Boundedness of quantum observables?


by A. Neumaier
Tags: boundedness, observables, quantum
Careful
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#37
Feb3-11, 04:31 PM
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Quote Quote by A. Neumaier View Post
Many concepts are not named after their originator.

For me the main purpose of naming is to ease communication rather than to honor the first who coined a concept. One doesn't want to rewrite articles and books each time a new historical fact turns up.
No need to rewrite, just use the correct terminology from the moment this fact becomes clear. I think we are obliged to honor the right persons in science, it is not only a matter of courtesy but of intellectual honesty too. And if this causes confusion in the beginning, then be it so; there are lots of confusing things in life.
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Feb3-11, 06:48 PM
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Quote Quote by Careful View Post
First of all, all natural observables in quantum theory (or QFT) correspond to unbounded (distributional) operators...
In response to the suggestion that C*-algebra is a reasonable starting point, you made an argument that C*-algebra is not a reasonable ending point. How is that helpful? It's like you completely ignored what I said in favor of what you wanted to say.


Quote Quote by Careful View Post
No need to rewrite, just use the correct terminology from the moment this fact becomes clear. I think we are obliged to honor the right persons in science, it is not only a matter of courtesy but of intellectual honesty too. And if this causes confusion in the beginning, then be it so; there are lots of confusing things in life.
AFAIK, the normal way is to add names -- e.g. to say "Krein-Nevanlinna space". To completely discard the common name, among other things, is obstructive to discussion.
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Feb3-11, 06:51 PM
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Quote Quote by yoda jedi View Post
and why just the same predictions ? why not that predictions, and more, beyond and broader predictions.
Because we were talking about alternatives to the algebraic approach to QM, not about attempts to find a better theory.

Quote Quote by yoda jedi View Post
then ?
contradicting yourself ?
I would have contradicted myself if I had called the theory obtained using the alternative approach "QM". I said "a theory that makes the same predictions as QM" to avoid the contradiction.
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Feb3-11, 07:06 PM
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Quote Quote by Hurkyl View Post
In response to the suggestion that C*-algebra is a reasonable starting point, you made an argument that C*-algebra is not a reasonable ending point. How is that helpful? It's like you completely ignored what I said in favor of what you wanted to say.
No, I did not. I said that C* algebra's are not natural at all for the reasons I mentioned. A mathematical object is only any good if you can formulate the physics directly into these terms. This never happens, neither in QM nor in QFT where unbounded operators enter the calculations. Normally, people might still be inclined to use the Weyl transformation to do what you suggested but this becomes completely problematic on Nevanlinna space. Actually your trick will fail there under any circumstances because any nontrivial analytic function blows up to infinity. For example, hermitian operators can have a complex spectrum (on the ''ghost'' states) which is totally unbounded. So an arctan, e^{ix} or something like that is not going to resolve anything.

What would be interesting from the point of view of ''C* algebra's'' is that you try to extend the GNS construction to non-positive states, so that you will get Nevanlinna space representations. This requires of course a change in the C* norm identities in the first place, but it might be good to define such generalized algebra's.
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#41
Feb3-11, 07:11 PM
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Quote Quote by Careful View Post
Right, there is no substance behind bounded operators. The best proof is that we never use them.
That hardly proves a claim as strong as "there's no substance behind bounded operators". I don't know the algebraic approach well, but it seems to me that we can never make a measurement that corresponds to the momentum operator (because there's no measuring device with infinite precision). In QM books at the level of Sakurai, it is often claimed that a measurement of an observable A that gives us the result a, leaves the system in a state represented by a vector in ker(A-a), i.e. the eigenspace corresponding to eigenvalue a. For an unbounded operator such as P, this would kick the state out of the Hilbert space entirely, and leave the system in a "state" |p> of perfectly well-defined momentum. I don't think a realistic measurement, which has finite precision, can do more than to confine the state vector to some subpace of the Hilbert space. So when we actually do a momentum measurement, the mathematical representation of the measuring device won't be P. It will be a member of that C*-algebra.

I do however agree that the fact that the unbounded operators are so prominent suggests that it would be desirable to start with some kind of algebra of unbounded operators instead. Perhaps there is such an approach, that gives us a rigged Hilbert space in a way that's similar to how the C*-algebra approach gives us a Hilbert space.
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Feb3-11, 07:26 PM
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Quote Quote by Fredrik View Post
That hardly proves a claim as strong as "there's no substance behind bounded operators". I don't know the algebraic approach well, but it seems to me that we can never make a measurement that corresponds to the momentum operator (because there's no measuring device with infinite precision). In QM books at the level of Sakurai, it is often claimed that a measurement of an observable A that gives us the result a, leaves the system in a state represented by a vector in ker(A-a), i.e. the eigenspace corresponding to eigenvalue a. For an unbounded operator such as P, this would kick the state out of the Hilbert space entirely, and leave the system in a "state" |p> of perfectly well-defined momentum. I don't think a realistic measurement, which has finite precision, can do more than to confine the state vector to some subpace of the Hilbert space. So when we actually do a momentum measurement, the mathematical representation of the measuring device won't be P. It will be a member of that C*-algebra.
Two quick reactions, Hilbert space is not only unsuitable because it has only positive norm but also because it cannot include distributional states. So rigged Hilbert spaces are much better. There is no problem in finding out a suitable probability interpretation for distributional states. It is not really important whether these states are measured or not, what matters is that they are the natural mathematical objects which show up in representation theory of the Poincare algebra. Since the generators of this algebra correspond to unbounded (distributional) operators (which even act well on the distributional states with a finite number of terms), the language of unbounded operators is the most natural thing. Whether energy or momentum can be measured sharply is a philosophical question (and one should not take comments of for example Sakurai too seriously); what matters is that dynamics is most naturally expressed in the unbounded (distributional) language.

