What is quantum measurement problem?
Tony012
Why do people say decoherence does not solve the quantum measurement
problem?
Thank your for any of your comment.
Arnold Neumaier
> What is quantum measurement problem?
> Why do people say decoherence does not solve the quantum measurement
> problem?
see the entry 'Does decoherence solve the measurement problem?'
(currently S11g) of the theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
Arnold Neumaier
we_p...@yahoo.com
> What is quantum measurement problem?
If you flip a classical coin the theory predicts
that you will get either heads or tails. It is no
problem for the theory to predict which one.
It does.
If you flip a quantum coin the theory predicts
that you will get either heads or tails. It is no
problem for the theory to predict which one.
It does not.
The measurement problem is:
"In case the quantum coin lands on heads, why (and not
tails instead) ? (and vice versa)"
>Why do people say decoherence does not solve the
>quantum measurement problem?
Googleit:
http://arxiv.org/PS_cache/quant-ph/pdf/0312/0312059.pdf
I think that to understand the paper requires relatively
strong quantum mechanics backround which I do not
yet have. I suppose that to say that decohorence solves
the measurement problem is like to say that it you poke
the quantum coin with a stick and take a photograph of
it you can be sure that it has landed on either side.
We Pretty
Cl.Massé
ka80n1pg6esesjoeq...@4ax.com...
> What is quantum measurement problem?
There is none quantum measurement problem, there are several quantum
measurement problems.
One of them is what makes the difference between a measurement and a mere
interaction. A measurement acts on one particle, an interaction acts on an
infinity of particles, that is, on the wave function. Problem solved, but
with no solution.
Second one, a measurement triggers a projection onto an eigenstate, it's a
discontinuous process, not time symmetric while quantum equations are so.
Third one. What makes the difference between a microscopic and a
macroscopic system? Why can't we measure superpositions we observe at the
microscopic level? Why are there macroscopic privileged states, like
localized states.
> Why do people say decoherence does not solve the quantum measurement
> problem?
Decoherence doesn't solve the third problem. It only says that a system is
spontaneously projected onto an eigenstate when it interacts with a
disordered surrounding, but doesn't explain why there are privileged
states. For example, why we don't observe plane waves of chairs.
Decoherence is in the framework of orthodox quantum mechanics, but quantum
mechanics has fundamental problem that aren't linked to the way we use it.
--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.
Eugene Stefanovich
> If you flip a quantum coin the theory predicts
> that you will get either heads or tails.
True.
> It is no
> problem for the theory to predict which one.
Not true.
Quantum mechanics never claimed to predict which side of
the quantum coin you'll get in each particular instance. QM can only
predict probabilities (i.e., 1/2 in your case)
> It does not.
QM never intended to do that. "Hidden variable" theories want
to do that, but they can't.
>
> The measurement problem is:
> "In case the quantum coin lands on heads, why (and not
> tails instead) ? (and vice versa)"
No answer. Our current theories are silent on this point.
You are allowed to think that each time you toss a
quantum coin, God throws a die and makes the choice
(metaphorically speaking, of course). There is a deep mystery
in the way our Universe works. At this time we are nowehere
close to resolving this mystery. Maybe it doesn't have a solution
at all. Perhaps quantum coins just fall as they pleased (obeying
quantum mechanical probabilities, of course).
Eugene
Marcel LeBel
> "Tony012" <diki...@gmail.com> a écrit dans le message de news:
> ka80n1pg6esesjoeq...@4ax.com...
>
>
>>What is quantum measurement problem?
>
>
> There is none quantum measurement problem, there are several quantum
> measurement problems.
>
> One of them is what makes the difference between a measurement and a mere
> interaction. A measurement acts on one particle, an interaction acts on an
> infinity of particles, that is, on the wave function. Problem solved, but
> with no solution.
to the degrees of freedom of a particle. Textbooks gives us the example
of a particle confined to a box. In that respect, could we say that
measurements apply a similar box-like constraint on the particle thereby
causing a temporary loss of freedom and quantization, entirely due to
the constraint of the measurement? Wouldn't this view be supported by
the uncertainty principle in that, measurements would necessarily cause
this temporary quantization and uncertainty?
Similarly, could we say that a particle interacting with another
particle is actually gaining degrees of freedom by sharing its box with
another particle's box? Both particles would now assume longer
wavelenghts or lower energy levels. This simplistic view make any sense?
