Constructing Domains for Unbounded Operators

An unbounded operator has no finite uniform bound on its output norm over unit vectors in its domain. Every individual output is still a Hilbert-space vector with finite norm. The difficulty is the absence of a uniform continuity estimate, which makes the domain and the treatment of limits essential.

Helpful background. Bounded Operators explains the continuity estimate that fails here. Self-Adjoint Operators supplies the distinction between an unbounded action and a quantum observable.

Unboundedness is a sequence test

A linear operator $A:D(A)\subseteq\mathcal H\to\mathcal H$ is unbounded precisely when there are unit vectors $\psi_n\in D(A)$ such that $\|A\psi_n\|\to\infty$. No vector with an “infinite output” is allowed into its domain.

For multiplication by position on $L^2(\mathbb R)$, take

$$(Q\psi)(x)=x\psi(x), \qquad D(Q)=\{\psi\in L^2:x\psi\in L^2\}.$$

Choose a normalized smooth function $g$ supported in $(0,1)$ and set $g_n(x)=g(x-n)$. Then $\|g_n\|=1$, while $\|Qg_n\|\geq n$. Translation of the packet proves unboundedness without any singular wavefunction.

Momentum admits a different sequence. Let $P=-i\hbar\,d/dx$ on $H^1(\mathbb R)$ and $\psi_k(x)=e^{ikx}g(x)$. Then

$$P\psi_k=e^{ikx}(\hbar k g-i\hbar g'), \qquad \|P\psi_k\|\geq\hbar|k|-\hbar\|g'\|.$$

The packet remains spatially localized, but its rapid oscillations make the momentum norm large. Position and momentum therefore probe different kinds of regularity.

Maximal domains and physical moments

A real measurable function $a$, finite almost everywhere, defines a maximal multiplication operator

$$D(M_a)=\{\psi:a\psi\in L^2\}, \qquad M_a\psi=a\psi.$$

The maximal-domain condition states exactly what is needed for the proposed output to exist in the Hilbert space. For position it is equivalent to a finite second position moment. For a self-adjoint $A$ with spectral probability measure $\mu_\psi$, the corresponding general statement is

$$\psi\in D(A) \quad\Longleftrightarrow\quad \int_{\mathbb R}\lambda^2\,d\mu_\psi(\lambda)<\infty.$$

A finite absolute first moment requires only $\psi\in D(|A|^{1/2})$. The integral defining the mean can therefore exist even when the vector expression $A\psi$ does not. Quadratic Forms explains how to handle such expectations without applying $A$.

For differentiation, the correct maximal domain is stated using weak derivatives. Momentum on the whole line has domain $H^1(\mathbb R)$, not merely the class of pointwise differentiable functions. The free Hamiltonian has domain $H^2(\mathbb R)$ in the absence of a potential. Boundary conditions become additional domain data on intervals and half-lines.

Why closed operators cannot act everywhere

An operator is closed when convergence of both inputs and outputs preserves the graph. The closed graph theorem says that an everywhere-defined closed linear map between Banach spaces is bounded. Consequently an unbounded closed quantum operator must have a proper domain.

The word “closed” cannot be omitted. Algebraically defined everywhere-domain discontinuous linear maps can exist, but they do not provide the stable limiting behavior needed for the spectral theory of observables. In particular, self-adjoint operators are closed, so an unbounded self-adjoint observable always has a proper domain.

This does not imply that states outside the operator domain are forbidden physical states. Their spectral probabilities and unitary time evolution still exist. What fails is the particular operator application, moment, or strong derivative being requested.

A diagonal model for domain calculations

On $\ell^2(\mathbb N)$ define

$$(Ac)_n=n c_n, \qquad D(A)=\left\{c:\sum_{n=1}^{\infty}n^2|c_n|^2<\infty\right\}.$$

Finite sequences form a dense subspace of this domain, and $Ae_n=ne_n$ proves unboundedness. The vector with coefficients $c_n=n^{-1}$ is in $\ell^2$ but outside $D(A)$. Its finite truncations converge in $\ell^2$, while their images do not converge in $\ell^2$. Thus Hilbert-space convergence by itself does not justify passing $A$ through a limit.

The square has a smaller domain:

$$D(A^2)=\left\{c:\sum_{n=1}^{\infty}n^4|c_n|^2<\infty\right\}.$$

For example $c_n=n^{-2}$ belongs to $D(A)$ but not to $D(A^2)$. The formal expression $\langle c,A^2c\rangle$ is undefined, although $\|Ac\|^2$ is finite and represents the second spectral moment. This distinction reappears in energy variances and uncertainty calculations.

Exercises

Use the diagonal model to find $c^{(N)}\to0$ in Hilbert norm for which $Ac^{(N)}$ does not tend to zero.

Solution

Set $c^{(N)}=e_N/N$. Then $\|c^{(N)}\|=1/N\to0$, but $Ac^{(N)}=e_N$ has norm one. Its images have no norm-convergent subsequence. This demonstrates discontinuity, while remaining compatible with closedness: closedness would impose a condition only if the images also converged.

For the free Hamiltonian, explain why finite kinetic-energy expectation need not imply a strong Schrödinger derivative.

Solution

In momentum representation, finite mean kinetic energy requires $\int p^2|\widehat\psi(p)|^2\,dp<\infty$, so $\psi\in H^1$. Membership in $D(H)$ requires $\int p^4|\widehat\psi(p)|^2\,dp<\infty$, or $\psi\in H^2$. The latter gives the strong derivative. A momentum amplitude decaying as $(1+p^2)^{-1}$ satisfies the first integral but not the second.

References

• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, doi:10.1007/978-94-007-4753-1 — closed operators and maximal domains.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014, author’s book page — multiplication operators and free quantum dynamics.