# Defining the square root of an unbounded linear operator

1. Dec 7, 2013

### AxiomOfChoice

I have started coming across square roots $(H+kI)^{\frac 12}$ of slight modifications of Schrodinger operators $H$ on $L^2(\mathbb R^d)$; that is, operators that look like this:
$$H = -\Delta + V(x),$$
where $\Delta$ is the $d$-dimensional Laplacian and $V$ corresponds to multiplication by some function. But how do we go about defining $(H+kI)^{1/2}$? My understanding was that we defined functions of self-adjoint operators by using the spectral theorem, but that only holds for bounded Borel functions, right? And $f(x) = \sqrt x$ certainly isn't bounded. And because of the $\Delta$, the operator $H$ isn't even bounded. So what do we do?

Last edited: Dec 7, 2013
2. Dec 8, 2013

### dextercioby

How did you get the square root ?

3. Dec 9, 2013

### AxiomOfChoice

What exactly do you mean? Are you asking how the square root is being used in the paper? From what I can tell, they are using the square root as a tool to prove a lemma that characterizes the operator $H$ itself. Not sure that answers your question though...

4. Dec 9, 2013

### dextercioby

Yes, I want to know the context they use the square root. Hopefully it's a free paper, I can't get in a journal behind a pay-wall.

5. Dec 9, 2013

### AxiomOfChoice

Ok, sure! I think it's on the arXiv: http://arxiv.org/pdf/math-ph/0304009.pdf. The relevant lemma is on pg. 15; it's Lemma 5.1.

6. Dec 9, 2013

### dextercioby

The powers of the operators can be defined through their spectral decomposition or by trans-Fourier-ing their (eigen)vectors, so that the power is transferred onto regular functions. These 2 'tricks' work irrespective whether the operator is bounded or not.