🚀 go-pugleaf

No Ramsay your comment is dead wrong in what follows as I show below. And
there is no need for you to adopt a hostile arrogant tone which is not good
manners e.g. "regressed" and "ugh" - let's keep a civilized high minded
tone of discourse that will not repress creative thought. I use "dead
wrong" because you use "ugh" tit for tat - but you really are very wrong
here as everyone else will recognize. You jumped too quick blinded by
emotion that diminished your normally competent reason.

From: ramsay@math.ubc.ca (Keith Ramsay)
Newsgroups: sci.physics,alt.sci.physics.new-theories,sci.optics
Subject: Re: Does unitary QM permit connection signals after all?
Followup-To: alt.sci.physics.new-theories
Date: 12 Dec 1992 23:19:27 GMT
Organization: University of British Columbia, Vancouver, B.C., Canada
Lines: 25
NNTP-Posting-Host: euclid.math.ubc.ca

Sarfatti has regressed.

In article <Bz4yut.83L@well.sf.ca.us> sarfatti@well.sf.ca.us
(Jack Sarfatti) writes:
|OK boys and girls. I may have found an error in my "retraction proof" that
|standard unitary reversible (forget dissipation) quantum mechanics does
not
|permit the quantum connection signal.

Ugh.

|Therefore, even though the "bare" inner product vanishes, i.e.,
|
|<1e+|1o-> = 0  (5)
|
|We cannot logically conclude that the "dressed" inner product (i.e.,matrix
|element of operator S(&)e^-i&|1&><1&| sandwiched between these orthogonal
|states) also vanishes.

Call this operator S(&)e^-i&|1&><1&| of yours "U". If U is unitary,
then <1e+|1o->=0 DOES logically imply that <1e+|U|1o->=0 also.

Follow-up to alt.sci.physics.new-theories.

Keith Ramsay
-
Sarfatti replies:

Ramsay this last statement of your is false. You have garbled things. What
you say is mathematically false as well as physically confused.

First for your mathematical error.

In general. Suppose we are given a bra <a t1| and a ket |b t1> representing
physical states at time t1 whose inner product <a t1|b t1> "bra-ket" is
some complex number.  Suppose there is a unitary time evolution operator
U(t1,t2) such that

<a t2| = <a t1|U*(t1,t2)

|b t2> = U(t1.t2)|a t1>

Therefore,

<a t2|b t2> = <a t1|b t1>

this is the proper invariance.  What you say is dead wrong! Thus you claim
that if

<a t1|b t1> = 0 that <a t1|U(t1,t2)|b t1> = 0 as well - which is evidently
an error and is totally unjustified mathematically.

Second - your error in physical visualization. Yes, it's even worse in
terms of the physics - I mean your remark

"Call this operator S(&)e^-i&|1&><1&| of yours "U". If U is unitary"

There is no reason in standard quantum mechanics to assert that the
operator

S(&)e^-i&|1&><1&|

must be unitary.  In fact it cannot be unitary - and even if it was your
conclusion would be wrong as I already showed. This operator is a modulated
sum of non-unitary projection operators corresponding to summing
distinguishable possible measurements of photon 1!

Thus, if you look back you will see that the unitary time evolution between
measurements is from |1,2> to  |1,2>' which introduces the relative phase
factor e^-i& between the terms in |1,2>' - (we might need the half-wave
plate after all I may have dropped it too hastily - let's put it back in -
but it makes no difference to the point at hand)

The expression with this "Fourier transform of projection operators" comes
from summing the nonlocal joint probabilities over all places where
transmitter photon 1 can be detected to get local receiver photon 2
probability at one receiver counter i.e.,

S(&)|<2e'+|<&|1,2>'|^2

Ramsay you have gotten so obsessed that everything in sight to you must be
"unitary" - not at this stage OK?