🚀 go-pugleaf

If there are any other objections to my claim that faster than light
communication on the quantum connection is allowed in standard quantum
mechanics please send them directly to me via e-mail as I have not
searched this conference (don't know how yet).

Response to Mike Gallis, Carlton Caves and Aephraim Steinberg on the
apparent
nonunitarity of the evolution in the superluminal communication
gedankenexperiment.

I shall summarize my arguments against your objections as clearly as I can
in terms of
Feynman's interpretation of quantum mechanics. You already know I do not
accept the
argument about the vanishing expectation of the commutator for space-like
separations
etc because of Schrodinger's 1930 theorem.

You all have essentially come up with the same objection independently which
I
summarize as follows:

Given
#1.  |E> -> |E>
#2.  |O> -> |E>
therefore
#3  |E> + |O> -> 2|E>
#4  |E>  - |O> ->  0

To quote Aephraim "a unitary transformation conserves probability for ANY
input state
.... this goes into zero. No good."

The point is that #1 and #2 do not describe the apparatus.
For the case that a single-photon state is incident, the
apparatus is the "transmitter" by itself consisting of a calcite C,
a phase plate P with translational relative phase shift phi in the
extraordinary path E, a
nonabsorbing mirror M with phase shift m that deflects E to half-silvered
beam
recombiner B with reflection phase shift b, a half wave plate H in ordinary
path O
positioned (at 45 degrees to plane of O polarization) so that it rotates O
to E before
going to B.

The two "inteferogram" outputs from B each go to a separate transmitter
counter I and
II. If a photon goes through B label the process B. If the photon is
reflected at B label
the process B*. The input to counter I consists of the Feynman history
CEPEMEBEI
added to the Feynman history COHEB*EI.  H and B* have "quantum erased" any
path
memory so that these two histories are indistinguishable interfering
alternatives in
Feynman's sense of paths.

Similarly for interferogram II we have CEPEMEB*EII and COHEBEII for the two
interfering alternatives. Note that, for example, CEPEMEBEI does not
interfere with
CEPEMEB*EII since they are detected by different counters. Nonunitarity of
the
measuring process is at this point creeping in to the Feynman way of doing
quantum
mechanics that I am using.

Let's take as input the dangerous normalized state

#4'  |-> = [|E> - |O>]/sqrt2

For interferogram I, CEPEMEBEI corresponds to

#5  |E> -> [e^i (phi + m)/sqrt2] |EI>

and COHEB*EI corresponds to

#6  |O> -> [e^ib/sqrt2] |EI>

Therefore, by linearity of quantum mechanical evolution

#7  |-> -> {[e^i (phi + m) - e^ib]/2}|EI>

This state is no longer normalized to unity because we are describing
everything
including the measurements.  Unitarity breaks down in a measurement. The
Feynman
path intepretation is nonlocal in both space and time.

If the Feynman interpretation of quantum mechanics gives answers to
experiments that
the traditional unitary matrix formalism does not give then I'll place my
bet on
Feynman. However, I think that a proper use of the formal matrix technique
will give
the same answers I am here explicitly computing.

Similarly, for interferogram II CEPEMEB*EII corresponds to

#8  |E> -> [e^i (phi + m +b)/sqrt2] |EII>

#9  |O> -> [1/sqrt2] |EII>

#10  |-> -> {[e^i (phi + m +b) - 1]/2}|EII>

Equations #7 and #10 describe the nonunitary transformation of the initial
pure state |->
to a statistical mixture of states or impure state.
The use of a unitary matrix is totally unnecessary in Feynman's
interpretation of
standard quantum mechanics based upon an organic nonlocal picture of quantum
reality.

