Does symmetry of the Green's function of some operator imply that the operator is hermitian?

Question

I know that if an operator $L$ is hermitian(self-adjoint), then its Green's function is symmetric, but is it true the other way?

In other words, is having a symmetric Green's function a necessary and sufficient condition for $L$ to be hermitian?

Answer 1

Nope, There are even finite dimensional counterexamples! So the definition of the Hermitian conjugate of an operator is $$\forall\vec x, \vec y\hspace{5mm}\langle L^\dagger x, y\rangle=\langle x, Ly\rangle$$ Let the inner product be defined via a positive definite matrix $G$ (not to be confused with Green's functions which are represented by $L$'s entries.) $$\langle x, y\rangle:=\vec x^{*T}G\vec{y}$$ A few lines of handwriting shows that the Hermitian conjugate is now given by $$L^\dagger=G^{-1}L^{*T}G$$ Now your question becomes: Does $L=L^T$ imply $L^\dagger=L$?

"Nope!" is justified by the counter example: $$G=\begin{pmatrix} 1 & 0& 0\\ 0 & 1 & \alpha\\ 0 & \alpha & 1 \end{pmatrix}\hspace{3mm}|\alpha|<1$$ $$L=\begin{pmatrix} 0 & 0& 1\\ 0 & 0 & 0\\ 1 & 0 & 0 \end{pmatrix}$$