Hermitian vs Self-Adjoint Operators

For a finite matrix, Hermitian and self-adjoint mean the same thing:

$$A^\dagger=A.$$

For an unbounded operator on an infinite-dimensional Hilbert space, the same symbols can hide different domains. A differential expression may be formally Hermitian, an operator built from it may be symmetric on one domain, and only a particular domain choice may make it self-adjoint.

The short rule is:

Symmetric means the inner-product identity holds on the chosen domain. Self-adjoint means the operator and its adjoint have the same action and the same domain.

Momentum on a finite interval gives a concrete example: setting the wavefunction to zero at both endpoints makes momentum symmetric, but phase-twisted endpoint conditions are needed for its self-adjoint realizations.

Helpful background. Operators explain domains and adjoints. Spectra explains what the domain choice changes about possible measurement values.

Finite-Dimensional Case

Let $\mathcal H=\mathbb C^d$. For a linear map from this whole space to itself, the domain is

$$\mathcal D(A)=\mathcal H.$$

Its adjoint is also defined on all of $\mathcal H$. Therefore there is no domain mismatch to track, and the conditions

$$\langle\phi|A\psi\rangle = \langle A\phi|\psi\rangle$$

for all vectors and

$$A^\dagger=A$$

are equivalent.

In an orthonormal basis,

$$[A^\dagger] = \left([A]^*\right)^{\mathsf T},$$

so the familiar matrix condition is

$$A_{mn}=A_{nm}^*.$$

A Hermitian matrix has:

• real eigenvalues;
• orthogonal eigenspaces for distinct eigenvalues;
• an orthonormal eigenbasis;
• a spectral decomposition by orthogonal projectors;
• real expectation values in every state.

Thus “observables are represented by Hermitian matrices” is correct for finite-level systems, spin matrices, and ordinary quantum-information models. See Hermitian Operators for the matrix theory.

Why Infinite Dimension Is Different

An operator is not merely a formula. It includes a domain:

$$A:\mathcal D(A)\subseteq\mathcal H \longrightarrow\mathcal H.$$

For an unbounded operator, $\mathcal D(A)$ is generally a proper dense subspace. Position, momentum, and differential Hamiltonians do not act on every square-integrable function.

For example, the formal expression

$$-i\hbar\frac{d}{dx}$$

does not specify:

• which interval or manifold $x$ belongs to;
• which square-integrable functions possess the required derivative;
• which endpoint or matching conditions hold;
• whether the output remains in the Hilbert space.

Changing any of these choices can change the operator, its adjoint, and its spectrum.

The Adjoint Has Its Own Domain

Let $A$ be densely defined. A vector $|\phi\rangle$ belongs to $\mathcal D(A^\dagger)$ when there is a vector $|\eta\rangle\in\mathcal H$ such that

$$\langle\phi|A\psi\rangle = \langle\eta|\psi\rangle$$

for every $|\psi\rangle\in\mathcal D(A)$. The adjoint is then defined by

$$A^\dagger|\phi\rangle=|\eta\rangle.$$

The representing vector $|\eta\rangle$ is unique because $\mathcal D(A)$ is dense. Equivalently, the map

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

must be continuous with respect to the Hilbert-space norm on $\mathcal D(A)$.

The key point is:

$$\mathcal D(A^\dagger) \quad\text{need not equal}\quad \mathcal D(A).$$

The adjoint is therefore another operator with its own domain, not merely a formal conjugate-transpose symbol. See Adjoint Operators for the construction.

Symmetric Operators

A densely defined operator $A$ is symmetric when

$$\langle\phi|A\psi\rangle = \langle A\phi|\psi\rangle$$

for all $|\phi\rangle,|\psi\rangle\in\mathcal D(A)$.

In operator-inclusion notation,

$$A\subseteq A^\dagger.$$

This means

$$\mathcal D(A)\subseteq\mathcal D(A^\dagger)$$

and

$$A^\dagger|\psi\rangle=A|\psi\rangle \qquad \text{for every }|\psi\rangle\in\mathcal D(A).$$

Symmetry implies real expectation values on the domain:

$$\begin{aligned} \langle\psi|A\psi\rangle^* &=\langle A\psi|\psi\rangle\\ &=\langle\psi|A\psi\rangle. \end{aligned}$$

This is necessary for a real-valued observable, but it is not sufficient for the full sharp-observable framework. A symmetric operator can have an adjoint defined on more vectors than the original operator.

