Unbounded Operators, Domains, Closure, and Adjoints

An unbounded operator must not be treated as a formula acting everywhere. It is a pair

$$\big(\mathcal D(A),A\big), \qquad A:\mathcal D(A)\subset H_1\longrightarrow H_2,$$

consisting of a linear rule and the vectors on which that rule is defined. Changing the domain can change the graph, the closure, the adjoint, the boundary conditions, and eventually the spectrum and dynamics, even when the displayed differential expression is unchanged.

The graph records simultaneous limits of inputs and outputs. A closed operator contains all such limits; a closable operator has a smallest closed extension. For a densely defined operator, the adjoint domain is not copied from the original domain. It consists of those vectors $\eta$ for which

$$\psi\longmapsto\langle\eta|A\psi\rangle$$

is bounded in the ambient Hilbert norm $\|\psi\|$. Riesz representation then determines $A^\dagger\eta$. These constructions explain why formal integration by parts is only the beginning of an operator analysis.

Required background. Banach and Hilbert Spaces, Completion, and Riesz Representation supplies completeness, orthogonal complements, and Riesz representation.

Operator domains and adjoint conventions

Throughout, $H$, $H_1$, and $H_2$ are complex Hilbert spaces; $H$ denotes the endomorphism case $H_1=H_2=H$. The bra slot is conjugate-linear and the ket slot is linear:

$$\langle\alpha\eta|A\psi\rangle = \alpha^*\langle\eta|A\psi\rangle, \qquad \langle\eta|A(\alpha\psi)\rangle = \alpha\langle\eta|A\psi\rangle.$$

The notation $\mathcal D(A)$ always means the operator domain, not merely a convenient set of test vectors suppressed from the definition. The adjoint is written $A^\dagger$. For a densely defined operator, the site convention defines it by

$$\eta\in\mathcal D(A^\dagger)\subset H_2 \quad\Longleftrightarrow\quad \text{there is a unique }w\in H_1\text{ such that } \langle\eta|A\psi\rangle=\langle w|\psi\rangle$$

for every $\psi\in\mathcal D(A)$, and then $A^\dagger\eta=w$.

This page develops domains, graphs, closures, cores, and adjoints. It distinguishes symmetric from self-adjoint operators but leaves extension theory, deficiency indices, boundary classifications, and unitary evolution to Self-Adjointness, Extensions, and Unitary Evolution. General spectra and resolvents come later at Spectra, Resolvents, Spectral Measures, and Functional Calculus.

The domain is part of the operator

Let $A$ and $B$ be linear operators from subspaces of $H_1$ into $H_2$. The notation

$$A\subset B$$

means

$$\mathcal D(A)\subseteq\mathcal D(B), \qquad B\psi=A\psi \quad\text{for every }\psi\in\mathcal D(A).$$

Thus $B$ is an extension of $A$, and $A$ is a restriction of $B$. Two operators are equal only when both their actions and their domains are equal. This convention prevents one differential expression from being silently identified with all of its possible boundary realizations.

The word unbounded means that there is no finite $C$ satisfying

$$\|A\psi\|_{H_2}\leq C\|\psi\|_{H_1} \qquad \text{for every }\psi\in\mathcal D(A).$$

It does not mean that $A\psi$ is an infinite vector. Every output is an element of $H_2$; what fails is one uniform estimate in the ambient norms. A proper domain alone is not evidence of unboundedness. If a linear rule on a dense domain does satisfy the displayed estimate, it extends uniquely to a bounded operator from $H_1$ to $H_2$.

For example, let $(e_n)_{n\geq1}$ be the standard basis of $\ell^2$ and set

$$Ne_n=ne_n.$$

The maximal natural domain is

$$\mathcal D(N) = \left\{ c=(c_n)\in\ell^2: \sum_{n=1}^{\infty}n^2|c_n|^2<\infty \right\}.$$

It contains the finite sequences and is therefore dense. Since $\|Ne_n\|=n$ while $\|e_n\|=1$, $N$ is unbounded. The rule cannot be extended continuously to all of $\ell^2$.

This last fact is not accidental. The Hellinger–Toeplitz theorem says that an everywhere-defined symmetric operator on a Hilbert space is bounded. More generally, the closed graph theorem says that an everywhere-defined closed operator between Banach spaces is bounded. A genuinely unbounded symmetric or closed operator must therefore have a proper domain.

