NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/alt/sci/physics/newtheo/2348

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Path: sparky!uunet!well!sarfatti
From: [e-mail] (Jack Sarfatti)
Subject: Sarfatti Lectures in Super-Physics 1
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Date: Thu, 19 Nov 1992 08:44:27 GMT
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I bring all the pieces on interferometry leading up to the quantum
connection communicator that can transmit messages backwards in time in one
place -for the convenience of students of the future:

The Sarfatti Lectures on Super Physics

Lecture 1

Clarification of Ramsay Sarfatti debate in simple case.

#1 Mach-Zehnder Interferometer (MZI) with ordinary unconnected one photon
input.

The MZI consists of two 50-50 non-absorbing beam splitters (\\) and two
nonabsorbing mirrors (\).

---->\\---------\
|a>  |    |t>   |
     |          - phase plate (phi)
     |          |
     \----------\\----- |tr+rt> = |I1> interferogram 1
          |r>   |
                |
                |
                 |tt+rr> = |I2> interferogram 2

Fig.1A MZI (orthogonal ket)

A. The first beam splitter resolves the input photon ket |a> into

|a> -> [|t> + i|r>]/sqrt2

The i is from reflection at the first beam splitter.

<t|r> = 0

<t|a> = 1/sqrt2

<r|a> = i/sqrt2

The effect of the phase plate (all optical paths otherwise equal) is

|t> -> e^iphi |t>

so that

[|t> + |r>]/sqrt2 -> [e^iphi|t> + |r>]/sqrt2

Interferogram 1 is then

[ie^iphi|t> + i|r>]/2 = |I1>

<I1|I2> = 1

The probability of counter 1 to "click" is then

p(1) = |<1|I1>|^2

= |[ie^iphi<1|t> + i<1|r>]/2|^2

= [|<1|t>|^2 + |<1|r>|^2 + 2|<1|t>||<1|r>|cos(phi + arg<1|t> - arg<1|r>)]/4

similarly, interferogram 2 is

[-|r> + e^iphi |t>]/2 = |I2>

<I2|I2> = 1

note that <I2|I1> = 0

and

p(2) = |<2|I2>|^2

= |[e^iphi<2|t> - <2|r>]/2|^2

= [|<2|t>|^2 + |<2|r>|^2 - 2|<2|t>||<2|r>|cos(phi + arg<2|t> - arg<2|r>)]/4


p(1) + p(2) = 1

is satisfied if

<1|t> = <1|r> = <2|t> = <2|r>

|<1|t>| = 1

so that

p(1) = [1 + cos(phi)]/2

p(2) = [1 - cos(phi)]/2

* Note that phi is definite. There is no reason to assume that phi is
random as Ramsay requires. There is no sum over phi required in an
individual experiment in order to conserve probability. This is a
significan difference from the classic 2-slit experiment. The MZ experiment
has a "discrete" screen of only two resolution-limited "points" in contrast
to the large number of resolution-limited points in the 2-slit experiment.

This model in which <t|r> = 0, requires 8 adhoc unobservable real
parameters which must be adjusted to get the right answer.

B. A much simpler model based upon the Feynman histories picture is:

---->\\---------\
|a>  |    +     |
     |          - phase plate (phi)
     |          |
     \----------\\----- |tr+rt> = |1> interferogram 1
          -     |
                |
                |
                 |tt+rr> = |2> interferogram 2

Fig.1B MZI (Feynman histories)

|a> = |1><1|a> + |2><2|a>

<1|a> = [a(+1) + a(-1)] = [ie^iphi + i]/2

<2|a> = [a(+2) + a(-2)] = [e^iphi -  1]/2

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

p(2) = |<2|a>|^2 = [1 - cos(phi)]/2

*Now we can see more clearly what I really mean when in the photon pair
case I write |a,e,+> parallel to |a,o,+>. That corresponds to the coherent
addition of the indistinguishable amplitudes a(+1) + a(-1) in which photon
a takes path + and it takes path - in both cases detected at counter 1.
I will develop this more for photon pair case in future postings.

It appears that Ramsay and I have been talking at cross purposes since
either picture gives the same answer in this simple case at least.

Note also that Wheeler's "delayed choice" experiment uses this
configuration. If one chooses (as Wheeler does choose) phi = 0, for
example, then

p(1) = 1 & p(2) = 0

*There is no question that this is not possible in a practical sense in the
real laboratory. There is no fundamental law that phi must be random so
that p(1) = p(2) = 1/2 - as Ramsay's picture seems to imply!


#2  Single counter MZI

The crux of the debate over quantum connection communication fact or
fantasy between me and Keith Ramsay has to do with the decision process
between interfering and non-interfering alternatives.  The following
example addresses the heart of the matter.  Ramsay professes that the two
alternative histories t and r for photon a are always non-interfering
because, he says, <t|r> = 0 and any unitary transformation must preserve
this bracket relation.

---->\\---------\
|a>  |    |t>   |
     |          - phase plate (phi)
     |          |
     \----------0 |3>
          |r>

Fig.2A MZI (single counter)

According to Ramsay:

|a> -> [e^iphi|t> + i|r>]/sqrt2 = |a>'

p(3) = |<3|a>'|^2 = |<3|t>|^2 + |<3|r>|^2

     + 2|<3|t>||<3|r>|cos(phi+arg<3|t>-arg<3|r> - pi/2)

At this point it is not clear what Ramsay would do. Ramsay what would you
do? One thing he might do is:

impose p(3) = 1 (ideal detectors in all these examples).

