(Bilge) wrote in message ...

Bilge
Actually the definition of a mixed state is a statistical mixture
obtained from independently prepared states.

Chaverondier
It is also used to model the state of a part S1 of an EPR correlated
system S=S1US2. This so called mixed state is characterized by a
reduced density operator (a weighted sum of rank 1 projectors instead
of a rank 1 projector characterazing a pure state). This reduced
density operator provides the statistics of quantum measurements on
part S1 when the state of S2 is not accounted for.

The reduced density operator modeling S1 alone together with the
reduced density operator modeling S2 alone provide an incomplete
knowledge of S1US2 quantum state. The state of this quantum whole is
more completely modeled by a pure state of S1US2 (or a density
operator modeling the mixed state of S1US2 if S1US2 is not isolated
hence not in a pure state).

Bilge
An epr pair is only an epr pair because the pair is described as a
single state which is completely unpolarized, so that the total spin
projection along any axis is S_z = 0. By contrast, if the photons were
independent, i.e., a statistical mixture,

Chaverondier
Photons can be independant and in a pure state (ie in a known spin
state).

Bill
I am having trouble understanding the context of some of the terminology
you use.

Bilge
Probably because he isn't using it in a very standard way and because he
appears to be attributing the correlations to the measurement.

Chaverondier
The decoherence process induced by a quantum measurement creates EPR
correlations between the observed system and the measuring apparatus,
then
between the measuring apparatus and the environment. When the
decoherence
process is achieved, the observed system is not any more in a pure
state. As any entangled system, the state of the system alone is
modeled by a reduced density operator which doesn't comprise the
information about the quantum correlation of the observed system with
its surrounding.

This knowledge is embedded in the quantum state of the quantum whole
comprising the observed system and the measuring apparatus and the
environment it is interacting with _when the measurment process has
taken place_ (before the measuring process has taken place, the
quantum state of the observed system is completely modeled by a pure
state if this sytem is isolated).

Bilge
[he thinks that] unknown information conspire at faster
than light speed in order to only make the outcome appear to be given by
the usual explanation.

Chaverondier
There is nothing unusual in the decoherence process. It gives rise to
the vanishing of the extradiagonal terms of the reduced density
operator of the observed system (caused by its entanglement with the
measuring apparatus when this reduced density operator is expressed in
the preferred Hilbert basis of the measuring apparatus). Neither is
there anything unusual or new in the entropy increase stemming out of
a quantum measurement. These are now well established facts.

Bilge
Don't be concerned. To the best I can tell, it means that the quantum
mechanical description of an epr pair as a single quantum state, in which
the state vector contains all there is to know about the state is wrong.

Chaverondier
No. It is right.

Bilge
and should be replaced by a mixed state in which there is more information

Chaverondier
This happens only AFTER one of the photons of the EPR correlated pair
has interacted with its polarizer, not before.

Bilge
I think the overall bottom line is that you can call a theory
deterministic by attributing the randomness to the unknowable and not
differentiating between probabilistic and statistical.

Chaverondier
A theory is deterministic when the evolution it predicts is
deterministic. No known facts (only hypotheses which are still
awaiting for an experimental proof) are up to justify that the
unitary, reversible and deterministic propagation of the Von Neumann
chain may be broken at some moment.

Bernard Chaverondier
http://perso.wanadoo.fr/lebigbang/epr.htm
Quantum determinism or relativist locality
http://perso.wanadoo.fr/lebigbang/no_communication.htm
The no-communication theorem

chaverondier:
(Bilge) wrote:

Bilge
Actually the definition of a mixed state is a statistical mixture
obtained from independently prepared states.

Chaverondier
It is also used to model the state of a part S1 of an EPR correlated
system S=S1US2. This so called mixed state is characterized by a
reduced density operator (a weighted sum of rank 1 projectors instead
of a rank 1 projector characterazing a pure state). This reduced
density operator provides the statistics of quantum measurements on
part S1 when the state of S2 is not accounted for.

