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.
  • #141
So J is unbounded because of L, irrespective of spin.
 
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  • #142
*** Deleted *** I see that I overlooked an essential part of what you said. I need to think for a minute...

OK, I've thought about it. Your [itex]J^2[/itex] is the total (squared) angular momentum operator. I agree that it's unbounded, because [itex]\vec L=\vec x\times \vec p[/itex], and both x and p are unbounded. The operator [itex]1\otimes S^2[/itex] is however bounded, with [itex]\|1\otimes S^2\|\leq s(s+1)[/itex].
 
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  • #143
Yes, but that's the spin operator; it's bounded, because the spin space is finite-dimensional.
 
  • #144
Fredrik said:
Your [itex]J^2[/itex] is the total (squared) angular momentum operator. I agree that it's unbounded, because [itex]\vec L=\vec x\times \vec p[/itex], and both x and p are unbounded.

No. It is unbounded because its spectrum consists of all j(j+1) with integral j (as can be seen by decomposing the wave functions into spherical harmonics) - other functions of x and p may well be bounded!

And for a scalar particle (which I had meant when I wrote ''bosonic particle'' - sorry for introducing ambiguity), J=L, so that spin need not be discussed to draw the conclusion.
 
  • #145
A. Neumaier said:
strangerep said:
But is your definition above equivalent to Hurkyl's?
He defines the spectrum, I defined the norm.

Actually, Hurkyl defined a norm as well:

Hurkyl said:
A complex number x is in the spectrum of T if and only if (T-x) is not
invertible. The norm is the supremum of the absolute values of the
numbers in the spectrum.

I was really just checking whether both his and your definitions
of norms are essentially the same norm, i.e., just different ways
of defining the same norm (meaning that both norms always give the
same result when applied to any arbitrary element of the algebra).

A. Neumaier said:
In C^*-algebras, both statements are theorems.

Which statements precisely? (Sorry, I just need this to be crystal-clear...)
 
  • #146
strangerep said:
Actually, Hurkyl defined a norm as well:

I was really just checking whether both his and your definitions
of norms are essentially the same norm, i.e., just different ways
of defining the same norm (meaning that both norms always give the
same result when applied to any arbitrary element of the algebra).

Yes. I didn't ''notice'' this although I read it. Yes, I have seen both versions in print, so that, at least, both definitions are provable from the axioms of a C^* algebra. Though it might not be trivial to prove this.
 
  • #147
A. Neumaier said:
strangerep said:
But the spectrum of [tex]J^2[/tex] is unbounded, [...]
I.e., 0, 1/2, 1, 3/2, 2, ...

Does this mean that even this elementary case can't be handled
in the context of C* algebras?

One handles unbounded self-adjoint operators A by considering instead
the bounded family of operators exp(isA), s real. [...]

I sense from your diminishing answers in this thread that you don't
have much interest in this subtopic, but I'll try to pursue it a
bit further anyway...

[To other readers: I want to concentrate just on the spectrum arising
from so(3), without the extra complication of position, linear momentum,
Hamiltonian, etc, i.e., without orbital angular momentum.]

I'll start by re-stating part of my earlier post:

strangerep said:
Thinking more about (un)boundedness of observables in a strictly
algebraic context (i.e., without Hilbert space), I'm wondering
how one would derive the usual angular momentum spectrum
in this context (if we didn't already know it).

In standard QM with Hilbert space, etc, one takes the usual two
commuting observables [tex]J^2 , J_z[/tex] and with the help of the
su(2) commutation relations and the Hermitian inner product, one
derives the spectrum in a page or two.

One does the latter by assuming common eigenstates [tex]|j,m\rangle[/tex]
of [tex]J^2 , J_z[/tex], and using the positive-definite inner
product, as well as properties of the so(3) algebra to show that the
spectrum of [tex]J^2[/tex] is given by j = 0, 1/2, 1, 3/2, 2, ...
and if a particular such j is specified, then m runs from
-j to +j in integer amounts.

So I pose this problem: if one didn't know these results, and wanted to
derive them using strictly C*-algebraic means, how does one proceed to
do so? Arnold suggested working with the exponentiated operators, but I
don't have a clue how to get started. Indeed, what does "joint spectrum
of [itex]J^2[/itex] and [itex]J_z[/itex]" mean in the algebraic context?
In the Hilbert space context we look for common eigenstates, but what does
one do in the algebraic context? If we introduce representations over the
algebra, that's essentially a retreat to Hilbert-space-like techniques, afaict.

