Spectra, Resolvents, Spectral Measures, and Functional Calculus

For a closed densely defined operator $A$ , the resolvent asks whether $A-zI$ has a bounded inverse on the whole Hilbert space. Its failures form the spectrum, and for a general operator they can occur because of eigenvectors, a dense but incomplete range, or even a nondense range. A self-adjoint operator is much more rigid: its spectrum is real, its residual spectrum is empty, its resolvent obeys a universal bound away from the real axis, and a unique projection-valued measure replaces the finite-dimensional idea of an eigenvector basis.

That measure is the input to the functional calculus. It defines $f(A)$ —including spectral projections, resolvents, and unitary evolution—while making the domain of an unbounded $f(A)$ explicit. In QFT, the same scalar-measure structure explains why an atom contributes a pole term and an absolutely continuous component contributes a boundary discontinuity. The full relativistic Källén–Lehmann construction requires additional physical hypotheses and remains with its Foundations treatment.

Required background. Unbounded Operators, Domains, Closure, and Adjoints supplies closed operators, adjoints, multiplication operators, and the range-orthogonality identity.

Helpful background. Self-Adjointness, Extensions, and Unitary Evolution supplies the domain analysis that selects self-adjoint realizations before their spectra are studied.

Closed operators and resolvent conventions

This page develops general closed-operator spectra, the residual-spectrum caveat, self-adjoint spectral measures, and the Borel functional calculus. It does not develop specialist operator frameworks, production software, or a physical spectral representation in full.

The Hilbert-space inner product is conjugate-linear in the bra and linear in the ket. Domains are part of every unbounded operator. This page fixes the resolvent sign convention

R_A(z)=(A-zI)^{-1}.

Changing to $(zI-A)^{-1}$ reverses several signs in resolvent and boundary formulas.

The resolvent records every failure of invertibility

Let $A:\mathcal D(A)\subset H\to H$ be densely defined and closed. Its resolvent set is

\rho(A) = \left\{ z\in\mathbb C: A-zI:\mathcal D(A)\longrightarrow H \text{ is bijective} \right\}.

For $z\in\rho(A)$ , the inverse has domain all of $H$ . It is a closed operator from $H$ to $H$ , so the closed graph theorem makes it bounded. Thus

R_A(z)=(A-zI)^{-1}\in\mathcal B(H).

The spectrum is the complement

\sigma(A)=\mathbb C\setminus\rho(A).

These definitions and the bounded-inverse consequence are the closed-operator formulation in Teschl 2014, §2.4, PDF. They show why “the inverse exists formally” is not enough: it must solve the equation for every vector and define a bounded operator.

For $z,w\in\rho(A)$ , elementary inverse algebra gives the first resolvent identity

\begin{aligned} R_A(z)-R_A(w) &=(z-w)R_A(z)R_A(w)\\ &=(z-w)R_A(w)R_A(z). \end{aligned}

If $z$ is sufficiently close to $z_0\in\rho(A)$ , the Neumann series for $I-(z-z_0)R_A(z_0)$ converges. Consequently $\rho(A)$ is open and $z\mapsto R_A(z)$ is operator-norm holomorphic there. These are local facts for every closed operator; they do not imply that the spectrum is real or that a self-adjoint resolvent bound holds.

Three spectral types for a general closed operator

The failure of $A-zI$ separates into three mutually exclusive cases:

$z$ is in the point spectrum $\sigma_{\mathrm p}(A)$ if $\ker(A-zI)\neq\{0\}$ ;
$z$ is in the continuous spectrum $\sigma_{\mathrm c}(A)$ if $A-zI$ is injective and has dense range, but is not onto; and
$z$ is in the residual spectrum $\sigma_{\mathrm r}(A)$ if $A-zI$ is injective but its range is not dense.

This taxonomy is the one recorded in NIST DLMF 2026, §1.18(ix). DLMF phrases the continuous case as an unbounded inverse on a dense range. For a closed $A$ , that is equivalent to the dense-but-not-onto formulation above: a bounded inverse on the range would make the range closed.

There is no fourth case for a closed operator. If $A-zI$ is injective and onto, its inverse is bounded by the closed graph theorem, so $z\in\rho(A)$ . The prerequisite page’s adjoint identity becomes

\operatorname{Ran}(A-zI)^\perp = \ker(A^\dagger-\overline z I).

It is therefore possible for $z$ not to be an eigenvalue of $A$ while $\overline z$ is an eigenvalue of $A^\dagger$ ; that is precisely the mechanism behind residual spectrum.

A nonnormal counterexample: the unilateral shift

Let $S$ be the right shift on $\ell^2(\mathbb N_0)$ ,

Se_n=e_{n+1}.

