bernard.chaverondier:
"Bilge" a écrit dans le message de
ue-al.net...

Chaverondier
The uncertainty principle applies to so called conjugate observables
as for instance position R and momentum P of a given particle.
Such observables are said to be conjugate because their
commutator [R, P] writes [R, P] = i hbar

Now, the question of Heisenberg uncertainty principle is more tricky
than it is generally believed. The commutation relations should not be
mistaken for an intrinsic cause of Heisenberg uncertainties relations.

Bilge
The uncertainty relations are defined by the commutation relations.
That's why the uncertainty relations quantize quantum mechanics.

Chaverondier
Heisenberg uncertainties show up because of the commutations
relation _and_ because of the uncertainties of quantum
measurements. The two conditions are needed to provide
these observed uncertainties.

Wrong. The uncertainty, \delta A\Delta B = hbar/2, occurs for any pair
of hermitian operators with a commutator equal to i\hbar.


Indeed, when you know perfectly the sate of the observed system in
representation position (or in momentum position, that's equivalent),
that's not enough to know perfectly the position measurement outcome you
will get. You can only predict the satistics of such position measurement
outcomes according to Born rules statistics.

Representation is irrelevant. You don't need to choose a representation
in which either p or x is diagonal.

That's because the quantum measurement outcomes depend not only on the
quantum state of the observed system. They depend also on the unknown
quantum state of the measuring apparatus and that of the environment that
interact with them.

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. Quantum mechanics might be weirdm but it's
not that weird.


Chaverondier
Indeed, when you know a given quantum wave in position
representation, you know its representation in momentum
representation too without any quantum indeterminacy.
The transformation between the two representations is
the unitary and deterministic Fourier transform. There
is no possibility to find any uncertainty there.

Bilge
That makes no sense. The wavefunction is not an hermitian operator,

Chaverondier
The wave function has nothing to do with an hermitian
operator of course. Why should it be otherwise ?

It shouldn't but I was responding to your paragraph above in which
you were talking about indeterminacy and various representations for
a wavefunction, which makes no sense. I was just pointing out that it
makes no sense.

(the Fourier transform is an unitary, deterministic process,
that has nothing to do with the unitary, deterministic
evolution operator Ut = exp(-iHt/hbar))

None of those has anything to do with a ``deterministic'' anything.
All those do is change repersentations.

Bilge
so it's non-sensical to talk about uncertainty in a representation.

Chaverondier
That's what I said. There is no intrinsic uncertainty to find
in the commutation relation (which rules the shift from
momentum to position representation).

Then why bring it up at all? You're using apples to explain oranges.

[...]
Bilge
It has everything to do with the conjugate nature of the
observables.

Chaverondier
This conjugate nature is necessary but not sufficient to give
rise to the impossibility of simultaneous measurements.

What's impossible about simultaneous measurements? Any two events
with a spacelike separation can be made simultaneous, by definition
of simultaneous. It would be impossible for any two events to _not_
be able to be made simultaneous if the events were spacelike.

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.

Nonsense. The uncertainty relations are defined by the commutation
relations. 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
The explanation seems so simple that it is somewhat puzzling.
This comes from our ignorance of the quantum state of the
measuring apparatus and that of the environment that interacts
with the measuring apparatus and the observed system.

Bilge
No, it doesn't otherwise that would be apparent in the entropy.

Chaverondier
And it is. When decoherence shows up, the system evolves from a pure
state rho = |psipsi| 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 choice of basis is irrelevant to the entropy. The entropy is the
entropy.

The entropy of the observed quantum system
S = -k ln(rhô) increases to S' = -k ln(rhô')

The entropy is S = k tr (\rho ln \rho). The trace is invariant under
a change of basis.

This entropy increase models the loss of information of the local
observer. He ignores the EPR correlations that are propagating from
the system to the measuring apparatus, then from the measuring
apparatus to the environnement according to the unitary, deterministic
and reversible propagation of the infinite Von Neumann Chain.

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.


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.

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.

"Bill Hobba" a écrit dans le message de
...
Bilge
The uncertainty relations are defined by the commutation relations.
That's why the uncertainty relations quantize quantum mechanics.

