From: sarfatti@well.sf.ca.us (Jack Sarfatti)
Subject: Re: Sarfatti Stuff FTL QM Connection Signal
Organization: Whole Earth 'Lectronic Link
Date: Sun, 8 Nov 1992 19:59:17 GMT


Sarfatti replies to:
From: ramsay@unixg.ubc.ca (Keith Ramsay)
Subject: Sarfatti stuff
Date: 8 Nov 1992 01:11:35 GMT
Organization: University of British Columbia, Vancouver, B.C., Canada
NNTP-Posting-Host: unixg.ubc.ca

I wrote:
||Therefore,
||
||p(V',OFF) = |<b(V')|<h|a.b>'|^2
|
|This is the probability that V' is observed *given that the photon is
|detected in state |h> at the transmitter end*. This is *not* what you
|want. How many times do we have to go over this particular point
|before you absorb it? You *must* count the other photon pairs, or you
|are not counting anything which can be determined at the "receiver"
|end of the apparatus.

In article <Bx632J.5Eo@well.sf.ca.us> sarfatti@well.sf.ca.us
(Jack Sarfatti) writes:
|Ramsay, I am saying there is not anything else if all the light goes into
|one detector symbolized by |h>.

This is precisely what does not make sense. There is not just one
state describing a photon hitting the detector. There are many
possible states a photon could have, while hitting the detector. (Many
different momentum states, for example.) So you cannot simply choose
to "symbolize" the detector by |h>.

*Sarfatti replies: Ramsay, you are confusing a pure state for a mixed state
and, also, perhaps, a eigenstate for pure superposition of eigenstates.  A
pure ket |h> can be a coherent superposition of many momentum eigenstates.
And if we do have a mixed state, I include them in the classical
statistical average,

Perhaps you don't realize that there is a Heisenberg uncertainty
relation involved here. If you reduce the uncertainty of the position
of your photons, the uncertainty of the conjugate momentum has to
increase. This prevents you from both focussing the beams on a small
spot, and having the phase vary slowly across that spot.

*Sarfatti replies:I am not contradicting any Heisenberg uncertainty
relation.  Ramsay look. Suppose we are doing a simple interferometer
experiment in which there are only two counters detecting outputs from a
beam splitter with complementary "fringes" say 1 + cos(phi)
and 1 - cos(phi) - if you were right these fringes could never be seen
because the phase phi would vary to fast across the spot! Yet, in fact,
this is not true. If you were right fringes could never be seen at all
because phi would aways be random cancelling itself out.  The conditions
for seeing the quantum connection signal are same as conditions for seeing
fringes in any optical interference phenomenon.  Also the issue is not
whether the phase varies fast or slow, but whether or not <cos(phi)> = 0
if there are many distinguishable spots contributing according to a
classical probability distribution p(phi) for the optical path differences
from the two beams to the different spots on the detector.

PS  How do you calculate the variation of phase "across that spot"?

What you say is also completely contrary to standard quantum
mechanics:

According to standard quantum mechanics, observations correspond to
self-adjoint operators. When an observation is carried out, the state
of the system is projected onto one of the eigenspaces of the
observable (and the projected vector is then normalized). The
probability of each such projection occurring (and the corresponding
eigenvalue being the observed value of the observable) is proportional
to the square of the magnitude of the projection.

For whatever observation you make, the probability of a state |f>
being projected to a fixed state |h> is at most |<h|f>|^2. You cannot
make both |<h|f>|^2 and |<h|g>|^2 be nearly 1, since |f> and |g> are
orthogonal.

*Sarfatti replies: Nonsense Ramsay, you are confusing formalism for
physics. |f>, |g> and |h> each consist of space and spin parts. Because of
the half wave plate all three kets have exactly the same spin states!  They
are not orthogonal in spin space! And because of the optics the space parts
of both |f> and |g> are guided right into the spot defining |h>, or the
finite set of distinguishable spots defining the density matrix that
replaces |h> with a classical probability distribution p(phi).  Since the
space parts of |f> and |g> go directly to |h> one can write above relations
without contradiction. Think Feynman paths rather than abstract kets and
you will see that what I am saying makes physical sense.

If a little is lost it should not make a
|big difference.  Almost everything collapses in a non-unitary process into
|the ket |h> - there are no other photon pairs.

This is definitely not standard quantum mechancics-- you cannot simply
assert that there is just one state involved and no other possibility.

*Sarfatti replies: Hogwash Ramsay! Just look physically! both beams enter
the single counter. And I said we can have a mixed state with a density
matrix replacing the pure key |h> - again your point that a pure |h> cannot
include many momentum eigenstates is false. You are raising a false issue.

