relation between quantum mechanics and classical mechanics
Ulf Klein
This post concerns the relation between quantum mechanics and
classical mechanics. One can find
many statements of the kind:
"Quantum mechanics is the covering theory of
classical mechanics",
or
"The classical limit of quantum mechanics is
classical mechanics".
in the literature. I have the following 2 problems
with these statements:
1.The classical limit of quantum mechanics is a
(uniquely defined) classical statistical theory.
(field theory for two not decoupled variables
\rho and S, one equation is Hamilton-Jacobi
the second is the continuity equation),
which is different from classical mechanics.
Would you agree that the second of the above
statements is wrong ? If not please tell me what
is wrong with my reasoning.
2.How can a statistical theory be a covering theory
for a deterministic theory ? The former is unable to
make predictions about individual events and
contains less information than the latter.
Best regards
Ulf Klein
ulf....@jku.at
Igor Khavkine
> 1.The classical limit of quantum mechanics is a
> (uniquely defined) classical statistical theory.
> (field theory for two not decoupled variables
> \rho and S, one equation is Hamilton-Jacobi
> the second is the continuity equation),
> which is different from classical mechanics.
> Would you agree that the second of the above
> statements is wrong ? If not please tell me what
> is wrong with my reasoning.
That actually sounds about right.
> 2.How can a statistical theory be a covering theory
> for a deterministic theory ? The former is unable to
> make predictions about individual events and
> contains less information than the latter.
There is a very simple way of turning a statistical theory into a
deterministic one. In usual statistical mechanics, the state of the
system is specified by a probability distribution on phase space. Take
this distribution to be a delta-function supported only on a single
point. Then you get a deterministic theory whose state is specified by
that point in phase space.
Hope this helps.
Igor
X-Phy
> Hi,
>
> This post concerns the relation between quantum mechanics and
> classical mechanics. One can find
> many statements of the kind:
> "Quantum mechanics is the covering theory of
> classical mechanics",
> or
> "The classical limit of quantum mechanics is
> classical mechanics".
> in the literature. I have the following 2 problems
> with these statements:
>
> 1.The classical limit of quantum mechanics is a
> (uniquely defined) classical statistical theory.
> (field theory for two not decoupled variables
> \rho and S, one equation is Hamilton-Jacobi
> the second is the continuity equation),
> which is different from classical mechanics.
> Would you agree that the second �of the above
> statements is wrong ? If not please tell me what
> is wrong with my reasoning.
A classical limit is obtained while letting some parameter tend to a
limit. This parameter is usually taken as the Plank constant tending
to zero, but it may also be the number of particles tending to
infinity. Actually, it isn't well understood, and that's the origin
of the difficulties in the interpretation of QM. Some claim that
decoherence explains everything, but it is not clear why the
"classical states" are favoured, since the formalism remains symmetric
with respect to the states, that is, nothing distinguishes a "quantum
mechanical" superposition from the superposed states.
I begin by this summary because I think that classical and quantum
mechanics are two different theories, which isn't a widespread
opinion. A statistical theory isn't the same theory as the evolution
of the mean values, even if the variance is comparatively small.
Thermodynamics isn't the same as the kinematical theory, even if one
can be deduced from the other. That is one example to see the
difference, but with quantum mechanics it is still worst.
> 2.How can a statistical theory be a covering theory
> for a deterministic theory ? The former is unable to
> make predictions about individual events and
> contains less information than the latter.
Classical and quantum mechanics have the same formal structure, for
instance comparing the Poisson bracket with the commutator. Quantum
mechanics involves more variables, since a point particle is replaced
by a wave function. It is in this sense that it may be called a
covering theory. But in quantum mechanics, not all of this variables
can be measured, that's why it contains less information. Though,
that is not really true if classical mechanics is taken as the limit
for an infinity of particles. It is therefore a collective
information, that is, less information since in quantum mechanics all
the particles can be measured individually.
Peter
> Hi,
>
> This post concerns the relation between quantum mechanics and
> classical mechanics. One can find
> many statements of the kind:
> "Quantum mechanics is the covering theory of
> classical mechanics",
> or
> "The classical limit of quantum mechanics is
> classical mechanics".
Indeed - but they grasp far too short as CM is also necessary to
interpret QM, so that there is the strange relationship that a theory
needs a limit case for understanding (cf Landau & Lifshits, Vol. III).
Similarly, Schr�dinger (1926) has required to justify the use of the
classical V(r) within QM. Moreover, he has required a foundation of
QM, where it follows uniquely that the energy and not the frequency is
discretized/quantized. And he has required a maths which is
appropriate to the quantum nature of quantum objects (eigenvalue maths
corresponds to classical waves).
A foundation of QM according these lines should answer your questions.
There are surely several such ones available; one is the axiomatic
deduction of the Schr�dinger eq. from Euler's representation of CM by
Suisky & myself (Int. J. Theor. Phys. 2006; Von der klassischen Physik
zur Quantenphysik, Springer 2006).
Best wishes,
Peter
Bob_for_short
>
> 2.How can a statistical theory be a covering theory
> for a deterministic theory ? The former is unable to
> make predictions about individual events and
> contains less information than the latter.
Statistical theory is always reacher than a deterministic theory: it
contains not only "deterministic" average values but also their
dispersions.
Look at the Ehrenfest equations - they are about average values. Isn't
it an answer?
Vladimir.
Ulf Klein
Thanks for your comment. I would like to reformulate
my (basically unchanged) question as follows
> > 2.How can a statistical theory be a covering theory
> > for a deterministic theory ? The former is unable to
> > make predictions about individual events and
> > contains less information than the latter.
> instance comparing the Poisson bracket with the commutator. Quantum
> mechanics involves more variables, since a point particle is replaced
> by a wave function.
why is the wave function used to describe a single particle
if (as apparently everybody agrees) quantum mechanics is a
statistical theory ?
Best regards
Ulf Klein
Bob_for_short
>
> why is the wave function used to describe a single particle
> if �(as apparently everybody agrees) quantum mechanics is a
> statistical theory ?
I does not describe "a single particle" but a single quantum state.
Think of an interference pattern. It consists of many points. The
whole pattern describes an electron/photon quantum state behind the
duble-slitted diaphragm. There is no interest in one point on the
screen. It is the pattern which is interesting to predict. The wave
function does it.
One point on a screen is like one letter in a text - it does not carry
much information. We are interested in the whole text which is an
ensemble of letters. This gives the description of a quantum state.
Vladimir.
X-Phy
First of all, the wave nature of particles is evidenced by
experiment. In comon interpretations, it is a wave of probabilities,
or better, of probability amplitudes. That is necessary to account for
the complexity of reality, which isn't as simple as the kinetic theory
of gases, and particularly for interference phenomena. It is not
possible to describe those phenomena with a mere classical statistical
theory.
Note that the Schr?dinger equation has a form similar to a diffusion
equation, which describes the diffusion of particles like for example
in the Brownian motion. But there is a factor i (complex unity) that
changes completely its nature. There are interpretations starting
from a stochastic theory, that is, similar to diffusion, but they need
something more to work, like a "quantum" potential.
Ulf Klein
>
> > why is the wave function used to describe a single particle
> > if �(as apparently everybody agrees) quantum mechanics is a
> > statistical theory ?
>
> I does not describe "a single particle" but a single quantum state.
> Think of an interference pattern. It consists of many points. The
> whole pattern describes an electron/photon quantum state behind the
> duble-slitted diaphragm. There is no interest in one point on the
> screen. It is the pattern which is interesting to predict. The wave
> function does it.
>
I agree that the wave function describes the interference pattern as
a whole. This is a good example to explain my question in more detail.
High-precision measurements (Tonomura) show beyond any doubt
that the pattern is due to the joint action of a huge number of
single
particles. This experiments indicate that quantum mechanics is
a theory describing the behavior of statistical ensembles of
particles.
So my question is: If quantum mechanics is a theory about ensembles
why do people believe that it can be used to explain or interpret the
behavior of the individual particles (forming the pattern). I found
answers
like "because they (the particles) are here" or "we have no better
theory"
not convincing.
Best
Ulf
Lester Welch
wrote:
Does one particle form a measureable interference pattern?
a student
> Hi,
>
> This post concerns the relation between quantum mechanics and
> classical mechanics. One can find
> many statements of the kind:
> "Quantum mechanics is the covering theory of
> classical mechanics",
> or
> "The classical limit of quantum mechanics is
> classical mechanics".
> in the literature. I have the following 2 problems
> with these statements:
>
> 1.The classical limit of quantum mechanics is a
> (uniquely defined) classical statistical theory.
> (field theory for two not decoupled variables
> \rho and S, one equation is Hamilton-Jacobi
> the second is the continuity equation),
> which is different from classical mechanics.
> Would you agree that the second �of the above
> statements is wrong ? If not please tell me what
> is wrong with my reasoning.
There is more than one 'classical limit'. First
there is the limit hbar -> 0. This is a purely
formal limit - hbar is a fixed physical constant.
In this limit there is indeed, as you suggest,
a classical statistical theory, described by
the fields rho and S on configuration space.
This doesn't really make the second statement
wrong, however, as this classical statistical
theory can describe individual trajectories
(where rho is a delta-function), and in this
sense includes deterministic classical
mechanics as a special case.
Second, one can look at more 'physical'
classical limits, where the effects of hbar
become 'small' so that description via
classical trajectories becomes a good
approximation. Such limits are reviewed,
for example, in
http://lanl.arxiv.org/abs/0705.2227
where the limit involves a weak coupling
to the environment that 'collapses' the
quantum state appropriately. The degree
of 'tuning' needed for such limits can be
a problem, however.
> 2.How can a statistical theory be a covering theory
> for a deterministic theory ? The former is unable to
> make predictions about individual events and
> contains less information than the latter.
Look at classical statistical mechanics on phase space,
where an ensemble described by rho(x,p) evolves via
the Liouville equation
d rho/dt = {H,rho}.
The set of possible densities includes, as it turns out,
deterministic 'point' densities
rho(x,p,t) = delta(x-x(t)) delta(p-p(t)),
where (x(t),p(t)) is a classical trajectory. In this case
one has an ensemble where every member of the
ensemble is identical to every other member. Thus,
the statistical theory is a covering theory for a
deterministic theory.
One interesting thing about Hilbert space quantum
mechanics, in the above regard, is that it does not
allow delta-functions for wave functions (they are not
square integrable) - the Schroedinger equation breaks
down in this case.
Phillip Helbig
> Hi,
>
> This post concerns the relation between quantum mechanics and classical
> mechanics. One can find
> many statements of the kind:
> "Quantum mechanics is the covering theory of classical mechanics", or
> "The classical limit of quantum mechanics is classical mechanics". in
> the literature. I have the following 2 problems with these statements:
>
> 1.The classical limit of quantum mechanics is a (uniquely defined)
> classical statistical theory. (field theory for two not decoupled
> variables \rho and S, one equation is Hamilton-Jacobi the second is the
> continuity equation), which is different from classical mechanics. Would
> you agree that the second of the above statements is wrong ? If not
> please tell me what is wrong with my reasoning.
Above statements are rather right. There is some important details to be
worked out (the problem of classicality is still open), however.
The approach I like more is to start from the basic quantum equation (the
Schr�dinger equation is a special case of this equation)
d(rho)/dt = L rho
next compute Wigner distributions f = f[rho] and then take the classical
limit of the resulting equation. For instance formally doing h --> 0.
The well-known result is the classical Liouville equation
d(f_cl)/dt = L_cl f_cl
Hamiltonian or Lagrangian equations of motion can be obtained from here in
the usual way doing f_cl = delta(p,q).
> 2.How can a statistical theory be a covering theory for a deterministic
> theory ? The former is unable to make predictions about individual
> events and contains less information than the latter.
Statistical theories deal with statistical variables. These can be
splinted into a deterministic part more a fluctuation, which has
associated a probability distribution.
A = <A> + delta(A)
Deterministic theories only deal with the deterministic part <A>.
--
http://www.canonicalscience.org/
BLOG:
http://www.canonicalscience.org/en/publicationzone/canonicalsciencetoday/canonicalsciencetoday.html
Juan R.
> Hi,
>
> This post concerns the relation between quantum mechanics and classical
> mechanics. One can find
> many statements of the kind:
> "Quantum mechanics is the covering theory of classical mechanics", or
> "The classical limit of quantum mechanics is classical mechanics". in
> the literature. I have the following 2 problems with these statements:
>
> 1.The classical limit of quantum mechanics is a (uniquely defined)
> classical statistical theory. (field theory for two not decoupled
> variables \rho and S, one equation is Hamilton-Jacobi the second is the
> continuity equation), which is different from classical mechanics. Would
> you agree that the second of the above statements is wrong ? If not
> please tell me what is wrong with my reasoning.