Chaverondier
Heisenberg uncertainties show up because of the commutations
relation _and_ because of the uncertainties of quantum
measurements. The two conditions are needed to provide
these observed uncertainties.

Indeed, when you know perfectly the sate of the observed
system in representation position (or in momentum position,
that's equivalent), that's not enough to know perfectly the
position measurement outcome you will get. You can only
predict the satistics of such position measurement outcomes
according to Born rules statistics.

Bill
Yea - but the state represents all that can be known.

Chaverondier
And in my opinion there is no more to know as far as
the system is in a shear state.

Some information is unknown from the local observer only if
the reduced density operator of the observed system indicates
that "the system is in a mixed state". Actually, the mixed nature
of the quantum state of the system should not be mistaken for
an intrinsic property of the observed system. Indeed, it is mainly
a property of the lack of knowlegde of the local observer.

It indicates that the system is, or has been, EPR correlated with
its surrounding so that the knowledge of the local observer (which
is encapsulated in the reduced density operator of the observed
system) has been somewhat deteriorated and is now incomplete.
A part of the information that would be necessary to predict
deteministically future evolutions of the observed system is
lies in the EPR correlations of the system with its surrounding.

These EPR correlations are not modeled in the reduced
density operator. The reduced density operator represents
the best knowledge the local observer can have as far as
he knows only this quantum part of the inseparable
quantum whole comprising the observed system and
the surrounding it is EPR correlated with.

The entropy encrease that follows a quantum measurement of a
system which is not in a pure state of the measuring apparatus
(as a measurement of position of a system which is nearly
in a shear quantum state of momentum for instance)
indicates this deterioration of the knowlege of the local
observer when the quantum collapse process occurs.

S = -k lnRhô increases when the density operator Rhô
shifts from a known shear state to a mixed state (ie a weighed
sum of shear states) providing the obsever whith only
probabilities that the system be in such or such shear state.

That's analog to a classical statistical physics situation when
the increase of the Boltzmann entropy is not and has not to be
interpreted as a fundamental irreversibility, but on the contrary
as a loss of knowledge of the macroscopic observer.

Chaverondier
That's because the quantum measurement outcomes depend
not only on the quantum state of the observed system.

Bill
Why that is no one really knows - different interpretations
different reasons.

Chaverondier
Here I provide my one. The hypothesis that there would be
some unknown fundamental irreversibility and indeterminism
in quantum measurement is (in my opinion) both superfluous
and incompatible with the unitary, deterministic and reversible
propagation of the infinite Von Neumann chain.

Chaverondier
They depend also on the unknown quantum state of the measuring
apparatus and that of the environment that interact with them.

Bill
In the Copenhagen interpretation the QM state of the measuring apparatus
is irrelevant because it is considered to operate along classical lines.

Chaverondier
In my opinion, there is no classical behaviour. This should be considered
only as a convenient macroscopic approximation. This additional
hypothesis has to be considered as an efficient manner to calculate,
not as an additional postulate that would be required in order to
complete quantum mechanics basics at a fundamental level.

That's as if one mistook the Biot and Savard law for an additional
law that would have to be added to the Maxwell equations to
complete the description of electromagnetism at a fundamental level.

Bill
One of the central issues with QM, if not the central issue,
is the point where we can draw that boundary is not defined
in the theory - leading to ideas such as it can be made at
chemical processes in the brain of a human observer.

Chaverondier
In my opioinion, there is no such boundary. All the world is a quantum
world and the classical boundary is only a convenient approximation.

Chaverondier
This conjugate nature is necessary but not sufficient to give rise
to the impossibility of simultaneous measurements. Measurement
uncertainties, following the Born rules, are necessary too.

Bill
Nope - only one rule is necessary. Namely if a system is in state p
and we conduct an experiment to see if it is in state q then if will give
a true result with a probability of |q|p|^2 and the system
will then be in state q

Chaverondier
Which is the Borne rule providing the statistics of quantum
measurements outcomes given the quantum state of the system
and the preferred basis of the measuring apparatus, ie the
eigen-vectors of the observable associated with this measuring
apparatus.

