No, the Hilbert space is *not* spanned by continuous "eigenfunctions" because they are not eigenfunctions at all!

A self-adjoint operator $T$ on $L^2(\mathbb{R})$ has a *point spectrum*, a *continuous spectrum* and a *singular spectrum*. The latter is physically irrelevant.

The *point spectrum* consists of the values $\lambda_i$ for which a true eigenvector $\psi_i\in L^2(\mathbb{R})$ with $T\psi_i = \lambda_i$ exists. The point spectrum is always discrete and countable because the Hilbert spaces of quantum mechanics are always separable, meaning they have countable bases, and so there are only countably many possible different eigenvectors.

The *continuous spectrum* is, as the name tells you, continuous, and apart from discrete intersections with the point spectrum, no elements of $L^2(\mathbb{R})$ are eigenvectors belonging to points of the continuous spectrum.

The quantum mechanical setting with bound *and* continuous states is described by a Gel'fand triple/rigged Hilbert space $\Phi \subset \mathcal{H} \subset \Phi^*$, where $\mathcal{H}$ is the physical Hilbert space and $\Phi^*$ is a larger space containing also eigenvectors for the elements of the continuous spectrum (e.g. for $L^2(\mathbb{R})$, this is frequently the space of tempered distributions containing the "eigenfunctions" $\delta(x-x_0)$ of the position operator). The large space of such a Gel'fand triple fails to be a Hilbert space - it has no proper inner product, for one. Therefore, there are indeed "eigenvectors" corresponding to the continuous spectrum in quantum mechanics, but they are not *elements of the Hilbert space*.

This post imported from StackExchange Physics at 2015-05-01 16:19 (UTC), posted by SE-user ACuriousMind