Are the eigenvectors of the Hamiltonian always in $L^2$?

In physics (in particular quantum mechanics), an important operator is the Hamiltonian operator. It plays an extremely important role, as it's spectrum gives the possible values for the energy of the system.

Now, in general, the eigenvectors of a self-adjoint operator defined on a dense subspace of an infinite dimensional Hilbert space are not always elements of the said Hilbert space (this spans the theory for rigged Hilbert spaces). However, in all the quantum systems I have encountered so far it seems that the eigenvectors of the Hamiltonian operator are always in the underlying Hilbert space. Is there a reason for this (like some special property that comes from the form of the Hamiltonian/the nice potential terms people generally use) or is it just pure coincidence (and maybe there are systems where this is not true and I just haven't had the chance to encounter them)?

Solution 1:

It is logically inconsistent to say that an eigenvector of the Hamiltonian $H$ is not in the space $X$. By definition, an eigenvector is a vector in the vector space $X$ under consideration. You can have approximate eigenvectors in the space. For example, you may well encounter a situation where $\{ \phi_{n} \}\subset X$ satisfies $$\|(H-\lambda I)\phi_n\| \le \frac{1}{n}\|\phi_n\|,\;\; \phi_n \ne 0,\;\; n=1,2,3,\cdots,$$ even though there may not be an eigenvector with eigenvalue $\lambda$. This is typical of the continuous spectrum of a self-adjoint operator $H$. The measurement of $H$ for $\phi_n$ is nearly $\lambda$, and the precision with which this is true is increasing with increasing $n$. So the value of $H$ can be made arbitrarily close to $\lambda$, even though it may never actually be $\lambda$.