bernard.chaverondier:
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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.

Bilge
That is completely incorrect.

Chaverondier
No.

Sorry. I thought you said the opposite of what you did say. My
apologies.

bernard.chaverondier:
"Bilge" a écrit dans le message de

A spin singlet is a pure state. It's not decomposable.
It does encompass all of the information of a pure j = 0 singlet.

Chaverondier
That's what I am stressing. state |psi = (|+-+|-+)/2^(1/2)
encompasses all the information about the quantum state
of a singlet like a pair of spin 1/2 particles (as a pair of
electrons in a maximum entangled state for instance).

That is certainly not what it appears you are stressing.

This quantum state can also be expressed equivalently,
without any loss of information, by a density operator
Rhô which represents perfectly and completely this
pure entangled state of pair of electrons.

Rhô = (1/2) (|+-+|-+)(+-|+-+|)

Of course. That's what an entangled state is.

There is no uncertainty about the pure state Rhô of the singlet.

So, you know the state is a singlet. So what?

All what has to be known about the state of the singlet is contained
in the wave vector or equivalently its density operator Rhô.

I believe I've already said this.

[...]
Bilge
If you want to say something about reduced density matrices,
write down the reduced density matrices. Then, it will be clear
what you mean and I won't assume it's meaningless jargon.

Chaverondier
OK.
So, as far as the two systems S1 and S2 don't interact,
S1 being the observed system and S2 the measuring
apparatus in the specific case when we are studying the
measurement problem, they are completely modeled
separately by their two density operators Rhô1 and Rhô2.

Indeed, the density operator Rhô of the whole system S
comprising S1 and S2 is simply the tensor product of the
density operators Rhô1 and Rhô2 modeling the quantum
states of each subsystem.

Now, when S1 and S2 begin to interact, the picture changes.
The density operator Rhô of the whole system S comprising the
observed system S1 and the measuring apparatus S2 accounts
for the EPR correlations between the two parts S1 and S2 (some
time later a third system S3 modeling the environment enters
the picture, but for the sake of simplicity let us first consider
only the observed system S1 and the measuring apparatus S2).

Now, your engaging in legerdemain. The epr correlations are the
result of the fact that the epr pair is emitted as a single quantum
state.

[...]

That's where the uncertainties in the measurement process
come from.

I have no idea what you are talking about. There is mo inherent
``uncertainty'' in the measurement process. The inherent uncertainty
is in the indeterminacy of what is being measured.

"Bilge" a écrit dans le message de
...
Chaverondier
When an observed system S1 and a measuring apparatus
S2 begin to interact. The density operator Rhô of the whole
system S comprising the observed system S1 and the measuring
apparatus S2 accounts for the EPR correlations between the
two parts S1 and S2.

Bilge
Now, your engaging in legerdemain. The epr correlations are the
result of the fact that the epr pair is emitted as a single quantum
state.

Chaverondier
No objection of course.

However, here I am not specifically speaking of an EPR correlated
pair of photons S1. I am speaking of any observed system S1 (which
can be an EPR pair or whatever else) and a measuring apparatus S2.

Chaverondier
That's where the uncertainties in the
measurement process come from.

Bilge
I have no idea what you are talking about. There is no inherent
``uncertainty'' in the measurement process. The inherent
uncertainty is in the indeterminacy of what is being measured.

Chaverondier
This indeterminacy is caused by the absence of knowledge
of the quantum state of the measuring apparatus S2 and that
of the environment S3 that interact with it.

Indeed, if the quantum state of the observed system S1
+ that of the apparatus S2 + that of the environment S3
were all known, the quantum evolution of this quantum
whole could be predicted deterministically.

