Self-Adjoint Operators

A quantum observable such as position, momentum, or energy is usually an unbounded operator. Such an operator is not specified by a formula alone: its domain is part of its definition. A densely defined operator $A:D(A)\subset\mathcal H\to\mathcal H$ is self-adjoint when

$$A=A^*,$$

meaning both that $A\psi=A^*\psi$ on their common vectors and that $D(A)=D(A^*)$. This equality of domains is the decisive condition. It is what supports a real spectral measure, a functional calculus, and—through Stone’s theorem—unitary dynamics.

Required background. State Vectors supplies Hilbert-space vectors, inner products, and norm convergence. This page also assumes introductory functional analysis: dense subspaces, continuous linear functionals, and closed graphs.

An operator includes its domain

Let $\mathcal H$ be a complex Hilbert space, with the inner product linear in its second argument. An operator is a pair consisting of a linear subspace $D(A)\subseteq\mathcal H$ and a linear map

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

For a bounded operator, continuity extends the action uniquely from a dense domain to all of $\mathcal H$. An unbounded operator cannot have an everywhere-defined closed extension: the closed graph theorem would make that extension bounded. Domain information is therefore unavoidable rather than a minor technicality.

Two operators with the same differential expression can be different operators. On $L^2([0,L])$, for example, the expression $-i\hbar\,d/dx$ becomes different momentum operators when paired with different boundary conditions. Those choices change the adjoint, spectrum, and admissible dynamics.

The graph

$$G(A)=\{(\psi,A\psi):\psi\in D(A)\} \subset\mathcal H\oplus\mathcal H$$

is useful for limiting arguments. The operator is closed when $G(A)$ is a closed subspace. Equivalently, if $\psi_n\to\psi$ and $A\psi_n\to\eta$ in norm, with every $\psi_n\in D(A)$, then $\psi\in D(A)$ and $A\psi=\eta$. A closable operator has a smallest closed extension, denoted $\overline A$. Every self-adjoint operator is closed.

The adjoint and its maximal domain

Suppose $D(A)$ is dense. A vector $\phi\in\mathcal H$ belongs to $D(A^*)$ if there is a vector $\eta\in\mathcal H$ such that

$$\langle\phi,A\psi\rangle = \langle\eta,\psi\rangle \qquad \text{for every }\psi\in D(A).$$

Density makes $\eta$ unique, and one defines $A^*\phi=\eta$. Thus

$$D(A^*) = \left\{ \phi\in\mathcal H: \psi\mapsto\langle\phi,A\psi\rangle \text{ is norm-continuous on }D(A) \right\},$$

where continuity is measured using the Hilbert-space norm, not the graph norm. The adjoint domain is maximal: integration by parts may suggest the formal adjoint, but $D(A^*)$ is determined by every vector for which the boundary pairing and distributional action define a Hilbert-space vector.

For densely defined operators $A$ and $B$, write $A\subseteq B$ when $D(A)\subseteq D(B)$ and $B\psi=A\psi$ on $D(A)$. The central distinctions are then

Property	Operator relation	Domain consequence
symmetric	$A\subseteq A^*$	$D(A)\subseteq D(A^*)$
self-adjoint	$A=A^*$	$D(A)=D(A^*)$
essentially self-adjoint	$\overline A=A^*$	the closure is self-adjoint

A symmetric operator satisfies

$$\langle\phi,A\psi\rangle = \langle A\phi,\psi\rangle, \qquad \phi,\psi\in D(A),$$

and therefore has real expectation values on its domain. The converse conclusion needed in quantum mechanics does not follow: real quadratic forms on a restricted domain do not make the operator self-adjoint.

In finite dimensions every linear operator is bounded and has the full vector space as its domain. There, a Hermitian matrix is self-adjoint, so the domain distinction disappears. Importing that matrix intuition unchanged into infinite-dimensional quantum mechanics is the common source of the mistake “symmetric equals self-adjoint.”

