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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.

Qmechanic
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Sofia
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    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 02 '15 at 22:06
  • Comment to the question (v2): OP seems to allow eigenfunctions outside the Hilbert space. Related: http://physics.stackexchange.com/q/68639/2451 , http://physics.stackexchange.com/q/90101/2451 and links therein. – Qmechanic Feb 02 '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 02 '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 02 '15 at 22:20
  • @WetSavannaAnimalakaRodVance , I know you are a professional, and I would be glad for a serious talk. – Sofia Feb 02 '15 at 22:21
  • For continuous spectra, eigenfunctions with real eigenvalues are Dirac-orthonormalisable and complete. See page 7 here. – lemon Feb 02 '15 at 22:25
  • @lemon , no, I am sorry, it doesn't solve my problem. You see, the Fourier functions are a special case. But I have functions that are *not* Fourier functions. My functions are not even defined on all the real $x$-axis, but only on the radial axis $r \ge 0$. – Sofia Feb 02 '15 at 22:37
  • @WetSavannaAnimalakaRodVance : do you have an email address, s.t. we could talk? – Sofia Feb 02 '15 at 22:40
  • With generalized eigenfunctions the notion of orthogonality simply has no meaning because these eigenfunctions do not belong, in general, to a space with an inner (scalar) product. – yuggib Feb 03 '15 at 00:07
  • @yuggib what about Fourier functions? Their scalar product is $\delta$ Dirac. It has some meaning, and we work with it a lot. Of course these functions don't have a bounded norm, but with the help of the concept of $\delta$ Dirac we manage not badly. – Sofia Feb 03 '15 at 00:11
  • This is not an orthogonality condition from a mathematical point of view. What do you want to know? you should define concepts better...A treatment of generalized eigenfunctions has been made from a mathematical standpoint and you should find the bibliography looking for rigged hilbert spaces. – yuggib Feb 03 '15 at 00:18
  • @yuggib which concepts to define better? The concepts with which I work are simple: Schrodinger equation? It has solutions in the point spectrum and in the continuous spectrum. Orthogonality? There exists normalization to $\delta$ Kronecker, or to $\delta$ Dirac, the 1st one for bound states, the other for continuous spectrum states. The problem is whether the two types of functions, regular and irregular are orthogonal to one another, as I said above, according to $\delta$ Dirac. About rigged Hilbert space, for Fourier functions you work with rigged Hilbert spaces? – Sofia Feb 03 '15 at 00:26
  • orthogonality is related to an Hilbert space. If you do not have it (and for generalized eigenfunctions you don't) then you do not have orthogonality. In addition, suppose you have an operator in $L^2$ with $e^{ikx}$ as a generalized eigenfunction, and $\psi\in L^2$ as a true eigenfunction. It seems plausible to me (actually every "real" eigenfunction will be in $L^2$). Then $\int e^{-ikx}\psi(x)dx=\hat{\psi}(k)$ is not a delta or zero, since the Fourier transform is a unitary operator on $L^2$. Hence the two are not "orthogonal" (with the meaning you give to the term). – yuggib Feb 03 '15 at 00:51
  • @yuggib I am afraid that we get into an argument about names of things, if two generalized Fourier functions can be named orthogonal or not. Call it under which names you prefer, but it won't address my problem. I stated it : the inner product of a regular solution and an irregular solution of the Schrodinger equation with energies in the continuous spectrum, *is zero or not? I.e. $\int _0^{\infty} \phi _{reg} (r; E) \ \phi _{irreg} (r; E') \ \text d \vec r = 0$ for any $E$ and $E'$ ? This is the question, and I am interested in an answer to this* question. – Sofia Feb 03 '15 at 01:31

2 Answers2

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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 $A$.

Phoenix87
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  • 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 02 '15 at 23:43
  • I guess that point spectrum yes is discrete. Am I right? – Sofia Feb 02 '15 at 23:45
  • I'm a physicist, not a mathematician! :P. 2. the point spectrum is roughly made of isolated points, so it is discrete.
    • – Phoenix87 Feb 02 '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 03 '15 at 00:03
  • 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 03 '15 at 13:14
  • I apologize, but An approximate eigenvector for λ is a sequence of vectors? An eigenvector is one, not a sequence. But, point spectrum is alien to my problem. As to the continuous spectrum (C.S.), people work currently with it using the tool of $\delta$ Dirac normalization. E.g., Breit-Wigner formula for resonances is a distribution of C.S. energies. What I ask is whether the irregular functions are normal to the regular, both corresponding to C.S. Iff you are sure that this issue is undecidable please tell me clearly, otherwise if you just doubt, I also doubt - what that gives me? – Sofia Feb 03 '15 at 14:08
  • @Phonix87 I am not lucky with you, and I am sorry. I wanted to give you points, and it didn't go. You are by far better at mathematics than I - I am a phenomenologist. Besides that, I learnt physics many years ago, while you learnt recently - this is why I ask questions about concepts with which I don't work currently. But, your answer was (probably) on the ground where you can be sure - the point spectrum. What shall I do, it's not there where I work. I's not simple. – Sofia Feb 03 '15 at 14:16
  • @Sofia. I'm sorry but I really thought that my answer was clear enough and addressed your question fully. You are asking whether the eigenfunctions of an operator are orthogonal. They are, but the notion of eigenfunction works for isolated points in the spectrum (discrete spectrum). If a point comes from the C.S. 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 approximated eigenvector. – Phoenix87 Feb 03 '15 at 14:34
  • Aha! I understand, I had also some look in Wikipedia. Oh, it's not that it is a long time since I learnt, no, actually I never learnt these details about which you talk. Well, I am very embarrassed to bother you so much. For my good feeling, (i.e. for allowing me to give you points as I wish) would you introduce your last statement in the answer? I.e. "If a point comes from the C.S. there is no vector $\psi$ such that $A \psi = λ \psi$. For any ϵ>0 though you can choose a vector $\psi _ϵ$ such that $∥(A−λI) \psi _ϵ∥<ϵ$, hence $\psi _ϵ$ is just an approximated eigenvector". – Sofia Feb 03 '15 at 15:49
  • I did it, you have my points. Just, please complete your answer, would you? – Sofia Feb 03 '15 at 15:53