# rigged Hilbert space, separable space, domain of CSCO, mapping

by Petro z sela
Tags: domain of csco, mapping, rigged hilbert space, separable space
Mentor
P: 16,690
 Quote by strangerep The Wiki page on "rigged Hilbert space" is sadly minimal. I've been wishing for several years for an expert to flesh it out some more. And the "best" textbook I know of, i.e., Gel'fand & Vilenkin vol 4, is quite difficult for the average physics type, including me. It's also growing a bit old. Cheers.
Hmm, I see. There is indeed more to a rigged Hilbert space than the wiki page shows. In particular, we want $\Phi$ to be reflexive. This implies that my example of the $\ell^2$-space would be wrong, since $\ell^1$ is not reflexive.

I don't have access to Antoine, sadly, but I'll look on the internet for a more complete definition.
P: 1,732
Are you sure? Try Google Scholar for Antoine "beyond hilbert space". You should be able to get a pdf from citeseer, or a ps from some other places.

I'll also dig out some other references and post them shortly.

 I am become the supreme onion, the saddener of worlds
Dunno why you say that. You always make me happier when you drop by...
P: 1,732
 Quote by me I'll also dig out some other references and post them shortly.
"On the Mathematical Basis of the Dirac Formulation of Quantum Mechanics",
IJTP, vol 42, no 10, 2003, p2225.
(If you can't access that, send me a PM... ;-)

"Eigenfunction Expansions and Transformation Theory",
Available as: math/0607548

and of course, there's always the reliable, but less rigorous:

"The role of the rigged Hilbert space in Quantum Mechanics",
Available as: quant-ph/0502053

Ref (3) is more physicist-oriented, but it does give a more intuitive picture of how the small space ##\Omega## is constructed in terms of an initial Hilbert space, one or more unbounded operators with continuous spectra, and the sequence of seminorms (and hence nested Hilbert spaces) induced by the latter.

[And now I need to go over your post #17 again more closely... ]
P: 1,732
 Quote by micromass We must have a map $i:\Phi\rightarrow H$ which is injective and a bounded linear operator.
So this corresponds to what others call the "canonical embedding map" ?
 [...] the injection $i:\ell^1\rightarrow \ell^2$ is bounded with norm (at most) 1.
So this means ##i## is also a Hilbert-Schmidt operator ?
 It turns out that $\ell^\infty$ is complete. However, we can prove that $\ell^\infty$ is not separable. The proof of this is essentially some kind of diagonal argument.
What do you mean by "diagonal argument" ?
P: 1,732
 Quote by micromass So, to my understanding (from wikipedia), we are looking for a couple $(\Phi,H)$, where H is a Hilbert space and $\Phi$ is a dense subset of H. Furthermore, we equip $\Phi$ with a topology (which is unrelated to the subspace topology of H).
In cases of physical interest, one usually performs the construction in a different sequence...

For a quantum theory, we must have a space with (+ve-definite) inner product which admits a probability measure. This is the Hilbert space ##H_0## (possibly after a completion in the norm induced by the inner product). The meaning of the "0" subscript will become clear below.

In a useful physical theory of dynamics, we essentially start from differential equations (of motion), and determine the maximal invariance group of this (system of) differential equation(s). This is the "dynamical group" -- it maps solutions of the dynamics among themselves. Depending on the degree of sophistication in the model, we might be content with restriction to a subgroup, rather than the maximal group.

So for a quantum-dynamical theory, this group must be represented as operators on the Hilbert space [note #1]. One might use group operators (i.e., in exponential form which are bounded), but more usually one finds the generators of the group and attempts to represent these on the Hilbert space. Typically, these generators are unbounded operators with continuous spectrum, hence cannot be defined everywhere on ##H_0##.

Take a typical such generator ##P## (ordinary momentum), represented by (e.g.,) ##-i\hbar\partial_x## on ##H_0##, being a Hilbert space of square-integrable wave functions. Now define
$$H_1 ~:=~ \left\{ \psi\in H_0 : P\psi \in H_0 \right\} ~.$$ Clearly ##H_1 \subset H_0##. Further, if the norm on ##H_0## is denoted ##\|\psi\|_0##, we can define another norm on ##H_1## :
$$\|\psi\|_1 ~:=~ \|\psi\|_0 + \|P\psi\|_0 ~,$$ so that ##\|\psi\|_1 \le \|\psi\|_0##. Clearly, both norms ##\|\cdot\|_0## and ##\|\cdot\|_1## are well-defined on ##H_1## and one can define a new topology ##\tau_1## on ##H_1## via the 1-norm.