It is bounded, closed, and injective, and it has no eigenvalues. For $|z|<1$ , however,

v_z=(1,\overline z,\overline z^{\,2},\ldots)\in\ell^2, \qquad S^\dagger v_z=\overline z\,v_z.

Hence $\operatorname{Ran}(S-zI)$ is not dense, and every point of the open unit disk lies in $\sigma_{\mathrm r}(S)$ . On $|z|=1$ , the adjoint vector is no longer square-integrable, so the range is dense. The normalized truncated vectors

u_N = \frac{1}{\sqrt{N+1}} \sum_{n=0}^{N}z^{-n}e_n

satisfy $\|(S-zI)u_N\|\to0$ , showing that $S-zI$ cannot have a bounded inverse. The unit circle is therefore continuous spectrum. For $|z|>1$ , the factorization $S-zI=-z(I-S/z)$ and its convergent Neumann series give a bounded inverse. Altogether,

\sigma(S)=\{z:|z|\le1\}, \qquad \sigma_{\mathrm p}(S)=\varnothing.

This example blocks two finite-dimensional habits at once: spectrum need not mean eigenvalues, and a general closed operator can have residual spectrum.

Self-adjointness sharpens the answer

Now let $A=A^\dagger$ . Write $z=x+iy$ with $y\neq0$ . For every $\psi\in\mathcal D(A)$ , symmetry makes $\langle\psi|(A-xI)\psi\rangle$ real, and the cross term cancels:

\|(A-zI)\psi\|^2 = \|(A-xI)\psi\|^2+y^2\|\psi\|^2.

Thus $A-zI$ is injective, its range is closed, and

\|(A-zI)\psi\|\ge |y|\,\|\psi\|.

The range is also dense because

\operatorname{Ran}(A-zI)^\perp = \ker(A-\overline zI) = \{0\}.

A closed, dense range is all of $H$ , so every nonreal $z$ belongs to $\rho(A)$ and

\|R_A(z)\|\le\frac{1}{|\operatorname{Im}z|}.

This proves both $\sigma(A)\subset\mathbb R$ and the standard nonreal resolvent estimate (Teschl 2014, Theorem 2.19, PDF).

The same range identity removes the residual spectrum on the real axis. If $\lambda\in\mathbb R$ and $A-\lambda I$ is injective, then

\operatorname{Ran}(A-\lambda I)^\perp = \ker(A-\lambda I) = \{0\}.

Its range is dense, so $\lambda$ can be in the continuous spectrum but not the residual spectrum. A self-adjoint operator may still have no normalizable eigenvectors at all; self-adjointness does not make the spectrum discrete.

Self-adjoint operators are not the only operators with a spectral theorem. Normal, non-self-adjoint operators admit an analogous projection-valued measure on $\mathbb C$ . That extension is outside this page’s real self-adjoint scope. The unilateral shift is not normal and does not inherit such a conclusion.

Spectral measures replace eigenvector lists

A projection-valued measure (PVM) on $\mathbb R$ is a map

E:\mathcal B(\mathbb R)\longrightarrow\mathcal B(H)

from Borel sets to orthogonal projections such that $E(\mathbb R)=I$ and, for pairwise disjoint sets $B_n$ ,

E\!\left(\bigcup_n B_n\right)\psi = \sum_n E(B_n)\psi

for every $\psi\in H$ , with convergence in Hilbert-space norm. It follows that $E(B_1)E(B_2)=E(B_1\cap B_2)$ .

Spectral theorem, cited form. For every self-adjoint operator $A$ there is a unique PVM $E_A$ on $\mathbb R$ such that

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

with operator domain

\mathcal D(A) = \left\{ \psi\in H: \int_{\mathbb R}\lambda^2\,d\mu_\psi(\lambda)<\infty \right\}, \qquad \mu_\psi(B)=\langle\psi|E_A(B)\psi\rangle.

The finite positive measure $\mu_\psi$ has total mass $\mu_\psi(\mathbb R)=\|\psi\|^2$ . The spectrum is the support of the PVM: $\lambda\in\sigma(A)$ exactly when every open interval $I$ containing $\lambda$ has $E_A(I)\neq0$ . This PVM statement, its uniqueness, and the domain formula are proved in Teschl 2014, Theorems 3.2, 3.6, and 3.7, PDF. Etingof’s Etingof 2023, §8.2.1 and Theorem 8.5, PDF give the complementary multiplication-operator realization.

The theorem is used here as a cited theorem. The examples below verify its formulas in concrete models, but they are not a proof for arbitrary self-adjoint operators.