Self-Adjoint Operators

An operator is self-adjoint when it equals its adjoint as an operator:

$$A=A^\dagger.$$

This statement contains two conditions:

$$\mathcal D(A)=\mathcal D(A^\dagger)$$

and, on that common domain,

$$A|\psi\rangle=A^\dagger|\psi\rangle.$$

Every self-adjoint operator is symmetric. The converse is false for unbounded operators.

The same momentum expression gives two different domain comparisons. Dirichlet endpoint conditions make the boundary form vanish but leave a larger adjoint domain. A fixed phase-twisted condition is also obeyed by every vector in the adjoint domain, making that realization self-adjoint. The spaces $H^1$ and $H^1_0$ are explained in the interval example below.

Formal Hermiticity

A formal adjoint is obtained by algebraic conjugation and integration by parts while temporarily suppressing boundary terms. A differential expression is often called formally Hermitian when it equals this formal adjoint.

For example,

$$-i\hbar\frac{d}{dx}$$

has the expected formal adjoint expression. But formal equality says nothing by itself about which boundary terms vanish or which domains agree.

The following terms answer related but different questions:

1. Formally Hermitian: the differential expression matches its formal adjoint.
2. Symmetric: the boundary form vanishes for every pair in the chosen operator domain.
3. Self-adjoint: the symmetric operator has the same domain as its Hilbert- space adjoint.
4. Essentially self-adjoint: the closure of a symmetric operator is self-adjoint.

Formal Hermiticity by itself is not yet an operator property because no domain has been supplied. Once a domain is chosen, the boundary form determines whether the resulting operator is symmetric; equality with the adjoint domain then determines self-adjointness. Skipping those steps is the common error.

Boundary Forms

Integration by parts exposes the domain dependence. For momentum on $[0,L]$,

$$P=-i\hbar\frac{d}{dx},$$

one finds for sufficiently regular functions

$$\begin{aligned} \langle\phi|P\psi\rangle -\langle P\phi|\psi\rangle &= -i\hbar \left[ \phi^*(x)\psi(x) \right]_0^L \\ &= -i\hbar\phi^*(L)\psi(L) +i\hbar\phi^*(0)\psi(0). \end{aligned}$$

The right-hand side is the boundary form. A proposed domain makes $P$ symmetric only when this form vanishes for every allowed pair $\phi,\psi$.

For a second-order expression such as

$$H_0 = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2},$$

the corresponding boundary form is

$$\langle\phi|H_0\psi\rangle -\langle H_0\phi|\psi\rangle = -\frac{\hbar^2}{2m} \left[ \phi^*\psi' -\phi'^*\psi \right]_0^L.$$

Dirichlet, Neumann, periodic, and Robin-type conditions can cancel this form in different ways. They need not define the same self-adjoint operator or the same spectrum.

Momentum on an Interval

Let

$$\mathcal H=L^2([0,L])$$

and begin with the momentum expression

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

Dirichlet domain: symmetric but not self-adjoint

Here $H^1([0,L])$ consists of square-integrable functions with a square-integrable weak first derivative. On a finite interval they have absolutely continuous representatives, so endpoint values are well-defined. Consider

$$\mathcal D_0(P) = \left\lbrace \psi\in H^1([0,L]) \mid \psi(0)=\psi(L)=0 \right\rbrace.$$

The boundary form vanishes because both endpoint values vanish. Therefore $P$ is symmetric on $\mathcal D_0(P)$.

However, testing against every Dirichlet $\psi$ imposes no endpoint condition on vectors in the adjoint domain. One obtains the larger adjoint domain

$$\mathcal D(P^\dagger)=H^1([0,L]).$$

Hence

$$\mathcal D_0(P) \subsetneq \mathcal D(P^\dagger),$$

so this momentum operator is not self-adjoint.

Phase-twisted domains: self-adjoint

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

$$\mathcal D_\theta(P) = \left\lbrace \psi\in H^1([0,L]) \mid \psi(L)=e^{i\theta}\psi(0) \right\rbrace.$$

If $\phi$ and $\psi$ satisfy the same condition, then

$$\begin{aligned} \phi^*(L)\psi(L) &= e^{-i\theta}\phi^*(0) e^{i\theta}\psi(0)\\ &= \phi^*(0)\psi(0), \end{aligned}$$

so the boundary form vanishes. To see why the adjoint has no extra vectors, testing against smooth functions supported away from the endpoints first gives $\phi\in H^1$. The remaining boundary form is proportional to

$$\bigl[e^{i\theta}\phi^*(L)-\phi^*(0)\bigr]\psi(0).$$

The endpoint value $\psi(0)$ is arbitrary within the twisted domain, so the bracket must vanish. Taking its complex conjugate gives $\phi(L)=e^{i\theta}\phi(0)$. Thus

$$\mathcal D(P^\dagger) = \mathcal D_\theta(P)$$

for this realization, and $P_\theta$ is self-adjoint.

