🚀 go-pugleaf

Sarfatti responds:
Article 11568 (986 more) in sci.physics:
From: svetlich@math.rutgers.edu (George Svetlichny)
Subject: Commuting observables are compatible. De la Torre et. al. are
wrong
Summary: Commuting observables behave like random variables thus
compatible.
Keywords: observables, commuting, incompatible, Heisenberg, uncertainty
Date: 9 Nov 92 18:14:16 GMT

*I want to respond specifically to this part:

"Sarfatti apparently doesn't want to change to another state but consider
subsamples of a run in a fixed quantum states to conclude that there is
a reciprocal variation in the "fluctuations" of the root-mean-square of
some variables. Here the situation is even worse as the inequalities
have nothing to do with "fluctuations". The inequality does not mean
that if in some subsample one has the sample value of Delta A halved
then in that very same subsample the sample value of Delta B is doubled.
The inequalities just don't say this, think of the A=B case again. Even
the Heisenberg inequalities do not say this, provided you can give
meaning to the allegation. Suppose that in some state Delta x Delta p is
just equal to hbar. Suppose you notice that in some subset of a run of
measured values of x the subsample value of Delta x is half the value
given by the quantum mechanical formula. This does not mean that if for
those instances you had measured p instead of x, the sample value of
Delta p would be twice the predicted value. In the first place, you did
not measure p, so the Copenhagen vision would not even allow you to talk
about the supposed contrafactual values of p, and even if you allow
yourself such pleasures by considering a hidden-variable theory
conforming to the Heisenberg inequality, the conclusion would still be
false. Fluctuations are not governed by these inequalities.

As a last note, standard quantum mechanics does not allow you to control
which subset of a run will exhibit a "fluctuation" from the predicted
root-mean square value of an observable. For simplicity's sake start
from a pure vector state F. Suppose you could by some physical process
guarantee that in the next run of N measurements of F, the root-mean
square Delta A of an observable is to a statistically significant extent
different from the predicted value of Delta A. Then you can use this
process as a preparation procedure for a new ensemble which differs in
some observable from the ensemble given by F, yet it would be a
subensemble of it. Since F is pure, by standard quantum theory, it has
no subensembles different from F. Thus such processes are disallowed by
standard quantum theory, it does not allow you to manipulate
"fluctuations" in any statistically significant way. Such putative
manipulations would form non-quantum ensembles, and then of course one
could send superluminal signals but not on the basis of standard quantum
theory."
George Svetlichny

My predictions for receiver counter probabilities

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

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

and for transmitter probability p = 1

are totally independent of the de la Torre (and Schrodinger 1930)
arguments.

My predictions come from the standard quantum formalism. The above results
are independent of fluctuations - fluctuations are only relevant if

Sum{cos(phi)} = 0 i.e. sum over phases corresponding to all distinguishable
Feynman histories in which transmitter photon lands in different macro-
spots vanishes.

Since the photo current in the +' counter, for example is proportional to
p(+'), it is elementary and obvious that the root mean square fluctuation
in that current will be proportional to sin(2theta) because root mean
square fluctuation in cos(phi) is not zero even if mean of cos(phi) is
zero.  The sin(2theta) is a coefficient - a nonlocal macroparameter having
to two with the relative orientations of the calcite crystals at the times
they scatter the two photons in the same pair.  If Dr. Svetlichny's fancy
math contradicts such an elementary consideration, then he is making some
sort of error in connecting his math to the physics.

The important point is that, in principle, there is no fundamental reason,
other than retarded causality itself - which we are testing - why

Sum{cos(phi)} should be zero under all possible experimental designs.  This
would require a conspiracy of vast proportions.  I do not see how the
Heisenberg momentum position uncertainty would require such a result. If it
did then Wheeler's famous delayed choice experiment in "Law Without Law"
would also be wrong because Wheeler's Mach-Zehnder interferometer equations
(for a single photon rather than a pair - so no sin(2theta) factor) i.e.

p(+') = [1 + Sum{cos(phi)}]/2

p(-') = [1 - Sum{cos(phi)}]/2

would be subject to the same criticism that Sum{cos(phi)} = 0 - which as
a matter of practical optics experience need not be true for real
interferometers and real counters - if Wheeler is right.