Sum of two essentially self-adjoint operators

Hi, I hope this question will make more sense than the one I posted yesterday.

I have two operators $p$ and $q$ which are essentially self-adjoint on a common domain $D$. Now I define $A = c_1 p + c_2 q$ with some real constants $c_1$, $c_2$. From this question I know that in general $A$ will not be essentially self-adjoint.

But in my case, I am working on $L^2(\mathbb R, dx)$. $q$ is the multiplication operator with $x$ and $p = -i \frac{d}{dx}$, both are essentially self-adjoint on the Schwartz space. I am pretty sure that the operator $A$ defined as above is again essentially self-adjoint on the Schwartz space.

By an explicit calculation, I think I would be able to show that $(A \pm i)D$ is dense in $L^2$, this would be sufficient. But this is a bit ugly and I am still hoping that there are theorems which prove the essential self-adjointness of $A$ in certain special cases. Does someone know of any such theorems?

The only one I know of / I could find is the Kato-Rellich theorem which requires one operator to be bounded (strongly) by the other. I don't think it is applicable here.

One can show that $P+cQ$ is unitarily equivalent to $P$ for every $c\in\mathbb R.$ This solves your problem.
Let $P=\frac{1}{i}\frac{d}{dx}$ and $Q=x$ be defined on the Schwartz space $S(\mathbb{R})$. Let $U$ be the unitary operator on $L^2(\mathbb{R})$ defined by $Uf(x)=e^{icx^2}f(x),\ c\in\mathbb{R}.$ Then $U$ is a bijection of $S(\mathbb{R})$ onto itself and we obviously have $U^*QU=Q.$ For $f\in S(\mathbb{R})$ we have $$(U^*PU)f=e^{-icx^2}\frac{1}i\left(e^{icx^2}f(x)\right)'=2cxf(x)+\frac{1}if'(x)=(P+2cQ)f.$$
• There is a lot of theorems indeed, and maybe I have missed something, but I couldn't find anything that would help me. The problem is that my operator $A$ is not a Schroedinger operator. – Paul Jun 24 '12 at 19:17