What is Hermitian operator: Definition and 59 Discussions

In mathematics, a self-adjoint operator on a finite-dimensional complex vector space V with inner product



⟨
⋅
,
⋅
⟩


{\displaystyle \langle \cdot ,\cdot \rangle }
(equivalently, a Hermitian operator in the finite-dimensional case) is a linear map A (from V to itself) that is its own adjoint:



⟨
A
v
,
w
⟩
=
⟨
v
,
A
w
⟩


{\displaystyle \langle Av,w\rangle =\langle v,Aw\rangle }
for all vectors v and w. If V is finite-dimensional with a given orthonormal basis, this is equivalent to the condition that the matrix of A is a Hermitian matrix, i.e., equal to its conjugate transpose A∗. By the finite-dimensional spectral theorem, V has an orthonormal basis such that the matrix of A relative to this basis is a diagonal matrix with entries in the real numbers. In this article, we consider generalizations of this concept to operators on Hilbert spaces of arbitrary dimension.
Self-adjoint operators are used in functional analysis and quantum mechanics. In quantum mechanics their importance lies in the Dirac–von Neumann formulation of quantum mechanics, in which physical observables such as position, momentum, angular momentum and spin are represented by self-adjoint operators on a Hilbert space. Of particular significance is the Hamiltonian operator






H
^





{\displaystyle {\hat {H}}}
defined by







H
^



ψ
=
−



ℏ

2



2
m




∇

2


ψ
+
V
ψ
,


{\displaystyle {\hat {H}}\psi =-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi +V\psi ,}
which as an observable corresponds to the total energy of a particle of mass m in a real potential field V. Differential operators are an important class of unbounded operators.
The structure of self-adjoint operators on infinite-dimensional Hilbert spaces essentially resembles the finite-dimensional case. That is to say, operators are self-adjoint if and only if they are unitarily equivalent to real-valued multiplication operators. With suitable modifications, this result can be extended to possibly unbounded operators on infinite-dimensional spaces. Since an everywhere-defined self-adjoint operator is necessarily bounded, one needs be more attentive to the domain issue in the unbounded case. This is explained below in more detail.

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