Its normalized eigenfunctions are

$$\psi_n(x) = \frac{1}{\sqrt L} \exp\left(i k_nx\right),$$

where

$$k_n = \frac{2\pi n+\theta}{L}, \qquad p_n = \hbar k_n, \qquad n\in\mathbb Z.$$

Different $\theta$ give different self-adjoint operators and shifted spectra, even though the local expression is unchanged.

A reflecting box has a different Hamiltonian

The Dirichlet kinetic-energy operator $-\hbar^2\psi^{\prime\prime}/(2m)$ on $H^2([0,L])$ with $\psi(0)=\psi(L)=0$ is self-adjoint. Here $H^2$ requires square-integrable weak derivatives through second order. This does not make the first-derivative Dirichlet momentum self-adjoint: the two operators have different adjoint domains. A reflecting box and a ring with joined endpoints describe different physical systems.

Spectral and Dynamical Consequences

It is tempting to treat a symmetric operator as a self-adjoint operator with a minor technical defect. That framing is misleading. The missing domain information controls whether:

• the spectrum is the real spectral set of a sharp observable;
• a projection-valued spectral measure exists;
• $f(A)$ is defined by the self-adjoint functional calculus;
• a Hamiltonian generates unitary evolution for all real times;
• boundary probability flux is conserved under the proposed dynamics.

Symmetry supplies real expectation values on a test domain. Self-adjointness supplies the complete operator needed by the spectral and dynamical theory.

Essentially Self-Adjoint Operators

A symmetric operator $A_0$ is essentially self-adjoint when its closure $\overline{A_0}$ is self-adjoint:

$$\overline{A_0} = A_0^\dagger.$$

Equivalently, the starting operator determines a unique self-adjoint extension. This is useful because differential operators are often introduced first on a convenient dense core such as smooth compactly supported functions.

If $A_0$ is essentially self-adjoint, no additional boundary parameter is needed to select the mathematical realization. If it is not, two mathematical possibilities must be distinguished:

• equal nonzero deficiency indices permit a family of self-adjoint extensions;
• unequal deficiency indices permit no self-adjoint extension.

Additional physical structure may select one member of an allowed family. The classification uses deficiency spaces and lies beyond this Core page. The mathematical entry point is Symmetric versus Self-Adjoint Operators.

Self-Adjoint Extensions

A self-adjoint extension $\widetilde A$ of a symmetric $A$ satisfies

$$A\subseteq\widetilde A = \widetilde A^\dagger.$$

The extension retains the original action on $\mathcal D(A)$ while enlarging the domain. For differential operators, extension parameters are often boundary conditions.

The interval momentum example has a one-parameter family $P_\theta$. The Dirichlet-domain momentum operator $P_{\mathrm D}$ is contained in each:

$$P_{\mathrm D} \subset P_\theta.$$

There is no contradiction in having several self-adjoint extensions. The starting symmetric operator did not yet encode a complete physical boundary condition. Choosing an extension completes the model.

Different extensions can have different spectra and dynamics. They are not different gauges or different matrix representations of one already-specified operator.

Why Self-Adjointness Matters for Observables

A standard sharp observable is represented by a self-adjoint operator. The spectral theorem provides a projection-valued measure

$$\Delta \longmapsto P_A(\Delta)$$

on real outcome sets. For a normalized state,

$$\Pr(A\in\Delta) = \langle\psi|P_A(\Delta)|\psi\rangle.$$

This framework handles discrete, continuous, and mixed spectra. A merely symmetric operator does not automatically carry the required self-adjoint spectral measure.

Real expectation values are therefore not the whole observable criterion. They are a local quadratic-form property; an observable needs a globally defined outcome measure.

See Observables and Spectral Theorem, Practical Version for the physical and mathematical viewpoints.

Why Self-Adjointness Matters for Dynamics

Stone’s theorem relates self-adjoint generators to strongly continuous one-parameter unitary groups. If $H$ is self-adjoint and time-independent, then

$$U(t) = e^{-iHt/\hbar}$$

is unitary on the whole Hilbert space for every real $t$ and satisfies

$$U(t+s)=U(t)U(s).$$

Conversely, a strongly continuous one-parameter unitary group has a unique self-adjoint generator, up to the conventional factor involving $\hbar$.

