Boundedness of quantum observables?

In summary, the C*-algebraic foundations of quantum mechanics assume that every observable must be bounded and self-adjoint, but this is not always the case.
  • #36
Fredrik said:
I don't think of Bohmian mechanics as QM either.


just a ψ-ontic hidden variable model (BM) of quantum theory.



Fredrik said:
about an approach that doesn't exist, or at least hasn't (yet) been developed to the point where it can be used to state the axioms of a theory that makes the same predictions as QM.

not so fast...

and why just the same predictions ? why not that predictions, and more, beyond and broader predictions.



Fredrik said:
I think of a "theory" as being defined by its axioms rather than by its predictions.

me too.

Fredrik said:
can be used to state the axioms of a theory that makes the same predictions as QM.

then ?
contradicting yourself ?
 
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  • #37
A. Neumaier said:
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.
 
  • #38
Careful said:
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. :frown:


Careful said:
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.
 
  • #39
yoda jedi said:
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.

yoda jedi said:
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.
 
  • #40
Hurkyl said:
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. :frown:
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.
 
  • #41
Careful said:
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.
 
  • #42
Fredrik said:
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.

Careful
 
  • #43
A. Neumaier said:
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.

A. Neumaier said:
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.

A. Neumaier said:
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://arnold-neumaier.at/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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  • #44
Fredrik said:
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.

Fredrik said:
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.
 
  • #45
Careful said:
A mathematical object is only any good if you can formulate the physics directly into these terms.
:confused: 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.

Careful said:
... 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​
 
  • #46
Careful said:
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.
 
  • #47
Hurkyl said:
:confused: 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.

Hurkyl said:
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.
:bugeye: 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.

Hurkyl said:
:
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.
 
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  • #48
bigubau said:
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.
 
  • #49
A. Neumaier said:
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 ? :rolleyes: Actually, I studied quantum physics rigorously from that point of view (my master education was in mathematical physics btw).
 
  • #50
Fredrik said:
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.

Fredrik said:
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.
 
  • #51
Careful said:
What is not rigorous about unbounded operators ? :rolleyes: 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.
 
  • #52
bigubau said:
.
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).
 
  • #53
A. Neumaier said:
My sentence was a response to a different statement of yours.
Which one then ? I am not going to bother about guessing...:smile: 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.

Careful
 
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  • #54
Careful said:
Which one then ? I am not going to bother about guessing...:smile:

You could have seen it by reading post #46 attentively.
 
  • #55
A. Neumaier said:
You could have seen it by reading post #46 attentively.
But again, I disagree here. All the calculations with unbounded operators are precise and one does not need to pass via bounded operators to show that. For example, to understand an easy Schroedinger equation, one first assigns these operators a densely defined domain, check that they are symmetric, compute the deficiency indices (and spaces) and construct (possibly unique) self-adjoint extensions. Everything is rigorous, there is nothing fuzzy about it.

You could have seen that answer coming by reading post 49 in detail. :-)
 
  • #56
Careful said:
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).

I guess the reasons for which the RHS formulation of the Copenhagian view of QM is not embraced by the community not only reside with the difficulties of the mathematical approach, but also with the conflict between RHS and the probabilistic view a\ la Born, which necessarily asks for Hilbert space and not for distributions on it.

If i better think about it, we've got conflicts in the Hilbert space axiomatization as well*. It turns out that, if one accepts/postulates that physical quantum states are described by unit rays in a complex separable Hilbert space, then the free massive Galilean particle doesn't exist, as it has no physical states, as follows from solving the Schroedinger equation (which is also postulated, of course). So the probabilistic interpretation a\ la Born of the free Galiean particle is not defined, as the probability to find this particle along the whole real axis is infinite.

* A way to circumvent that is to acknowledge that the (probably) commonly accepted axioms apply only to a very restricted class of physical systems, which, of course, is not desired for a theory.
 
  • #57
bigubau said:
I guess the reasons for which the RHS formulation of the Copenhagian view of QM is not embraced by the community not only reside with the difficulties of the mathematical approach, but also with the conflict between RHS and the probabilistic view a\ la Born, which necessarily asks for Hilbert space and not for distributions on it.

If i better think about it, we've got conflicts in the Hilbert space axiomatization as well*. It turns out that, if one accepts/postulates that physical quantum states are described by unit rays in a complex separable Hilbert space, then the free massive Galilean particle doesn't exist, as it has no physical states, as follows from solving the Schroedinger equation (which is also postulated, of course). So the probabilistic interpretation a\ la Born of the free Galiean particle is not defined, as the probability to find this particle along the whole real axis is infinite.

