Self-Adjointness, Extensions, and Unitary Evolution

Let $S$ be a densely defined symmetric operator on a complex Hilbert space. Symmetry gives only

$$S\subset S^\dagger.$$

After closing $S$, choose a scale $\mu>0$ with the same physical dimension as $S$ and form the deficiency spaces

$$\mathcal N_+ = \ker(S^\dagger-i\mu I), \qquad \mathcal N_- = \ker(S^\dagger+i\mu I).$$

Their dimensions do not depend on the chosen positive $\mu$. The operator has a self-adjoint extension exactly when

$$\dim\mathcal N_+=\dim\mathcal N_-.$$

It is essentially self-adjoint exactly when both dimensions vanish, in which case its closure is the unique self-adjoint extension. When the dimensions are equal and nonzero, each unitary map $W:\mathcal N_+\to\mathcal N_-$ selects one extension. For differential operators, that unitary choice appears as boundary conditions.

Self-adjointness is also the exact static-generator condition in Stone’s theorem. A self-adjoint $H$ determines the strongly continuous unitary group

$$U(t)=e^{-itH},$$

and every strongly continuous one-parameter unitary group has one unique self-adjoint generator in this convention. The evolution exists for every Hilbert-space vector, but the strong derivative

$$\frac{d}{dt}U(t)\psi=-iHU(t)\psi$$

exists precisely for $\psi\in\mathcal D(H)$.

Required background. Unbounded Operators, Domains, Closure, and Adjoints supplies operator inclusion, graph closure, adjoint domains, graph norms, cores, and the distinction between symmetry and self-adjointness.

Symmetric operators and self-adjoint realizations

The distinction used throughout is

$$S\subset S^\dagger \qquad\text{versus}\qquad S=S^\dagger.$$

This page works on a complex Hilbert space $H$ with the bra slot conjugate-linear and the ket slot linear, and uses the site unit convention $\hbar=1$. Thus

$$\langle\eta|S\xi\rangle = \langle S^\dagger\eta|\xi\rangle$$

for $\xi\in\mathcal D(S)$ and $\eta\in\mathcal D(S^\dagger)$.

This page develops the reusable extension criteria, the boundary-form test, and Stone’s theorem. General spectral measures and Borel functional calculus continue at Spectra, Resolvents, Spectral Measures, and Functional Calculus. Perturbation theorems for sums of unbounded operators, time-dependent Hamiltonians, and semigroup generators are not developed here.

The exact physical continuation is Hilbert Positivity and Unitary Evolution. That page develops probability conservation, physical positivity, and the qualification needed for gauge-fixed auxiliary spaces.

Symmetric, self-adjoint, and essentially self-adjoint

Let $S:\mathcal D(S)\subset H\to H$ be densely defined. It is symmetric when

$$\langle\eta|S\xi\rangle = \langle S\eta|\xi\rangle \qquad \text{for every }\eta,\xi\in\mathcal D(S),$$

equivalently $S\subset S^\dagger$. It is self-adjoint when the inclusion is equality as operators:

$$\mathcal D(S)=\mathcal D(S^\dagger), \qquad S\xi=S^\dagger\xi \quad \text{on that domain}.$$

It is essentially self-adjoint when its closure is self-adjoint:

$$\overline S=(\overline S)^\dagger.$$

An essentially self-adjoint operator may be presented on a small convenient core, while its closure supplies the unique self-adjoint operator. A merely symmetric operator may have no self-adjoint extension or many of them. Every self-adjoint operator is maximal symmetric: it has no proper symmetric extension. The converse is false; the half-line momentum below supplies a maximal symmetric operator that is not self-adjoint.

The expectation value of a symmetric operator is real on its domain:

$$\langle\xi|S\xi\rangle = \langle S\xi|\xi\rangle = \langle\xi|S\xi\rangle^*.$$

That calculation does not compare $\mathcal D(S)$ with $\mathcal D(S^\dagger)$ and therefore does not prove self-adjointness.

