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- From: svetlich@math.rutgers.edu (George Svetlichny)
- Newsgroups: sci.physics,alt.sci.physics.new-theories
- Subject: No FTL signal proofs still good.
- Summary: Sarfatti's objections to no FTL signal proofs do not stand.
- Keywords: superluminal, signal, quantum, correlation, Heisenberg, uncertainty
- Message-ID: <Nov.5.12.47.40.1992.23831@math.rutgers.edu>
- Date: 5 Nov 92 17:47:41 GMT
- Followup-To: sci.physics
- Organization: Rutgers Univ., New Brunswick, N.J.
- Lines: 135
-
- Sarfatti's argument against the validity of the contention that standard
- quantum mechanics (SQM) forbids superluminal quantum-correlation signals
- is apparently based on two allegations:
-
- 1. Besides dynamical equations there are boundary conditions and
- Sarfatti uses novel boundary conditions.
-
- 2. Commuting observables are nevertheless incompatible due to state
- entanglement as was shown by Schroedinger in 1930 and by de la Torre in
- the past two years.
-
- Neither of these allegations stand up as an argument for SQM implying
- superluminal signals.
-
- In relation to the first one should call attention to the fact that SQM
- _does_ impose restrictions on boundary conditions, usually through
- precise domain requirements on unbounded operators in hilbert space.
- Such requirements are usually to the effect that any putative operator
- (generally differential) that is to represent an observable should be
- essentially self adjoint, and this translates itself into boundary
- conditions on the vectors in its domain. Imposing any other "novel"
- boundary conditions is as much a deviation from SQM as is using
- non-unitary evolution (and in particular cases can amount to the same
- thing).
-
- More to the point though is that considerations of evolution for the
- no-signal theorem is secondary. The theorem is basically a consequence
- of two main assumptions:
-
- a) The mean value of any observable is given as a matrix element
- (F,AF) of a self-adjoint operator A in the state F.
-
- b) Operators corresponding to observations carried out at space-like
- separation commute.
-
- Now an observation can be carried out by a complex arrangement of
- physical object which may interact among themselves and with the system
- being observed. The functioning of such an arrangement can then be
- described by appropriate dynamical laws, evolution equations, boundary
- conditions, etc. If these do not follow SQM rules one may end up with an
- arrangement that doesn't satisfy a) above. But then such an arrangement
- by definition does not realize a SQM observable. The role of evolution
- in the no-signal theorem is thus indirect and the argument does not
- essentially depend on it. One just has to ask if the arrangements on
- both sides of Sarfatti's experiment satisfy a) above or not. If not, SQM
- has been abandoned, if yes, all arguments concerning evolution and
- boundary conditions are irrelevant.
-
- If Sarfatti admits that his arrangements satisfy a) he must now face b).
- Since he apparently doesn't deny this, the no-signal theorem follows and
- his device can't work.
-
- Well, what about Schroedinger and de la Torre? Sarfatti doesn't supply us
- with references and I've not located Schroedinger's argument, but for
- de la Torre (where one finds some mention of Schroedinger's expression)
- he probably means:
-
- A. C. del la Torre, P. Catuogno and S. Ferrando
-
- "Uncertainty and nonseparability"
- Found. Phys. Lett. 2, 235 (89)
-
- and
-
- "Nonseparability and noncommutativity in quantum systems"
- Found. Phys. Lett. 4, 49 (1991)
-
- Now these papers are certainly within the realm of SQM and though the
- authors seem to derive some of their motivation from philosophical
- considerations usually found in hidden-variable arguments, their
- calculations are within SQM. Their calculation is nothing more than to
- derive what appears to be a Heisenberg-type uncertainty relation for
- commuting observables in a given state, that is if A, B are quantum
- mechanical observables (commuting or not) then in a state F one can
- derive that
-
- Delta A Delta B >= |T(A,B,F)|
-
- where Delta A is the root mean square of the observed values of A in F
- and similarly for B. (Delta A)^2 = (F, A^2F) - (F, AF)^2, and T(A,B,F)
- = (F, ABF)-(F, AF)(F, BF).
-
- This is nothing more than the Cauchy Schwartz inequality. Similarly if X
- and Y are two _classical_ random variables then
-
- s(X)s(Y) >= |C(X,Y)|
-
- where s(X) is the standard deviation of X, likewise for Y. and C(X, Y)
- is the covariance. Thus _classical_ random variables satisfy (shall we
- say it?) a Heisenberg-type inequality.
-
- De la Torre et al. have a curious phrase in their second paper:
-
- "In the last section we have shown that there are physical states where
- commuting observables are incompatible due to the nonseparability or
- non-PI."
-
- (non-PI translates to T(A, B, F) not vanishing). This is a very strange
- idea of incompatibility. They seem to feel a Heisenberg-type inequality
- implies incompatibility. They missed a few essential points about
- Heisenberg uncertainty relations (like being able to give a
- state-independent lower bound to the right hand side). One cannot derive
- from their inequality for commuting observables the same type of
- conclusions that one derives from the Heisenberg uncertainty relations.
- If one could, all statisticians would have to study quantum mechanics
- as classical random variable would then behave like quantum
- observables.. For instance, knowing that incidence of lung cancer and
- smoking are correlated one deduces an a-priori inequality relating the
- standard deviations of the number of cigarettes smoked and of the
- severity of lung lesions, making their product bounded below by the
- modulus of the covariance. Does does this means these two variables are
- incompatible? Does it mean reducing the one number (holding smoking to
- zero, hence zero standard deviation) will make the other standard
- deviation (severity of lung lesions) go infinite? Do people with perfect
- lungs invariably smoke a widely varying number of cigarettes a day?
-
- The cited results (and probably Schroedinger's as well) have nothing to
- do with superluminal signals as the same reasoning would apply to states
- and experimental arrangements conforming to Bell's inequalities for
- which a common cause explanation for the correlations can be given, and
- no one believes that common-cause correlations can be used for
- superluminal signals. Otherwise one could build a superluminal telegraph
- utilizing a "source" that randomly picks matched pairs of socks and
- sends each one in opposite directions. There would be an observer
- catching one of the socks, and by analyzing it, would determine if the
- space-like separated observer was looking at the color or at the
- weave-pattern of the other mate. If a state-dependent seemingly
- Heisenberg-like inequality was all that was needed to send signals, this
- would work.
-
- --
- George Svetlichny /\ On leave from:
- Department of Mathematics /***| Departamento de Matematica
- Hill Center, Rutgers University /****| Pontificia Universidade Catolica
- New Brunswick, 08903 NJ /*****| Rio de Janeiro, Brazil
-