Rigged Hilbert Spaces, First Look

A rigged Hilbert space is a Hilbert space together with a dense space of well-behaved test vectors and a larger space of distributions. It is the clean mathematical setting behind statements such as “the position eigenket $\lvert x\rangle$ is not normalizable, but it can still be used under an integral.”

The practical slogan is:

Normalizable states live in the Hilbert space. Continuous-spectrum kets live on the distribution side of a larger triple.

This page is a first look. The main measurement formalism is still the Spectral Theorem, Practical Version. Rigged Hilbert spaces explain why the corresponding Dirac notation often works so well.

For the practical rules behind individual continuous-spectrum kets, see Generalized Eigenvectors.

Why Quantum Mechanics Uses Them

The Hilbert space $\mathcal H$ is the home of physical pure states with finite norm. For a particle on the line, the standard example is $L^2(\mathbb R)$.

Yet many useful objects in wave mechanics are not elements of $L^2(\mathbb R)$:

• position eigenstates $\lvert x\rangle$;
• momentum plane waves $\lvert p\rangle$;
• delta functions;
• scattering states normalized to delta functions.

These objects are indispensable for calculations, but treating them as ordinary Hilbert-space vectors causes contradictions. Their “norms” are not finite, their inner products involve delta distributions, and they usually represent exact values in a continuous spectrum rather than physically normalizable wave packets.

A rigged Hilbert space separates the roles:

• use $\mathcal H$ for normalizable states and probabilities;
• use a test-vector space $\Phi$ for vectors on which singular operations behave well;
• use a distribution space $\Phi'$ for generalized eigenvectors.

The Gelfand Triple

The basic structure is a Gelfand triple:

$$\Phi \subset \mathcal H \subset \Phi'.$$

Here:

• $\Phi$ is a dense subspace of $\mathcal H$ with a stronger topology than the Hilbert norm;
• $\mathcal H$ is the Hilbert space of square-normalizable states;
• $\Phi'$ is the space of continuous linear functionals on $\Phi$.

The word “topology” matters. It says what convergence means in $\Phi$, and it determines which linear functionals count as continuous distributions. The same Hilbert space can support different useful riggings depending on the operators and representations under study.

With the site’s inner-product convention, $\langle\phi\vert\psi\rangle$ is conjugate-linear in $\phi$ and linear in $\psi$. Thus an ordinary Hilbert-space vector $\psi\in\mathcal H$ defines a linear functional on $\Phi$ by

$$\eta_\psi(\varphi) = \langle\psi\vert\varphi\rangle_{\mathcal H}, \qquad \varphi\in\Phi.$$

This embeds $\mathcal H$ into $\Phi'$. Some mathematical texts use the opposite inner-product convention and write an antidual, often denoted $\Phi^\times$, instead of the linear dual. The underlying idea is the same: Hilbert-space vectors sit inside a larger distributional space.

Standard Example on the Line

For a particle on the real line, the standard example is

$$\mathcal S(\mathbb R) \subset L^2(\mathbb R) \subset \mathcal S'(\mathbb R).$$

Here $\mathcal S(\mathbb R)$ is the Schwartz space: smooth functions whose derivatives decay faster than any power. One way to express this is that, for all nonnegative integers $m,n$,

$$p_{m,n}(\varphi) = \sup_{x\in\mathbb R} \lvert x^m\partial_x^n\varphi(x)\rvert < \infty.$$

The space $\mathcal S'(\mathbb R)$ is the space of tempered distributions: continuous linear functionals on $\mathcal S(\mathbb R)$. It contains ordinary square-integrable functions, but it also contains delta distributions and plane waves.

This triple is especially useful because the Fourier transform maps $\mathcal S(\mathbb R)$ to itself and extends naturally to $\mathcal S'(\mathbb R)$. That is why it fits position and momentum representations so well.

Test Functions

Test functions are the vectors on which singular or limiting expressions are tested. They are chosen to be nice enough that distributional pairings, derivatives, Fourier transforms, and integrations by parts make sense.

In the Schwartz triple, a test function $\varphi\in\mathcal S(\mathbb R)$ is not merely square-integrable. It is smooth and rapidly decreasing. This extra regularity is stronger than physical normalizability, but it gives a controlled core for calculations.

The point is not that every physical state must be a Schwartz function. The point is that many formal identities are first honest on $\Phi$, then extended to larger classes of states through Hilbert-space completion or spectral theory.

Distributions

A distribution $T\in\Phi'$ is a continuous linear map

$$T:\Phi\to\mathbb C.$$

Linearity means

$$T(a\varphi+b\chi) = aT(\varphi)+bT(\chi).$$

Continuity is the condition that rules out pathological linear functionals and makes limits compatible with the action of $T$.

