I understand that quantum mechanics asserts that position and momentum
(velocity?) cannot be simultaneously determined. Yet, what happens
when you measure the position of an electron as precisely as possible?

Suppose you shoot an electron at a phosphorus screen. At the moment
of impact, you have an exact measure of the position of the electron,
relative to the screen. You know the momentum/velocity of the screen
relative to itself (i.e., 0). That must mean that you have no
information whatever about the momentum of the electron. But you do.
You know that the absolute magnitude of the velocity of the electron
is surely less than or equal to what it was before the moment of
impact. Or can electrons, when striking an object at relative rest,
ricochet off that object at a greater velocity than when the electron
was approaching the object that it will strike? Are there laws of
thermodynamics that atomic particles obey, and some they don't obey?

wrote in message
om...
I understand that quantum mechanics asserts that position and momentum
(velocity?) cannot be simultaneously determined. Yet, what happens
when you measure the position of an electron as precisely as possible?

Yes, that is known as the Heisenberg uncertainty principle.


Suppose you shoot an electron at a phosphorus screen. At the moment
of impact, you have an exact measure of the position of the electron,
relative to the screen. You know the momentum/velocity of the screen
relative to itself (i.e., 0). That must mean that you have no
information whatever about the momentum of the electron. But you do.
You know that the absolute magnitude of the velocity of the electron
is surely less than or equal to what it was before the moment of
impact. Or can electrons, when striking an object at relative rest,
ricochet off that object at a greater velocity than when the electron
was approaching the object that it will strike? Are there laws of
thermodynamics that atomic particles obey, and some they don't obey?

The key to understanding this is that you cannot observe the electron until
AFTER it has hit the screen, when it is too late. If you attempt to observe
it before it hits the screen, you affect its path. In order to observe it,
you
need to hit it with a photon (that disturbs it) and observe what happens
to the photon.
It is rather like measuring the velocity of a muon. To do that, use a pair
of
scintillators to provide a measured distance over which the velocity of the
muon can be timed. The problem is the muon has to pass through the
glass of the scintillator, then out again, then through the glass of the
second
scintillator. This is somewhat like placing a pair of tollgates on an
highway
to measure the speed of road traffic. It slows to pass through the
tollgates.
All you get is the velocity of the muon in glass.
Tame muons in the laboratory always have a velocity less than c.
Wild, feral muons have no restriction on their velocity.
Therefore it is unreasonable to conclude that a wild muon has the same
velocity as a tame one.
Androcles.

Dear richardconers:

wrote in message
om...
I understand that quantum mechanics asserts that position and momentum
(velocity?)

momentum is correct, so that it can apply to light as well.

cannot be simultaneously determined. Yet, what happens
when you measure the position of an electron as precisely as possible?

You know "next to nothing" about its momentum.

Suppose you shoot an electron at a phosphorus screen. At the moment
of impact, you have an exact measure of the position of the electron,
relative to the screen.

How exact is a phosphor dot. About 1 part in 10,000 to the size of an
atom. Not all that accurate.

You know the momentum/velocity of the screen
relative to itself (i.e., 0). That must mean that you have no
information whatever about the momentum of the electron.

Not much, but since you really haven't localized the electron, and the
"phosphor illumination" yields information about *energy* not momentum.
All vector information is lost, although something could be assumed about
path of the *host* of electrons being sent at the pixel.

But you do.
You know that the absolute magnitude of the velocity of the electron
is surely less than or equal to what it was before the moment of
impact. Or can electrons, when striking an object at relative rest,
ricochet off that object at a greater velocity than when the electron
was approaching the object that it will strike?

Actually the energy they carry can tunnel past the phosphor, through the
glass, and end up displacing electrons onto the outside of the monitor
screen. The better the tube, the better the grounding, so some screens
have little residual "static electricity". You'll recognize the ones that
don't, because they always collect lots of dust...

Are there laws of
thermodynamics that atomic particles obey, and some they don't obey?

The laws of thermodynamics don't apply to individual particles. Only to
large statistical populations, like objects, streams of particles (even
"one at a time"), etc.

David A. Smith

wrote:
I understand that quantum mechanics asserts that position and momentum
(velocity?) cannot be simultaneously determined. Yet, what happens
when you measure the position of an electron as precisely as possible?

Momentum. This is the Heisenberg unceretainty principle, and not only
does it say that position and momentum cannot be simultaneously
measured, it gives a lower bound on how accurately they can be measured.


Suppose you shoot an electron at a phosphorus screen. At the moment
of impact, you have an exact measure of the position of the electron,
relative to the screen.

No, you do not have an "exact measure". No matter what you do or how you
measure the location on the screen, you will be limited to the size of a
scintillating molecule. Note that to be able to identify which molecule
was hit by the electron, the electron must interact with the molecule,
transferring both energy and momentum to the molecule. The molecule will
recoil from that interaction, and you do not know in which direction
this recoil is, giving a lower bound on the accuracy with which you know
the momentum of the electron. There are more unknowns in this process,
but this is the major one....


You know that the absolute magnitude of the velocity of the electron
is surely less than or equal to what it was before the moment of
impact.

