Is Schrodinger operator with potential self adjoint Consider real-valued one-dimensional Schrodinger operator with potential $V(x)$, s.t. $L: H^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$
$$L(u)=-u''+V(x)u$$ with $V(x)$ bounded
Is this operator self-adjoint?
 A: One way to show this, assuming you know $-\frac{d^2}{dx^2}$ is self adjoint on $H^2(\mathbb{R})$, is the Kato-Rellich theorem. The theorem can be found on page 162 of Reed-Simon volume 2.
Definition: Let $A$ and $B$ be densely defined operators on a Hilbert space with $D(A)\subseteq D(B)$, and suppose there are constants $a,b\in \mathbb{R}$ with
$$
||B\phi||\leq a||A\phi||+b||\phi||
$$
Then $B$ is $A$ bounded with relative bound the infimum of all such $a$.
Then, the theorem states that if $A$ is a self adjoint, and $B$ is symmetric with $A$ relative bound $a<1$, then $A+B$ on $D(A)$ is self adjoint. 
In our setting, taking $A=-\frac{d^2}{dx^2}$, which is self adjoint on $D(A)=H^2(\mathbb{R})$ and $B=T_V$, multiplication by $V$, which is symmetric and defined on $L^2(\mathbb{R})\supseteq H^2(\mathbb{R})$, then it's easy to check that the relative bound is $a=0$, taking $b=||V||_{L^\infty(\mathbb{R})}$. So the hypotheses of the theorem are satisfied, and the operator is self adjoint on $H^2(\mathbb{R})$.
A: If $u\in\mathcal{S},$ then $$(Lu,v)=\int (-u''\bar{v} +V u\bar{v})=\int \left(u(\overline{-v''})+u\overline{(\bar{V}v)}\right),$$ via integration by parts. If $V$ is real-valued, then $V=\bar{V},$ making the right-hand side of the above equal to $(u,Lv)$. Hence, it's formally self-adjoint. 
$V$ is just a multiplication operator, so it won't mess up (formal) self-adjointness if it's real-valued. I think it’s essentially self-adjoint, but I’m not sure for a non-compactly-supported real-valued potential. 
A: Theorem: If $A : \mathcal{D}(A)\subseteq\mathcal{H}\rightarrow\mathcal{H}$ is selfadjoint on the Hilbert space $\mathcal{H}$, and if $B$ is a bounded selfadjoint operator on $\mathcal{H}$, then $A+B$ is self-adjoint on $\mathcal{H}$ and $(A+B)^*=A^*+B^*$.
Proof: Suppose $A$ and $B$ are as stated. Then
$$
        \Phi(x)=\langle (A+B)x,y\rangle
$$
is a bounded linear functional on $\mathcal{D}(A)$ iff $\Psi(x)=\langle Ax,y\rangle$ is a bounded linear functional on $\mathcal{D}(A)$. The proof follows from this. $\blacksquare$
