Re: a question on incompatibility of properties in a one particle system

From: Peter Kinane (pkinane_at_iol.ie)
Date: 10/19/04 

Date: 19 Oct 2004 07:58:06 -0700

Sorry about threadlet.

"bernard.chaverondier" <bernard.chaverondier@wanadoo.fr> wrote in message news:417500b1$0$4031$8fcfb975@news.wanadoo.fr...
> "Bilge" <dubious@radioactivex.lebesque-al.net> a écrit dans le message de
> news:slrncn9kad.1qe.dubious@radioactivex.lebesque-al.net...
>
> Bilge
> > The uncertainty, \delta A\Delta B >= hbar/2, occurs for any
> > pair of hermitian operators with a commutator equal to i\hbar.
>
> Chaverondier
> Yes and the issue we are discussing about is that the interpretation
> of this result in terms of uncertainties relies on the interpretation of
> quantum measurement uncertainties. The question is to know if
> such measurement induced uncertainties are fundamental or result
> from the lack of knowlege ot the observer about the quantum state
> of the measuring apparatus and that of the environement.
>
> I believe into this second interpretation which is a deterministic,
> contextual hidden variables, explicitly non local interpretation
> of quantum mechanics and I provide in this post some of the
> reasons why I favor this deterministic interpretation of QM.
>
> (see also the sub-quantum (deterministic) theory of Micho Durdevich,
> Universidad Nacional Autonoma de Mexico, "Physics Beyond the
> Limits of Uncertainty Relations". A picture of physical reality which
> is based on individual physical systems, completely causal,
> and statistically compatible with quantum mechanics.
> http://www.matem.unam.mx/~micho/subq.html)
>
> Bilge
> > Representation is irrelevant. You don't need to choose
> > a representation in which either p or x is diagonal.
>
> Chaverondier
> Of course. That is not the issue. The issue is about Heisenberg
> uncertainties and the fact that these uncertainties pertains actually
> to the uncertainties showing up in the quantum measurement process
>
> Now, the manner you transform a position representation into a
> momentum one stems from the commutation relation. This relation
> amounts to indicate that the two representations are Fourier
> transforms one of each other.
>
> The fact that the two representation are Fourier transform of each other
> (ie the commutation relation) is not intrinsically the cause of Heisenberg
> uncertainties. Indeed, the fact that two observables cannot be diagonalized
> in the same Hilbert basis wouldn't give rise two the indeterminacy of the
> measurement of one of them just after the other one has been measured
> if the quantum measurement uncertainties were not to show up.
>
> Bilge
> > You are reading too much into quantum mechanics. The uncertainty
> > in your apparatus also obeys quantum mechanics, unless you got the
> > equipment from a different universe.
>
> Chaverondier
> In don't believe that there were any uncertainty in quantum
> mechanics itself.
>
> The uncertainty is a property indicating the lack of knowledge
> of the observer, not a lack of cause deterministically ruling quantum
> dynamics of isolated quantum systems. One of the idea that plague
> the understanding of quantum mechanics is the idea that the state
> of a given particle for instance would be ill known if we are not
> up to provide one unique value of position and one unique value
> of momentum at the same time for instance.
>
> That's because of this classical belief that the Heisenberg
> uncertainties are misleadingly named uncertainties relations.
>
> This stems from a classical interpretation of what should be a complete
> knowledge of the state of the particle. This classical point of view
> cannot fit quantum world. In quantum world, the state of a particle
> is perfectly known as soon as its wave function is known. Its being
> in a superposition state should not to be considered as a lack of
> knowledge of the state of the system, but a lack of knowledge of
> the result of a quantum measurement.
>
> This lack of knowledge is a consequence of the lack of knowledge
> of the quantum state of the measuring apparatus and its environment
> and should not be mistaken for a lack of knowledge of the quantum
> state of the observed system itself.
>
> Chaverondier
> > >(the Fourier transform is an unitary, deterministic process,
> > >that has nothing to do with the unitary, deterministic
> > >evolution operator Ut = exp(-iHt/hbar))
>
> Bilge
> > None of those has anything to do with a ``deterministic'' anything.
> > All those do is change representations.
>
> Chaverondier
> The best knowledge of the momentum of a quantum system you can
> have is, for instance, its momentum representation. Now, you have
> absolutely no uncertainty or incomplete knowledge of this momentum
> representation if you know the position representation.
>
> Chaverondier
> > >This conjugate nature is necessary but not sufficient to give
> > >rise to the impossibility of simultaneous measurements.
>
> Bilge
