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Questions about Rigged Hilbert Space

by lugita15
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lugita15
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Jan9-12, 10:31 PM
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Dirac's bra-ket formalism implicitly assumed that there was a Hilbert space of ket vectors representing quantum states, that there were self-adjoint linear operators defined everywhere on that space representing observables, and that the eigenvectors of any such operator formed an orthogonal basis of the ket space. But this runs into a few problems: first of all, some operators like position and momentum are unbounded, so they cannot be defined everywhere on the ket space due to the Hellinger-Toeplitz theorem. Second, the criterion that you have an orthogonal basis seems to require the ket space to be seperable, but seperable Hilbert spaces can only have countably many dimensions, which is incompatible with the idea that position eigenstates form an uncountable basis.

The RHS construction tries to alleviate these concerns as follows (this is mostly gleamed from reading old PF threads on the topic). We start with the standard Hilbert space [itex]H=L^{2}(ℝ^{3})[/itex], and we pick out a so-called nuclear subspace [itex]\Phi[/itex] which is dense in [itex]H[/itex] and on which any continuous function [itex]f(P,Q,H)[/itex] (defined via Taylor series) of the position, momentum, and Hamiltonian operators is defined. We then basically use [itex]\Phi[/itex] as the space of test functions for a theory of Schwartz distributions - the space [itex]\Phi^{\times}[/itex] of continuous anti-linear functionals will be our ket space, and the space [itex]\Phi'[/itex] of continuous linear functionals will be our bra space.

My question is, does RHS sucessfully implement all of Dirac's formalism unscathed, so that we can completely forget about the original Hilbert space once we've set up the rigged Hilbert space? I've heard that the Gelfand-Maurin nuclear spectral theorem proves that every self-adjoint linear operator acting on [itex]H[/itex] has an eigenbasis in [itex]\Phi^{\times}[/itex], but can the same be said of all self-adjoint linear operators acting on [itex]\Phi^{\times}[/itex]? Is it possible to define a distribution-valued inner product on [itex]\Phi^{\times}[/itex] (or do we run into the age-old problem of multiplication of distributions?), and does this make [itex]\Phi^{\times}[/itex] an uncountable-dimensional non-seperable Hilbert space? Can [itex]\Phi'[/itex] also be considered as the dual space of [itex]\Phi^{\times}[/itex], and is there a one-to-one correspondence between these two spaces? Finally, can things like delta-function potentials be rigorously justified in RHS framework?

Sorry for asking so many questions, but I only recently discovered Rigged Hilbert Spaces, and I'm absolutely fascinated by them. Can anyone recommend a good book on the subject? (I've already read the couple pages Ballentine has to say on the subject, and I wish there was a whole quantum mechanics text using this approach.)

Any help would be greatly appreciated.

Thank You in Advance.
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Fredrik
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Jan9-12, 11:42 PM
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Every Hilbert space has an orthormal basis. It has a countable orthonormal basis if and only if the space is separable.

I don't know this stuff well, but I think the references in this thread will be a good place to start.
strangerep
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Jan10-12, 01:56 AM
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Quote Quote by lugita15 View Post
Dirac's bra-ket formalism implicitly assumed that there was a Hilbert space of ket vectors representing quantum states, that there were self-adjoint linear operators defined everywhere on that space representing observables, and that the eigenvectors of any such operator formed an orthogonal basis of the ket space. But this runs into a few problems: first of all, some operators like position and momentum are unbounded, so they cannot be defined everywhere on the ket space due to the Hellinger-Toeplitz theorem. Second, the criterion that you have an orthogonal basis seems to require the ket space to be seperable, but seperable Hilbert spaces can only have countably many dimensions, which is incompatible with the idea that position eigenstates form an uncountable basis.
Try this paper first:

Rafael de la Madrid,
"The role of the rigged Hilbert space in Quantum",
Available as: arXiv:quant-ph/0502053

Hilbert spaces have a scalar-valued inner product. But Dirac's formalism has a distribution-valued inner product in general. Rigged Hilbert space can be regarded as a version of Dirac's formalism made rigorous and generalized.