Bill
This is called the collapse of the wave function and is probably the
central issue with QM. Different interpretations have different takes on
exactly what this means eg in the consistent histories interpretation they
try to do away with the idea of observation entirely so the collapse is
just something a theorist calculates - not something that actually occurs.
All the rest follows from this one assertion as was shown by
Von Neumann in Mathematical Foundations of QM.

Chaverondier
The collapse causes off-diagonal terms of the reduced density
operator of the observed system to vanish (when expressed in the
preferred Hilbert basis of the measuring apparatus). This is a real
phenomenon that stems from a progressive (though very fast)
entanglement of the observed system with the measuring
apparatus followed by the entanglement of the measuring
apparatus with its environment. That is not any more a working
hypothesis. It has been observed experimentally and it is now
studied by numerous physics laboratories.

Chaverondier
Without these measurements uncertainties the conjugate
nature of the observables wouldn't be enough to give rise
to the uncertainties relations.

Bill
I have no idea what you are trying to say.

Chaverondier
You just answered this question above. The uncertainties
that link conjugate observables stems from the uncertainties
of the measuring process (which statistics are given
by the Born rules).

In my opinion, there is no fundamental indeterminacy here.
There is only a lack of knowlege of the local observer on
the quantum state of the apparatus and the environnement
that prevent the observer to know what outcome will
be found out of the quantum measurement process.

Except in the special case when the system is in an
eigen state of the measured observable, the knowledge
of the quantum state of the system isn't enough to predict
the outcome that will stem from the deterministic quantum
evolution of the inseparable quantum whole comprising
the observed system the measuring apparatus and the
environmement that interact with them.

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 ?

"Bilge" a écrit dans le message de
...

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 = |psipsi| 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 ?

Sorry about threadlet.

"bernard.chaverondier" wrote in message ...
"Bilge" a écrit dans le message de
...

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 = |psipsi| 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.

om...

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

"bernard.chaverondier" wrote in message
...
"Bill Hobba" a écrit dans le message de
...
Bilge
The uncertainty relations are defined by the commutation relations.
That's why the uncertainty relations quantize quantum mechanics.

Chaverondier
Heisenberg uncertainties show up because of the commutations
relation _and_ because of the uncertainties of quantum
measurements. The two conditions are needed to provide
these observed uncertainties.

Indeed, when you know perfectly the sate of the observed
system in representation position (or in momentum position,
that's equivalent), that's not enough to know perfectly the
position measurement outcome you will get. You can only
predict the satistics of such position measurement outcomes
according to Born rules statistics.

Bill
Yea - but the state represents all that can be known.

Chaverondier
And in my opinion there is no more to know as far as
the system is in a shear state.

Shear state? Do you mean pure state? If so see below.


Some information is unknown from the local observer only if
the reduced density operator of the observed system indicates
that "the system is in a mixed state". Actually, the mixed nature
of the quantum state of the system should not be mistaken for
an intrinsic property of the observed system. Indeed, it is mainly
a property of the lack of knowlegde of the local observer.

Mixed state means a state that is a superposition of states of definite
position say. But such only has a meaning in regard to a predefined
experimental setup. In principle any state is 'pure' in the sense it is
assumed some experiment can be designed to detect it. I am having trouble
understanding the context of some of the terminology you use. Can you confi
rm you accept the terminology of the following presentation
http://www.physics.utoronto.ca/~stei...enna_Lect2.ppt. If not please
let me know because that details my understanding of it eg we have no idea
whether a collapse really occurs as demonstrated by the fact that
interpretations that demand it exist and other that do not also exist - and
they are all perfectly compatible with QM. What I find difficult in your
writing is you seem to be taking a certain predetermined view of things that
is known to yourself but is not spelt out clearly from the start and mixing
it up with concepts that we simply do not know for sure. The same for non
locality - we simply do not know if QM has non locality or not - it is
entirely interpretation dependant. Nothing a-priori demands that QM is
determined by hidden variables as assumed in a bell type analysis.


