Self-Adjoint Operator

(redirected from Essentially self-adjoint)

self-adjoint operator

[¦self ə¦jȯint ′äp·ə‚rād·ər]
(mathematics)
A linear operator which is identical with its adjoint operator.

Self-Adjoint Operator

 

(or Hermitian operator), an operator coincident with its adjoint. The theory of self-adjoint operators arose as a generalization of the theory of, for example, integral equations with a symmetric kernel, self-adjoint differential equations, and symmetric matrices. Examples of self-adjoint operators are (1) the operator of multiplication by the independent variable in the space of functions that are defined on the entire number axis and are square integrable and (2) the differentiation operator

in the same space.

If the function K(x,y) is continuous in the square a ≤ x ≤ b,a ≤ y ≤ b and if K(x,y) = K(y, x), then the integral operator

is self-adjoint. The spectrum of a self-adjoint operator lies on the real axis. In quantum mechanics physical quantities have corresponding self-adjoint operators whose spectra give the possible values of these quantities. A self-adjoint operator can be represented as an integral that is the limit of linear combinations of pairwise orthogonal projection operators with real coefficients.

References in periodicals archive ?
A densely defined symmetric operator is called essentially self-adjoint if its closure is self-adjoint.
infinity]] Riemannian manifold is essentially self-adjoint on the dense domain [C.
infinity]] Riemannian manifold, we will let [DELTA]' denote the essentially self-adjoint Laplacian whose domain is [C.
Since, by Theorem (2), [DELTA]' is essentially self-adjoint on [C.
But since [DELTA]' is essentially self-adjoint, [DELTA] = [([DELTA]').