Self-adjoint Operator

A self-adjoint operator is a linear operator on a Hilbert space that equals its adjoint. For an unbounded operator, this equality concerns both its action and its domain. Self-adjoint operators generalize real symmetric and complex Hermitian matrices; their spectral representations make them central to functional analysis and the mathematical formulation of quantum mechanics. (math.ubc.ca)

Definition and the role of the domain

Let $H$ be a complex Hilbert space, with inner product $\langle x,y\rangle$ taken to be linear in its first argument. Let

be a linear operator whose domain $D(A)$ is a dense linear subspace. The domain of the adjoint operator $A^*$ consists of all $y\in H$ for which there exists $z\in H$ satisfying

Density makes $z$ unique, and one defines $A^*y=z$. The operator $A$ is self-adjoint when

For a bounded operator defined on all of $H$, both domains are automatically $H$, so the condition reduces to $A=A^*$. (math.ubc.ca)

A densely defined operator is symmetric if

Equivalently, $A\subseteq A^*$: the adjoint agrees with $A$ on $D(A)$, but may have a larger domain. Thus self-adjointness implies symmetry, whereas symmetry alone does not imply self-adjointness for unbounded operators. Every self-adjoint operator is closed, meaning that its graph is closed in $H\oplus H$. (math.toronto.edu)

The word Hermitian is commonly synonymous with self-adjoint in finite dimensions. In discussions of unbounded operators, however, it is sometimes used for symmetry alone. The explicit domain condition avoids this terminological ambiguity. An operator is essentially self-adjoint when its closure is self-adjoint; it need not itself be closed or self-adjoint on its original domain. (math.brown.edu)

Finite-dimensional operators and matrices

In an orthonormal basis, a self-adjoint operator is represented by a matrix $M$ satisfying

where $M^*$ is the conjugate transpose. Thus $m_{ij}=\overline{m_{ji}}$, and diagonal entries are real. Over a real inner-product space, the condition becomes $M=M^T$: the matrix is symmetric. Complex symmetry $M=M^T$, without conjugation, is a different condition. (math.brown.edu)

The finite-dimensional spectral theorem gives

where $U$ is a unitary matrix and $\Lambda$ is diagonal with real entries. Consequently, all eigenvalues are real, eigenvectors belonging to distinct eigenvalues are orthogonal, and there is an orthonormal basis of eigenvectors. For real symmetric matrices, $U$ can be chosen orthogonal. (math.brown.edu)

Infinite-dimensional spectral theory

For a self-adjoint operator on a complex Hilbert space, its spectrum lies on the real axis. Nevertheless, spectral values need not be eigenvalues, and an orthonormal basis of eigenvectors need not exist. Infinite-dimensional spectral theory replaces diagonal matrices with multiplication operators or spectral integrals. (math.toronto.edu)

The general spectral theorem associates with $A$ a unique projection-valued measure $E$ on the Borel subsets of $\mathbb R$, such that

Each $E(B)$ is an orthogonal projection, and

These projections describe the portions of the Hilbert space associated with different spectral regions. (math.ubc.ca)

Writing $\mu_x(B)=\langle E(B)x,x\rangle$, the domain is recovered from the spectral measure:

The associated functional calculus defines

Bounded measurable functions give bounded operators; unbounded functions require corresponding domain conditions. This framework gives precise meanings to functions of operators, including exponentials and square roots of nonnegative operators. (math.ubc.ca)

For a compact self-adjoint operator on a separable Hilbert space, the orthogonal complement of its kernel has an orthonormal basis of eigenvectors with nonzero real eigenvalues. Nonzero eigenvalues have finite multiplicity, and any infinite sequence of them can accumulate only at zero. Adding an orthonormal basis of the kernel gives a basis for the whole space. (ocw.mit.edu)

Examples and boundary conditions

A fundamental example is multiplication by a real-valued measurable function $m$ on a measure space:

On this maximal domain, $M_m$ is self-adjoint. It is bounded precisely when $m$ is essentially bounded. The multiplication-operator formulation of the spectral theorem makes this example a model for general self-adjoint operators. (ocw.mit.edu)

Differential operators illustrate why domains matter. On $L^2(0,1)$, consider

where $H^1$ is a Sobolev space and $H_0^1(0,1)$ imposes zero endpoint values. Integration by parts shows that $P$ is symmetric. Its adjoint has the same differential expression but domain $H^1(0,1)$, without endpoint restrictions. Hence $P$ is not self-adjoint. (faculty.washington.edu)

Self-adjoint realizations instead arise from boundary conditions such as

These form a one-parameter family of self-adjoint extensions. The periodic case $\theta=0$ has exponential eigenfunctions. Thus specifying the differential expression $-i\,d/dx$ alone does not specify the operator. (ocw.mit.edu)

By contrast, the second-order operator

is self-adjoint: its Dirichlet boundary conditions also impose the necessary endpoint restrictions on the adjoint domain. Boundary conditions must therefore be analyzed together with the order and form of the differential expression. (faculty.washington.edu)

Essential self-adjointness and extensions

For a densely defined symmetric operator $S$, define its deficiency indices by

The extension theory associated with John von Neumann distinguishes three cases:

• If $n_+=n_-=0$, $S$ is essentially self-adjoint.
• If $n_+=n_-$, self-adjoint extensions exist.
• If $n_+\ne n_-$, no self-adjoint extension exists in the same Hilbert space.

For a closed symmetric operator with equal indices, its self-adjoint extensions are parametrized by unitary maps between the two deficiency spaces. (math.ucr.edu)

Essential self-adjointness supplies a unique self-adjoint extension, namely the closure. Otherwise, choosing an extension can amount to choosing additional boundary data. A densely defined symmetric operator bounded below also admits a self-adjoint extension preserving that lower bound. (ocw.mit.edu)

Quantum mechanics and unitary evolution

In the standard Hilbert-space formulation of quantum mechanics, sharp observables are represented by self-adjoint operators. For a normalized state $\psi$, the Born rule expresses the probability of a measurement result in a Borel set $B$ as

When $\psi\in D(A)$, the expectation value is $\langle A\psi,\psi\rangle$, which is real. Spectral measures accommodate both discrete and continuous measurement distributions. (math.ucr.edu)

Stone's theorem connects self-adjointness to dynamics: every self-adjoint $A$ generates a strongly continuous one-parameter unitary group $e^{itA}$, and every such group has a unique self-adjoint generator. For a time-independent Hamiltonian operator $H$, this yields

the norm-preserving evolution underlying the Schrödinger equation. Symmetry alone does not provide this correspondence. (math.ucr.edu)

Limits of formal operator manipulations

Domain control remains essential when combining unbounded operators. Even if two operators are self-adjoint, their sum or product is not automatically self-adjoint. Likewise, agreement of $AB$ and $BA$ on a common dense domain does not by itself establish that their spectral projections commute. The stronger spectral notion of commutation is needed for a joint spectral representation. (math.ucr.edu)

References

1. Review of Unbounded Operatorsmath.ubc.ca
2. Chapter 6 Structure of operators in inner product spaces — Linear Algebra Done Wrongmath.brown.edu
3. S13.4 — Self-adjoint theorymath.toronto.edu
4. Quantum Theory I, Recitation 1 Notesocw.mit.edu
5. The Spectral Theorem for Self-Adjoint and Unitary Operatorsmtaylor.web.unc.edu
6. Lecture 20 — Spectral theorem for compact self-adjoint operatorsocw.mit.edu
7. Mathematics for Physicspeople.physics.illinois.edu
8. Quantum Theory and Analysismath.ucr.edu