Maybe my problem here is that in the Hilbert space context we're really
attacking a different physical question, i.e., finding the unirreps -
which means finding all subspaces which are invariant under SO(3)
rotations and determining their respective dimensions and other
properties. In contrast, finding all values u,v such that neither
[tex](J^2 - u)[/tex] nor [tex](J_z - v)[/tex] have inverses in the
so(3) universal enveloping algebra seems like a distinctly different
problem. So maybe I should be thinking in terms of automorphisms of
the algebra, finding operators A(u,v) such that

[tex]
e^{i(aJ^2 + bJ_z)} \; A \; e^{-i(aJ^2 + bJ_z)} ~=~ A ~,~~~~~~~~~ (1)
[/tex]

where A is a polynomial expression in the so(3) U.E.A. and
a,b are real parameters. Is that how one should start?
Presumably one should reach the outcome that the set of
such polynomials A is characterized by two values that
correspond to the usual j,m of the standard treatment?

[Edit: I just realized (1) can't be the right approach -- because [itex]J^2[/itex]
commutes with everything. Therefore the [itex]J^2[/itex] parts of the exponentials
cancel without effect.]

I hope someone can help disentangle my obvious confusion about all this.
 
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  • #148
What would be the problem with this approach other than that it mentions Hilbert spaces?

Let's say that we have a C*-algebra [itex]\mathcal A[/itex] that's equal to the subalgebra generated by three of its members, [itex]E_1,E_2,E_3[/itex], which satisfy [itex][E_i,E_j]=i\varepsilon_{ijk}E_k[/itex]. Suppose that [itex]\pi:\mathcal A\rightarrow\mathcal B(\mathcal H)[/itex] is a representation. (B(H)=the set of bounded linear operators on a Hilibert space H). Define [itex]J_i=\pi(E_i)[/tex]. Proceed the way you're familiar with. Now you can conclude a bunch of things about what the irreducible representations must look like.
 
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  • #149
Fredrik said:
What would be the problem with this approach other than that it mentions Hilbert spaces?

There's no problem if you bring in Hilbert spaces. That's (equivalent to) the standard
approach. But I'd always understood that part of the "benefit" of the algebraic approach
was that one could do without Hilbert spaces (i.e., find spectra, etc). I'm trying to discover
whether that's really correct or not, and precisely why.
 
  • #150
strangerep said:
There's no problem if you bring in Hilbert spaces. That's (equivalent to) the standard
approach. But I'd always understood that part of the "benefit" of the algebraic approach
was that one could do without Hilbert spaces (i.e., find spectra, etc). I'm trying to discover
whether that's really correct or not, and precisely why.
The C*-algebra approach is not really focused on doing without Hilbert spaces, but really it is about being able to move between them.
For example if I create the abstract C*-algebra of a scalar quantum field theory, states become linear functionals on that algebra. By the GNS theorem, each state is equivalent to a vector in a Hilbert space and the C*-algebra the set of bounded operators on that Hilbert space. The Hilbert space of the state is called its GNS space.
Commonly, many states give the same GNS space, they just turn out to be different vectors in it. This is what is treated as a Hilbert space in the standard approach and is called a folium in the algebraic approach.

The advantage of the algebraic approach is that you can easily use a state from any folium. For example I could use a thermal state for a free QFT, which isn't a Fock state, I simple choose a state from a thermal folium of the right temperature.

So all states, even if they are in different Hilbert spaces, belong to the same space of linear functionals on the algebra. This allows you to analyse difficult problems. For instance free scalar field theory and [tex]\phi^{4}[/tex] have the same C*-algebra in two dimensions. However the automorphisms on that algebra are different (different time evolution or different Hamiltonian in conventional language). It turns out that these two automorphisms never have a well-defined representation in one folium simultaneously. Hence the folium the free theory is defined in (the Fock folium) is different from the folium of [tex]\phi^{4}[/tex]. This is the algebraic version of Haag's theorem.

In summary, you do not dispose of Hilbert space, you just become detached from it and you are able to treat all states in a unified way.
 
  • #151
DarMM said:
For example if I create the abstract C*-algebra of a scalar quantum field theory, states become linear functionals on that algebra.
I would be interested in seeing what the abstract C*-algebra of a scalar QFT looks like. I would also appreciate if you (or someone) can tell me what the three "types" mentioned below are?

DarMM said:
In my opinion one of the major advantages of C*-algebra approach is that it allows one to see a tower of probability theories. For every type of algebra one gets a probability theory:
Type I:
Abelian: Discrete Kolmogorov probability
NonAbelian: Quantum Mechanics
Type II:
Abelian: Statistical Mechanics
NonAbelian: Quantum Statistical Mechanics
Type III:
NonAbelian: Quantum Field Theory
 
  • #152
Fredrik said:
I would be interested in seeing what the abstract C*-algebra of a scalar QFT looks like. I would also appreciate if you (or someone) can tell me what the three "types" mentioned below are?
The abstract C*-algebra of a QFT is basically the algebra generated by the exponential forms of the smeared fields [tex]e^{\iota \phi(f)}[/tex]. Its properties are quite subtle.