The delta distribution at $x_0$ is the evaluation functional

$$\delta_{x_0}(\varphi) = \varphi(x_0).$$

This is not a vector in $L^2(\mathbb R)$. It is a distribution acting on test functions.

For the standalone Fourier-analysis introduction to this language, see Distributions.

Similarly, a momentum plane wave defines a tempered distribution by

$$F_p(\varphi) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar}\varphi(x)\,dx.$$

This integral is well behaved for Schwartz test functions. It is the mathematical object behind the formal bra $\langle p\vert$.

Position and Momentum Kets

In Dirac notation, physicists write

$$\langle x_0\vert\psi\rangle = \psi(x_0).$$

The rigged-Hilbert-space reading is that $\langle x_0\vert$ is an evaluation distribution on the test space. For $\varphi\in\mathcal S(\mathbb R)$,

$$\langle x_0\vert\varphi\rangle = \delta_{x_0}(\varphi) = \varphi(x_0).$$

It satisfies the generalized eigenvalue relation

$$\langle x_0\vert X\varphi\rangle = x_0\langle x_0\vert\varphi\rangle,$$

where $(X\varphi)(x)=x\varphi(x)$.

For momentum, define

$$\langle p\vert\varphi\rangle = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar}\varphi(x)\,dx.$$

With $P=-i\hbar\,d/dx$, integration by parts gives

$$\langle p\vert P\varphi\rangle = p\langle p\vert\varphi\rangle.$$

Thus $\langle p\vert$ is a generalized momentum eigenbra. The corresponding ket $\lvert p\rangle$ is the Dirac notation for the same distributional spectral label, not a finite-norm vector in $\mathcal H$.

Why This Makes Dirac Notation More Honest

The formal completeness relation

$$I = \int_{\mathbb R} \lvert x\rangle\langle x\rvert\,dx$$

should not be read as an ordinary sum of projectors onto Hilbert-space vectors. On suitable test vectors, it means that inner products can be reconstructed from the position representation:

$$\langle\chi\vert\varphi\rangle = \int_{\mathbb R} \langle\chi\vert x\rangle \langle x\vert\varphi\rangle\,dx.$$

The same idea holds in momentum representation:

$$\langle\chi\vert\varphi\rangle = \int_{\mathbb R} \langle\chi\vert p\rangle \langle p\vert\varphi\rangle\,dp.$$

These equations are the continuous analogues of expanding a vector in an orthonormal basis. The difference is that the labels $x$ and $p$ correspond to generalized eigenvectors in $\Phi'$, not to countably many normalizable basis vectors in $\mathcal H$.

This is the disciplined meaning of expressions such as

$$\langle x\vert x'\rangle = \delta(x-x'), \qquad \langle p\vert p'\rangle = \delta(p-p').$$

The delta function is not a finite inner product. It is the distributional kernel that makes the corresponding integral transform work.

Relation to the Spectral Theorem

Rigged Hilbert spaces do not replace the spectral theorem. For observables, the rigorous object that assigns probabilities is still the projection-valued measure $E_A(\Delta)$ of a self-adjoint operator $A$:

$$\Pr(A\in\Delta) = \langle\psi\vert E_A(\Delta)\psi\rangle.$$

The rigged-Hilbert-space viewpoint supplements this by giving a controlled setting for generalized eigenvectors. In favorable cases, one can regard the formal kets $\lvert\lambda\rangle$ as elements of $\Phi'$ satisfying

$$A\lvert\lambda\rangle = \lambda\lvert\lambda\rangle$$

in a distributional sense.

The phrase “in favorable cases” is important. The choice of $\Phi$ must be compatible with the operator and the representation. One does not prove self-adjointness, domains, or spectral properties merely by writing a Gelfand triple.

Worked Example: Evaluation Is Not an L2 Functional

The position distribution $\langle x_0\vert$ evaluates a function at a point. This is continuous on $\mathcal S(\mathbb R)$, but it is not a continuous functional on $L^2(\mathbb R)$ with the $L^2$ norm.

To see why, choose a smooth function $f$ with $f(0)\ne0$ and define

$$f_n(x) = \sqrt n\,f(nx).$$

Its $L^2$ norm is independent of $n$:

$$\lVert f_n\rVert_2^2 = \int_{-\infty}^{\infty} n\lvert f(nx)\rvert^2\,dx = \int_{-\infty}^{\infty} \lvert f(u)\rvert^2\,du.$$

But its value at the origin grows:

$$f_n(0) = \sqrt n\,f(0).$$

So no constant $C$ can make

$$\lvert f(0)\rvert \le C\lVert f\rVert_2$$

hold for all test functions. Point evaluation is not controlled by the Hilbert norm alone. This is exactly why the test-space topology is part of the rigged Hilbert space.

