Talk:Rigged Hilbert space

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Why is the word "rigged" used in the name? - Gauge 07:23, 4 September 2005 (UTC)

Term "rigged" should be interpreted as "equipped and ready for action", in analogy with the rigging of a sailing ship. - Kongruencja 22:53, 22 November 2005 (UTC)

I assumed it meant "artificially constructed", as in, "this poker game is rigged".  :-) linas 05:06, 23 November 2005 (UTC)

I don't really know what a rigged Hilbert space is. It seems to be some extension of a Hilbert space, but that is all I got from reading this article. A lot of important mathematical concepts are mentioned quite casually in the article, but a true formal definition is missing. When I see the words "a rigged Hilbert space is," or "Formally, a rigged Hilbert space consists of ..." I expect to see a formal mathematical definition, not a bunch of advanced mathematical concepts casually thrown together.

This article seems to have two definitions of a rigged Hilbert space, neither of them formal. It would be better to say something on the order of: "A rigged Hilbert space is a Hilbert space H together with a subspace Φ ... such that:" followed by a formal list of conditions which together are necessary and sufficient. This shouldn't be too hard for somebody reasonably knowledgeable in the subject. It is completely unnecessary and in fact needlessly confusing to define here what is meant by "finer topology" when there is a link to that already. And please don't ramble on in the middle of a formal definition as to why or how this or that concept or theorem in connected. (All that's important and should remain part of the article, but not as part of a definition.) And don't "consider" things or use vague language such as "of some sort" in the middle of a definition, either.

Sorry if this sounds like a rant, but I am a mathematical type person, not a physicist, and I really do have trouble understanding any kind of mathematical concept without a formal definition. Perhaps someone will be so kind as to include a formal definition in this article or put a little box on this article that says "in need of expert attention." (I don't know how to do either one.) But I do know a formal mathematical definition when I see one, and this article doesn't have one, and it really needs one.

130.94.162.64 22:45, 27 October 2005 (UTC)

I don't see a nice way of inserting the definition into the article without restructuring it significantly, and the technical definition is actually there among all the admittedly long winded verbiage. The definition is, a rigged Hilbert space is a triple (H,S,S') of a Hilbert space, a dense subspace (in the strong top), and its dual, viewed as a superset of H by means of Riesz. HTH -Lethe | Talk 08:05, 28 October 2005 (UTC)
The topology on the subspace makes it into a locally convex space.--CSTAR 12:53, 28 October 2005 (UTC)
OK TVS.--CSTAR 12:54, 28 October 2005 (UTC)

I don't get the point.

Let ${\displaystyle \Phi =H}$, and every Hilbert space is canonically self-rigged. So what's special about a rigged Hilbert space that makes it worth a name and definition of its own? 84.160.220.238 16:17, 8 January 2006 (UTC)

Well, it's actually more interesting about H that its elements can represent, say square-integrable holomorphic functions on the unit disc, than just to know about a definition in terms of square-summable sequences. Equally it is interesting to know about the elements of H as generalized functions, not just functions. This concept is capable of expressing that. It pays dividends in spectral theory, for example, where it can be more interesting to know more about eigenvectors than just that they are vectors. They are typically functions or distributions, with smoothness and other properties. Charles Matthews 17:13, 8 January 2006 (UTC)

So the focus is not the Hilbert space H but the dense subspace ${\displaystyle \Phi }$ it admits? 84.160.244.63 20:15, 8 January 2006 (UTC)
Exactly. In the Gelfand-style approach there is the extra 'degree of freedom' this allows (choose Φ to fit the problem). Connected certainly with the ideas of doing representation theory in infinite-dimensional spaces. With a Lie group acting, say in a unitary representation on H, the Lie algebra ought to be some differential operators acting on H, or rather a dense. In simple examples you might be able to get away with a domain for those which was trigonometric polynomials or suchlike. In general you want something a bit bigger and more like test functions, complete therefore with respect to a bunch of seminorms. Charles Matthews 20:26, 8 January 2006 (UTC)

I have a Ph.D. in mathematics (analysis) and I find this article hard to read. I think it simply needs to be longer with longer explanations. I have seen a better explanation in a quantum mechanics book. Such phrases as "Phi 'carries' a finer topology", and "linear functionals on the subspace Phi of type phi -> <v, \phi> for v in H are 'faithfully represented', etc." may cause trouble even for mathematicians. The "Formal definition (Gelfand triple)" section is better, but there should be at least a link with "the adjoint to i'. The assertion that "this isomorphism is not the same and the composition of the inclusion i with its adjoint i^*..." should be backed up with a reference or counterexample. Gsspradlin (talk) 10:29, 24 August 2013 (UTC)

Identification of the Hilbert space and it's dual

I read the identification ${\displaystyle {\mathcal {H}}={\mathcal {H}}^{\ast }}$ several times now, though I am quite sure those are not canonically isomorphic, while still isomorphic as vector spaces. I'd suggest to change the notation to ${\displaystyle {\mathcal {H}}\cong {\mathcal {H}}^{\ast }}$ and drop the confusing/wrong ${\displaystyle H\subseteq \Phi ^{\ast }}$. This may be useful if you care about applications and calculations only (as physicists usually do) but confuses the reader who thinks about the given explanation. --Clebor42 24 May 2014

But they are canonically isomorphic, because ${\displaystyle {\mathcal {H}}}$ has an inner product. Note that the isomorphism is only over the reals. It is complex antilinear if ${\displaystyle {\mathcal {H}}}$ is complex (with Hermitian inner product). So I agree, when you see ${\displaystyle H\subseteq \Phi ^{\ast }}$ it is potentially confusing: it is given by a canonical injection (monomorphism), not an actual inclusion, and furthermore, you have to remember that it is only real-linear, not complex-linear.