Self-adjoint operator

'''Definition: In a finite dimensional space, given any linear operator A defined on the whole space, there exists an operator A †  such that '''

 ⟨ϕ, Aχ⟩ = ⟨A † ϕ, χ⟩,  ϕ, χ ∈ H.
A †  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,  D(A), is dense in  H and only for those vectors ϕ for which  ⟨ϕ, Aχ is a continuous function of χ. The set of these vectors is a vectorial subset of  H and by definition it will be the domain,  D(A † ), of A † .

Definition: A linear operator A is said to be:

  1. a Hermitian operator if A ⊆ A † , i.e. if

 ⟨ϕ, Aϕ⟩ = ⟨Aϕ, ϕ⟩,  ϕ ∈ D(A)

  1. a symmetric operator if A ⊆ A †  and $\; \overline{\mathcal{D}(A) }=\mathcal{H}$, where $\; \overline{\mathcal{D}(A) }$ is the complement of  D(A);
  2. a self-adjoint operator if A = A †  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 '''

 ⟨ϕ, Aχ⟩ = ⟨Aϕ, χ⟩,  ϕ, χ ∈ 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⟨ϕ, Aχ⟩ = ⟨ϕ + χ, A(ϕ + χ)⟩⟨ϕ − χ, A(ϕ − χ)⟩ − iϕ + iχ, A(ϕ + iχ)⟩ + iϕ − iχ, A(ϕ − iχ)⟩.

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:

Am, n *  = ⟨ξm, Aξn *  = ⟨A † ξm, ξn *  = ⟨ξn, A † ξm⟩ = An, m † 

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

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Monday, October 26, 2015 - 17:56