This is the precise content behind the statement that a closed-system Hamiltonian must generate unitary time evolution. Mere symmetry of a formal differential expression is not the hypothesis of the theorem.

For initial states in $\mathcal D(H)$, this evolution also solves $i\hbar\dot\psi=H\psi$ as a Hilbert-space derivative. States outside that domain still evolve unitarily, although that strong derivative need not exist. The dynamics-facing treatment is Unitary Time Evolution.

Probability Current and Boundary Conditions

For the Schrödinger Hamiltonian on an interval, self-adjoint boundary conditions are closely related to vanishing net boundary form and conservation of probability. In one dimension, the probability current is

$$j(x) = \frac{\hbar}{m} \operatorname{Im} \left( \psi^*(x)\psi'(x) \right).$$

For a sufficiently regular solution,

$$\frac{d}{dt} \int_0^L|\psi(x,t)|^2\,dx = j(0,t)-j(L,t).$$

Boundary conditions compatible with a self-adjoint Hamiltonian ensure that the boundary contribution is controlled so the unitary dynamics preserves total probability. Different self-adjoint boundary conditions can encode different physical connections of the endpoints or different reflecting boundaries.

This current argument is a useful physical diagnostic, but it is not a substitute for proving equality of operator domains.

Common Physics Shorthand

Physicists often say “Hermitian operator” in all dimensions. The intended meaning varies:

• For a finite matrix, it unambiguously means $A^\dagger=A$ and is equivalent to self-adjointness.
• For a bounded everywhere-defined operator, symmetry implies self-adjointness, so the shorthand is normally safe.
• For an unbounded differential operator, it may mean formally Hermitian, symmetric on an understood domain, essentially self-adjoint on a core, or genuinely self-adjoint.

The shorthand is acceptable when the domain-sensitive statement has already been established and the context is stable. When domains, boundary conditions, singular potentials, spectra, or generated dynamics are at issue, use the precise term.

Common Pitfalls

Vanishing boundary terms prove symmetry first. Self-adjointness requires testing which additional functions the adjoint accepts. The interval example shows how a boundary condition can make the first test pass while the second fails.

A self-adjoint extension changes the model’s domain. It is not merely a basis change. Extension parameters can alter both spectra and dynamics, so geometry and physical boundary conditions must select the appropriate realization.

Real expectations are insufficient. Symmetric operators have real expectations on their domains, yet may have several self-adjoint extensions or none. A sharp spectral measure and a unitary group require a self-adjoint realization.

For proofs of the extension criteria, continue with Self-Adjoint Operators. For concrete wavefunctions and currents, see Boundary Conditions.

Exercises

Exercise 1: Finite-dimensional equivalence

Let $A$ be a linear operator on $\mathbb C^d$. Suppose

$$\langle\phi|A\psi\rangle = \langle A\phi|\psi\rangle$$

for all $\phi,\psi$. Prove that $A^\dagger=A$.

Solution

By the definition of the adjoint,

$$\langle\phi|A\psi\rangle = \langle A^\dagger\phi|\psi\rangle.$$

Combining the two identities gives

$$\langle(A^\dagger-A)\phi|\psi\rangle=0$$

for every $\psi$. The only vector orthogonal to every vector is zero, so

$$(A^\dagger-A)\phi=0$$

for every $\phi$. Hence $A^\dagger=A$.

Exercise 2: The momentum boundary form

For $P=-i\hbar\,d/dx$ on $[0,L]$, derive

$$\langle\phi|P\psi\rangle -\langle P\phi|\psi\rangle = -i\hbar \left[ \phi^*(x)\psi(x) \right]_0^L.$$

Solution

Starting from

$$\langle\phi|P\psi\rangle = \int_0^L \phi^*(x) \left(-i\hbar\psi'(x)\right)\,dx,$$

integration by parts gives

$$\langle\phi|P\psi\rangle = -i\hbar \left[\phi^*\psi\right]_0^L +\int_0^L \left(-i\hbar\phi'(x)\right)^* \psi(x)\,dx.$$

The integral is $\langle P\phi|\psi\rangle$. Subtracting it yields the stated boundary form.

Exercise 3: Dirichlet momentum

Show that the momentum operator on

$$\mathcal D_0 = \left\lbrace \psi\in H^1([0,L]) \mid \psi(0)=\psi(L)=0 \right\rbrace$$

is symmetric. Explain why this calculation alone does not prove self-adjointness.