Boundary forms reveal the missing domain data

Consider the formal momentum expression

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

on $\mathcal H=L^2([0,L])$. For sufficiently regular $\phi$ and $\psi$, integration by parts gives the boundary form

$$\langle\phi,P\psi\rangle - \langle P\phi,\psi\rangle = -i\hbar \left[ \phi(x)^*\psi(x) \right]_{0}^{L}.$$

The action alone does not decide whether this term vanishes. For each $\theta\in[0,2\pi)$, define

$$D(P_\theta) = \left\{ \psi\in H^1([0,L]): \psi(L)=e^{i\theta}\psi(0) \right\}.$$

The boundary form vanishes for pairs of vectors in this domain, and the same condition emerges when the adjoint domain is computed. Hence $P_\theta$ is self-adjoint. Changing $\theta$ leaves the differential expression unchanged but shifts the spectrum. The detailed interval calculation belongs to the physicist-facing Hermitian versus Self-Adjoint Operators page; here the family is evidence that a formal action does not select one operator. By contrast, imposing both $\psi(0)=0$ and $\psi(L)=0$ produces a symmetric restriction whose adjoint has a larger domain.

This example is a model for more complicated Hamiltonians. Boundary conditions at a finite endpoint, at a singularity, or at infinity are part of the operator. Physically admissible choices are constrained by self-adjointness, but self-adjointness alone need not select a unique choice.

The half-line gives a sharper warning. Start with $-i\hbar\,d/dx$ on $C_c^\infty(0,\infty)$. Its closure is symmetric, but the two deficiency equations have respectively one and zero square-integrable solutions. The deficiency indices are unequal, so this momentum operator has no self-adjoint extension on $L^2(0,\infty)$. Not every symmetric operator can be repaired by choosing a boundary condition.

Deficiency spaces and extension tests

Let $A$ be densely defined, closed, and symmetric. Its deficiency spaces are

$$\mathcal N_+ = \ker(A^*-iI), \qquad \mathcal N_- = \ker(A^*+iI),$$

with deficiency indices $n_\pm=\dim\mathcal N_\pm$. The labels depend on the displayed sign convention; the dimensions, not the labels, carry the criterion.

• $A$ is self-adjoint exactly when $n_+=n_-=0$.
• $A$ has self-adjoint extensions exactly when $n_+=n_-$.
• When the common finite value is $n$, the extensions are parametrized by unitary maps from $\mathcal N_+$ to $\mathcal N_-$.

An equivalent range test says that a densely defined symmetric operator is self-adjoint precisely when

$$\operatorname{Ran}(A-iI) = \operatorname{Ran}(A+iI) = \mathcal H.$$

For essential self-adjointness, the corresponding ranges need only be dense. These tests convert domain equality into solvability properties of shifted operators.

An operator first defined on a convenient small domain—smooth functions of compact support, for example—is often not closed. Calling that initial operator “self-adjoint” is usually false. The useful statement is that it is essentially self-adjoint on that domain: it has one self-adjoint closure, so the convenient domain is a core that determines the physical operator.

Spectral and dynamical consequences

Self-adjointness has consequences that symmetry alone does not guarantee.

First, the spectrum is real. More quantitatively, for $z\in\mathbb C\setminus\mathbb R$,

$$\|(A-zI)\psi\| \geq |\operatorname{Im}z|\,\|\psi\|,$$

and the self-adjoint range property makes $A-zI$ invertible with

$$\|(A-zI)^{-1}\| \leq \frac{1}{|\operatorname{Im}z|}.$$

Second, the spectral theorem assigns a unique projection-valued measure $E_A$ on $\mathbb R$ such that

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

This is not merely a symbolic diagonalization. It defines bounded functions $f(A)$ for bounded Borel functions $f$, and it fixes the domains of unbounded functions through integrability conditions. The full construction belongs to the unbounded spectral theorem.

Third, a self-adjoint $A$ defines unitary operators

$$U(t)=e^{-itA}$$

for all real $t$. They form a strongly continuous one-parameter group. The converse—every such group has a unique self-adjoint generator—is Stone’s theorem. For a Hamiltonian $H$, the physical convention is $U(t)=e^{-itH/\hbar}$.

These facts explain the observable postulate more precisely. A self-adjoint operator supplies real-valued sharp events through $E_A(\Delta)$, while a self-adjoint Hamiltonian supplies norm-preserving time evolution. The two roles use different parts of the same operator theory.

Standard examples and nonexamples

Multiplication by position

On $L^2(\mathbb R)$, define

$$(Q\psi)(x)=x\psi(x), \qquad D(Q)=\{\psi\in L^2(\mathbb R):x\psi(x)\in L^2(\mathbb R)\}.$$

This maximal multiplication operator is self-adjoint. Its spectrum is all of $\mathbb R$, yet it has no normalizable eigenvector. The spectral projections act by multiplication with indicator functions:

$$(E_Q(\Delta)\psi)(x) = \mathbf 1_\Delta(x)\psi(x).$$

Thus a self-adjoint operator need not possess an orthonormal basis of ordinary eigenvectors.