Above statements are rather right. There is some important details to be
worked out (the problem of classicality is still open), however.
The approach I like more is to start from the basic quantum equation (the
Schr�dinger equation is a special case of this equation)
d(rho)/dt = L rho
next compute Wigner distributions f = f[rho] and then take the classical
limit of the resulting equation. For instance formally doing h --> 0.
The well-known result is the classical Liouville equation
d(f_cl)/dt = L_cl f_cl
Hamiltonian or Lagrangian equations of motion can be obtained from here in
the usual way doing f_cl = delta(p,q).
> 2.How can a statistical theory be a covering theory for a deterministic
> theory ? The former is unable to make predictions about individual
> events and contains less information than the latter.
Statistical theories deal with statistical variables. These can be
Gerry Quinn
@b2g2000yqi.googlegroups.com>, end...@dekasges.de says...
> On 18 Nov., 07:16, Ulf Klein <klein....@gmail.com> wrote:
> > Hi,
> >
> > This post concerns the relation between quantum mechanics and
> > classical mechanics. One can find
> > many statements of the kind:
> > "Quantum mechanics is the covering theory of
> > classical mechanics",
> > or
> > "The classical limit of quantum mechanics is
> > classical mechanics".
>
> Indeed - but they grasp far too short as CM is also necessary to
> interpret QM, so that there is the strange relationship that a theory
> needs a limit case for understanding (cf Landau & Lifshits, Vol. III).
Can't we bootstrap QM, by replacing the classical limit with, so to
speak, a "fully decohered environment" which behaves classically?
- Gerry Quinn
Bob_for_short
>
> So my question is: If quantum mechanics is a theory about ensembles
> why do people believe that it can be used to explain or interpret the
> behavior of the individual particles (forming the pattern). I found
> answers
> like "because they (the particles) are here" or "we have no better
> theory" not convincing.
You know, the Classical Mechanics also describes ensembles and is a
statistical theory - it describes the average measured positions of
heavy bodies. Experimentally one observes a body with many-photon
exchange, just like in a double-slit experiment, but then one makes an
average, assigns this value to the body (geometric) center and
considers the equations for this average variable. It is deterministic
because it is average over many points. The dispersion of points
describes the body size. Similarly the quantum state of the photon/
electron has a dispersion that describes its (state) size. The size is
an additional data testifying that the studied system is not point-
like: the total information is not reduced to three numbers
(coordinates of center of mass). One-point measurement is a pixel of
the total image in both CM and QM.
Peter
> ...
> One interesting thing about Hilbert space quantum
> mechanics, in the above regard, is that it does not
> allow delta-functions for wave functions (they are not
> square integrable) - the Schroedinger equation breaks
> down in this case.
Not at all.
Recall that quantum particles may move in regions, where
total energy, E < classical potential energy, V(r)
Consequently, the contribution of the configuration {r} to the total
energy is not equal to V(r), but equals
V_noncl(r) = F(r) V(r), F(r) >= 0
Analogously, the contribution of the 'momentum configuration' {p} to
the total energy is not equal to T(p), but equals
T_noncl(p) = G(p) T(r), G(p) >= 0
The "weight functions" (Schr?dinger 1926, 4th Commun.) F(r) and T(p)
limit the values of V_noncl(r) and T_noncl(p) such that E remains
finite (in particular, this is necessary for T(p) for p->oo). All
possible orbits contribute to a given stationary state (Schr?dinger,
loc. cit.), therefore,
E = $ F(r) V(r) dr / $ F(r) dr
+ $ G(p) T(p) dp / $ G(p) dp
For a free particle of momentum p_0,
G(p) = delta(p - p_0)
The wave functions follow from
G(p) = |phi(p)|^2, F(r) = |psi(r)|^2
They are generalized functions (distributions), but square-integrable,
and the obey the ordinary Schr?dinger eq. in space and momentum space,
respectively.
Best wishes,
Peter
Ulf Klein
Hi a student,
I agree partly with your last point concerning the structure of
statistical mechanics (We are talking about the classical phase space
theory which is the microscopic counterpart of thermodynamics). There
is in fact a deterministic element contained in this theory since
particles (with x and p as observable properties) exist and the
trajectories of these particles can be, or could be, calculated by
means of classical laws. The uncertainty is not in the laws but in
the initial values. This situation has been called
"cryptodeterministic"
(Moyal). Probably, one can consider a limiting situation where a
probability 1 is assigned to initiial values and then obtain
the delta solutions you wrote down.
Nevertheless I would - even in this case - not say that statistical
mechanics is the covering theory of classical mechanics. Rather
I would say that a limiting situation exists where both theories
touch. If theory A is the covering theory of theory B than each
solution of A should correspond to a solution of B. This is not
the case here. Also, a statistical theory must contain uncertainty,
otherwise it is not a statistical theory but a unnecessary
complicated
version of a deterministic theory.
Let us discuss quantum mechanics. I think that there is only a single
classical limit for quantum theory, namely the limit \hbar -> 0. This
parameter has a fixed value in nature. So, classical mechanics is a
nonexistent theory (whenever quantum theory applies) and the limit is
formal. But we are comparing theories, as given by clearly defined
sets
of equations (If classical mechanics behaviour is obtained if the
environment is taken into account is a different question). Thus,
there is no other (less 'formal') way to perform the classical limit
than \hbar -> 0.
Of course, we know that no vestiges of quantum effects can be seen in
our macroscopic world. But this is not necessarily a consequence (or
a proof) of the fact that the classical limit of quantum theory is
classical mechanics. Complexity may do the same job of suppressing
the true microscopic features.
Quantum mechanics contains even more uncertainty than statistical
physics,
because there are no laws for particle movement and such things as
initial
values for particle trajectories do not exist any more. The above
mentioned
singular point where statistical physics touches classical mechanics,
does
not exist for quantum mechanics and cannot be added by hand. I think
that
there is even less justification to consider quantum mechanics as a
covering theory of classical mechanics. A interesting paper on
randomness
in quantum mechanics as compared to classical statistics has been
written
by Torre (Eur.J.Phys. vol.29 (2008) 567)
Best
Ulf
Arnold Neumaier
> why is the wave function used to describe a single particle
> if (as apparently everybody agrees) quantum mechanics is a
> statistical theory ?
The wave function describes a pure state of the system.
To see the relation with classical mechanics, one needs to consider
- as in statistical mechanics - mixed states, which correspond to
density matrices.
A density matrix describes the stochastic behavior of a quantum system
in the same way as a density function describes the stochastic behavior
of a classical system. In both cases, if the system is nice enough that
the stochastic uncertainties in the quantities of interest are much
smaller than the quantities themselves, one can form a deterministic
approximation.
This deterministic approximation is given by a classical dynamical
system for the (expectations of the) quantities of interest.
Thus, in a sense, classical variables are simply expectations of
relevant quantum variables with small uncertainty. The small uncertainty
makes these variables approximately predictable in each individual
event, and hence classical.
For more details, see, e.g., Sections 7.3-7.5 of
A. Neumaier and D. Westra,
Classical and Quantum Mechanics via Lie algebras
http://de.arxiv.org/abs/0810.1019
Igor Khavkine
> Nevertheless I would - even in this case - not say that statistical
> mechanics is the covering theory of classical mechanics. Rather
> I would say that a limiting situation exists where both theories
> touch. If theory A is the covering theory of theory B than each
> solution of A should correspond to a solution of B. This is not
> the case here. Also, a statistical theory must contain uncertainty,
> otherwise it is not a statistical theory but a unnecessary
> complicated
> version of a deterministic theory.
This is a very strange way to define "covering theory". If it is taken
seriously, neither of these theories is covered by any other one:
statistical mechanics, Boltzmann's equation, hydrodynamics,
thermodynamics. To be honest, I have not see anywhere the phrase
"covering theory" used as a technical term before your post. But
intuitively, I would have imagined the following sequence of
coverings:
stat -> boltz -> hydro -> thermo.
If your terminology is taken seriously, then you are looking for some
sort of mathematical equivalence. However, it is quite clear that
quantum mechanics and classical mechanics are different theories and
are not mathematically equivalent in any way.
Igor
X-Phy
> > why is the wave function used to describe a single particle
> > if (as apparently everybody agrees) quantum mechanics is a
> > statistical theory ?
> The wave function describes a pure state of the system.
On 25 nov, 10:40, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
wrote:
Yes, and the system can be a single particle.
> To see the relation with classical mechanics, one needs to consider
> - as in statistical mechanics - mixed states, which correspond to
> density matrices.
Mixed states are an incoherent superposition, that is, described by
classical statistical mechanics. It is no longer QM.
> A density matrix describes the stochastic behavior of a quantum system
> in the same way as a density function describes the stochastic behavior
> of a classical system.
Yet, a density matrix is also able to describe a pure state.
Actually, the density matrix conceals the distinction between pure and
mixed states. So, using it to reason about the relationship between
pure (fully quantum) and mixed (classical statistical) states is a bit
acrobatic.
> In both cases, if the system is nice enough that
> the stochastic uncertainties in the quantities of interest are much
> smaller than the quantities themselves, one can form a deterministic
> approximation.
Remember that in QM, the uncertainties aren't bounded. If we are
talking about the position uncertainty in order to define a
trajectory, and make precise position measurements, then the momentum
uncertainty becomes very large, and the trajectory concept (actually
defined in the phase space) loses sense.
> This deterministic approximation is given by a classical dynamical
> system for the (expectations of the) quantities of interest.
>
> Thus, in a sense, classical variables are simply expectations of
> relevant quantum variables with small uncertainty. The small uncertainty
> makes these variables approximately predictable in each individual
> event, and hence classical.
For that to make sense, a "nice" enough system (or rather state) must
be taken from the begining (for exemple coherent states,) and some
measurements not allowed (or macroscopically not feasible.) That may
happen in our macroscopical world, but it is a contingency that
doesn't enters into the discussion. Some sort of second principle of
QM is perhaps necessary for CM to emerge from QM.
Arnold Neumaier
> Ulf Klein wrote:
>
>>> why is the wave function used to describe a single particle
>>> if (as apparently everybody agrees) quantum mechanics is a
>>> statistical theory ?
>
>> The wave function describes a pure state of the system.
>
> On 25 nov, 10:40, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
> wrote:
>
> Yes, and the system can be a single particle.
>
>> To see the relation with classical mechanics, one needs to consider
>> - as in statistical mechanics - mixed states, which correspond to
>> density matrices.
>
> Mixed states are an incoherent superposition, that is, described by
> classical statistical mechanics. It is no longer QM.
You are the first whom I saw claiming that statistical mechanics with
density matrices is no longer quantum mechanics.
In fact, almost all quantum mechanics applied to real systems not in
the ground state needs density matrices, since pure states are very
difficult to create and propagate unless a system is in the ground
state. Pure states describe only an idealized version of quantum
reality.
>> A density matrix describes the stochastic behavior of a quantum system
>> in the same way as a density function describes the stochastic behavior
>> of a classical system.
>
> Yet, a density matrix is also able to describe a pure state.
> Actually, the density matrix conceals the distinction between pure and
> mixed states.
In practice, this distinction is blurred anyway since one cannot
determine of a real system whether it is in a pure state or whether
it has a slight admixture of other states.
> So, using it to reason about the relationship between
> pure (fully quantum) and mixed (classical statistical) states is a bit
> acrobatic.
There is nothing acrobatic here. Things get more elegant with density
matrices, and intuition is easier since the classical analogy fits
much better.
In fact, one can set up the whole machinery of quantum mechanics without
ever introducing pure states. The emphasis on pure states in textbooks
is only a historical accident.
>> In both cases, if the system is nice enough that
>> the stochastic uncertainties in the quantities of interest are much
>> smaller than the quantities themselves, one can form a deterministic
>> approximation.
>
> Remember that in QM, the uncertainties aren't bounded.
Neither are they in classical statistical mechanics.
This doesn't affect deterministic approximations, which only require
that variances are small.
> If we are
> talking about the position uncertainty in order to define a
> trajectory, and make precise position measurements, then the momentum
> uncertainty becomes very large, and the trajectory concept (actually
> defined in the phase space) loses sense.
I did not say that deterministic approximations are _always_ adequate.