My question is, does RHS sucessfully implement all of Dirac's formalism unscathed, so that we can completely forget about the original Hilbert space once we've set up the rigged Hilbert space?
Don't forget that the Hilbert space is the middle space in the Gel'fand triple (RHS). So it's still there.

I've heard that the Gelfand-Maurin nuclear spectral theorem proves that every self-adjoint linear operator acting on [itex]H[/itex] has an eigenbasis in [itex]\Phi^{\times}[/itex], but can the same be said of all self-adjoint linear operators acting on [itex]\Phi^{\times}[/itex]?
The statement of the theorem is a bit trickier than that. One constructs a spectral measure in terms of bras and kets living in [itex]\Phi^{\times}[/itex] and [itex]\Phi'[/itex].

If you really want to learn the maths, try Gelfand & Vilenkin vol 4. :-)

Can [itex]\Phi'[/itex] also be considered as the dual space of [itex]\Phi^{\times}[/itex],
No. [itex]\Phi'[/itex] is the dual of [itex]\Phi[/itex]. That's the basis for how everything is set up: find a space [itex]\Phi[/itex] on which arbitrary powers of the observables are defined everywhere, and then consider its dual [itex]\Phi'[/itex] (and anti-dual [itex]\Phi^{\times}[/itex]). Observables acting on [itex]\Phi[/itex] are then extended to operators acting on the larger space [itex]\Phi'[/itex] via the dual pairing. (This is kinda critical to the whole construction.)

and is there a one-to-one correspondence between these two spaces?
Afaik, one is the conjugate space of the other (interpreted appropriately).

Finally, can things like delta-function potentials be rigorously justified in RHS framework?
As mentioned in quant-ph/0502053 (near the end),

Quote Quote by Rafa
it is not clear that one can always
find a nuclear space Φ that remains invariant under the action of the observables.
Nevertheless, Roberts has shown that such Φ exists when the potential is infinitely
often differentiable except for a closed set of zero Lebesgue measure [...].
...which seems to cover delta-function potentials.

Sorry for asking so many questions, but I only recently discovered Rigged Hilbert Spaces, and I'm absolutely fascinated by them. Can anyone recommend a good book on the subject?
I still haven't found one. In ordinary QM one can use Hilbert space without being totally familiar with the vast body of functional analysis. Same thing with RHS.

(I've already read the couple pages Ballentine has to say on the subject, and I wish there was a whole quantum mechanics text using this approach.)
Most already do -- if they use distribution-valued inner products (i.e., Dirac deltas).

Another (more mathmatical) paper is:

M. Gadella, F. Gomez,
"On the Mathematical Basis of the Dirac Formulation of Quantum Mechanics",
IJTP, vl. 42, No. 10, October 2003, p2225.

Also do a search on Google Scholar for papers/books by J-P Antoine.

lugita15
#4
Jan10-12, 01:44 PM
P: 1,583
Questions about Rigged Hilbert Space

Quote Quote by strangerep View Post
Don't forget that the Hilbert space is the middle space in the Gel'fand triple (RHS). So it's still there.
I'm aware that that's how the RHS is constructed, but having constructed it can you discard the scaffolding and only deal with the spaces [itex]\Phi^{\times}[/itex] and [itex]\Phi'[/itex]? Or can they be defined without reference to [itex]\Phi[/itex] and [itex]H[/itex]?
The statement of the theorem is a bit trickier than that. One constructs a spectral measure in terms of bras and kets living in [itex]\Phi^{\times}[/itex] and [itex]\Phi'[/itex].
By spectral measure do you mean doing a Riemann-Stieltjes integral with respect to an operator, as shown here in Kreyszig's Introductory Functional Analysis? (I wish there was a book at the level of Kreyszig that tackled rigged Hilbert spaces.)
If you really want to learn the maths, try Gelfand & Vilenkin vol 4. :-)
Is this stuff scattered across the various volumes of Gelfand? Because I see some websites that say that the notion of symmetric operators possessing complete sets of eigenfunctionals is developed in volumes 2 and 3 (which are by Gelfand and Shilov; looks like there was a change in coauthors).
No. [itex]\Phi'[/itex] is the dual of [itex]\Phi[/itex]. That's the basis for how everything is set up: find a space [itex]\Phi[/itex] on which arbitrary powers of the observables are defined everywhere, and then consider its dual [itex]\Phi'[/itex] (and anti-dual [itex]\Phi^{\times}[/itex]). Observables acting on [itex]\Phi[/itex] are then extended to operators acting on the larger space [itex]\Phi'[/itex] via the dual pairing. (This is kinda critical to the whole construction.)
So then do we lose the beauty of the Riesz representation theorem? Can we no longer view the bras as inner products with fixed kets? Or are the bras only inner products with fixed kets that happen to be part of the nuclear space? In that case, do the intuitive notions underpinning the bra-ket formalism go out the window, like <x| acting on |ψ> being equal to the inner product of |x> and |ψ>?
Afaik, one is the conjugate space of the other (interpreted appropriately).
Do you mean this?
As mentioned in quant-ph/0502053 (near the end),


