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 a dense linear domain is necessarily the full vector space, and every linear map on it is bounded. 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

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

Here $H^1(0,L)$ consists of $L^2$ functions with a weak first derivative in $L^2$. Such functions have continuous representatives and well-defined endpoint traces. Begin with the closed symmetric realization $P_{\mathrm D}$ on $H_0^1(0,L)$, where both traces vanish. Its adjoint domain can be computed without guessing a boundary condition:

1. Test the adjoint relation against $C_c^\infty(0,L)\subset H_0^1(0,L)$. If $\phi\in D(P_{\mathrm D}^*)$, the representing vector is $-i\hbar\phi'$ in the distributional sense. Thus $\phi'\in L^2$ and $\phi\in H^1(0,L)$.
2. Conversely, every $\phi\in H^1(0,L)$ satisfies the adjoint relation: integration by parts has no boundary term because the input $\psi$ has zero traces. No endpoint restriction on $\phi$ is needed.

Consequently,

$$D(P_{\mathrm D}^*)=H^1(0,L), \qquad P_{\mathrm D}^*\phi=-i\hbar\phi'.$$

The constant function lies in this adjoint domain but not in $H_0^1(0,L)$. This is an explicit failure of self-adjointness, even though the boundary form vanishes for every pair in $D(P_{\mathrm D})$. Adjoints of Unbounded Operators develops the general continuity and restriction arguments used in this test.

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

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

Compactly supported tests again force an adjoint vector into $H^1(0,L)$. The remaining boundary pairing is

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

The endpoint value $\psi(0)$ is arbitrary within $D(P_\theta)$, so the bracket must vanish. Its conjugate gives $\phi(L)=e^{i\theta}\phi(0)$. Conversely that condition cancels the pairing for every input. Therefore $D(P_\theta^*)=D(P_\theta)$: the test proves maximality as well as symmetry. Bonneau, Faraut, and Valent, Appendix A.3 and Section 5.3, give the corresponding extension classification.

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 spectrum and a state outside the momentum domain

Solving $-i\hbar u'=pu$ and imposing the boundary condition gives

$$u_n^{(\theta)}(x)=\frac{1}{\sqrt L}e^{i(2\pi n+\theta)x/L}, \qquad p_n^{(\theta)}=\frac{\hbar}{L}(2\pi n+\theta), \qquad n\in\mathbb Z.$$

These functions form an orthonormal basis: multiplication by $e^{i\theta x/L}$ is unitary on $L^2(0,L)$ and maps the ordinary Fourier basis onto this one. It also maps the periodic $H^1$ domain onto $D(P_\theta)$, with the transformed action $P_0+\hbar\theta/L$. Thus the displayed eigenvalues give the full spectrum, and Fourier coefficients $c_n=\langle u_n^{(\theta)},\psi\rangle$ characterize the domain by

$$\psi\in D(P_\theta) \quad\Longleftrightarrow\quad \sum_n \bigl(p_n^{(\theta)}\bigr)^2|c_n|^2<\infty.$$

For a concrete distinction between a state and an operator-domain vector, take the normalized constant $\psi(x)=L^{-1/2}$. With periodic momentum it has momentum zero with certainty. With antiperiodic momentum ($\theta=\pi$), direct integration instead gives

$$c_n=\int_0^1 e^{-i(2n+1)\pi y}\,dy =\frac{2}{i\pi(2n+1)}, \qquad \mathbb P_\psi\bigl(P_\pi=p_n^{(\pi)}\bigr) =\frac{4}{\pi^2(2n+1)^2}.$$

The probabilities sum to one over $n\in\mathbb Z$, but $\sum_n(p_n^{(\pi)})^2|c_n|^2$ diverges. The constant is a valid state with a normalized spectral probability law, although it is outside $D(P_\pi)$ and applying $P_\pi$ to it is undefined. Even the absolute first moment diverges: its summands are proportional to $1/|2n+1|$. Pairwise cancellation of positive and negative momenta gives a principal value zero, not an expectation value. The two domain choices describe different observables on the same Hilbert space.

The example also explains why a finite Fourier truncation can conceal a domain failure: every truncation has finite moments, while the limiting state need not. For the separate question of approximating an operator on its domain, see the boundary-layer counterexample in Dense Domains.

Half-line momentum has no self-adjoint extension

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. In fixed units, 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. Continuous translations without a derivative

For $T_\theta(a)=e^{-iaP_\theta/\hbar}$, where $a$ is a displacement, derive the translation across the joined endpoints and find $T_\theta(L)$. For the normalized constant state and $\theta=\pi$, compute $\|T_\pi(a)\psi-\psi\|^2$ for $0<a<L$. Does its strong derivative at $a=0$ exist?

Solution

Extend a function quasiperiodically by $\psi(x+L)=e^{i\theta}\psi(x)$. The spectral phases show that $T_\theta(a)\psi(x)=\psi(x-a)$ with this extension. For $0<a<L$,

$$(T_\theta(a)\psi)(x)= \begin{cases} e^{-i\theta}\psi(x-a+L),&0<x<a,\\ \psi(x-a),&a<x<L. \end{cases}$$

In particular $T_\theta(L)=e^{-i\theta}I$; this is the same phase obtained from every eigenvalue. For the constant and $\theta=\pi$, the translated function changes sign only on a segment of length $a$. Therefore

$$\|T_\pi(a)\psi-\psi\|^2=\frac{4a}{L}, \qquad \left\|\frac{T_\pi(a)\psi-\psi}{a}\right\|^2=\frac{4}{aL}.$$

The first quantity tends to zero, as strong continuity requires. The second diverges, excluding a strong derivative. This agrees with the spectral domain test in the main example; it does not invalidate unitary evolution of that state. Reality of expectations on the domain is proved separately in Symmetric Operators.

2. The interval momentum family

Find the operator domain and spectrum of $H_\theta=P_\theta^2/(2m)$. Does any member of this family equal the Dirichlet well Hamiltonian?

Solution

The product requires $\psi\in D(P_\theta)$ and $P_\theta\psi\in D(P_\theta)$. Hence

$$\begin{aligned} D(H_\theta)=\{\psi\in H^2(0,L):\ &\psi(L)=e^{i\theta}\psi(0),\\ &\psi'(L)=e^{i\theta}\psi'(0)\}. \end{aligned}$$

The eigenfunctions remain $u_n^{(\theta)}$ and the energies are $\hbar^2(2\pi n+\theta)^2/(2mL^2)$, with degeneracies when different momenta have equal squares. These operators are self-adjoint by spectral calculus. The Dirichlet Hamiltonian instead has domain $H^2(0,L)\cap H_0^1(0,L)$ and energies $\hbar^2\pi^2j^2/(2mL^2)$ for $j=1,2,\ldots$. None of the domains coincides: each $u_n^{(\theta)}$ belongs to $D(H_\theta)$ but has nonzero endpoint values. Squaring the differential expression is not enough to identify the Hamiltonian. For the physical distinction between a reflecting box and joined endpoints, see Hermitian versus Self-Adjoint Operators.

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, author manuscript, Sections 5.3 and 7.3 and Appendix A.3 — interval momentum, its adjoint domain, and the domain of its square.
• 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.