Let us consider the Hilbert space $L^2([0,\infty))$ and operator $H=-\frac{d^2}{dx^2} + \frac{1}{x}$ on the domain of $C^{\infty}\_0((0,\infty))$ (smooth functions with compact support away from $0$). 


Is the operator H essentially self-adjoint? What is the domain of its self-adjoint extension?



Selfadjointness of hamiltonian with 1/x potential
The answer is classical and negative. It is a particular instance of Thm X.11 of Reed-Simon [here](http://rads.stackoverflow.com/amzn/click/0125850026).


Let $V(x)$ be a continous potential on $(0,+\infty)$. If
\begin{equation}
V(x) \geq \frac{3}{4x^2}
\end{equation}
Then $-\partial\_x^2 +V$ it is essentially self-adjoint. On the other hand, if for some $\varepsilon >0$
\begin{equation}
0\leq V(x) \leq \left(\frac{3}{4} - \varepsilon\right)\frac{1}{x^2}
\end{equation}
Then $-\partial\_x^2 +V$ is not essentially self-adjoint