...which seems to cover delta-function potentials.
I'm not so sure about that. I think that's about when your Hamiltonian is made up of regular functions, not distributions. I think the range [itex]R(H)[/itex] of the Hamiltonian operator corresponding to a delta function potential will be a subspace of [itex]\Phi^{\times}[/itex] whose intersection with the Hilbert space is {0}. In other words Hψ(x), where ψ(x)≠0 is in the Hilbert space, will not be in the Hilbert space. So I would expect a trivial nuclear subspace and thus problems for the RHS framework.
I still haven't found one.
I've been looking for the book "Dirac Kets, Gamow Vectors, and Gelfand Triplets" by Arno Bohm, which is out of print.
Another (more mathmatical) paper is:

M. Gadella, F. Gomez,
"On the Mathematical Basis of the Dirac Formulation of Quantum Mechanics",
IJTP, vl. 42, No. 10, October 2003, p2225.

Also do a search on Google Scholar for papers/books by J-P Antoine.
Thanks for the references.
lugita15
#5
Jan10-12, 02:20 PM
P: 1,583
I've found a nice article from the Stanford Encyclopedia of Philosophy which among other things discusses Rigged Hilbert Space. But it has a few surprising assertions:
It has been argued in (Kronz 1999), for example, that separable Hilbert space is not a suitable framework for quantum mechanics under Bohm's ontological interpretation (also known as Bohmian mechanics).
I thought interpretations of quantum mechanics agreed on the math, but disagreed on what the math means.
It was later demonstrated that the rigged Hilbert space formulation of quantum mechanics can handle a broader range of phenomena than the separable Hilbert space formulation.
So RHS formulation differs on actual experimental results? Then how can it be a formulation of QM at all?
It is worth noting that neither [itex]\Phi[/itex] nor [itex]\Phi^{\times}[/itex] is a Hilbert space in that each lacks an inner product that induces a metric with respect to which the space is complete, though for each space there is a topology with respect to which the space is complete.
Does this hold true even if you endow [itex]\Phi^{\times}[/itex] with a distribution-valued inner product?
Fredrik
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Jan10-12, 05:13 PM
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Quote Quote by lugita15 View Post
By spectral measure do you mean doing a Riemann-Stieltjes integral with respect to an operator, as shown here in Kreyszig's Introductory Functional Analysis?
I think that's a Lebesgue-Stieltjes integral. A spectral measure is also called a projection-valued measure because it's a function that's just like a measure, except that its codomain is the set of projection operators on a Hilbert space instead of the set [0,∞] of non-negative extended real numbers.

Quote Quote by lugita15 View Post
(I wish there was a book at the level of Kreyszig that tackled rigged Hilbert spaces.)
I don't think you need one. The (extremely) hard part is to get to the point where you understand the spectral theorem for not necessarily bounded normal operators on a Hilbert space. (A "normal" operator is a linear operator that commutes with its adjoint, so self-adjoint operators are normal). Once you're there (I'm still not), I think it's going to be very easy compared to what you've already done to understand rigged Hilbert spaces and the generalized spectral theorem. The article I linked to in the other thread looks like it's going to be very useful at that point.