The multiplication model

Let $(X,\Sigma,\nu)$ be a $\sigma$ -finite measure space and let $m:X\to\mathbb R$ be measurable. The maximal multiplication operator

\begin{aligned} (M_m\psi)(x)&=m(x)\psi(x),\\ \mathcal D(M_m) &=\{\psi\in L^2(X,\nu):m\psi\in L^2(X,\nu)\} \end{aligned}

is self-adjoint. Its spectral measure is visible pointwise:

(E_{M_m}(B)\psi)(x) = \mathbf 1_{m^{-1}(B)}(x)\psi(x).

Its spectrum is the essential range of $m$ :

\lambda\in\sigma(M_m) \quad\Longleftrightarrow\quad \nu\{x:|m(x)-\lambda|<\varepsilon\}>0 \text{ for every }\varepsilon>0.

An eigenspace consists of the square-integrable functions supported on the level set $m^{-1}(\{\lambda\})$ . For $M_x$ on $L^2(\mathbb R,dx)$ , every level set has measure zero. Consequently

\sigma(M_x)=\mathbb R, \qquad \sigma_{\mathrm p}(M_x)=\varnothing,

and the whole spectrum is continuous. This is the canonical check that a real, self-adjoint spectrum need not provide a Hilbert basis of eigenvectors.

Functional calculus keeps domains visible

Let $f:\mathbb R\to\mathbb C$ be Borel measurable. The spectral theorem defines

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

on the domain

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

If $f$ is bounded, $f(A)$ is bounded on all of $H$ and $\|f(A)\|\le\|f\|_\infty$ ; functions equal outside an $E_A$ -null set define the same operator. If $f$ is real $E_A$ -almost everywhere, $f(A)$ is self-adjoint. Important special cases are

\begin{aligned} \mathbf 1_B(A)&=E_A(B),\\ R_A(z)&=\int_{\mathbb R}\frac{1}{\lambda-z}\,dE_A(\lambda), \qquad z\notin\mathbb R,\\ e^{-itA}&=\int_{\mathbb R}e^{-it\lambda}\,dE_A(\lambda). \end{aligned}

The last line reconstructs the unitary group discussed on the recommended self-adjointness page. The multiplication model checks the whole calculus:

f(M_m)=M_{f\circ m}.

Domains cannot be discarded when $f$ is unbounded. For two unbounded Borel functions, $f(A)g(A)$ agrees with $(fg)(A)$ only on

\mathcal D(g(A)) \cap \{\psi:g(A)\psi\in\mathcal D(f(A))\},

which can be smaller than $\mathcal D((fg)(A))$ . Likewise, the natural domain of a sum is the intersection of the two operator domains. The precise inclusions are part of the Borel functional calculus in Teschl 2014, Theorem 3.2, PDF.

Atomic and continuous measures give different resolvents

For a diagonal self-adjoint operator on $\ell^2$ ,

H_{\mathrm d}e_n=E_ne_n, \qquad \mathcal D(H_{\mathrm d}) = \left\{ c:\sum_n E_n^2|c_n|^2<\infty \right\},

the PVM and scalar spectral measure are atomic:

E_{H_{\mathrm d}}(B) = \sum_{E_n\in B}|e_n\rangle\langle e_n|, \qquad \mu_\psi = \sum_n|c_n|^2\delta_{E_n}.

By contrast, for $M_x$ and a vector $\psi\in L^2(\mathbb R)$ ,

d\mu_\psi(E)=|\psi(E)|^2\,dE.

These are not different definitions of spectral measure. They are two measure types allowed by the same PVM theorem. Singular continuous measures are possible as well, so “not discrete” must not be silently replaced by “has a smooth density.”

Controlled QFT bridge: poles versus continua

Let $H$ be self-adjoint and let $\psi\in H$ . The scalar resolvent is

G_\psi(z) = \langle\psi|(H-zI)^{-1}\psi\rangle = \int_{\mathbb R}\frac{d\mu_\psi(E)}{E-z}, \qquad z\in\mathbb C\setminus\mathbb R.

This is the Borel, or Stieltjes, transform of the positive measure $\mu_\psi$ (Teschl 2014, §§3.1 and 3.4, PDF). If $\mu_\psi$ has an atom of weight $w>0$ at $E_0$ , then

G_\psi(z) = \frac{w}{E_0-z} +\text{the transform of the remaining measure}.

The atomic term appears only if $\psi$ has nonzero spectral weight at $E_0$ . It is a genuine isolated pole of $G_\psi$ when $E_0$ is isolated from the support of the remaining measure; with this page’s sign convention its residue as a function of $z$ is $-w$ . If an atom is embedded in continuous support, the displayed term remains present but the full transform need not be meromorphic in a punctured neighborhood. An eigenvalue invisible to this vector produces no atomic term in this particular matrix element.

