0
$\begingroup$

Consider the eigenvalue equation:

$$\hat{Q}\Psi = q\Psi$$

where $q$ and $\Psi$ are eigenvalues and eigenfunctions of the hermitian operator $\hat{Q}$. If the spectrum of the hermitian operator is discrete, then the eigenfunctions lie in the Hilbert Space and constitute physically realizable states.

Why do discrete eigenvalues imply that their associated eigenfunctions are square-integrable (and hence live in the Hilbert Space)?

$\endgroup$
2
  • 1
    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/68639/9887 $\endgroup$ – Alfred Centauri Nov 9 '14 at 19:48
  • $\begingroup$ This is almost by definition: the point spectrum of an operator $Q$(i.e. the set of all eigenvalues) is the set of $\lambda\in\mathbb{C}$($\mathbb{R}$ for hermitian operators) such that it exists one $0\neq x\in \mathscr{H}$ such that $Qx=\lambda x$. Usually the point spectrum of a self-adjoint operator is a set of isolated points in the real line, but this is not always the case. For example an eigenvalue may be the accumulation point of other eigenvalues, or imbedded in the continuum spectrum. $\endgroup$ – yuggib Nov 9 '14 at 19:56