Range criteria

Fix $\mu>0$. Symmetry makes the cross term vanish in

$$\|(S\pm i\mu I)\xi\|^2 = \|S\xi\|^2+\mu^2\|\xi\|^2.$$

In particular, $S\pm i\mu I$ are injective and bounded below by $\mu$. The adjoint relation gives

$$\begin{aligned} \operatorname{Ran}(S+i\mu I)^\perp &= \ker(S^\dagger-i\mu I),\\ \operatorname{Ran}(S-i\mu I)^\perp &= \ker(S^\dagger+i\mu I). \end{aligned}$$

Consequently,

$$\boxed{ S\text{ is essentially self-adjoint} \quad\Longleftrightarrow\quad \operatorname{Ran}(S\pm i\mu I)\text{ are both dense}. }$$

If $S$ is closed and symmetric, the displayed norm identity makes those ranges closed. In that case density can be replaced by surjectivity, and

$$\boxed{ S\text{ is self-adjoint} \quad\Longleftrightarrow\quad \operatorname{Ran}(S\pm i\mu I)=H. }$$

Both signs are required. One dense or surjective range controls only one deficiency space.

The boundary form locates admissible domains

For a densely defined symmetric $S$, define on $\mathcal D(S^\dagger)$ the boundary form

$$\mathfrak b_S(\eta,\xi) = \langle\eta|S^\dagger\xi\rangle -\langle S^\dagger\eta|\xi\rangle.$$

It is conjugate-linear in $\eta$, linear in $\xi$, and anti-Hermitian:

$$\mathfrak b_S(\eta,\xi) = -\mathfrak b_S(\xi,\eta)^*.$$

Suppose

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

Then $A$ is symmetric exactly when

$$\mathfrak b_S(\eta,\xi)=0 \qquad \text{for all }\eta,\xi\in\mathcal D(A).$$

The adjoint domain is the boundary-form orthogonal complement:

$$\mathcal D(A^\dagger) = \left\{ \eta\in\mathcal D(S^\dagger): \mathfrak b_S(\eta,\xi)=0 \text{ for every }\xi\in\mathcal D(A) \right\}.$$

Therefore $A$ is self-adjoint precisely when its domain equals this orthogonal complement. Vanishing of the boundary form proves symmetry; self-adjointness requires the domain to equal its full boundary-form orthogonal complement.

The form vanishes whenever either entry lies in $\mathcal D(\overline S)$, so the nontrivial boundary data live in the quotient

$$\mathcal D(S^\dagger)/\mathcal D(\overline S).$$

Self-adjoint extensions correspond to self-orthogonal, often called Lagrangian, subspaces of this boundary-data space. Merely being maximal among isotropic subspaces is not enough when the two deficiency dimensions are unequal. For ordinary differential operators, integration by parts turns $\mathfrak b_S$ into endpoint values.

Deficiency indices classify the extensions

Every densely defined symmetric operator is closable, and $S^\dagger=(\overline S)^\dagger$. Replace $S$ by $\overline S$ for the classification, so assume in this section that $S$ is closed.

For $\mu>0$ with the same physical dimension as $S$, define

$$\mathcal N_+ = \ker(S^\dagger-i\mu I), \qquad \mathcal N_- = \ker(S^\dagger+i\mu I),$$

and

$$n_+=\dim\mathcal N_+, \qquad n_-=\dim\mathcal N_-.$$

The numbers $n_\pm$ are independent of the chosen point in the corresponding open half-plane. They may be finite or infinite. Von Neumann’s decomposition is the direct sum

$$\mathcal D(S^\dagger) = \mathcal D(S) \mathbin{\dotplus} \mathcal N_+ \mathbin{\dotplus} \mathcal N_-.$$

The boundary form has opposite signs on the two deficiency spaces:

$$\begin{aligned} \mathfrak b_S(\eta_+,\eta_+) &= 2i\mu\|\eta_+\|^2,\\ \mathfrak b_S(\eta_-,\eta_-) &= -2i\mu\|\eta_-\|^2. \end{aligned}$$

Cancellation is possible exactly when the positive and negative deficiency spaces have the same Hilbert dimension.