Note my qualification that ``the system is nice enough that
the stochastic uncertainties in the quantities of interest are much
smaller than the quantities themselves''
>> This deterministic approximation is given by a classical dynamical
>> system for the (expectations of the) quantities of interest.
>>
>> Thus, in a sense, classical variables are simply expectations of
>> relevant quantum variables with small uncertainty. The small uncertainty
>> makes these variables approximately predictable in each individual
>> event, and hence classical.
>
> For that to make sense, a "nice" enough system (or rather state) must
> be taken from the begining (for exemple coherent states,) and some
> measurements not allowed (or macroscopically not feasible.) That may
> happen in our macroscopical world,
This happens in many microscopic systems. Laser light is usually in a
coherent state.
Classicality develops whenever the variances of the quantities of
interest become small compared to their expectations. Of course,
there is significant interst in quantum systems that are decidedly
non-classical, so that this does not happen; but this does not change
the general principle.
Arnold Neumaier
p.ki...@ic.ac.uk
> Does one particle form a measureable interference pattern?
Yes -- but only if your detectors somehow measure that pattern
directly. I suggest something equivalent to a time-reversed
version of whatever is providing the interference pattern you
want to measure :-).
If you want to use an array of pointlike detectors, then you
have to wait for the statistics to build up.
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (Photonics) (ph) +44-20-759-47734 (fax) 47714
Imperial College London, Dr.Paul...@physics.org
SW7 2AZ, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/
X-Phy
wrote:
> You know, the Classical Mechanics also describes ensembles and is a
> statistical theory - it describes the average measured positions of
> heavy bodies. Experimentally one observes a body with many-photon
> exchange, just like in a double-slit experiment, but then one makes an
> average, assigns this value to the body (geometric) center and
> considers the equations for this average variable. It is deterministic
> because it is average over many points. The dispersion of points
> describes the body size. Similarly the quantum state of the photon/
> electron has a dispersion that describes its (state) size. The size is
> an additional data testifying that the studied system is not point-
> like: the total information is not reduced to three numbers
> (coordinates of center of mass). One-point measurement is a pixel of
> the total image in both CM and QM.
You seem to forget the uncertainty relation. For a single particle,
the position can be measured with any precision, even if its quantum
state is dispersed. Yet, the quantum state is afterward changed,
projected onto an eigenfunction, and thus it is impossible the measure
the initial dispersion. That is what makes QM very different from a
classical statistical theory, especially when considering a single
particle. That is one of the reasons why I think CM isn't a limiting
case of QM.
X-Phy
wrote:
> > Mixed states are an incoherent superposition, that is, described by
> > classical statistical mechanics. It is no longer QM.
> You are the first whom I saw claiming that statistical mechanics with
> density matrices is no longer quantum mechanics.
I was too radical effectively. But it remains that incoherent
superposition occurs in classical mechanics, and is formalized by
plain classical probabilities.
> In fact, almost all quantum mechanics applied to real systems not in
> the ground state needs density matrices, since pure states are very
> difficult to create and propagate unless a system is in the ground
> state. Pure states describe only an idealized version of quantum
> reality.
That is a contingency I spoke about. We were discussing the theories
per se.
> >> A density matrix describes the stochastic behavior of a quantum system
> >> in the same way as a density function describes the stochastic behavior
> >> of a classical system.
> > Yet, a density matrix is also able to describe a pure state.
> > Actually, the density matrix conceals the distinction between pure and
> > mixed states.
> In practice, this distinction is blurred anyway since one cannot
> determine of a real system whether it is in a pure state or whether
> it has a slight admixture of other states.
> > So, using it to reason about the relationship between
> > pure (fully quantum) and mixed (classical statistical) states is a bit
> > acrobatic.
> There is nothing acrobatic here. Things get more elegant with density
> matrices, and intuition is easier since the classical analogy fits
> much better.
QM isn't intuitive, and that's why it's a bit acrobatic (or risky if
you prefer.) Elegance isn't correctness, and precisely a classical
analogy might insinuate while the theories don't reduce to one
another.
> In fact, one can set up the whole machinery of quantum mechanics without
> ever introducing pure states. The emphasis on pure states in textbooks
> is only a historical accident.
That can also be done without operators, without Hibert space and so
on. But a theory that isn't able to describe pure states, in a way or
another, couldn't be QM. An interference pattern can occur only with a
pure state.
> > Remember that in QM, the uncertainties aren't bounded.
> Neither are they in classical statistical mechanics.
> This doesn't affect deterministic approximations, which only require
> that variances are small.
If the variances are small, they are bounded, and they play the role
of uncertainties in QM. To the contrary, in QM the uncertainties can
alway be made greather than the mean, provided a precise enough
measurement is feasible (and done.)
> > If we are
> > talking about the position uncertainty in order to define a
> > trajectory, and make precise position measurements, then the momentum
> > uncertainty becomes very large, and the trajectory concept (actually
> > defined in the phase space) loses sense.
> I did not say that deterministic approximations are _always_ adequate.
> Note my qualification that ``the system is nice enough that
> the stochastic uncertainties in the quantities of interest are much
> smaller than the quantities themselves''
Your qualification means "the system is classical." Otherwise you
need another qualification for the measurements performed. And even
not, see below.
> >> This deterministic approximation is given by a classical dynamical
> >> system for the (expectations of the) quantities of interest.
> >> Thus, in a sense, classical variables are simply expectations of
> >> relevant quantum variables with small uncertainty. The small uncertainty
> >> makes these variables approximately predictable in each individual
> >> event, and hence classical.
> > For that to make sense, a "nice" enough system (or rather state) must
> > be taken from the begining (for exemple coherent states,) and some
> > measurements not allowed (or macroscopically not feasible.) That may
> > happen in our macroscopical world,
> This happens in many microscopic systems. Laser light is usually in a
> coherent state.
Here a "coherent state" is a special state where the product dp.dx is
minimum, that is, a "wave packet" whose mean values behave
classically. But we know that even for those states there is
dispersion, so that the position uncertainty (variance of position)
grows indefinitely.
> Classicality develops whenever the variances of the quantities of
> interest become small compared to their expectations. Of course,
> there is significant interst in quantum systems that are decidedly
> non-classical, so that this does not happen; but this does not change
> the general principle.
Assuming there are situations in which classicality so defined
develops, is it the same as to make h tend to 0? Apparently, it isn't
sufficient, and classicality can be lost even once it has developped.
p.ki...@ic.ac.uk
> Second, one can look at more 'physical'
I always liked Milburn's alternative approach using a "random sequence
of unitary phase changes" to provide the decoherence, i.e. in
Phys. Rev. A 44, 5401 - 5406 (1991)
http://prola.aps.org/abstract/PRA/v44/i9/p5401_1
a student
> On 22 Nov., 21:14, a student <of_1001_nig...@hotmail.com> wrote:
>
> > ...
> > One interesting thing about Hilbert space quantum
> > mechanics, in the above regard, is that it does not
> > allow delta-functions for wave functions (they are not
> > square integrable) - the Schroedinger equation breaks
> > down in this case.
>
> Not at all.
[snip]
> For a free particle of momentum p_0,
>
> � �G(p) = delta(p - p_0)
>
> The wave functions follow from
>
> � �G(p) = |phi(p)|^2, � F(r) = |psi(r)|^2
>
> They are generalized functions (distributions), but square-integrable,
> and the obey the ordinary Schr?dinger eq. in space and momentum space,
> respectively.
OK, let me point out this is not a counterexample,
and clarify my original 'throwaway' remark (and
expand on it a little).
First, your phi(p) above is not a wave-function that
is a delta-function (the latter is not square-integrable
and hence not in Hilbert space) - hence, my point
still stands. Your phi(p) is instead, the 'square root
of a delta-function' - which is an interesting idea,
and which can probably be defined rigorously.
Second, I should have made it clear I was
thinking of the position representation, and
with time evolution in mind.
In particular, take a free-particle. One can
extend Hilbert space QM (eg, via C*-algebras
or Bohmian mechanics) to make sense of
an initial state that has a definite momentum.
The evolution of such a state is trivial -
momentum is conserved. OK.
However, consider the same free particle,
and suppose it can somehow be
described as initially having a definite
position. How does it evolve? I don't
believe any sensible evolution can be
defined. For example, in Bohmian
mechanics the so-called quantum
potential is not well-defined in such a
case, and if one tries to define it via
a sequence of distributions one still
gets nonsense (intuitively, the
particle experiences an infinite
acceleration and the probability
distribution spreads instantaneously
everywhere). This is very
different from classical mechanics!
Note that, for a non-free particle,
one has a similar problem even for
intial states of definite momentum
(unless one goes to periodic
boundary conditions, eg, a
particle on a ring).
Peter
> On Nov 25, 5:53�pm, Peter <end...@dekasges.de> wrote:
>
> > On 22 Nov., 21:14, a student <of_1001_nig...@hotmail.com> wrote:
>
> > > ...
> > > One interesting thing about Hilbert space quantum
> > > mechanics, in the above regard, is that it does not
> > > allow delta-functions for wave functions (they are not
> > > square integrable) - the Schroedinger equation breaks
> > > down in this case.
>
> > Not at all.
>
> [snip]
>
> > For a free particle of momentum p_0,
>
> > � �G(p) = delta(p - p_0)
>
> > The wave functions follow from
>
> > � �G(p) = |phi(p)|^2, � F(r) = |psi(r)|^2
>
> > They are generalized functions (distributions), but square-integrable,
> > and the obey the ordinary Schr?dinger eq. in space and momentum space,
> > respectively.
>
> OK, let me point out this is not a counterexample,
> and clarify my original 'throwaway' remark (and
> expand on it a little).
>
> First, your phi(p) above is not a wave-function that
> is a delta-function (the latter is not square-integrable
> and hence not in Hilbert space) - hence, my point
phi(p) = delta(p - p_0)
is an unphysical solution to the stationary Schr�dinger equation in
momentum space, because |phi(p)|^2=delta(p - p_0)^2 has no classical
counterpart (mass density of classical point-like particle in momentum
space)
============== Moderator's note ===============================
delta(p - p_0)^2 has not only no classical counterpart, but it is not
defined at all. It's a meaningless expression. It's clear that generalized
eigensolutions of essentially adjoint operators with a continuous spectral
value are no L**2 functions but distributions, defined as linear forms acting
on the much smaller nuclear space of the rigged-Hilbert space construction
(Gelfand triple).
================================================================
BTW, the Schr�dinger eq. does *not* "break down"
Thus, the exclusion of functions like delta(p - p_0) from Hilbert
space is physically correct.
> Your phi(p) is instead, the 'square root
> of a delta-function' - which is an interesting idea,
> and which can probably be defined rigorously.
It's not my idea, I have read it in Eugene Stefanovich's book, but the
approach to QM via the weight functions F(r) and G(p) provides them,
IMHO, with a deeper physical reasoning.
> Second, I should have made it clear I was
> thinking of the position representation, and
> with time evolution in mind.
psi(r,t) is the Fourier transform of phi(p,t) as usual
> In particular, take a free-particle. �One can
> extend Hilbert space QM (eg, via C*-algebras
> or Bohmian mechanics) to make sense of
> an initial state that has a definite momentum.
No extension necessary, see above
> The evolution of such a state is trivial -
> momentum is conserved. � OK.
>
> However, consider the same free particle,
> and suppose it can somehow be
> described as initially having a definite
> position. �
Within QM, this is impossible. For in free space, no configuration is
distinguished =>
F(r,t) = |psi(r,t)|^2 = const
> How does it evolve? �I don't
> believe any sensible evolution can be
> defined. �For example, in Bohmian
> mechanics the so-called quantum
> potential is not well-defined in such a
> case, and if one tries to define it via
> a sequence of distributions one still
> gets nonsense (intuitively, the
> particle experiences an infinite
> acceleration and the probability
> distribution spreads instantaneously
> everywhere). �This is very
> different from classical mechanics!
BTW, for a free particle, one cannot strictly say whether it is
quantum or classical. For the description of the classical stationary
state by means of time-independent functions holds as well that p=p_0,
while the position is undefined.
> Note that, for a non-free particle,
> one has a similar problem even for
> intial states of definite momentum
> (unless one goes to periodic
> boundary conditions, eg, a
> particle on a ring).
I don't know to which extend one can here superpose stationary wave
functions such that the initial wave function is localized in
(momentum) space...
Best wishes,
Peter
Arnold Neumaier
> On 26 nov, 19:32, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
> wrote:
>
>>> Mixed states are an incoherent superposition, that is, described by
>>> classical statistical mechanics. It is no longer QM.