Quote Quote by lugita15 View Post
Is this stuff scattered across the various volumes of Gelfand?
I haven't read the book, but I just looked at the foreword of vol. 4, and it says this:
The material of this fourth volume represents a complete unit in itself, and, as we have said, the exposition is practically independent of the preceding volumes. In spite of the relation of one chapter to another, one can begin a reading of this book with the first chapter, which contains the general theory of nuclear and rigged Hilbert spaces, or with the second chapter, which discusses the more elementary theory of positive definite generalized functions.
So it looks like it's just chapter 1 of volume 4.

Quote Quote by lugita15 View Post
So RHS formulation differs on actual experimental results? Then how can it be a formulation of QM at all?
I've been told (by A. Neumaier*) that separable Hilbert spaces also can't handle some quantum field theory stuff. So if we view "QM" as the framework in which quantum theories are defined, and quantum field theories with interactions as quantum theories, then the separability requirement should be dropped from the axioms of QM.

*) He supported his claim by linking to an article that apparently explains the details, but I didn't read it.

Quote Quote by lugita15 View Post
Does this hold true even if you endow [itex]\Phi^{\times}[/itex] with a distribution-valued inner product?
I'm not sure I understand the question. A "distribution-valued inner product" is not an inner product, since an inner product is complex-valued. A space that doesn't have an inner product isn't an inner product space, and therefore not a Hilbert space (=complete inner product space).
lugita15
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Jan10-12, 10:53 PM
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Quote Quote by Fredrik View Post
I think that's a Lebesgue-Stieltjes integral.
Well, Kreyszig doesn't discuss Lebesgue measure or Lebesgue integration during any of this, but are you saying that what he is doing in taking the limit with respect to the operator norm is equivalent to a Lebesgue-Stieltjes integral? (This conscious omission of any discussion of the Lebesgue integral goes so far that Kreyszig has to define the space L2[a,b] as the completion of C[a,b] with respect to a norm defined by a Riemann integral.)
A spectral measure is also called a projection-valued measure because it's a function that's just like a measure, except that its codomain is the set of projection operators on a Hilbert space instead of the set [0,∞] of non-negative extended real numbers.
Just as a regular measure must always equal nonnegative number, must the projection-valued measure always equal a positive operator?
I don't think you need one. The (extremely) hard part is to get to the point where you understand the spectral theorem for not necessarily bounded normal operators on a Hilbert space. (A "normal" operator is a linear operator that commutes with its adjoint, so self-adjoint operators are normal). Once you're there (I'm still not), I think it's going to be very easy compared to what you've already done to understand rigged Hilbert spaces and the generalized spectral theorem. The article I linked to in the other thread looks like it's going to be very useful at that point.
Hmm, Kreyszig just takes a chapter to discuss the spectral theory of unbounded operators. I haven't gotten to that part, so I can't judge whether he does the whole thing or leaves out the more advanced pieces.
I haven't read the book, but I just looked at the foreword of vol. 4, and it says this:
The material of this fourth volume represents a complete unit in itself, and, as we have said, the exposition is practically independent of the preceding volumes. In spite of the relation of one chapter to another, one can begin a reading of this book with the first chapter, which contains the general theory of nuclear and rigged Hilbert spaces, or with the second chapter, which discusses the more elementary theory of positive definite generalized functions.
So it looks like it's just chapter 1 of volume 4.
OK, that's good to know. I was afraid there were 5 volumes of Gelfand standing between me and the nuclear spectral theorem
I've been told (by A. Neumaier*) that separable Hilbert spaces also can't handle some quantum field theory stuff. So if we view "QM" as the framework in which quantum theories are defined, and quantum field theories with interactions as quantum theories, then the separability requirement should be dropped from the axioms of QM.
That sounds interesting. I wonder, what was the original physical motivation for separability of the Hilbert space? There must have been a reason it was included as an axiom.
I'm not sure I understand the question. A "distribution-valued inner product" is not an inner product, since an inner product is complex-valued. A space that doesn't have an inner product isn't an inner product space, and therefore not a Hilbert space (=complete inner product space).
Admittedly the language is a bit imprecise: strictly speaking the value of the inner product is not a distribution, just the value of a distribution, but intuitively it's clear. In ordinary inner products, for a fixed vector u we have <u,v>=f(v) where f is a function. But in this case we want to have <u,v>=f(v) where f is a distribution. Once the "distribution-valued inner product" is defined (and I'm still unclear on how exactly this is done), we can make sense of things like <x|x'>=δ(x-x'). So we are generalizing the notion of inner products, so that they can equal not only ordinary numbers but also stranger beasts like δ(0). (Of course, distributions are usually defined under an integral sign, so I don't think we deal directly with δ(0) and the like).
strangerep
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Jan10-12, 11:18 PM
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Quote Quote by lugita15 View Post
I'm aware that that's how the RHS is constructed, but having constructed it can you discard the scaffolding and only deal with the spaces [itex]\Phi^{\times}[/itex] and [itex]\Phi'[/itex]? Or can they be defined without reference to [itex]\Phi[/itex] and [itex]H[/itex]?
The dual and antidual spaces [itex]\Phi^{\times}[/itex] and [itex]\Phi'[/itex] need [itex]\Phi[/itex] in order to make useful sense. For one thing, the action of operators on [itex]\Phi'[/itex] is only defined in terms of their action on [itex]\Phi[/itex]. E.g., think about how one defines the derivative of a delta function: it's a weak derivative -- only defined via its action on test functions.
Cf. http://en.wikipedia.org/wiki/Weak_derivative