I have no idea what SU1US2 issupposed to mean. Physics has a standard
language. If you expect to be understood, use it. If it's too much
effort to write, it's too much effort for me to try and decipher.
The reduced density matrix applies to systems in which one or more
components is _not_ detected.

The reduced density operator modeling S1 alone together with the
reduced density operator modeling S2 alone provide an incomplete
knowledge of S1US2 quantum state. The state of this quantum whole is
more completely modeled by a pure state of S1US2 (or a density
operator modeling the mixed state of S1US2 if S1US2 is not isolated
hence not in a pure state).

A spin singlet is a pure state. It's not decomposable. Stop
posting a lot of meaningless babble. If you have a point,
write out a real equation.

Bilge
An epr pair is only an epr pair because the pair is described as a
single state which is completely unpolarized, so that the total spin
projection along any axis is S_z = 0. By contrast, if the photons were
independent, i.e., a statistical mixture,

Chaverondier
Photons can be independant and in a pure state (ie in a known spin
state).

No, they cannot. But, don't take my word for it. Let me provide you
with some statements taken verbatim from, ``Density Matrix Theory and
Applications'', Blum, Karl:

``It is not possible to characterize a a mixture by a single
state vector.''

``In general, {\it a beam of photons is said to be in a mixed state
if it is not possible to describe the beam in terms of a single
state vector}.''

``The system is a coherent superposition of basis states, \phi_m,
if its density matrix is not diagonal in the \phi_n representation.
If, in addition, the system is in a pure state, it is said to be
completely coherent.''

``If \rho is diagonal, the system is said to be an incoherent
superposition of the basis states states (provided there is
more than one non-vanishing element.''

[...]
Bilge
and should be replaced by a mixed state in which there is more information

Chaverondier
This happens only AFTER one of the photons of the EPR correlated pair
has interacted with its polarizer, not before.

The detection occurs on a spacelike interval. That means the detection
can be made simultaneous. There is _no_ absolute time ordering.

Bilge
I think the overall bottom line is that you can call a theory
deterministic by attributing the randomness to the unknowable and not
differentiating between probabilistic and statistical.

Chaverondier
A theory is deterministic when the evolution it predicts is
deterministic.

A theory is determimistic if it is completely specified by the
initial conditions, i.e., it's classical. Deterministic systems
are also generally chaotic.

No known facts (only hypotheses which are still
awaiting for an experimental proof) are up to justify that the
unitary, reversible and deterministic propagation of the Von Neumann
chain may be broken at some moment.

I have no idea what you are trying to say.

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

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 )

That's a dead issue.


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.

Bill Hobba:
"Bilge" wrote in message
[...]

Don't be concerned. To the best I can tell, it means that the quantum
mechanical description of an epr pair as a single quantum state, in which
the state vector contains all there is to know about the state is wrong
and should be replaced by a mixed state in which there is more information
which is simply inaccessible and the environment somehow manages to be
involved in the conspiracy. It's an elaborate and rather obtuse way of
saying there are hidden variables that conspire to look just like quantum
mechanics. I think the overall bottom line is that you can call a theory
deterministic by attributing the randomness to the unknowable and not
differentiating between probabilistic and statistical.

BTW on another matter I must admit defeat with those lectures on QFT you
gave me the link to. I did get something out of them but its mathematical
complexity is beyond me; and functional analysis - especially Hilbert Spaces
and Distribution Theory - was what I specialized in when I was into math.

As I mentioned, the material is rather dense. Since the time I mentioned
the link to those lectures, the author of the lectures on renormalization
(the one in which you were interested) was given a nobel prize for his
work in quantum field theory, so if you found things less obvious than he
did, (and a lot of the material in those volumes is like that), you have a
good reason. On the plus side, I'd say it's a rather comprehensive and
compact treatment of field theory with a good outline of string theory,
even if terse. Anytime I think it might have been more fun to have been a
theorist, I can peruse those volumes and realize I had a better aptitude
for experimental physics. People like ed witten have more mathematics
at their immediate disposal than I know exists.

As I said at the time be careful what you wish for - you may get it.
When I got Griffiths book I also got Zee's book on QFT in a Nutshell.
Its light an breezy style I find quite refreshing highly recommended
to anyone.

