From: sarfatti@well.sf.ca.us (Jack Sarfatti)
Subject: Sarfatti to Ramsay NO! 10^3XNO& Wheeler's delayed choice.
Organization: Whole Earth 'Lectronic Link
Date: Mon, 9 Nov 1992 03:11:19 GMT


Sarfatti answers Ramsay NO! A thousand times NO!

From: ramsay@unixg.ubc.ca (Keith Ramsay)
Subject: Re: Sarfatti Stuff FTL QM Connection Signal
Date: 9 Nov 1992 00:02:12 GMT
Organization: University of British Columbia, Vancouver, B.C., Canada
NNTP-Posting-Host: unixg.ubc.ca


Just one simple question, Sarfatti:

Are you claiming that, according to standard quantum mechanics, it is
possible for there to be two (pure) states represented by |f> and |g>,
which are orthogonal, and an observation which can be performed, which
has the property that both |f> and |g> will reliably be collapsed into
a single state |h> by performing that observation? Let |h> be pure or
mixed-- whatever you like. Let us just be clear: are you claiming that
this is consistent with standard quantum mechanics?

Keith Ramsay
ramsay@unixg.ubc.ca

Sarfatti's answer is NO! I do not say that. Here is what I do say:
The distorted photon pair state |a,b>' after the half-wave plate acts in
the o path , and the beam splitter acts on both e and o paths (the splitter
output ports are f and g) is:

|a,b>'=

{|b+>[|f><f|a,e,+>+|g><g|a,e,+>]+|b->[|f><f|a,o,+>+|g><g|a,e,+>]}/sqrt2

<b+'|<h|a,b>' = {<b+'|b+>[<h|f><f|a,e,+>+<h|g><g|a,e,+>]

+<b+'|b->[<h|f><f|a,o,+>+<h|g><g|a,e,+>]}/sqrt2

= {cos(theta)[<h|f><f|a,e,+>+<h|g><g|a,e,+>]

+sin(theta)[<h|f><f|a,o,+>+<h|g><g|a,e,+>]}/sqrt2

<b-'|<h|a,b>' = {<b-'|b+>[<h|f><f|a,e,+>+<h|g><g|a,e,+>]

+<b-'|b->[<h|f><f|a,o,+>+<h|g><g|a,e,+>]}/sqrt2

= {-sin(theta)[<h|f><f|a,e,+>+<h|g><g|a,e,+>]

+cos(theta)[<h|f><f|a,o,+>+<h|g><g|a,e,+>]}/sqrt2

If |f><f| + |g><g| = 1

therefore, the nonlocal joint probabilities are:

<b+'|<h|a,b>' = {cos(theta)<h|a,e,+> + sin(theta)<h|a,o,+>}/sqrt2

<b-'|<h|a,b>' = {-sin(theta)<h|a,e,+> + cos(theta)[<h|a,o,+>}/sqrt2

|<b+'|<h|a,b>'|^2 = |{cos(theta)<h|a,e,+> + sin(theta)<h|a,o,+>}/sqrt2|^2

= {cos^2(theta)<a,e,+|h><h|a,e,+> + sin^2(theta)<a,o,+|h><h|a,o,+>

+ sin(theta)cos(theta)[<a,e,+|h><h|a,o,+> + <a,o,+|h><h|a,e,+>]}/2

= p(b+'|a,h)

|<b-'|<h|a,b>'|^2 = |{-sin(theta)<h|a,e,+> + cos(theta)<h|a,o,+>}/sqrt2|^2

= {sin^2(theta)<a,e,+|h><h|a,e,+> + cos^2(theta)<a,o,+|h><h|a,o,+>

- sin(theta)cos(theta)[<a,e,+|h><h|a,o,+> + <a,o,+|h><h|a,e,+>]}/2

= p(b-'|a,h)

