About the orthogonality of the Hamiltonian eigenstates for the the continuous energy spectrum

I would like first to describe a strange case that I encountered. $\ \ -$ I solved the Schrodinger equation with a potential barrier (a potential well limited by a finite height wall which decrease with the distance $r$ from the center of the well). Relevant for me was the continuous spectrum of energies. I selected the set of solutions ${\phi (r, E)}$ regular at $r=0$ - see definition in end of the text. Then, I picked a certain function, $S(r, t)$, which is not an eigenfunction, but is regular at $r=0$.

I thought that $S(r, t)$ can be fully developed as a superposition of the regular eigenfunctions, i.e. $S(r, t) = \sum _E A(E, t) \ \phi (r; E)$. But, I discovered that $S(r, t)$ has a non-null projection on the irregular eigenfunctions.

Now, my question: is there some general proof that the eigenfunctions of a Hamiltonian, in the continuous energy spectrum are mutually orthogonal? Could it be that they are not?

I mention that the spectral theorem doesn't seem helpful for the continuous spectra.

DEFINITIONS : The Schrodinger equation may have (as in my case), two types of solutions, finite at $r=0$ which we call regular, and infinite at $r=0$ which we call irregular. The regular solutions are physical, while the irregular are non-physical.

• Could you detail your solution a little more? This sounds like a good question, but one can take any linear combination of bound and unbound eigenfunctions, and the result "solves" the SE and by definition has a nonzero projection on the irregular eigenfunction subspace. Superpositions of irregular eigenfunctions can be normalisable - witness a finite energy wavepacket decomposed by Fourier transform into a superposition o plane waves. – Selene Routley Feb 2 '15 at 22:06
• Comment to the question (v2): OP seems to allow eigenfunctions outside the Hilbert space. Related: physics.stackexchange.com/q/68639/2451 , physics.stackexchange.com/q/90101/2451 and links therein. – Qmechanic Feb 2 '15 at 22:07
• @Qmechanic of course that the eigenfunctions corresponding to the continuous spectrum have infinite norm? Yes, they are supposed to be normalized to $\delta (E' - E)$. But this is not the issue. The problem is that I am not convinced that the regular eigenfunctions are orthogonal on the irregular. This I why I told the strange situation that I found. As to the spectral theorem, it isn't good for these states (so it seems to me - if you'd ask I would tell you more). – Sofia Feb 2 '15 at 22:18
• @WetSavannaAnimalakaRodVance : I would be very glad to tell you more details. Just, I don't see a possibility to have a professional discussion by comments, not even in a private chat room. Do you have an email? – Sofia Feb 2 '15 at 22:20
• @WetSavannaAnimalakaRodVance , I know you are a professional, and I would be glad for a serious talk. – Sofia Feb 2 '15 at 22:21

Eigenvectors exist only for the point spectrum of an operator. For any other point of the spectrum one can only find a sequence of vectors for which $(A-\lambda I)u_n\to0$, where $A$ is said operator, and $\lambda$ is a point in the spectrum which is not an isolated point. So in this case there is a sequence of approximate eigenvectors. With a bit of extra details, if a point $\lambda$ comes from the continuous spectrum of the operator, there is no vector $\psi$ such that $A\psi=\lambda\psi$. For any $\epsilon>0$, though, you can choose a vector $\psi_\epsilon$ such that $\Vert(A-\lambda I)\psi_\epsilon\Vert<\epsilon$, hence $\psi_\epsilon$ is just an approximate eigenvector.

This can also be observed from the spectral theory for $A$, where there exists a projection-valued measure $E$ supported by the spectrum of such that $$A=\int\limits_{\sigma(A)}\lambda\text dE(\lambda).$$ If $\lambda$ is an isolated point, then $E(\{\lambda\})$ is a non-zero projection which projects onto the $\lambda$-eigenspace, but if $\lambda$ is not isolated then such a projection is simply zero.

As for the decomposition of the function $S$, given a othonormal system of functions (say the eigenfunctions for the point spectrum of $A$), this can be completed to an orthonormal basis of the Hilbert space. Any function $S$ can then be decomposed w.r.t to such a basis. The only problem is that there is no clear meaning of the rest of the vectors in the basis for the operator $A$.

• you are a mathematician and I am not. Some concepts to which you refer I don't know their meaning. For understanding you, try to tell me the definitions of the concepts with which I am not familiar. Would you? Then, what is point-spectrum? Is it a discrete spectrum? (Let me just mention that I proved somehow the $\delta$ Dirac normalization of the regular eigenfunctions, and, separately, of the irregular eigenfunctions. But the normality of one category on the other, I can't say that I proved.) But, please don't be influenced by this information. – Sofia Feb 2 '15 at 23:43
• I guess that point spectrum yes is discrete. Am I right? – Sofia Feb 2 '15 at 23:45
• 1. I'm a physicist, not a mathematician! :P. 2. the point spectrum is roughly made of isolated points, so it is discrete. – Phoenix87 Feb 2 '15 at 23:51
• For the problem that I am treating, it is the continuous spectrum that matters. As you say that you are a physicist, you of course heard of open systems. I have to do with resonances, and as Feshbach explained in his reaction theory, these are generated by the coupling of some bound state with the scattering continuum. But, tell me, what you mean by approximate eigenvectors? – Sofia Feb 3 '15 at 0:03
• Just what it is said in the answer: an approximate eigenvector for the spectral value $\lambda$ is a sequence of vectors $u_n$ in the (separable) Hilbert space such that $(A-\lambda I)u_n\to 0$. – Phoenix87 Feb 3 '15 at 13:14