Algebra also has domains

For endomorphisms $A$ and $B$ in $H$, their formal sum and product have the natural domains

$$\begin{aligned} \mathcal D(A+B) &= \mathcal D(A)\cap\mathcal D(B),\\ \mathcal D(AB) &= \{\psi\in\mathcal D(B):B\psi\in\mathcal D(A)\}. \end{aligned}$$

The second line is ordered: $B$ must act first and must produce a vector on which $A$ can act. Consequently $AB$ and $BA$ can have different domains, and the commutator

$$[A,B]\psi=(AB-BA)\psi$$

is initially meaningful only on $\mathcal D(AB)\cap\mathcal D(BA)$. An algebraic identity verified on a common invariant test domain is an identity there; it does not automatically extend to all of $H$ or to the maximal domains of the displayed products.

Closure does not distribute naively over these operations. Even when $A$ and $B$ are closable, a sum or product can require additional hypotheses before it is closable or before its closure has a simple formula. This is one place where bounded perturbations are much safer than arbitrary unbounded ones.

Graphs, closed operators, and closures

The graph of $A$ is the linear subspace

$$\Gamma(A) = \{(\psi,A\psi):\psi\in\mathcal D(A)\} \subset H_1\oplus H_2.$$

The operator is closed when $\Gamma(A)$ is closed in the product Hilbert space. Equivalently,

$$\psi_n\longrightarrow\psi, \qquad A\psi_n\longrightarrow\eta$$

with every $\psi_n\in\mathcal D(A)$ implies

$$\psi\in\mathcal D(A), \qquad A\psi=\eta.$$

Both limits matter. Closedness does not say that $A\psi_n$ converges whenever $\psi_n$ does; that would be continuity. It says that if the input and output limits both exist, the limiting pair still lies in the graph.

The graph norm on $\mathcal D(A)$ is

$$\|\psi\|_A = \left(\|\psi\|_{H_1}^2+\|A\psi\|_{H_2}^2\right)^{1/2}.$$

The map

$$\psi\longmapsto(\psi,A\psi)$$

is an isometry from $(\mathcal D(A),\|\cdot\|_A)$ onto $\Gamma(A)$. Hence

$$\boxed{ A\text{ is closed} \quad\Longleftrightarrow\quad (\mathcal D(A),\|\cdot\|_A)\text{ is complete}. }$$

The ambient norm and graph norm have different jobs. The graph norm controls both a vector and the result of applying $A$. The adjoint domain below, however, is defined by boundedness with respect to the ambient norm, not the graph norm.

Closability and the sequence test

An operator $A$ is closable when the closure $\overline{\Gamma(A)}$ is itself the graph of an operator. That operator is the closure $\overline A$. It is the smallest closed extension of $A$:

$$A\subset\overline A\subset B$$

for every closed extension $B$ of $A$.

The obstruction to closability is a vertical vector in the closed graph. Equivalently,

$$\boxed{ A\text{ is closable} \quad\Longleftrightarrow\quad \left. \begin{aligned} \psi_n&\longrightarrow0,\\ A\psi_n&\longrightarrow\eta \end{aligned} \right\} \Longrightarrow\eta=0. }$$

If a nonzero $\eta$ survived, the closed graph would contain both $(0,0)$ and $(0,\eta)$ and could not be the graph of a single-valued map. Conversely, if no such vertical vector exists, the first component of a point in $\overline{\Gamma(A)}$ determines its second component uniquely.

A subspace $\mathcal C\subset\mathcal D(A)$ is a core for a closed operator $A$ when it is dense in $\mathcal D(A)$ in the graph norm. The restriction $A|_{\mathcal C}$ is then closable and

$$\overline{A|_{\mathcal C}}=A.$$

Thus a core is more than an ambient-dense set. It lets every domain vector be approximated together with its image.

A densely defined operator that is not closable

Density of the domain does not by itself imply closability. On $\ell^2$, let $c_{00}$ be the finite sequences and define

$$Ax = \left(\sum_{n=1}^{\infty}n x_n\right)e_1, \qquad \mathcal D(A)=c_{00}.$$

The sum is finite on this domain. For

$$x^{(n)}=\frac1n e_n,$$

one has

$$x^{(n)}\longrightarrow0, \qquad Ax^{(n)}=e_1.$$

The sequence test fails, so $A$ is not closable. This example separates three questions that are often conflated: the rule is well defined on each domain vector, the domain is dense, but simultaneous graph limits still fail to define an operator.