A bounded Hermitian operator

If $B$ is bounded and defined on all of $\mathcal H$, then

$$\langle\phi,B\psi\rangle = \langle B\phi,\psi\rangle \quad\text{for all }\phi,\psi\in\mathcal H$$

already implies $B=B^*$. This is the regime in which “Hermitian” and “self-adjoint” are harmlessly used as synonyms.

A merely symmetric restriction

Restricting a self-adjoint operator to a smaller dense domain usually preserves symmetry but destroys self-adjointness. The adjoint remembers the larger set of vectors on which the boundary pairing is meaningful. A formal calculation that checks only vectors in the restricted domain cannot detect this mismatch.

Common pitfalls

Checking only the differential expression. Integration by parts identifies a boundary form, not a self-adjoint operator. State the domain and compare it with the adjoint domain.

Equating real expectations with self-adjointness. A symmetric operator has real expectations on its domain. It can still lack a spectral resolution or a unitary group generated on the whole Hilbert space.

Calling a test-function operator self-adjoint. A differential operator on $C_c^\infty$ is often a convenient symmetric seed. The correct result may be essential self-adjointness of that seed, or the existence of several self-adjoint extensions.

Assuming every symmetric operator has an extension. Unequal deficiency indices rule out self-adjoint extensions on the same Hilbert space. Equality, not symmetry alone, is the extension criterion.

Expecting only eigenvalues. Continuous spectrum is fully compatible with self-adjointness. Spectral projections and generalized representations replace an ordinary eigenvector sum.

Exercises

1. Reality of expectation values

Show that if $A$ is symmetric and $\psi\in D(A)$, then $\langle\psi,A\psi\rangle$ is real. Explain why this does not prove that $A$ is self-adjoint.

Solution

Symmetry gives

$$\langle\psi,A\psi\rangle = \langle A\psi,\psi\rangle = \langle\psi,A\psi\rangle^*.$$

Hence the expectation value is real. The calculation uses only vectors in $D(A)$ and shows $A\subseteq A^*$; it says nothing about whether $D(A)=D(A^*)$.

2. The interval momentum family

Show that the boundary form vanishes on $D(P_\theta)$ and derive the spectrum of $P_\theta$.

Solution

If $\phi$ and $\psi$ obey the same quasiperiodic condition, then $\phi(L)^*\psi(L)=\phi(0)^*\psi(0)$, so the boundary form vanishes. Solving $-i\hbar\psi'=p\psi$ gives $\psi(x)=C e^{ipx/\hbar}$. The condition $\psi(L)=e^{i\theta}\psi(0)$ requires

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

so $p=\hbar(2\pi n+\theta)/L$. Normalization gives $|C|=L^{-1/2}$.

3. Resolvent estimate

Let $A$ be self-adjoint and $z=a+ib$ with $b\ne0$. Prove $\|(A-zI)\psi\|\geq |b|\|\psi\|$.

Solution

Because $A-aI$ is symmetric, $\langle\psi,(A-aI)\psi\rangle$ is real. Expanding the squared norm gives

$$\|(A-aI-ibI)\psi\|^2 = \|(A-aI)\psi\|^2+b^2\|\psi\|^2,$$

because the cross terms cancel. Taking square roots yields the estimate. The surjectivity part needed for a bounded inverse uses self-adjointness, not only this inequality.

4. Domain of the position operator

Give an $L^2(\mathbb R)$ function that does not belong to $D(Q)$, where $Q\psi=x\psi$.

Solution

For example,

$$\psi(x)=\frac{1}{1+|x|}$$

belongs to $L^2(\mathbb R)$, but $x\psi(x)$ approaches a nonzero constant in magnitude as $|x|\to\infty$ and is not square-integrable. Thus $D(Q)$ is a proper dense subspace of $L^2(\mathbb R)$.

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.
• J. B. Conway, A Course in Functional Analysis, 2nd ed., Springer, 1990.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• T. Kato, Perturbation Theory for Linear Operators, 2nd ed., Springer, 1976.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• K. Schmüdgen, Unbounded Self-Adjoint Operators on Hilbert Space, Springer, 2012.
• G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., American Mathematical Society, 2014.