Spectral Theorem for Bounded Self-Adjoint Operators
Every bounded self-adjoint operator can be represented as multiplication by a bounded real function. The representation need not be an eigenvector expansion: a bounded observable can have entirely continuous spectrum. The bounded theorem is also the starting point for constructions of the unbounded spectral theorem.
Helpful background. Bounded Operators supplies norm continuity and adjoints. Familiarity with continuous functions on compact sets, finite Borel measures, and completion is assumed.
The bounded spectral representation
Section titled “The bounded spectral representation”Let . There is a unique projection-valued measure supported on the compact set such that
The integral is defined on every vector. For every bounded Borel , spectral integration defines , and continuous satisfy
The measure’s additivity is in the strong operator topology. Its full measure-theoretic construction and event interpretation belong to Projection-Valued Measures. This page explains the route from elementary bounded-operator algebra to a multiplication representation.
The theorem extends to bounded normal operators, with a PVM on a compact subset of . General bounded operators need not have this representation; a nontrivial Jordan block is already a finite-dimensional counterexample.
From polynomials to continuous functions
Section titled “From polynomials to continuous functions”Polynomials are initially unambiguous:
A key lemma is the norm identity . One route proves it from polynomial spectral mapping together with the equality of norm and spectral radius for normal operators. The latter follows from the identity and the spectral-radius formula, rather than assuming an eigenbasis.
By uniform polynomial approximation on the compact real spectrum, each continuous can be approximated by polynomials . The norm identity makes a Cauchy sequence and defines
in operator norm. The result is independent of the approximating sequence. Multiplication and complex conjugation pass to the limit, producing a unital isometric -homomorphism from into .
Positivity follows because a nonnegative continuous has a continuous square root: . This is the bridge from operator algebra to measures.
A cyclic vector turns the operator into multiplication
Section titled “A cyclic vector turns the operator into multiplication”A vector is cyclic for when vectors span a dense subspace. For normalized , the functional
is positive and satisfies . The Riesz representation theorem gives a probability measure on with .
Define initially
This is well-defined in because
Polynomial functions are dense in , and cyclicity supplies density on the operator side. Thus extends to a unitary map . Multiplication of polynomials shows
The spectral projections now arise concretely as . Their strong additivity follows from dominated convergence in .
This proves the cyclic case once the functional-calculus lemmas and the two approximation theorems have been supplied. It is not an argument that an arbitrary vector is cyclic.
General operators require several cyclic pieces
Section titled “General operators require several cyclic pieces”For any , the closed span of is invariant under . Since is self-adjoint, its orthogonal complement is invariant too, so the subspace is reducing.
Decomposing into orthogonal cyclic reducing subspaces and applying the construction to each yields a direct sum of multiplication representations. In a separable Hilbert space only countably many nonzero orthogonal summands are needed. The repeated spectral channels carry multiplicity.
Uniqueness of the PVM follows because its integrals against continuous functions must equal the already determined continuous calculus. Uniqueness of scalar measures, followed by polarization of matrix elements, fixes every projection.
This is a proof architecture with its substantive ingredients identified. It shows why compactness of the spectrum, positivity, and cyclic decomposition enter the theorem, rather than treating the theorem as formal diagonalization.
A bounded observable with no eigenvectors
Section titled “A bounded observable with no eigenvectors”On take . The constant function is cyclic because polynomials are dense in this space. The preceding construction returns Lebesgue measure and the original multiplication representation.
An eigenvector at would be supported on , a null set, so none exists. Nevertheless . Compactness of the spectrum is not compactness of the operator: this is bounded but not compact.
Exercises
Section titled “Exercises”Let be a Hermitian matrix with an eigenvalue of multiplicity two. Can one vector be cyclic for ?
Solution
No. In that eigenspace, every is a scalar multiple of the single projection of onto the eigenspace. Those vectors cannot span a two-dimensional eigenspace. More cyclic channels, or additional commuting observables, are needed to resolve the multiplicity.
For bounded self-adjoint and continuous , explain why the construction is stable under uniform approximation but not automatically under pointwise approximation.
Solution
The norm identity gives . Uniform convergence therefore controls the operator norm. Pointwise convergence lacks that bound; with a uniform bound it gives strong convergence through spectral dominated convergence, but need not give norm convergence.
References
Section titled “References”- B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5, Chapters 7–8 — statements and proofs of the bounded self-adjoint spectral theorem.
- R. B. Melrose, Functional Analysis, MIT lecture notes, 2020, course reading page — bounded-operator and spectral preliminaries.
- G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014, author’s book page — spectral representations and functional calculus.