Here's another way to see it: Suppose that we call the two possible
values of the variable observed by the detector "detect" (+) and "does
not detect" (-). You have not mentioned the detector observing
anything else about the photon, just whether it strikes the given
spot. If you can (almost) guarantee that a photon in state |f> will be
detected, then |f> must be (almost) in the (+)-eigenspace of the
operator corresponding to the observable. Therefore when (if) it is
detected, it will remain the same (be changed by only a little bit).
Likewise for |g>. So |f> and |g> cannot both be (almost) guaranteed to
be collapsed to the same state |h> unless they are (almost) the same
as |h> to begin with, since neither one can be changed (by more than a
small deviation-- one of small norm). *That* is what standard quantum
mechanics says.

*Sarfatti replies: Ramsay if |f> is photon in extraordinary path and |g> is
photon in ordinary path, and |h> is mixed state (in general - I mean impure
density matrix) then , including unitary evolution tacitly , from the
geometry it is certain that no matter with path transmitter photon is in it
will hit set of spots represented by density matrix for h.

If there are other photon pairs, they are not counted at the transmitter
nor at the receiver counters.

*Sarfatti replies, the photons are connected, and their detections are
connected. However, it is a moot point, because I show, that for ideal
detectors  the probabilities balance perfectly of each side. For
transmitter p = 1 and for receivers the sum of the two counter
probabilities also add to 1. So anythoing you say about "other photon
pairs" is ad-hoc and not part of sqm.

Why not? This seems totally arbitrary.

*Sarfatti: On the contrary, what you say seems totally arbitrary.

|So that what is counted will show the "fringe modulation".  Now
|if there is a density matrix so we do a classical average over several
|phases we may get zero signal in the mean- but we can pick it up in a
|Hanbury Brown Twiss correlation of intensity fluctuations from photo-
|currents of the two receiver counters. This is simple math and I have
|already done it more than once.

The math does not say what you are claiming that it says.

Here again is an piece of text where you do this "simple math":

|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.

So? This is not the "root mean square" variation of anything you can
detect at the receiver.

*Sarfatti replies: Of course it is!

I think you must be misapplying some statistical formula, but since you're
not specifying what formula it is, it is hard to tell why you think this is
obviously correct.

*Sarfatti replies: I am simply making the standard ansatz that the photo-
current in the counter outputs are proportional to the quantum
probabilities of detection of a photon that I have computed. That's all. I
simply don't get your point Ramsay. Do you deny what I just wrote?

|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

Because you do not give a solid justification.

*Sarfatti replies: This is elementary physics of the photo-electric effect.
For quantum optics formalsim see Glauber;s 1964 Les Houches Lecture. You
are on very weak ground here Ramsay.  My basic point is that you are giving
formal arguments that , if correct, would preclude ever measuring any
coherent phase effects locally - because you inmply that they will always
wash out - certainly in many kinds of everyday interferometers as descirved
in detail in Born & Wolf for example.

Elsewhere, you describe how you are going to measure these "fluctuations":

|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.

So let N be the total number of photons which arrive over the interval
(83%)of time required to determine the count rate. The probability of
counting m of those photons at the V' counter is C(N,m) p(V')^m
(1-p(V'))^{N-m}, since the probabilities of each photon being counted
V' are mutually independent. Thus the likelihood of getting a certain
rate m/N of detection over that period of time is dependent only on
p(V'). Correlations with this extra variable |h> are simply
irrelevant. So how can "fluctuations" possibly enter into it, unless
you are able to affect p(V') in the first place?

*Sarfatti replies - yes p(V') is affected in the first place because it
depends on sin(2theta)cos(phi) - but the really important point is whether
<cos(phi)> really must be zero to begin with. There is no reason from
conservation of energy or probability for it to be zero. There is no
principle of physics, other than causality which we are testing, for it to
be zero. If it's not zero then the local receiver probabilities are

p(V') = [1 + sin(2theta)<cos(PHI)>]/2

p(H') = [1 - sin(2theta)<cos(PHI)>]/2

so p(V') + p(H') = 1

and at the transmitter p(h) = 1  This is all "unitary" in the practical
significant experimental sense.

Remember <cos(PHI)> depends on the classical probability distribution
p(PHI) which can be engineered to be non-uniform I think using phase
compensation plates - this is classical optical engineering problem.

It *appears* that you want to calculate these deviations in a more
devious way as follows:

Calculate the variance of the number of photons which arrive over the
interval (out of N) by partitioning them up according to the
associated values of |h>. Take the mean of each group. Calculate the
variance of the resulting distribution of means. This corresponds to
what you have calculated above. Unfortunately for you, this procedure
is simply wrong. So what are you in fact thinking?

*Sarfatti replies: I am predicting that if <cos(PHI)> is not zero then the
mean photo-currents at the receiver will vary as sin(2theta) and possibly
as cos<PHI> as well. And if, <cos(PHI)> is zero then we will see only a
sin(2theta) dependence in the root mean square fluctuations of the receiver
photo-currents about the mean. Theta is controlled from the transmitter.
Let experiment decide. Those are my predictions, and I say my model is sqm
and there is no contradiction of Heisenberg, or Hermitian observables etc.

                       "In no way, shape or form did Kevin represent
Keith Ramsay            a viable alternative to mental illness."

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