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Feb3-11, 07:37 PM
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Quote Quote by A. Neumaier View Post
Yes, though the axioms are not always clearly or fully spelled out. (I don't know clear axioms for the standard model.)
Agreed. A complete definition of a theory would have to include a lot more than the stuff we can list on a page of a book, including a specification of e.g. what measuring devices or procedures we should think of as measuring "energy". If we can't come up with a good axiom scheme that specifies how to identify all (or a sufficiently large class of) measuring devices with self-adjoint operators (or whatever represents them mathematically in the theory we're considering), we need a separate axiom for each type of measuring device.

Quote Quote by A. Neumaier View Post
Then two versions of SR that use opposite conventions for the signature would be different theories (since the axioms differ), although they make the same predictions.
Yes, this is pretty annoying. What I've been doing when I have only been thinking about these things, is to allow myself to use sloppy terminology, and call the opposite signature SR "SR" even though this contradicts my definition of "theory". This isn't very different from how mathematicians define a group as a pair (X,b) or a 4-tuple (X,b,u,e) with certain properties...and then start referring to X as a "group" the moment they're done with the definition. It doesn't confuse me, but I think I need a better system for when I explain these things to other people.

Quote Quote by A. Neumaier View Post
The approach outlined is fully developped though not widely publicized; it is the basis of my thermal interpretation of quantum mechanics. It agrees with how one does measurements in thermodynamics (the macroscopic part of QM (derived via statistical mechanics), and therefore explains naturally the classical properties of our quantum world. It is outlined in my slides
http://www.mat.univie.ac.at/~neum/ms/optslides.pdf
and described in detail in Chapter 7 of my book
Classical and Quantum Mechanics via Lie algebras
http://lanl.arxiv.org/abs/0810.1019
Cool, I'll check it out, but not right now. I need to get some sleep. I didn't realize that you're one of the authors of that book. I downloaded it in May, but haven't gotten around to reading it yet. I don't think I will the next few months either, because I'm trying to learn functional analysis, and it takes an absurd amount of time.
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Feb3-11, 07:41 PM
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Quote Quote by Fredrik View Post
In QM books at the level of Sakurai, it is often claimed that a measurement of an observable A that gives us the result a, leaves the system in a state represented by a vector in ker(A-a), i.e. the eigenspace corresponding to eigenvalue a. For an unbounded operator such as P, this would kick the state out of the Hilbert space entirely, and leave the system in a "state" |p> of perfectly well-defined momentum.
.

(bolding by me). That's the von Neumann's projection postulate which pertains to the Copenhagian view of things. It's not universally accepted and other interpretations and axiomatizations of QM completely disregard it.

Quote Quote by Fredrik View Post
I do however agree that the fact that the unbounded operators are so prominent suggests that it would be desirable to start with some kind of algebra of unbounded operators instead. Perhaps there is such an approach, that gives us a rigged Hilbert space in a way that's similar to how the C*-algebra approach gives us a Hilbert space.
This part I agree with. I haven't seen the bolded part yet, theories of unbounded operator algebras use a Hilbert space as an environment, not an RHS.

As far as I recall (but if I'm wrong, please correct me), putting an RHS into a QM problem with unbounded operators turns these operators into bounded ones, but of course, not in the original topology of the H-space, but in the topology of the antidual space in which the original operators will find their eigenvectors.
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Feb3-11, 08:08 PM
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Quote Quote by Careful View Post
A mathematical object is only any good if you can formulate the physics directly into these terms.
At face value, this comment seems utterly absurd.

For example, hermitian operators can have a complex spectrum (on the ''ghost'' states) which is totally unbounded.
Er, so? Splitting it into the sum of real part and imaginary parts is an even more standard trick than taking the arctangent to make a real variable bounded.

Quote Quote by Careful View Post
... what matters is that dynamics is most naturally expressed in the unbounded (distributional) language.
Again, that hardly proves a claim as strong as "there's no substance behind bounded operators". A laundry list of reasons why you want an unbounded / distributional language does not constitute a denial of the hypothesis
The language of unbounded / distributional operators can be constructed using foundations built from of bounded operators
A. Neumaier
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#46
Feb4-11, 02:44 AM
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Quote Quote by Careful View Post
Mathematical physicists often investigate things which are not useful for physics. Give me one application which hinges upon them.
But this thread is about rigorous quantum mechanics - giving logical impeccable justiifications for what theoretical physicists commonly do.
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Feb4-11, 02:47 AM
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Quote Quote by Hurkyl View Post
At face value, this comment seems utterly absurd.
It is clear you are a mathematician; there is nothing absurd about this comment. I would even go much further and state that mathematicians only develop the easy language and the hard one, which is useful, is left entirely to the physicists. But yeah, you need to do theoretical physics to understand why this is true.