By spectral measure do you mean doing a Riemann-Stieltjes integral with respect to an operator, as shown here in Kreyszig's Introductory Functional Analysis?
That's the general idea -- though it's a Lebesgue measure as Fredrik mentioned. The [itex]dE_\lambda[/itex] in your example can be expressed more explicitly as a projection-valued measure, i.e.,
[tex]
\def\<{\langle}
\def\>{\rangle}
|E\>\<E| \, dE
[/tex]
where E ranges over the (continuous) range of energies in the spectrum of the self-adjoint operator.

I wish there was a book at the level of Kreyszig that tackled rigged Hilbert spaces.
Me too! Kreyszig is always my go-to guy when I need to learn something for the first time. G+V is hard work. I've tried a few times to look for other books on generalized functions, but they either seem too engineering-oriented or too puremath-oriented.

Is this stuff scattered across the various volumes of Gelfand? [...]
I only have vols 4 & 5. Sometimes, it was (and still is) a challenge to grok their notation and concepts, but it's doable.

So then do we lose the beauty of the Riesz representation theorem? Can we no longer view the bras as inner products with fixed kets? Or are the bras only inner products with fixed kets that happen to be part of the nuclear space? In that case, do the intuitive notions underpinning the bra-ket formalism go out the window, like <x| acting on |ψ> being equal to the inner product of |x> and |ψ>?
Afaik, the RR thm only works as it does because a Hilbert space is isomorphic to its dual.
(That's why a Hilbert space -- the completion of [itex]\Phi[/itex] -- sits in the middle of a Gel'fand triple.)

But we're dealing with a case where [itex]\Phi^\times[/itex] is typically much larger than [itex]\Phi[/itex].

In that case, do the intuitive notions underpinning the bra-ket formalism go out the window, like <x| acting on |ψ> being equal to the inner product of |x> and |ψ>?
Not really -- the notion of the inner product is simply generalized to a larger context that pretty much conforms with Dirac's bra-ket formalism.

[...conjugate space...] Do you mean this?
Yes.

Quote Quote by strangerep
[... delta function potentials...]
I'm not so sure about that. I think that's about when your Hamiltonian is made up of regular functions, not distributions. [...]
Argh. Yes -- I was thinking of something else. I've seen frameworks were one works with a generalized class of Hamiltonian H such that
[tex]
H ~\in~ Lin(\Phi, \Phi')
[/tex]
but this gets tricky since [itex]H^2[/itex] might not be defined everywhere, and one must retreat to [itex]\exp(itH)[/itex] and take a t derivative to get the generator. I vaguely remember that Thirring treats things this way, but I don't have a copy of his book here.
J-P Antoine's stuff on PIP-spaces (partial inner product spaces) is also relevant here. PIP-spaces allow a finer grained treatment than the (rather coarse) RHS -- but I haven't yet studied it deeply.

