Schrodinger with real potential is self-adjoint? Suppose I define the operator
$$
-\frac{d^2}{dx^2}+V(x)
$$
on the space of Schwartz class functions $\mathcal{S}(\mathbb{R})$ and take its closure to form the operator $H$  acting on a domain in $L^2(\mathbb{R})$. Is there an example of a real valued polynomial $V$ such that the closed (densely defined) operator is not self-adjoint?
If I consider two dimensions and the potential $x_1^4+x_2^4-\lambda x_1^2x_2^2$ is this not self adjoint for $\lambda>2$? 
 A: From Dunford and Schwartz, Vol III, p. 1409, with $p=1$:

Theorem: Let $\tau f = -\frac{d^2}{dt^2} f -q(t)f$ be a second order formal differential operator defined on the interval $[a,b)$ ($a < b \le \infty$.) Assume that 
$\;\;\;$ (a) $q(t) > 0$ for $t$ sufficiently near to $b$.
$\;\;\;$ (b) $\int_{a}^{b}\left|\left[\frac{q'(t)}{q(t)^{3/2}}\right]'+\frac{1}{4}\frac{q'(t)^2}{q(t)^{5/2}}\right|dt < \infty$

Then
$\;\;\;$ (a) if $\int_{x}^{b}\frac{1}{|q(t)|^{1/2}}dt = \infty$ for all $x$,
then $\tau$ has no boundary values at $b$;
$\;\;\;$ (b) if for sufficiently large $x$, $\int_{x}^{b}\frac{1}{|q(t)|^{1/2}}dt < \infty$, then $\tau$ has two boundary values at $b$.

You only need to apply the theorem to $[0,\infty)$ in order to conclude that the operator $\tau$ is not essentially selfadjoint on $(-\infty,\infty)$ if $\tau$ has a boundary value at $\infty$ on $[0,\infty)$.
This theorem sharply divides the cases $n = 2$ and $n > 2$, and you only have to look at the interval $[0,\infty)$. If you let $q(t)=t^2$, then (a) applies, and there are no boundary values at $\infty$, which leads to an essentially selfadjoint operator. However, if $q(t)=t^n$ for any higher power, the operator $\tau$ is no longer essentially selfadjoint.