Von Neumann extension theorem. The closed symmetric operator $S$ has a self-adjoint extension if and only if

$$n_+=n_-.$$

For every unitary map

$$W:\mathcal N_+\longrightarrow\mathcal N_-,$$

define

$$\mathcal D(S_W) = \left\{ \phi+\eta_++W\eta_+: \phi\in\mathcal D(S),\ \eta_+\in\mathcal N_+ \right\}$$

and

$$S_W(\phi+\eta_++W\eta_+) = S\phi+i\mu\eta_+-i\mu W\eta_+.$$

Then $S_W$ is self-adjoint, and every self-adjoint extension of $S$ arises from exactly one such $W$. Thus:

• $n_+=n_-=0$ means that $S$ is self-adjoint; for an unclosed initial operator, this means essential self-adjointness and a unique closure;
• $n_+=n_->0$ gives a family of self-adjoint extensions; and
• $n_+\neq n_-$ means that no self-adjoint extension exists in the given Hilbert space.

The theorem classifies mathematical realizations. Additional physics may select among them through locality, symmetries, positivity, or a specified boundary interaction, but those criteria are not consequences of the deficiency dimensions alone.

Proof status. The full von Neumann domain decomposition is taken here as a standard theorem. Once that decomposition is known, the extension formula has a short proof. The boundary form has coefficients $+2i\mu$ on $\mathcal N_+$ and $-2i\mu$ on $\mathcal N_-$ with equal magnitude. Its restriction to

$$\{\eta_++W\eta_+:\eta_+\in\mathcal N_+\}$$

vanishes exactly when $W$ is an isometry, and self-orthogonality requires it to be onto. The displayed action is simply the restriction of $S^\dagger$, using $S^\dagger\eta_+=i\mu\eta_+$ and $S^\dagger W\eta_+=-i\mu W\eta_+$. This proves the parametrization once the decomposition theorem is granted. Teschl 2014, §2.6, PDF proves the equivalent extension classification via the Cayley transform; Bonneau, Faraut, and Valent 2001, §4, PDF state the deficiency-index trichotomy used in the examples.

A semibounded operator has a distinguished extension

There is an important sufficient existence theorem that does not require computing deficiency vectors. Suppose the densely defined symmetric operator $S$ is bounded below:

$$\langle\psi|S\psi\rangle \geq \gamma\|\psi\|^2 \qquad \text{for every }\psi\in\mathcal D(S)$$

for some $\gamma\in\mathbb R$. The Friedrichs extension theorem produces a self-adjoint extension $S_{\mathrm F}$ with the same lower bound,

$$S_{\mathrm F}\geq\gamma I.$$

The construction shifts the quadratic form by $\gamma$, closes the resulting nonnegative form, and recovers its associated self-adjoint operator. The full closed-form representation theorem is cited rather than developed here (Teschl 2014, §2.3, PDF).

This result guarantees a distinguished semibounded extension. It does not say that $S$ is essentially self-adjoint, that the extension is the only self-adjoint one, or that every other extension preserves the same lower bound. For example, a minimal second-order operator on a domain with boundary can have many realizations even though its Friedrichs extension is distinguished by the closed energy form.

One derivative, three outcomes

Take the formal momentum expression

$$P=-i\frac{d}{dx}.$$

For each interval below, begin on $C_c^\infty$ and then close the minimal operator. Its adjoint is the maximal weak derivative on $H^1$.

A finite interval: many extensions

On $L^2(0,\ell)$,

$$\begin{aligned} \mathcal D(S)&=H_0^1(0,\ell),\\ \mathcal D(S^\dagger)&=H^1(0,\ell),\\ S^\dagger f&=-if'. \end{aligned}$$

The deficiency equations give

$$\mathcal N_+ = \operatorname{span}\{e^{-\mu x}\}, \qquad \mathcal N_- = \operatorname{span}\{e^{\mu x}\}.$$

Here $\mu$ has inverse-length units. Both functions are square-integrable on the finite interval, so the indices are $(1,1)$. The boundary form is

$$\mathfrak b_S(g,f) = -i\,[g(x)^*f(x)]_0^\ell.$$

For each $\theta\in[0,2\pi)$, define

$$\begin{aligned} \mathcal D(P_\theta) &= \{f\in H^1(0,\ell): f(\ell)=e^{i\theta}f(0)\},\\ P_\theta f&=-if'. \end{aligned}$$

If $f$ and $g$ obey the same phase-twisted boundary condition, the endpoint term vanishes, so $P_\theta$ is symmetric. Conversely, if $g\in\mathcal D(P_\theta^\dagger)$, the boundary form must vanish for every admissible $f$. Varying $f(0)$ gives

$$g(\ell)=e^{i\theta}g(0).$$

Thus the adjoint has exactly the same domain:

$$P_\theta^\dagger=P_\theta.$$

The two-endpoint condition $f(0)=f(\ell)=0$ defines the closed minimal symmetric operator, not a self-adjoint momentum operator. It imposes too many conditions for this first-order expression.