The adjoint determines its own domain

Assume from now on that $A:\mathcal D(A)\subset H_1\to H_2$ has domain dense in $H_1$. For fixed $\eta\in H_2$, consider

$$F_\eta(\psi)=\langle\eta|A\psi\rangle, \qquad \psi\in\mathcal D(A).$$

The vector $\eta$ belongs to $\mathcal D(A^\dagger)$ exactly when there is a constant $C_\eta<\infty$ such that

$$|\langle\eta|A\psi\rangle| \leq C_\eta\|\psi\|_{H_1} \qquad \text{for every }\psi\in\mathcal D(A).$$

This estimate allows $F_\eta$ to extend uniquely from the dense subspace $\mathcal D(A)$ to a bounded linear functional on all of $H_1$. Riesz representation gives a unique $w\in H_1$ with

$$\langle\eta|A\psi\rangle = \langle w|\psi\rangle \qquad \text{for every }\psi\in\mathcal D(A),$$

and the definition is $A^\dagger\eta=w$.

Density is what makes $w$ unique. If $\mathcal D(A)$ were not dense in $H_1$, adding any vector in $\mathcal D(A)^\perp\subset H_1$ would give the same pairings. One can develop adjoint relations for nondense domains, but not the single-valued Hilbert-space adjoint used here.

Several structural facts now follow:

• $A^\dagger$ is always closed;
• if $A\subset B$, then $B^\dagger\subset A^\dagger$;
• $A$ is closable exactly when $\mathcal D(A^\dagger)$ is dense in $H_2$;
• in that case, $\overline A=A^{\dagger\dagger}$ and $(\overline A)^\dagger=A^\dagger$.

For example, closedness of $A^\dagger$ follows directly. If $\eta_n\to\eta$ and $A^\dagger\eta_n\to w$, then for every $\psi\in\mathcal D(A)$,

$$\begin{aligned} \langle\eta|A\psi\rangle &= \lim_{n\to\infty}\langle\eta_n|A\psi\rangle\\ &= \lim_{n\to\infty}\langle A^\dagger\eta_n|\psi\rangle = \langle w|\psi\rangle. \end{aligned}$$

Therefore $\eta\in\mathcal D(A^\dagger)$ and $A^\dagger\eta=w$.

The nonclosable example above makes the criterion concrete. A coefficient calculation gives

$$\mathcal D(A^\dagger) = \{y\in\ell^2:y_1=0\}, \qquad A^\dagger y=0.$$

This codimension-one subspace is closed and not dense, exactly as the closability theorem predicts.

Symmetric is an inclusion, not an equality

A densely defined endomorphism $A:\mathcal D(A)\subset H\to H$ is symmetric when

$$\langle\eta|A\psi\rangle = \langle A\eta|\psi\rangle \qquad \text{for all }\eta,\psi\in\mathcal D(A).$$

In operator notation, this is

$$A\subset A^\dagger.$$

It implies that every densely defined symmetric operator is closable. It does not imply equality of domains. A self-adjoint operator satisfies the stronger condition

$$A=A^\dagger, \qquad \mathcal D(A)=\mathcal D(A^\dagger).$$

The next page develops why this distinction controls extensions and unitary evolution. On the present page, the important lesson is already visible: checking a formal identity on $\mathcal D(A)$ establishes at most the inclusion. The adjoint-domain calculation decides whether equality holds.

Multiplication gives a model closed operator

Let $(X,\mu)$ be a measure space and let $m:X\to\mathbb C$ be measurable and finite almost everywhere. Define the maximal multiplication operator on $L^2(X,\mu)$ by

$$(M_m f)(x)=m(x)f(x),$$

with

$$\mathcal D(M_m) = \{f\in L^2(X,\mu):mf\in L^2(X,\mu)\}.$$

The functions

$$f^{(n)}=\mathbf 1_{\{|m|\leq n\}}f$$

belong to $\mathcal D(M_m)$ and converge to $f$ for every $f\in L^2$, so the domain is dense. If $h_n\to h$ and $m h_n\to g$ in $L^2$, subsequences converge almost everywhere, so $g=mh$ almost everywhere. Thus $M_m$ is closed.

The adjoint is

$$M_m^\dagger=M_{m^*}, \qquad \mathcal D(M_m^\dagger)=\mathcal D(M_{m^*}).$$

The inclusion follows from

$$\langle g|M_mf\rangle = \langle M_{m^*}g|f\rangle.$$

For the reverse inclusion, testing on functions supported where $|m|$ is bounded forces the representing vector to equal $m^*g$ and hence to lie in $L^2$. When $m$ is real almost everywhere, the maximal multiplication operator has equal operator and adjoint domains. When $m(x)=x$ on $L^2(\mathbb R)$, it is nevertheless unbounded.