I've been looking for the book "Dirac Kets, Gamow Vectors, and Gelfand Triplets" by Arno Bohm, which is out of print.
I have a copy, but it doesn't go into the heavy math detail. I think you can find the same material (and in a more modern form) in other papers and books by those authors and those already cited earlier.
strangerep
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Jan10-12, 11:40 PM
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Quote Quote by Fredrik View Post
I don't think you need [a Kreyszig-level book on RHS]. The (extremely) hard part is to get to the point where you understand the spectral theorem for not necessarily bounded normal operators on a Hilbert space.
I found lots of stuff in G+V that could surely have been explained more clearly,
especially the details for handling nuclear spaces and the various theorems.

[...] separable Hilbert spaces also can't handle some quantum field theory stuff.
I think it was Kibble's stuff that Arnold mentioned. But it's easier to get an overview from Haag's textbook, sect II.1.1 on "the representation problem". All the unitarily inequivalent representations that one encounters in advanced QFT can be considered to live in an mostly-intractable nonseparable space.
strangerep
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Jan10-12, 11:47 PM
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Quote Quote by lugita15 View Post
Well, Kreyszig doesn't discuss Lebesgue measure or Lebesgue integration during any of this, but are you saying that what he is doing in taking the limit with respect to the operator norm is equivalent to a Lebesgue-Stieltjes integral? (This conscious omission of any discussion of the Lebesgue integral goes so far that Kreyszig has to define the space L2[a,b] as the completion of C[a,b] with respect to a norm defined by a Riemann integral.)
Try Lax, or Reed & Simon vol 1, as the next step up from Kreyszig.

There must have been a reason [separability] was included as an axiom.
Mainly because nonseparability is just too darned difficult to work with directly.

Once the "distribution-valued inner product" is defined (and I'm still unclear on how exactly this is done),
As with most things distributional, it's done in terms of its action on test functions.

we can make sense of things like <x|x'>=δ(x-x'). So we are generalizing the notion of inner products,
Yes, except that we should probably call it a "partial inner product", because...

so that they can equal not only ordinary numbers but also stranger beasts like δ(0).
No, [itex]\delta(0)[/itex] is still not "defined".
Fredrik
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Jan11-12, 12:58 AM
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Quote Quote by lugita15 View Post
Well, Kreyszig doesn't discuss Lebesgue measure or Lebesgue integration during any of this, but are you saying that what he is doing in taking the limit with respect to the operator norm is equivalent to a Lebesgue-Stieltjes integral? (This conscious omission of any discussion of the Lebesgue integral goes so far that Kreyszig has to define the space L2[a,b] as the completion of C[a,b] with respect to a norm defined by a Riemann integral.)
Kreyszig certainly knows this stuff better than I do, so if he says that you can avoid Lebesgue integrals, then you probably can. I haven't studied all the details of the spectral theorem. I don't even know the exact definition of the integral.

Quote Quote by lugita15 View Post
Just as a regular measure must always equal nonnegative number, must the projection-valued measure always equal a positive operator?
A projection valued measure takes sets to projection operators, and projection operators are positive operators, so yes. But there's also something called positive operator valued measure (POVM). I guess spectral measures are a special case.

Quote Quote by lugita15 View Post
Hmm, Kreyszig just takes a chapter to discuss the spectral theory of unbounded operators. I haven't gotten to that part, so I can't judge whether he does the whole thing or leaves out the more advanced pieces.
I don't know either. I would guess that he states and proves all the relevant theorems, but avoids talking about related topics that the authors of more recent textbooks like to include. They seem to think that the best way to understand the spectral theorem is as a corollary to a theorem about representations of C*-algebras. That's the approach I'm trying to follow, but it's hard as ****. I suspect that Kreyszig's approach completely ignores all of that, and just tries to get to the theorems with as little effort as possible. I've been wondering if I made the wrong choice, but at least I know that if I finish it, I'll understand not only the spectral theorems, but also the algebraic approach to QM.

