TITLE: Addition of Unbounded Operators
QUESTION [5 upvotes]: Let $H$ be a (separable complex) Hilbert space and let $A$ and $B$ be two densely-defined, maximally-defined linear operators on $H$ with domains $D(A)$ and $D(B)$ respectively.  (By maximall-defined, I mean that $A$ and $B$ admit no extensions).  Then, we can define the operator $A+B$ on $D(A)\cap D(B)$, however, in general, the operator $\left( D(A)\cap D(B),A+B\right)$ will not be maximally-defined.  The question is:  does this operator admit a unique maximal extension?
My conjecture is that the answer is no, but I would absolutely love for the answer to be yes.
Any ideas?
Thanks again!

REPLY [5 votes]: There are no maximally-defined operators except those defined everywhere.  That is, if $D(A)$ is any proper subspace of $H$, you can take any $u \notin D(A)$ and define an extension $\tilde{A}$ on $D(\tilde{A}) = \text{span}(D(A),u)$ by $\tilde{A}(x+cu) = Ax$ for $x \in D(A)$ and scalars $c$.  
Perhaps you might be interested in operators related to self-adjoint operators.  There I have a little result that you might find interesting.  Let $A$ be any self-adjoint unbounded linear operator with purely discrete spectrum.  Consider $T = U A$ and $T^* = A U^*$ where $U$ is a unitary operator.  Then the set of $U$ for which ${\cal D}(T) \cap {\cal D}(T^*) = \{0\}$ is a dense $G_\delta$ in the unitary operators on $H$.