Spectral Theorem for Unbounded Self-Adjoint Operators

The spectral theorem says that every self-adjoint operator is a real measurable multiplication operator in a suitable representation. In its coordinate-free form, a self-adjoint operator $A$ has a unique projection-valued measure $E_A$ on the Borel subsets of $\mathbb R$ such that

$$A = \int_{\mathbb R}\lambda\,dE_A(\lambda).$$

For an unbounded $A$, this display is incomplete without its domain. If

$$\mu_\psi^A(\Delta) = \langle\psi,E_A(\Delta)\psi\rangle,$$

then

$$D(A) = \left\{ \psi\in\mathcal H: \int_{\mathbb R}\lambda^2\,d\mu_\psi^A(\lambda)<\infty \right\}.$$

The theorem therefore supplies the spectral projections, the action, and the domain in one structure. It does not assert that $A$ has an orthonormal basis of ordinary eigenvectors.

Required background. Self-Adjoint Operators supplies the domain-sensitive operator concept. The external measure-theory assumption is Borel sets, countable additivity, measurable functions, and Lebesgue integration.

Helpful background. Projectors supplies the sharp-subspace geometry used to interpret spectral events.

The unbounded spectral representation

Let $A:D(A)\subset\mathcal H\to\mathcal H$ be self-adjoint. There is a unique map

$$E_A:\mathcal B(\mathbb R) \longrightarrow \{\text{orthogonal projections on }\mathcal H\}$$

with the following properties:

1. $E_A(\varnothing)=0$ and $E_A(\mathbb R)=I$;

2. $E_A(\Delta_1\cap\Delta_2)=E_A(\Delta_1)E_A(\Delta_2)$;

3. for pairwise disjoint Borel sets $\Delta_n$,

   $$E_A\!\left(\bigcup_{n=1}^{\infty}\Delta_n\right)\psi = \sum_{n=1}^{\infty}E_A(\Delta_n)\psi$$

   for every $\psi\in\mathcal H$, with convergence in Hilbert-space norm;

4. $A$ is the spectral integral of the identity function, on the domain determined by its square-integrability condition.

These are the essential axioms of a projection-valued measure. Strong countable additivity, set operations, scalar measures, and the distinction from POVMs are developed on the dedicated Projection-Valued Measures page.

For each $\psi$, the scalar measure $\mu_\psi^A$ is finite and positive, with

$$\mu_\psi^A(\mathbb R)=\|\psi\|^2.$$

For a normalized state it is a probability measure. In quantum mechanics,

$$\Pr_\psi(A\in\Delta) = \langle\psi,E_A(\Delta)\psi\rangle$$

is the Born probability for a sharp measurement result in $\Delta$. The theorem constructs the event operators; it does not specify a measurement instrument or post-measurement state.

How the integral defines an unbounded operator

For bounded simple functions

$$s(\lambda) = \sum_{j=1}^{m}c_j\mathbf 1_{\Delta_j}(\lambda),$$

the spectral integral is the bounded operator

$$\int s(\lambda)\,dE_A(\lambda) = \sum_{j=1}^{m}c_jE_A(\Delta_j).$$

Bounded measurable functions are obtained by controlled limits of simple functions. The identity function $\lambda\mapsto\lambda$ is generally unbounded, so its integral cannot be an everywhere-defined bounded operator. Instead, truncate it:

$$A_n = \int_{[-n,n]}\lambda\,dE_A(\lambda).$$

Each $A_n$ is bounded. For precisely those $\psi$ satisfying

$$\int_{\mathbb R}\lambda^2\,d\mu_\psi^A(\lambda)<\infty,$$

the sequence $A_n\psi$ converges in $\mathcal H$, and its limit defines $A\psi$. The identity

$$\|A\psi\|^2 = \int_{\mathbb R}\lambda^2\,d\mu_\psi^A(\lambda)$$

shows why the second moment, rather than merely the first absolute moment, determines the operator domain.

The notation $A=\int\lambda\,dE_A(\lambda)$ therefore abbreviates a strong limit on a declared domain. It is not an improper Riemann integral converging in operator norm.

