Rigged Hilbert Spaces — Mathematics Knowledge System

Overview

A rigged Hilbert space (or Gelfand triple) is a triple Phi subset H subset Phi* where H is a Hilbert space, Phi is a dense nuclear space continuously embedded in H, and Phi* is the topological dual of Phi (containing H as a subspace via the Riesz identification). This construction allows continuous spectra and generalised eigenvectors (distributional eigenfunctions) for self-adjoint operators that have no eigenvectors in H itself -- the key example being the position and momentum operators in quantum mechanics, whose 'eigenstates' are delta functions and plane waves.

Intuition

The Hilbert space L^2(R) is too small for position eigenstates (delta functions) and too large in some sense -- it lacks the differentiability properties needed to define operators like d/dx on all elements. The test function space S(R) (Schwartz functions) sits inside L^2(R) and consists of very smooth, rapidly decaying functions. Its dual S*(R) (tempered distributions) extends outward, containing delta functions and plane waves e^{ikx}. The triple S(R) subset L^2(R) subset S*(R) is the standard Gelfand triple for quantum mechanics.

Formal Definition

Definition

A Gelfand triple is a continuous dense embedding Phi -> H -> Phi* where Phi is a nuclear locally convex space, H is a Hilbert space (identified with its dual via Riesz), and Phi* is the strong dual of Phi. The nuclear spectral theorem states: for any self-adjoint operator A on H that maps Phi into Phi continuously, there exist generalised eigenvectors xi in Phi* (with A*xi = lambda*xi in Phi*) forming a complete spectral resolution of A.

\Phi \subset H \subset \Phi^*\quad (\text{Gelfand triple})

Gelfand triple / rigged Hilbert space

\langle \xi, \phi \rangle_{\Phi^* \times \Phi} = (\xi, \phi)_H \text{ for } \xi \in H \subset \Phi^*

Compatibility of pairings

A^* \xi = \lambda \xi,\quad \xi \in \Phi^*\setminus H \text{ (generalised eigenvector)}

Generalised eigenvalue equation

f = \int_\sigma \langle f, \xi_\lambda \rangle\, \xi_\lambda\, d\mu(\lambda),\quad f \in \Phi

Spectral decomposition via generalised eigenvectors

Notation

Notation	Meaning
$\Phi \subset H \subset \Phi^*$	Gelfand triple (rigged Hilbert space)
$\mathcal{S}(\mathbb{R})$	Schwartz space of rapidly decreasing test functions
$\mathcal{S}'(\mathbb{R})$	Tempered distributions, dual of Schwartz space
$\delta_x$	Dirac delta at x, a generalised eigenvector of the position operator

Theorems

Theorem 1: Nuclear Spectral Theorem (Gelfand-Maurin)

Let A: Phi -> Phi be a continuous self-adjoint operator on a Gelfand triple Phi subset H subset Phi*. Then there exists a measure mu on the spectrum sigma(A) and generalised eigenvectors xi_lambda in Phi* for mu-almost every lambda, such that A*xi_lambda = lambda*xi_lambda and every f in Phi admits the expansion f = integral xi_lambda * <xi_lambda, f> dmu(lambda) (convergence in Phi*).

Theorem 2: Dirac's Completeness Relation

For the position operator Q on L^2(R) with Gelfand triple S(R) subset L^2(R) subset S'(R), the generalised eigenvectors delta_x (x in R) satisfy <delta_x, delta_y> = delta(x-y) and the completeness relation integral |x><x| dx = I (as an operator on S(R) extended to S'(R)).

Worked Examples

1
The position operator acts as Q*phi(x) = x*phi(x) for phi in S(R).
2
The adjoint Q* on S'(R): (Q*xi)(phi) = xi(Q*phi) = xi(x*phi) for xi in S'(R).
3
For xi = delta_0: (Q*delta_0)(phi) = delta_0(x*phi) = [x*phi(x)]_{x=0} = 0*phi(0) = 0 = 0*(delta_0(phi)).
$(Q^* \delta_0)(\phi) = \delta_0(x\phi) = 0 = 0 \cdot \delta_0(\phi)$
4
Hence Q*delta_0 = 0*delta_0, confirming delta_0 is a generalised eigenvector with eigenvalue 0.

✓ Answer

delta_0 satisfies Q*delta_0 = 0*delta_0 in S'(R), making it a generalised eigenvector with eigenvalue 0, even though it is not in L^2(R).

Practice Problems

Hardfree response

Explain why the position operator Q on L^2(R) has no eigenvectors in L^2(R) but has generalised eigenvectors in S'(R).

Common Mistakes

Common Mistake

Thinking that elements of Phi* are ordinary vectors in H.

Phi* contains H (via the Riesz map), but also distributions like delta functions that are NOT in H. They are functionals on Phi, not elements of the Hilbert space.

Quiz

The purpose of a Gelfand triple Phi subset H subset Phi* is to:

Replace the Hilbert space with a larger oneProvide generalised eigenvectors for operators with continuous spectrumMake all operators boundedEmbed H into a finite-dimensional space

Historical Background

The Gelfand triple was introduced by Gelfand and Shilov in the 1950s as part of their systematic study of generalised functions (distributions). Maurin (1955) proved the nuclear spectral theorem. The framework provides mathematical rigour to Dirac's bra-ket notation in quantum mechanics, where physicists routinely use delta-function eigenstates of the position operator that lie outside the Hilbert space L^2(R).

1950s
Gelfand and Shilov develop the theory of generalised functions and nuclear spaces
Israel Gelfand, Georgiy Shilov
1955
Maurin proves the nuclear spectral theorem for rigged Hilbert spaces
Krzysztof Maurin
1930
Dirac introduces bra-ket notation and delta-function eigenstates (informal precursor)
Paul Dirac

Summary

A Gelfand triple Phi subset H subset Phi* provides a framework for generalised eigenvectors of self-adjoint operators.
The nuclear spectral theorem guarantees a complete spectral resolution using distributional eigenvectors in Phi*.
The canonical example is S(R) subset L^2(R) subset S'(R), which rigorises Dirac's bra-ket formalism.

References

BookGelfand, I.M. and Vilenkin, N.Ya. Generalised Functions Vol. 4. Academic Press, 1964.
BookBohm, A. Quantum Mechanics: Foundations and Applications. Springer, 1993.
WebsiteWikipedia -- Rigged Hilbert space