For a discussion of the types of C*-algebras read:
http://arxiv.org/abs/quant-ph/0601158" [Broken]
It will explain it better than I can.
 
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  • #153
DarMM said:
The C*-algebra approach is not really focused on doing without Hilbert spaces, but really it is about being able to move between them. [...]

This allows you to analyse difficult problems. For instance free scalar field theory and [tex]\phi^{4}[/tex] have the same C*-algebra in two dimensions. However the automorphisms on that algebra are different (different time evolution or different Hamiltonian in conventional language). It turns out that these two automorphisms never have a well-defined representation in one folium simultaneously. Hence the folium the free theory is defined in (the Fock folium) is different from the folium of [tex]\phi^{4}[/tex]. This is the algebraic version of Haag's theorem.

A very nice summary!


strangerep said:
I'd always understood that part of the "benefit" of the algebraic approach was that one could do without Hilbert spaces (i.e., find spectra, etc). I'm trying to discover whether that's really correct or not, and precisely why.

One doesn't do without Hilbert spaces but creates the latter as needed.

Usually, the C^*-algebra doesn't contain a Lie algebra such as so(3), but the Lie group SO(3) it generates. To find the spectrum of an operator in some Lie algebra generating a Lie group inside the C^* algebra, one usually finds all unitary representations of that Lie group. Then one has to check which of these irreps extend to a rep of the C^*-algebra. For those that extend, one then diagonalizes the operator in that irrep.
 
  • #154
DarMM said:
The C*-algebra approach is not really focused on doing without Hilbert spaces, but really it is about being able to move between them.
[...]
In summary, you do not dispose of Hilbert space, you just become detached from it and you are able to treat all states in a unified way.

A. Neumaier said:
strangerep said:
[...]
One doesn't do without Hilbert spaces but creates the latter as needed.
[...]

Thank you both. I think I can now move past that particular mental roadblock.
 
  • #155
Returning to the central theme of this thread, I'll extract and
summarize part of the conversation between Arnold and DarMM near the
beginning...

You (Arnold) opened by saying:

A. Neumaier said:
I don't like the C^*-algebraic foundations of quantum mechanics since it
assumes that every observable must be bounded and self-adjoint.

But most physical observables are not bounded.

but, in post #135, you appear to retreat from this, (though in a vague way):

A. Neumaier said:
One handles unbounded self-adjoint operators A by considering instead the
bounded family of operators exp(isA), s real. In each representation, their
spectrum is in 1-to-1 correspondence with each other. Thus, in _some_ sense,
one doesn't need unbounded operators. But many things become quite awkward
when expressed in terms of the exponentials rather than their generators.

and I guess your emphasis in the above relates back to another quote
from your post #3:

A. Neumaier said:
My primary requirement is that the usage [of the term "observable"]
should not be too different from what physicists in general call
observables. Of course it entails always an idealization since nobody
can measure all values in an unbounded domain of results, or even all
real numbers in a bounded domain. But before the advent of axiomatic
field theory, basic observables were position, momentum, angular
momentum, energy, force, torque, etc., all of them unbounded. The
axiomatists changed the terminology for purely technical reasons, which
is unacceptable and will never find general agreement, I think.

Most of physics is phrased in terms of differential equations involving
unbounded observables. Most of physics become unexpressible or clumsy to
express when phrased in terms of bounded operators only. No commutation
relations, no continuity equation, no field equations, no Noether theorem

and echoed later in post #3:

A. Neumaier said:
[...] one can represent the unbounded observables on the nuclear space,
take its completion if the Hilbert space is needed, and the dual space
if more singular objects are encountered.

So I think the right mathematical setting should be an inner product space
on which all observables of interest are defined (this common domain exists
in all applications I am aware of), and its closure in various topologies
depending on what one wants to do on the technical level.

Similarly in post #6:

A. Neumaier said:
[...] it seems to me that the relevant unbounded observables always
have a common domain on which they are true linear self-mappings, so
that one could work instead with the algebra of linear self-mappings of
this domain. Resolvents and exponentials would then live in closures of
dense subalgebras of this algebra under appropriate topologies. [...]

Thus far, you seem to be advocating the rigged Hilbert space approach,
in which one defines a nuclear space as an inductive limit of a
sequence of (dense) subspaces of a Hilbert space such that arbitrary
powers of (generally unbounded) observables are well-defined on this
nuclear space. Then one considers the dual of this space, and so on.

