Functional Calculus

Functional calculus assigns an operator $f(A)$ to a function of a self-adjoint observable. Its main practical advantage is that it defines exponentials, square roots, inverses, and cutoffs without assuming an eigenbasis or a convergent power series. For unbounded functions, the associated integrability domain is part of the answer.

Helpful background. The Unbounded Spectral Theorem constructs the spectral measure and integral. This page focuses on how to calculate with that integral; Functions of Observables develops the observable interpretation.

Borel functions and their domains

Let $A$ be self-adjoint with projection-valued measure $E_A$. For $\psi\in\mathcal H$, write $\mu_\psi(\Delta)=\langle\psi,E_A(\Delta)\psi\rangle$. For a finite-valued Borel function $f:\mathbb R\to\mathbb C$, define

$$f(A)=\int f(\lambda)\,dE_A(\lambda),$$

with domain

$$D(f(A)) =\left\{\psi:\int |f(\lambda)|^2\,d\mu_\psi(\lambda)<\infty\right\}.$$

Its norm on a domain vector is the square root of that integral. The spectral theorem constructs a closed densely defined normal operator; when $f$ is real-valued it is self-adjoint.

If $f$ is bounded, the domain is all of $\mathcal H$ and $\|f(A)\|\leq\sup_{\lambda\in\sigma(A)}|f(\lambda)|$. For bounded Borel functions the exact norm is the essential supremum relative to the projection-valued measure. For continuous $f$ on the spectrum, this equals the ordinary supremum.

The qualification matters: changing $f$ at a point whose spectral projection is zero does not change $f(A)$, even if the point belongs to continuous spectrum. A scalar supremum over such a point can therefore overestimate the operator norm.

Examples: phases, moments, and inverses

The phase function $f_t(\lambda)=e^{-it\lambda}$ has modulus one, so $f_t(A)$ is unitary on every vector. It is not necessary that $\psi$ belong to all domains $D(A^n)$, and expanding the phase as an operator power series on an arbitrary vector is unnecessary.

The function $f(\lambda)=|\lambda|^{1/2}$ gives the form domain

$$D(|A|^{1/2}) =\left\{\psi:\int|\lambda|\,d\mu_\psi<\infty\right\}.$$

This is the domain for a finite absolute first moment, whereas applying $A$ requires a finite second moment. Powers distinguish higher regularity and higher moments.

For $z\in\rho(A)$, $f(\lambda)=(\lambda-z)^{-1}$ gives the bounded resolvent. In contrast, an inverse at zero is bounded only if zero is outside the spectrum. If $A$ is injective but zero lies in continuous spectrum, the reciprocal defines an unbounded inverse on $\operatorname{Ran}(A)$, with the zero point omitted or assigned any value because its spectral projection vanishes.

Products require the intermediate domain

For bounded $f$ and $g$, the familiar rule is

$$f(A)g(A)=(fg)(A).$$

When functions are unbounded, the natural product domain is

$$D(f(A)g(A)) =D(g(A))\cap D((fg)(A)),$$

and the product agrees with $(fg)(A)$ there. The second condition controls the final output, while the first ensures that the intermediate vector exists. The maximal operator $(fg)(A)$ can have a larger domain.

A particularly useful counterexample takes $f(\lambda)=\lambda$ and $g(\lambda)=0$. Then $A\,0$ is defined on every vector, while $0\,A$ is defined only on $D(A)$. Both actions vanish wherever defined, but the operators are unequal because their domains differ.

Similarly, $f(A)+g(A)$ agrees with $(f+g)(A)$ on the intersection of the two domains; cancellation can enlarge the domain of the latter. The correct closure of a restricted sum or product is another question, not a license to ignore its starting domain.

Spectral truncation controls limits

Define $f_n=f\,\mathbf1_{\{|f|\leq n\}}$. Each $f_n(A)$ is bounded, and for $\psi\in D(f(A))$,

$$\|(f(A)-f_n(A))\psi\|^2 =\int_{\{|f|>n\}}|f|^2\,d\mu_\psi\longrightarrow0.$$

This constructs the unbounded action by a controlled limit. The vectors $\psi_n=E_A(\{|f|\leq n\})\psi$ also converge to $\psi$ in the graph norm of $f(A)$, so the union of these spectral subspaces is a core.

For uniformly bounded Borel functions $f_n$ converging pointwise to $f$, dominated convergence gives strong convergence $f_n(A)\psi\to f(A)\psi$ for every vector. Uniform convergence gives the stronger conclusion of operator-norm convergence.

Pointwise convergence without a common bound is not enough for convergence on every vector. A spectral integral must provide the domination or integrability needed for the chosen limit.

Exercises

Let $A$ be position multiplication on $L^2(\mathbb R)$ and let $f$ be the indicator of the singleton $\{0\}$. Find $f(A)$ and its norm.

Solution

The projection acts by multiplication with $\mathbf1_{\{0\}}$, which vanishes almost everywhere. Therefore $f(A)=0$ and its norm is zero, although $\sup_{\lambda\in\sigma(A)}|f(\lambda)|=1$. The projection-valued essential supremum is the correct Borel-calculus norm.

For $Ae_n=n e_n$, compare $A A^{-1}$ and $A^{-1}A$ as operators.

Solution

Here $A^{-1}e_n=n^{-1}e_n$ is bounded and maps all of $\ell^2$ into $D(A)$. Hence $AA^{-1}=I$ on the whole space. The reverse product is identity only on $D(A)$, because $A$ must act first. Its closure is the full identity, but the product itself has the smaller domain.

References

• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — spectral theorem and functions of operators.
• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, doi:10.1007/978-94-007-4753-1 — spectral integrals and unbounded functional calculus.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014, author’s book page — Borel calculus and spectral approximation.