Self adjoint operator

(also self-adjoint operator or Hermitian operator)

Introduction

A coordinate free expression (and an infinite dimensional generalization) of self adjoint matrixs.

I think that in finite dimensional case, their matrix with respect to an orthonormal basis is a self adjoint matrix.

By the way, observe that given an operator $A$ , the sum or product of $A$ with its conjugate transpose $A^{†}$ is a Hermitian operator (easily checked by computation).

There is a version of spectral theorem for these operators.
One important property of such operators is that the eigenvalues of a self-adjoint operator are necessarily real. Indeed, if $k$ is any eigenvalue with corresponding (normalized) eigenvector $v$ , we see

k = k ⟨ v, v ⟩ = ⟨ k v, v ⟩ = ⟨ S v, v ⟩ = ⟨ v, S v ⟩ = ⟨ v, k v ⟩ = \overset{―}{k} ⟨ v, v ⟩ = \overset{―}{k}

showing that $k$ is real.

If we have a basis of eigenvectors we can construct the associated resolution of identity.

In Quantum Mechanics

They correspond to observables in Quantum Mechanics.
If we multiply them by the complex unit $i$ we obtain antiHermitian operators, which constitute the "Lie algebra" of the "group" of unitary operators. That is, given a self-adjoint operator $A$ in a Hilbert space then $e^{i A}$ is an unitary operator. In fact, its Hermitian adjoint is $e^{- i A^{†}} = e^{- i A}$ because $A$ is Hermitian ( $A = A^{†}$ ). Now, we compute $e^{i A} e^{- i A}$ and $e^{- i A} e^{i A}$ :

e^{i A} e^{- i A} = e^{i A - i A} = e^{0} = I

e^{- i A} e^{i A} = e^{- i A + i A} = e^{0} = I

where we use the fact that $[A, - A] = 0$ .