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Discrete or continuous spectrum?

  1. Mar 15, 2013 #1

    ShayanJ

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    Consider an unbounded self-adjoint operator defined in a hilbert space(its domain isn't the entire hilbert space,of course).Can its spectrum have discrete and continuous parts simultaneity?Does it have eigenvectors with finite norm?
    Thanks
     
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  3. Mar 15, 2013 #2

    micromass

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    Take a "nice" (for example: piecewise continuous) function [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex]. We can consider the multiplication operator

    [tex]M:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R}): g\rightarrow fg[/tex]

    An element ##\lambda\in \mathbb{C}## is not in the spectrum of ##M## if and only if ##M-\lambda I## is not invertible. But we have

    [tex](M-\lambda I)(g)(x) = f(x)g(x) - \lambda g(x) = ( f(x) - \lambda 1) g(x)[/tex]

    We can see from this (and we can prove), that ##M-\lambda I## is invertible only if ##f(x)\neq \lambda## for any ##x\in \mathbb{R}##. So the spectrum of ##M## is ##\overline(f(\mathbb{R}))##.

    Now take for example, the function

    [tex]f(x) = \left\{\begin{array}{l}x-1 ~\text{if}~ x\leq 0\\ 0~\text{if} ~x>0\end{array}\right.[/tex]

    Then the spectrum will be ##(-\infty,-1]\cup \{0\}##, which consists of "continuous" and "discrete parts".

    Now, ##g## is an eigenvalue of ##M## if ##Mg = \lambda g## for a certain ##\lambda\in \mathbb{C}##. So ##f(x)g(x) = \lambda g(x)## for all ##x\in \mathbb{R}##.

    If we take the function ##f##as above, then we see that

    [tex]g(x) = \left\{\begin{array}{l} 0~\text{if} ~x\leq 0\\ 1~\text{if}~ 0<x<1\\ 0~\text{if}~ x\geq 1\end{array}\right.[/tex]

    is an eigenvector. Furthermore, this has finite norm.
     
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