From: sarfatti@well.sf.ca.us (Jack Sarfatti)
Subject: UNITARITY WITHOUT UNITARITY
Organization: Whole Earth 'Lectronic Link
Date: Sat, 15 Aug 1992 09:54:39 GMT


UNITARITY WITHOUT UNITARITY

IN SUPERLUMINAL QUANTUM CONNECTION COMMUNICATION

by Jack Sarfatti

It is shown that the cosine of the difference between the reflection and
transmission phase shifts of a beam splitter need not vanish for
resonant thicknesses of the glass plate tuned to the phase shift of an
optical impurity. This removes the last objection of Aephraim Steinberg
to my gedankenexperiment for faster than light and backwards in time
communication on the quantum connection. If this device works it will
mean the demise of the retarded principle of causality in physics -
effects can happen before their causes under certain conditions. It also
provides a new mechanism for parity violation.


#1 Surface excitations distort reflection shift?
The Fresnel equations express the electromagnetic wave boundary
conditions that tangential components of E and H and normal components
of D and B field vectors are continuous at the faces of the beam
recombiner. They determine the pi phase difference between the external
and internal reflections. Is it possible to have some kind of localized
surface elementary excitation like a high frequency plasmon or some kind
of frequency-matched surface wave that will shift the phase difference
away from pi? If so, that would make cosb not zero.

#2 Localized optical doping inside glass?
The localized optical inhomgeneity inside the glass plate of the beam
recombiner positioned to intercept the return path for the internal
reflection from back face to front face (giving it an extra shift of
theta) should work in principle because the return path need not overlap
with the incident path from front face to back face. These two paths
inside the glass are separated by twice the angle of refraction. This
would also make cosb not zero because the phase shift of the transmitted
beam would be kd (k = 2pi/wavelength) but the phase shift of the
reflected beam would be kd + theta/2 + pi/2.

#3. Pair experiment.
If you look at my previous equations (7a&b)

|E> -> e^i(phi+m)2^-1/2|I> +  e^i(phi+m+b)2^-1/2 |II> (7a)

|O> -> e^ib 2^-1/2|I> + 2^-1/2 |II>                   (7b)

I noted that this transformation is not unitary in general if we assume
that |I> and |II> are orthonormal. It is only unitary when b = pi/2
which corresponds to Aephraim's remarks. Hoever, for the pair correlated
state this transformation being only a part, an intermediate local
process, need not be unitary as long as the whole nonlocal process is
unitary conserving locally observable probabilities directly measured by
the photon counters.

Amazingly enough, even when b is not equal to pi/2, nevertheless, the
whole process is unitary - recall that the initial pair state |a,b> of
equation (1) becomes

|a,b> = [|e>|E> + |o>|O>]cosx 2^-1/2  + [|e>|O> - |o>|E>]sinx 2^-1/2

-> |a,b>' = [|e> {e^i(phi+m)2^-1/2|I> +  e^i(phi+m+b)2^-1/2 |II>}  +

|o>{e^ib 2^-1/2|I> + 2^-1/2 |II>}]cosx 2^-1/2 +

[|e>{e^ib 2^-1/2|I> + 2^-1/2 |II>}

- |o>{e^i(phi+m)2^-1/2|I> + e^i(phi+m+b)2^-1/2 |II>}]sinx 2^-1/2

= |e>|I>[e^i(phi+m)cosx + e^ib sinx]/2 + |e>|II>[e^i(phi+m+b)cosx +
sinx]/2

+ |o>|I>[e^ib cosx - e^i(phi+m)sinx]/2 + |o>|II>[cosx -
e^i(phi+m+b)sinx]/2

(8)

Therefore, at the receiver


p(e) = |[e^i(phi+m)cosx + e^ib sinx]/2|^2 + |[e^i(phi+m+b)cosx +
sinx]/2|^2

= [1 + sin2x cos(phi+m)cosb]/2
(9a)


p(o) = |[e^ib cosx - e^i(phi+m)sinx]/2|^2 + |[cosx -
e^i(phi+m+b)sinx]/2|^2

= [1 - sin2x cos(phi+m)cosb]/2
(9b)

So that the locally observable probabilities to measure the receiver
photon polarization eigenvalues are conserved and we get the connection
communication equation

p(e) - p(o) = sin2x cos(phi+m)cosb
(6)


This is the complete solution so we can see what happens at the
transmitter


p(I)=|[e^i(phi+m)cosx+e^ib sinx]/2|^2 +|[e^ib cosx-e^i(phi+m)sinx]/2|^2

= 1/2
(10)

Similarly p(II) = 1/2 so that probability is conserved on the
transmitter side and there is no trace of the coherent phase parameters
phi,m and b as we would expect for ordinary light on the same apparatus!
The coherent phase parameters have teleported to the other receiver
side.

