From: sarfatti@well.sf.ca.us (Jack Sarfatti)
Newsgroups: sci.physics
Subject: Round 3 FTL Debate Sarfatti vs Ramsay
Date: 5 Nov 92 06:29:54 GMT
Article-I.D.: well.Bx8Cpu.LL
Organization: Whole Earth 'Lectronic Link


Round 3 The Great Superluminal Debate: Sarfatti vs. Ramsay
Article 11172 (962 more) in sci.physics:
From: ramsay@unixg.ubc.ca (Keith Ramsay)
Subject: Re: "irreversible" beam splitter - correction
Date: 5 Nov 1992 03:29:53 GMT
Organization: University of British Columbia, Vancouver, B.C., Canada
NNTP-Posting-Host: unixg.ubc.ca
Summary: More Sarfatti stuff

-In article <Bx7Cot.4D7@well.sf.ca.us> sarfatti@well.sf.ca.us
(Jack Sarfatti) writes:
|I woke up this morning realizing that, of course, if the two beams into
the splitter are coherent and 90 degrees out of phase then one output port
will have no signal due to constructive interference because of extra 90
degree shift of splitter. In this special circumstance splitter is not
|"irreversible". But his trick can be used to get a single |h> in my
|experiment.  Put a 90 degree shift phase plate in path of one output of
|first recombiner then use second beamsplitter so that all of the signal
|goes to only one counter.

Will it, in your experiment, really always be 90 degrees? Won't it
rather have to be by an angle which will complement the phase shift
introduced by your earlier phase plate, and hence prevent you from
modulating the phase difference at the "transmitter"?

[Ramsay, I do not think that matters (not that I agree it is even true). I
think we can throw away the beam splitter all together!

OK Ramsay- let me give a simpler more transparent (I hope) formal picture
so that we can see just where the problem is.  Also, it solves the missing
factor of 2 mystery.

Let the initial photon pair state be

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

e(o) = extraordinary (ordinary) space path of "transmitter" calcite
+(-) linear polarization state relative to local basis specified by actual
orientation of crystal at time photon locally interacts with it.

Note that the "entanglement" structure demands that action at a distance
has already happened because we are in the basis (e(o)(+,-))defined by
local interaction of transmitter photon a with transmitter crystal, but the
faraway receiver photon b is nonlocally prepared in that same basis!  The
basis (e'(o')(+',-')) of the receiver crystal that locally interacts with
photon b analyses what the transmitter crystal prepared. That is a simple
physical picture of the actually observed correlations (e.g., p(e+,e'+') =
p(o-,o'-') = cos^2(theta)/2 etc.)

We must think of the spin basis (frame) as a kind of fiber frame attached
to the space paths.  Thus, the e path has a (+,-) frame different from the
o path (+.-) frame etc.

So for example, the action of the half-wave plate in the o (transmitter)
path is the local off-diagonal unitary matrix acting in the spin fiber o-
space that does|a,o+> -> |a,o-> AND  |a,o-> -> |a,o+>.  The fact that |a,b>
does not have |a,o+> does not matter.  Similarly, the phase plate action is
another diagonal local unitary matrix in the spin fiber e-space that does
|a,e+(-)> -> e^iphi|a,e+(-)>.  Again |a,b> does not have |a,e->. Not
realizing that there are two little spin fiber vector spaces attached to
each space path has been a source of confusion about unitarity of the
reversible dynamical time evolution in absence of irreversible measurement.

First consider the simplest possible "transmitter". Forget the half wave
plate. Just use mirrors (assume zero external reflection phase shift) and
dump both e and o beams into a single counter C. The process photon a hits
C and it is counted is described by ket |C>.

C is an extended blob it is not a mathematical point. Kets can describe
blobs, otherwise we could never see fringes of the form <PSI|x><x|PSI> at
blob x on screen of double slit experiment.

The "counting" is an irreversible making of the record. It is a non-unitary
process according to the Von-Neumann interpretation which I assume.

Let's be clear what is unitary reversible and what is non-unitary collapse!

The local actions of the phase plate and the half wave plate in different
space paths e and o respectively are reversible unitary and they result in
the "erasable" dynamical evolution:

|a,b> -> |a,b>' = {e^iphi|a,e,+>|b,e,+> + |a,o,+>|b,o,->}/sqrt2

Let's compute the nonlocal joint probability that receiver photon b is
detected in state e'+' while its twin photon a is detected in counter C.

*Note, there is no other place but C for photon a to be detected! This is
very important!  It is the main point of our disagreement I think.

The non-unitary irreversible double -measurement collapse is the Von-
Neumann projection:

p(b,e'+'|a,C) ='<a,b|b,e'+'>|C><C|<b,e'+'|a,b>'

= |{e^iphi<C|a,e,+><b,e'+'|b,e,+> + <C|a,o,+><b,e'+'|b,o,->}/sqrt2|^@

<b,e'+'|b,e,+> = cos(theta)

<b,e'+'|b,o,-> = sin(theta)

Assuming "equal illumination" take special case

<C|a,e,+> = |C|e^iphi(e)

<C|a,o,+> = |C|e^iphi(o)

Therefore,

p(b,e'+'|a,C) =

|{e^iphi|C|e^iphi(e)cos(theta) + |C|e^iphi(o)sin(theta)}/sqrt2|^2

= |C|^2[1 + sin(2theta)cos(phi + phi(e) - phi(o))]/2

Similarly,

p(b,o'-'|a,C) = |C|^2[1 - sin(2theta)cos(phi + phi(e) - phi(o))]/2

The local response probability of ideal transmitter counter C detecting
photon a is the sum of the above two nonlocal joint probabilities since the
two possible detections of photon b are distinguishable (non-interfering)
Feynman histories. Therefore,

p(a,C) = |C|^2 = 1

Now don't squawk that it has to be 1/2. It doesn't! The reason is subtle
and we all missed it.  There is nothing non-unitary or contradictory in
writing both

<C|a,e,+> = e^iphi(e)

<C|a,o,+> = e^iphi(o)

because |a,e,+> and |a,o,+> are kets in two different vector spaces which
become indistinguishable upon measurement by C.  They are not orthogonal
kets in the same vector space! We need not assume an innconsistent unitary
evolution of two orthogonal kets into the same ket |C>.  All these two
equations say is that if photon a is in path e (o) it will hit blob C with
certainty!  That is very sensible and physical.

Since there is |C> and only |C>, and we do not invoke a unitary evolution
of two orthogonal kets in the same space into it, the nonlocal joint
probabilities are the same as the local receiver probabilities. So that,
the quantum connection signal locally observable at the receiver end, with
no comparison with transmitter counter data required, is

p(b,e',+') - p(b,o',-') = sin(2theta)cos(phi + phi(e) - phi(o))

The photo-current from a subtractor will be proportional to it.  This is
best case.  Worst case is there are several classically distinct blobs so
that phi(e) - phi(o) is a random variable.  When we make a classical
statistical average <...> over the blobs suppose

<cos(phi + phi(e) - phi(o))> = 0

But then

<cos^2(phi + phi(e) - phi(o))> is not zero -but the phi dependence drops
out at the receiver. Nevertheless the root mean square fluctuation of the
receiver photo-current will be proportional to sin(2theta). Keep receiver
crystal fixed and rotate transmitter crystal to see quantum connection
signal!

Now I will go back and check to see if I can keep phi dependence in the
fluctuations if I put back the beam splitter.