However, for the many-particle case, one must presumably take an
infinite tensor product of the above? Does the Schwartz fundamental
theorem generalize satisfactorily for this case? [I.e., the theorem
that basically says [tex](S \otimes S)^* = S^* \otimes S^*[/tex], where
S is the nuclear space for 1 particle.]

I guess this question is related to what DarMM said in post #9:

DarMM said:
The existence of such a domain basically follows from the Wightman axioms,
I'll call it [tex]D[/tex]. I have never seen a study of the mappings of this
domain itself. Getting a workable specification of this domain is an open
problem in axiomatic quantum field theory. A subset of this domain and its
mappings are well studied. This domain are the states generated by polynomials
of the fields acting on the vacuum and a study of its mappings is found in
papers by Borchers, Ruelle, Dixmier and Haag. It is called [tex]D_0[/tex] .

It is a long standing conjecture that:
[tex] D = D_0 [/tex]
However this is not proven. [...]

So is the problem that, in the field theoretic case, one cannot
(rigorously) take an inductive limit to obtain a nuclear space like one
can in the finite-particle case? And that we cannot proceed to a
generalization of the Schwartz fundamental theorem?
 
  • #156
DarMM said:
The abstract C*-algebra of a QFT is basically the algebra generated by the exponential forms of the smeared fields [tex]e^{\iota \phi(f)}[/tex]. Its properties are quite subtle.

For a discussion of the types of C*-algebras read:
http://arxiv.org/abs/quant-ph/0601158" [Broken]
It will explain it better than I can.
Thanks. I was hoping that there would be an article like that. This one looks good.
 
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  • #157
DarMM said:
For a discussion of the types of C*-algebras read:
http://arxiv.org/abs/quant-ph/0601158" [Broken]
It will explain it better than I can.

This is the first paper on Type I-III I saw that clearly relates the classification to physical content without getting too abstract. Still, it seems to me that the paper mainly makes statements of the form: Given a useful C^* algebra, one can construct factors from it that have a certain type.

But what I'd like to know is: If I know that a useful C^* algebra has factors of a certain type, how can i use this information to conclude something of physical interest? Or how does the classification help me in constructing useful C^*-algebras?
 
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  • #158
I interpolate into your comments summaries of what I had written, starting with [AN - and ending with ].
strangerep said:
Returning to the central theme of this thread, I'll extract and
summarize part of the conversation between Arnold and DarMM near the
beginning...

You (Arnold) opened by saying:
[AN - C^* algebras do not contain unbounded operators]
but, in post #135, you appear to retreat from this, (though in a vague way):
[AN - No retreat, but explaining that they are treated indirectly, replaced by their exponentials]
and I guess your emphasis in the above relates back to another quote
from your post #3:
[AN - where I say that the observables dominant in physics are unbounded,
physics is phrased in terms of differential equations relating these, while when working in a C^* algebra, all this must be given up.]
and echoed later in post #3:
[AN - saying that one can consider the full algebra, if one let's it act on a smaller spacer than the Hilbert space]
Similarly in post #6:
[AN - saying the same again]

Thus far, you seem to be advocating the rigged Hilbert space approach,
in which one defines a nuclear space as an inductive limit of a
sequence of (dense) subspaces of a Hilbert space such that arbitrary
powers of (generally unbounded) observables are well-defined on this
nuclear space. Then one considers the dual of this space, and so on.

In contrast to the rigged Hilbert space approach, where the base space is nuclear. I want to leave open what precisely must be required from the base space (since I am not clear about that myself). Thus I just want a vector space with an inner product, on which the algebra acts, its closure under the induced norm (which is a Hilbert space), and its dual under the induced topology. This seems to be enough to do everything needed in quantum mechanics.

strangerep said:
However, for the many-particle case, one must presumably take an
infinite tensor product of the above? Does the Schwartz fundamental
theorem generalize satisfactorily for this case? [I.e., the theorem
that basically says [tex](S \otimes S)^* = S^* \otimes S^*[/tex], where
S is the nuclear space for 1 particle.]

I guess this question is related to what DarMM said in post #9:
[AN - where he says that a common domain exists.]

Maybe his open question is related to the question of nuclearity. I believe that D_0 suffices, and no physics depends on whether this equals D, though I don't understand the situation well enough to claim this as a fact.

strangerep said:
So is the problem that, in the field theoretic case, one cannot
(rigorously) take an inductive limit to obtain a nuclear space like one
can in the finite-particle case? And that we cannot proceed to a
generalization of the Schwartz fundamental theorem?

