🚀 go-pugleaf

The standard NO FTL signal proofs base on unitarity fail to include the
|possibility of interference using a ket |h> that is not orthogonal to
|either |f> or |g>.

No, they don't make such an assumption. You just think they do. (Ramsay
writes)

Ramsey, then show that they don't. JS

 Those proofs are trivial. They are the analog of the
|idea that if you observe which slit the photon passes there are no fringes
|on the screen!

No, they don't make that assumption. (Ramsay writesw)

Ramsay I did not say they made that assumption, only that they are an
'analog'.

They allow a quite arbitrary choice of observation at the "transmitter"
end. (Ramsay writes)

Ramsay, the basic points of difference are two: .1. can we collapse all (or
almost all) of the transmitter light into a single ket |h> with an
uncertainty in the phase difference arg<h|g> - arg<h|f> << 2pi. If we can,
then we get a connection signal in the mean at the receiver.  Now this
collapse to a single |h> is not a unitary evolution it is a non-unitary
collapse of two indistinguishable Feynman histories for the photon pair.
The "fringes" are seen for the receiver photon not for the transmitter
photon. Perhaps with clever optics and phase compensation plates we can
focus almost 100% of light from both recombiner output ports f and g into a
single diffraction limited spot (one cell in Gabor's photon phase space).

2. If the recombiner output relative phase noise is too big so we sum over
many spots (classical probability distribution for several |h> - one for
each diffraction limited spot - then we lose the connection signal in the
mean but can recover it by correlating intensity fluctuations from the two
receiver counters.

Again: the V' receiver counter photo-current I(V') will be proportional to

I(V') = x + ysin(2theta)cos(phi)<cos(arg<h|g> - arg<h|f>)>

the H' counter photo-current proportional to

I(H') = x - ysin(2theta)cos(phi)<cos(arg<h|g> - arg<h|f>)>

where <  > is a classical statistical average over set of spots or set of
kets |h>. The different macro spots of finite resolution each represent
distinguishable Feynman non-interfering histories.

Even if

<cos(arg<h|g> - arg<h|f>)> = 0

<cos^2(arg<h|g> - arg<h|f>)> is not zero.

Therefore the root mean square fluctuations in the receiver counter
electric currents will be modulated by sin(2theta)cos(phi).

This is a simple argument. Why don't you get it? Sarfatti

From ramsay@unixg.ubc.ca Tue Nov  3 15:43:34 1992
Date: Tue, 3 Nov 92 15:43:58 PST
From: Keith Ramsay <ramsay@unixg.ubc.ca>
To: sarfatti@well.sf.ca.us
Subject: Problems with your |h>.

|This is where we disagree. All the light goes to a single |h>.

Okay, how do you justify saying that it does? I see no justification
of this at all. Standard QM does not allow one to assert such
identities of states out of nowhere, especially for states that
originally were orthogonal!

Ramsay it's simple physics. If |f(g)> represent a photon leaving an blob
f(g), and if |h> represents a photon arriving at blob h - you are saying
that one cannot focus and guide two narrow (f and g) beams into a single
spot of finite area. I am not talking mathematical points but finite blobs.
I am assuming a single ket represents a blob not a point.

If you were right Ramsay, beam splitters would be impossible. Suppose we
have a photon in ket |i> incident on a beam splitter. The output is |r> and
|t> for reflected and transmitted paths.  One could always put in the
virtual zero-pont photon for the second input - but then we are no longer
in the Fock sub-space of 1-photon states.  But in the space of 1-photon
states the beam splitter is non-unitary.

There is a certain interesting irreversibility intrinsic to a beam
splitter.  Thus, I can shoot a single beam into a half-silvered mirror, and
out comes two beams.  But I can not shoot two beams into the beam splitter
and have only one beam come out - can I?  You could tell if a movie was
being run backwards - not for light (too fast) but, say,for slow moving
waves in a medium.

On the other hand, I say it is possible to get almost all the light from
two output ports of a beam splitter into a single counter - a single
counter means in ideal limit only one |h> - but the process is not unitary
because there is an irreversible making of a record - a measurement.
Unitary time evolution is reversible!  The quantum connection signal
depends both on coherence and irreversibility that go beyond the unitarity
prrofs. Unitarity is not all of quantum mechanics - it does not include
measurement.

|h> does not represent a mathematical point. At best it represents a single
|diffraction-limited spot on photo-sensive surace- at worst a finite set
|of them- and if the phases tend to cancel look at the fluctuations to
|recover the information.