If, on an interval, the measure has density $d\mu_\psi(E)=\rho_\psi(E)\,dE$ , then at almost every Lebesgue point where the boundary values exist,

\begin{aligned} \operatorname{Im}G_\psi(E+i0)&=\pi\rho_\psi(E),\\ G_\psi(E+i0)-G_\psi(E-i0)&=2\pi i\,\rho_\psi(E). \end{aligned}

Thus an absolutely continuous component produces the displayed boundary discontinuity at almost every Lebesgue point where the boundary values exist; the jump can vanish at support points where the density vanishes. Calling the support a branch cut further assumes an analytic continuation and enough regularity to define the chosen branches. A singular continuous measure need not admit an ordinary density, so the simple pole-versus-smooth-cut picture is not exhaustive.

A free one-particle Hamiltonian in a finite spatial box has a discrete energy measure and a sum of pole terms. In an infinite-volume multiplication model, the energy variable becomes continuous and the scalar resolvent becomes an integral with boundary values. This is the controlled mathematical content behind spectral representations and propagator poles versus continua.

It is not yet the relativistic Källén–Lehmann derivation. That result uses a translation-invariant vacuum, Poincaré covariance, completeness of physical states, the spectrum condition, positivity, and operator-valued distributions to produce a measure in invariant mass squared. It also has its own Fourier and $i0$ conventions. In this page’s convention,

G(z)=\int\frac{d\mu(s)}{s-z},

so $G$ maps the upper half-plane to itself and an isolated atom has $z$ -residue $-w$ . Schwartz instead writes the Fourier coefficient as $i\Pi(p^2)$ and uses

\Pi(p^2)=\int_0^\infty\frac{\rho(s)\,ds}{p^2-s+i0}.

For the analytic functions away from the real-axis prescription, the schematic identification is $\Pi(z)=-G(z)$ , so the corresponding particle-pole residue of $\Pi$ is $+w$ . This is a convention translation, not a disagreement.

The physical assumptions and normalizations are developed at The Källén–Lehmann Representation; the pole and multiparticle-threshold discussion is supported by Schwartz 2014, §24.2.1, printed pp. 467–470, Eqs. (24.67)–(24.77). Its corrected support condition is $p^2\ge0$ and $p^0>0$ .

One must not infer that every momentum-space propagator is literally a positive-Hilbert-space Hamiltonian resolvent. In gauge-fixed descriptions, gauge-variant fields can act in a state space where the positivity argument for a scalar spectral density does not apply. The Foundations page develops that physical qualification.

What the contrast decides

The principal distinction can now be stated compactly:

Question	General closed $A$	Self-adjoint $A$
Where can $\sigma(A)$ lie?	In $\mathbb C$	In $\mathbb R$
Can residual spectrum occur?	Yes	No
What controls $R_A(z)$ ?	Local identities and operator-specific estimates	$\lVert R_A(z)\rVert\le\lvert\operatorname{Im}z\rvert^{-1}$ off the real axis
Is there a real PVM?	Not in general	A unique $E_A$ on $\mathbb R$
How is $f(A)$ defined?	No self-adjoint Borel calculus follows from closedness alone	By $f(A)=\int f\,dE_A$ , with an explicit domain

The most consequential misuse is to import the right column merely because a formal differential expression looks real. The domain and self-adjoint realization must be established first.

Common pitfalls

The spectrum is the set of eigenvalues. It is only the point spectrum in finite-dimensional language. $M_x$ has spectrum $\mathbb R$ and no $L^2$ eigenvectors, while the unilateral shift has spectrum without any eigenvalues at all.

A dense range means a resolvent point. The range must be all of $H$ , and the inverse must be bounded. For a closed operator, bijectivity supplies boundedness. Injectivity together with a dense, non-surjective range describes the continuous-spectrum case.

Residual spectrum is impossible for a densely defined operator. It is impossible for a self-adjoint operator, not for a general densely defined closed one. The unilateral shift gives an explicit bounded counterexample.

The spectral theorem is an eigenvector expansion. A PVM includes point, absolutely continuous, and singular continuous parts. Generalized eigenfunctions can be useful representations, but they need not be vectors in the Hilbert space.

A PVM and a scalar spectral measure are the same object. $E_A$ is operator-valued and independent of the test vector. The positive measure $\mu_\psi(B)=\langle\psi|E_A(B)\psi\rangle$ depends on $\psi$ .