> > What's impossible about simultaneous measurements?
>
> Chaverondier
> It's impossible to know with certainty the position measurement
> outcome of a quantum particle (for instance) even if a momentum
> measurement has provided a precise representation of its
> quantum state (in momentum representation).
>
> The uncertainty that shows up is because the outcome of a given
> measurement doesn't depend only on the quantum state of the system.
> It depends also on the quantum state of the measuring apparatus
> and that of the environment.
>
> Chaverondier
> > >Measurement uncertainties, following the Born rules, are necessary
> > >too. Without these measurements uncertainties the conjugate nature
> > >of the observables wouldn't be enough to give rise to the
> > >uncertainties relations.
>
> Bilge
> > Nonsense. The uncertainty relations
> > are defined by the commutation relations.
>
> Chaverondier
> No. Not only and you provide the reason.
>
> Bilge
> > The uncertainty is defined as the rms deviation from
> > the expectation value of an operator,
> > \Delta A^2 = <A - <A>>^2 = <A^2> - <A>^2
> > Find \Delta A\Delta B for any pair of hermitian operators which
> > have a commutator [A,B] = i.
>
> Chaverondier
> Yes, and the statistical interpretation of what is \Delta A^2
> as well as the interpretation of the inequalities between rms
> of conjugate observables as uncertainty relations rely on the
> Born rule devoted to predict the statistics of quantum
> measurements uncertainties.
>
> Chaverondier
> > >And it is. When decoherence shows up, the system evolves from a pure
> > >state rho = |psi><psi| to a so called mixed state rhô' (a weighed sum
> > >of rank 1 projectors connected to the preferred Hilbert basis of the
> > >measuring apparatus)
> > >The entropy of the observed quantum system
> > >S = -k <ln(rhô)> increases to S' = -k <ln(rhô')>
>
> Bilge
> > The choice of basis is irrelevant to the entropy. The entropy
> > is the entropy. The entropy is S = k tr (\rho ln \rho).
> > The trace is invariant under a change of basis.
>
> Chaverondier
> Why should it be otherwise ?
>
> Bilge
> > EPR correlations don't constitute information. When you actually
> > perform an experiment demonstrating otherwise, then you can say that.
> > Until then, you're trying to pass off your idea on how to fix something
> > that isn't broken as fact.
>
> Chaverondier
> The proof of the impossibility to transfer information thanks to
> EPR effect relies on the interpretation of quantum indeterminacy
> as a fundamental one. This interpretation conflicts (in my opinion)
> with the unitary, deterministic and reversible propagation of the
> infinite Von Neumann chain, but let us come back to the issue
> under discussion (which is not about FTL but about quantum
> mechanics determinism).
>
> The knowledge about EPR correlations of the system with its
> surrounding is encapsulated in the knowledge of the quantum
> state of the quantum whole comprising the observed system,
> the measuring appararus and the environmement.
>
> The reduced density operator of the observed system
> lacks the information about EPR correlations that are
> modeled in the quantum state of the quantum whole
> comprising the observed system, the measuring
> apparatus and its environement.
>
> Chaverondier
> > >As in classical physics, this apparent irreversibility and
> > >indeterminacy is a consequence of the loss of information
> > >of the local observer and (in my opinion) should not be
> > >interpreted as a fundamental indeterminacy.
>
> Bilge
> > That's your opinion. On the other hand, what quantum mechanics
> > does say is that indeterminacy is fundamental. If it weren't, it
> > would be classical mechanics.
>
> Chaverondier
> No. There is no superposition of state in classical mechanics.
> There is no quantification of the energy spectrum in classical systems.
> There are no refraction and interference effects of classical particles.
> There is no quantum inseparability in classical mechanics and
> there is no indistinguishability of particles in classical mechanics
>
> The quantum measurement indeterminacy is a consequence of
> quantum inseparability that prevents the observer to predict
> deterministically the evolution of the observed system knowing
> only the reduced density operator modeling the system
> alone (instead of the quantum state of the quantum whole
> comprising the observed system, the measuring apparatus
> and its environement).
>
> Bernard Chaverondier
> http://perso.wanadoo.fr/lebigbang/transformation.htm
> Derivation of Lorentz transforms and definition of inertial
> systems of coordinates in the framework of Aristotle space-time.
> http://perso.wanadoo.fr/lebigbang/epr.htm
> Quantum determinism or Relativist locality ?
>
>

Suggestion: Consider dropping the term "deterministic" and use
"effectuationist" instead.

news:d8097fcc.0410190642.2aa584e9@posting.google.com...

Peter Kinane
http://www.effectuationism.com/

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