Marcel, le...@muontailpig.com remove particle
Chris H. Fleming
> "Tony012" <diki...@gmail.com> a écrit dans le message de news:
> ka80n1pg6esesjoeq...@4ax.com...
>
> > What is quantum measurement problem?
>
> There is none quantum measurement problem, there are several quantum
> measurement problems.
>
> One of them is what makes the difference between a measurement and a mere
> interaction. A measurement acts on one particle, an interaction acts on an
> infinity of particles, that is, on the wave function. Problem solved, but
> with no solution.
>
> Second one, a measurement triggers a projection onto an eigenstate, it's a
> discontinuous process, not time symmetric while quantum equations are so.
This one has been solved in the framework of quantum open systems. In a
large system, the Poincare Recurrence time becomes large. Measurements
involve interactions with many degrees of freedom. In a quantum open
system, evolution is not unitary nor time symmetric. Interaction
between the microscopic system and the reservoir can cause irreversable
localization.
Cl.Massé
> > Second one, a measurement triggers a projection onto an eigenstate, it's
> > a discontinuous process, not time symmetric while quantum equations are
> > so.
"Chris H. Fleming" <chris_h...@yahoo.com> a écrit dans le message de
news: 1131663467.3...@f14g2000cwb.googlegroups.com...
> This one has been solved in the framework of quantum open systems. In a
> large system, the Poincare Recurrence time becomes large. Measurements
> involve interactions with many degrees of freedom. In a quantum open
> system, evolution is not unitary nor time symmetric. Interaction
> between the microscopic system and the reservoir can cause irreversable
> localization.
That implies the outcome of a measurement depends on the state of the
apparatus, thus isn't determined by the measured system and probabilities
alone like according to quantum mechanics. There must already be a
discontinuous process in the system alone to drive the apparatus
(irreversibly) to the configuration corresponding to the eigenvalue. I
guess there is no detailed description of such a measurement process, and
thus this concept is more of a hope than of a discovery. I wouldn't invest
my time in it.
Cl.Massé
mMydnT_rJ6N...@rogers.com...
> I see some example of quantization as the result of constraints applied
> to the degrees of freedom of a particle.
In this statement, we can see two meaning of quantization:
1 - Quantization of the energy levels, that is, discretization. It's a
physical process.
2 - Path between the classical description of a system, and its description
in quantum mechanics. It's a full theoretical process.
The two concepts must not be confused.
> Textbooks gives us the example of a particle confined to a box.
The system "particle" is first quantized. Then the wave has a continuous
energy spectrum. Placed in a box, the particle-wave get a discreet
spectrum, but its state can be a superposition of states with different
energies.
> In that respect, could we say that
> measurements apply a similar box-like constraint on the particle thereby
> causing a temporary loss of freedom and quantization, entirely due to
> the constraint of the measurement?
No. There is discretization or there is not. In both case, the
measurement project the particle-wave onto a state of definite energy
(assuming the energy or a function thereof is measured.) The energy values
measured may be continuous or discrete.
> Wouldn't this view be supported by
> the uncertainty principle in that, measurements would necessarily cause
> this temporary quantization and uncertainty?
>
> Similarly, could we say that a particle interacting with another
> particle is actually gaining degrees of freedom by sharing its box with
> another particle's box? Both particles would now assume longer
> wavelenghts or lower energy levels. This simplistic view make any sense?
In a mere interaction, even if the energy of the particles are discretized,
they remain superpositions of states of different energy, to the contrary of
a measure. Put simply, each state of the superposition behave as though it
were alone.
p.ki...@imperial.ac.uk
> Decoherence doesn't solve the third problem. It only says that a system is
> spontaneously projected onto an eigenstate when it interacts with a
> disordered surrounding, but doesn't explain why there are privileged
> states. For example, why we don't observe plane waves of chairs.
It's the particular coupling of the system to the reservior (your
"disordered surroundings") that determines the eigenstate.
If you could find a mechanism to couple the chairs in a system to and
from a suitable reservoir of chairs, without other chair-destroying
dymanics, then indeed it might seem as if planes-waves-of-chairs
were privileged states.
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (QOLS) (ph) +44-20-759-47520 (fax) 47714
Imperial College London, Dr.Paul...@physics.org
SW7 2BW, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/
Chris H. Fleming
> I wrote:
>
> > > Second one, a measurement triggers a projection onto an eigenstate, it's
> > > a discontinuous process, not time symmetric while quantum equations are
> > > so.