Consequently the incertainty of the quantum measurement
outcomes and the final quantum sate of observed system S1
is a consequence of the lack of knowledge of the observer
about the quantum state of the measuring apparatus S2
(and that of the environment S3 that interact with the
apparatus S2 and with the observed system S1).

http://perso.wanadoo.fr/lebigbang
Compatibility of Alain Aspect experiment interpretation
as an action at a distance with a formulation of relativist
invariance of phenomena that satisfy this invariance in
the framework of Aristotle space-time SE(1)xSE(3)/SO(3)
and the compatibility of possible instantaneous transfer
of information thanks to EPR effect with an explicitly non
local and deterministic interpretation of quantum measurement.

bernard.chaverondier:
"Bilge" a écrit dans le message de
ue-al.net...
Chaverondier
When an observed system S1 and a measuring apparatus
S2 begin to interact. The density operator Rhô of the whole
system S comprising the observed system S1 and the measuring
apparatus S2 accounts for the EPR correlations between the
two parts S1 and S2.

Bilge
Now, your engaging in legerdemain. The epr correlations are the
result of the fact that the epr pair is emitted as a single quantum
state.

Chaverondier
No objection of course.

However, here I am not specifically speaking of an EPR correlated
pair of photons S1. I am speaking of any observed system S1 (which
can be an EPR pair or whatever else) and a measuring apparatus S2.

Chaverondier
That's where the uncertainties in the
measurement process come from.

Bilge
I have no idea what you are talking about. There is no inherent
``uncertainty'' in the measurement process. The inherent
uncertainty is in the indeterminacy of what is being measured.

Chaverondier
This indeterminacy is caused by the absence of knowledge
of the quantum state of the measuring apparatus S2 and that
of the environment S3 that interact with it.

Not in the least. The indeterminacy is stated explicitly by
the theory and the phenomena which exhibit that indeterminacy
are the pure states, i.e., the ones which are completely coherent.
An interference pattern occurs because the phases are completely
indeterminate, i.e., there is a non-zero commutator between the
phase and the photon number operator:

[N, \phi] = -i

That occurs for a state of _maximum_ knowledge, i.e., a
pure state in which the paths of the interfering photons
are indeterminate, not uknown or uncertain by virtue of
an interaction.

Indeed, if the quantum state of the observed system S1
+ that of the apparatus S2 + that of the environment S3
were all known, the quantum evolution of this quantum
whole could be predicted deterministically.

That simply isn't the case. The fact that [p,x] = -i\hbar means
that position and momentum are intrinsically not simultaneously
determinate quantities, even in principle. It doesn't matter
what you include in the wave function. If you include the environment,
all you get is a different wavefunction for which the same commutator
applies.

Consequently the incertainty of the quantum measurement
outcomes and the final quantum sate of observed system S1
is a consequence of the lack of knowledge of the observer
about the quantum state of the measuring apparatus S2
(and that of the environment S3 that interact with the
apparatus S2 and with the observed system S1).

It's not a lack of knowledge. From the viewpoint of quantum mechanics,
there is nothing more to know. You can't lack knowledge of what doesn't
exist in the first place.

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

Bilge
I have no idea what you are talking about. There is no inherent
``uncertainty'' in the measurement process. The inherent
uncertainty is in the indeterminacy of what is being measured.

Chaverondier
Quantum indeterminacy is caused by the absence of knowledge
of the quantum state of the measuring apparatus S2 and that
of the environment S3 that interact with it.

Bilge
Not in the least. The indeterminacy is stated explicitly
by the theory and the phenomena which exhibit that
indeterminacy are the pure states.

Chaverondier
There is no indeterminacy in the quantum evolution of pure
states. Indeterminacy shows up only when the system in a
known pure state interacts with a measuring apparatus
which quantum state is not known (or incompletely known).

Bilge
An interference pattern occurs because the phases are
completely indeterminate, i.e., there is a non-zero commutator
between the phase and the photon number operator:
[N, \phi] = -i

Chaverondier
This indeterminacy shows up only if you perform a measurement
of photon number (by a photographic plate for instance).
This implies that your wave interacts with a measuring apparatus
which quantum state is unknown (or intractable deterministically
even if known). Now, you have a very good statistical model
of the indeterminacy that stems from your lack of knowledge.