* A way to circumvent that is to acknowledge that the (probably) commonly accepted axioms apply only to a very restricted class of physical systems, which, of course, is not desired for a theory.
I have never given the Galileian limit any thought, but I agree with what you say for the rest (I guess you simply say that a stationary state for a free particle corresponds to an eigenstate outside Hilbert space, but this is also the case in the relativistic theory). I would go even further and say that even RHS are too limited... and indeed, the Born rule is in for a huge generalization. The problem with QT does not only reside in a technical framework which is too limited, but also in a too simplistic probability interpretation.

Careful
 
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  • #58
bigubau said:
* A way to circumvent that is to acknowledge that the (probably) commonly accepted axioms apply only to a very restricted class of physical systems, which, of course, is not desired for a theory.

But Born's rule _does_ apply only to a very restricted class of physical systems!

1. The interpretation cannot apply unrestrictedly. Suppose a Hamiltonian has a discrete spectrum with zero ground state energy and irrational eigenvalues otherwise. (There are plenty of these; e.g., the anharmonic oscillator with H=a^*a+g(a^*a)^2 and nonzero g.) Then Born's rule claims that each measurement of H results in one of the eigenvalues. This is ridiculous since no measurement can give the value of an irrational number exactly.

2. In quantum optics and quantum information theory, one frequently has measurements that have no associated self-adjoint operator to which Born's rule could be applied.Instead, one represents observables by so-called positive operator valued measures (POVMs); see http://en.wikipedia.org/wiki/POVM. The Born rule applies only in the special case where the POVM is actually projection-valued.

3. Even POVMs are adequate only for measurements in the form of clicks, flashes or events (particle tracks) in scattering experiments. They do not cover measurements of, say, the Lamb shift, or of particle form factors.

One therefore needs to be even more general: Any function of the model parameters defining a quantum system (i.e., the Hamiltonian and the state) may be an observable. Clearly, anything computable in quantum mechanics belongs there.

Indeed, whenever we are able to compute something from raw measurements according to the rules of the theory, and it agrees with something derivable from quantum mechanics, we call the result of that computation a measurement of the latter. This correctly describes the practice of measurement.

See Chapter 7 of my book ''Classical and Quantum Mechanics via Lie algebras'' http://lanl.arxiv.org/abs/0810.1019
 
  • #59
Careful said:
So, in my opinion, bounded operators are dead. Heisenberg thought about it in the same way and as far as my knowledge of history goes, he was playing with Krein spaces later on in his life.
I don't understand this, ignoring C*-algebras, Spin is a bounded operator in the quantum mechanical theory of a single fermion. Some Hamiltonians are also bounded.
 
  • #60
bigubau said:
If i better think about it, we've got conflicts in the Hilbert space axiomatization as well*. It turns out that, if one accepts/postulates that physical quantum states are described by unit rays in a complex separable Hilbert space, then the free massive Galilean particle doesn't exist, as it has no physical states,
Could you explain this?
 
  • #61
DarMM said:
I don't understand this, ignoring C*-algebras, Spin is a bounded operator in the quantum mechanical theory of a single fermion. Some Hamiltonians are also bounded.
Spin is only a part of the total angular momentum and does not make physical sense as an operator by itself. It is the orbital angular momentum part which is unbounded. Further, you miss the entire point, one can construct plenty of bounded Hamiltonians but the natural variables from which they are constructed correspond to unbounded operators.

This is not a matter of bad variables, but it is a deep consequence of Lorentz covariance.
 
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  • #62
DarMM said:
Could you explain this?

The physical states, as commonly axiomatized, are described by unit norm eigenvectors for time-independent Hamiltonians in the Schroedinger picture. But there are no unit norm eigenvectors for the Hamiltonian of a free non-relativistic particle moving freely in R^3. All of them lie in the Schwartz space S'(R^3, dx). So no physical states.
 
  • #63
bigubau said:
The physical states, as commonly axiomatized, are described by unit norm eigenvectors for time-independent Hamiltonians in the Schroedinger picture. But there are no unit norm eigenvectors for the Hamiltonian of a free non-relativistic particle moving freely in R^3. All of them lie in the Schwartz space S'(R^3, dx). So no physical states.
That's what I thought you meant, but it is not correct. There are physical states, the whole Hilbert space [tex]\mathcal{L}^{2}(\mathbb{R}^{3})[/tex] is full of them. It's just that there are no states which are eigenstates of the Hamiltonian. There are still states which evolve unitarily under the Hamiltonian.
 
  • #64
DarMM said:
That's what I thought you meant, but it is not correct. There are physical states, the whole Hilbert space [tex]\mathcal{L}^{2}(\mathbb{R}^{3})[/tex] is full of them. It's just that there are no states which are eigenstates of the Hamiltonian. There are still states which evolve unitarily under the Hamiltonian.
You are falling over semantics and miss the point again. What people call physical states is a matter of agreement, it is a social construct without any deeper meaning. Even at this level, you constantly use the distributional states by means of the Fourier transform, still you wish to expell them to the margins. Let me turn the game around and you show us a Hilbert space construction which is adequate for QFT. I have given plenty of positive arguments against Hilbert space, so you show now a positive argument pro Hilbert space. Then, we will talk.
 
