Hilbert spaces are mostly assumed to be separable. A Hilbert space is separable if and only if it admits a countable orthonormal basis. How does this fit together with the possible existence of the continuous part of the spectrum, e.g. in case of the hydrogen atom?

  • 1
    $\begingroup$ It fits perfectly together. A separable space has not to be countable itself (the reals with the usual topology are a separable space). In addition, the continuous "eigenvectors" does not belong to the space (hence are not really eigenvectors), but belong a larger space that contains the Hilbert space (the space of distributions). $\endgroup$
    – yuggib
    Jun 24, 2015 at 23:04
  • $\begingroup$ Related: physics.stackexchange.com/q/90004/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 25, 2015 at 18:09

1 Answer 1


By definition, a value is in the continuous spectrum of $A$ if it not an eigenvalue, but the range of $A-\lambda I$ is a proper dense subset of the Hilbert space. There is nothing in this definition distinguishing separable spaces, or precluding operators in them from having continuous spectrum, and indeed some do. There is an equivalent definition in terms of "generalized eigenstates" that do not belong to the original space.

Perhaps the puzzlement comes from implicitly reversing the definition of separable. A space is separable if it admits a countable orthonormal basis, this does not mean that it can not also admit an uncountable "basis" of "generalized eigenstates" (caution: this can not happen with an honest basis, any orthonormal basis in a separable space is countable). For example, $L_2(\mathbb{R})$ admits a countable orthonormal basis of suitably normalized Hermite functions (the eigenfunctions of the quantum oscillator), but it also admits a continuous "basis" of $\delta$ functions. They happen to be the "generalized eigenfunctions" of the position operator, which is why its spectrum is continuous.

  • 1
    $\begingroup$ @wondering If this answer was helpful, consider 'accepting' it by clicking on the checkmark to the left. $\endgroup$ Jun 29, 2015 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.