I haven't seen either of those, but if I come across them, I'll
give them a look-through. The only book of griffiths' that I've
seen was the E&M textbook I used as an undergraduate, which I
think is pretty good, especially for the physical intuition (although
I didn't really appreciate that fact at the time.)

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

Chaverondier
a mixed state is also used to model the state of a part S1
of an EPR correlated system S = S1 U S2. This so called
mixed state is characterized by a reduced density operator
(a weighted sum of rank 1 projectors instead of a rank 1
projector characterizing a pure state). This reduced density
operator provides the statistics of quantum measurements
on subsystem S1 of system S when the state of S2 is not
accounted for.

Bilge
I have no idea what S1 U S2 is supposed to mean.

Chaverondier
A system S comprising a subsystem S1 and a subsystem S2

Bilge
Physics has a standard language. If you expect to be
understood, use it. If it's too much effort to write, it's too
much effort for me to try and decipher.

Chaverondier
When some problem of translation or notation arise,
I try to correct it or provide a definition if it is needed
(when somebody points out the problem).

Bilge
The reduced density matrix applies to systems in
which one or more components is _not_ detected.

Chaverondier
Nevertheless, it has a precise physical meaning and can be calculated
on any subsystem of a given quantum system by the partial trace
operation (applied to the density operator of the whole system).

Chaverondier
The reduced density operator modeling S1 alone together with
the reduced density operator modeling S2 alone provide an
incomplete knowledge of S = S1 U S2 quantum state. The state
of this quantum whole is more completely modeled by a pure
state of system S (or a density operator modeling the mixed state
of system S if system S is not isolated hence not in a pure state).

Bilge
A spin singlet is a pure state.

Chaverondier
Yes.

Bilge
It's not decomposable.

Chaverondier
What do you mean by decomposable ? I precisely point out that
the reduced density matrix of each part S1 and S2 of a system
S comprising subsystems S1 and S2 (S be a singlet or not, that's
not my point) doesn't encompass all the information modeled by
the pure state of the whole system S as soon as S1 and S2 are
EPR correlated.

Bilge
Stop posting a lot of meaningless babble.
If you have a point, write out a real equation.

Chaverondier
Before writing any equation, it is necessary to
agree on what we are writing equation about.

Bilge
An epr pair is only an epr pair because the pair is described
as a single state which is completely unpolarized, so that the
total spin projection along any axis is S_z = 0. By contrast,
if the photons were independent, i.e., a statistical mixture,

Chaverondier
Photons can be independant and in a pure
state (ie in a known polarization state).

Bilge
No, they cannot.

Chaverondier
Yes they can.

Bilge
But, don't take my word for it. Let me provide you with some
statements taken verbatim from, ``Density Matrix Theory and
Applications'', Blum, Karl:

``It is not possible to characterize a mixture by a single
state vector.''

Chaverondier
Agreed. That's precisely one of the points I am stressing.

Bilge
``In general, {\it a beam of photons is said to be in a mixed state
if it is not possible to describe the beam in terms of a single
state vector}.''

Chaverondier
No objection up to here.

Bilge
``The system is a coherent superposition of basis states, \phi_n,
if its density matrix is not diagonal in the \phi_n representation.
If, in addition, the system is in a pure state, it is said to be
completely coherent.''

Chaverondier
Still no objection here.

Bilge
``If \rho is diagonal, the system is said to be an incoherent
superposition of the basis states states (provided there is
more than one non-vanishing element.''

Chaverondier
And I still completely agree with that last statement.

Bilge
[system state] should be replaced by a mixed state
in which there is more information

Chaverondier
This happens only AFTER one of the photons of the EPR
correlated pair has interacted with its polarizer, not before.

Bilge
The detection occurs on a spacelike interval. That means the detection
can be made simultaneous. There is _no_ absolute time ordering.

Chaverondier
I was not adressing the question of time ordering of spacelike
events here (this question depends on the interpretation of
quantum indeterminacy). You can make only one photon of
the pair interact. The entanglement between the EPR correlated
photons pair and one polarizer occurs as soon as one polarizer
interacts with one photon of the pair (the interaction of a second
photon with its polarizer is not needed for the entanglement of
the EPR pair with one polarizer to take place).

