As e.g. Griffiths says (p. 103, Introduction to Quantum Mechanics, 2nd ed.), if a spectrum of a linear operator is continuous, the eigenfunctions are not normalizable, therefore it has no eigenfunctions in the Hilbert space.

On the other hand, both bound and continuous eigenfunctions are required to have a complete set, to be able to expand an arbitrary wave function in terms of the eigenfunc­tions (Landau&Lifshitz, Quantum Mechanics, p.19). How are these results connected, how to explain the apparent contradiction?

Is a formulation that the Hilbert space is spanned by both bound and continuous hydrogen atom eigenfunctions correct?

Update: I just found Scattering states of Hydrogen atom in non-relativistic perturbation theory which is related (but only partially answers this question).


1 Answer 1


If a self-adjoint operator has pure point spectrum, then (by one definition of pure point spectrum) its eigenfunctions form a complete basis for the Hilbert space.

However, an operator may also have continuous spectrum, in which case to get a decomposition of the identity one must use the spectral measure of the self-adjoint operator.

The usual thing physicists are used to in quantum mechanics is a decomposition of the identity when the operator has only pure point spectrum: $$ I = \sum_{n}|\psi_n\rangle\langle\psi_n|\,. $$

$|\psi_n\rangle$ are the eigenfunctions of the operator: $H|\psi_n\rangle=E_n|\psi_n\rangle$.

The way to generalize this relation to operators that also have other types of spectrum is as follows: for fixed $n$, think of $|\psi_n\rangle\langle\psi_n|$ like a projection so that the above displayed equation can just as well be written as: $$ I = \sum_{\lambda \in \text{spectrum of }H}P_\lambda$$ where $P_{\lambda_n}=|\psi_n\rangle\langle\psi_n|$ is the projection operator onto the particular eigenstate that has the corresponding eigenvalue. Now, operators that have continuous rather than discrete spectrum call for an integral, and so we get $$ I = \int_{\lambda \in \text{spectrum of }H} \mathrm{d}P_\lambda\,. $$

This might seem a bit intimidating: Now $P$ is what is called the projection-valued spectral measure of the operator $H$. It plays the same role that was played before by the same symbol, but now it's not limited to be an atomic measure. If the kind of continuous spectrum we have is only absolutely continuous (which is often the case) then this may be further decomposed as $$ I = \sum_{\lambda_n \in \text{point spectrum of }H}|\psi_n\rangle\langle\psi_n|+\int_{\lambda \in \text{continous spectrum of }H} F(\lambda)\mathrm{d}\lambda\,. $$

$F$ is called the projection-valued density of the spectral measure of $H$ w.r.t. the Lebesgue measure (it is the Radon-Nikodym derivative of the spectral measure $P$ w.r.t. the Lebesgue measure).

The point is you really need both parts to get a decomposition of the identity, and the hydrogen atom (as well as a particle in a finite box) has both types of spectrum indeed.

If you want more details about the construction of these spectral measures I recommend Rudin's book on functional analysis.


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