Is $\Delta$ self adjoint? Put $T=\Delta = \frac{d^2}{dx^2}.$ We put $D=\{f\in L^{2}(\mathbb R):Tf\in L^{2}(\mathbb R) \}.$ 

My Questions: (1) Is $D$ is dense in $L^{2}(\mathbb R)$. If yes, how to justify.  (2) Can we show $T:D\subset L^{2}\to L^{2}$ is self-adjoint operator? (3) Put $ H=-\Delta + |x|^2$, ($x\in \mathbb R$). Is $D=\{f\in L^{2}: Hf\in L^{2}\}$ is dense in $L^{2}$? Is the operator $H:D\subset L^{2} \to L^{2} $ is self- adjoint operator? 

(See  this also for $ H$.)
Note: Just for curiosity: Why at various places authors consider $-\Delta$ instead of $\Delta$? 
 A: If $Hf = -\frac{d^2}{dx^2}f+x^2f$, then $H$ is densely-defined and selfadjoint on the domain $\mathcal{D}(H)$ consisting of twice absolutely continuous functions $f$ on $\mathbb{R}$ such that $f\in L^2$ and $-f''+x^2f \in L^2$.
Differentiability isn't quite the right condition. Absolutely continuity is required so that $f$ is locally the integral of its derivative, and $f'$ is the integral of its derivative. More precisely, the equivalence class of $f\in L^2$ contains an element that is locally absolutely continuous on $\mathbb{R}$. And one does not have to assume that $f' \in L^2$. However, $f'$ must be equal almost everywhere to a locally absolutely continuous function such that $-f''+x^2f \in L^2$. With these minimal assumptions, $H$ as given is closed, densely-defined and selfadjoint.
What this implies from a Physics point of view is that there are no other conditions needed for a well-posed problem, which is good because if $H$ were symmetric on this natural domain, but not selfadjoint, further conditions at $\infty$ would have to be imposed in order to have a well-posed selfadjoint problem. And that would basically mean this equation would have no definite Physical meaning. Subtle details like this working out so perfectly help validate the Physics model as a tight Mathematical model.