It indicates that the system is, or has been, EPR correlated with
its surrounding so that the knowledge of the local observer (which
is encapsulated in the reduced density operator of the observed
system) has been somewhat deteriorated and is now incomplete.
A part of the information that would be necessary to predict
deteministically future evolutions of the observed system is
lies in the EPR correlations of the system with its surrounding.

I have no idea what your are trying to say. In QM the state tells us
everything we can know about a system - EPR or no EPR (which is just and
example of strange correlations inherent in the idea that two particles are
described by the one state in certain circumstances).


These EPR correlations are not modeled in the reduced
density operator. The reduced density operator represents
the best knowledge the local observer can have as far as
he knows only this quantum part of the inseparable
quantum whole comprising the observed system and
the surrounding it is EPR correlated with.

EPR correlations are modeled completely in QM and totally consistent with
it.


The entropy encrease that follows a quantum measurement of a
system which is not in a pure state of the measuring apparatus
(as a measurement of position of a system which is nearly
in a shear quantum state of momentum for instance)
indicates this deterioration of the knowlege of the local
observer when the quantum collapse process occurs.

Again I have no idea what you are trying to say.


S = -k lnRhô increases when the density operator Rhô
shifts from a known shear state to a mixed state (ie a weighed
sum of shear states) providing the obsever whith only
probabilities that the system be in such or such shear state.

That's analog to a classical statistical physics situation when
the increase of the Boltzmann entropy is not and has not to be
interpreted as a fundamental irreversibility, but on the contrary
as a loss of knowledge of the macroscopic observer.

Chaverondier
That's because the quantum measurement outcomes depend
not only on the quantum state of the observed system.

Bill
Why that is no one really knows - different interpretations
different reasons.

Chaverondier
Here I provide my one. The hypothesis that there would be
some unknown fundamental irreversibility and indeterminism
in quantum measurement is (in my opinion) both superfluous
and incompatible with the unitary, deterministic and reversible
propagation of the infinite Von Neumann chain.

Then please provide the full mathematical details of the interpretation. An
example of such would be primary state diffusion as proposed by Ian
Percival - http://arxiv.org/abs/quant-ph/9508021. I would appreciate it if
your ideas were expressed in a similar way because I have great difficulty
understanding what you are on about. Please do not take what am about to
say next in a bad way - it is not meant to be. But what I see from you
writing is ideas jumbled together - what I would like to see in then
coherently presented such as what Griffith has done in the Consistent
Histories interpretation.


Chaverondier
They depend also on the unknown quantum state of the measuring
apparatus and that of the environment that interact with them.

Bill
In the Copenhagen interpretation the QM state of the measuring apparatus
is irrelevant because it is considered to operate along classical lines.

Chaverondier
In my opinion, there is no classical behaviour.

What we need is less opinion and more fact. The fact is interpretations
exist in which classical behavior is assumed to exist eg the Copenhagen
interpretation. The problem with the idea of no classical behavior is what
is to count as an observation. One possibility is the consistent histories
interpretation that does not require the concept from the outset.

This should be considered
only as a convenient macroscopic approximation. This additional
hypothesis has to be considered as an efficient manner to calculate,
not as an additional postulate that would be required in order to
complete quantum mechanics basics at a fundamental level.

That's as if one mistook the Biot and Savard law for an additional
law that would have to be added to the Maxwell equations to
complete the description of electromagnetism at a fundamental level.

Ok when you have the full detail of such worked out repost.


Bill
One of the central issues with QM, if not the central issue,
is the point where we can draw that boundary is not defined
in the theory - leading to ideas such as it can be made at
chemical processes in the brain of a human observer.

Chaverondier
In my opioinion, there is no such boundary. All the world is a quantum
world and the classical boundary is only a convenient approximation.

Then what is an observation? PSD solves it by introducing random processes
at about the plank time and that is what collapses the wave function.
Consistant histories resolves it by not introducing it into the description
in the first place. What you need to do is detail your ideas, including
exactly what you mean by terms like 'shear state' so we can see how it all
fits together.


Chaverondier
This conjugate nature is necessary but not sufficient to give rise
to the impossibility of simultaneous measurements. Measurement
uncertainties, following the Born rules, are necessary too.