Quote Quote by lugita15 View Post
That sounds interesting. I wonder, what was the original physical motivation for separability of the Hilbert space? There must have been a reason it was included as an axiom.
I don't think it needs to be motivated by anything other than that separable spaces are easier to work with. (Why not keep things as simple as possible until there is evidence that we have simplified things too much?) I think that von Neumann found that Heisenberg's and Schrödinger's theories are statements about two different Hilbert spaces, both of which are separable. They weren't called "Hilbert spaces" at the time. That term was invented by von Neumann. His definition included the separability axiom. Other mathematicians later dropped it from the definition of Hilbert space.
lugita15
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Jan11-12, 10:47 AM
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Quote Quote by strangerep View Post
The dual and antidual spaces [itex]\Phi^{\times}[/itex] and [itex]\Phi'[/itex] need [itex]\Phi[/itex] in order to make useful sense. For one thing, the action of operators on [itex]\Phi'[/itex] is only defined in terms of their action on [itex]\Phi[/itex]. E.g., think about how one defines the derivative of a delta function: it's a weak derivative -- only defined via its action on test functions.
Cf. http://en.wikipedia.org/wiki/Weak_derivative
Yes, I forgot how dependent distributions are on the space of test functions.
That's the general idea -- though it's a Lebesgue measure as Fredrik mentioned. The [itex]dE_\lambda[/itex] in your example can be expressed more explicitly as a projection-valued measure, i.e.,
[tex]
\def\<{\langle}
\def\>{\rangle}
|E\>\<E| \, dE
[/tex]
where E ranges over the (continuous) range of energies in the spectrum of the self-adjoint operator.
I think I may need to read more about abstract integration with respect to arbitrary measures. I'm only familiar with integration with respect to the Lebesgue measure; that's what is mainly discussed Wheeden and Zygmund.
EDIT: On second thought I think I may be good with abstract integration: if I understand correctly you just have to replace the Lebesgue measure with another measure, and the rest is pretty straightforward. What I'm not familiar with might be more of a notational thing: I'm not used to treating what looks like a differential form, like [tex]
\def\<{\langle}
\def\>{\rangle}
|E\>\<E| \, dE
[/tex], as a measure. I've seen things like [itex]∫fd\mu[/itex], where [itex]\mu[/itex] is the Lebesgue measure, but I thought that was just fancy notation.
Afaik, the RR thm only works as it does because a Hilbert space is isomorphic to its dual.
(That's why a Hilbert space -- the completion of [itex]\Phi[/itex] -- sits in the middle of a Gel'fand triple.)