Chaverondier
A theory is deterministic when the
evolution it predicts is deterministic.

Bilge
A theory is determimistic if it is completely
specified by the initial conditions, i.e., it's classical.

Chaverondier
Though deterministic (quantum evolution of an isolated quantum
system is completely specified by the initial value of its state vector
and its Hamiltonian), the unitary evolution of isolated quantum
systems is generally not considered as a classical process.

Bilge
Deterministic systems are also generally chaotic.

Chaverondier
Yes. And it's probably extremely difficult to extract some specific
deterministic feature out of a chaotically deterministic dynamic
(here is the conspiracy you were evocating in a previous post).

Chaverondier
No known facts (only hypotheses which are still
awaiting for an experimental proof) are up to justify that
the unitary, reversible and deterministic propagation of
the Von Neumann chain may be broken at some moment.

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

Chaverondier
Quantum dynamics of isolated systems is deterministic.
Presently, as far as I know, no known physics observation
have provided any proof that an isolated quantum system
might, under certain circumstances, exhibit an indeterminist
or irreversible behavior.

Quantum irreversibility and indeterminacy show up only
when the observed system interacts with an other system
in an incompletely specified quantum state (or too
complex to handle other than classically).

Bernard Chaverondier
http://perso.wanadoo.fr/lebigbang/transformation.htm
Derivation of Lorentz transforms and inertial system
of coordinates in the framework of Aristotle space-time.
http://perso.wanadoo.fr/lebigbang/epr.htm
Quantum determinism or Relativist locality ?

"Bilge" wrote in message
...
Bill Hobba:
"Bilge" wrote in message
[...]

Don't be concerned. To the best I can tell, it means that the
quantum
mechanical description of an epr pair as a single quantum state, in
which
the state vector contains all there is to know about the state is
wrong
and should be replaced by a mixed state in which there is more
information
which is simply inaccessible and the environment somehow manages to be
involved in the conspiracy. It's an elaborate and rather obtuse way
of
saying there are hidden variables that conspire to look just like
quantum
mechanics. I think the overall bottom line is that you can call a
theory
deterministic by attributing the randomness to the unknowable and not
differentiating between probabilistic and statistical.

BTW on another matter I must admit defeat with those lectures on QFT you
gave me the link to. I did get something out of them but its
mathematical
complexity is beyond me; and functional analysis - especially Hilbert
Spaces
and Distribution Theory - was what I specialized in when I was into
math.

As I mentioned, the material is rather dense. Since the time I mentioned
the link to those lectures, the author of the lectures on renormalization
(the one in which you were interested) was given a nobel prize for his
work in quantum field theory, so if you found things less obvious than he
did, (and a lot of the material in those volumes is like that), you have a
good reason. On the plus side, I'd say it's a rather comprehensive and
compact treatment of field theory with a good outline of string theory,
even if terse. Anytime I think it might have been more fun to have been a
theorist, I can peruse those volumes and realize I had a better aptitude
for experimental physics. People like ed witten have more mathematics
at their immediate disposal than I know exists.

One does not win a Fields medal for no reason. And one does not decide on
physics over math without reason. Asimov mentioned all but the greatest
mathematicians eventually reach a point where is becomes too hard - that is
when you realize either you love math for moths sake and you persevere or
you go looking for something else - I went looking for something else - and
never regretted it. Greats like Feynman and Witten never did reach that
point of it becoming too hard - they choose physics because that is what
they liked.


As I said at the time be careful what you wish for - you may get it.
When I got Griffiths book I also got Zee's book on QFT in a Nutshell.
Its light an breezy style I find quite refreshing highly recommended
to anyone.

I haven't seen either of those, but if I come across them, I'll
give them a look-through. The only book of griffiths' that I've
seen was the E&M textbook I used as an undergraduate, which I
think is pretty good, especially for the physical intuition (although
I didn't really appreciate that fact at the time.)