Momentum is a vector, and you don't know in which direction the electron
was scattered. So the momentum is uncertain.


Or can electrons, when striking an object at relative rest,
ricochet off that object at a greater velocity than when the electron
was approaching the object that it will strike?

In certain cases this can happen (the object transfers momentum and
energy to the electon). But that's not necessary here.


Are there laws of
thermodynamics that atomic particles obey, and some they don't obey?

Until this sentence you were discussing quantum mechanics, not
thermodynamics. The laws of both are probabilistic. The QM model of the
world is quite good within its domain of applicability, and does not
apply "sometimes" within that domain. the laws of thermodynamics arae
statistical, and do not apply to individual atoms or particles, only
ensembles or large numbers of such objects.


Tom Roberts

:
I understand that quantum mechanics asserts that position and momentum
(velocity?) cannot be simultaneously determined. Yet, what happens
when you measure the position of an electron as precisely as possible?

The momentum is completely indeterminate.

Suppose you shoot an electron at a phosphorus screen. At the moment
of impact, you have an exact measure of the position of the electron,
relative to the screen.

No, you only think that's the case. We're talking about really
small numbers here. Remember that we can pin down the position
and momentum of an electron to that of an atomic orbital.

You know the momentum/velocity of the screen
relative to itself (i.e., 0). That must mean that you have no
information whatever about the momentum of the electron. But you do.
You know that the absolute magnitude of the velocity of the electron
is surely less than or equal to what it was before the moment of
impact. Or can electrons, when striking an object at relative rest,
ricochet off that object at a greater velocity than when the electron
was approaching the object that it will strike?

The first thing is that velocity is not a quantum observable, momentum
is and the momentum is -i\hbar\grad\Psi, not mv. The second thing is
you shouldnt think of the indeterminacy as a lack of precision. Think
of it as literally meaning what it implies. A precise position measurement
implies the momentum was really indeterminate, not that it really had
a value but you can't measure the value.

Are there laws of thermodynamics that atomic particles obey, and
some they don't obey?

Atoms and other particles not only obey the laws of thermodynamics,
but the quantum theory that describes them resolved a long standing
problem with classical physics called gibb's paradox. Because of the
uncertainty relations, identical particles really are indistinguishable
particles, e.g., two electrons in a single state have to be treated
as a single state, not two electrons. Gibb's paradox was the result
of a _classical_ counting argument in which interchanging two particles
defined a different state of the system and overcounted the number of
states by N!, giving an entropy which was too large.

a écrit dans le message de
om...

I understand that quantum mechanics asserts that position and
momentum (velocity?) cannot be simultaneously determined.

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.

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.

Actually, the so called quantum uncertainties shows up
only when quantum measurement processes are involved.

Indeed, when you perform a position measurement, even
if you know perfectly the representation of the observed
system in its position representation, you aren't up to predict
the outcomes of your position measurement better than according
to the statistics of the Born rules. That's to say, the frequency f
that your position outcome be x0 +/-deltax/2 writes

f(x0,deltax) = || |psi(x0) ||^2 deltax

Where |psi denotes the repesentation of the quantum
state of your particle in the position repesentation.

This indeterminacy has nothing to do with the conjugate
nature of position and momentum observables.
This uncertainty is completely and uniquely defined by
the characteristic uncertainty of quantum measurements
as they are defined by the statistic Born rules.

Now, if you know perfectly the representation of the observed
system in momentum representation (which is equivalent to know
it perfectly in position representation) you can assert a similar
conclusion : the frequency that your momentum outcome
be p0 +/-deltap/2 writes

f(p0, deltap) = || |psi'(p0)||^2 delta p

Where |psi' denotes the repesentation of the quantum
state of your particle in its momentum repesentation.

This stressing of the quantum measurement origin of Heisenberg
uncertainties is very important because it shows that the interpretation
of quantum indeterminacy relies completely and uniquely on the
interpetation of quantum measurement indeterminacy.
There is no other source of quantum indeterminacy.

Now, where does this quantum
measurement indeterminacy come from ?

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.

If the quantum state of this quantum whole were to be perfectly
known a complete knowledge of the dynamic quantum evolution
of this quantum whole would be possible according to the unitary,
reversible and deterministic evolution of quantum processes.
In my opinion, the phenomenological quantum indeterminacy
stemming from the measurement process should not mistaken
for a fundamental law of nature that should have to be added
to the other postulates to get a complete description of quantum
physics at a fundamental level.

That's as if the laws ruling the behaviour of elastic solids were
mistaken for fundamental laws of nature that should be added
to ground physics of condensed matter completely at a
fundamental level.

Indeed, when a quantum measurement is performed, one
consider only the observed system. Unhappily, owing to
quantum non separability, the behaviour of the quantum whole
comprising the observed system, the measuring apparatus
and the environment interacting with them cannot be modeled
in a complete manner thanks to a model of the _reduced_
density operators modeling separately each one of these
three EPR entangled components of this quantum whole.

