My caricature

 

 

Ark's Homepage

Publications

Curriculum Vitae

What's New

Physics and the Mysterious

Event Enhanced Quantum Physics (EEQT)

Quantum Future

My Kaluza-Klein pages

Links to my other online papers dealing with hyperdimensional physics

QFG Site Map

My posts to the newsgroup sci.physics.research - Part V

Full threads can be found at the URL http://groups.google.com/groups?hl=en&group=sci.physics.research

and also http://www.lns.cornell.edu/spr/

On Tue, 9 Jul 2002 02:30:24 GMT, Anselmi_Fabio@hotmail.com (fabio) wrote:

>Hi everybody!
>
>A quantum channel can be represented by a strictly contractive
>operator.
>So suppose you have two channels : S , T. If at least one of them is
>not unital
>(doesn't transform the identity operator into itself) the tensor
>product could be a non strictly contractive operator.
>Can anyone show me an example of this?
>Or , better, a family of operators (strictly contractive) whose
>tensor product is not?
>
>Thanks.
>Fabio.

It would be nice if you could be more specific. Strictly contractive operator acting on what space? I guess on some space of operators,
but which one? Prigogine invented the term "supeoperator" in order to avoid a confusion.

If you like to adress youe question also to those who do not necessarily know your definition of a "quantum channel" but may know something about
tensor products of Hilbert spaces and operator algebras - you would better provide a precise definition or a reference.

ark


On Sat, 13 Jul 2002 20:42:44 +0000 (UTC), Pol <xpol@interfree.it> wrote:

>Can you please point me at relevant studies
>of problems of that kind?
>How can i algebraically simplify eigenvalues?
>
>Yor suggestions will be very welcome

Looking at the algebraic structure of eigenvalues
you will probably be able to find relations between parameters when
there are changes in behavior. Then numerically solving your evolution equation (for instance for some observable of interest) for several
values of the parameters in each parameter domain should give
you agood idea of what is going on.

Usually it is not necessary to have a nice general formula that
gives you an explicit solution. Numerical solution in a certain
parameter range, with a plot that tells the story in pictures, is often
enough.

ark


On Sun, 14 Jul 2002 15:51:44 GMT, Ilja Schmelzer <schmelzer@wias-berlin.de> wrote:

>Arkadiusz Jadczyk <ark@cassiopaea.com> writes:
>> <schmelzer@wias-berlin.de> wrote:
>
>>>> Today perhaps we need to start thinking on our own again. What if
>>>> quantum theory - the next version of it - version 2.0, say, is not
>>>> gonna be probabilistic?
>>>
>>> Then QM 2.0 would be probably assigned another name, not QM 2.0.
>>> It would be too different from QM.
>>
>> Who says so?
>
>According to the quotes, Ilja Schmelzer says so.
>

Then Ilya Schmelzer may easily be wrong. Ark says so.
And why? Because Ilya Schmelzer says things that he believes
to be true without giving reasons.

>>>> ... because Bell's theorem in a popular edition (not Bell's original
>>>> one) is based on oversimplified concepts ...
>
>>> Who cares about properties of popular editions?
>
>> Those who read them. And there are many.
>
>So what? It doesn't invalidate Bell's theorem.

That depends on what Ilya Schmelzer calls "Bell's
theorem" and he understands by "invalidating".
It is always good to be explicit and precise when
there is a danger that personal understanding may be
different from understanding of fellow physicists.

>>>> Do not forget that Bell wrote his paper in 1964 - it is about
>>>> "measurements" and "experiments". More than 20 years later he wrote
>>>> another paper "Against measurement" where he realized and pointed it
>>>> out to the rest of the physics community that these concepts are not
>>>> even defined! Few people really listened.
>
>>> Bell's theorem certainly does not depend on particular interpretations
>>> of QM and moreover not on the choice of an appropriate name for the
>>> expression <psi|A|psi> .
>
>> Of course it does. Without interpretation there is no quantum mechanics
>> at all.
>
>Yep. But the known interpretations of QM have sufficient things in
>common. The common part is known as "minimal interpretation".

Again we see "known" interpretations, whereas it shoul be "known to Ilya
Schmelzer". And "the common part" shoul read as "common part to
those selected by I.S. to prove his point."

