Sum of two self-adjoint unbounded operators

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

$$ D(T)=\Bigl\lbrace u\in H;\;\int_0^1 \bigl\vert u'(t)\bigr\vert^2 dt <\infty,\;;\;x(0)=x(1)=0\;\Bigr\rbrace, $$

$$ D(S)=\Bigl\lbrace u\in H;\;\int_0^1 \bigl|u'(t)\bigr|^2 dt <\infty,\;;\;y(0)=y(1)=0\;\Bigr\rbrace. $$

Denote by $J:\bR^2\to\bR^2$ the linear operator $(x,y)\mapsto (-y,x)$. For $u\in D(T)$ defines

$$ Tu=J\frac{du}{dt}, $$

while for $ u\in D(S)$ define

$$ Su=J\frac{du}{dt}. $$

Both operators $S$ and $T$ are selfadjoint. Note that

$$ D(T)\cap D(S)= \Bigl\lbrace u\in H;\;\int_0^1 \bigl|u'(t)\bigr|^2 dt <\infty,\;\;u(0)=u(1)=0\,\Bigr\rbrace. $$

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 $v\in D(A^*)\setminus D(A)$.

Let $T:D(T)\to H$ be a self-adjoint unbounded operator.

Take $S=-T$ with domain $D(S)=D(T)$.

Then $S$ is a self-adjoint unbounded operator but the sum $T+S$ is the null operator $0:D(T)\to H$, which is not self-adjoint if $D(T)\neq H$.

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