If I have an unbounded operator $A$ with domain $D$ on a hilbert space, I can define the sum of $A$ and its adjoint $A^\ast$ on $D$. I know that in general, $A + A^\ast$ will not be self-adjoint, because we only have $(A + B)^\ast \subset A^\ast + B^\ast$.


But for the same reason, $A + A^\ast$ certainly is symmetric. I am wondering if $A + A^\ast$ is also essentially self-adjoint on $D$.


It seems like a simple question but I'm stuck. I also couldn't find anything in books (Reed/Simon, Teschl and whatever I could find on the internet), and the problem is hard to google.
I hope someone can help me out here, or just point me to some book that has information about this topic.


(Additional info: In my specific problem, $A$ is minus i times the annihilation operator of the 1D quantum harmonic oscillator (with $D$ taken to be the Schwartz space). But I would be more interested in a general solution ;))



Sum of operator and adjoint operator
For a closed densely-defined operator $A$ the domains of $A$ and $A^\star$ in general are different, and their intersection may be $\{0\}$.  

In fact, let $H$ be any self-adjoint operator on a separable infinite-dimensional 
Hilbert space with purely discrete spectrum.
 Let $A = UH$ where $U$ is a unitary operator.  

Then the set of $U$ for which ${\cal D}(A) \cap {\cal D}(A^\*)
= \{0\}$ is a dense $G\_\delta$.
See my "Some Generic Results in Mathematical Physics", Markov Processes and Related Fields 10 (2004), 517-521, <http://www.math.ubc.ca/~israel/papers/is.ps>