>
> "Chris H. Fleming" <chris_h...@yahoo.com> a écrit dans le message de
> news: 1131663467.3...@f14g2000cwb.googlegroups.com...
>
> > This one has been solved in the framework of quantum open systems. In a
> > large system, the Poincare Recurrence time becomes large. Measurements
> > involve interactions with many degrees of freedom. In a quantum open
> > system, evolution is not unitary nor time symmetric. Interaction
> > between the microscopic system and the reservoir can cause irreversable
> > localization.
>
> That implies the outcome of a measurement depends on the state of the
> apparatus, thus isn't determined by the measured system and probabilities
> alone like according to quantum mechanics.
Probabilities come from Born interpretation. The Schrödinger equation
is fully deterministic. If you use the Schrödinger equation alone and
model the entire apparatus and measurement process as interaction, then
yes.
> There must already be a
> discontinuous process in the system alone to drive the apparatus
> (irreversibly) to the configuration corresponding to the eigenvalue.
It is as discontinuous as the decoherence time is small.
It is as irreversable as the Poincare recurrence time is long.
Both these things are generally overwhelming as the measuring aparatus
and environment contain many, many degrees of freedom.
Newtonian mechanics is time reversable, but that does not prevent
irreversable processes from happening in statistical mechanics. You can
let the air out of a balloon, but you cannot let it back in.
> I
> guess there is no detailed description of such a measurement process, and
> thus this concept is more of a hope than of a discovery. I wouldn't invest
> my time in it.
See Halliwell and Hu-Paz-Zhang. Search for decoherence and quantum
brownian motion.
Measurements are not magic, the measurement and the thing to be
measured all need to be described by the same fundamental theory. The
Schrödinger equation alone is already fundamental, complete, and
consistent. We might as well attempt to see if it works before throwing
our hands in the air.
Marcel LeBel
>>In that respect, could we say that
>>measurements apply a similar box-like constraint on the particle thereby
>>causing a temporary loss of freedom and quantization, entirely due to
>>the constraint of the measurement?
>
> Cl.Massé wrote:
> No. There is discretization or there is not. In both case, the
> measurement project the particle-wave onto a state of definite energy
> (assuming the energy or a function thereof is measured.) The energy values
> measured may be continuous or discrete.
Marcel LeBel
Hum! Lets take a photon. A photon can take any/any direction. So
"direction" is a continuous variable. Send the photon across a slit or
pinhole and from then on, "direction" is discrete. The observation here
includes both the constraint applied by the slit or pinhole just as is
the fact of watching the interference bands on the screen. I know that,
in a sense, we like to separate the two of them as experience (slit) and
results (screen measurements) but the later is not possible without the
former.
My question is still; can the constraint of observation impart on an
otherwise continuous variable its discretization thereafter? I know
about waves interference and wavelength equations... I am just stepping
back and trying to put in words what it is that we see.
le...@muontailpig.com remove particle
Cl.Massé
> > That implies the outcome of a measurement depends on the state of the
> > apparatus, thus isn't determined by the measured system and
> > probabilities alone like according to quantum mechanics.
"Chris H. Fleming" <chris_h...@yahoo.com> a écrit dans le message de
news: 1132180817.2...@f14g2000cwb.googlegroups.com...
> Probabilities come from Born interpretation. The Schrödinger equation
> is fully deterministic. If you use the Schrödinger equation alone and
> model the entire apparatus and measurement process as interaction, then
> yes.
The Born (rather Copenhagen) interpretation implies a projection onto the
eignestate.
> > There must already be a
> > discontinuous process in the system alone to drive the apparatus
> > (irreversibly) to the configuration corresponding to the eigenvalue.
> It is as discontinuous as the decoherence time is small.
Therefore not discontinous, see the definition. And decoherence depends on
the state of the surrounding. Both approaches aren't reconciliables.
> We might as well attempt to see if it works before throwing our hands in
> the air.
I don't forbid you to try, it's your time.
Chris H. Fleming
> I wrote:
>
> > > That implies the outcome of a measurement depends on the state of the
> > > apparatus, thus isn't determined by the measured system and
> > > probabilities alone like according to quantum mechanics.
>
> "Chris H. Fleming" <chris_h...@yahoo.com> a écrit dans le message de
> news: 1132180817.2...@f14g2000cwb.googlegroups.com...