Therefore,

p(b+'|a,h) + p(b-'|a,h) = {<a,e,+|h><h|a,e,+> + <a,o,+|h><h|a,o,+>}/2

* The important point is that we must not assume that

|a,e,+><a,e,+| + |a,o,+><a,o,+| = 1

That is the mistake that led to the bogus "missing factor of 2" and which
is still hanging you up Ramsay.  Indeed, |a,e,+> and |a,o,+> do not live in
the same spin vector spaces at all. Together they do not form an
orthonormal basis.  The space e and o q. nos. put them in different spaces.
they are like adding apples to oranges if you want to make a unitary
evolution. They each evolve according to different unitary operators U(e)
and U(o).  This is what you do not understand! (You're not the only one -
Caves, even Brian Josephson didn't get it - neither did I at first.) What
is true is that

|a,e,+><a,e,+| + |a,e,-><a,e,-| = 1

|a,o,+><o,e,+| + |o,e,-><o,e,-| = 1

It is contingent that |a,b>' has zero projection on either |a,e,-> or
|a,o,->.

In fact, it is no contradiction to say that

|<a,e,+|h>| = |<a,o,+|h>| = 1

because the geometry is such that both e and o beams go to h with
certainty.  Therefore, everything is unitary locally on each end!  Thus,

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

p(b-'|a,h) = = {1 - sin[2theta]cos[phi(e) - phi(o)]}/2

The local receiver probabilities, that do not require data from the
transmitter to measure, are:

p(b+') = = 1/2 + sin[2theta] Sum{cos[phi(e) - phi(o)]}/2

p(b-') = = 1/2 - sin[2theta] Sum{cos[phi(e) - phi(o)]}/2

The sum is over any classically distinguishable "spots" h that the
transmitter photon a can land in, if there are more than one.  A classical
phase difference probability distribution will determine this sum or
statistical average. Classical diffraction theory describes it and we can
use phase compensation plates to engineer a non uniform phase distribution.

The local transmitter probability, from Feynman's quantum amplitude rules
of adding squares before adding for distinguishable histories, is:

p(h) = p(b+') + p(b-') = 1

Quantum connection signal is

S = p(b+') - p(b-') = sin[2theta] Sum{cos[phi(e) - phi(o)]}

There is nothing in the quantum formalism that says that

Sum{cos[phi(e) - phi(o)]} = 0

And even if it is in a special case, nevertheless,

Sum{cos^2[phi(e) - phi(o)]} is not zero which implies that we can see a
sin[2theta] dependence in the root mean square fluctuations of the receiver
counter photo currents.

PS John Wheeler's article "LAW WITHOUT LAW" in QUANTUM THEORY AND
MEASUREMENT (i.e. Wheeler & Zurek, Princeton,1983) shows that your main
physical objection that

Sum{cos[phi(e) - phi(o)]} = 0

because of a specious alleged Heisenberg-dictated rapid phase variation is
wrong!  Or, if you are right Wheeler has made a very big boo boo! I mean
his "delayed choice" experiment "Fig. 4 p.183" which is physically
equivalent to what I propose - the fact that I use photon pair and Wheeler
uses one photon is not an essential difference to the issue at hand.
Wheeler's interferometer has only two counters getting outputs from a beam
splitter. One counter sees a signal {1 + cos(phi)}/2, the other sees the
complementary signal {1 - cos(phi)}/2. The argument, you have given, would
imply that <cos(phi)> = 0 in Wheeler's experiment as well as mine! This
would mean that all kinds of practical interferometers would not work. But
they do Your random phase argument is a red herring! I will go into more
detail on this if you like. The important point is that Wheeler's two
counters in his delayed choice experiment play the same role as my two
receiver counters do in the pair experiment. Indeed, my pair experiment is
equivalent to a double slit experiment! It is better than Wheeler's
experiment because in my experiment I can delay the choice until after the
(receiver) photon is counted!  In Wheeler's experiment he can only delay
the choice of whether the photon passed two slits or one up until the time
that the photon is counted - but after it passes the slits.

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