# 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.

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## 2 Answers

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 $$(U^*PU)f=e^{-icx^2}\frac{1}i\left(e^{icx^2}f(x)\right)'=2cxf(x)+\frac{1}if'(x)=(P+2cQ)f.$$

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Thank you a lot! – Paul Jul 3 '12 at 15:23

There is a big complex of theorems used for proving essential selfadjointness of the Schroedinger operators under quite different assumptions on potentials. It would not be reasonable to try to give an exposition here. See M. Reed and B. Simon, Methods of Modern Mathematical Physics, II: Fourier Analysis, Self-Adjointness, Academic Press, 1975.

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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
I meant a variety of methods which can be helpful in other situations too. – Anatoly Kochubei Jun 25 '12 at 4:47