What good are rigged hilbert spaces?

1. Apr 1, 2013

jostpuur

I have seen the definition of a rigged Hilbert space many times, but I have never seen rigged Hilbert spaces actually being used for anything. Like for proving something. Has anyone else ever seen rigged Hilbert spaces being used for proving something?

2. Apr 1, 2013

MathematicalPhysicist

It works in the background, did you read the motivation for it in Wiki?

3. Apr 1, 2013

jostpuur

I am mostly motivated by proving stuff.

4. Apr 1, 2013

5. Apr 3, 2013

strangerep

It's a framework for working with unbounded operators and continuous spectra. RHS is not absolutely essential (afaict), and many things can be done while remaining strictly inHilbert space (and I've noticed mathematicans tend to prefer that -- see Reed & Simon for an extensive treatise). But the Dirac bra-ket formalism used in physics sits more naturally (and rigorously) in rigged Hilbert space. Both formalisms have spectral theorems, which is one of the most important features in any quantum theory.

But did you mean "proving stuff" in quantum theory, or more generally?

6. Apr 3, 2013

strangerep

It's a bit more than that. Sobolev started working with spaces more general than Hilbert spaces, since the latter seemed not quite suitable when working with certain classes of differential equation. Then came Laurent Schwartz and his theory of distributions (which is essentially a prototype of RHS) in which Fourier transforms find a nice home. There's also a more general theory called "Gel'fand transforms".

7. Apr 4, 2013

dextercioby

Mathematicians didn't exploit the work by Gel'fand, Kostyuchenko and later Berezanskii and Maurin to use the concept any further. Basically the mathematical reasearch into this subject stops at the end of the 1960's. Main result: the spectral theorem for unitary and self-adjoint operators on a Hilbert space.

Fortunately the physicists took over and now we have a pretty decent chapter in the mathematical foundations of quantum theories.

8. Apr 4, 2013

jostpuur

When new students are confused about the delta function formulation of the Fourier transforms, they are often told: "You can make this delta function rigorous with rigged Hilbert spaces."

However, I have never seen the anyone actually proving that Fourier inverse transform works, by somehow using the definition of the rigged Hilbert spaces. If you actually look some proof of the Fourier inverse transform, it has never anything to do with rigged Hilbert spaces. The proof is something completely different.

Doesn't this mean, that the advice given to new students is actually false? The advice implies, that if the delta function can be made rigorous, then also the formal delta function proof of the Fourier inverse transform could be made rigorous. But the truth is that the formal computation is never made rigorous with rigged Hilbert spaces.

9. Apr 7, 2013

strangerep

It would be better if they were told: "this delta function is not really a function -- it's a distribution". Like I said above: the framework involved in Schwartz's theory of distributions is an example of rigged Hilbert space.

Please give reference(s) to the proofs you have in mind. (It's impossible to have a constructive discussion in a vacuum.)