You haven't shown why "fluctuations" recognizable at the "receiver"
should carry information. [Yes, I did and repeated it above. JS]
And you haven't answered my explanation for
why, if there are no variations in the mean, that there also won't be
any special "fluctuations" to detect either. [Nonsense!]

| I am told that we can capture essentially 100%
|of light- even if some leaks away- a small percentage will not destroy the
|entire effect.

I am not worried about such details either.

The problem is that there are two possible things you could have in
mind, and you're not describing the detector consistently enough for
us to know which exactly you mean:

(a) Placing a detector with a small enough aperture that the phase
differences between the two paths can be treated as essentially
constant over it.

(b) Placing a detector with a wide enough aperture that it captures
nearly all the light.

You can get (a) at the expense of missing a lot of the light.

You can get (b) at the expense of having no single state "|h>" describing
the state of a photon at the detector. To make your latest calculation
make sense, you need both.

No I don't, if (a) I get signal in the mean, provided the receiver counters
also miss corresponding parts of their light!  If (b) I get no signal in
the mean but get it in the fluctuations as I showed above. JS

And then to cover yourself you make this
terribly handwavy claim about "fluctuations", which is also flawed.

In article <Bx3AFr.9IF@well.sf.ca.us> sarfatti@well.sf.ca.us
(Jack Sarfatti) writes:
|The one's basedon unitary time evolutions leave out shifts in global
|boundary conditions.

Please specify these "global boundary conditions", and how they
supposedly invalidate the argument.

|Furthermore, they only apply to means and not to fluctuations.

No, you are mistaken. You have presented no actual argument for this.
If the "means" remain constant, the "fluctuations" are unaffected as
well, as I argued before:

|Let J be flux of photons arriving at a receiver V' counter. The count rate
|(clicks per unit time) will be proportional to J times p(V'). Plot the
|count rate as a function of time over scale bigger than duration required
|to determine count rate. This time series will fluctuate about a mean.
Now
|I do not say that such an analysis is easy- other fluctuations of
equipment
|processes must be subtracted etc. but in principle it is possible.

The probability of making a given observation with one photon pair and
of making a given observation with another photon pair are independent
of one another. So the only way to affect the probability distribution
of measurements, and hence of any statistical measure on the data, is
by way of changing the probability of observing results for
*individual* pairs.

You have accused the rest of us of clutching at straws in order to
hold on to cherished beliefs- consider that you may be doing that
yourself with these shifting claims. I would not be upset if it turned
out that FTL communication were possible- on the contrary. But it
seems you are psychologically unprepared seriously to consider the
possibility that quantum mechanics does not imply it.

||RMS [p(V') - p(H')] = sin(2theta)cos(phi)RMS[cos[arg<3|1> - arg<3|2>]>]/2
|
|"What is the variable over which you are taking the "RMS"?"
|
|The variable is cos[arg<3|1> - arg<3|2>] or arg<3|1> - arg<3|2> if you
|like.

Which, if you are defining |3> to be the state corresponding to where
the particle hits the detector, is also random, independent for each
observation, and is not known to the people at the "detector" end
during the "reception". This "RMS" it is not something they can detect
until they get to compare notes with the people at the "transmitter"
end afterward.

HOGWASH! the equation shows that the local fluctuations at the receiver
have a nonlocal functional dependence on the local fluctuations in phase at
the transmitter.  Ramsay- you are really off the wall in your remarks here.
Sarfatti

|See de la Torre in Foundations of Physics Letters - at
|least two papers in last 2 years. I keep saying this but it passes
|people by. Vanishing spacelike commutators are necessary if there
|is no ftl correlation signal but not sufficient! Schrodinger showed
|this in 1930!

I have examined these papers of de la Torre et al (the Schrodinger
reference was not available in the library). They do not say what you
imply that they say.

Then what do they say? De la Torres shows that the product of uncertainties
in the spin fluctuations for each photon in the pair obey a Heisenberg
inequality even though the operators commute. That's what he shows, and
that's what I said. Sarfatti

|You can have a Heisenberg uncertainty principle i.e. incompatibility
|or product of root mean square fluctuations greater than non-zero
|c-number even if the corresponding operators commute!  What does the
|trick is entanglement! Because of nonlcoal entanglement, measurements
|of commuting local operators outside each other's light cones can
|interfere with each other!

This "incompatibility" is a distinct property from the property
required to contradict a "no-signal" result.

Oh yeah? Why? What's the difference?

You are confusing the two. The "incompatibility" described in de la Torre
et al. is something completely different, which you have wishfully read as
implying what you want to be the case. There is a possibility of
correlation between what is observed at either end, but this does not
imply any communication. It is not possible to manipulate it in order
to send a signal.

Where is the proof that it is not possible to manipulate it. I have shown
that it is possible! JS