I don't see any problem if one takes as the base space the space of wave functions
psi=(psi_0,psi_1,...) with arbitrarily often differentiable functions psi_N(p_1,...,p_N) of an indefinite number N of momentum arguments, decaying rapidly enough, so that in particular the inner product
[tex]\langle\phi|\psi\rangle:=
\sum_N \frac{1}{N!} \int dp_{1:N} \phi_N(p_1,...,p_N)^* \psi_N(p_1,...,p_N)
[/tex]
is absolutely convergent. This space is canonically isomorphic to a subspace of Fock space,
via
[tex]\psi \to
\sum_N \frac{1}{\sqrt{N!}} \int dp_{1:N} \psi_N(p_1,...,p_N)|p_1,...,p_N\rangle.
[/tex]
 
  • #159
A. Neumaier said:
In contrast to the rigged Hilbert space approach, where the base space is nuclear. I want to leave open what precisely must be required from the base space (since I am not clear about that myself). Thus I just want a vector space with an inner product, on which the algebra acts, its closure under the induced norm (which is a Hilbert space), and its dual under the induced topology. This seems to be enough to do everything needed in quantum mechanics.

After pondering this for a while, I don't really see the significant difference between
standard RHS and what you're looking for. With RHS, we start with a Hilbert space "H"
and set of particular candidate observables, then discover that they're not defined
everywhere on the Hilbert space, then construct smaller and smaller (nested, dense)
subspaces of H where successively higher orders of polynomials in the observables
are well-defined. The inductive limit of this sequence of nested subspaces becomes
the "small" space in a Gel'fand triple. These are all nuclear spaces because each
step in the sequence is constructed via a seminorm which ensures that the next
higher order of polynomial in the observables is well-defined there.

In contrast you want a vector space with an inner product on which the observables
act, and then find a larger Hilbert space by taking closure. You skip mention of the
nuclearity details involved in going the other way, but I wonder what this really buys?

What is wrong with using the standard RHS construction and the body of theory
that accompanies it? (It's puzzling to me that you would not use it up to the
point where it proves to be deficient in some specific way.)
 
  • #160
strangerep said:
After pondering this for a while, I don't really see the significant difference between standard RHS and what you're looking for. [...]

In contrast you want a vector space with an inner product on which the observables
act, and then find a larger Hilbert space by taking closure. You skip mention of the
nuclearity details involved in going the other way, but I wonder what this really buys?

What is wrong with using the standard RHS construction and the body of theory
that accompanies it? (It's puzzling to me that you would not use it up to the
point where it proves to be deficient in some specific way.)
There is nothing wrong with it. But proceeding the other way, one has an explicit description of the small space, and one doesn't have to worry about whether this space is or isn't identical with the nuclear space. There are lots of concrete quantum systems for which one can write down such a small space explicitly, and I don't know how to check easily whether these spaces are nuclear.

But perhaps you know a simple criterion for checking that?
 
  • #161
A. Neumaier said:
There is nothing wrong with it. But proceeding the other way, one has an explicit description of the small space, and one doesn't have to worry about whether this space is or isn't identical with the nuclear space. There are lots of concrete quantum systems for which one can write down such a small space explicitly, and I don't know how to check easily whether these spaces are nuclear.

But perhaps you know a simple criterion for checking that?

:uhh: Danger, danger. This sounds like a rhetorical teaching device, likely to end
with me embarrassing myself in public. But ok, I'll play...

First, would you please give one or two of the concrete quantum systems you
have in mind, and their explicit small spaces, to focus this exercise?
 
  • #162
strangerep said:
:uhh: Danger, danger. This sounds like a rhetorical teaching device
It asn't intended that way. I don't want you to actually recognize it; just though that you might have come across theorems or techniques that I could look up, which give sufficient conditions for nuclearity.

strangerep said:
First, would you please give one or two of the concrete quantum systems you
have in mind, and their explicit small spaces, to focus this exercise?

The prime examples that I'd like to recognize as nuclear using a simple, general recipe are

(i) the space of Schwartz function on R, with the algebra of linear combinations of
[tex]x^k \exp(i u x) p^l \exp(i vp)[/tex] acting on it.

(ii) the space of phi=(phi_),phi_1,phi_2,...) where phi_n is a square integrable symmetrized n-particle C^inf wave function on R^n (regarded as coefficient functions of the corresponding Fock space, with the algebra generated by a(x), a(f) for smooth f, their adjoints and their exponentials.
 
  • #163
Does not observation in itself change the observation like in the quantum chip where the light bands would show up in one of two ports unless third port checked first this is done after the light has been recorded. Best theory given was the light knew where it was and knew where the other light were?
 
