Self-Adjoint Operator (Observable)

Let $H$ be a complex Hilbert space and let $A:H\to H$ be a bounded operator (see Bounded Operator Hilbert). The operator $A$ is self-adjoint (or Hermitian) if

A = A^\ast,

where $A^\ast$ is the adjoint defined by

\langle x,Ay\rangle = \langle A^\ast x,y\rangle \quad \text{for all } x,y\in H.

In (finite-dimensional) quantum mechanics, self-adjoint operators are used to represent observables (measurable quantities).

Equivalent condition

$A$ is self-adjoint if and only if

\langle x,Ay\rangle = \langle Ax,y\rangle \quad \text{for all } x,y\in H.

Spectral properties in finite dimension

If $H$ is finite-dimensional (Complex Hilbert Space Finite), then:

All eigenvalues of $A$ are real.
Eigenvectors corresponding to distinct eigenvalues are orthogonal.
$A$ is unitarily diagonalizable: there exists an orthonormal basis of eigenvectors.

Equivalently, $A$ admits a spectral decomposition as in Spectrum Self Adjoint Finite.

Expectation values and measurement

Given a quantum state $\rho$ (a Density Operator), the expectation value of the observable $A$ is

\mathbb{E}_\rho[A] = \operatorname{Tr}(\rho A),

where $\operatorname{Tr}$ is the operator trace (Trace Operator).

In a pure state $|\psi\rangle$ (unit vector), the corresponding density operator is $\rho = |\psi\rangle\langle \psi|$ , and the expectation becomes

\mathbb{E}_\psi[A] = \langle \psi, A\psi\rangle.

Note on unbounded observables

In infinite-dimensional quantum theory, important observables (like position and momentum) are often unbounded. Then “self-adjoint” requires careful domain conditions. This knowl focuses on the bounded (and in particular finite-dimensional) setting.