Sorry for the confusion - it is not the same Griffith's (although he also
wrote a book on introductory QM). This is the Griffith that along with
Gell-Mann and Hartel developed the Consistent Quantum Histories
interpretation - http://info.phys.cmu.edu/people/faculty/griffiths_bob/

Thanks
Bill

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

Chaverondier
a mixed state is also used to model the state of a part S1
of an EPR correlated system S = S1 U S2. This so called
mixed state is characterized by a reduced density operator
(a weighted sum of rank 1 projectors instead of a rank 1
projector characterizing a pure state). This reduced density
operator provides the statistics of quantum measurements
on subsystem S1 of system S when the state of S2 is not
accounted for.

Bilge
I have no idea what S1 U S2 is supposed to mean.

Chaverondier
A system S comprising a subsystem S1 and a subsystem S2

Bilge
Physics has a standard language. If you expect to be
understood, use it. If it's too much effort to write, it's too
much effort for me to try and decipher.

Chaverondier
When some problem of translation or notation arise,
I try to correct it or provide a definition if it is needed
(when somebody points out the problem).

It's not a translation issue. I appreciate the difficulty in having
to communicate in a different language and do the best I can to figure
out something stated awkwardly. What I'm referring to are things like
S1US2, which is not a word in any language nor a standard way of writing
anything.

[...]
Bilge
A spin singlet is a pure state.

Chaverondier
Yes.

Bilge
It's not decomposable.

Chaverondier
What do you mean by decomposable ?

I mean precisely that. It's a single wavefunction, not two photons
propagating with independent identities.

I precisely point out that
the reduced density matrix of each part S1 and S2 of a system
S comprising subsystems S1 and S2 (S be a singlet or not, that's
not my point) doesn't encompass all the information modeled by
the pure state of the whole system S as soon as S1 and S2 are
EPR correlated.

It does encompass all of the information of a pure j = 0 singlet.
If you don't like that, you'll need to find a wavefunction that
includes whatever else you want to include.

Bilge
Stop posting a lot of meaningless babble.
If you have a point, write out a real equation.

Chaverondier
Before writing any equation, it is necessary to
agree on what we are writing equation about.

Then there isn't much point in responding, since the only way
I can differentiate between meaningless babble and something
concrete is the terms are defined by equations. 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 assune it's meaningless jargon.

Chaverondier
Photons can be independant and in a pure
state (ie in a known polarization state).

Bilge
No, they cannot.

Chaverondier
Yes they can.

Then you aren't talking about quantum mechanics. A singlet state
is _ONE_ wavefunction, |j,j_z = |0,0. It's not decomposable into
two spin 1 particles. That is a basic part of quantum theory. You
can discard quantum theory and say anything you want, however.


Bilge
But, don't take my word for it. Let me provide you with some
statements taken verbatim from, ``Density Matrix Theory and
Applications'', Blum, Karl:

``It is not possible to characterize a mixture by a single
state vector.''

Chaverondier
Agreed. That's precisely one of the points I am stressing.

Then why do you also accept the epr pair as being a singlet? A
singlet is a single state vector which is indecomposable into
two state vectors. Make up your mind.


Bilge
``In general, {\it a beam of photons is said to be in a mixed state
if it is not possible to describe the beam in terms of a single
state vector}.''

Chaverondier
No objection up to here.

Then you won't object if I assume you don't consider the photons
to be in a singlet state described by j=0.

[...]
Bilge
``The system is a coherent superposition of basis states, \phi_n,
if its density matrix is not diagonal in the \phi_n representation.
If, in addition, the system is in a pure state, it is said to be
completely coherent.''

Chaverondier
Still no objection here.

You aren't being consistent, since you also assume the photons are
correlated quantum mechanically. If you're going to argue about something,
try to be consistent and not adopt different positions on the same thing.


Bilge
``If \rho is diagonal, the system is said to be an incoherent
superposition of the basis states states (provided there is
more than one non-vanishing element.''

Chaverondier
And I still completely agree with that last statement.

OK, then you'll have to agree that you don't believe the epr correlations
exist, since you are saying that the density matrix is mixed above you
agreed that pire states are the ones which are coherent.