This example is a prototype for the spectral representation of a self-adjoint operator, but the general spectral theorem belongs to the later functional-calculus page.

A derivative exposes the boundary data

Consider

$$P_0=-i\frac{d}{dx}, \qquad \mathcal D(P_0)=C_c^\infty(0,\ell) \subset L^2(0,\ell).$$

The domain is dense. For $f,g\in C_c^\infty(0,\ell)$, integration by parts gives

$$\langle g|P_0f\rangle -\langle P_0g|f\rangle = -i\,[g(x)^*f(x)]_{0}^{\ell} =0.$$

Therefore $P_0$ is symmetric and closable. To determine its adjoint, however, one must allow $g$ outside the test domain. The result is

$$\begin{aligned} \mathcal D(P_0^\dagger)&=H^1(0,\ell),\\ P_0^\dagger g&=-ig', \end{aligned}$$

where $g'$ is the weak derivative. Indeed, the adjoint boundedness condition is precisely the distributional statement that $g$ has a weak derivative in $L^2$. The closure is

$$\begin{aligned} \mathcal D(\overline P_0)&=H_0^1(0,\ell),\\ \overline P_0f&=-if'. \end{aligned}$$

Here $H_0^1(0,\ell)$ is the closure of $C_c^\infty(0,\ell)$ in the $H^1$ norm; in one dimension its elements have zero trace at both endpoints. Thus

$$P_0\subset\overline P_0 \subset P_0^\dagger,$$

and all three operators use the same differential expression on different domains.

For arbitrary $f,g\in H^1(0,\ell)$, the boundary form is

$$\langle g|-if'\rangle -\langle-ig'|f\rangle = -i\,[g(x)^*f(x)]_0^\ell.$$

Periodic, phase-twisted, or endpoint conditions select different restrictions of the maximal derivative. Vanishing of this form on a proposed domain tests symmetry. It does not by itself prove equality with the adjoint domain. Classifying the self-adjoint restrictions is deliberately deferred to the next page.

The same warning applies to $-d^2/dx^2$. The Dirichlet domain

$$\{f\in H^2(0,\ell):f(0)=f(\ell)=0\}$$

and the Neumann domain

$$\{f\in H^2(0,\ell):f'(0)=f'(\ell)=0\}$$

define different operators, despite sharing the displayed differential expression.

Hamiltonians and differential operators: a free-field example

For the page’s QFT-facing example, put a massive free scalar field in a spatial box with periodic boundary conditions. Let

$$\Lambda_L=\frac{2\pi}{L}\mathbb Z^s, \qquad \omega_{\mathbf k} = \sqrt{|\mathbf k|^2+m^2}, \qquad L>0,\quad m>0,\quad s\in\mathbb N,\quad s\geq1.$$

Each momentum mode is a harmonic oscillator. Let $\mathcal N_{\mathrm{fin}}$ be the set of occupation functions

$$\mathbf n:\Lambda_L\to\mathbb N_0$$

with finite support. The bosonic Fock space has an orthonormal occupation basis $|\mathbf n\rangle$ and can be represented as

$$\mathcal F \cong \ell^2(\mathcal N_{\mathrm{fin}}).$$

Begin on the algebraic span

$$\mathcal D_{\mathrm{fin}} = \operatorname{span}_{\mathrm{alg}} \{|\mathbf n\rangle:\mathbf n\in\mathcal N_{\mathrm{fin}}\}.$$

It is dense in $\mathcal F$. Assign the vacuum zero excitation energy and initially define

$$H_{0,\mathrm{alg}}|\mathbf n\rangle = E_{\mathbf n}|\mathbf n\rangle, \qquad E_{\mathbf n} = \sum_{\mathbf k\in\Lambda_L} \omega_{\mathbf k}n_{\mathbf k}.$$

The sum in $E_{\mathbf n}$ is finite for each basis vector. Yet the operator is unbounded: choose one-particle states with $|\mathbf k_j|\to\infty$. They have norm one while $\|H_{0,\mathrm{alg}}|\mathbf 1_{\mathbf k_j}\rangle\| =\omega_{\mathbf k_j}\to\infty$.