Solution

For $\phi,\psi\in\mathcal D_0$, both endpoint products vanish:

$$\phi^*(L)\psi(L) = \phi^*(0)\psi(0) = 0.$$

The boundary form is therefore zero, which proves symmetry.

Self-adjointness additionally requires

$$\mathcal D(P^\dagger)=\mathcal D_0.$$

The integration-by-parts calculation with Dirichlet test functions does not force an adjoint-domain vector to vanish at the endpoints. The adjoint domain is larger, so the equality fails.

Exercise 4: Twisted boundary condition

Let

$$\psi(L)=e^{i\theta}\psi(0), \qquad \phi(L)=e^{i\theta}\phi(0).$$

Show that the momentum boundary form vanishes.

Solution

Complex conjugation gives

$$\phi^*(L)=e^{-i\theta}\phi^*(0).$$

Therefore,

$$\begin{aligned} \phi^*(L)\psi(L) &= e^{-i\theta}\phi^*(0) e^{i\theta}\psi(0)\\ &= \phi^*(0)\psi(0). \end{aligned}$$

The endpoint difference in the boundary form is zero. This proves symmetry on the twisted domain; the adjoint-domain argument then shows this realization is self-adjoint.

Exercise 5: Twisted momentum spectrum

Solve the momentum eigenvalue equation on $[0,L]$ with

$$\psi(L)=e^{i\theta}\psi(0).$$

Find the allowed momenta.

Solution

The eigenvalue equation

$$-i\hbar\psi'(x)=p\psi(x)$$

has solutions

$$\psi(x)=Ce^{ipx/\hbar}.$$

The boundary condition requires

$$e^{ipL/\hbar}=e^{i\theta}.$$

Thus

$$\frac{pL}{\hbar} = \theta+2\pi n, \qquad n\in\mathbb Z,$$

and

$$p_n = \frac{\hbar}{L} \left(2\pi n+\theta\right).$$

Changing the extension parameter shifts the spectrum.

Exercise 6: Second-derivative boundary form

For

$$H_0 = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2},$$

integrate by parts twice to derive

$$\langle\phi|H_0\psi\rangle -\langle H_0\phi|\psi\rangle = -\frac{\hbar^2}{2m} \left[ \phi^*\psi' -\phi'^*\psi \right]_0^L.$$

Solution

Begin with

$$\langle\phi|H_0\psi\rangle = -\frac{\hbar^2}{2m} \int_0^L\phi^*\psi''\,dx.$$

Integrating once,

$$\int_0^L\phi^*\psi''\,dx = \left[\phi^*\psi'\right]_0^L -\int_0^L\phi'^*\psi'\,dx.$$

Integrating the remaining integral once more,

$$\int_0^L\phi'^*\psi'\,dx = \left[\phi'^*\psi\right]_0^L -\int_0^L\phi''^*\psi\,dx.$$

Combining terms leaves the stated boundary expression plus $\langle H_0\phi|\psi\rangle$.

Exercise 7: Real expectation values are not enough

Explain why the fact that

$$\langle\psi|A\psi\rangle\in\mathbb R$$

for every $\psi$ in a dense test domain does not by itself prove that $A$ is a self-adjoint observable.

Solution

Real quadratic expectations establish, under the usual polarization assumptions, symmetry on the chosen domain. They do not show that the adjoint has the same domain. A symmetric operator may have a larger adjoint domain, several self-adjoint extensions, or no self-adjoint extension. The projection-valued spectral measure for a sharp observable requires a specified self-adjoint realization, not only real expectations on a test domain.

Exercise 8: Why the Hamiltonian condition is stronger

State why symmetry alone does not guarantee that the formal expression

$$U(t)=e^{-iHt/\hbar}$$

defines unitary evolution for all real $t$.

Solution

The theorem connecting generators to strongly continuous unitary groups uses a self-adjoint generator. A symmetric $H$ may not possess the spectral calculus needed to define a unitary exponential on the whole Hilbert space, and different self-adjoint extensions can generate different dynamics. One must identify a self-adjoint $H$, or prove essential self-adjointness of a chosen core, before invoking the standard unitary evolution.

References

• G. Bonneau, J. Faraut, and G. Valent, “Self-adjoint extensions of operators and the teaching of quantum mechanics,” American Journal of Physics 69, 322–331 (2001). doi:10.1119/1.1328351; author manuscript.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013. Publisher.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014, Chapters 2–5. Author’s book page and online edition.