Bilge
The detection occurs on a spacelike interval. That means the detection
can be made simultaneous. There is _no_ absolute time ordering.

Chaverondier
I was not adressing the question of time ordering of spacelike
events here (this question depends on the interpretation of
quantum indeterminacy). You can make only one photon of
the pair interact. The entanglement between the EPR correlated
photons pair and one polarizer occurs as soon as one polarizer
interacts with one photon of the pair (the interaction of a second
photon with its polarizer is not needed for the entanglement of
the EPR pair with one polarizer to take place).

You have completely misunderstood relativity. The points have a spacelike
separation. ``As soon as'' has no meaning. The two events that constitute
the mesurement have no intrinsic temporal relationshop to each other.
``As soon as'' implies some reality which contridicts relativity.

[...]
Bilge
A theory is determimistic if it is completely
specified by the initial conditions, i.e., it's classical.

Chaverondier
Though deterministic (quantum evolution of an isolated quantum
system is completely specified by the initial value of its state vector
and its Hamiltonian), the unitary evolution of isolated quantum
systems is generally not considered as a classical process.

Again, you are trying to convolute the meaning of deterministic
to suit your argument. I have one 22Na nucleus. Give me an equation
that tells me how long before it decays. Not a half-life or mean
lifetime. I want nn equation that let's me point to each nucleus
in any group of nuclei and know when each decays.

If you can't do that, then stop making that ridiculous statement you
keep making.

Bilge
Deterministic systems are also generally chaotic.

Chaverondier
Yes. And it's probably extremely difficult to extract some specific
deterministic feature out of a chaotically deterministic dynamic
(here is the conspiracy you were evocating in a previous post).

Chaotic systems are deterministic, Quantum mechanics is not chaotic.

[...]

Chaverondier
Quantum dynamics of isolated systems is deterministic.
Presently, as far as I know, no known physics observation
have provided any proof that an isolated quantum system
might, under certain circumstances, exhibit an indeterminist
or irreversible behavior.

Then you haven't looked very hard. The decay time for a simple
2p-1s transition is only specified by a probability. In fact,
transition rates, are by definition, the inverse of the expectation
value, i.e., 1/t is proportional to f|A|i. Give me the exact
expression that probability represents or stop being ridiculous.

bernard.chaverondier:
"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.

That is completely incorrect. A mixed state is _incoherent_. The
probabilities add up _classically_, i.e., |psi_1|^2 + |psi_2|^2 ...
You do NOT get cross terms, since the density matrix is diagonal.

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

Bilge
Things like S1US2, is not a word in any language
nor a standard way of writing anything.

Chaverondier
OK. I used S1 U S2 to mean the system S which is the Union of
subsystem S1 and subsystem S2. My intention was to ascribe a
subscript 1 to the first subsystem denoted S1 and a subscript 2 to the
second one denoted S2. The whole system is denoted S and U stands
for the union symbol often used in classical mechanics when a system
S encompasses two subsystems respectively denoted S1 and S2.

Bilge
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).

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) (|+-+|-+)(+-|+-+|)

There is no uncertainty about the pure state Rhô of the singlet.
All what has to be known about the state of the singlet is contained
in the wave vector or equivalently its density operator Rhô.

Now, on the contrary, the _reduced_ density operators
of each electron are the density matrixes

Rhô1 = (|++| +|--|)/2 for S1
(and the same for Rhô2 the density operator of electron S2)

The singlet state is incompletely described by the mixed state
of each photon modeled separately by there reduced density
operator of each subsystem S1 and S2 of the quantum whole
S because the EPR correlations between S1 and S2 are not
present in this incomplete statistical description.

Now, my point was not about the pure state of a singlet but
about the more general case concerning two EPR correlated
systems like an observed system S1 interacting with a measuring
apparatus S2. I come to that below.

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).

This density operator Rhô is not any more the tensor product
of the reduced density operators Rhô1 and Rhô2 of each part.
The reduced density operators Rhô1 and Rhô2 of each part
of this entangled quantum system S don't provide a complete
description of the quantum whole S (union of S1 and S2).