But we're dealing with a case where [itex]\Phi^\times[/itex] is typically much larger than [itex]\Phi[/itex].
But is [itex]\Phi^\times[/itex] complete with respect to the distribution-valued inner product? If so, can we get a result analogous to the Riesz representation theorem?
Not really -- the notion of the inner product is simply generalized to a larger context that pretty much conforms with Dirac's bra-ket formalism.
I'm a bit confused. Doesn't the fact that the action of <x| on |ψ> is equal to the inner product of |x> and |ψ> for all |ψ> imply that <x| has a Riesz representation? So are you saying other bras in [itex]\Phi'[/itex] do not have a Riesz representation?
Argh. Yes -- I was thinking of something else. I've seen frameworks were one works with a generalized class of Hamiltonian H such that
[tex]
H ~\in~ Lin(\Phi, \Phi')
[/tex]
but this gets tricky since [itex]H^2[/itex] might not be defined everywhere, and one must retreat to [itex]\exp(itH)[/itex] and take a t derivative to get the generator. I vaguely remember that Thirring treats things this way, but I don't have a copy of his book here.
Which book of Thirring are you referring to? This one?
lugita15
#13
Jan11-12, 10:51 AM
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As with most things distributional, it's done in terms of its action on test functions.
Where can I read the details of how the distribution-valued inner product is defined? Is it in one of the papers you told me about?
lugita15
#14
Jan11-12, 11:02 AM
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Quote Quote by Fredrik View Post
I don't know either. I would guess that he states and proves all the relevant theorems, but avoids talking about related topics that the authors of more recent textbooks like to include. They seem to think that the best way to understand the spectral theorem is as a corollary to a theorem about representations of C*-algebras. That's the approach I'm trying to follow, but it's hard as ****. I suspect that Kreyszig's approach completely ignores all of that, and just tries to get to the theorems with as little effort as possible. I've been wondering if I made the wrong choice, but at least I know that if I finish it, I'll understand not only the spectral theorems, but also the algebraic approach to QM.
I just took a peek at the Kreyszig chapter on unbounded linear operators, and it seems relatively straightforward. First he discusses the spectral theory of unitary operators, which is doable because they're bounded. Then, for any self-adjoint operator T, whether bounded or unbounded, he defines the Cayley transform [itex]U=(T-iI)(T+iI)^{-1}[/itex], which is always unitary. So he uses the spectral representation of U to find the spectral representation of T.
Fredrik
#15
Jan11-12, 11:59 AM
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Quote Quote by lugita15 View Post
I just took a peek at the Kreyszig chapter on unbounded linear operators, and it seems relatively straightforward. First he discusses the spectral theory of unitary operators, which is doable because they're bounded. Then, for any self-adjoint operator T, whether bounded or unbounded, he defines the Cayley transform [itex]U=(T-iI)(T+iI)^{-1}[/itex], which is always unitary. So he uses the spectral representation of U to find the spectral representation of T.
That sounds good. I think I might switch to Kreyszig when I'm done with the spectral theorem for bounded normal operators. (I have come too far along this path to just abandon it).
dextercioby
#16
Jan12-12, 01:55 AM
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There's been some history here over this subject. I remember sending people who have questions on this to first make a solid read of the most informative source on RHS I know of: Rafael de la Madrid's PhD thesis, which should still be freely available online. So our OP is kindly invited to perform some further reading.
Fredrik
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Jan12-12, 02:33 AM
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I never looked at his thesis before. The table of contents certainly looks good. Here's a link. http://physics.lamar.edu/rafa/webdis.pdf
strangerep
#18
Jan12-12, 04:11 AM
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Quote Quote by lugita15 View Post
[...] On second thought I think I may be good with abstract integration: if I understand correctly you just have to replace the Lebesgue measure with another measure, and the rest is pretty straightforward. What I'm not familiar with might be more of a notational thing: I'm not used to treating what looks like a differential form, like
[tex]
\def\<{\langle}
\def\>{\rangle}
|E\>\<E| \, dE
[/tex], as a measure. I've seen things like [itex]∫fd\mu[/itex], where [itex]\mu[/itex] is the Lebesgue measure, but I thought that was just fancy notation.
I believe it's still a Lebesgue measure, since one defines it over a [itex]\sigma[/itex]-algebra (iiuc). The only unusual thing is that it's an operator-valued measure. Think of the usual resolution of unity:
[tex]
1 ~=~ \int |E\>\<E| \, dE
[/tex]

But is [itex]\Phi^\times[/itex] complete with respect to the distribution-valued inner product?
That "inner product" is not defined on all pairs of vectors (else we'd have [itex]\delta(0)[/itex]). Normally, one doesn't think of the topology on [itex]\Phi^\times[/itex] in those terms, but only as a weak or weak-* topology (which are defined in terms of the standard norm topology on the small space [itex]\Phi[/itex] via the dual pairing).

If so, can we get a result analogous to the Riesz representation theorem?
I don't think it works that way. Hilbert spaces are very special (completion of [itex]\Phi[/itex] in norm topology, and self-dual). I suspect the RR thm only works because of the self-dual property -- or maybe one could regard it as a proof of the latter.

I'm a bit confused.
Only a bit? That's good. :-)

Doesn't the fact that the action of <x| on |ψ> is equal to the inner product of |x> and |ψ> for all |ψ> imply that <x| has a Riesz representation? [...]
Depends what spaces each lives in.

Which book of Thirring are you referring to? This one?
I think so -- it's been a while and I never really looked into this thoroughly.


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