The Borel functional calculus

The same spectral measure defines $f(A)$ for every complex Borel function $f$. Its maximal domain is

$$D(f(A)) = \left\{ \psi\in\mathcal H: \int_{\mathbb R}|f(\lambda)|^2 \,d\mu_\psi^A(\lambda)<\infty \right\},$$

and

$$f(A) = \int_{\mathbb R}f(\lambda)\,dE_A(\lambda).$$

When $f$ is bounded on the spectrum, $f(A)$ is bounded on all of $\mathcal H$, and

$$\|f(A)\| = \sup_{\lambda\in\sigma(A)}|f(\lambda)|$$

for continuous $f$; for bounded Borel functions the corresponding statement uses the essential supremum relative to the spectral measure. Important special cases are

$$\mathbf 1_\Delta(A)=E_A(\Delta),$$ $$e^{-itA} = \int_{\mathbb R}e^{-it\lambda}\,dE_A(\lambda),$$

and, for $z\notin\mathbb R$,

$$(A-zI)^{-1} = \int_{\mathbb R} \frac{1}{\lambda-z} \,dE_A(\lambda).$$

The exponential and resolvent are bounded even when $A$ is unbounded. This is why spectral calculus can construct unitary dynamics on every vector while the derivative generator acts only on $D(A)$.

Algebraic rules such as

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

are automatic for bounded functions. For unbounded functions, equality of actions must be accompanied by the appropriate product domains; formal function manipulation does not erase domain questions.

Equivalent multiplication-operator form

There is an equivalent representation in which spectral integration becomes ordinary multiplication. More precisely, one can find a measure space with a multiplicity field, a real measurable function $a(x)$, and a unitary map $W$ from $\mathcal H$ to the associated $L^2$ direct integral such that

$$(WAW^{-1}f)(x)=a(x)f(x)$$

on

$$D(WAW^{-1}) = \left\{ f: \int |a(x)|^2\|f(x)\|^2\,d\nu(x)<\infty \right\}.$$

In this representation,

$$(WE_A(\Delta)W^{-1}f)(x) = \mathbf 1_{a^{-1}(\Delta)}(x)f(x),$$

and

$$(Wf(A)W^{-1}g)(x)=f(a(x))g(x).$$

The multiplicity data matters: the theorem does not say that every self-adjoint operator is multiplication on one scalar $L^2$ space with multiplicity one. A cyclic self-adjoint operator admits that simpler scalar form; a general operator may require several spectral channels or a direct integral of fibers.

Eigenvalues, atoms, and continuous spectrum

For any $\lambda\in\mathbb R$,

$$\operatorname{Ran}E_A(\{\lambda\}) = \ker(A-\lambda I).$$

Thus $\lambda$ is an eigenvalue exactly when the singleton projection is nonzero. The point spectrum is the atomic part of the spectral measure.

But $\lambda\in\sigma(A)$ need not be an eigenvalue. For the position operator $Q$ on $L^2(\mathbb R)$,

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

the spectral projections are

$$(E_Q(\Delta)\psi)(x) = \mathbf 1_\Delta(x)\psi(x).$$

Every singleton has Lebesgue measure zero, so $E_Q(\{\lambda\})=0$ for every $\lambda$. Nevertheless $\sigma(Q)=\mathbb R$. Position values belong to a continuous spectrum without normalizable position eigenvectors.

More generally, the support of $E_A$ is the spectrum in the sense that an open set disjoint from $\sigma(A)$ has zero spectral projection, while every open neighborhood of a spectral point has nonzero projection. The spectral theorem contains point, absolutely continuous, and singular continuous behavior; it does not force a purely discrete or purely absolutely continuous spectrum.

Proof architecture and boundary

A full construction is a substantial theorem of functional analysis. One standard proof proceeds as follows:

1. form the Cayley transform

   $$U=(A-iI)(A+iI)^{-1},$$

   which is unitary for self-adjoint $A$;

2. apply the bounded spectral theorem to the unitary operator $U$; its spectral measure satisfies $E_U(\{1\})=0$ because $1$ is not an eigenvalue of this Cayley transform;

3. transport the projection-valued measure from the unit circle to $\mathbb R$ through $\lambda=i(1+u)/(1-u)$; the inverse map is undefined and unbounded at $u=1$, but it may be assigned arbitrarily on the null spectral projection $E_U(\{1\})$;

4. define the unbounded spectral integral by truncation and verify its exact domain;

5. prove uniqueness from the resolvent or from the bounded functional calculus.

Other proofs use the resolvent family and the Riesz representation theorem, or the commutative operator algebra generated by bounded functions of $A$. This page proves the elementary consequences of the resulting measure but does not compress those construction theorems into a misleading short proof.

The self-adjoint hypothesis is essential. A merely symmetric operator may lack a self-adjoint extension, may have nonreal spectral behavior in the relevant operator sense, and need not possess a normalized real PVM whose identity integral recovers that operator. One may first prove essential self-adjointness or choose a self-adjoint extension, but that is additional work.

Scope relative to practical treatments

This route owns the exact unbounded theorem, its domain, multiplicity-aware representation, and proof architecture. Practical matrix calculations remain supporting methods, while the Spectral Theorem card retains compact lookup scope. The continuous-outcome Born rule owns detector regions, probability densities, and experimental interpretation rather than the operator-theory construction.