When I model the observed system by its reduced density
operator, I loose information about the EPR link that propagate
unitarily, reversibly and deterministically from the observed system
to the measuring apparatus and then from the measuring apparatus
to its environment during the measurement process according to
the unitary, reversible and deterministic quantum dynamics of
propagation of the so called Von Neumann chain.

Actually, the only fondamental phenomenon that explains
this loss of information is the quantum non-separability.

The entropy S = -kln rhô where rhô denotes the reduced density
operator of the observed system is a measure of the information
accessible to a local observer of the system. The increase of entropy
caused by a quantum measurement stems from a loss of information
_of the local observer_ because the model rhô of the quantum state
of the system is incomplete. It doesn't encompass the EPR
correlations of the observed system with its surrounding.

Hence, in my opinion, the loss of information of the local
observer should not be mistaken for a fundamental surge of
quantum indeterminacy but should be considered as a local
observer dependant one.

This is a very important point because, if this interpretation
of quantum indeterminacy happens to be the correct one,
the no_communication theorem, that is alleged to prove the
impossibility to send information thanks to quantum
measurements of EPR intricated parts of a quantum system,
relies on a wrong interpretation of quantum indeterminacy and
cannot any more be considered as providing a valid conclusion
(see http://perso.wanadoo.fr/lebigbang/no_communication.htm )

In such a case, the quantum measurement process has to be
interpreted as a not Lorentz covariant process implicitly
grounding an universal simultaneity (ie an observer
independant simultaneity). Moreover this provides at a principle
level a possibly observable motion with regard to the medium
where quantum waves propagate in violation of the principle
of relativity of motion (but in accordance with the principle that
there are no effects and no correlations that would have no cause).

There are only (as in classical thermodynamics) informations
that escape the possibility to be gained by the observer. Hence
entropy increase models the loss of information stemming from
this inability of the local observer to get a complete information
about the quantum whole comprising the observed system, the
measuring apparatus and the environment interacting with them.

In Quantum mechanics, this inability stems from
quantum unseparability. This interpretation gives rise to a
deterministic interpretation of quantum measurement,
hence it requires to consider quantum phenomena as
explicitly non local (owing to Bells inequalities violation
together with this deterministic interpretation of quantum
measurement).

(see 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

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 ?

wrote:

I understand that quantum mechanics asserts that position and momentum
(velocity?) cannot be simultaneously determined. Yet, what happens
when you measure the position of an electron as precisely as possible?

Suppose you shoot an electron at a phosphorus screen. At the moment
of impact, you have an exact measure of the position of the electron,
relative to the screen. You know the momentum/velocity of the screen
relative to itself (i.e., 0). That must mean that you have no
information whatever about the momentum of the electron. But you do.
You know that the absolute magnitude of the velocity of the electron
is surely less than or equal to what it was before the moment of
impact. Or can electrons, when striking an object at relative rest,
ricochet off that object at a greater velocity than when the electron
was approaching the object that it will strike? Are there laws of
thermodynamics that atomic particles obey, and some they don't obey?

The point of the uncertainty principle is that you don't know the
momentum of the electron before it struck the screen.

John Anderson

bernard.chaverondier:
a écrit dans le message de
. com...

I understand that quantum mechanics asserts that position and
momentum (velocity?) cannot be simultaneously determined.

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.

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


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.


That makes no sense. The wavefunction is not an hermitian operator,
so it's non-sensical to talk about uncertainty in a representation.
That entire paragraph is a non-sequiter.

[...]
This indeterminacy has nothing to do with the conjugate
nature of position and momentum observables.
This uncertainty is completely and uniquely defined by
the characteristic uncertainty of quantum measurements
as they are defined by the statistic Born rules.

It has everything to do with the conjugate nature of the
observables. It makes no difference _what_ you choose for a
representation.

[...]

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.

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

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

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.

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.

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.

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

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).

Chaverondier
This indeterminacy has nothing to do with the conjugate
nature of position and momentum observables.
This uncertainty is completely and uniquely defined by
the characteristic uncertainty of quantum measurements
as they are defined by the statistic Born rules.

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.
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
It makes no difference _what_ you choose for a
representation.

Chaverondier
Agreed (the issue is not that one).

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

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.

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.

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

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.

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.

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


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

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

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


In the Copenhagen interpretation the QM state of the measuring apparatus is
irrelevant because it is considered to operate along classical lines. 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
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 ?
(the Fourier transform is an unitary, deterministic process,
that has nothing to do with the unitary, deterministic
evolution operator Ut = exp(-iHt/hbar))

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).

Chaverondier
This indeterminacy has nothing to do with the conjugate
nature of position and momentum observables.
This uncertainty is completely and uniquely defined by
the characteristic uncertainty of quantum measurements
as they are defined by the statistic Born rules.

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.
Measurement uncertainties, following the Born rules, are necessary
too.

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. 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 form this one assertion as was shown by Von Neumann in
Mathematical Foundations of QM.

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

I have no idea what you are trying to say.

Thnaks
Bill


Bilge
It makes no difference _what_ you choose for a
representation.

Chaverondier
Agreed (the issue is not that one).

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

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.

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.

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 ?