  • #164
JEDIGnome said:
Does not observation in itself change the observation like in the quantum chip where the light bands would show up in one of two ports unless third port checked first this is done after the light has been recorded. Best theory given was the light knew where it was and knew where the other light were?
http://www.photonics.com/Article.aspx?AID=45161 Inserting web link sorry if this is a double entry.
 
  • #166
You have to be kidding me?
http://www.photonics.com/Article.aspx?AID=45161

"Such an interconnected quantum system could function as a quantum computer, or, as proposed by the late Caltech physicist Richard Feynman in the 1980s, as a "quantum simulator" for studying complex problems in physics."
 
  • #167
JEDIGnome said:
You have to be kidding me?
http://www.photonics.com/Article.aspx?AID=45161

"Such an interconnected quantum system could function as a quantum computer, or, as proposed by the late Caltech physicist Richard Feynman in the 1980s, as a "quantum simulator" for studying complex problems in physics."
Again you forgot to tell us the connection to the topic of the thread, ''Boundedness of quantum observables''.
 
  • #168
[I had intended to defer posting again in this thread until I'd chased
down some references and digested the issues more thoroughly. But as
that will take ages, Arnold suggested privately that I should at least
post some references -- in case others are interested.]

First, regarding criteria for nuclearity of a space...

Gel'fand & Vilenkin (vol 4) do indeed mention some techniques for
checking nuclearity of a space (their section 3.4).

0) A nuclear space is also perfect (no isolated points: every point
can be approximated arbitrarily well by other points),

which implies other properties:

1) Both in a nuclear space and its adjoint (dual), strong and weak
convergence coincide.

2) A closed bounded set in the dual of a nuclear space is compact
relative to weak and strong convergence.

3) A nuclear space is separable.

4) A nuclear space is complete relative to weak convergence.

It's also the case that:

5) A Banach space is not nuclear.

Regarding item (3) above (that a nonseparable space is not nuclear), in
view of Kibble's work on the IR problem in QED involving a nonseparable
space, this no-go sounds disturbing. Then I found out (partly from G&V,
but also from references following below) that it's not strictly
necessary for the small space in the triple to be nuclear. One can use
merely a locally convex TVS, provided the embedding map into its
closure is nuclear (i.e., trace-class).

The following two papers by Antoine appear to be an early investigation
of precisely the question that Arnold mentioned earlier about
the minimally necessary conditions that the small space of the triple
be stable under the action of a set of observables.

[Antoine, 1969a]
J-P Antoine,
"Dirac Formalism and Symmetry Problems in QM I. General Dirac Formalism",
J. Math. Phys., Vol. 10, (1969), p.53-69

[Antoine, 1969b]
J-P Antoine,
"Dirac Formalism and Symmetry Problems in QM II. Symmetry Problems",
J. Math. Phys., Vol. 10, (1969), p.2276-2290

Antoine mentions that the small space need not be nuclear, it being
sufficient that its embedding map into its closure (the Hilbert space)
be nuclear (i.e., trace-class). However, he then seems to back away
from this to work with nuclear spaces anyway. It's not clear to me on a
single reading whether this is essential, or just convenient.

He also notes that the small space must be Frechet (metrizable topology)
for the Schwarz kernel theorem to hold (the one which says
[tex](V \otimes V)' = V' \otimes V'[/tex]. (But it's unclear to me whether
this is really an essential property for quantum theory.)

He also opens the can of worms of marrying such spaces with probability
satisfactorily. He seems to emphasize relative probability, but I'm not
yet clear about what he's suggesting.

Both papers are worth reading if you're into this sort of stuff,
provided one overlooks some questionable material. (He rejects
eigenvalues which are not in the Hilbert space spectrum as unphysical
which I think is too harsh.)
The theme (of defining the small space by means of the observables it
must support) is also explored in this more recent paper:

[Wickramasekara+Bohm, 2003]
S. Wickramasekara & A. Bohm,
"Symmetry Representations in the RHS Formulation of QM",
math-ph/0302018

Their definition of rigged Hilbert space does not insist on nuclearity
of the small space.