Continuing in this way we can define a sequence of nested spaces, e.g.,
$$H_{n+1} ~:=~ \left\{ \psi\in H_n : P\psi \in H_n \right\} ~,$$ and corresponding ##(n+1)##--norms similarly. Thus, one has an infinite sequence of nested spaces with a corresponding sequence of seminorms and topologies. Gel'fand & Vilenkin call this setup a "countably-Hilbert" space. Then, one puts
$$\Omega ~:=~ \bigcap_{n=0}^\infty H_n ~,$$ thus arriving at the "small" space in the triple (which hopefully is not trivially empty).

Anyway,... the point of posting all this is because I'm not sure whether the extra structure implied by the above might affect what you've been saying about duals...

[Edit: I also think this construction guarantees that each canonical embedding map from ##H_{n+1}## into ##H_n## is nuclear (Hilbert-Schmidt). It appears (eg Wiki) that "nuclear" requires only trace-class embedding operators, but the nuclear spectral theorem seems to require Hilbert-Schmidt embedding operators. Can you shed any more light on this?]

[Note #1: Algebraic quantum theory handles this differently: one constructs abstract functionals over a ##C^*##--algebra related to the dynamical group.]
Mentor
P: 16,690
 Quote by strangerep So this corresponds to what others call the "canonical embedding map" ?
Yes.

 So this means ##i## is also a Hilbert-Schmidt operator ?
The only Hilbert-Schmidt operators I ever encountered were of the form $T:H\rightarrow H$ between Hilbert spaces. I don't see how i is of this form. But I'm sure that generalizations exist.

 What do you mean by "diagonal argument" ?
The idea is the following. Assume that there is a countably dense subset
$$\{\mathbf{x}_1,\mathbf{x}_2,...\}$$ of $\ell^\infty$. Then we can write this as

$$\mathbf{x}_1: ~x_1^1 ~ x_1^2~ x_1^3 ~ ...$$
$$\mathbf{x}_2: ~x_2^1 ~ x_2^2~ x_2^3 ~ ...$$
$$\mathbf{x}_3: ~x_3^1 ~ x_3^2~ x_3^3 ~ ...$$
and so on
But consider the following diagonal sequence: $(x_n^n +1)_n$. This is an element of $\ell^\infty$, but it has distance at least 1 from any of the $\mathbf{x}_k$. So our set isn't dense.

 Quote by strangerep In cases of physical interest, one usually performs the construction in a different sequence... For a quantum theory, we must have a space with (+ve-definite) inner product which admits a probability measure. This is the Hilbert space ##H_0## (possibly after a completion in the norm induced by the inner product). The meaning of the "0" subscript will become clear below. In a useful physical theory of dynamics, we essentially start from differential equations (of motion), and determine the maximal invariance group of this (system of) differential equation(s). This is the "dynamical group" -- it maps solutions of the dynamics among themselves. Depending on the degree of sophistication in the model, we might be content with restriction to a subgroup, rather than the maximal group. So for a quantum-dynamical theory, this group must be represented as operators on the Hilbert space [note #1]. One might use group operators (i.e., in exponential form which are bounded), but more usually one finds the generators of the group and attempts to represent these on the Hilbert space. Typically, these generators are unbounded operators with continuous spectrum, hence cannot be defined everywhere on ##H_0##. Take a typical such generator ##P## (ordinary momentum), represented by (e.g.,) ##-i\hbar\partial_x## on ##H_0##, being a Hilbert space of square-integrable wave functions. Now define $$H_1 ~:=~ \left\{ \psi\in H_0 : P\psi \in H_0 \right\} ~.$$ Clearly ##H_1 \subset H_0##. Further, if the norm on ##H_0## is denoted ##\|\psi\|_0##, we can define another norm on ##H_1## : $$\|\psi\|_1 ~:=~ \|\psi\|_0 + \|P\psi\|_0 ~,$$ so that ##\|\psi\|_1 \le \|\psi\|_0##. Clearly, both norms ##\|\cdot\|_0## and ##\|\cdot\|_1## are well-defined on ##H_1## and one can define a new topology ##\tau_1## on ##H_1## via the 1-norm. Continuing in this way we can define a sequence of nested spaces, e.g., $$H_{n+1} ~:=~ \left\{ \psi\in H_n : P\psi \in H_n \right\} ~,$$ and corresponding ##(n+1)##--norms similarly. Thus, one has an infinite sequence of nested spaces with a corresponding sequence of seminorms and topologies. Gel'fand & Vilenkin call this setup a "countably-Hilbert" space. Then, one puts $$\Omega ~:=~ \bigcap_{n=0}^\infty H_n ~,$$ thus arriving at the "small" space in the triple (which hopefully is not trivially empty). Anyway,... the point of posting all this is because I'm not sure whether the extra structure implied by the above might affect what you've been saying about duals... [Edit: I also think this construction guarantees that each canonical embedding map from ##H_{n+1}## into ##H_n## is nuclear (Hilbert-Schmidt). It appears (eg Wiki) that "nuclear" requires only trace-class embedding operators, but the nuclear spectral theorem seems to require Hilbert-Schmidt embedding operators. Can you shed any more light on this?] [Note #1: Algebraic quantum theory handles this differently: one constructs abstract functionals over a ##C^*##--algebra related to the dynamical group.]
I have to admit that this entire post by you is very fascinating. But unfortunately, I am very much out of my comfort zone here, so I can't really help. Gelfand & Vilenkin looks a really nice book, so I will go through it. Maybe I can answer some more after I have read up on the book and the other references you posted.
 Mentor P: 16,690 Gelfand-Vilenkin seem to define a rigged Hilbert space as a nuclear countably Hilbert space $\Phi$ together with a continuous, injective, linear map $T:\Phi\rightarrow H$ into a Hilbert space. However, since $\Phi$ is nuclear countably Hilbert, it is separable. This seems very much nontrivial, but Gelfand and Vilenkin do prove it (in an earlier volume). Furthermore, any nuclear countably Hilbert space is reflexive. This means that $\Phi^\times$ is separable as well. So if we use the definition from Gelfand-Vilenkin, then we can answer (1) and (2) with yes.
P: 1,732
 Quote by micromass [...] So if we use the definition from Gelfand-Vilenkin, then we can answer (1) and (2) with yes.
Cool!

