(A,a) is operator that projects onto the subset a; for example...
|
|P(A,a) = Sum ai in a[|ai><ai|]                     (6.2.5)

In your case, "a" is allegedly the whole spectrum of A-- leaving no
photons uncounted at the "transmitter" end. What is P(A,a) in such a
case? It's the identity operator.

|If the measurement of A yields a result lying in a, any further predictions
|must be made using the density matrix
|
|p(a) = {P(A,a,t1)p(0)P(A,a,t1)}/tr[P(A,a,t1)p(o)]  (6.2.6)

If P(A,a,t1) is the identity, then p(a)=p(0).

|"Now let the system evolve until time t2 when a measurement of observable B
|is made..... the probability of finding B in a range b, ...conditional on A
|.. found to be in a at time t1, is
|
|Prob(B in b, t2|A ina,t1;p(0)) = tr(P(B,b,t2)p(a)) (6.2.8)"

If p(a)=p(0), then having made the (trivial) observation of A has no
effect upon the likelihood of observing a given value of the
observable B. It's just ridiculous to say that making a trivial
observation, one whose answer you know beforehand, can affect the
result of another observation, by rendering the alternatives
"indistinguishable".

I wrote:
|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.

Sarfatti:
|*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! 

As I explained in another post, it is enough that it be possible to
distinguish the states reliably. Of course the actual measurement may
be of momentum, or of point-of-impact, and measuring the one can
prevent one from going back to measure the other. But failing to
measure one or both of them doesn't make the alternatives
"indistinguishable".

Write coherently next time, or I'll not reply.

Keith Ramsay
ramsay@unixg.ubc.ca
Newsgroups: sci.physics