>
> > Probabilities come from Born interpretation. The Schrödinger equation
> > is fully deterministic. If you use the Schrödinger equation alone and
> > model the entire apparatus and measurement process as interaction, then
> > yes.
>
> The Born (rather Copenhagen) interpretation implies a projection onto the
> eignestate.
yes it certainly does...?
> > > There must already be a
> > > discontinuous process in the system alone to drive the apparatus
> > > (irreversibly) to the configuration corresponding to the eigenvalue.
>
> > It is as discontinuous as the decoherence time is small.
>
> Therefore not discontinous, see the definition.
Ah but can you prove that the actual event is more discontinuous than
the decoherence time is small. You are trusting too much in the theory
and not the experiment.
> And decoherence depends on
> the state of the surrounding.
So what.
> Both approaches aren't reconciliables.
Yes and if Decoherence is true and complete, then the Born
Interpretation is some kind of an approximation.
> > We might as well attempt to see if it works before throwing our hands in
> > the air.
>
> I don't forbid you to try, it's your time.
Consistency is the first requirement for truth. What other consistent
theories are there that aren't some kind of BS local variables or
something? I am too green to know.
mark...@yahoo.com
http://groups.google.com/group/sci.physics.research/msg/ea90208e4340eea7
>What is quantum measurement problem?
>Why do people say decoherence does not solve the quantum measurement
>problem?
It's the problem of determining why and how a pure state seems to turn
into a mixed state upon "measurement".
Decoherence brings you much of the way by not only showing how
"mixtures" can arise in a world of "pure" states (i.e., states which
describe part of an overall system can be in what are called "improper
mixtures" even if the total system is in a pure state), but also why
"measurements" seem to delineate a specific frame of eigenmodes for the
states to decohere into, even though any basis serves just as well as
another to describe the space of pure states.
There is an additional feature described in the FAQ of that of
determining why or how a mixed state turns into one of the pure states
comprising its mixture. But I don't see any reason for recognizing that
as anything but a null operation, since "mixed state" *already* means
probabilistic combination of pure states, and there's no reason for
interpreting it any other way. So that, when you say, "state W becomes
the mixture p W1 + q W2 with p and q > 0 and p + q = 1" this will be
nothing more than a fancy way of saying "state W becomes either W1
(with probability p) or W2 (with probability q)".
For instance, a thermal state is a mixed state, even though a system
lying in that state is really in one of its pure states, with
probabilities given by the coefficients of the mixture. The
probabilities reflect an absence of information in the mind of the
perceiver, not something out there.
So, in reality, the difference between "mixed" and "random pure" is all
in the mind of the person seeing it and depends entirely on who much
they know of the system. The mixture represents nothing more than an
absence of information.
Moreover, the latter feature of "mixed -> one of the pure state at
random", which I deem a null operation, would have nothing to do with
quantum physics, per se, even if it were something. Mixed states are
not quantum theoretic in the first place. They're a generic feature of
all classical and quantum physics.
Quentum theory left the discussion as soon as you quit talking about
the superpositions of the original pure state. You could just as well
discuss the question of why a thermal mixed state "became" one of its
pure states after being looked at through a microscope, yet you don't
hear anybody pretending here that this "becoming" was anything more
than a change of perception in the mind of the observer, rather than
something that actually happened out there.
The real issue is not the latter step (or, more appropriately,
non-step), but the former one. Decoherence gets you 99% of the way to
resolving a pure state into one looking very much like a mixed state.
But not all the way.
It still leaves the universe in a pure state. So, all you end up doing
is shifting the issue over to a "Wigner's friend" problem and resolving
nothing.
The more fundamental question is this: why does there appear to be a
non-trivial superselection structure in the universe? Superselection is
just a fancy way of describing this situation. It is what underlies
decoherent superpositions.
To be more precise, a superselection structure consists of what is
called a partition of unity:
1 = P1 + P2 + ... + Pn (+ ..., possibly infinite
sum or even integral)
with Pi Pj = 0 if i != j and Pi' (adjoint) = Pi Pi = Pi ... such that
every observable commutes with every Pi. Each of these operators
identifies what is known as a superselection sector. States from two
different sectors will only combine decoherently.
But in reality, the whole question (and whole problem) is based on a
false premise -- namely, that the universe is a purely quantum system.
Nobody ever said that all degrees of freedom or all modes in all
systems had to be quantum. It is entirely consistent to discuss hybrid
classico-quantum systems in which some modes are classical and some
quantum.