The whole line: a unique extension

On $L^2(\mathbb R)$, neither $e^{-\mu x}$ nor $e^{\mu x}$ is square-integrable at both ends. Hence

$$(n_+,n_-)=(0,0).$$

The momentum operator on $C_c^\infty(\mathbb R)$ is essentially self-adjoint, and its unique closure has domain $H^1(\mathbb R)$.

The half-line: no extension

On $L^2(0,\infty)$, $e^{-\mu x}$ is square-integrable and $e^{\mu x}$ is not. Thus

$$(n_+,n_-)=(1,0),$$

and the minimal half-line momentum has no self-adjoint extension in $L^2(0,\infty)$. A plausible-looking boundary condition cannot overcome unequal deficiency indices. It is also maximal symmetric. Indeed, because $\mathcal N_-=\{0\}$, every vector in $\mathcal D(S^\dagger)\setminus\mathcal D(S)$ has a nonzero component $\eta_+\in\mathcal N_+$. The decomposition above gives

$$\mathfrak b_S(\eta_+,\eta_+) = 2i\mu\|\eta_+\|^2 \neq0,$$

so adjoining any such vector would destroy symmetry. There is therefore no proper symmetric extension, even though $S\neq S^\dagger$.

These three cases share the same formal expression. The geometry and domain change the operator-theoretic answer (Bonneau, Faraut, and Valent 2001, §§3–5, PDF).

Stone’s theorem turns self-adjointness into dynamics

A strongly continuous one-parameter unitary group is a map

$$U:\mathbb R\longrightarrow\mathcal B(H)$$

such that

$$\begin{aligned} U(0)&=I,\\ U(t+s)&=U(t)U(s),\\ U(t)^\dagger U(t)&=U(t)U(t)^\dagger=I,\\ \lim_{t\to0}\|U(t)\psi-\psi\|&=0 \quad\text{for every }\psi\in H. \end{aligned}$$

Strong continuity is an essential hypothesis. Pointwise algebraic group identities alone do not supply a densely defined generator.

Stone’s theorem. For every strongly continuous one-parameter unitary group there is a unique self-adjoint operator $H$ such that

$$U(t)=e^{-itH}.$$

Its domain and action are recovered from the strong derivative:

$$\begin{aligned} \mathcal D(H) &= \left\{ \psi\in H: \lim_{t\to0} \frac{U(t)\psi-\psi}{t} \text{ exists in }H \right\},\\ H\psi &= i\lim_{t\to0} \frac{U(t)\psi-\psi}{t}. \end{aligned}$$

Conversely, every self-adjoint $H$ defines such a group. For $\psi\in\mathcal D(H)$,

$$U(t)\psi\in\mathcal D(H), \qquad HU(t)\psi=U(t)H\psi,$$

and

$$i\frac{d}{dt}U(t)\psi = HU(t)\psi.$$

For a general $\psi\in H$, $U(t)\psi$ remains norm-continuous and norm-preserving, but it need not be differentiable. Stone’s theorem concerns an autonomous group. A time-dependent family $H(t)$ requires separate existence, domain-stability, and propagator results (Teschl 2014, §5.1, PDF).

Proof status. Stone’s theorem is used here as a cited theorem. The self-adjoint-generator-to-unitary-group direction uses spectral functional calculus, whose general construction is deferred to the next page; the converse generator theorem is likewise cited rather than reproved. The diagonal Fock calculation below is an independent special-case check, not a proof of the general theorem.

