NetNews Usenet Archive 1992 #26/NN_1992_26.iso/spool/sci/physics/18746

Internet Message Format | 1992-11-11 | 7.9 KB

Path: sparky!uunet!gumby!destroyer!cs.ubc.ca!unixg.ubc.ca!ramsay
From: [e-mail] (Keith Ramsay)
Newsgroups: sci.physics
Subject: Sarfatti errata
Date: 12 Nov 1992 06:08:28 GMT
Organization: University of British Columbia, Vancouver, B.C., Canada
Lines: 183
Distribution: world
Message-ID: <[e-mail]>
NNTP-Posting-Host: unixg.ubc.ca

I am open to suggestions about this discussion with Sarfatti, as to
whether this is the best way to deal with him. What I want most is to
avoid anyone being convinced by mistaken arguments here. Sarfatti
regularly states conclusions based on his arguments as fact, and I
don't want anyone misled. On the other hand, it may be that I am
really addressing myself to an audience of just one, i.e., Sarfatti
himself.

Sarfatti writes:
|Now let's apply this Feynman lesson to my quantum connection communicator.
|On the receiver side there are only two counters which catch all the light
|to a good approximation. Therefore x' is not a continuous variable but a
|discrete one x' = +1 or x' = -1, and they are exclusive alternatives.  On
|the opposite transmitter side, there is only one counter that catches all
|the light to a good approximation. So there is only on alternative, say, x
|= 0.

No, there is a continuous spectrum of possibilities, corresponding to
the various points in your detector. For that matter, there are
multiple independent paths corresponding to x'=+-1 as well. To be "the
same path" they have to be the *same path*, not just paths which
resemble each other in coarse outline. So your invocation of Feynman
methods is faulty.

|Now the skeptics will say that the Delta function model is impossible in
|principle. That the interference terms will always be zero in the sum over
|exclusive alternatives on the transmitter side. But this is adhoc there is
|nothing in quantum mechanics to demand that.

First, I am not a "skeptic". I don't have some pre-planned bias
against you. I examine what you are saying and find mistakes in it.
That is the only reason I don't believe you.

Second, no, it is not ad-hoc. It is required by the orthogonality of
the two states, corresponding to photons arriving from distinguishable
sources.

To many a crackpot, the fact that the amount of work performed by
going from on state of a system to another is independent of the path
taken is regarded as "ad hoc", or as something that would require "a
conspiracy of massive proportions" to work out exactly in all cases.
They just don't see it as a natural requirement. So they waste a lot
of time trying to circumvent it.

Recall how in classical electromagnetism the energy of a
electromagnetic field at a time t is quadratic in the magnitude of the
field. It can be regarded as a sum of the integrals of squared norms
of the electric and magnetic fields. Suppose then that there are two
electromagnetic waves, initially separate (sources far apart from each
other), which then propagate so as to overlap. What is it that
guarantees that the energy of the resulting field is the same as the
sum of the energies of the two fields separately? It is an
orthogonality of the two fields-- a natural though non-obvious
consequence of their "independent" origin.

The exactly corresponding fact in quantum mechanics is that the
states-- the wave functions-- of two photons which are initially
separate (and hence orthogonal) remain orthogonal under evolution.
When they arrive at the detector, their wave functions are orthogonal
still. It implies that the two wave functions are not simply multiples
of on another by some phase factor.

You are wasting a lot of time trying to circumvent this.

|Trivial, but false, formal proof that quantum connection communication is
|impossible.
...
|1) |a,b> can have a parametric dependence on phi so that it is
|mathematically incorrect to use the completeness relation in isolation. So
|that it does not cancel out of the problem destroying the controllable
|nonlocal coherence effect.

The state |a,b> is the one in which the photon pair was initially
prepared; there is nothing in QM which permits us to prepare a photon
pair in a state which depends upon later emerging (unpredictable)
variables.

|However, if
|
|p(phi) = Dirac delta [phi - phi']

I.e., if it is possible to focus a beam of light on a single point, so
as to evade all questions of coherence....

|Ramsay writes:
|
|"In order for the evolution to be unitary, we must have <a,e,+|a,o,+> = 0."
|
|Why?

Because they are the result of evolving two orthogonal states (the two
polarizations for the "a" photon emerging from the source) by a
unitary evolution-- the one given by the experimental apparatus in
between the source of the photons and the detector on the "a" side.

|Both |a,e,+> and |a,o,+> are identical in spin space and they both
|evolve to |h>.

No, they evolve to different ones.

If they both *evolved* unitarily to |h>, then the evolution would have
to be non-unitary. In fact, you yourself have asserted vehemently that
saying |a,e,+> --> |h> and |a,o,+> --> |h> is a "red herring" which
"has nothing to do with your proposal". So the only other alternative
is that the two (still distinct) states can be reliably observed to be
the one same |h>. But to this you have said "a thousand times no"!

Shall we now go again in a circle? And having you object that you're
being misunderstood again? It is not possible to evolve two distinct
states into just one state with certainty, by any combination of
unitary evolution and observation.

|Ramsay is confused. The proper orthogonality relations are:
|
|<a,e,+|a,e,-> = 0
|
|<a,o,+|a,o,-> = 0

These hold as well-- there is not just one set of "the" orthogonality
relations. |a,e,+>, |a,e,->, |a,o,+>, and |a,o,-> as you have
described them are all orthogonal to each other. No confusion.

|How do you know you can't combine the beams coherently.

They are distinguishable at their sources. With a different sort of
detector in place of yours, one could determine from which direction
the photons were arriving, so as to be able to measure which beam they
were on.

|If they can't be
|coherent the scheme won't work.

Indeed.

|But it is obvious that they can be coherent.

No its not, or else we would not be disagreeing.

|There are analogous experiments with Stern Gerlach's and with neutron
|interferometers.

This analogy is vague, and does not hold up under close scrutiny.

|Also if you have two calcite rhombs back to back you can
|recombine the extraordinary and ordinary beams from the first rhomb
|coherently to reproduces the incident 45 degrre linear polarized beam.

Certainly. If the photons are initially in a state {|+> + |->}/sqrt2,
where |+> and |-> represent the states of vertically and horizontally
polarized photons, then one can split the components apart and
recombine them.

However, if you feed *just* the |+> portion of the beam into the
second crystal, blocking the other, say, you will *not* get diagonally
polarized light. You will get only vertically polarized light. In
order to do what you want to, you have to be able to do what is
analogous to arranging that |+> and |-> both give rise to diagonally
polarized light when fed into the second crystal.

|I will now lay to rest forever the bogus objection made by Mike Gallis,
|Carlton Caves, Keith Ramsay et-al that my gedankenexperiment requires
|violation of local unitarity in which a finite norm superposition is
|transformed into a zero-norm superposition.

You can stop pretending that each pronouncement of yours is
irrefutable and correct.

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

If this is unitary, then <a,e,-|a,e,+> calculated using the states
referred to in the first line has to equal <a,e,+|a,o,+> calculated
using the states referred to in the second line.

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

False. And the whole rest of the argument depends upon this false
premise, which is based on a faulty visualization of the situation,
not backed up by any other physics.

Keith Ramsay
[e-mail]