Antoine has obviously spent decades extending and refining these ideas.
There's this paper:

[Antoine, 1998]
J-P Antoine, "QM beyond Hilbert Space",
in "Irreversibility and Causality Semigroups and Rigged Hilbert Spaces",
Lecture Notes in Physics, 1998, Volume 504, Springer-Verlag, p3.
(Also available as a PDF from psu.edu).

and this one:

[Antoine+Trapani, 2010]
J-P Antoine, C. Trapani,
"The Partial Inner Product Space Method: A Quick Overview",
Adv. Math. Phys, vol 2010, Art ID 457635,
http://www.hindawi.com/journals/amp/2010/457635/
(PDF freely downloadable)

which is a "quick" (37 page) overview of this book:

[Antoine+Trapani, 2009]
J-P Antoine, C. Trapani,
"Partial Inner Product Spaces: Theory & Applications",
Springer, 2009, ISBN-13 978-3642051357

I haven't yet got the book, but the overview paper opened my eyes a bit
more to other sophisticated approaches beyond orthodox rigged Hilbert
spaces, (which are only a special case of PIP spaces).---------------------

On the question of algebraic frameworks for working with unbounded
operators, there's this paper in which generalized GNS representation
ideas used rigged Hilbert spaces (I think -- this paper is highly
mathematical and rather impenetrable for me). I'd be interested to
hear what the more mathematically gifted make of it.

S. Iguri, M. Castagnino,
"The Formulation of Quantum Mechanics in Terms of Nuclear Algebras"
IJTP, vol38, no1, (1999), pp143-164

Abstract:

In this work we analyze the convenience of nuclear barreled b*-algebras
as a better mathematical framework for the formulation of quantum
principles than the usual algebraic formalism in terms of C*-algebras.
Unbounded operators on Hilbert spaces have an abstract counterpart in
our approach. The main results of the C*-algebra theory remain valid.
We demonstrate an extremal decomposition theorem, an adequate
functional representation theorem, and an extension of the classical
GNS theorem.

---------------------

Then there's this book which I haven't seen -- beyond what Amazon
allows one to read. To me it looks like pretty heavy stuff.

[Antoine+Inoue+Trapani, 2002]
J-P Antoine, I. Inoue, C. Trapani,
"Partial *-Algebras and Their Operator Realizations",
Springer, 2002, ISBN-13: 978-1402010255

Product Description:

Algebras of bounded operators are familiar, either as C*-algebras or as
von Neumann algebras. A first generalization is the notion of algebras
of unbounded operators (O*-algebras), mostly developed by the Leipzig
school and Japanese mathematicians. This is the first textbook to go
one step further by considering systematically partial *-algebras of
unbounded operators (partial O*-algebras) and the underlying algebraic
structure, namely, partial *-algebras. The first part of the text
begins with partial O*-algebras covering basic properties and
topologies with many examples and accumulates in the generalization to
this new framework of the celebrated modular theory of Tomita-Takesaki,
one of the cornerstones for the applications to statistical physics.
The text then focuses on abstract partial *-algebras and their
representation theory, again obtaining generalizations of familiar
theorems, for example Radon-Nikodym and Lebesgue. Partial *-algebras of
operators on Rigged Hilbert Spaces are also considered. The last
chapter discusses some applications in mathematical physics, for
example quantum field theory and spin systems. This book will be of
interest to graduate students or researchers in pure mathematics as
well as mathematical physicists.

---------------------

I recall Bert Schroer often mentions "Tomita-Takesaki modular theory",
and I've often wonder what the heck it is. Can anyone give a
physicist-friendly overview?
 
  • #169
strangerep said:
[Arnold suggested privately that I should at least
post some references -- in case others are interested.]
Thanks for posting it!

strangerep said:
Gel'fand & Vilenkin (vol 4) do indeed mention some techniques for
checking nuclearity of a space (their section 3.4).
Unfortunately, what you listed are only necessary conditions - they can be used to exclude nuclearity but not to prove it.
strangerep said:
Regarding item (3) above (that a nonseparable space is not nuclear), in
view of Kibble's work on the IR problem in QED involving a nonseparable
space, this no-go sounds disturbing.
While in earlier times, nonseparable Hilbert spaces were considered to be unphysical curiosities, they were shown more recently to be unavoidable in some cases of great physical interest: All infrared problems in QFT are related to nonseparable spaces defined by nonregular representations of Weyl groups - as described e.g. in
Acerbi, F. and Morchio, G. and Strocchi, F.,
Infrared singular fields and nonregular representations of canonical commutation relation algebras,
J. Math. Phys. 34 (1993), 889.
strangerep said:
The following two papers by Antoine appear to be an early investigation
of precisely the question that Arnold mentioned earlier
Especially his recent book (of which I only learned through your notice) is relevant in this respect!
strangerep said:
He also notes that the small space must be Frechet (metrizable topology)
for the Schwarz kernel theorem to hold (the one which says
[tex](V \otimes V)' = V' \otimes V'[/tex]. (But it's unclear to me whether
this is really an essential property for quantum theory.)
It is important since topologies that are not Frechet are hardly usable in the analysis.
strangerep said:
I'd be interested to
hear what the more mathematically gifted make of it.