Quote Quote by Hurkyl View Post
Er, so? Splitting it into the sum of real part and imaginary parts is an even more standard trick than taking the arctangent to make a real variable bounded.
Errr, the only natural splitting which is allowed is by means of the natural involution dagger. There is nothing you can do here, because the operator is self-adjoint. Actually on Nevanlinna space, there is no natural algebraic criterion which gives only operators with a real spectrum. So what you propose is even bad mathematics; simply accept that your point -which any student can make- is only valid in Hilbert space.

Quote Quote by Hurkyl View Post
:
Again, that hardly proves a claim as strong as "there's no substance behind bounded operators". A laundry list of reasons why you want an unbounded / distributional language does not constitute a denial of the hypothesis
The language of unbounded / distributional operators can be constructed using foundations built from of bounded operators
Again, this is only true on Hilbert space!! You seem to be trapped here in some irrational wish for bounded operators and are willing to go through all unnatural constructions possible to save their ***. It is possible of course to define bounded operators on Krein space, but it is not the natural class of operators (since their very definition requires a Hilbert space construction!) and there would be no reason for me to even consider Krein space if I would stick to these simple animals.
A. Neumaier
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#48
Feb4-11, 02:48 AM
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Quote Quote by bigubau View Post
Apparently there's some work in the field of <algebras of unbounded operators> as this review article (and the quoted bibliography) shows:

http://arxiv.org/abs/0903.5446
Thanks. This is a nice paper that I didn't know before. I need to read it more carefully.
Careful
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#49
Feb4-11, 02:51 AM
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Quote Quote by A. Neumaier View Post
But this thread is about rigorous quantum mechanics - giving logical impeccable justiifications for what theoretical physicists commonly do.
What is not rigorous about unbounded operators ? Actually, I studied quantum physics rigorously from that point of view (my master education was in mathematical physics btw).
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Feb4-11, 02:56 AM
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Quote Quote by Fredrik View Post
A complete definition of a theory would have to include a lot more than the stuff we can list on a page of a book, including a specification of e.g. what measuring devices or procedures we should think of as measuring "energy". If we can't come up with a good axiom scheme that specifies how to identify all (or a sufficiently large class of) measuring devices with self-adjoint operators (or whatever represents them mathematically in the theory we're considering), we need a separate axiom for each type of measuring device.
This is why measurement (and the whole interpretational stuff) doesn't belong to the axioms. Imagine we'd have to start classical theoretical mechanics with a discussion of the classical measurement problem (it is not well settled - there are lots of unresolved issues in classical statistical mechanics).

Instead, one starts with a clean slate figuring a configuration space with an action, or a phase space with a Hamiltonian. Nobody cares there about how it relates to reality - the theory stands for itself though it is inspired by reality. And the examples used are heavily idealized compared to the real thing - they illustrate the math and physics but would get really complicated if one would have to discuss them in the context of reality.

Quote Quote by Fredrik View Post
Cool, I'll check it out, but not right now. I need to get some sleep. I didn't realize that you're one of the authors of that book. I downloaded it in May, but haven't gotten around to reading it yet. I don't think I will the next few months either, because I'm trying to learn functional analysis, and it takes an absurd amount of time.
The slides should be an easy read, though, and give the main idea of the thermal interpretation. However, that should be discussed in a new thread.
A. Neumaier
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Feb4-11, 02:59 AM
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Quote Quote by Careful View Post
What is not rigorous about unbounded operators ? Actually, I studied quantum physics rigorously from that point of view (my master education was in mathematical physics btw).
My sentence was a response to a different statement of yours.
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#52
Feb4-11, 03:02 AM
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Quote Quote by bigubau View Post
.
As far as I recall (but if I'm wrong, please correct me), putting an RHS into a QM problem with unbounded operators turns these operators into bounded ones, but of course, not in the original topology of the H-space, but in the topology of the antidual space in which the original operators will find their eigenvectors.
That looks right... but there are subtleties as far as I understand. The space of distributions is not a Hilbert space, but a locally convex Hausdorff space generated by semi-norms defined by smearing functions of compact support. And for each of these seminorms, the unbounded operator is indeed bounded (which basically comes down to truncating a divergent series after any finite number of terms).
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#53
Feb4-11, 03:04 AM
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Quote Quote by A. Neumaier View Post
My sentence was a response to a different statement of yours.
Which one then ? I am not going to bother about guessing... If you point to the sociological fact that most physicists are not ready yet to leave Hilbert space; well yes, I never cared about such things. All I am pointing out is that from where I stand and how I know quantum gravity to work out, bounded operators have evaporated.

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A. Neumaier
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#54
Feb4-11, 03:27 AM
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Quote Quote by Careful View Post
Which one then ? I am not going to bother about guessing...
You could have seen it by reading post #46 attentively.


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