Every continuum automatically gives a branch cut. A continuous measure gives boundary behavior of its transform. A conventional branch cut also requires a specified analytic continuation, while a singular continuous part may not have a density at all.

Functional-calculus algebra ignores domains. The familiar sum and product rules hold without qualification for bounded functions. With unbounded functions, operator inclusions and domain intersections are part of the statement.

Exercises

Residual-spectrum check. For $|z|<1$ , show directly why $z$ is not an eigenvalue of the unilateral shift but is in its residual spectrum.

Solution

If $Sx=zx$ , the zeroth component gives $zx_0=0$ and the remaining recurrence forces every component to vanish; $S$ is also injective at $z=0$ . Thus $z$ is not an eigenvalue. But $v_z=(1,\overline z,\overline z^{\,2},\ldots)$ belongs to $\ell^2$ and obeys $S^\dagger v_z=\overline z\,v_z$ . Hence $v_z\perp\operatorname{Ran}(S-zI)$ , so that range is not dense.

Resolvent-bound check. Let $A=M_x$ on $L^2(\mathbb R)$ . Compute $R_A(z)$ for $\operatorname{Im}z\neq0$ and verify the self-adjoint bound.

Solution

The inverse is multiplication by $(x-z)^{-1}$ . Therefore

\|R_A(z)\| = \operatorname*{ess\,sup}_{x\in\mathbb R}\frac{1}{|x-z|} = \frac{1}{|\operatorname{Im}z|}.

The general estimate is saturated.

Domain check. For $A=M_x$ , what is the domain of $A^2$ obtained from the functional calculus, and why is it smaller than $\mathcal D(A)$ ?

Solution

Taking $f(\lambda)=\lambda^2$ gives

\mathcal D(A^2) = \left\{ \psi\in L^2(\mathbb R): \int_{\mathbb R}x^4|\psi(x)|^2\,dx<\infty \right\}.

The domain of $A$ requires only the second moment. For example, a tail can make the second moment finite while the fourth diverges: $\psi(x)=(1+|x|)^{-2}$ has exactly this behavior.

QFT-transfer check. Suppose $d\mu_\psi(E)=w\delta(E-E_0)+\rho(E)\,dE$ . Identify what the scalar resolvent can reveal and what still requires the physical handoff.

Solution

The atom gives $w/(E_0-z)$ , while the absolutely continuous part gives $\int \rho(E)/(E-z)\,dE$ and, where boundary values exist, a discontinuity $2\pi i\rho(E)$ . This identifies the operator-theoretic atomic and continuum contributions; the atom is an isolated pole only when separated from the remaining support. Interpreting $E_0$ as a particle mass, fixing relativistic normalization, locating multiparticle thresholds, and justifying positivity for a particular field require the Källén–Lehmann hypotheses.

Where to continue

Use Self-Adjointness, Extensions, and Unitary Evolution to establish the realization whose spectrum is being studied. Continue to The Källén–Lehmann Representation for the exact physical treatment of spectral densities, poles, thresholds, and positivity.

References

Pavel Etingof, Mathematical Ideas and Notions of Quantum Field Theory, PDF, MIT 18.238 lecture notes, 2023, §8.2.1 and Theorem 8.5 (printed pp. 103 and 105–106). These results develop the multiplication-operator form of the bounded and unbounded self-adjoint spectral theorem. Its inner product is conjugate-linear in the first argument, matching this page.
NIST Digital Library of Mathematical Functions (accessed August 11, 2026), §1.18(ix), “Spectrum of an Operator”. This is the independent structural reference for the point, continuous, and residual spectrum taxonomy and for the self-adjoint specialization. The shift classification on this page follows from the displayed calculation rather than being attributed to DLMF; the site’s inner-product convention is used in the adjoint calculation.
Matthew D. Schwartz, Quantum Field Theory and the Standard Model, §24.2.1, printed pp. 467–470, especially Eqs. (24.67)–(24.77), Cambridge University Press, 2014. This section derives the Källén–Lehmann measure, isolated poles, and multiparticle continua; it is not used as the authority for unbounded-operator domains. The author’s first-printing corrections were checked. The corrected sentence on printed p. 467 gives support at $p^2\ge0$ with $p^0>0$ ; material in §24.2.2 is not used here.
Gerald Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, PDF, second edition, §§2.4, 3.1, and 3.4, American Mathematical Society, 2014. These sections establish resolvents, the PVM theorem, Borel functional calculus, and scalar resolvent transforms. The author’s errata, PDF, updated March 18, 2026, were checked. In particular, the multiplication example above uses the corrected eigenspace formulation rather than treating one characteristic function as automatically square-integrable.