S. Iguri, M. Castagnino,
"The Formulation of Quantum Mechanics in Terms of Nuclear Algebras"
IJTP, vol38, no1, (1999), pp143-164
Far too technical (in contrast to Antoine's work) to make good foundations - and it depends heavily on nuclearity, which is too strong in the IR regime.
strangerep said:
[Antoine+Inoue+Trapani, 2002]
J-P Antoine, I. Inoue, C. Trapani,
"Partial *-Algebras and Their Operator Realizations",
Springer, 2002, ISBN-13: 978-1402010255
Probably obsolete given his new book. But I haven't seen it.
strangerep said:
I recall Bert Schroer often mentions "Tomita-Takesaki modular theory",
and I've often wonder what the heck it is. Can anyone give a
physicist-friendly overview?
It is a generalization of the KMS state conditions for equilibrium states. But modular theory has always been too abstract for me to relate it to useful physics; so I can't help.
 
Last edited:
  • #170
A. Neumaier said:
strangerep said:
Gel'fand & Vilenkin (vol 4) do indeed mention some techniques for
checking nuclearity of a space (their section 3.4).
[...]

Unfortunately, what you listed are only necessary conditions - they can be
used to exclude nuclearity but not to prove it.

Oh, I knew you would say that!

There's also some other techniques for cases where each member of the
family of (semi)norms can be bounded by an order set of functions with
certain properties. Proving nuclearity then seems straightforward
(which G&V show for the case of Schwartz space).

A. Neumaier said:
[Frechet] is important since topologies that are not Frechet are
hardly usable in the analysis.

I meant: it's not clear to me whether the property
[tex](V \otimes V)' = V' \otimes V'[/tex] is really
essential for quantum theory.
 
  • #171
strangerep said:
Oh, I knew you would say that!
I added it only because you didn't say it.
strangerep said:
There's also some other techniques for cases where each member of the
family of (semi)norms can be bounded by an order set of functions with
certain properties. Proving nuclearity then seems straightforward
(which G&V show for the case of Schwartz space).
Thanks, that's better. But as I recently learnt, nuclearity fails in simple examples with nontrivial infrared behavior; so this seems now less relevant.
strangerep said:
I meant: it's not clear to me whether the property [tex](V \otimes V)' = V' \otimes V'[/tex] is really essential for quantum theory.
This is needed so that one knows how the duals of subspaces of Fock space look like.
 
  • #172
A. Neumaier said:
But as I recently learnt, nuclearity fails in simple examples with nontrivial infrared behavior; so [nuclearity of a space] seems now less relevant.

If one relaxes the requirement so that only the embedding map of the small space
into its closure is nuclear (trace-class), I'm wondering whether this helps in
the case of nonseparable spaces. For that matter, how does one take a trace in
a nonseparable space anyway? I guess one needs a tractable example of a
nonseparable space to study this? Or maybe the correspondence between "nuclear
mapping" and "trace-class mapping" only applies for separable spaces?
 
  • #173
strangerep said:
If one relaxes the requirement so that only the embedding map of the small space
into its closure is nuclear (trace-class), I'm wondering whether this helps in
the case of nonseparable spaces. For that matter, how does one take a trace in
a nonseparable space anyway? I guess one needs a tractable example of a
nonseparable space to study this? Or maybe the correspondence between "nuclear
mapping" and "trace-class mapping" only applies for separable spaces?
The trace is undefined. A simple nonseparable Hilbert space is obtained by taking functions f(x,s) of two real arguments x and s, which hare square integrable with respect to x and vanish uniformly in x for all but a countable number of values of s, such that
[tex] ||f||^2:= \sum_x \int dx |f(x,s)|^2[/tex]
is finite, and defining the inner product to be
[tex]\langle f,g\rangle := \sum_s \int dx f(x,s)^*g(x,s).[/tex]
Maybe this is enough for you to play with the weakening of the nuclear property.
 
  • #174
A. Neumaier said:
The trace is undefined. A simple nonseparable Hilbert space is obtained by taking functions f(x,s) of two real arguments x and s, which are square integrable with respect to x and vanish uniformly in x for all but a countable number of values of s,

Sadly, I failed to discover the rigorous meaning of "vanish uniformly in x"
applicable in this context.

[tex] ||f||^2:= \sum_x \int dx |f(x,s)|^2[/tex]

I guess that summation should have been over s.
 
  • #175
strangerep said:
Sadly, I failed to discover the rigorous meaning of "vanish uniformly in x" applicable in this context.
It means that the sum only has countably many nonzero terms (else it wouldn't make sense).
strangerep said:
I guess that summation should have been over s.
Yes.
 
<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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