Defining the square root of an unbounded linear operator

[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:
<br /> H = -\Delta + V(x),<br />
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?

 

[dextercioby]

How did you get the square root ?

 

[AxiomOfChoice]

How did you get the square root ?

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...

 

[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.

 

[AxiomOfChoice]

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.

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.

 

[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.