  • #65
DarMM said:
That's what I thought you meant, but it is not correct. There are physical states, the whole Hilbert space [tex]\mathcal{L}^{2}(\mathbb{R}^{3})[/tex] is full of them. It's just that there are no states which are eigenstates of the Hamiltonian. There are still states which evolve unitarily under the Hamiltonian.

Then what you say totally disagrees with what's axiomatized through the Schroedinger equation. The physical states are solutions of the SE. If the Hamiltonian is time-independent (true for the free particle in the Schroedinger picture), then the physical state must have the form

[tex] \Psi_{\mbox{physical}}(t) = \psi_E \exp{\frac{t p^2}{2mi\hbar}} [/tex]

where \psi_E is an eigenfunction of the free-particle hamiltonian, a member of S'(R^3), so the whole physical state becomes a tempered distribution, thus contradicting the physical state postulate.
 
  • #66
bigubau said:
Then what you say totally disagrees with what's axiomatized through the Schroedinger equation. The physical states are solutions of the SE. If the Hamiltonian is time-independent (true for the free particle in the Schroedinger picture), then the physical state must have the form

[tex] \Psi_{\mbox{physical}}(t) = \psi_E \exp{\frac{t p^2}{2mi\hbar}} [/tex]

where \psi_E is an eigenfunction of the free-particle hamiltonian, a member of S'(R^3), so the whole physical state becomes a tempered distribution, thus contradicting the physical state postulate.

DarMM is correct. The Schroedinger equation i hbar d ps(t)i/dt = H psi(t), which describes the dynamics of states in time has a unique solution for arbitrary initial conditions psi(0) in the Hilbert space.

Most states are not eigenstates - the latter are the very specialized solutions that lead to harmonically oscillating phases. If all states had to be eigenstates, nothing ever would happen in our universe, since eigenstates are stationary.
 
  • #67
A. Neumaier said:
Most states are not eigenstates - the latter are the very specialized solutions that lead to harmonically oscillating phases. If all states had to be eigenstates, nothing ever would happen in our universe, since eigenstates are stationary.
That is YOUR problem, NOT mine. My theory deals with spacetime dependent local Hamiltonians and there is no issue with Haag's theorem and describing preferred states. Your comments here are negative, since you have no good way to define what a real observable is (you only utter that you do not believe it is connected to some global Hamiltonian because that would give you trouble indeed in the standard formalism; moreover you have an ''irrational'' wish to preserve Hilbert space). All you can do is spit into your hands, guess how a measurement apparatus ''thinks'' and pray to God you guessed right (at least at asymptotic infinity). You should not confuse the shortcomings of standard QFT with a valid comment regarding the interpretation of distributional states.

Moreover, I don't know if you are aware of this, but the correct states in the canonical quantization of gravity are stationary states with zero energy (and guess what? they are distributional, like the Kodama state). People call this the problem of time while I prefer ''the end of standard QFT''. :-)
 
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  • #68
A. Neumaier said:
DarMM is correct. The Schroedinger equation i hbar d ps(t)i/dt = H psi(t), which describes the dynamics of states in time has a unique solution for arbitrary initial conditions psi(0) in the Hilbert space..

(bolded by me). I think you're not really seeing my point, though it seems you're disputing it. What I emphasized by bold characters holds true iff the Hamiltonian (thus the dynamics), when properly defined and self-adjoint, has a purely discrete spectrum. Else, the whole equation (the psi, its derivative and the H) would no longer be living in a Hilbert space, but into extension of it obtained by the Gelfand - Maurin spectral theorem.

So your claim has only restricted validity and in no way refutes any of my arguments.

A. Neumaier said:
Most states are not eigenstates - the latter are the very specialized solutions that lead to harmonically oscillating phases. If all states had to be eigenstates, nothing ever would happen in our universe, since eigenstates are stationary.

Of course that the whole quantum dynamics and everything going around in the (quantum) world is driven by time-dependent Hamiltonians. I wasn't questioning that. In QM (and also in QFT for what is worth) for time-depending Hamiltonians the SE cannot be solved completely and one is forced to make approximations. If you say that <most states are not eigenstates>, I'll go further <most states are actually unknown, but we can very well do without them, because we've been smart enough and invented perturbation theory>.
 