Bill
Nope - only one rule is necessary. Namely if a system is in state p
and we conduct an experiment to see if it is in state q then if will
give
a true result with a probability of |q|p|^2 and the system
will then be in state q

Chaverondier
Which is the Borne rule providing the statistics of quantum
measurements outcomes given the quantum state of the system
and the preferred basis of the measuring apparatus, ie the
eigen-vectors of the observable associated with this measuring
apparatus.

If you call the above the Born rule then nothing else is required -
indeterminacy of momentum and position follows immediately from it.


Bill
This is called the collapse of the wave function and is probably the
central issue with QM. Different interpretations have different takes on
exactly what this means eg in the consistent histories interpretation
they
try to do away with the idea of observation entirely so the collapse is
just something a theorist calculates - not something that actually
occurs.
All the rest follows from this one assertion as was shown by
Von Neumann in Mathematical Foundations of QM.

Chaverondier
The collapse causes off-diagonal terms of the reduced density
operator of the observed system to vanish (when expressed in the
preferred Hilbert basis of the measuring apparatus). This is a real
phenomenon that stems from a progressive (though very fast)
entanglement of the observed system with the measuring
apparatus followed by the entanglement of the measuring
apparatus with its environment. That is not any more a working
hypothesis. It has been observed experimentally and it is now
studied by numerous physics laboratories.

Correct - Quantum State Diffusion and entanglement are now well known
phenomena associated with wave function collapse. But they in no way
resolve the fundamental issue - what causes the collapse or even is such
really exists. Different interpretations handle it differently. The above
sounds a lot like the consistent histories interpretation.


Chaverondier
Without these measurements uncertainties the conjugate
nature of the observables wouldn't be enough to give rise
to the uncertainties relations.

Bill
I have no idea what you are trying to say.

Chaverondier
You just answered this question above. The uncertainties
that link conjugate observables stems from the uncertainties
of the measuring process (which statistics are given
by the Born rules).

I do not know what you mean by 'Born rules'. There is one and only one
statistical assertion of QM - the Von Neumann statistical assertion that I
detailed above from which all statistical assertions of QM follow. If that
is what you mean by 'Born rules' please use the singular form - Born rule
because only one rule is necessary.


In my opinion, there is no fundamental indeterminacy here.
There is only a lack of knowlege of the local observer on
the quantum state of the apparatus and the environnement
that prevent the observer to know what outcome will
be found out of the quantum measurement process.

Fine - then write up the detail of your interpretation similar to the paper
I gave on PSD and the book by Griffith on Consistent histories.


Except in the special case when the system is in an
eigen state of the measured observable, the knowledge
of the quantum state of the system isn't enough to predict
the outcome that will stem from the deterministic quantum
evolution of the inseparable quantum whole comprising
the observed system the measuring apparatus and the
environmement that interact with them.

Thanks
Bill


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 ?

bernard.chaverondier:
"Bilge" a écrit dans le message de
ue-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 think it's fairly straight forward to determine. In order to construct
an experiment, you have to choose in advance what it is you plan to
measure so that you don't try to measure two incommensurate quantities.
This doesn't happen in classical physics and in scattering experiments,
the actual interactions in the detectors is irrelevant. Measuring an
anguar distribution that distinguishes between spin observables almost
never is done by analyzing the spin of the scattered particles. That is
too hard. What is usually done is just a simple measurement of an
asymmetry in the direction of the scattered particles or a difference in
total cross section using a polarized beam or polarized target or possibly
both. A tyipical analyzing power measurement does nothing but count
particles in two detectors which are located symmetrically about the
target. You get counts in the right side and left side detectors and take
the analyzing power to be something like A = (L-R)/(L+R). There's no
mysterious epr correlations involved. You seem to forget that quantum
mechanics gets the most use in situations that require quantum mechanics,
but have nothing at all to do with epr measurements. You still don't
have a classical experiment, since something like a tensor polarized
deuteron beam is not classical.

What you're doing is attempting to explain a particular experiment
with an unnecessary and elaborate interpretation and completely
ignoring the zillions of other, more mundane examples of commutation
relations and the uncertainty principle appear for exactly the same
reason.