Boundary conditions and quantum time evolution

Return to the finite-volume massive free-scalar excitation Hamiltonian from the prerequisite page. The mode-Hamiltonian interpretation follows Schwartz 2014, §§2.2–2.3; the operator-domain argument below is the page’s controlled mathematical specialization. In the occupation basis,

$$H_0\Psi = \sum_{\mathbf n} E_{\mathbf n}c_{\mathbf n}|\mathbf n\rangle,$$

with

$$\mathcal D(H_0) = \left\{ \Psi: \sum_{\mathbf n} E_{\mathbf n}^2|c_{\mathbf n}|^2<\infty \right\}.$$

There, $H_{0,\mathrm{alg}}$ was the restriction to finite occupation-basis sums, and the adjoint and closure calculations gave

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

It follows that

$$H_0^\dagger = (\overline{H_{0,\mathrm{alg}}})^\dagger = H_{0,\mathrm{alg}}^\dagger = H_0.$$

Thus the maximal diagonal Hamiltonian is self-adjoint, while the algebraic restriction is essentially self-adjoint.

Stone’s unitary group is explicit:

$$U_0(t)\Psi = \sum_{\mathbf n} e^{-itE_{\mathbf n}} c_{\mathbf n}|\mathbf n\rangle.$$

It preserves the norm because every multiplier has absolute value one. It is strongly continuous for every $\Psi\in\mathcal F$, since

$$\|U_0(t)\Psi-\Psi\|^2 = \sum_{\mathbf n} |e^{-itE_{\mathbf n}}-1|^2 |c_{\mathbf n}|^2 \longrightarrow0$$

by dominated convergence, using the summable majorant $4|c_{\mathbf n}|^2$.

The derivative requires the stronger weighted condition:

$$\lim_{t\to0} \frac{U_0(t)\Psi-\Psi}{t} = -iH_0\Psi \quad\Longleftrightarrow\quad \Psi\in\mathcal D(H_0).$$

In particular, the normalized state constructed on the prerequisite page with square-summable coefficients but divergent energy-weighted norm evolves unitarily, yet has no strong Schrödinger derivative at $t=0$.

This is the page’s controlled QFT application. It does not construct an interacting Hamiltonian, prove positivity of a physical gauge-theory state space, or handle a time-dependent background. Those physical questions continue at Hilbert Positivity and Unitary Evolution.

Decision path

For a proposed Hamiltonian or differential realization:

1. state the dense domain and close the symmetric restriction;
2. calculate the adjoint domain and boundary form;
3. compute both deficiency spaces or invoke a theorem that proves essential self-adjointness;
4. if the indices agree but do not vanish, declare the additional principle selecting a self-adjoint extension; and
5. only after a self-adjoint realization is fixed, invoke Stone’s theorem for its autonomous two-sided unitary group.

Different extension choices can produce different dynamics. The differential expression alone does not select one.

Common pitfalls

A real formal expression is self-adjoint. A differential expression has no adjoint domain by itself. One must specify a dense domain, calculate the Hilbert adjoint, and compare domains.

A vanishing boundary term proves self-adjointness. It proves symmetry on the proposed domain. Self-adjointness requires the domain to equal its full boundary-form orthogonal complement.

Every symmetric operator has a self-adjoint extension. Equal deficiency indices are necessary and sufficient. The half-line momentum has indices $(1,0)$ and no self-adjoint extension in the stated Hilbert space.

The Friedrichs extension proves uniqueness. Semiboundedness guarantees a distinguished self-adjoint extension that preserves the lower bound. It does not force the deficiency indices to vanish or exclude other extensions.

Maximal symmetric means self-adjoint. Self-adjoint operators are maximal symmetric, but the converse fails. Unequal deficiency indices can leave no room for a proper symmetric extension while still preventing equality with the adjoint.

Essentially self-adjoint means already closed. It means that the closure is self-adjoint. A small core such as $C_c^\infty(\mathbb R)$ need not itself be complete in the graph norm.

A symmetric expression has the evolution $e^{-itS}$. Symmetry alone does not produce a unique global unitary group. If self-adjoint extensions exist but are nonunique, the evolution depends on the selected extension.

Self-adjoint means positive. Self-adjointness makes the spectrum real and licenses unitary generation; it does not imply positivity or a lower bound. The momentum operators above are self-adjoint without being positive.