>
>> You are the first whom I saw claiming that statistical mechanics with
>> density matrices is no longer quantum mechanics.
>
> I was too radical effectively. But it remains that incoherent
> superposition occurs in classical mechanics, and is formalized by
> plain classical probabilities.
But density matrices are intrinsically quantum mechanical.
Nevertheless they exhibit very close analogies to classical densities.
Therefore everyone interested in the relations between classical and
quantum mechanics is well-advised to look at both in the statistical
mechanics version, where the analogies are obvious, and the transition
from quantum to classical takes the form of a simple approximation.
>> In fact, almost all quantum mechanics applied to real systems not in
>> the ground state needs density matrices, since pure states are very
>> difficult to create and propagate unless a system is in the ground
>> state. Pure states describe only an idealized version of quantum
>> reality.
>
> That is a contingency I spoke about. We were discussing the theories
> per se.
>
>
>> Things get more elegant with density
>> matrices, and intuition is easier since the classical analogy fits
>> much better.
>
> QM isn't intuitive, and that's why it's a bit acrobatic (or risky if
> you prefer.) Elegance isn't correctness, and precisely a classical
> analogy might insinuate while the theories don't reduce to one
> another.
QM in the statistical mechanics version is almost as intuitive as
classical statistical mechanics. The only somewhat nonintuitive part
is in both cases how to interpret probability. (This is already a
severe problem in classical statistical mechanics, as the book by
Laurence Sklar, Physics and Chance, explains in detail.)
>> In fact, one can set up the whole machinery of quantum mechanics without
>> ever introducing pure states. The emphasis on pure states in textbooks
>> is only a historical accident.
>
> That can also be done without operators, without Hibert space and so
> on. But a theory that isn't able to describe pure states, in a way or
> another, couldn't be QM. An interference pattern can occur only with a
> pure state.
No. Interference patterns even arise in classical electrodynamics.
And interference patterns occur in quantum mechanics also with
approximately pure states. Most real states are only approximately
pure; nevertheless, interference is often observed.
Finally, as you observed, pure states are special mixed states,
so have a place in statistical mechanics (where they typically arise
as approximations in the cold limits T-->0.)
>>> Remember that in QM, the uncertainties aren't bounded.
>
>> Neither are they in classical statistical mechanics.
>> This doesn't affect deterministic approximations, which only require
>> that variances are small.
>
> If the variances are small, they are bounded, and they play the role
> of uncertainties in QM. To the contrary, in QM the uncertainties can
> alway be made greather than the mean, provided a precise enough
> measurement is feasible (and done.)
In classical statistical mechanics, the uncertainties can also be made
greater than the mean, so this is not a distinctive feature of quantum
mechanics.
But in many important cases, the variances are much smaller than the
mean. Then (and only then) is a deterministic approximation adequate.
>>> If we are
>>> talking about the position uncertainty in order to define a
>>> trajectory, and make precise position measurements, then the momentum
>>> uncertainty becomes very large, and the trajectory concept (actually
>>> defined in the phase space) loses sense.
>
>> I did not say that deterministic approximations are _always_ adequate.
>> Note my qualification that ``the system is nice enough that
>> the stochastic uncertainties in the quantities of interest are much
>> smaller than the quantities themselves''
>
> Your qualification means "the system is classical."
Yes; precisely this was my claim. A quantum system is essentially
classical if its relevant quantities have uncertainties that are small
compared to their expectations.
>>>> This deterministic approximation is given by a classical dynamical
>>>> system for the (expectations of the) quantities of interest.
>
>>>> Thus, in a sense, classical variables are simply expectations of
>>>> relevant quantum variables with small uncertainty. The small uncertainty
>>>> makes these variables approximately predictable in each individual
>>>> event, and hence classical.
>
>>> For that to make sense, a "nice" enough system (or rather state) must
>>> be taken from the begining (for exemple coherent states,) and some
>>> measurements not allowed (or macroscopically not feasible.) That may
>>> happen in our macroscopical world,
>
>> This happens in many microscopic systems. Laser light is usually in a
>> coherent state.
>
> Here a "coherent state" is a special state where the product dp.dx is
> minimum, that is, a "wave packet" whose mean values behave
> classically. But we know that even for those states there is
> dispersion, so that the position uncertainty (variance of position)
> grows indefinitely.
No. Usually the position uncertainty is quite well bounded.
Otherwise one could not even say that the system under study is
confined to a region behind a polarizer, on the desk of the
experimenter, or in the domain of vision of a telescope.
>> Classicality develops whenever the variances of the quantities of
>> interest become small compared to their expectations. Of course,
>> there is significant interst in quantum systems that are decidedly
>> non-classical, so that this does not happen; but this does not change
>> the general principle.
>
> Assuming there are situations in which classicality so defined
> develops, is it the same as to make h tend to 0?
No. h is a fixed constant of Nature, which is often even set to
one to have convenient units. The classical limit is the limit of large
quantum numbers M (typically of mass, number of particles, or size of
angular momentum).
But in many situations, the effect is similar to taking the limit
hbar --> 0. In these cases the relative uncertainties scale with
sqrt(hbar/M), which becomes small if either hbar is made formally
tiny or if M is large.
> Apparently, it isn't
> sufficient, and classicality can be lost even once it has developped.
And it can be gained if it was not there initially. Both is in
the nature of quantum systems, and no argument against what I said.
Classical and nonclassical quantum systems coexist in Nature,
and the border line between them is quite fuzzy.
Decoherence only handles a small part of this border line, using
much technical machinery, while statistical mechanics handles it
all in an easily understood manner.
I added two sections
S1s. The classical limit of quantum mechanics
S1t. The classical limit via coherent states
to my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
containing the essence of the arguments in this thread, and some
additional (more technical) explanation of the classical limit
in terms of coherent states.
Arnold Neumaier
Peter
> ============== Moderator's note ===============================
> delta(p - p_0)^2 has not only no classical counterpart, but it is not
> defined at all. It's a meaningless expression. It's clear that generalized
> eigensolutions of essentially adjoint operators with a continuous spectral
> value are no L**2 functions but distributions, defined as linear forms acting
> on the much smaller nuclear space of the rigged-Hilbert space construction
> (Gelfand triple).
> ================================================================
Please consult L. Berg, Products of the Dirac delta distribution and
their derivatives, Nova Acta Leopoldina 44 (1976) Suppl. 8, pp.69-77
(in German)
Without questioning the invention of Gelfand's mathematical genius,
let me ask whether we really need the rigged Hilbert space for a
mathematically sound foundation of the Schr�dinger wave functions for
free particles?
For your wording "eigensolutions" indicates that you may not follow
Schr�dinger's, Heisenberg's and others' insight that the maths of
eigenvalue problems is appropriate for standing classical waves
(strings, pipes), but not for quantum systems.
I agree, of course, that my posting(s) aim(s) at physical insight
rather than maths rigour, without discarding that the latter should
finally (not from the very beginning!) be reached in theoretical
physics :-)
Best wishes,
Peter
X-Phy
> > superposition occurs in classical mechanics, and is formalized by
> > plain classical probabilities.
On 30 nov, 18:30, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
wrote:
> But density matrices are intrinsically quantum mechanical.
> Nevertheless they exhibit very close analogies to classical densities.
>
> Therefore everyone interested in the relations between classical and
> quantum mechanics is well-advised to look at both in the statistical
> mechanics version, where the analogies are obvious, and the transition
> from quantum to classical takes the form of a simple approximation.
Whatever all the virtues of the density matrix, it contains reduced
informations. I could also say I have five fruits in my basket, so
apples are an approximation of oranges, and the fruit concept make
obvious the transition between apples and oranges.
> >> Things get more elegant with density
> >> matrices, and intuition is easier since the classical analogy fits
> >> much better.
> > QM isn't intuitive, and that's why it's a bit acrobatic (or risky if
> > you prefer.) �Elegance isn't correctness, and precisely a classical
> > analogy might insinuate while the theories don't reduce to one
> > another.
> QM in the statistical mechanics version is almost as intuitive as
> classical statistical mechanics. The only somewhat nonintuitive part
> is in both cases how to interpret probability.
Not the probabilities. QM deals with probability amplitudes, which
are nonintuitive (unanschaulich.) There have been many attemps to
understand QM intuitively, and all have failed. I challenge you to
make an intuitive account of QM that neither is inconsistent nor hide
the difficulty under the carpet. No hidden variable or signal faster
than light can be used, which makes the task impossible as soon as the
collapse intervene, since intuitively we have either of them. It is
sort of a mathematical miracle that a so "ill founded" theory gives
consistent predictions.
> >> In fact, one can set up the whole machinery of quantum mechanics without
> >> ever introducing pure states. The emphasis on pure states in textbooks
> >> is only a historical accident.
> > That can also be done without operators, without Hibert space and so
> > on. But a theory that isn't able to describe pure states, in a way or
> > another, couldn't be QM. An interference pattern can occur only with a
> > pure state.
> No. Interference patterns even arise in classical electrodynamics.
You are playing with words. Without pure states, interference
patterns with electrons couldn't occur, and that is what we are
discussing about.
> And interference patterns occur in quantum mechanics also with
> approximately pure states. Most real states are only approximately
> pure; nevertheless, interference is often observed.
Pure or approximately pure must be described by the theory, that's the
same. There is as well no pure water, although thermodynamics of pure
bodies can be done.
> Finally, as you observed, pure states are special mixed states,
> so have a place in statistical mechanics (where they typically arise
> as approximations in the cold limits T-->0.)
It is the converse we are discussing.
> > If the variances are small, they are bounded, and they play the role
> > of uncertainties in QM. �To the contrary, in QM the uncertainties can
> > alway be made greather than the mean, provided a precise enough
> > measurement is feasible (and done.)
> In classical statistical mechanics, the uncertainties can also be made
> greater than the mean, so this is not a distinctive feature of quantum
> mechanics.
Yes, but if it is the case, it isn't deterministic approximations, and
therefore we are no longer addressing the classical limit. In
constrast, QM remains QM even with large uncertainties.
> >> I did not say that deterministic approximations are _always_ adequate.
> >> Note my qualification that ``the system is nice enough that
> >> the stochastic uncertainties in the quantities of interest are much
> >> smaller than the quantities themselves''
> > Your qualification means "the system is classical." �
> Yes; precisely this was my claim. A quantum system is essentially
> classical if its relevant quantities have uncertainties that are small
> compared to their expectations.
So you say, the classical limit isn't obtained from the only
requirement that h tends formaly to 0 (or that every action value
characteristic of the system is >> h.) A further requirement is that
some measurements, especially too precise ones, aren't allowed,
otherwise your qualifacation should be "described by CM", and thus
irreal.
> >>>> Thus, in a sense, classical variables are simply expectations of
> >>>> relevant quantum variables with small uncertainty. The small uncertainty
> >>>> makes these variables approximately predictable in each individual
> >>>> event, and hence classical.
> >>> For that to make sense, a "nice" enough system (or rather state) must
> >>> be taken from the begining (for exemple coherent states,) and some
> >>> measurements not allowed (or macroscopically not feasible.) �That may
> >>> happen in our macroscopical world,
> >> This happens in many microscopic systems. Laser light is usually in a
> >> coherent state.
> > Here a "coherent state" is a special state where the product dp.dx is
> > minimum, that is, a "wave packet" whose mean values behave
> > classically. �But we know that even for those states there is
> > dispersion, so that the position uncertainty (variance of position)
> > grows indefinitely.
> No. Usually the position uncertainty is quite well bounded.
> Otherwise one could not even say that the system under study is
> confined to a region behind a polarizer, on the desk of the
> experimenter, or in the domain of vision of a telescope.
It is empirically bounded, but that is inconsistent with naive QM
predictions. There is no reason either that decoherence should favour
localized states. It is a great difficulty for trying to squeeze QM
in the little box of CM. Usually, before an experiment, a
comparatively localized state is prepared, so that it isn't really a
problem for the cases you talk about.
> >> Classicality develops whenever the variances of the quantities of
> >> interest become small compared to their expectations. Of course,
> >> there is significant interst in quantum systems that are decidedly
> >> non-classical, so that this does not happen; but this does not change
> >> the general principle.
>
> > Assuming there are situations in which classicality so defined
> > develops, is it the same as to make h tend to 0? �
> No. h is a fixed constant of Nature, which is often even set to
> one to have convenient units. The classical limit is the limit of large
> quantum numbers M (typically of mass, number of particles, or size of
> angular momentum).