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.

You can believe whatever you wish, but unless you can find an
experiment that gives what you believe some additional reality,
all you're doing is adding philosophical baggage that at best
serves no purpose and at worst is ligically incompatible with
what you are trying to explain with it.

(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)

OK, what that boils down to is yet another attempt to attribute the
uncertainty principle to something else in order to say some even
more abstract quantity is deterministic (in a weird sort of way),
but for some rather nebulous reason, we can't observe a particle
such that it doesm't appear to behave just like quantum mechanics.
Sorry, I don't buy it without an example of an experiment which can
distinguish between the interpretations. Attributing reality to
something with no real existence is silly.

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

So far, they show up just like the standard theory says they will.
As far as I know, the epr experiment did not require any novel
explanations to construct. It was constructed from standard quantum
theory, in part, because einstein thought what quantum mechanics
predicted was ridiculous on its face.

Now, the manner you transform a position representation into a
momentum one stems from the commutation relation.

That's a rather content-free statement. What you just said is
equivalent to saying the manner in which you perform a canonical
transforms stems from the poisson bracket. Well, that's only true because
a canonical transformation preserves poisson brackets by definition. The
reason canonical transformations are important is because they preserve
possoin brackets. All that means is that the poisson brackets are
important to the theory and the particular choice of variables that
preserve them is irrelevant to any physics, apart from what it happens
to be convenient to measure.

This relation amounts to indicate that the two representations are Fourier
transforms one of each other.

So what? How you transform between representations is irrelevant.
What's relevant is that the commutator bracket isn't changed in
fundamental way. What's the commutator [p,x] in the number basis,
|n? It's simple. It's -i\hbar. You can check it by performing a
canonical transformation and obtaining x = a + a(+), p = i(a - a(+)).
That quantizes the harmonic oscillator, which is quantized in whatever
basis you choose.

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.

Buy a quantum mechanics book. What you are saying is complete non-sense.

[...]
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).

Again, you demonstrate that you don't know what you are talking about.
Even conjugate operators commute on spacelike intervals.

[*snip*]

"Bill Hobba" a écrit dans le message de
...

"bernard.chaverondier" wrote in message
...

Bill
Mixed state means a state that is a superposition
of states of definite position say.

Chaverondier
No. Mixed states should not be mistaken for superposition states.
A superposition state is still a pure state because the interference
between the components of the state of the system is possible.

Bill
In principle any state is 'pure'

Chaverondier
No. If a system S1 is entangled with a system S2, then
even if S=S1US2 is in a pure state |psi, you cannot
separate system S1 from system S2 without loss
of information. This is explained on the slide
"What happens if you don't look at part of your system"
of the link you provided in this thread
http://www.physics.utoronto.ca/~stei...enna_Lect2.ppt. )

When you look only the part S1 of a system S = S1US2 in a pure
but entangled state |psi, the part S1 of S is not any more in a pure
state. It is in a mixed state that is obtained by the so called partial
trace operation rhô1 = somme_k psi2_k| rhô |psi2_k
* where rhô denotes the density operator
of system S = S1US2, rhô = |psipsi|
* where rhô1 denotes the reduced density operator of system S1
* where the |psi2_k denote an Hilbert basis of the
Hilbert space state of system S2

You loose information when you try to define an entangled
part S1 of an inseparable quantum whole S=S1US2 separately
from the rest of this quantum whole.

See for instance, the slides "decoherence arises from throwing
away information" and "decoherence party line" of the link
http://www.physics.utoronto.ca/~stei...enna_Lect2.ppt
you provided to me.

I quote the slide "decoherence party line" because it's very important
"coherence is never lost, as _unitary_ evolution preserves the purity
of states. In principle, the measurement interaction is _reversible_.
In practice, once the system interacts with the environment, ie
anything with too many degrees of freedom for us to handle,
we cannot reverse it. Just as in classical mechanics, it is the
_approximation_ of an open system which leads to effective
_irreversibility_ and _loss of information_ (increase in entropy)

loss of information = loss of coherence"

Bill
I am having trouble understanding the context
of some of the terminology you use. Can you
confirm you accept the terminology of the following presentation
http://www.physics.utoronto.ca/~stei...enna_Lect2.ppt

Chaverondier
First "shear state" is a translation error of mine. I wanted to say pure
state. Second, what I call the statistics Born rule is what you call the
Von Neumann statistical assertion.