The adjoint-domain definition now performs the essential completion. If

$$\Psi=\sum_{\mathbf n}c_{\mathbf n}|\mathbf n\rangle,$$

then the functional $\Phi\mapsto\langle\Psi|H_{0,\mathrm{alg}}\Phi\rangle$ on $\mathcal D_{\mathrm{fin}}$ is ambient-norm bounded exactly when

$$\sum_{\mathbf n} E_{\mathbf n}^2|c_{\mathbf n}|^2<\infty.$$

Therefore

$$\begin{aligned} \mathcal D(H_{0,\mathrm{alg}}^\dagger) &= \left\{ \Psi\in\mathcal F: \sum_{\mathbf n} E_{\mathbf n}^2|c_{\mathbf n}|^2<\infty \right\},\\ H_{0,\mathrm{alg}}^\dagger\Psi &= \sum_{\mathbf n} E_{\mathbf n}c_{\mathbf n}|\mathbf n\rangle. \end{aligned}$$

Write

$$H_0:=H_{0,\mathrm{alg}}^\dagger$$

for this maximal diagonal operator. It is closed because every adjoint is closed. Finite occupation-basis truncations converge in its graph norm because, for $\Psi\in\mathcal D(H_0)$,

$$\|\Psi\|^2+\|H_0\Psi\|^2 = \sum_{\mathbf n} (1+E_{\mathbf n}^2)|c_{\mathbf n}|^2.$$

Hence $\mathcal D_{\mathrm{fin}}$ is a core for $H_0$, and

$$\overline{H_{0,\mathrm{alg}}}=H_0.$$

The domain is proper. Choose distinct one-particle modes with $\omega_{\mathbf k_j}\geq j$ and set

$$\Psi=C\sum_{j=1}^{\infty}\frac1j |\mathbf 1_{\mathbf k_j}\rangle,$$

where $C$ normalizes the vector. Then $\Psi\in\mathcal F$, but

$$\sum_{j=1}^{\infty} \omega_{\mathbf k_j}^2\left|\frac Cj\right|^2 \geq |C|^2\sum_{j=1}^{\infty}1 =\infty.$$

Therefore $\Psi\notin\mathcal D(H_0)$. A Hilbert-space state need not be in the Hamiltonian domain.

Only the excitation-energy diagonal operator is being analyzed here. The choice of representation, creation and annihilation operators, their common domains and canonical commutation relations, the additive vacuum-energy term, and its physical treatment belong to the Foundations page below. This example establishes neither a general interacting Hamiltonian nor a continuum construction of pointlike fields.

Canonical Quantization: Algebra, Representation, and State develops the canonical algebra, representation, vacuum, and free-field Hamiltonian. The present page supplies the domain-and-closure discipline needed for that treatment.

What each construction changes

Construction	Input	Output	Decisive test
Restriction	an operator and a smaller domain	a generally different operator	action agrees on the smaller domain
Closedness	one graph	no enlargement	simultaneous input-output limits stay in the graph
Closure	a closable graph	the smallest closed extension	no nonzero vertical graph limit
Core	a subspace of a closed operator domain	a reconstructing restriction	density in the graph norm
Adjoint	a densely defined operator	a closed operator with a new domain	$\psi\mapsto\langle\eta\mid A\psi\rangle$ is ambient-norm bounded
Symmetry	one densely defined operator	an operator inclusion	$A\subset A^\dagger$
Self-adjointness	an operator and its adjoint	exact equality	action and domains both agree

The central dependencies can be summarized as

$$\begin{gathered} A\text{ densely defined} \quad\Longrightarrow\quad A^\dagger\text{ exists and is closed},\\ A\text{ closable} \quad\Longleftrightarrow\quad \overline{\mathcal D(A^\dagger)}=H_2,\\ A\text{ closable} \quad\Longrightarrow\quad \overline A=A^{\dagger\dagger}. \end{gathered}$$

These statements do not say that every densely defined operator is closable, every closed operator is bounded, every symmetric operator is self-adjoint, or every formal product has a useful closure.

Common pitfalls

“Unbounded” means some output is infinite. Every $A\psi$ is an element of the Hilbert space. Unboundedness means that the ratio $\|A\psi\|/\|\psi\|$ has no finite uniform upper bound on the domain.

A dense test domain is automatically a core. Density in $\|\cdot\|$ is not enough. A core must be dense in the graph norm, so both $\psi_n\to\psi$ and $A\psi_n\to A\psi$ are controlled.

