NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/sci/physics/19118

Text File | 1992-11-17 | 13.8 KB | 347 lines

Newsgroups: sci.physics
Path: sparky!uunet!well!sarfatti
From: [e-mail] (Jack Sarfatti)
Subject: re:Ramsay's confusions on FTL signal expt.
Message-ID: <[e-mail]>
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Organization: Whole Earth 'Lectronic Link
Date: Wed, 18 Nov 1992 04:27:51 GMT
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Sarfatti responds to:
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
Lines: 215
Message-ID: <[e-mail]>
References: <[e-mail]>
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 <[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.

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

In article <[e-mail]> [e-mail]
(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.

*I don't deny it. What point are you trying to make. This appears to be
irrelevant rambling.*

|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.

*I don't deny it. There is no reason why, in the one-photon problem with
zero connection that arg<a,e,+|a,o,+> cannot be different for every
different superposition in order to preserve unitarity. Why do you
automatically assume invariance of arg<a,e,+|a,o,+> - what law of physics
requires such invariance?*

|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.

*OK:

|psi> = [|a,e,+> + |a,o,->]/sqrt2

|psi>' = U(e,phi)U(o,1/2)|psi> = [e^iphi|a,e,+> + |a,o,+>]/sqrt2

We are using a Mach-Zehnder interferometer, so |psi> is sent to a beam
splitter giving

|psi>'' = U(splitter)|psi>' = {[e^i(phi+argr) + e^iargt]|a,1>

                              + [e^i(phi+argt) + e^iargr]|a,2>}/2

The responses of the two counters 1 and 2 are

p(1) = |<a,1|psi>''|^2 = [1 + cos(phi + argr - argt)]/2

     = [1 + cos(phi + pi/2)]/2 = [1 - sin(phi)]/2


p(2) = |<a,2|psi>''|^2 = [1 + cos(phi + argt - argr)]/2

     = [1 + cos(phi - pi/2)]/2 = [1 + sin(phi)]/2

so, now, Ramsy, if your argument was right we would have to conclude that
phi was a random variable so that <sin(phi)> = 0 and "fringes" would never
be seen.

AHA! I think I understand your confusion - your physical confusion! You
keep referring to f(x) and g(x) in double slit experiment! The difference
is that here, in the Mach-Zehnder x is phi! so that f(phi) and g(phi) - and
summing over different x is like resetting the phase plate phi and summing.
But, in one double slit experiment we see many x all at once. This is
different from a Mach-Zehnder in each experiment we see only one phi - not
all phi over 2pi. That is why

p(1) + p(2) = 1 for a fixed phi! We do not need to integrate over phi to
conserve probability. True, there is an analogy between 2-slit and Mach-
Zehnder but you are garbling the analogy. In the Mach-Zehnder you pick a
fixed phi by setting the phase plate - then you detect relative count rates
of the two detectors. Then, you re-set the phase plate, measure count rates
again, and so on until you cover 2pi. Doing all those separate MZ runs (all
phi) is equivalent to looking at entire screen (all x) in 2-slit. An
ensemble of MZ experiment is equivalent to one 2-slit experiment. This is
the seat of the miscommunication between us.*


|*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,

*Yes, but that does not mean that the result of the action of the operators
is the same on all states of the state space. In particular the result of
the action of U(e,phase plate)U(o,1/2 wave plate) on z1|a,e,+> + z2|a,o,->
is not the same as its result on [|a,e,+>|b,e,+> + |a,o,->|b,o.->}/sqrt2.*

which means talking about states which you're not planning to use or see in
the experiment.

*May be so, but these unused states do not influence results of my
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.

*Why? You just can't claim it without reason. You have given no reason.
Even your argument for one-photon superposition is not mathematically
correct, and even if it were, you cannot apply it to the two-photon
problem. Your argument here, Ramsay, is irrational, not at all demanded by
the principles of standard quantum mechanics.*

(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.

* It is supported by simple physical reasoning using Feynman histories
approach. First, the half-wave plate erases the polarization distinction
between the two paths, and the recombination of the paths erases the space
distinction. Therefore, there is interference between the two corresponding
pair amplitudes detectable locally for photon b, though not for photon a.*

|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.

*We keep going round in circles. Where else is the photon detected? There
is only one counter. It is detected in that counter. Are you talking about
summing over different points of surface of counter? If so, that will not
destroy the phase information - if it did, then Wheeler's "delayed choice"
experiment would not work either.

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.

*But, I say it is not possible. Why do you say it is possible?*


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.
*I ask you to do the same.*

PS on the density matrix formalism proof of no-go FTL - the key is
sum(a) - yes in a two slit experiment where a = x position on screen
each two slit experiment sums over all x. this is very different from
a Mach-Zehnder type experiment in which each experiment is for a fixed phi
(now a is pectrum of phi) - the sum over phi is for an ensemble of MZ expts.
Therefore, the density matrix proof is the right answer to the wrong
question. Yes, if one uses a pair photon version of Mach-Zehnder - which I
do-
then for a fixed phi at transmitter one will get phi dependent responses
at the receiver - if averages the data over all phi one will erase the
signal - but that is a stupid unnecessary thing to do/ You can garble any
signal if you're stupid.