That's where the uncertainties in the measurement process
come from. When you know only the reduced density
operator Rhô1 of the observed system S1 and not the density
operator Rhô of the quantum whole comprising the observed
system S1 and the measuring apparatus S2, your knowledge
is dincomplete (as reflected by the increase of entropy induced
by the entanglement process caused by a quantum measurement).

Because of your absence of knowledge of the quantum state
of the measuring apparatus S2 (and that of the environment S3),
you can only predict the statistics of quantum measurement
outcomes in agreement with Von Neumann statistics.

Chaverondier
Photons can be independent and in a pure
state (ie in a known polarization state).

Bilge
Then you aren't talking about quantum mechanics.
A singlet state is _ONE_ wavefunction, |j,j_z = |0,0.
It's not decomposable into two spin 1 particles.

Chaverondier
That's precisely what I am stressing (when EPR correlated
systems are considered). I hope the above details will
clear up any ambiguity on what I am pointing out.

Bilge
``If \rho is diagonal, the system is said to be an incoherent
superposition of the basis states (provided there is
more than one non-vanishing element.''

Chaverondier
And I still completely agree with that last statement.

Bilge
OK, then you'll have to agree that you don't believe the epr
correlations exist, since you are saying that the density matrix
is mixed.

Chaverondier
What density matrix ?
* If you are evocating the density matrix of an EPR correlated pair
of photons (for instance) before it has interacted, it is a pure state.

Rhô = (1/2) (|+-+|-+)(+-|+-+|)

* if you are evocating the reduced density matrix of _one_ photon
of the EPR correlated pair, this density matrix is in a mixed state.

Rhô1 = (1/2) |++| + (1/2) |--|

* if you are evocating the density matrix of the system comprising
one polarizer and the EPR pair this quantum whole is in a pure state
(as far as the environment has not begun to interact).

* if you are evocating the reduced density matrix of the EPR pair alone
when it has finished to interact with one polarizer, this is a mixed state.

Rhô = (1/2) |+-+-| + (1/2) |-+-+|

A mixed state is a model of the incomplete knowledge of the observer.
Any isolated system S1 which quantum state is perfectly known is in
a pure state and its quantum evolution is deterministic as soon as he
doesn't interact with any system S2 which quantum state is ill known
or too difficult to handle other than approximately (and approximately
modeled as a classical measuring apparatus for instance when a
measurement process is considered).

Chaverondier
I was not addressing the question of time ordering of spacelike
events here (this question depends on the interpretation of
quantum indeterminacy). You can make only one photon of
the pair interact. The entanglement between the EPR correlated
photons pair and one polarizer occurs as soon as one polarizer
interacts with one photon of the pair (the interaction of a second
photon with its polarizer is not needed for the entanglement of
the EPR pair with one polarizer to take place).

Bilge
You have completely misunderstood relativity. The points have
a spacelike separation. ``As soon as'' has no meaning.

Chaverondier
First, even in a classical presentation of relativity, "as soon as"
has a precise meaning when you define what inertial frame you
are considering (when it is not explicitly stated, it means that
there is no ambiguity on the implicitly chosen frame).

Second, There is a possibility to provide a precise meaning to "as soon"
in a manner that doesn't depend on the motion of the observation frame
provided you work in the appropriate framework, ie a geometrical
framework that doesn't assume that all physics phenomena would
satisfy the principle of relativity of motion (without any exception).

You have then to derive the Lorentz transforms in the framework of
Aristotle spacetime (endowed with 3D Euclidean slices of simultaneity
and 1D Euclidean motionless world lines that are the characteristic
foliations of the rank 3 space metric and the rank 1 time metric of
Aristotle space-time A_4 = SE(1)xSE(3)/SO(3) generated by the
Aristotle strict subgroup SE(1)xSE(3) of the restricted Poincaré
group SE(1,3). See http://perso.wanadoo.fr/lebigbang
http://perso.wanadoo.fr/lebigbang/epr.htm and
http://perso.wanadoo.fr/lebigbang/transformation.htm
to get more mathematical and physical details)

Bilge
The two events that constitute the measurement have no
intrinsic temporal relationship to each other. ``As soon as''
implies some reality which contradicts relativity.