>
> But in many situations, the effect is similar to taking the limit
> hbar --> 0. In these cases �the relative uncertainties scale with
> sqrt(hbar/M), which becomes small if either hbar is made formally
> tiny or if M is large.
> > Apparently, it isn't sufficient,
You didn't answer the question. We saw that large quantum numbers
don't make sure that the uncertainties get lower than the mean values,
unless full buckets of holy water are used. We aren't in the case of
thermodynamics where large deviations become very unlikely from the
foundations of the theory themselves. Even if they are possible, that
doesn't make it analogous to QM, since it is rather a likelihood than
a possibility. We lack a second principle of QM which would free it
from any mystery and make it intuitive and compatible with CM.
In the case of large numbers of particles, there is a thermodynamics
limit along with the classical one, both must no be mixed up.
Arnold Neumaier
>>> I was too radical effectively. But it remains that incoherent
>>> superposition occurs in classical mechanics, and is formalized by
>>> plain classical probabilities.
>
> On 30 nov, 18:30, Arnold Neumaier <Arnold.Neuma...@univie.ac.at>
> wrote:
>
>> But density matrices are intrinsically quantum mechanical.
>> Nevertheless they exhibit very close analogies to classical densities.
>>
>> Therefore everyone interested in the relations between classical and
>> quantum mechanics is well-advised to look at both in the statistical
>> mechanics version, where the analogies are obvious, and the transition
>> from quantum to classical takes the form of a simple approximation.
>
> Whatever all the virtues of the density matrix, it contains reduced
> informations.
How do you know? There is no experimental way to decompose a mixed
state into its pure components. This tells us that the mixed state
is something irreducible, not something reduced.
If one writes QM from the start in terms of density matrices,
one gets a consistent theory that is more general than the
wave-function-based QM, fits reality better than the latter,
and is closer to the classical version of mechanics.
All this suggests that this is the correct formulation,
and that the preferred wave-function-based approach is a
purely historical accident.
>>>> Things get more elegant with density
>>>> matrices, and intuition is easier since the classical analogy fits
>>>> much better.
>
>>> QM isn't intuitive, and that's why it's a bit acrobatic (or risky if
>>> you prefer.) Elegance isn't correctness, and precisely a classical
>>> analogy might insinuate while the theories don't reduce to one
>>> another.
>
>> QM in the statistical mechanics version is almost as intuitive as
>> classical statistical mechanics. The only somewhat nonintuitive part
>> is in both cases how to interpret probability.
>
> Not the probabilities. QM deals with probability amplitudes, which
> are nonintuitive (unanschaulich.)
In the density matrix version, no probability amoplitudes appear,
only true probabilities. This adds to the intuitiveness of the density
matrix version.
> There have been many attemps to
> understand QM intuitively, and all have failed. I challenge you to
> make an intuitive account of QM that neither is inconsistent nor hide
> the difficulty under the carpet.
See Chapter 7 of
Arnold Neumaier and Dennis Westra,
Classical and Quantum Mechanics via Lie algebras,
http://www.mat.univie.ac.at/~neum/papers/physpapers.html#QML
http://de.arxiv.org/abs/0810.1019
> It is
> sort of a mathematical miracle that a so "ill founded" theory gives
> consistent predictions.
No miracles are needed in the above Chapter 7.
>>>> In fact, one can set up the whole machinery of quantum mechanics without
>>>> ever introducing pure states. The emphasis on pure states in textbooks
>>>> is only a historical accident.
>
>>> That can also be done without operators, without Hibert space and so
>>> on. But a theory that isn't able to describe pure states, in a way or
>>> another, couldn't be QM. An interference pattern can occur only with a
>>> pure state.
>
>> No. Interference patterns even arise in classical electrodynamics.
>
> You are playing with words. Without pure states, interference
> patterns with electrons couldn't occur, and that is what we are
> discussing about.
According to the thread title, we are discussing the relation between
quantum mechanics and classical mechanics
>> And interference patterns occur in quantum mechanics also with
>> approximately pure states. Most real states are only approximately
>> pure; nevertheless, interference is often observed.
>
> Pure or approximately pure must be described by the theory, that's the
> same. There is as well no pure water, although thermodynamics of pure
> bodies can be done.
You misunderstood. Approximately pure states are described by density
matrices, not by pure states. And one can predict with density matrices
the extent to which an interference pattern is blurred. Thus
interference is not an argument for pure states.
>> Finally, as you observed, pure states are special mixed states,
>> so have a place in statistical mechanics (where they typically arise
>> as approximations in the cold limits T-->0.)
>
> It is the converse we are discussing.
We are discussing that, typically, observed states are not pure,
since we always work at finite temperature.
>>> If the variances are small, they are bounded, and they play the role
>>> of uncertainties in QM. To the contrary, in QM the uncertainties can
>>> alway be made greather than the mean, provided a precise enough
>>> measurement is feasible (and done.)
>
>> In classical statistical mechanics, the uncertainties can also be made
>> greater than the mean, so this is not a distinctive feature of quantum
>> mechanics.
>
> Yes, but if it is the case, it isn't deterministic approximations, and
> therefore we are no longer addressing the classical limit. In
> constrast, QM remains QM even with large uncertainties.
Classical statistical mechanics also ramains classical statistical
mechanics when uncertainties are large. But with large uncertainties,
the classical limit becomes blurred.
>>>> I did not say that deterministic approximations are _always_ adequate.
>>>> Note my qualification that ``the system is nice enough that
>>>> the stochastic uncertainties in the quantities of interest are much
>>>> smaller than the quantities themselves''
>
>>> Your qualification means "the system is classical."
>
>> Yes; precisely this was my claim. A quantum system is essentially
>> classical if its relevant quantities have uncertainties that are small
>> compared to their expectations.
>
> So you say, the classical limit isn't obtained from the only
> requirement that h tends formaly to 0 (or that every action value
> characteristic of the system is >> h.)
h is a constant, hence never goes to zero.
One calls the limit h to 0 also the classical limit, but this
is a purely theoretical construct. The only experimentally
meaningful classical limit is that of small uncertainties.
> A further requirement is that
> some measurements, especially too precise ones, aren't allowed,
> otherwise your qualifacation should be "described by CM", and thus
> irreal.
If one is interested in the dynamics, one cannot make too precise
measurements, since one always needs position and momentum
(or position at many times, which allows to reconstruct the momentum).
>
>> No. Usually the position uncertainty is quite well bounded.
>> Otherwise one could not even say that the system under study is
>> confined to a region behind a polarizer, on the desk of the
>> experimenter, or in the domain of vision of a telescope.
>
> It is empirically bounded, but that is inconsistent with naive QM
> predictions.
No. Momentum measurements in a real experiment on a desk are never
so precise that the uncertainty in position is bigger than the size of
the desk.
>>>> Classicality develops whenever the variances of the quantities of
>>>> interest become small compared to their expectations. Of course,
>>>> there is significant interst in quantum systems that are decidedly
>>>> non-classical, so that this does not happen; but this does not change
>>>> the general principle.
>>> Assuming there are situations in which classicality so defined
>>> develops, is it the same as to make h tend to 0?
>
>> No. h is a fixed constant of Nature, which is often even set to
>> one to have convenient units. The classical limit is the limit of large
>> quantum numbers M (typically of mass, number of particles, or size of
>> angular momentum).
>>
>> But in many situations, the effect is similar to taking the limit
>> hbar --> 0. In these cases the relative uncertainties scale with
>> sqrt(hbar/M), which becomes small if either hbar is made formally
>> tiny or if M is large.
>
>>> Apparently, it isn't sufficient,
In many situations it is. I didn't claim in all.
Arnold Neumaier
Hendrik van Hees
> On 30 Nov., 10:56, Peter <end...@dekasges.de> wrote:
>
>> ============== Moderator's note =================================
>> delta(p - p_0)^2 has not only no classical counterpart, but it is not
>> defined at all. It's a meaningless expression. It's clear that
>> generalized eigensolutions of essentially adjoint operators with a
>> continuous spectral value are no L**2 functions but distributions,
>> defined as linear forms acting on the much smaller nuclear space of
>> the rigged-Hilbert space construction (Gelfand triple).
>> ==================================================================
>
> Please consult L. Berg, Products of the Dirac delta distribution and
> their derivatives, Nova Acta Leopoldina 44 (1976) Suppl. 8, pp.69-77
> (in German)
I'm not sure whether I can get this journal at my university. Can you
briefly summarize what might be a sensible meaning of a squared delta
function? In my experience, whenever this appears (and to my surprise it
once even appeared in a scientific paper, I had to review!) it's just a
bad handling of the mathematics by the author.
>
> Without questioning the invention of Gelfand's mathematical genius,
> let me ask whether we really need the rigged Hilbert space for a
> mathematically sound foundation of the Schr�dinger wave functions for
> free particles?
Of course not. You can do it like von Neumann in the good oldfashioned
way, but it's so much simpler with the rigged Hilbert space. I guess
that's why it's called rigged!
>
> For your wording "eigensolutions" indicates that you may not follow
> Schr�dinger's, Heisenberg's and others' insight that the maths of
> eigenvalue problems is appropriate for standing classical waves
> (strings, pipes), but not for quantum systems.
I'm perhaps not aware about newer history-of-science studies, but where
have Schr�dinger and Heisnberg stated this "insight"? To the contrary,
I think that the spectral theory of essentially adjoint operators is at
the heart of the very foundation of quantum theory (particularly since
Schr�dinger, Heisenberg, Born, Jordan, and Dirac).
>
> I agree, of course, that my posting(s) aim(s) at physical insight
> rather than maths rigour, without discarding that the latter should
> finally (not from the very beginning!) be reached in theoretical
> physics :-)
I'm also more towards physics than mathematical rigor, as you can see in
my brief remark above, where I refer to generalized eigensolutions which
is more a physicist's language than a rigoros mathematical expression!
--
Hendrik van Hees
Institut f�r Theoretische Physik
Justus-Liebig-Universit�t Gie�en
http://theorie.physik.uni-giessen.de/~hees/
Peter
giessen.de> wrote:
> >> ============== Moderator's note =================================
> >> delta(p - p_0)^2 has not only no classical counterpart, but it is not
> >> defined at all. It's a meaningless expression. It's clear that
> >> generalized eigensolutions of essentially adjoint operators with a
> >> continuous spectral value are no L**2 functions but distributions,
> >> defined as linear forms acting on the much smaller nuclear space of
> >> the rigged-Hilbert space construction (Gelfand triple).
> >> ==================================================================
>
> > Please consult L. Berg, Products of the Dirac delta distribution and
> > their derivatives, Nova Acta Leopoldina 44 (1976) Suppl. 8, pp.69-77
> > (in German)
>
> I'm not sure whether I can get this journal at my university. Can you
> briefly summarize what might be a sensible meaning of a squared delta
> function? In my experience, whenever this appears (and to my surprise it
> once even appeared in a scientific paper, I had to review!) it's just a
> bad handling of the mathematics by the author.
Hello Hendrik,
Berg considers
- Dirac's delta function d(t)
- Heaviside's jump function h(t)
- the break ("Knick") function k(t) = t h(t)
It holds
h = k', d = h' = k''
h = h^2 (1)
k = h k = k h (2)
d = d h + h d (3)
d' = d' h+ 2d^2 + h d'
d'' = d'' h + 3d' d + 3d d' + h d'' etc.
Here and in further expression obtained in a similar manner, d^2
occurs; btw,
d^3 = 0
There are subtleties mentioned in that contribution, more details are
in papers by Berg in Math. Nachr. and ZAMM
> > Without questioning the invention of Gelfand's mathematical genius,
> > let me ask whether we really need the rigged Hilbert space for a
> > mathematically sound foundation of the Schr�dinger wave functions for
> > free particles?
> Of course not. You can do it like von Neumann in the good oldfashioned
> way, but it's so much simpler with the rigged Hilbert space. I guess
> that's why it's called rigged!
Babylon say about 'rig'.
set up, make ready for use; equip, furnish with supplies or equipment;
falsify, tamper with, manipulate fraudulently; dress, clothe
(Informal); fit or install sails on a sailboat
;-)
> > For your wording "eigensolutions" indicates that you may not follow
> > Schr�dinger's, Heisenberg's and others' insight that the maths of
> > eigenvalue problems is appropriate for standing classical waves
> > (strings, pipes), but not for quantum systems.
> I'm perhaps not aware about newer history-of-science studies, but where
> have Schr�dinger and Heisnberg stated this "insight"?
