Self-adjoint operator

Fri, 23/10/2015 - 17:53 by Anonymous

'''Definition: In a finite dimensional space, given any linear operator \; A defined on the whole space, there exists an operator \; A^\dagger such that ''' :\; \langle \phi, A\chi \rangle = \langle A^\dagger\phi, \chi \rangle, \quad \phi, \chi \in \mathcal{H}. '''\; A^\dagger is uniquely determined and is called the ''adjoint'' of \; A.''' It must be noticed that for bounded operators the adjoint operator can be defined naturally, whereas for unbounded operators this can be done only if the domain of \; A, \; \mathcal{D}(A), is dense in \; \mathcal{H} and only for those vectors \; \phi for which \; \langle \phi, A\chi \rangle is a continuous function of \; \chi. The set of these vectors is a vectorial subset of \; \mathcal{H} and by definition it will be the domain, \; \mathcal{D}(A^\dagger), of \; A^\dagger. '''Definition:''' A linear operator \; A is said to be: # a ''Hermitian'' operator if \; A \subseteq A^\dagger, i.e. if :\; \langle \phi, A\phi \rangle = \langle A\phi, \phi \rangle, \quad \phi \in \mathcal{D}(A) ; # a ''[[symmetric operator]]'' if \; A \subseteq A^\dagger and \; \overline{\mathcal{D}(A) }=\mathcal{H}, where \; \overline{\mathcal{D}(A) } is the complement of \; {\mathcal{D}(A) }; # a '''self-adjoint''' operator if \; A = A^\dagger and \; \overline{\mathcal{D}(A) }=\mathcal{H}. For bounded linear operators, and in particular for linear operators in finite dimensional Hilbert spaces, the three definitions coincide. '''Proposition: A necessary and sufficient condition for the linear operator \; A to be ''Hermitian'' is that ''' :\; \langle \phi, A\chi \rangle = \langle A\phi, \chi \rangle, \quad \phi, \chi \in \mathcal{H}. '''Proof:''' This equality certainly implies the definition 1. of Hermitian operator. The converse is true because of the following identity, which can be easily verified: :\; 4\langle \phi, A\chi \rangle = \langle \phi+\chi, A(\phi+\chi) \rangle \langle \phi-\chi, A(\phi-\chi) \rangle -i\langle \phi+i\chi, A(\phi+i\chi) \rangle + i\langle \phi-i\chi, A(\phi-i\chi) \rangle. The matrix which represents the ''adjoint'' of a linear operator \; A in any orthonormal basis is the Hermitian conjugate of the matrix representing \; A. Indeed its matrix elements are given by: :\; A_{m,n}^* = \langle \xi_m, A\xi_n\rangle^* = \langle A^\dagger\xi_m, \xi_n\rangle^* = \langle \xi_n, A^\dagger\xi_m\rangle = A_{n,m}^\dagger. So, if a bounded linear operator is '''self-adjoint''', it is represented by a [[Hermitian matrix]] in any orthonormal basis. [[Category:Linear Algebra]] [[category:Handbook of Quantum Information]]