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 ;))

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    $\begingroup$ You need some assumptions: for example if $A$ is not closable, then $A^*$ is not densely defined. $\endgroup$ – András Bátkai Jun 22 '12 at 16:30

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

  • $\begingroup$ Of course, all of you are right, I was confusing something (I had $D(A) \subset D(A^\ast)$ in my mind, but that is only true for symmetric $A$). Now my question doesn't make much sense anymore... Thanks for your help nevertheless, I think I'll just ask another question as soon as I'm certain what my actual question is. $\endgroup$ – Paul Jun 22 '12 at 17:51

It is a non-trivial question even for a weighted shift $Ae_k=\lambda_ke_{k+1},\ k\in\mathbb N.$ Depending on the sequence of $\lambda_k\in\mathbb C$ the operator $A+A^*$ defined on the linear hull of $e_k,\ k\in\mathbb N,$ can be essentially self adjoint or have defect indices $(1,1).$


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