Schr�dinger in his 1926 pioneering papers, Heisenberg in his 1929
Chicago lecture
> To the contrary,
> I think that the spectral theory of essentially adjoint operators is at
> the heart of the very foundation of quantum theory (particularly since
> Schr�dinger, Heisenberg, Born, Jordan, and Dirac).
Schr�dinger knew it better :-)
For Schr�dinger remarked what one furthermore has to postulate when
founding QM that way :-)
Looking forward,
Peter
Igor Khavkine
> On 3 Dez., 11:25, Hendrik van Hees <Hendrik.vanH...@physik.uni-
> > > Please consult L. Berg, Products of the Dirac delta distribution and
> > > their derivatives, Nova Acta Leopoldina 44 (1976) Suppl. 8, pp.69-77
> > > (in German)
>
> > I'm not sure whether I can get this journal at my university. Can you
> > briefly summarize what might be a sensible meaning of a squared delta
> > function? In my experience, whenever this appears (and to my surprise it
> > once even appeared in a scientific paper, I had to review!) it's just a
> > bad handling of the mathematics by the author.
> Berg considers
> - Dirac's delta function d(t)
> - Heaviside's jump function h(t)
> - the break ("Knick") function k(t) = t h(t)
>
> It holds
>
> h = k', d = h' = k''
>
> h = h^2 (1)
This formula makes sense because h is an honest function (which d is
not).
> k = h k = k h (2)
>
> d = d h + h d (3)
Does d h != h d? Why the non-commutativity?
> d' = d' h+ 2d^2 + h d'
>
> d'' = d'' h + 3d' d + 3d d' + h d'' etc.
Is the point here that d^2 is defined in terms of d', d' h, and h d'?
How are are d' h and h d' any better defined than d^2?
> Here and in further expression obtained in a similar manner, d^2
> occurs; btw,
>
> d^3 = 0
>
> There are subtleties mentioned in that contribution, more details are
> in papers by Berg in Math. Nachr. and ZAMM
This smells fishy. Even with subtleties, I don't see how this approach
would lead to anything sensible.
AFAIK, there are two contexts where multiplication of distributions
makes mathematical sense: microlocal analysis and Colombeaux algebras.
Both have a substantial history of mathematical development.
The basic idea in the microlocal analysis context is that two
distributions can be multiplied if one is smooth where the other is
singular. When this notion is analyzed in terms of "local" Fourier
transforms, it becomes clear that this mutual non-singularity
condition can be generalized to some cases when neither distribution
is smooth at the same point, but only in some cases.
The basic idea in the Colombeaux algebra context is that that any
distribution can be smeared and thus become non-singular in a very
nice way. All the algebraic and differential operations are well
defined on smeared distributions. Formally taking the limit where the
smearing is removed preserves the algebraic and differential
structure. The price is that the resulting algebra is much larger than
just the smooth functions together with linear distributions.
I do have to add the disclaimer that my understanding of these
mathematical ideas is fairly rudimentary. So I'd be happy if someone
could correct any of the above.
In any case, all of the above is waaaay outside the usual L^2 where
the Schroedinger equation and ordinary quantum mechanics are very well
defined and understood.
Igor
X-Phy
>> informations.
On 3 d�c, 11:25, Arnold Neumaier <Arnold.Neuma...@univie.ac.at> wrote:
> How do you know?
Because it is a sum.
> There is no experimental way to decompose a mixed
> state into its pure components. This tells us that the mixed state
> is something irreducible, not something reduced.
There no experimental way either to decompose a dice into six faces up,
nor to decompose a plane into length and width.
> If one writes QM from the start in terms of density matrices,
> one gets a consistent theory
A reduced theory is always consistent if the initial theory is so.
Arithmetic with fruits is as well valid as arithmetic with apples or
oranges.
> that is more general than the
> wave-function-based QM, fits reality better than the latter,
> and is closer to the classical version of mechanics.
>
> All this suggests that this is the correct formulation,
Correct? We are no longer talking about physics. Moderator!
Any discussion about which formulation is "correct", "fundamental",
"true", "basic"... is sterile from the beginning.
> and that the preferred wave-function-based approach is a
> purely historical accident.
All yours arguments miss the point. But now, it has been overlooked
that the density matrix contains more variables than the wave function
in the case of a pure state. Therefore no accident. Describing
mixtures of sates, when they are numerous, is essentially a forced
classical situation. But if the meaning of the classical probabilities
is literally taken, there is only one pure state, but we don't know
which it is. Physics has no sleeves (or no carpet if you prefer.)
>>> QM in the statistical mechanics version is almost as intuitive as
>>> classical statistical mechanics. The only somewhat non intuitive part
>>> is in both cases how to interpret probability.
>> Not the probabilities. QM deals with probability amplitudes, which
>> are nonintuitive (unanschaulich.)
> In the density matrix version, no probability amoplitudes appear,
> only true probabilities. This adds to the intuitiveness of the density
> matrix version.
Are you saying that the density matrix is unable to describe a quantum
system? If not, you are inconsistent. True probabilities can't be used
to describe entanglement, for instance, since they are equivalent to
local hidden variables.
>> There have been many attemps to
>> understand QM intuitively, and all have failed. I challenge you to
>> make an intuitive account of QM that neither is inconsistent nor hide
>> the difficulty under the carpet.
> See Chapter 7 of
> Arnold Neumaier and Dennis Westra,
> Classical and Quantum Mechanics via Lie algebras,
> http://www.mat.univie.ac.at/~neum/papers/physpapers.html#QML
> http://de.arxiv.org/abs/0810.1019
What? This chapter is about thermal theory. What has it to do with the
EPR paradox for example? you use a very thick carpet of equations and
formula of all sorts, that's all I can see. Intuition is a very direct
way of looking at things. If you can't make an account in a few lines
for somebody who is familiar with quantum theory, forget it.
>> It is
>> sort of a mathematical miracle that a so "ill founded" theory gives
>> consistent predictions.
> No miracles are needed in the above Chapter 7.
Of course, it is a quaint expression. Nevertheless, there is nothing
intuitive at all, and nobody ever have understood why it is so. That's
like a magician cutting a woman, separating arms and legs in A and B
corners, then bringing them together in the same whole woman. Yet we
don't know the trick. You may use the mathematics you want to describe
the motion of legs and arms, that don't make you understand what happen.
>>>> That can also be done without operators, without Hibert space and so
>>>> on. But a theory that isn't able to describe pure states, in a way or
>>>> another, couldn't be QM. An interference pattern can occur only with a
>>>> pure state.
>>> No. Interference patterns even arise in classical electrodynamics.
>> You are playing with words. Without pure states, interference
>> patterns with electrons couldn't occur, and that is what we are
>> discussing about.
> According to the thread title, we are discussing the relation between
> quantum mechanics and classical mechanics
Included in QM is the quantum theory of the electron. You can't put
aside what doesn't fit your ideas.
>>> And interference patterns occur in quantum mechanics also with
>>> approximately pure states. Most real states are only approximately
>>> pure; nevertheless, interference is often observed.
>> Pure or approximately pure must be described by the theory, that's the
>> same. There is as well no pure water, although thermodynamics of pure
>> bodies can be done.
> You misunderstood. Approximately pure states are described by density
> matrices, not by pure states.
No, that's plain wrong, sorry. They are also described by pure states
with there respective probabilities, even if that decomposition isn't
unique, it doesn't matter. There isn't either a unique frame of
reference.
> And one can predict with density matrices
> the extent to which an interference pattern is blurred. Thus
> interference is not an argument for pure states.
You just proved the contrary.
>>> Finally, as you observed, pure states are special mixed states,
>>> so have a place in statistical mechanics (where they typically arise
>>> as approximations in the cold limits T-->0.)
>> It is the converse we are discussing.
> We are discussing that, typically, observed states are not pure,
> since we always work at finite temperature.
You are mixing QM with thermodynamics. We aren't discussing the
thermodynamical limit of classical statistical theory.
>>> In classical statistical mechanics, the uncertainties can also be made
>>> greater than the mean, so this is not a distinctive feature of quantum
>>> mechanics.
>> Yes, but if it is the case, it isn't deterministic approximations, and
>> therefore we are no longer addressing the classical limit. In
>> constrast, QM remains QM even with large uncertainties.
> Classical statistical mechanics also ramains classical statistical
> mechanics when uncertainties are large.
> But with large uncertainties, the classical limit becomes blurred.
We are talking about *classical mechanics*, not statistical mechanics.
>>> Yes; precisely this was my claim. A quantum system is essentially
>>> classical if its relevant quantities have uncertainties that are small
>>> compared to their expectations.
>> So you say, the classical limit isn't obtained from the only
>> requirement that h tends formaly to 0 (or that every action value
>> characteristic of the system is >> h.)
> h is a constant, hence never goes to zero.
>
> One calls the limit h to 0 also the classical limit, but this
> is a purely theoretical construct. The only experimentally
> meaningful classical limit is that of small uncertainties.
You didn't answer. The formal limit, be it only theoretical, exists on
its own right. The question is whether this limit is CM. I argue it
isn't. Consequently, any limit that is mathematically equivalent to it
isn't CM.
>> A further requirement is that
>> some measurements, especially too precise ones, aren't allowed,
>> otherwise your qualifacation should be "described by CM", and thus
>> irreal.
> If one is interested in the dynamics, one cannot make too precise
> measurements, since one always needs position and momentum
> (or position at many times, which allows to reconstruct the momentum).
Being interested in the dynamics doesn't prevent to make precise
measurements. You have the same type of argument as Goethe who said the
prism experiment of Newton wasn't valid because it didn't observe Nature
in itself but a nature modified by the slit. I'm interested in the
dynamics *while* precise measurements are done, because I am a physicist
and want to know everything, and because I think precise measurements
belong to Nature.
>>> No. Usually the position uncertainty is quite well bounded.
>>> Otherwise one could not even say that the system under study is
>>> confined to a region behind a polarizer, on the desk of the
>>> experimenter, or in the domain of vision of a telescope.
>>
>> It is empirically bounded, but that is inconsistent with naive QM
>> predictions.
>
> No. Momentum measurements in a real experiment on a desk are never
> so precise that the uncertainty in position is bigger than the size of
> the desk.
Which means exactly what I said, it is *empirically* bounded. But QM
allows unlocalized states. If you don't explain why those states aren't
empirically observed, you don't explain why CM emerges from QM.
>>> But in many situations, the effect is similar to taking the limit
>>> hbar --> 0. In these cases the relative uncertainties scale with
>>> sqrt(hbar/M), which becomes small if either hbar is made formally
>>> tiny or if M is large.
>>>> Apparently, it isn't sufficient,
> In many situations it is. I didn't claim in all.
So you don't address why in the macroscopical world, quantum effects
aren't observed. So you didn't address the relation between QM and CM
at all.
Let me synthesize and formalize all that.
We have invoked a parallel between two limiting models. First, the
transition between (classical or quantum) mechanics (Me) and
thermodynamics (Th), and second, between quantum mechanics (QM) and
classical mechanics (CM).
In Me -> Th, the domain of validity is given by the parameter N, the
number of particles. If N is large enough, then Me and Th are identical
models, that is, they give indistinguishable predictions up to the
measurement precision. Making a prediction from Me is as usual more
complex since it involves the use of the classical equations plus
probability theory, the so called statistical mechanics.
In QM -> CM, it seems that analogously taking the limit M (quantum
numbers) large will work, but that's not so simple. In Me -> Th, taking
N large (through probability theory) implies the uncertainties of Th
variables become very small compared to their mean values. But it is
not true in QM. A stronger condition is necessary, that is, (all) the
uncertainties must be small compared to the mean values (or the
measurement uncertainties in some case like for the position,) which
also entails the condition that the quantum numbers are large.
Now there is still a big difference. While in Me the condition only
need to be satisfied at a time, then it is (likely) satisfied at a later
time, for the reasons I have given that isn't true for QM. That why I
think we can't say QM and CM aren't identical models.
If we look further, we see that even the large N condition isn't so
simple. Systems occur in Nature where N is large but the uncertainties
aren't small, like for example a vessel where all the molecules are
located in the same half. Such systems can be created by man, or may
happen naturally, for example when a membrane is broken. But they
(likely) evolve toward a state where the condition is fulfilled, through
what is called an irreversible process. Above we silently used the
second principle of Th, with the concept of entropy. This principle is
necessary if Me and Th should remain identical.
