# Isomorphism of rigged Hilbert spaces

In connection with the statement that QM can be formulated in terms of separable complex (rigged) Hilbert spaces, the fact that all infinite dimensional separable complex Hilbert spaces are isomorphic to one another is sometimes emphasized.

I understand why some authors like to add "(rigged)" in order hint how to make this slightly simplified statement more rigorous, but is this modification also compatible the "additional hint" that all "physically relevant" Hilbert spaces are isomorphic? For a "physically relevant" rigged Hilbert space, the ket space has to be nuclear:

A nuclear space is a topological vector space with a topology defined by a family of Hilbert seminorms, such that for any Hilbert seminorm $p$ we can find a larger Hilbert seminorm $q$ so that the natural map from $V_q$ to $V_p$ is trace class.

So I have some questions:

1. I have the impression that rigged Hilbert spaces don't add anything for finite dimensional Hilbert spaces, especially all finite dimensional rigged Hilbert spaces of the same dimension should be isomorphic. Is this correct?
2. Are there simple examples of non-isomorphic infinite dimensional separable rigged Hilbert spaces where the ket space is nuclear?
3. Are the most common examples of the $\mathbb R$- and $\mathbb R^3$-Schwartz-space (i.e. $\mathscr{S}(\mathbb{R}) \subset L^2(\mathbb{R}) \subset \mathscr{S'}(\mathbb{R})$ and $\mathscr{S}(\mathbb{R}^3) \subset L^2(\mathbb{R}^3) \subset \mathscr{S'}(\mathbb{R}^3)$) examples of non-isomorphic rigged Hilbert spaces?
4. Is the $\mathbb R^3$-Schwartz-space at least isomorphic to a subspace of the $\mathbb R$-Schwartz-space?
• Keep in mind that neither the Schwarz space $\mathscr{S}$ nor its dual $\mathscr{S}'$ are Hilbert spaces. – yuggib Feb 13 '15 at 10:15
• @yuggib What I mean is the rigged Hilbert space given by $\mathscr{S}(\mathbb{R}) \subset L^2(\mathbb{R}) \subset \mathscr{S'}(\mathbb{R})$. I edited the question to clarify this. I also added a link to "rigged Hilbert space" for readers who are unfamiliar with the definition of rigged Hilbert space in terms of a sandwich $S\subset H \subset S'$. – Thomas Klimpel Feb 13 '15 at 10:52
• 1. The RHS is relevant for continuous spectrum only, so it is relevant for infinite dimensions only; 2. I'd need to investigate on this; 3. My feeling is that they are isomorphic: the $L^2(\mathbb R^n)$ spaces surely are, as they are all separable; moreover $\mathscr S(\mathbb R^n)$ must be infinite-dimensional and at most separable, so that it should be possible to construct isomorphisms between them (but I haven't checked this!) 4. Related to the previous point. – Phoenix87 Feb 13 '15 at 11:23
• I know what the rigging means...but I don't clearly understand your question. The isomorphism of Hilbert spaces is due, in simple terms, to the fact that for each separable Hilbert space it is possible to choose a countable orthonormal basis. This does not give, however any information on the topology you choose on dense sets. – yuggib Feb 13 '15 at 11:29
• @Phoenix87 You are right, the Schwartz spaces are indeed isomorphic. Yurii Savchuk provided a proof that the obvious candidate mapping is an isomorphism. I added a community wiki answer with a link and a short summary. – Thomas Klimpel Feb 16 '15 at 23:27

To specify a Gel'fand triple $(\Phi^*,\mathscr{H},\Phi)$ it is sufficient to specify the Hilbert space $\mathscr{H}$ and the topological vector space $\Phi\subset \mathscr{H}$. The necessary requirement is that the imbedding of $\Phi$ into $\mathscr{H}$ is continuous with respect to the topology of $\Phi$, so this gives the imbedding of $\mathscr{H}^*=\mathscr{H}\subset \Phi^*$.

Therefore, we consider the triple $(\mathscr{H},\Phi,\mathscr{T})$, where $\mathscr{T}$ is the topology of $\Phi$, as defining the Gel'fand triple $(\Phi^*,\mathscr{H},\Phi)$.