What the theorem does not say

• It does not guarantee an ordinary eigenvector basis.
• It does not turn generalized eigenvectors such as $|x\rangle$ into elements of $\mathcal H$.
• It does not make an unbounded function $f(A)$ everywhere defined.
• It does not identify a merely symmetric operator with an observable.
• It does not specify state update, apparatus dynamics, or a POVM for an unsharp measurement.
• It does not remove spectral multiplicity or choose a preferred coordinate representation.

Common pitfalls

Omitting the domain. The integral formula without the square-integrability domain describes only part of the unbounded theorem.

Treating the integral as operator-norm convergent. Truncated integrals converge on $D(A)$ vector by vector. If they converged in operator norm to $A$, the limit would be bounded.

Equating spectrum with eigenvalues. Singleton spectral projections detect eigenvalues. Continuous spectrum can remain even when every singleton projection vanishes.

Using bounded functional-calculus rules for unbounded functions. Products and compositions of unbounded operators require domain checks.

Forgetting uniqueness. The spectral measure is determined by $A$; it is not an arbitrary resolution chosen after the fact.

Exercises

1. Recover the finite-dimensional theorem

Let $A=\sum_j\lambda_jP_j$ be the spectral decomposition of a Hermitian matrix, with distinct eigenvalues $\lambda_j$. Construct $E_A(\Delta)$ and show that the spectral integral recovers $A$.

Solution

Define

$$E_A(\Delta) = \sum_{\lambda_j\in\Delta}P_j.$$

The projectors are mutually orthogonal and sum to $I$, so this is a PVM. The identity function is constant with value $\lambda_j$ on each atom, giving

$$\int\lambda\,dE_A(\lambda) = \sum_j\lambda_jP_j=A.$$

Every vector lies in the domain because the finite spectrum is bounded.

2. A domain from the spectral measure

For the position operator on $L^2(\mathbb R)$, use its PVM to recover the domain $D(Q)=\{\psi:x\psi\in L^2\}$.

Solution

The scalar spectral measure is

$$d\mu_\psi^Q(x)=|\psi(x)|^2\,dx.$$

Therefore

$$\int_{\mathbb R}x^2\,d\mu_\psi^Q(x) = \int_{\mathbb R}x^2|\psi(x)|^2\,dx.$$

This is finite exactly when $x\psi(x)\in L^2(\mathbb R)$.

3. Atoms are eigenvalues

Show that $E_A(\{\lambda\})\psi=\psi$ implies $A\psi=\lambda\psi$, and that an eigenvector with eigenvalue $\lambda$ lies in the range of this projection.

Solution

If the spectral measure of $\psi$ is concentrated at $\lambda$, then

$$A\psi = \int x\,dE_A(x)\psi = \lambda\psi.$$

Conversely, if $A\psi=\lambda\psi$, then for any bounded Borel function $f$, $f(A)\psi=f(\lambda)\psi$. Taking $f=\mathbf 1_{\{\lambda\}}$ gives $E_A(\{\lambda\})\psi=\psi$.

4. Boundedness from spectral support

Suppose $E_A([-M,M])=I$ for some finite $M$. Prove that $A$ is bounded and $\|A\|\leq M$. Prove the converse inclusion $E_A([-\|A\|,\|A\|])=I$ when $A$ is bounded and self-adjoint.

Solution

For every $\psi$,

$$\|A\psi\|^2 = \int_{[-M,M]}\lambda^2\,d\mu_\psi^A(\lambda) \leq M^2\|\psi\|^2.$$

Thus $D(A)=\mathcal H$ and $\|A\|\leq M$. Conversely, the spectrum of a bounded self-adjoint operator lies in $[-\|A\|,\|A\|]$, so its spectral measure assigns the complement projection zero.

5. Resolvent bound

Use the functional calculus to show that, for $z\notin\mathbb R$,

$$\|(A-zI)^{-1}\| \leq \frac{1}{|\operatorname{Im}z|}.$$

Solution

The resolvent is $f_z(A)$ for $f_z(\lambda)=(\lambda-z)^{-1}$. Since $|\lambda-z|\geq|\operatorname{Im}z|$ on the real line,

$$\sup_{\lambda\in\sigma(A)}|f_z(\lambda)| \leq \frac{1}{|\operatorname{Im}z|}.$$

The bounded functional calculus gives the claimed operator-norm estimate.

References

• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• K. Schmüdgen, Unbounded Self-Adjoint Operators on Hilbert Space, Springer, 2012.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014.