Similarly, QM and CM can't be identical unless a second principle of QM
is introduced. And if that isn't possible, we must conclude that QM is
false. The analogy of the half full vessel is the Schr�dinger's cat. A
two states system (vs empty/full) is coupled to a large M system (vs
large N.) There should be an irreversible process which gives a
transition (dead+alive) -> (dead|alive), where the latter state is
distinguished from the former by a higher value of a quantum mechanical
entropy, or QM must be modified in order that states like (dead+alive)
can't exist (and not only are unlikely.)
Arnold Neumaier
>
> AFAIK, there are two contexts where multiplication of distributions
> makes mathematical sense: microlocal analysis and Colombeaux algebras.
> Both have a substantial history of mathematical development.
Actually, in Colombeau algebras one still cannot multiply two dirac
deltas. Colombeau algebras allow the associative multiplication of
arbitrary generalized functions, but the latter are not distributions,
but only _equivalent_ to distributions.
Thus each delta distribution is represented by infinitely many different
generalized functions, and the square depends on which of these you
take. In this way, Colombeau algebras are able to avoid the no-go
theorems for algebras of distributions.
It is easy to show that there cannot be an algebra of distributions
satisfying the usual arithmetic rules: The Heaviside function h
satisfies h=h^2=h^3, hence h'=2hh'=3h^2h. Now h^2=h implies that
h'=2hh'=2(3h^2h'-2hh')=0, But d=h' is the Dirac delta function,
contradiction. With similar arguments, one can find a contradiction
from assuming the existence of d^2.
Arnold Neumaier
Nicolaas Vroom
news:2e8038c9-8f41-4ad0...@m16g2000yqc.googlegroups.com...
> Hi,
>
> This post concerns the relation between quantum mechanics and
> classical mechanics. One can find
> many statements of the kind:
> "Quantum mechanics is the covering theory of
> classical mechanics",
> or
> "The classical limit of quantum mechanics is
> classical mechanics".
> in the literature. I have the following 2 problems
> with these statements:
>
> 1.The classical limit of quantum mechanics is a
> (uniquely defined) classical statistical theory.
> (field theory for two not decoupled variables
> \rho and S, one equation is Hamilton-Jacobi
> the second is the continuity equation),
> which is different from classical mechanics.
> Would you agree that the second of the above
> statements is wrong ? If not please tell me what
> is wrong with my reasoning.
>
> for a deterministic theory ? The former is unable to
> make predictions about individual events and
> contains less information than the latter.
>
>
> Ulf Klein
> ulf....@jku.at
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html
Is it not correct to state that gravity and
the electromagnetic forces belong to clasical mechanics
And that the weak and strong forces
belong to quantum mechanics ?
That means there is a clear dividing line between the two.
Or is the whole more complicated ?
Does it make sense to speak about deterministic theories
and statistical theories ?
Is Newton's Law a deterministic theory a statistical theory or both ?
Newton's Law is based on measurements and act of measuring
is a statistical process. Does it make sense to call Newto's Law
a statistical theory?
IMO the answer is No.
IMO the same is true for all laws.
Newton's Law is a mathematical description of the reality.
The mathematics pure is a deterministic abstraction.
Does it make sense to call Newton's law deterministic.
IMO the answer is No.
IMO the same is true for all laws.
Nicolaas Vroom
http://users.pandora.be/nicvroom/
Arnold Neumaier
>>> Whatever all the virtues of the density matrix, it contains reduced
>>> informations.
>
>
>> How do you know?
>
> Because it is a sum.
No. A density matrix is a positive semidefinite Hermitian operator of
trace one. The representation as a sum is artificial and has no
experimental basis.
It is well-known that one cannot distinguish experimentally the
contributions to a density matrix when written as a weighted sum of
pure density matrices.
>> There is no experimental way to decompose a mixed
>> state into its pure components. This tells us that the mixed state
>> is something irreducible, not something reduced.
One cannot prove experimentally that the density matrix
rho = 1/2 |up><up| + 1/2 |down><down| is composed of a pure up state and
a pure down state. If it were so, it would also be composed of pure
states in arbitrary superposition, since these can be observed to
be seen in corresponding detection experiments.
Thus the state rho would be composed of all pure states
although it is written only as a sum of two - arbitrarily chosen and
orthogonal, or of none. This shows that your argument is meaningless.
Thus a density matrix is irreducible.
> There no experimental way either to decompose a dice into six faces up,
> nor to decompose a plane into length and width.
One can prove experimentally that a die has 6 faces with six different
marks on them. This makes your analogy faulty.
>> If one writes QM from the start in terms of density matrices,
>> one gets a consistent theory
>
> A reduced theory is always consistent if the initial theory is so.
> Arithmetic with fruits is as well valid as arithmetic with apples or
> oranges.
QM with density matrices is like arithmetics, without reduction.
But if you restrict to pure states (apples) you get a reduction like for
arithmetics with apples only.
>> and that the preferred wave-function-based approach is a
>> purely historical accident.
>
> All yours arguments miss the point. But now, it has been overlooked
> that the density matrix contains more variables than the wave function
> in the case of a pure state.
All components of the density matrix are observable,
while the components of a wave function (the amplitudes) are not.
This shows the priority of the density matrix.
That there are more components to a density matrix reflects the physical
fact that most observed states are not pure and pure states do not have
enough degrees of freedom to account for experimental facts.
>>>> QM in the statistical mechanics version is almost as intuitive as
>>>> classical statistical mechanics. The only somewhat non intuitive part
>>>> is in both cases how to interpret probability.
>
>>> Not the probabilities. QM deals with probability amplitudes, which
>>> are nonintuitive (unanschaulich.)
>
>> In the density matrix version, no probability amoplitudes appear,
>> only true probabilities. This adds to the intuitiveness of the density
>> matrix version.
>
> Are you saying that the density matrix is unable to describe a quantum
> system?
No. I am saying that
- pure states and probability amplitudes are unable to describe
most observed quantum systems.
- density matrices do not use probability amplitudes.
- density matrices fully describe the state of a quantum system.
> If not, you are inconsistent. True probabilities can't be used
> to describe entanglement, for instance, since they are equivalent to
> local hidden variables.
A density matrix for a qubit is more than probabilities for up and down,
which only determine its diagonal entries. There is also the
off-diagonal element that describes the nonclassical part.
Similarly, entanglement is fully encoded in the 4x4 density matrix
of a pair of qubits.
>>> There have been many attemps to
>>> understand QM intuitively, and all have failed. I challenge you to
>>> make an intuitive account of QM that neither is inconsistent nor hide
>>> the difficulty under the carpet.
>
>> See Chapter 7 of
>> Arnold Neumaier and Dennis Westra,
>> Classical and Quantum Mechanics via Lie algebras,
>> http://www.mat.univie.ac.at/~neum/papers/physpapers.html#QML
>> http://de.arxiv.org/abs/0810.1019
>
> What? This chapter is about thermal theory. What has it to do with the
> EPR paradox for example? you use a very thick carpet of equations and
> formula of all sorts, that's all I can see.
That's only because you looked at the pattern of characters rather
than read the chapter. The book is written for people who are willing
to read. They can easily find there the desired intuitive account of QM.
> Intuition is a very direct
> way of looking at things. If you can't make an account in a few lines
> for somebody who is familiar with quantum theory, forget it.
I am not interested in giving simplistic intuitions for people who think
that quantum mechanics can be understood in a few lines.
>>>> And interference patterns occur in quantum mechanics also with
>>>> approximately pure states. Most real states are only approximately
>>>> pure; nevertheless, interference is often observed.
>
>>> Pure or approximately pure must be described by the theory, that's the
>>> same. There is as well no pure water, although thermodynamics of pure
>>> bodies can be done.
>
>> You misunderstood. Approximately pure states are described by density
>> matrices, not by pure states.
>
> No, that's plain wrong, sorry. They are also described by pure states
No. An approximately pure state is only approximately described by a
pure state but exactly described by a density matrix.
> with there respective probabilities, even if that decomposition isn't
> unique, it doesn't matter. There isn't either a unique frame of
> reference.
The decomposition doesn't matter, it is as spurious as the coordinates
of a vector field without the frame of reference.
The vector field is the physical invariant; the coordinates only a
view of it. Similarly, the density matrix is the physical invariant,
while the pure states it ``contains'' (in a very vague sense, since
every state is contained in a positive definite mixed state) only give
a view of it.
This view is never used when actually working with density matrices.
For pure states give a very strange point of view for quantum dynamics.
If you look at the standard decomposition of a density matrix rho(t)
into its singular vectors psi(t), which presumably are the pure states
you think rho(t) contained of. and apply the dynamics to the density
matrix rho(t) and to the individual states psi(t), you get completely
different results at any other time....
>> And one can predict with density matrices
>> the extent to which an interference pattern is blurred. Thus
>> interference is not an argument for pure states.
>
> You just proved the contrary.
No. Interference can be completely described by density matrices,
and in fact needs the latter for a full discussion, e.g., in the
case of partially polarized photons. You can find this discussion
in the bible on quantum optics,
L. Mandel and E. Wolf,
Optical Coherence and Quantum Optics,
Cambridge University Press, 1995.
But it takes more than a few lines to understand....
>>>> Finally, as you observed, pure states are special mixed states,
>>>> so have a place in statistical mechanics (where they typically arise
>>>> as approximations in the cold limits T-->0.)
>
>>> It is the converse we are discussing.
>
>> We are discussing that, typically, observed states are not pure,
>> since we always work at finite temperature.
>
> You are mixing QM with thermodynamics. We aren't discussing the
> thermodynamical limit of classical statistical theory.
Classical mechanics is an intuitive limit of quantum mechnaics only
if one takes thermal effects into account. Quantum mechanics without
thermal effects is an idealization that does not apply to real life.
>>>> In classical statistical mechanics, the uncertainties can also be made
>>>> greater than the mean, so this is not a distinctive feature of quantum
>>>> mechanics.
>
>>> Yes, but if it is the case, it isn't deterministic approximations, and
>>> therefore we are no longer addressing the classical limit. In
>>> constrast, QM remains QM even with large uncertainties.
>
>> Classical statistical mechanics also ramains classical statistical
>> mechanics when uncertainties are large.
>> But with large uncertainties, the classical limit becomes blurred.
>
> We are talking about *classical mechanics*, not statistical mechanics.
statistical mechanics is as much part of classical mechanics
as quantum mechnaics is part of statistical mechanics.
Limiting one's attention to pure states makes the relation between
classical and quantum mechanics much harder to understand than
when both are viewed in the right context.
>>>> Yes; precisely this was my claim. A quantum system is essentially
>>>> classical if its relevant quantities have uncertainties that are small
>>>> compared to their expectations.
>
>>> So you say, the classical limit isn't obtained from the only
>>> requirement that h tends formaly to 0 (or that every action value
>>> characteristic of the system is >> h.)
>
>> h is a constant, hence never goes to zero.
>>
>> One calls the limit h to 0 also the classical limit, but this
>> is a purely theoretical construct. The only experimentally
>> meaningful classical limit is that of small uncertainties.
>
> You didn't answer. The formal limit, be it only theoretical, exists on
> its own right. The question is whether this limit is CM. I argue it
> isn't. Consequently, any limit that is mathematically equivalent to it
> isn't CM.
For the simple systems where the formal limit exists, the result is
formal classical mechanics. This needs no argument but is obvious.
But I am interested in explaining the deeper reasons and the
relation to real physics, and there this formal limit must be relpaced
by the sort of limit I was discussing.
>>> A further requirement is that
>>> some measurements, especially too precise ones, aren't allowed,
>>> otherwise your qualifacation should be "described by CM", and thus
>>> irreal.
>
>> If one is interested in the dynamics, one cannot make too precise
>> measurements, since one always needs position and momentum
>> (or position at many times, which allows to reconstruct the momentum).
>
> Being interested in the dynamics doesn't prevent to make precise
> measurements.
The uncertainty relation prevents you from making too precise
measurements. if you know that your system is under your atomic
microsope or behind your polarizer then it is impossible to measure
its momentum to an accuracy bigger than what the uncertainty relation
declares possible based on the knowledge of position.
Thus one cannot make too precise measurements.
> But QM
> allows unlocalized states. If you don't explain why those states aren't
> empirically observed, you don't explain why CM emerges from QM.
They are not empirically observed since they are not localized.
Everything we observe is necessarily localized at the instrument
of observation (though we often infer something about their sources,
which may be far away).