In fact, at one extreme, every single degree of freedom in a system
will be its own superselection sector. In that case, there are no
coherent superpositions at all -- you're in a pure classical world.
At the other extreme, there will be no superselection at all -- then
you're in a pure quantum world.
But there is a whole continuum of possibilities in between these two
extremes.
The appearance of the world is actually that of one which has a few
classical degrees of freedom; e.g. classical degees that come from the
"outside", with the entire universe, itself, being effectively an open
system, and weave into every system to create a striation of
superselection sectors which provide the framework for decoherent
superposition and fill out the remaining 10% of the gap that
decoherence, itself, cannot close; ... and quantum degrees of freedom.
So it is natural to assume that the world actually is what it appears
to be -- a hybrid classico-quantum universe, which possesses some
classical degrees of freedom and perhaps a large majority of degrees of
freedom that are quantum.
Again, nobody ever said the quantum hypothesis had to be universal or
that the universe had to be entirely quantum.
This occurs, in particular, if (a) the universe, itself, is NOT in a
pure quantum state, e.g. a thermal state at a temperature of 3 degrees
Kelvin, or (for that matter), (b) if there doesn't even exist a
universal state space (as Smolin has argued), but only state spaces
associated with finite regions. In the latter case, there is ALWAYS an
outside and you have no recourse but to prescribe "improper mixtures",
since there will be no universal pure state, to begin with.
Case (a) is imminent (give the foregoing) from the fact that all
observers are finite. So they will all cut off the world at a boundary
and see the universe as being an (improper) mixture. But case (b) goes
even further and brings up the possibility that even from God's
perspective, the universe is STILL a mixture, and that there are no
proper mixtures at all (so that mixture means improper mixture) because
there is no universal state space in the first place!
Case (b) must also occur if the Universe is not globally hyperbolic.
What global hyperbolicity means is that you can layer the universe into
a sequence of "snapshots" such that every maximal worldline passes
through each one exactly once. That is, global hyperbolic universes are
those in which you can regard time as "flowing" globally along the
sequencing of snapshots comprising any of its layerings.
To define a universal state space requires, almost as a prerequisite,
that the universe be globally hyperbolic.
If there are time-like loops, or in the presence of time travel routes,
you need not have global hyperbolicity and you're then forced to resort
to the notion of time given in relativity as being "all there", and not
"flowing" since there would be no consistent global flow in such a
universe.
In that case, you can't even do the basic things required to set up a
universal state space (e.g. formulate the Cauchy problem), so that the
conclusion Smolin hypothesized would blindside even him from an oblique
angle, coming home to bear -- no universal state space.
Aaron Denney
> There is an additional feature described in the FAQ of that of
> determining why or how a mixed state turns into one of the pure states
> comprising its mixture. But I don't see any reason for recognizing that
> as anything but a null operation, since "mixed state" *already* means
> probabilistic combination of pure states, and there's no reason for
> interpreting it any other way. So that, when you say, "state W becomes
> the mixture p W1 + q W2 with p and q > 0 and p + q = 1" this will be
> nothing more than a fancy way of saying "state W becomes either W1
> (with probability p) or W2 (with probability q)".
>
> in the mind of the person seeing it and depends entirely on who much
> they know of the system. The mixture represents nothing more than an
> absence of information.
>
> Moreover, the latter feature of "mixed -> one of the pure state at
> random", which I deem a null operation, would have nothing to do with
> quantum physics, per se, even if it were something. Mixed states are
> not quantum theoretic in the first place. They're a generic feature of
> all classical and quantum physics.
Yes, so far as it goes -- it's a perfectly valid stance to take, yet at
the same time it misses something fundamental. Mixed states can be
resolved into different mixtures of the pure states. As such, a mixture
of pure states contains "more information" than that of the
corresponding mixture, yet the statistics of any possible measurement
are the same. Given that, it doesn't seem to operationally add anything
to assert that there is some underlying unobservable mixture, and
computations using a mixed density matrix are often easier...
--
Aaron Denney
-><-
Cl.Massé
1135812069.8...@g44g2000cwa.googlegroups.com
> It's the problem of determining why and how a pure state seems to turn
> into a mixed state upon "measurement".
Who say that? Because that seems the problem tailored to be solved by
decoherence. There are other problems that, if not officially called
"measurement", aren't solved at all by decoherence. That's what I would
call the *muleta* approach to quantum mechanics interpretation.