# What to do with symmetric but not essentially self-adjoint operators

If we have a densely defined and symmetric operator on a Hilbert space, which is not even essentially self-adjoint, what can we do with it? It is at least closable, but i cannot really find/think of any other nice things we can say about it.

• You can find many self-adjoint extensions of this operator. Mar 7, 2022 at 19:54
• @RyszardSzwarc : You may not have self-adjoint extensions of a closed densely-defined symmetric $A$. You have self-adjoint extensions iff there is a unitary map between $\mathcal{N}(A^*+iI)$ and $\mathcal{N}(A^*-iI)$. Mar 8, 2022 at 14:46
• @Disintegrating By Parts: You are right. I forgot to mention the defect indices should be equal. This holds for example when the operator commutes with a conjugation operator. Mar 8, 2022 at 18:03

A canonical case of a symmetric operator is $$A=\frac{1}{i}\frac{d}{dx}$$ on the domain $$\mathcal{D}(A)$$ consisting of all absolutely continuous functions $$f\in L^2[0,\infty)$$ for which $$f'\in L^2[0,\infty)$$ and $$f(0)=0$$. This operator $$A : \mathcal{D}\left(A\right)\subset L^2[0,\infty]\rightarrow L^2[0,\infty)$$ is closed. And it is symmetric operator because $$\langle Lf,g\rangle-\langle f,Lg\rangle=\frac{1}{i}\int_0^{\infty}f'\,\overline{g}+f\overline{g'}dx=\left.\frac{1}{i}f\overline{g}\right|_{0}^{\infty}=0,\;\; f,g\in\mathcal{D}(L).$$ The theory of this operator involves the Laplace transform and holomorphic functions on the upper half plane, which includes index theory, and other interesting aspects of holomorphic function theory. The associated holomorphic function theory can be studied on the unit disk instead, but it really came from here.
General symmetric operators can be decomposed into multiple copies of the operator $$A$$ and a self-adjoint operator, which may be $$0$$.