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  • #69
bigubau said:
(bolded by me). I think you're not really seeing my point, though it seems you're disputing it. What I emphasized by bold characters holds true iff the Hamiltonian (thus the dynamics), when properly defined and self-adjoint, has a purely discrete spectrum. Else, the whole equation (the psi, its derivative and the H) would no longer be living in a Hilbert space, but into extension of it by obtained by the Gelfand - Maurin spectral theorem.
Just to add that a discrete spectrum is not even enough. If the spectrum isn't bounded, domain issues will arise and not any initial conditions can be chosen. But I guess what Arnold wanted to say is that for those states in the domain of H (which of course remain in the domain of H under evolution), the Hilbert space picture actually works.
But what I want to do is pull this discussion away from some silly textbook prejudices people have to situations where it really matters. For example to QFT or quantum gravity: that is where these issues really show their theeth, not in standard QM.
 
  • #70
bigubau said:
A. Neumaier said:
The Schroedinger equation i hbar d ps(t)i/dt = H psi(t), which describes the dynamics of states in time has a unique solution for arbitrary initial conditions psi(0) in the Hilbert space.
What I emphasized by bold characters holds true iff the Hamiltonian (thus the dynamics), when properly defined and self-adjoint, has a purely discrete spectrum.
No. What you emphasized by bold characters holds iff the Hamiltonian is self-adjoint.

For in this case it is the infinitesimal generator of a 1-parameter group exp(itH), which is a bounded operator defined on the full Hilbert space, and psi(t)=exp(-it/hbar H)psi(0) is a well-defined solution of the Schroedinger equation for every psi(0) in the Hilbert space.

Thus the statement holds in _all_ well-defined and time-reversal invariant quantum theories.
bigubau said:
Of course that the whole quantum dynamics and everything going around in the (quantum) world is driven by time-dependent Hamiltonians.

No. Most of quantum mechanics is done with time-independent Hamiltonians - for example the whole of quantum chemistry. If all states were eigenstates, chemical reactions would be impossible!
 
<h2>1. What is the concept of boundedness in quantum observables?</h2><p>The concept of boundedness in quantum observables refers to the idea that certain physical quantities, such as position or momentum, have a finite range of possible values. This means that the observable can only take on values within a certain interval, and cannot exceed that range.</p><h2>2. How is boundedness related to uncertainty in quantum mechanics?</h2><p>Boundedness is closely related to the uncertainty principle in quantum mechanics. According to this principle, it is impossible to know the exact value of certain pairs of observables, such as position and momentum, simultaneously. This is because the more precisely one observable is measured, the less precise the measurement of the other observable becomes, due to the inherent uncertainty in the quantum world.</p><h2>3. Can all quantum observables be considered bounded?</h2><p>No, not all quantum observables are bounded. Some observables, such as energy, can take on an infinite range of values. This is because energy is a continuous quantity, and there is no limit to how high or low it can be. However, other observables, such as angular momentum, are bounded and can only take on discrete values.</p><h2>4. How does boundedness affect the behavior of particles in quantum systems?</h2><p>The concept of boundedness has a significant impact on the behavior of particles in quantum systems. It means that the particles can only exist in certain energy states, and cannot exceed the bounds set by the observable. This leads to the quantization of energy levels and the discrete nature of quantum systems.</p><h2>5. What are the implications of boundedness for the measurement of quantum observables?</h2><p>The concept of boundedness has important implications for the measurement of quantum observables. It means that the observable can only be measured within a certain range of values, and any measurement outside of this range is not physically meaningful. This also highlights the importance of the uncertainty principle in quantum measurements, as it sets limits on the precision of these measurements.</p>

1. What is the concept of boundedness in quantum observables?

The concept of boundedness in quantum observables refers to the idea that certain physical quantities, such as position or momentum, have a finite range of possible values. This means that the observable can only take on values within a certain interval, and cannot exceed that range.

2. How is boundedness related to uncertainty in quantum mechanics?

Boundedness is closely related to the uncertainty principle in quantum mechanics. According to this principle, it is impossible to know the exact value of certain pairs of observables, such as position and momentum, simultaneously. This is because the more precisely one observable is measured, the less precise the measurement of the other observable becomes, due to the inherent uncertainty in the quantum world.

3. Can all quantum observables be considered bounded?

No, not all quantum observables are bounded. Some observables, such as energy, can take on an infinite range of values. This is because energy is a continuous quantity, and there is no limit to how high or low it can be. However, other observables, such as angular momentum, are bounded and can only take on discrete values.

4. How does boundedness affect the behavior of particles in quantum systems?

The concept of boundedness has a significant impact on the behavior of particles in quantum systems. It means that the particles can only exist in certain energy states, and cannot exceed the bounds set by the observable. This leads to the quantization of energy levels and the discrete nature of quantum systems.

5. What are the implications of boundedness for the measurement of quantum observables?

The concept of boundedness has important implications for the measurement of quantum observables. It means that the observable can only be measured within a certain range of values, and any measurement outside of this range is not physically meaningful. This also highlights the importance of the uncertainty principle in quantum measurements, as it sets limits on the precision of these measurements.

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