#4 Single photon experiment.
The apparatus is different and there is no reason to assume apriori that
the beam recombiner will respond to a single particle photon state in
the same way it responds to an entangled photon pair state. This is
intuitively clear in Bohm's "super quantum potential" description of
field theory. Indeed, going back to equations (7a&b) we can get
conservation of probability and energy for any value of b by not making
the assumption that |I> and |II> are normalized to 1. Specifically for
any input single photon state of the form

|1> = cosx|E> + sinx|O>  (11)

putting in an extra phasor e^i chi factor in front of sinx will merely
renormalize phi - so that the equations below are general for any input
polarization on the transmitter interferometer.

<1|1> = 1    (12)

the output from the beam recombiner integrated as a part of the whole
"transmitter" apparatus is unitary provided that

<I|II> = 0    (13a)

<I|I> = <II|II> = 1/2[1 + sin2xcos(phi + m)cosb]    (13b)

Therefore,

|1> -> |1>' =  cosx{e^i(phi+m)2^-1/2|I> +  e^i(phi+m+b)2^-1/2 |II>}

            +  sinx{e^ib 2^-1/2|I> + 2^-1/2 |II>}

            =  {cosx e^i(phi+m)2^-1/2 + sinx e^ib 2^-1/2}|I>

            +  {cosx e^i(phi+m+b)2^-1/2 + sinx 2^-1/2} |II>     (14)

Therefore, the probabilities of response of the two ideal counters I and
II are

p(I)='<1|I><I|1>' =  [1 + sin2x cos(phi+m-b)]/2[1 + sin2x
cos(phi+m)cosb]
(15a)

p(II)='<1|II><II|1>'=[1 + sin2x cos(phi+m+b)]/2[1 + sin2x
cos(phi+m)cosb] (15b)

p(I) + p(II) = 1  (15c)

so that energy and probability are conserved for an arbitrary single
photon input state - not to the isolated beam recombiner but to the
entire apparatus of calcite, phase plate in E path, half-wave plate in O
path, mirror in E path and beam recombiner whose inputs are the E and O
paths and whose outputs are the I and II paths. If b = pi/2 then

p(I) ->  [1 + sin2x sin(phi+m)]/2 (16a)

p(II) -> [1 - sin2x sin(phi+m)]/2 (16b)

If, further, m = pi the usual value for external reflection

p(I) ->  [1 - sin2x sin(phi)]/2    (17a)

p(II) -> [1 + sin2x sin(phi)]/2    (17b)

This explicit computation shows that there is no violation of
conservation of probability or energy in either experiment if some way
can be found to make cosb not equal to zero in a beam recombiner.

#5 Time reversal invariance and energy conservation at beam recombiner.
Unless the device coheres zero point vacuum fluctuation energy (which
appears far fetched) then

t*r + r*t = 0     (18)

but that energy conservation condition need not imply that

b = phi(r) - phi(t) = pi/2  (19).

Here's why,suppose the optical doping (i.e. theta in return kd path back
face to front face) idea works. The external reflection phasor (TE wave)
is e^ipi. The internal reflection phasor is e^i(2kd + theta). The
transmisson phasor is e^ikd. Add the internal reflection phasor to the
external reflection phasor to get

e^ipi+e^i(2kd+theta)=

e^{i(2kd+theta+pi)/2}[e^-{i(2kd+theta-pi)/2}+e^{i(2kd+theta-pi)/2}

= e^{i(2kd+theta+pi)/2} 2cos{(2kd+theta-pi)/2}  (20)    proportional to
r

Aephraim you forgot the loophole provided by the periodic modulation of
the reflection amplitude! Therefore,

t*r + r*t = |r||t|4cos{kd+theta/2-pi/2} cos{(theta + pi)/2}    (21)

in which

b = (theta + pi)/2       (22)

Therefore, energy conservation is consistent with any b provided that we
make the thickness of the beam splitter "resonant" such that

cos{kd+theta/2-pi/2} = 0         (23a)

or at resonance

kd+theta/2-pi/2 = pi/2

kd + theta/2 = pi                (23b)

theta = 2(pi - kd)               (23c)

b = pi - kd + pi/2 = 3pi/2 - kd  (23d)

cosb = cos(3pi/2 - kd) = -sin(kd)

In this case the superluminal (FTL) and backward-in-time (BIT) quantum
connection communication equation is (for m = pi)

p(e) - p(o) = sin2x cos(phi)sinkd  (24)

Do the single photon experiment on the transmitter alone in equations 15
as a homework problem for m = pi