Chaverondier
Yes. It contradicts the hypothesis that any phenomenon
would satisfy the principle of relativity of motion. This
hypothesis isn't satisfied if some interaction propagate
Faster Than Light.

Nevertheless, the proof of the impossibility of interactions
propagating Faster Than Light thanks to EPR effect relies
on the hypothesis that quantum indeterminacy would be
fundamental, ie that quantum uncertainties would not be
the consequence of the lack of knowledge of the observer,
hypothesis which I don't believe (to get more details, see
http://perso.wanadoo.fr/lebigbang/no_communication.htm )

Anyhow this additional hypothesis about quantum indeterminacy
nature conflicts (in my opinion) with the unitary, deterministic
and reversible dynamics of quantum evolutions.

Bilge
You are trying to convolute the meaning of deterministic to suit
your argument. I have one 22Na nucleus. Give me an equation that
tells me how long before it decays. Not a half-life or mean lifetime.

Chaverondier
You can't. You don't know the quantum state of the quantum
whole that is involved in the decay's by products observation.
You know only the quantum state of a part of this quantum whole :
the observed 22Na nucleus alone not the EPR correlated quantum
whole comprising the 22Na nucleus, the detectors and the environment
that interact with these detectors.

When this 22Na decays, this quantum system is not anymore
isolated from its surrounding. The by products of this decay have
to be detected. This detection is a position measurement involving
the searched for source of quantum indeterminacy.

This all comes from the lack of knowledge of the quantum state of the
detectors (and of the quantum state of the environment with which these
detectors interact). Here is the source of quantum indeterminacy. In My
opinion, there is no source of quantum indeterminacy that couldn't be
interpreted as a lack of knowledge of the observer.

Bilge
Chaotic systems are deterministic, Quantum mechanics is not chaotic.

Chaverondier
This has to be proven, more particularly when a measurement process
is involved. My guess is that a deterministically chaotic process
is involved as Paul Budnik seems to argue on his web site.

Chaverondier
Quantum dynamics of isolated systems is deterministic.
Presently, as far as I know, no known physics observation
have provided any proof that an isolated quantum system
might, under certain circumstances, exhibit an indeterminist
or irreversible behavior.

Bilge
Then you haven't looked very hard. Stop being ridiculous. The decay
time for a simple 2p-1s transition is only specified by a probability.

Chaverondier
Indeed, there is a quantum measurement indeterminacy (ie an
observer's loss of information) induced by the detection process
of the by products emerging out of any quantum diffusion process.

Bernard Chaverondier
http://perso.wanadoo.fr/lebigbang/transformation.htm
Derivation of Lorentz transforms and inertial system
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
...

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.

Bilge
A mixed state is _incoherent_. The probabilities add up _classically_,
i.e., |psi_1|^2 + |psi_2|^2 ... You do NOT get cross terms,
since the density matrix is diagonal.

Chaverondier
Of course.

Except if there were some problem of terminology a superposition
state (ie implicitly a _coherent_ superposition state) is a pure state

For instance the (coherent it's implicit when not explicitly stated)
superposition |psi = (|+ + |-)/2^(1/2) is a pure state

Its density operator
rhô = (1/2) (|+ + |-)( +| +-|)
is that of a pure state, ie a rank 1 projector

The _incoherent_ superposition (it has to be explicitly stated
if an incoherent "superposition" is considered) of the
two states |+ and |- is modeled by the density operator

rhô = (1/2) |+ +| + (1/2) |--|

This _incoherent_ superposition is the weighted sum of two
rank 1 projectors. It's not a pure state. It's a mixed state.

Bernard Chaverondier
http://perso.wanadoo.fr/lebigbang/transformation.htm
Derivation of Lorentz transforms and inertial system
of coordinates in the framework of Aristotle space-time.
http://perso.wanadoo.fr/lebigbang/epr.htm
Quantum determinism or Relativist locality ?