>>>> But in many situations, the effect is similar to taking the limit
>>>> hbar --> 0. In these cases the relative uncertainties scale with
>>>> sqrt(hbar/M), which becomes small if either hbar is made formally
>>>> tiny or if M is large.
>
>>>>> Apparently, it isn't sufficient,
>
>> In many situations it is. I didn't claim in all.
>
> So you don't address why in the macroscopical world, quantum effects
> aren't observed. So you didn't address the relation between QM and CM
> at all.
See Chapter 7 of the book mentioned.
There is no need to repeat things here.
Arnold Neumaier
X-Phy
On 6 d?c, 17:54, Arnold Neumaier <Arnold.Neuma...@univie.ac.at> wrote:
> No. A density matrix is a positive semidefinite Hermitian operator of
> trace one. The representation as a sum is artificial and has no
> experimental basis.
A density matrix is a statistical expectation value, that is, all the
elements that are summed have *no experimental basis*, they are only
potentialities, possibilities that may or may not occur according to a
probability. The real reason why it isn't possible to empirically
decompose
it is that *isn't possible to experimentally know the wave function*
of a
pure states. You think you can conceal the latter point by
constructing
another objet and shifting the responsibility to the construction, but
you
haven't made advance the understanding a sole millimeter. It is a sum
and
thus a reduced description, and theoretically can't bring anything.
In the
same way, talking about the mean position of a wave function is a dead
end,
since the dispersion can take any large value, especially for a plane
wave,
and indeed the dispersion grows with time.
> It is well-known that one cannot distinguish experimentally the
> contributions to a density matrix when written as a weighted sum of
> pure density matrices.
>>> There is no experimental way to decompose a mixed
>>> state into its pure components. This tells us that the mixed state
>>> is something irreducible, not something reduced.
> One cannot prove experimentally that the density matrix
> rho = 1/2 |up><up| + 1/2 |down><down| is composed of a pure up state and
> a pure down state. If it were so, it would also be composed of pure
> states in arbitrary superposition, since these can be observed to
> be seen in corresponding detection experiments.
Ditto. It is because |up> can be written as a composition of
eigenfunctions
in another basis and that *a measurement can be done in this other
basis.*
Your density matrix is actually |up> *or* |down>, which I can write
1/2 |left> + 1/2 |right> *or* 1/2 |left> - 1/2 |right>. You have done
nothing, except reducing the description. The state can still be
measured
left or right, and that only because of the *probability amplitude*,
the
classical probability playing no role at all, save reducing the
description.
There is no empirical difference between + and - when the left-right
quantum
number is measured, and that from purely quantum theoretical reasons.
Since
the impossibility of an empirical decomposition is ascribable to
quantum
theory *only*, you can't use that argument to say that it isn't a
reduction,
but as it is a sum, *it is a reduction.*
> Thus the state rho would be composed of all pure states
> although it is written only as a sum of two - arbitrarily chosen and
> orthogonal, or of none. This shows that your argument is meaningless.
>
> Thus a density matrix is irreducible.
>> There no experimental way either to decompose a dice into six faces up,
>> nor to decompose a plane into length and width.
> One can prove experimentally that a die has 6 faces with six different
> marks on them. This makes your analogy faulty.
You haven't understood the analogy. A dice has only one face up at a
time.
>> All yours arguments miss the point. But now, it has been overlooked
>> that the density matrix contains more variables than the wave function
>> in the case of a pure state.
> All components of the density matrix are observable,
> while the components of a wave function (the amplitudes) are not.
> This shows the priority of the density matrix.
But that is wrong. An observable is an operator, not the modulus of
an
amplitude. You can't measure the components of the density matrix.
If you
mean measuring the probabilities by repeated experiments, that's
exactly
the same for a wave function. The density matrix isn't attached to a
particular interpretation, so I don't know how you can call it
priority, one
more physically vacuous term. The wave function is /smarter/ than the
density matrix.
> That there are more components to a density matrix reflects the physical
> fact that most observed states are not pure and pure states do not have
> enough degrees of freedom to account for experimental facts.
Every *observed* state is pure, since if it is observed, that means it
is
the actual real state, and not one of the other possibilities. If we
can't
decide which possibility it is, that is linked, as I have shown, to
the
impossibility of a complete knowledge of the wave function. For
example,
when I observe a fruit that is round and sweet, I can't know if it is
an
apple or an orange, but I know it is either of them, and that's not
because
it must be both at the same time.
We can imagine a setup where a different pure state is prepared
according to
the outcome of a dice. We don't need more variables to describe the
prepared state, since we must include the dice in the system, and the
supplementary variables are for it only. The dice can be described
with a
pure state, and so also the whole system, which confirms that the
density
matrix is a reduced description.
The dice could be a spin 1/2, and the outcome be the measurement of
the spin
along a given axis. Then according to it, a particle with a spin
state up
or down could be prepared. That is just the example you gave. When
we
measure the spin projection of this particle along the up-down axis,
we know
in which state it is, so we achieve an experimental decomposition of
the
density matrix. This is possible because of the knowledge of the
whole
system. Anew, this shows that the density matrix is a reduced
description,
and precisely because it can't be experimentally decomposed. A
smaller
system is chosen, and the degrees of freedom of the remaining is
summed
over. You only have put the measurement problem outside the system,
and you
make believe that both possibilities are still therein, while only one
still
exist.
The setup can be simplified, without loss of generality, by measuring
directly the spin of the dice. There is no longer a projection before
the
measurement, and that is just an example of the Heisenberg cut. So,
we have
a pure state, which shows that mixtures aren't necessary to describe
reality, just like ensembles. They are necessary only to a reduced
description, like thermodynamics.
>>> In the density matrix version, no probability amoplitudes appear,
>>> only true probabilities. This adds to the intuitiveness of the density
>>> matrix version.
>>
>> Are you saying that the density matrix is unable to describe a quantum
>> system?
>
> No. I am saying that
> - pure states and probability amplitudes are unable to describe
> most observed quantum systems.
Most *reduced* quantum systems, or ensembles of quantum systems.
> - density matrices do not use probability amplitudes.
> - density matrices fully describe the state of a quantum system.
Which is wrong. If you haven't the phases of the amplitudes, you
haven't a
full description of the state. The phases are contained in the off-
diagonal
elements, up to an overall phase.
>> If not, you are inconsistent. True probabilities can't be used
>> to describe entanglement, for instance, since they are equivalent to
>> local hidden variables.
> A density matrix for a qubit is more than probabilities for up and down,
> which only determine its diagonal entries. There is also the
> off-diagonal element that describes the nonclassical part.
Ok, which contradict your previous statement: "In the density matrix
version, no probability amplitudes appear"
> Similarly, entanglement is fully encoded in the 4x4 density matrix
> of a pair of qubits.
>>> See Chapter 7 of
>>> Arnold Neumaier and Dennis Westra,
>>> Classical and Quantum Mechanics via Lie algebras,
>>> http://www.mat.univie.ac.at/~neum/papers/physpapers.html#QML
>>> http://de.arxiv.org/abs/0810.1019
>> What? This chapter is about thermal theory. What has it to do with the
>> EPR paradox for example? you use a very thick carpet of equations and
>> formula of all sorts, that's all I can see.
> That's only because you looked at the pattern of characters rather
> than read the chapter. The book is written for people who are willing
> to read. They can easily find there the desired intuitive account of QM.
You're kicking to touch. Sorry, I don't need a introductory physics
course,
and my time isn't elastic.
>> Intuition is a very direct
>> way of looking at things. If you can't make an account in a few lines
>> for somebody who is familiar with quantum theory, forget it.
> I am not interested in giving simplistic intuitions for people who think
> that quantum mechanics can be understood in a few lines.
If you have the impression I'm a big newbie in QM, I'm sorry for you.
I've
always some defiance to people that are unable to make a simple
account and
hide themselves behind a thick screen of obfuscated terms.
>>> You misunderstood. Approximately pure states are described by density
>>> matrices, not by pure states.
>> No, that's plain wrong, sorry. They are also described by pure states
> No. An approximately pure state is only approximately described by a
> pure state but exactly described by a density matrix.
You didn't notice the 's' at the end of 'state'. I could point you to
a big
book about grammar, but I think you don't need it. In most tongues,
the
adjective also agree, and it is not a bad thing.
>> with there respective probabilities, even if that decomposition isn't
>> unique, it doesn't matter. There isn't either a unique frame of
>> reference.
> The decomposition doesn't matter, it is as spurious as the coordinates
> of a vector field without the frame of reference.
>
> The vector field is the physical invariant; the coordinates only a
> view of it. Similarly, the density matrix is the physical invariant,
> while the pure states it ``contains'' (in a very vague sense, since
> every state is contained in a positive definite mixed state) only give
> a view of it.
I didn't spoke about a field, but about space.
Anew, you make the same error, a wave function can't give a view of
anything, since it can't be known completely.
> This view is never used when actually working with density matrices.
> For pure states give a very strange point of view for quantum dynamics.
> If you look at the standard decomposition of a density matrix rho(t)
> into its singular vectors psi(t), which presumably are the pure states
> you think rho(t) contained of. and apply the dynamics to the density
> matrix rho(t) and to the individual states psi(t), you get completely
> different results at any other time....
>>> And one can predict with density matrices
>>> the extent to which an interference pattern is blurred. Thus
>>> interference is not an argument for pure states.
>> You just proved the contrary.
> No. Interference can be completely described by density matrices,
> and in fact needs the latter for a full discussion, e.g., in the
> case of partially polarized photons. You can find this discussion
> in the bible on quantum optics,
> L. Mandel and E. Wolf,
> Optical Coherence and Quantum Optics,
> Cambridge University Press, 1995.
> But it takes more than a few lines to understand....
I'm not your pupil. Anew kicking to touch, perhaps because you don't
know
what to answer. I advice you to RTFM, and that time, trying and
understand.
In the matter at hand, it takes actually only a few lines. Since only
the
pure states give interferences, the resulting interference pattern is
simply
the superposition (in the common meaning) of the patterns for each
pure
state. That it be partially polarized photons, that is, a mixture of
photons with different polarization states, is only a useless
complication.
Therefore, interference phenomena can only be explained in the
framework of
pure states.
>> You are mixing QM with thermodynamics. We aren't discussing the
>> thermodynamical limit of classical statistical theory.
> Classical mechanics is an intuitive limit of quantum mechnaics only
> if one takes thermal effects into account. Quantum mechanics without
> thermal effects is an idealization that does not apply to real life.
We are going in circles. CM is an intuitive limit of QM only if we
replace
probability amplitudes by classical probabilities? Hmmm. - Then it
is very
simple? - Of course, take the rabbit off the hat, and it is an
ordinary
rabbit. - Elementary dear Watson. - You said it!
Nothing more will emerge from that discussion, I stand by my
conclusion.
Arnold Neumaier
This is amply contradicted in the literature.
But I get tired of preaching deaf ears, and end this discussion
with the present mail.
>> No. I am saying that
>> - pure states and probability amplitudes are unable to describe
>> most observed quantum systems.
>
> Most *reduced* quantum systems, or ensembles of quantum systems.
All observed quantum s=ystems are reduced, since they neglect the
effect of the environment to which they are necessarily coupled
(else they couldn't be observed).
>> - density matrices do not use probability amplitudes.
>> - density matrices fully describe the state of a quantum system.
>
> Which is wrong. If you haven't the phases of the amplitudes, you
> haven't a
> full description of the state. The phases are contained in the off-
> diagonal
> elements, up to an overall phase.
You cannot describe unpolarized light better than with a diagonal
density matrix. This contains the full description of the state.
There is no description of unpolarized light in terms of a pure state.
>>> If not, you are inconsistent. True probabilities can't be used
>>> to describe entanglement, for instance, since they are equivalent to
>>> local hidden variables.
>
>> A density matrix for a qubit is more than probabilities for up and down,
>> which only determine its diagonal entries. There is also the
>> off-diagonal element that describes the nonclassical part.
>
> Ok, which contradict your previous statement: "In the density matrix
> version, no probability amplitudes appear"
No. The off-diagoaanl elements of a density matrix are not probability
amplitudes but express correlations.
> Nothing more will emerge from that discussion, I stand by my
> conclusion.
So do I. The discussion is closed.
============ Moderator's note ================================
It's clear that density-matrix elements are measurable quantities.
In fact they are what you have to measure to determine the quantum
state of a system. A good introductory discussion about how to
determine the state by measurements on an ensemble of equally prepared
systems (state determination) can be found in
L. Ballentine, Quantum Mechanics, A Modern Development,
World Scientific