Not all self-adjoint operators are hermitian

We show this via a counterexample.

To find a counter-example we need to go to infinite dimensional spaces, and find an Operator with mismatching Domains. Take the Position Operator $X$ as an example, defined with support on the Schwarz space with support on ${\mathbb{R}}_{\ge 1}$. $X$ has full support in this space as $\Vert Xf{\Vert }_2={\int }_{{\mathbb{R}}_{\ge 1}}{x}^2\Vert f(x)\Vert dx$, and any function $f$ in the Schwarz space decays faster than any polynomial.

However,

• ${\mathbb{I}}_{x\le 2}\notin Dom(X)$ since $f$ is not a smooth function, hence it does not below to the Schwarz space
• ${\mathbb{I}}_{x\le 2}\in Dom(X^{\ast })$ since $\forall g\in Dom(X)$ , we have $⟨{\mathbb{I}}_{x\le 2},Xg⟩={\int }_{{\mathbb{R}}_{\ge 1}}x{\mathbb{I}}_{x\le 2}(x)g(x)dx={\int }_1^{2}xg(x)dx=⟨x{\mathbb{I}}_{x\le 2},g⟩$, which is well-defined for all $g\in Dom(X)$.

Since we have a domain mismatch, $X\ne X^{\ast }$, so $X$ is not a Self-Adjoint Operator