BTW, I'm really impressed that you were able to absorb that much from G+V in such a disgustingly short time.
Mentor
P: 16,690
 Quote by strangerep Edit: I also think this construction guarantees that each canonical embedding map from ##H_{n+1}## into ##H_n## is nuclear (Hilbert-Schmidt). It appears (eg Wiki) that "nuclear" requires only trace-class embedding operators, but the nuclear spectral theorem seems to require Hilbert-Schmidt embedding operators. Can you shed any more light on this?
Could you provide me a reference to the nuclear spectral theorem?

But in any case, your language seems to imply that trace-class is a weaker property that being of Hilbert-Schmidt type. But it is actually a stronger property, isn't it? It seems to me that trace-class operators are always of Hilbert-Schmidt type. So if you say that the spectral theorem requires Hilbert-Schmidt operators, then it holds for nuclear operators as well.
 Sci Advisor HW Helper P: 11,863 Yes, the trace-class property is stronger than Hilbert-Schmidt. From page 61 of R.de la Madrid PhD thesis: Definition A compact operator A = U]A] is called Hilbert-Schmidt if ynC__l/2 n  OD, where the An are the eigenvalues of the operator ]A]. Therefore, an operator is of Hilbert-Schmidt type if[ admits a decomposition of the form (2.5.18) such that the series ynC__l/2 n converges. One can also see that in order an operator A be of Hilbert-Schmidt type, it is necessary and sufficient that the series ynC__l ]]Aen]] 2 converge for at least one orthonormal basis ˝1, ˝2,... in 7-/. An even more restrictive requirement that the operator A be Hilbert-Schmidt is that it be a nuclear operator. Definition A compact operator is called nuclear (or trace class) if ynřC=l/n < O(), where the An are the eigenvalues of the operator IAI appearing in the decomposition A = UIAI. Since the convergence of the series Y]i A} follows from the convergence of Y]i An, every nuclear operator is of Hilbert-Schmidt type.
P: 1,732
 Quote by micromass Could you provide me a reference to the nuclear spectral theorem?
Gelfand & Vilenkin vol 4. They build it up in several versions in section 4.5.

 But in any case, your language seems to imply that trace-class is a weaker property that being of Hilbert-Schmidt type. But it is actually a stronger property, isn't it? It seems to me that trace-class operators are always of Hilbert-Schmidt type. So if you say that the spectral theorem requires Hilbert-Schmidt operators, then it holds for nuclear operators as well.
Aaark! It seems I am closer to the edge of my comfort zone than I realized.