>> Let me quote from Bell:
>> "... a hidden variable interpreation of elementary quantum theory
>> has been expliocitly constructed. That particular interpretation has
>> indeed a grossly non-local structure. This is characteristic, according
>> to the result to be proved here, of any such theory, which reproduces
>> exactly the quantum mechanical predictions."
>
>Nice quote. So what?

So that you (and other people who read it) can see what John Bell himself had in mind. Once we discuss Bell's theorem, it is good to go
back to the source, if only in order to stress that "my understanding
of what Bells' theorem theorem is about is not the same, and not even close, to what Bell himself could have in mind. And here is why:...."


>> We can see the word "interpretation" twice.
>
>interpretation, interpretation, interpretation, interpretation.

I will not comment on the above.

>In this line you can see the word "interpretation" four times. And
>now?

I will not comment on the above too.

>> Without "interpretation" Bell's theorem is a mathematical tautology
>> (as ANY "theorem" that can be proven from assumptions.)
>
>The way Bell's theorem depends on interpretation is minimal. The
>disagreement between different QM interpretations is completely
>irrelevant for the validity of Bell's theorem.

Again, it depends on ill undefined terms "Bell's theorem" and
"validity". If you want, you can DEFINE Bell's theorem and "validity" in
such a way that the above statement of yours is true. But, and that
is what I am trying to convey, there is also another view according
to which Bell's theorem can be of no consequence because it is based on assumptions which will not apply in certain alternative formulations of quantum theory. Bell's theorem deals with expectation values. What is
the relation of expectation values (mathematical concepts) to real "measurements" (finite time series of data) depends on interpretation.

>Of course, Bell's theorem is interesting only for realistic
>interpretations of QM. For non-realistic interpretations it is simply
>irrelevant.
>
>Ilja

It would be nice to have a definition of "realistic" and non-realistic
interpretation. It would also be nice to have a definition of
"measurement" that is precise enough and is a part of a
rigorous formalism - otherwise there will be no end to
this discussion due to the use of undefined concepts.

It seems to me that something is wrong with our understanding
of QM and that is what John Bell was trying to convey in
his paper "Agains measurement". It seems to me that is does not
good to think that nothing is wrong, that John Bell simply
"over-reacted". Why do I think so? I consider EEQT - the framework and interpretation I was active in developing, is a realistic one, nevertheless I see no way in which Bell's theorem applies to it.
The "standard interpretation" or "minimal interpretation" can
be indeed derived, but only in some limit, and as an approximation.
EEQT descrinbes individual quantum system under observation.
There are no "observables" and no "expectation values" - except
in a limit and as an idealization. Of course EEQT is, perhaps, wrong.
But, on the other hand, perhaps, it is right?

ark


On Mon, 15 Jul 2002 22:27:31 +0000 (UTC), George Jones <george_llew_jones@yahoo.com> wrote:

(snip)

>In either case, the space of quantum states is a representation space
>for a unitary representation of the appropriate group. For the Poincare
>group, the appropriate group is its universal cover, i.e., replace the
>rotation group by SU(2). For the Galilean group, the appropriate group
>is the central extension of its universal cover.


>Now, why does the position operator generate boosts?
>Now let U = exp(-i*a*X).
(,,,)
>U P U^(-1) = exp(-i*a*X) P exp(i*a*X)
>
> = exp(-i*a*X)( a exp(i*a*X) + exp(i*a*X) P)
>
> = P + a.
>
>Thus, X generates Galilean boosts.
>
>I don't guarantee any of the +/- signs.
>
>Regards,
>George


Hi,

Let me just add few comments. As a reference I will use a paper

"LOGICS GENERATED BY CAUSALITY STRUCTURES.
COVARIANT REPRESENTATIONS OF THE GALILEAN LOGIC "

available form

http://www.cassiopaea.com//quantum_future/jadpub.htm#ceja75

The "central extension" is indeed important. This is how mass parameter
enters into Galilei invariant quantum theory (through the parameter alpha in the paper) . Formally it comes from study of central extensions - see work of V. Bargman quoted in the above paper.

