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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?

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  • $\begingroup$ The recent edit seemed completely gratuitous, and it is debatable if the changes made any substantive improvement $\endgroup$
    – Yemon Choi
    Commented May 11, 2016 at 2:23

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The answer is classical and negative. It is a particular instance of Thm X.11 of Reed-Simon here.

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

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  • $\begingroup$ The inequality $0\leq V(x) \leq \left(\frac{3}{4} - \varepsilon\right)\frac{1}{x^2}$ does not hold everywhere for $V(x)=1/x$. Is it not a problem? $\endgroup$
    – user72829
    Commented May 8, 2016 at 14:19
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    $\begingroup$ No, it is not a problem. If you have that inequality only near 0, then the operator is in the "limit circle" case near 0. By Weyl's criterion (R&S Theorem X.7), it is not essentially self-adjoint. $\endgroup$
    – Raziel
    Commented May 8, 2016 at 14:26
  • $\begingroup$ Are there any self-adjoint extensions of the considered operator. E.g. is the operator H self-adjoint on the domain $\{ f\in L^2([0,\infty)): f'' \in L^2([0,\infty)), f(0)=0\}$? $\endgroup$
    – user72829
    Commented May 8, 2016 at 14:41
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    $\begingroup$ Yes, this is one possible self-adjoint extension (with Dirichlet boundary condition). Characterizing all the infinite self-adjoint extension is not really straightforward, but they are listed, e.g., here: arxiv.org/abs/0806.2764 $\endgroup$
    – Raziel
    Commented May 8, 2016 at 15:12

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