Now, I read your link and I didn't notice any point of disagreement
except on the slide "what are the effect of measurement ?" where
there is a big mistake about the explanation of the collapse of a
system S1US2 in the state (|+-+|-+)/2^(1/2) when measuring
one of the two entangled parts.

The explanation which is provided on this slide amounts
to a _local_ hidden variable interpretation which has been
discarded by the experimental verification of Bells inequalities
violations. Otherwise, your link is OK.

Bill
What I find difficult in your writing is you seem to be taking
a certain predetermined view of things that is known to
yourself but is not spelt out clearly from the start and mixing
it up with concepts that we simply do not know for sure.

Chaverondier
Presently; the only thing where my point of view seems to differ
from the most widely accepted one is that I don't believe in the
hypothesis that the so called collapse would be something different
from the unitary, deterministic and reversible evolution of the quantum
whole comprising all parts involved in the measurement process, ie
* the observed system
* the measuring apparatus
* the environment interacting with them.

As in classical statistical mechanics, when an apparent
irreversibility and an apparent indeterminacy show up, that's
(in my opinion) a consequence of a loss of information of the
observer.

Surprisingly, your link seems to agree with this point of view
(which suggests that the views on that topic may be changing).

Bill
The same for non locality - we simply do not know if QM has
non locality or not - it is entirely interpretation dependant.

Chaverondier
You know that the result of a measurement on one part S2
depends on the measurement that has been performed on
The part S1 of an entangled system S=S1US2 (ie the state
of the measuring apparatus of S1 when the measurement
on S1 has been performed)

The best illustration is the Greenberg Horn Zeilinger
thought experiment that is even more striking that the
EPR experiment because only equalities are involved.
(see "do we really understand quantum mechanics ?" by Frank Laloe
http://www.phys.ens.fr/cours/notes-d...mq-anglais.pdf
8 Mo unhappily, chapter 4.3 GHZ equality )

Now, the interpretation of EPR effect as an action at a distance
depends on the interpretation of quantum measurement indeterminacy.
The principle of relativity of motion is preserved if quantum
measurement indeterminacy is assumed to be fundamental.
In the case when quantum indeterminacy is interpreted as a
loss of knowledge of the local observer, then the principle
of relativity of motion is lost.

Bill
Nothing a-priori demands that QM is determined
by hidden variables as assumed in a Bell type analysis.

Chaverondier
The Bell experiment and the Bell's inequalities violation
that have been proven don't demand hidden variables.
They are incompatible with _local_ hidden variables
and are compatible with _contextual_ hidden variables.

If we stick to the idea that the unitary, deterministic, reversible
quantum formalism that applies to isolated quantum systems
works fine and that no strange and unknown process breaking
this reversibility would show up, I don't see how it is possible
to escape the interpretation according to which the quantum
measurement indeterminacy stems from a loss of knowledge
of the contextual hidden variables (forgotten variables instead
of hidden would be better in my opinion) ie (in my opinion)
the quantum state of the measuring apparatus and its
environment

(see "Hidden Variables and Nonlocality
in Quantum Mechanics" Douglas Hemmick"
http://www.intercom.net/~tarababe/DissertPage.html )

and 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

Incidentally, the deterministic interpretation of the measurement
process seems to be shared by the author of the link you have provided.

Chaverondier
A mixed state indicates that the system is, or has been, EPR correlated
with its surrounding so that the knowledge of the local observer (which
is encapsulated in the reduced density operator of the observed
system) has been somewhat deteriorated and is now incomplete.
A part of the information that would be necessary to predict
deterministically future evolutions of the observed system is
lies in the EPR correlations of the system with its surrounding.

Bill
I have no idea what your are trying to say.