The adjoint uses the same domain. The adjoint domain is derived from an ambient-norm boundedness condition. For the minimal derivative it is $H^1(0,\ell)$, not $C_c^\infty(0,\ell)$.

Integration by parts proves self-adjointness. Vanishing boundary terms on the proposed domain proves symmetry. Self-adjointness also requires that the proposed domain equal the full adjoint domain.

A formal commutator is an operator identity on all states. Both products must first be defined. A calculation on a common invariant core remains a calculation on that core unless an extension theorem supplies more.

Exercises

Domain comparison. Let $A e_n=n e_n$ on $c_{00}\subset\ell^2$, and let $N$ be the maximal diagonal operator defined earlier. Are $A$ and $N$ the same operator?

Solution

No. They have the same action on $c_{00}$ but different domains, so $A\subset N$. Finite truncations converge in the $N$ graph norm; therefore $c_{00}$ is a core for $N$ and $\overline A=N$. Equality holds only after taking the closure.

Closability test. On $c_{00}$ define $Ax=(\sum_n n x_n)e_1$. Diagnose the obstruction to closure and compare it with the adjoint-domain criterion.

Solution

For $x^{(n)}=e_n/n$, one has $x^{(n)}\to0$ but $Ax^{(n)}=e_1\not\to0$. Hence the graph closure contains the nonzero vertical vector $(0,e_1)$ and is not the graph of an operator. Directly, $\mathcal D(A^\dagger)=\{y:y_1=0\}$, which is not dense. The sequence and adjoint criteria agree.

Boundary calculation. For $P=-i\,d/dx$ on $H^1(0,\ell)$, derive the boundary form. What does it prove on $C_c^\infty(0,\ell)$, and what does it not prove?

Solution

Integration by parts gives

$$\langle g|Pf\rangle-\langle Pg|f\rangle =-i[g^*f]_0^\ell.$$

The term vanishes for $f,g\in C_c^\infty(0,\ell)$, proving that the minimal operator is symmetric. It does not prove self-adjointness: its closure has domain $H_0^1(0,\ell)$ while its adjoint has domain $H^1(0,\ell)$.

Free-field transfer. Why can a normalized vector belong to Fock space but fail to belong to the domain of the free Hamiltonian?

Solution

Fock-space membership asks for $\sum_{\mathbf n}|c_{\mathbf n}|^2<\infty$. Hamiltonian-domain membership asks for the stronger weighted condition $\sum_{\mathbf n}E_{\mathbf n}^2|c_{\mathbf n}|^2<\infty$. Coefficients can therefore be square-summable while their energy-weighted coefficients are not. The explicit one-particle series with coefficients $1/j$ and energies at least $j$ gives such a vector.

Etingof 2023, §§8.2.2–8.2.5, pp. 102–107, PDF and Teschl 2014, §2.2, PDF give complementary treatments of domains, graphs, closures, adjoints, and the symmetric/self-adjoint distinction. The harmonic-oscillator and free-field application can be compared with Schwartz 2014, §§2.2–2.3.

Where to continue

Continue to Self-Adjointness, Extensions, and Unitary Evolution for boundary-form classifications, essential self-adjointness, deficiency indices, and Stone’s theorem. Continue to Spectra, Resolvents, Spectral Measures, and Functional Calculus for the resolvent set and spectral calculus of closed and self-adjoint operators.

References

• Pavel Etingof, Mathematical Ideas and Notions of Quantum Field Theory, PDF, §§8.2.2–8.2.5, pp. 102–107, MIT OpenCourseWare lecture notes, 2023. These sections develop domains, graphs, closures, adjoints, boundary forms, and the symmetric/self-adjoint distinction. Its inner product is conjugate-linear in the first slot, matching the site convention.
• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, §§2.2–2.3, Cambridge University Press, 2014. These sections derive harmonic oscillators, free-field modes, the free Hamiltonian, and creation and annihilation operators. The author’s first-printing corrections were checked. The finite-volume occupation-basis realization and the explicit maximal-domain and core argument on this page are the controlled mathematical specialization added here; they do not claim a construction of an interacting theory.
• Gerald Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, PDF, second edition, §2.2, American Mathematical Society, 2014. This is the structural source for operator inclusion, graph norms, adjoints, closability, double adjoints, the closed graph theorem, and the Hellinger–Toeplitz theorem. The author’s errata, PDF, updated March 18, 2026, were checked; in particular, this page does not use the unqualified closure-of-sums formula corrected there.