That's not that different from the case when I play head and tail by
flipping a coin. I can provide a mathematical statistical model which
will provide me with an excellent agreement with observed statistics.
Nevertheless, this mathematical model doesn't oppose the interpretation
that this indeterminacy stems from my lack of knowledge of the little
details that cause the coin to fall on one face or the other one.

Bilge
That occurs for a state of _maximum_ knowledge,

Chaverondier
Of maximum knowledge of _a part_ of the quantum whole
comprising the observed system, the measuring apparatus
+the environment that interact with them. The part comprising
only the observed system.

The maximum knowledge of the observed system is not enough
to predict deterministically what happens to the inseparable
quantum whole comprising the observed system, the measuring
apparatus + the environment that interact with them

Bilge
i.e., a pure state in which the paths of the interfering photons
are indeterminate, not unknown or uncertain by virtue of
an interaction.

Chaverondier
There are no paths for the photon whatever the experiment you
consider. That's a classical concept which is inappropriate.

Now, when you measure the position of a photon you cannot
say that the photon was where you have measured it to be before
you performed the quantum measurement. The measuring apparatus
(hence its exact quantum sate) participates strongly to the creation
of this interaction event.

You cannot predict where the photon's position will be observed
because the pure state of your photon is perfectly known but this
outcome depends also on the quantum state of your detector
which is not known.

Chaverondier
Indeed, if the quantum state of the observed system S1
+ that of the apparatus S2 + that of the environment S3
were all known, the quantum evolution of this quantum
whole could be predicted deterministically.

Bilge
That simply isn't the case. The fact that [p,x] = -i\hbar means
that position and momentum are intrinsically not simultaneously
determinate quantities, even in principle.

Chaverondier
This doesn't oppose the physics interpretation I propose for
the uncertainties that can be ascribed to this commutator.

Bilge
It doesn't matter what you include in the wave function.
If you include the environment, all you get is a different
wavefunction for which the same commutator applies.

Chaverondier
Your commutator has no intrinsic statistical meaning.
The uncertainties associated to these commutators
enter the play only when a measurement situation is
considered and the same interpretation of the
measurement indeterminacy can be provided.

You cannot predict the quantum evolution of the quantum whole
Comprising the quantum system + the measuring apparatus +
the environment having only a maximal knowledge of the state
of the observed quantum system. My interpretation doesn't oppose
the Heisenberg uncertainties. It provides a physics interpretation
which is inherently linked to quantum measurement uncertainty
(which stems form the lack of knowledge of the quantum state
of the measuring apparatus).

Chaverondier
Consequently the uncertainty of the quantum measurement
outcomes and the final quantum sate of observed system S1
is a consequence of the lack of knowledge of the observer
about the quantum state of the measuring apparatus S2
(and that of the environment S3 that interact with the
apparatus S2 and with the observed system S1).

Bilge
It's not a lack of knowledge. From the viewpoint of quantum
mechanics, there is nothing more to know. You can't lack knowledge
of what doesn't exist in the first place.

Chaverondier
Whatever the way you perform quantum measurements, the observer
will always lack some knowledge of a part of the inseparable quantum
whole involved in a measuring process. I don't see the need to assume
any additional source of quantum indeterminacy which (in my opinion)
would conflict with the deterministic behavior of isolated quantum
systems.

http://perso.wanadoo.fr/lebigbang
Compatibility of Alain Aspect experiment interpretation
as an action at a distance with a formulation of relativist
invariance of phenomena that satisfy this invariance in
the framework of Aristotle space-time SE(1)xSE(3)/SO(3)
and the compatibility of possible instantaneous transfer of
information thanks to EPR effect with an explicitly non local
and deterministic interpretation of quantum measurement.