Chaverondier
It is explained in the link you have provided.
I quote the last slide "summary"

"the reduced density matrix of an entangled sub-system appears
mixed because the discarded part of the system carry away
information. This is the origin of decoherence of the measured
subsystem."

Bill
In QM the state tells us everything we can know about
a system - EPR or no EPR

Chaverondier
When you consider the measurement process, the system
becomes EPR correlated with the measuring apparatus.
The whole comprising the system + the measuring apparatus
can still be represented by a pure state (as far as it is not
entangled with the environment) but the observed system
cannot be anymore modeled by a pure state. It is modeled
by a mixed state, ie by its reduced density operator
(which doesn't provide information about the EPR
correlation of the observed system with the environment).
This is explained on the link you have provided.

Chaverondier
The entropy increase that follows a quantum measurement of a
system which is not in a pure state of the measuring apparatus
(as a measurement of position of a system which is nearly
in a shear quantum state of momentum for instance)
indicates this deterioration of the knowledge of the local
observer when the quantum collapse process occurs.

Bill
Again I have no idea what you are trying to say.

Chaverondier
See the slide "decoherence party line" of the
link you have provided for instance.

Chaverondier
Here I provide my one. The hypothesis that there would be
some unknown fundamental irreversibility and indeterminism
in quantum measurement is (in my opinion) both superfluous
and incompatible with the unitary, deterministic and reversible
propagation of the infinite Von Neumann chain.

Bill
Then please provide the full mathematical details of the
interpretation. An example of such would be primary state
diffusion as proposed by Ian Percival
http://arxiv.org/abs/quant-ph/9508021
I would appreciate it if your ideas were expressed in a similar
way because I have great difficulty understanding what you are
on about.

Chaverondier
I am still working the bibliography and what you are asking
amounts to a work of several years (at least). Presently, I
am working about all that and use the net as a mean to
get a lot of valuable hints and links on the topic.

The point where I am presently is that one.

1/ I have stated the compatibility of faster than light propagating
interactions with an appropriate formulation of relativist
invariance of phenomena that actually satisfy this symmetry
in the framework of Aristotle space-time
see http://perso.wanadoo.fr/lebigbang/epr.htm
(unhappily written in French. Up to now, I have not
taken the time to translate it in English) and in English
http://perso.wanadoo.fr/lebigbang/transformation.htm
derivation of the Lorentz transforms and definition of
inertial frames in the framework of Aristotle space-time.

2/ I have pointed out that the proof on the no communication
theorem relies on the dubious hypothesis that quantum
measurement indeterminacy would be of fundamental nature.
see http://perso.wanadoo.fr/lebigbang/no_communication.htm

Now, I cannot see how this assumed fundamental indeterminacy
could be interpreted as compatible with the unitary, reversible and
determinist propagation of the Von Neumann chain. So, I rather interpret
the Copenhagen interpretation to be an efficient tool as for instance is
the Biot and Savart law of electromagnetism. I believe the quantum
measurement process to be a deterministic process as soon as the
quantum whole encompassing the observed system, the measuring
apparatus and the environment apparatus are accounted for.

Bill
Then what is an observation?

Chaverondier
Of course, a model of quantum measurement is needed to support
a deterministic, explicitly non local interpretation of quantum
measurement and that's where I would like to go.

Presently, I just want to discuss these ideas to get all the objections
that can oppose this point of view and get valuable hints and links on
the topic. Indeed, I think it would not be a good idea to do a lot of
work without knowing in the first place all the objections that have
to be considered and the present "state of the art" on the topic.

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 ?

"Bilge" a écrit dans le message de
...

Chaverondier
I believe into this second interpretation which is a deterministic,
contextual hidden variables, explicitly non local interpretation
of quantum mechanics.

(see "Hidden Variables and Nonlocality in Quantum Mechanics" by
Douglas Hemmick http://www.intercom.net/~tarababe/DissertPage.html )

Bilge
You can believe whatever you wish, but unless you can find an
experiment that gives what you believe some additional reality,
all you're doing is adding philosophical baggage that at best
serves no purpose and at worst is logically incompatible with
what you are trying to explain with it.

Chaverondier
Agreed. That's what I am interested in.