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

From: bernard.chaverondier (bernard.chaverondier_at_wanadoo.fr)
Date: 10/19/04 

Date: Tue, 19 Oct 2004 13:55:27 +0200

"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 ?

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