The exponential solves Schrödinger’s equation on every vector. The unitary group acts on every vector, but the strong differential equation holds exactly on the generator domain.

Stone’s theorem covers $H(t)$. The theorem classifies autonomous one-parameter unitary groups. A time-dependent Hamiltonian is a different evolution problem.

Exercises

Three levels. State the operator equalities that distinguish symmetric, self-adjoint, and essentially self-adjoint operators.

Solution

Symmetry is the inclusion $S\subset S^\dagger$. Self-adjointness is the operator equality $S=S^\dagger$, including equality of domains. Essential self-adjointness is $\overline S=(\overline S)^\dagger$; it says that the closure is the unique self-adjoint extension.

Missing-extension test. Compute the deficiency indices of $-i\,d/dx$ initially defined on $C_c^\infty(0,\infty)$. Why can no endpoint phase repair the result?

Solution

The adjoint acts as $-i\,d/dx$ on $H^1(0,\infty)$. The equations $S^\dagger f=\pm i\mu f$ give $e^{-\mu x}$ and $e^{\mu x}$. Only the first is square-integrable, so $(n_+,n_-)=(1,0)$. Von Neumann’s theorem requires equal indices; no unitary map can exist from a one-dimensional deficiency space to a zero-dimensional one.

Boundary derivation. Prove directly that

$$\mathcal D(P_\theta) = \{f\in H^1(0,\ell):f(\ell)=e^{i\theta}f(0)\}$$

makes $-i\,d/dx$ self-adjoint.

Solution

Integration by parts gives $\mathfrak b(g,f)=-i[g^*f]_0^\ell$. If $f$ and $g$ satisfy the same twisted condition, the endpoint products agree, proving symmetry. If $g$ belongs to the adjoint domain, the form must vanish for all such $f$. Since $f(0)$ is arbitrary and $f(\ell)=e^{i\theta}f(0)$, this forces $g(\ell)=e^{i\theta}g(0)$. The adjoint domain is therefore exactly $\mathcal D(P_\theta)$.

QFT transfer. A Fock vector $\Psi$ lies in $\mathcal F$ but not in $\mathcal D(H_0)$. Which parts of unitary evolution still exist?

Solution

The vector $U_0(t)\Psi$ exists for every $t$, depends continuously on $t$ in norm, and satisfies $\|U_0(t)\Psi\|=\|\Psi\|$. What fails is the strong derivative at $t=0$, so $H_0\Psi$ and the strong Schrödinger equation are not defined for that initial vector.

Where to continue

Continue to Spectra, Resolvents, Spectral Measures, and Functional Calculus for the spectral theorem, projection-valued measures, resolvents, and functions of the selected self-adjoint realization. For the physical role of positivity, probability conservation, free-scalar evolution, and the gauge-fixed qualification, continue to Hilbert Positivity and Unitary Evolution.

References

• Guy Bonneau, Jacques Faraut, and Galliano Valent (2001), extended arXiv version of “Self-adjoint extensions of operators and the teaching of quantum mechanics”, PDF, §§3–5 (pp. 5–9), American Journal of Physics 69 (2001), 322–331, doi:10.1119/1.1328351. This is the specialist anchor for deficiency indices and the momentum operator on the line, half-line, and finite interval. This page sets the paper’s explicit $\hbar$ to one; for the momentum examples, $\mu$ is the inverse of the paper’s positive length scale after that specialization. The remaining notation is translated to the site’s adjoint and inner-product convention.
• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, §§2.2–2.3, Cambridge University Press, 2014. These sections derive oscillator modes and the free-field Hamiltonian. The author’s first-printing corrections were checked. The finite-volume diagonal domain and strong-continuity argument are the controlled mathematical specialization on this page.
• Gerald Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, PDF, second edition, §§2.2–2.3, 2.6, and 5.1, American Mathematical Society, 2014. These sections establish self-adjointness, the Friedrichs extension, defect spaces, boundary realizations, and Stone’s theorem. The author’s errata, PDF, updated March 18, 2026, were checked, including the corrections on printed pp. 93–94 and to the time-evolution arguments on pp. 146 and 148. This page does not reproduce the corrected conjugation or resolvent-difference proof text.