Now, even if X (called Q in the paper )commutes with P as a boost operator (up to a multiplicative, mass-dependent constant, that does necessarily mean
that X can serve the same role as boost general for all purposes.

In the paper above it is shown that it is possible to consider
quantum mechanics of "extended" Gaileli particles. In this case
there is internal space, and this space is infinite-dimensional.
In that case the difference between X and K (it is called lower case "k": k= K-alpha Q) is a nontrivial operator acting on this internal space. It can be seen that k gives rise to a nontrivial internal
angular momentum of the extended object (a "quantized" rigid rod")

ark


On Wed, 17 Jul 2002 15:48:39 +0000 (UTC), "Peter Grieve" <karenlyn@bridgernet.com> wrote:

> Many elementary texts on fiber bundles present the Moebius band as a
>bundle with base S1 and typical fiber [0,1]. They then say that "the
>structure group of the Moebius band is Z2".

(snip)

> Peter Grieve


I would say this is misleading. It would be good if you provide
the exact reference, so that other people can be warned.

Usually, when we speak about "structure group" we think either
of principal or of associated bundle. Moebius band with [0,1] as
a fiber is not a principal bundle. It would be a principle bundle
if the fibre was {0,1} - just two points, rather than the whole
interval. You CAN make an interval [0,1] a representation space
for Z2, sure, but that must be stated explicitly, as there are many
ways of doing so. Without that Mobius band does not carry
a unique structure of an associated bundle and its structure group
is not clear.

ark


On Fri, 26 Jul 2002 16:18:58 +0000 (UTC), dgoncz@aol.comp.mil ( Doug Goncz ) wrote:

>I have three ways available to get over my block on eigenstates and
>eigenvectors. One, take Linear Algebra. I don't have the prerequisite
>skills, although I do have the credits.

Taking linear algebra course may be not so bad idea. One interesting and important application, easier than Principal Component Analysis and Factor Analysis are Least Square Fit and Linear Regression methods.
There is a nice little book "Data Reduction and Error Analysis for The Physical Sciences" by Philip R. Bevington and D.Keith Robinson, which
introduces the subject, with all necessary calculation and shows how
linear algebra is important in solving the very general question: how to
choose parameterst so as to best fit my data, and how to estimate
"goodness of the fit". You do not need eigenvectors yet, but you need
to know how to invert a hermitean matrix, and you get some idea
about how matrix elements, in this particular application, are related to "correlations". You will understand the need for "scalar product."

Understanding eigenvalues and eigenvectors in quantum theory
needs more algebra. You reaaly need to learn "spectral theorem"
- at least its matrix version. Otherwise you will be told to believe
certain "axioms of qm" and you will be asking "why?".

Again there is a little book that tells all about qm in the matrix
form:

"Introduction to quantum mechanics" by P. T. Matthews. It is one
of the easiest one to read, and yet it has lot of interesting and
originally presented stuff.

ark


On Sat, 31 Aug 2002 07:17:37 +0000 (UTC), naughty_sonny@yahoo.com (Craig Joyce) wrote:

>Hi!
>
>I'd like to know if a set of operators on a state could themselves be
>classified as a state. For example, when you apply the operators Q'
>and Q". We can have two possible combinations Q"Q' and Q'Q" on a state
>|a>
>
>Does this mean, we could consider the operation Q"Q' as a new state
>|x> and the operation Q'Q" as a state |y> ?
>
>What is the alternate option to this?
>
>-Craig

In a sense the answer can be "yes", but not for the reason you are giving.

In algebraic quantum theory we have what is known as GNS (Gelfand -Naimark - Segal) construction. Let A be a C* algebra unit and let f be a normalized positive functional on A. Then a Hilbert space (a representation space for A) is being constructed. The first step in this construction is to define a left ideal N of A as follows:

y is in N if and only if f(x*y)=0 for all x in A.

Then H is defined as the quotient linear space A/N. Thus the non-zero elements of H (representing the vector states of the representation of A) indeed consists of "sets of operators".

Any book on C* star algebras describes the construction in details.

ark

 

 

This site is a member of WebRing.
To browse visit Here.

 

Last modified on: June 27, 2005.

.