An alternative thing he might do is to make a formal sum over all detection
points 3 and assume random phases to get

Sum{p(3)} = |<3|t>|^2 + |<3|r>|^2
3

and make reasonable ansatz that

|<3|t>| = |<3|r>| = 1/sqrt2

This way of doing things is unesthetic having all these ad-hoc parameters
like epicycles that must be adjusted to get what we know by intuition is
the right answer.

The Feynman history way of doing this problem is simple to write:

|a> -> [e^iphi a(t) + a(r)]|3> = |a>'

where a(t) is amplitude for photon to take t(r) path without intervening
phase plates etc.. We add these ampltudes coherently since the single
counter (clamped) makes them indistinguishable (i.e. no 1-1 correlation to
orthogonal kets of some measuring system like a Heisenberg microscope in
"on" mode).

Therefore,

p(3) = |<3|a>'|^2 = |e^iphi a(t) + a(r)|^2 = 1

but,

a(t) = e^iarga(t)/sqrt2

a(r) = e^iarga(r)/sqrt2

Therefore, conservation of probability requires

cos[phi + arg(t) - arg(r)] = 0

or

phi + arg(t) - arg(r) = pi/2

is a solution.

Note that the relative phase shift of the beam splitter is not frozen in
stone like the Ten Commandments but adjusts itself globally to the boundary
conditions or "total experimental arrangement" - in this case the setting
on the variable phase plate.

The point is that there is nothing in the quantum formalism that prohibits
"indistinguishable" single counter detection as shown in Fig.1c.  Ramsay's
comments on different momentum components is adhoc nonsense.  I will go on
to develop a sequence of examples leading to final quantum connection
communicator in future posts.


#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?

Preamble:Ramsay's remarks about summing over phi are clearly wrong as he
has not conceptually distinguished between a 2-slit experiment and a Mach-
Zehnder experiment.

*Yes, if my transmitter used two slits and a screen, then Ramsay would be
right. But his point is not relevant for the Mach-Zehnder case.




#6 pair-correlated light (single transmitter counter |1>)


   |3>     _                   _
   -|-|-|-| |                 | |-|-|-|-|-|-\
          | |                 | |    |e+>   |
          | |                 | |           - (phi)
   |4>    | |                 | |    _      |
   -o-o-o-|_| <-    O       ->|_|-o-|_|-|-|-0 |1>

                                    1/2



   receiver                     transmitter


Fig. 6A

As before in #5:

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>'

The recombination at counter 1 induces

|a,b>' -> |a,+>|1>{e^iphi|b,+> + |b,->}/sqrt2 = |a,b>'''

Ramsay and others object that no unitary transformation can make two
orthogonal states go to a single state in the sense that an orthogonal
basis cannot be collapsed to one vector in the basis. That is true, but
that is not what is happening here. What we have here is the time-reverse
of a beam splitter. For example in #1 an initial state |a> was split into
two orthogonal states |t> and |r>. That is:

|a> -> [|t> + i|r>]/sqrt2

clearly the time reverse is also possible

[|t> + i|r>]/sqrt2 -> |a>

This is not the same as |t> -> |t> and |r> -> |t> which is what I was
construed to have meant - partly my own fault, perhaps. But this is what I
mean now! In the case at hand, it not that

|a,e> -> |a,e> and |a,o> -> |a,e> but rather |a,b>' -> |a,b>'''

note that

<a,b|a,b> = '<a,b|a,b>' = '''<a,b|a,b>''' = 1

so these transformations are unitary.

Let us review:

|a,b> is entangled both in space and spin spaces. The half-wave plate
disentangles |a,b> in spin space to |a,b>'. The single counter arrangement,
which may require some clever optical engineering to construct properly,
further disentangles in space |a,b>' to |a,b>'''.

Indeed,

|a,b>''' = |a>|b> where

|a> = |a,+>|1>

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


Therefore, the phase information at the transmitter has been "teleported"
to the receiver on the quantum connection by the quantum erasure of both
the spin and space information at the transmitter. This is a new law of
physics buried in standard quantum mechanics.

p(3) = |<3|b>|^2 = |(e^iphi cos@ + sin@)/sqrt2|^2 =  [1 + sin2@ cos(phi)]/2

p(4) = |<4|b>|^2 = |(-e^iphi sin@ + cos@)/sqrt2|^2 = [1 - sin2@ cos(phi)]/2

p(3) + p(4) = 1

The quantum connection signal is

p(3) - p(4) = sin2@ cos(phi)

where @ is the nonlocal parameter - the difference in orientations of the
two crystals at the spacetime events in which each photon in the same pair
locally interacts with its crystal. The space-time interval between these
two events is irrelevant( which Budnik fails to understand in his wrong
"collapse or collapse" usenet posting).

Ramsay's remarks about summing over phi are clearly wrong as he has not
conceptually distinguished between a 2-slit experiment and a Mach-Zehnder
experiment.

*Yes, if my transmitter used two slits and a screen, then Ramsay would be
right. But his point is not relevant for the Mach-Zehnder case.