What It Does Not Mean

A rigged Hilbert space does not mean that delta functions become physical pure states. Physical wavefunctions are still normalized in $\mathcal H$, and probabilities are still assigned to spectral sets, not to exact points in an absolutely continuous spectrum.

It also does not mean that all formal eigenfunction expansions are automatically valid. Theorems such as the Gelfand–Maurin nuclear spectral theorem require hypotheses on the test space and operator. For most practical quantum mechanics, the safe hierarchy is:

1. establish the Hilbert space and domains;
2. identify self-adjoint observables;
3. use the spectral theorem for probabilities and time evolution;
4. use generalized kets as distributional notation when the rigging supports them.

Common Mistakes

• Treating $\delta(x-x_0)$ as a square-integrable wavefunction.
• Saying $\lvert x\rangle$ is a Hilbert-space basis vector.
• Interpreting $\langle x\vert x\rangle$ as an infinite physical norm rather than as failed Hilbert-space language.
• Forgetting that $\Phi$ has extra topology, not merely extra smooth elements.
• Assuming there is a unique rigged Hilbert space for every quantum system.
• Using generalized eigenvectors to avoid checking self-adjointness or domains.

Cross-Links

• Hilbert Spaces
• Continuous Spectra
• Generalized Eigenvectors
• Spectral Theorem, Practical Version
• Position and Momentum Representations
• Delta Function
• Distributions
• Fourier Transform
• Wavefunctions as Representations
• Discrete and Continuous Spectra
• Scattering Amplitude

References

• I. M. Gel’fand and N. Ya. Vilenkin, Generalized Functions, Volume 4: Applications of Harmonic Analysis, Academic Press, 1964.
• K. Maurin, Generalized Eigenfunction Expansions and Unitary Representations of Topological Groups, Polish Scientific Publishers, 1968.
• A. Bohm and M. Gadella, Dirac Kets, Gamow Vectors and Gelfand Triplets, Springer, 1989.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, Academic Press, 1980.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.

Exercises

1. In the triple $\mathcal S(\mathbb R)\subset L^2(\mathbb R)\subset\mathcal S'(\mathbb R)$, classify each object as a test vector, Hilbert-space vector, or distribution: a normalized Gaussian, the delta distribution $\delta_{x_0}$, and the plane wave $e^{ipx/\hbar}$.

Solution

A normalized Gaussian is a Schwartz function, hence also an $L^2$ vector. The delta distribution $\delta_{x_0}$ is in $\mathcal S'(\mathbb R)$ but not in $L^2(\mathbb R)$. The plane wave is not square-integrable on the full line, but it defines a tempered distribution by integration against Schwartz test functions.

2. Show that the delta distribution is a generalized eigenvector of the position operator.

Solution

For $\varphi\in\mathcal S(\mathbb R)$,

$$\langle x_0\vert X\varphi\rangle = (X\varphi)(x_0) = x_0\varphi(x_0) = x_0\langle x_0\vert\varphi\rangle.$$

So $\langle x_0\vert$ satisfies the eigenvalue equation distributionally.

3. Verify the generalized momentum eigenvalue equation for plane waves.

Solution

For $\varphi\in\mathcal S(\mathbb R)$,

$$\langle p\vert P\varphi\rangle = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar} \left(-i\hbar\frac{d\varphi}{dx}\right)\,dx.$$

Integrating by parts gives no boundary term because $\varphi$ is rapidly decreasing. Since

$$\frac{d}{dx}e^{-ipx/\hbar} = -\frac{ip}{\hbar}e^{-ipx/\hbar},$$

one obtains

$$\langle p\vert P\varphi\rangle = p\langle p\vert\varphi\rangle.$$

4. Why is the formal equation $I=\int\lvert x\rangle\langle x\rvert\,dx$ not an ordinary Hilbert-space sum of projectors?

Solution

The labels $x$ refer to generalized eigenvectors, not normalizable Hilbert-space vectors. The expression is shorthand for how the position representation reconstructs inner products and wavefunctions:

$$\langle\chi\vert\varphi\rangle = \int_{\mathbb R} \chi(x)^*\varphi(x)\,dx.$$

The rigorous measurement object is the position spectral measure, whose projector onto a set $\Delta$ multiplies by $\mathbf 1_\Delta(x)$.