Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Defining the square root of an unbounded linear operator

  1. Dec 7, 2013 #1
    I have started coming across square roots [itex](H+kI)^{\frac 12}[/itex] of slight modifications of Schrodinger operators [itex]H[/itex] on [itex]L^2(\mathbb R^d)[/itex]; that is, operators that look like this:
    [tex]
    H = -\Delta + V(x),
    [/tex]
    where [itex]\Delta[/itex] is the [itex]d[/itex]-dimensional Laplacian and [itex]V[/itex] corresponds to multiplication by some function. But how do we go about defining [itex](H+kI)^{1/2}[/itex]? 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 [itex]f(x) = \sqrt x[/itex] certainly isn't bounded. And because of the [itex]\Delta[/itex], the operator [itex]H[/itex] isn't even bounded. So what do we do?
     
    Last edited: Dec 7, 2013
  2. jcsd
  3. Dec 8, 2013 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    How did you get the square root ?
     
  4. Dec 9, 2013 #3
    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 [itex]H[/itex] itself. Not sure that answers your question though...
     
  5. Dec 9, 2013 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  6. Dec 9, 2013 #5
    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.
     
  7. Dec 9, 2013 #6

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Defining the square root of an unbounded linear operator
Loading...