From: sarfatti@well.sf.ca.us (Jack Sarfatti)
Subject: Quantum Spin Mach-Zehnder Interferometry III
Organization: Whole Earth 'Lectronic Link
Date: Thu, 19 Nov 1992 05:04:22 GMT


#3 Calcite MZI

    _
   | |-|-|-|-|-|\
   | |    |e+>  |
   | |          - phase plate (phi)
   | |          |
/ /|_|-o-o-o-o--\\----- |e(+)r+o(-)t> = |1> (interferogram 1)
|a>             |
          |o->      |
                |o(-)r+e(+)t> = |2> (interferogram 2)

Fig. 3A

Note polarization kets: | = |+> & o = |-> and <+|-> = 0

|e+> = |e>|+>

|e> is a path ket etc.

Fig.2 CMZI

The first beam splitter is replaced by a calcite crystal

because of orthogonality of the polarization states

p(1) = p(2) = 1/2

independent of phi because (i.e, using quick and dirty "Feynman
history"picture)

|a> -> [e^iphi|e>|+> + |o>|->]/sqrt2

    -> |1> (ie^iphi|+> + |->)/2 + |2>(e^iphi|+> + i|->)/2 = |a>'


p(1) = |<1|a>'|^2 = |(ie^iphi|+> + |->)/2|^2 = 1/2

p(2) = |<2|a>'|^2 = |(e^iphi|+> + i|->)/2|^2 = 1/2

#4 Now insert a half-waveplate in the o-space path. The unitary operator
U(o:1/2) only acts locally and unitarily in the o-path because that is
physically where the plate is. There is no half-wave plate in the e-path.
Ramsay seems to get confused on this, falsely applying the unitary operator
to the entire state - which is wrong standard quantum mechanics (SQM).
The half-wave plate "quantum erases" the polarization distinguishability
between the two space paths. That is

U(o:1/2)|-> = |+>
    _
   | |-|-|-|-|--\
   | |    |e+>   |
   | |           - phase plate
phi)
   | |    _      |
/ /|_|-o-|_|-|-|-\\----- |e(+)r+o(+)t> = |1> (interferogram 1)
|a>              |
      |o->  |o+> |  |o(+)r+e(+)t> = |2> (interferogram 2)

Fig. 4A

Now we have, using the Feynman histories picture to avoid all the extra
confusing "epicycles" of the Ramsayian "Dirac ket" picture (artifacts of a
bad notation - perhaps Ramsay can show a more efficient method?):

|a> -> |a>' = {|1>(ie^iphi + 1)/2 + |2>(e^iphi + i)/2}|+>

p(1) = |<1|a>'|^2 = |(ie^iphi + 1)/2|^2 = [1 + cos(phi+pi/2)]/2

     = [1 - sin(phi)]/2

p(2) = |<2|a>'|^2 = |(e^iphi + i)/2|^2 = [1 + cos(phi-pi/2)]/2

     = [1 + sin(phi)]/2

so that

p(1) + p(2) = 1

The local "fringe signal" is

p(1) - p(2) = sin(phi)

for a fixed definite phi. There is no need to sum over random phi's. It
makes no sense physically! Ther is a similarity with the 2-slit experiment,
but this is a significant difference.  There are only two "spots" on this
"screen" (i.e. counters 1 and 2) that is everything - a fact that Ramsay
does not appear to appreciate sufficiently. The "fringes" are seen in a
sequence or ensemble of distinct macro-experiments corresponding to
changing the setting on the phase-meter (i.e. phase plate).

#5 pair-correlated light


   |3>     _                   _
   -|-|-|-| |                 | |-|-|-|-|-|-\
          | |                 | |    |e+>   |
          | |                 | |           - (phi)
   |4>    | |                 | |    _      |
   -o-o-o-|_| <-    O       ->|_|-o-|_|-|-|-\\-|- |1>
                                    1/2    _|_
                                            |
                                             |2>

Fig. 5A


The initial "entangled" (i.e. quantum-connected) photon-pair state is:

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

local unitary action of both phi-meter in e-path and 1/2 wave plate in o
path of photon a is "polarization disentanglement":

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

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

local unitary action of the beam splitter \\ on photon a is

|a,b>' -> |a,+>{|1>(ie^iphi|b,+> + |b,->)/2 + |2>(e^iphi|b,+> + i|b,->)/2}

        = |a,b>''

Consider the nonlocal joint probabilities:

p(3,1) = |<3|<1|<a,+||a,b>|^2

=|(ie^iphi cos@ + sin@)/2|^2 = [1 + sin2@ cos(phi+pi/2)]/4

= [1 - sin2@ sin(phi)]/4


p(3,2) = |<3|<2|<a,+||a,b>|^2

=|(e^iphi cos@ + isin@)/2|^2 = [1 + sin2@ cos(phi-pi/2)]/4

= [1 + sin2@ sin(phi)]/4

Therfore, the local "receiver" probability for counter 3 of photon b is

p(3) = p(3,1) + p(3,2) = 1/2

similarly, p(4) = 1/2

Therefore, there is no quantum connection communication signal in this
experimental arrangement because of the i phase factor of the transmitter
beam splitter.  The i phase factor is required by local unitarity of the
beam splitter in the one-photon problem.  Is it required for the photon
pair problem as well? What about Fig.2A. Is that a clue?

to be continued




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