Let $H$ be a Hilbert space, and $T:D(T)\subset H\rightarrow H$ and $S:D(S)\subset H\rightarrow H$ be unbounded selfadjoint operators.
Is $T+S:D(T)\cap D(S)\rightarrow H$ selfadjoint?
Let $H$ be a Hilbert space, and $T:D(T)\subset H\rightarrow H$ and $S:D(S)\subset H\rightarrow H$ be unbounded selfadjoint operators. Is $T+S:D(T)\cap D(S)\rightarrow H$ selfadjoint? 


The answer is no. $\newcommand{\bR}{\mathbb{R}}$ Consider the Hilbert space of $L^2$functions $$ u:[0,1]\to \bR^2,\;\;u(t)=(x(t), y(t)) . $$ Define $$ D(T)=\Bigl\lbrace u\in H;\;\int_0^1 \bigl\vert u'(t)\bigr\vert^2 dt <\infty,\;;\;x(0)=x(1)=0\;\Bigr\rbrace, $$ $$ D(S)=\Bigl\lbrace u\in H;\;\int_0^1 \biglu'(t)\bigr^2 dt <\infty,\;;\;y(0)=y(1)=0\;\Bigr\rbrace. $$ Denote by $J:\bR^2\to\bR^2$ the linear operator $(x,y)\mapsto (y,x)$. For $u\in D(T)$ defines $$ Tu=\frac{du}{dt}, $$ while for $ u\in D(S)$ define $$ Su=\frac{du}{dt}. $$ Both operators $S$ and $T$ are selfadjoint. Note that $$ D(T)\cap D(S)= \Bigl\lbrace u\in H;\;\int_0^1 \biglu'(t)\bigr^2 dt <\infty,\;\;u(0)=u(1)=0\,\Bigr\rbrace. $$ The operator $A:=T+S:D(A)= D(T)\cap D(S)\to H$ is symmetric, but not selfadjoint. Indeed, $ v\in D(A^*)$ if and only if, there exists $C>0$ such that $$ \bigl\vert\;(Au, v)_{H} \;\bigr\vert \leq C\Vert u\Vert_H,\;\;\forall u\in D(A). $$ The constant function $t\mapsto v(t)= (1,1)$ satisfies this condition because for any $u\in D(A)$ we have $$ (Au, v)_H=\int_0^1 \bigl(\; 2Ju'(t), v(t)\;\bigr) dt = 2\int_0^1\frac{d}{dt} \bigl(\; Ju(t), v(t)\;\bigr) =0. $$ We have produced a function $v\in D(A^*)\setminus D(A)$. 


This is one of the central problems of modern mathematical physics. One could fill libraries with the mathematics generated by the most important special casethe Schrödinger operator. Mathematically, this is the question of when the sum of the Laplace operator and a multiplication operator on $L^2$ is selfadjoint (more precisely, essentially selfadjoint). The writings of Barry Simon, many of which are available online, are a good place to start. 