Consider the class of Gel'fand triples; we may define a morphism $f$ between objects of the class $(\mathscr{H},\Phi,\mathscr{T})$ and $(\mathscr{K},\Psi,\mathscr{R})$ as following:

• $f(\mathscr{H},\Phi,\mathscr{T})$ is a triple of sets $(A,B,C)$;
• $f\rvert_{Hilb}:\mathscr{H}\to \mathscr{K}$ is a morphism of Hilbert spaces, $f\lvert_{Hilb}(\mathscr{H})=A$;
• $f\rvert_{Top}:(\Phi,\mathscr{T})\to (\Psi,\mathscr{R})$ is a morphism of topological vector spaces, $f\lvert_{Top}(\Phi,\mathscr{T})=(B,C)$;
• $(\Psi,\mathscr{R})$ is continuously imbedded in $\mathscr{K}$ (i.e. $(\mathscr{K},\Psi,\mathscr{R})$ is a triple).

The map is an isomorphism if each of the reduced maps is an isomorphism (i.e. if it is one-to-one and $A=\mathscr{K}$, $B=\Phi$, $C=\mathscr{T}$). [The inverse map $f^{-1}=(\,f\rvert_{Hilb}^{-1}\,,\,f\rvert_{Top}^{-1}\,)$]

Obviously there are non-isomorphic triples, simply choose a triple $(\mathscr{H},\Phi,\mathscr{T}_1)$ and the triple $(\mathscr{H},\Phi,\mathscr{T}_2)$ with $\mathscr{T}_1\subset \mathscr{T}_2$ (i.e. $\mathscr{T}_2$ a topology strictly finer than $\mathscr{T}_1$) and $(\Phi,\mathscr{T}_1)$, $(\Phi,\mathscr{T}_2)$ not homeomorphic (I suppose that e.g. different cardinality of $\mathscr{T}_1$ and $\mathscr{T}_2$ would do the trick).

Given $L^2(\mathbb{R}^d)$ and the spaces $\mathscr{S}(\mathbb{R}^d)$ (rapid decrease smooth functions) and $\mathscr{D}(\mathbb{R}^d)$ (compactly supported smooth functions) we have that $(\mathscr{S}',L^2,\mathscr{S})$ and $(\mathscr{D}',L^2,\mathscr{D})$ are not isomorphic w.r.t. the definition above ($\mathscr{D}$ and $\mathscr{S}$ are not homeomorphic, because $\mathscr{S}$ is metrizable while $\mathscr{D}$ is not).

• your definition of isomorphism doesn't really convince me, as it seems that it lacks some "symmetry" between both sides. – Phoenix87 Feb 13 '15 at 14:58
• @Phoenix87 clarified the definition of the morphism a little bit ;-) – yuggib Feb 13 '15 at 15:19
• "Obviously there are non-isomorphic triples, simply choose a triple $(\mathscr{H},\Phi,\mathscr{T}_1)$ and the triple $(\mathscr{H},\Phi,\mathscr{T}_2)$ with $\mathscr{T}_1\subset \mathscr{T}_2$ (i.e. $\mathscr{T}_2$ a topology strictly finer than $\mathscr{T}_1$)." I agree that I can choose such triples (your example below), but are you sure that this is enough to ensure that the triples are non-isomorphic? If yes, can you explain to me why you are sure, such that I can also be sure? – Thomas Klimpel Feb 13 '15 at 16:09
• It would be nice if you could support your arguments with some references :) – Phoenix87 Feb 13 '15 at 22:27
• @Phoenix87 It would be very nice indeed...also to be sure I am not saying something wrong ;-) unluckily I don't know references on these triples, apart from the Gel'fand book on generalized eigenfunctions. I know they are still used to some extent in spectral theory to study the properties of continuous spectrum (someone in my department told me so) but I am not expert of the field. – yuggib Feb 14 '15 at 10:04

The answers to item 3. and 4. have been given without proof by Phoenix87 in the comments, namely that the $\mathbb R$- and $\mathbb R^3$-Schwartz-space are isomorphic as rigged Hilbert spaces. A nice proof of this fact is given in this answer to a simplified versions of item 3. and 4. from this question. It is the "obvious" candidate isomorphism, which maps the Hermite functions basis of $\mathscr{S}(\mathbb{R}^3)$ to the Hermite functions basis of $\mathscr{S}(\mathbb{R})$.
It seems that the question has been answered completely. But what about the motivation behind this question: "..., but is this modification also compatible the "additional hint" that all "physically relevant" Hilbert spaces are isomorphic?" As rigged Hilbert space, $\mathscr{D}(\mathbb{R}^d)$ is not really "physically relevant". My impression is that indeed many physically relevant rigged Hilbert spaces are isomorphic, i.e. the "additional hint" is not necessarily wrong.