Gelfand Triples

A Gelfand triple equips a Hilbert space with a dense test space carrying a finer topology and a continuous dual or antidual containing generalized vectors. The extra topology makes functionals such as point evaluation continuous even when they are not bounded in the Hilbert norm. It also supplies a precise meaning for extending an operator to such functionals; it does not make them normalizable states.

Required background. Distribution Theory Minimum distinguishes test-function continuity from Hilbert-norm continuity. Dense Domains explains why an operator includes its domain. The first look gives the elementary motivation.

A rigging begins with a continuous dense inclusion

Let $\Phi$ be a locally convex complex vector space and let $j:\Phi\to\mathcal H$ be a continuous injective linear map with dense range. Usually $\Phi$ is identified with its image. Its topology must be specified, not inferred from the set of vectors. For many spectral applications one chooses a nuclear Fréchet space; nuclearity is additional structure, not a consequence of density.

Write $\Phi^\times$ for the continuous antilinear functionals on $\Phi$. An ordinary vector $\psi\in\mathcal H$ defines

$$(J\psi)(\phi)=\langle j\phi,\psi\rangle, \qquad \phi\in\Phi.$$

The map $J$ is linear in $\psi$. Cauchy–Schwarz gives

$$|(J\psi)(\phi)|\le\|j\phi\|\,\|\psi\|,$$

so the continuity of $j$ makes $J\psi$ a continuous functional on $\Phi$. Density proves injectivity: if this pairing vanishes for every $\phi$, then $\psi=0$. This constructs the chain

$$\Phi\xrightarrow{\ j\ }\mathcal H \xrightarrow{\ J\ }\Phi^\times.$$

For the weak antidual topology, continuity of $J$ follows from each evaluation bound. It also holds for the topology of uniform convergence on bounded subsets of $\Phi$, because a continuous $j$ maps each such set to a norm-bounded set in $\mathcal H$.

The linear dual describes bras with a different embedding

The linear dual $\Phi'$ is equally useful. The functional

$$B_\psi(\phi)=\langle\psi,j\phi\rangle$$

is linear in $\phi$, but the map $\psi\mapsto B_\psi$ is antilinear. Thus the notation $\mathcal H\subset\Phi'$ needs this qualification under the conjugate-linear-first inner product convention. One can instead linearly embed the conjugate Hilbert space into $\Phi'$.

Both conventions describe the same pairing structure. They should not be mixed inside a calculation. A generalized ket is an antilinear functional of the test vector $\phi$, whereas its associated bra is a linear functional of that vector. For example, at a real position $x_0$,

$$\langle x_0|\phi\rangle=\phi(x_0), \qquad \langle\phi|x_0\rangle=\overline{\phi(x_0)}.$$

These are a linear functional and an antilinear functional, respectively; neither is evaluation on an arbitrary $L^2$ equivalence class.

The Schwartz rigging supports position and momentum

Take $\Phi=\mathcal S(\mathbb R)$ with its Schwartz seminorms and $\mathcal H=L^2(\mathbb R,dx)$. The inclusion is continuous and dense. Multiplication $X\phi=x\phi$ and differentiation $P\phi=-i\hbar\phi'$ map $\mathcal S$ continuously into itself. Their polynomial combinations can therefore be applied repeatedly on this common test space.

This invariant space is especially useful for the canonical commutation relation and Fourier transformation. It does not replace the larger self-adjoint domains $D(X)$ and $D(P)$. The distinction between an invariant test space and a full operator domain is developed in Commutation Relations on Domains.

A different Hamiltonian may require a different rigging. For example, a nonsmooth potential need not preserve $\mathcal S$. Choosing this space solely because it works for a free particle does not establish invariance for the interacting model.

Extending a symmetric operator to generalized kets

Suppose $A$ is symmetric on $\mathcal H$, $\Phi\subset D(A)$, and $A\Phi\subset\Phi$ continuously. Define

$$(A^\times F)(\phi)=F(A\phi), \qquad F\in\Phi^\times.$$

Continuity of $A$ ensures that $A^\times F$ remains in the antidual. Symmetry ensures consistency with Hilbert-space action: for $\psi\in D(A)$,

$$A^\times J\psi=J(A\psi).$$

For a general nonsymmetric operator, an extension agreeing with its action uses its adjoint on the test side and requires the corresponding invariance conditions. The transpose of an arbitrary operator should not silently be identified with its Hilbert adjoint.

A generalized eigenvector satisfies $A^\times F=\lambda F$. Existence of a complete eigenfunction expansion requires an appropriate spectral theorem and compatible rigging. The triple alone proves no completeness claim, and it need not produce useful eigenfunctionals for every selected spectral value. See Generalized Eigenfunction Expansions.

Exercise: three different convergence tests

Use dimensionless $x$ and set $\psi_n(x)=\pi^{-1/4}e^{-x^2/2}e^{inx}$. Show that $J\psi_n$ tends to zero in the weak antidual topology, although $\psi_n$ tends to zero neither in Hilbert norm nor in the Schwartz topology.

Solution

For each Schwartz $\phi$, $\langle\phi,\psi_n\rangle$ is the Fourier transform at increasing frequency of the integrable function $\pi^{-1/4}\overline{\phi(x)}e^{-x^2/2}$. It tends to zero by the Riemann–Lebesgue lemma. But $\|\psi_n\|_2=1$, and $|\psi_n'(0)|=n\pi^{-1/4}$ makes the Schwartz derivative seminorm unbounded. Weak convergence of the embedded functionals does not imply convergence in either stronger topology.

References

• de la Madrid, Rafael, “The role of the rigged Hilbert space in quantum mechanics,” European Journal of Physics 26, 287–312 (2005), doi:10.1088/0143-0807/26/2/008, arXiv:quant-ph/0502053 — explicit dual and antidual constructions for quantum models.
• Gel’fand, I. M., and N. Ya. Vilenkin, Generalized Functions, Vol. 4: Applications of Harmonic Analysis, translated by Amiel Feinstein, Academic Press, 1964, publisher edition — nuclear spaces and spectral analysis in rigged Hilbert spaces.
• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — Hilbert-space domains and spectral representations.