🚀 go-pugleaf

Sarfatti responses in [..]
ramsay@unixg.ubc.ca Wed Nov 11 23:18:52 1992
Date: Wed, 11 Nov 92 22:09:46 PST
From: Keith Ramsay <ramsay@unixg.ubc.ca>
To: sarfatti@well.sf.ca.us

As usual:

Newsgroups: sci.physics
Subject: Sarfatti errata
Organization: University of British Columbia, Vancouver, B.C., Canada

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.

[This is a very arrogant remark - not at all relevant to an honest
disagreement over a scientific issue of fundamental significance - if the
quantum connection is a communication channel then that is very significant
for physics. Indeed, it is a paradigm shift that gets people upset because
it insults their prejudices about how the world is. Look what they did to
Giordano Bruno and to Galileo.]

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.

[Yes, there may be a continuous spectrum - but they are all operationally
equivalent. When Feynman discusses simple examples like the two-slit
experiment he does not use a continuous spectrum. The main point is whether
the continuous spectrum destroys phase information. If it always does then,
for example, Wheeler's delayed thoice experiment, which you have ignored,
would not work because the cos(phi) factors in his detector probabilities
would wash out. I suspect that other interferometers would also not work if
you were right - but they do work.]

|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".

[There iws nothing wrong in being a "skeptic".]

I don't have some pre-planned bias against you. I examine what you are
saying and find mistakes in it. [Sometimes you do, and sometimes you think
you do but don't. Fine no problem.]
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.

***[ WHAT? What "distinguishable sources" are you talking about.  A photon
is incident on a calcite rhomb. It can take essentially two alternative
paths the ordinary and the extraordinary. This is precisely equivalen to a
a photon having the option of passing one slit or the other. (The
"continuous spectrum" you refer to above is not operationally significant
at this stage.). Ramsay you seem to have a fundamental physical
misconception in your sentence "It is required by the orthogonality of
the two states, corresponding to photons arriving from distinguishable
sources." What are you imagining when you say "distinguishable sources"?]

To many a crackpot,[Ramsay - you are losing it. You just went off the wall
with "distinguishable sources" which do not exist in my dedankenexperiment
and no you are stooping to a guttersnipe remark by insinuating that I am a
crackpot - like Joe Mc Carthy calling someone a communist.  If you thought
I was a crackpot why bother in the first place. Why debase yourself by
calling insulting names? It only adds emotional noise to the rational
scientific signal. I may be wrong, but I am not a crackpot. So don't be
disrespectful - you are better than that.]

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.

[Ramsay, you lose me here in theis irrelevant tirade. I never challenge
path independence in conservative mechanical systems - what universe have
you slipped into?]

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.

[OK you seem to be back on the "distinguishable sources" obsession. There
are no distinguishable sources - unless you mean that the e and o rays are
intrinsically distinguishable so that you cannot construct reconstruct in
incident polarized ray by using a second calcite - that is false to fact,
but, if it were true it would destroy the effect I predict in a testable
quantitative way.]

The exactly corresponding fact in quantum mechanics is that the
states-- the wave functions-- of two photons

[two photons? you mean transmitter photon a and receiver photon b.
The half wave plate only acts on transmitter photon a. If photon a is on
the o path the plate rotates its plane of polarization by 90 degrees to
line it up in same direction as it is on the e path - so the polarizations
on the two alternative paths for single photon a are no longer orthogonal!
This is obvious and elementary.]

which are initially separate (and hence orthogonal) remain orthogonal under
evolution. hen they arrive at the detector, their wave functions are
orthogonal till. It implies that the two wave functions are not simply
multiples of on another by some phase factor.

***[Gottcha Ramsay! - you are really confused on the physics here. It is
now obvious that by "two photons" you do not mean photons a and b. You
falsely picture one photon for state |a,e,+> and another photon for state
|a,0,-> - but there is only one photon here not two! these states are
alternatives for the same photon!  The transmitter photon a can take either
the e path or the o path just like a photon can pass one slit or another.
The half wave plate makes the two polarizations on the two paths parallel
to each other. Therefore, the two states only differ by a phase factor
because of the rotation of the plane of polarization of one of them by the
half wave plate.]

You are wasting a lot of time trying to circumvent this. [You physical
picture is all garbled- I am not trying to circumvent what you say I am
trying to circumvent. Your picture is not what I am picturing.]

|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....

[No - the delta function is an idealization - physically it must be that
the rms deviation in phi is small compared to 2pi.]

|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.

[So what does the half-wave plate do?  For ordinary uncorrelated light if
you use the half wave plate in one beam and then superpose them you will
get first order fringes - they are no longer orthogonal.]

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

No, they evolve to different ones.

[By |h> I mean photon a will hit the same detector no matter if it is on o
space path or e space path.]

If they both *evolved* unitarily to |h>, then the evolution would have
to be non-unitary.

[Only because you deny the existence of half-wave plates and assume they
are orthogonal when they are parallel.]

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.

[It's obvious that a half-wave plate will transform the ordinary ray's
polarization into an extraordinary ray's polarization and vice versa. Also
you can, as a matter of fact, coherently recombine the extraordinary ray
with the ordinary ray to reconstruct the original incident polarized ray -
so what you say is physically wrong.]

|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.
[Again you invoke these obscure ineffable "sources"! Please define what
"sources" you are talking about. Does anyone else out there know what
Ramsay means by "sources" in this context?]

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.

[Yes, and with that detector, there would be no signal - just like
switching Heisenberg microscope on and destroying fringes. So what!
That would be a different experiment! Read Bohr's account of his
discussions with Einstein in Wheeler and Zurek's Measurement book. So your
remark is not relevant.]

|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.

[Yes it does.]

|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.

[Now you have contradicted yourself!]

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.

[I never implied that it would!]

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
ramsay@unixg.ubc.ca

&