One can use the density matrix formalism. The initial density matrix will be
an
idempotent projection operator whose square equals itself.
A similarity transformation will take the initial density matrix to the
final one implicit in equations #7 and #10 which is not idempotent.
Nevertheless ther is no violation of conservation of probability. The
experimental
prediction is that the probability for counter I to click is

#11 p(I) =  |{[e^i (phi + m) - e^ib]/2}|^2/{ |{[e^i (phi + m) -

e^ib]/2}|^2 + |{[e^i (phi + m +b) - 1]/2}|^2}

= [1 - cos (phi + m -b)]/ { [1 - cos (phi + m -b)] +

[1 - cos (phi + m +b)]

or

#12  p(I) =  [1 - cos (phi + m -b)]/ 2[ 1 - cos (phi + m) cos(b)]

Similarly,

#13  p(II) =  [1 - cos (phi + m +b)]/ 2[ 1 - cos (phi + m) cos(b)]

Probability is conserved

#14 p(I) + p(II) = 1.

This device does not have any output going to zero violating conservation of
probability.

We can apply the same picture to the initial photon pair case emitted at
space-time event
S.  Photon 1 is the transmitter photon and photon 2 is the receiver photon.
In terms of
the base states (E.O) of the transmitter calcite in its actual orientation
at space-time
event T when photon 1 locally interacts with it

#15  |1,2> = [ |E1> |E2> + |O1>|O2> ]/sqrt2

For interferogram I

#16  |1,2> -> { [e^i (phi + m)/sqrt2] |E1I>  |E2> + [e^ib/sqrt2] |E1I> |O2>
}/sqrt2

=  {e^i (phi + m)  |E2> + e^ib |O2>} |E1I>/2

For interferogram II

#17  |1,2> -> { [e^i (phi + m +b)/sqrt2] |E1II>  |E2> + [1/sqrt2] |E1II>
|O2> }/sqrt2

=  {e^i (phi + m +b)  |E2> + |O2>} |E1II>/2

Note that the local transmitter photon detection probabilities p(I) and
p(II) are each 1/2
because of the orthogonality of |E2> and |O2>.  The phase information phi
disappears
from event T and reappears at event R if the phase shift b is selected
properly.

Now resolve the receiver photon 2 states into the base states (e,o) of the
receiver calcite
at spacetime event R when photon 2 locally interacts with it.  The receiver
is simply a
calcite with two counters behind it to catch the e and o paths.

#18   |E2> = cos(theta) |e2> + sin(theta) |o2>

#19   |O2> = -sin(theta)|e2> + cos(theta) |o2>

#20  |1,2> -> =  {e^i (phi + m) {cos(theta) |e2> + sin(theta) |o2>}

+ e^ib { -sin(theta)|e2> + cos(theta) |o2>} |E1I>/2

The nonlocal joint probability to detect receiver photon in the e path at
event R and its
twin transmitter photon in the I path at event T is then

#21  p[e(R), I(T)] = [1 - sin(2theta)cos(phi+m -b)]/4

Similarly

#22  p[o(R), I(T)] = [1 + sin(2theta)cos(phi+m -b)]/4

For interferogram II the corresponding results are

#23  p[e(R), II(T)] = [1 - sin(2theta)cos(phi+m+b)]/4

#24  p[o(R), II(T)] = [1 + sin(2theta)cos(phi+m +b)]/4

Since the different transmitter interferograms are distinguishable non
interfering
alternatives, Feynman's nonunitary rule tells us

#25  p[e(R)] = p[e(R), I(T)] + p[e(R), II(T)] = [1 - sin(2theta) cos(phi +m)
cos (b)]/2

#26  p[o(R)] = p[o(R), I(T)] + p[o(R), II(T)] = [1+ sin(2theta) cos(phi +m)
cos (b)]/2

Therefore, the local probabilities on the receiver side are conserved

#27  p[e(R)] + p[o(R)]  = 1

The quantum spin connection signal which is independent of the space-time
interval
between the transmitting event T and the receiving event R is

#28  p[e(R)] - p[o(R)]  = - sin(2theta) cos(phi +m) cos (b)

* Note to the reader: Professor Robert Solovay of the UCB Mathematics
Department
says I do not understand quantum mechanics but he can't show specifically
where this
model goes wrong.