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I am wondering if it is possible to have a continuum of bound states in a finite system, for example a molecular system with a fixed number of nuclei and electrons. As a chemist, I'm used to discussions that are limited to discrete bound states, so I never thought about the possibility of a continuous spectrum of bound states besides unbound states above the ionization limit.

So my question is is whether there can be a continuous spectrum of bound states in an operator of the type

$$ \hat H = \hat T + \hat V $$ where $\hat T$ is the sum of all kinetic energy operators for all particles and $\hat V$ is a Coulomb potential operator for all involved particles. The system is not allowed to be a crystal with translational symmetry. It is a finite system. Let us also assume that we disregard the translation of the center of mass and the associated states. Consider this "total translation" to be separated such that only internal degrees of freedom remain.

This question was sparked by a review of the Born-Oppenheimer approximation, where one often encounters the claim that any molecular wavefunction can be expanded into a sum of the form $$ \psi(r,R) = \sum_n\theta_n(R)\varphi_n(r;R), $$ where $\theta_n(R)$ is the nuclear wavefunction and $\varphi_n(r;R)$ is an electronic eigenfunction of the associated electronic clamped-nuclei Hamiltonian. It is not clear why there shouldn't be continuous states involved in such an expansion, but in all the sources, that are known to me, I have never encountered a discussion of this point.

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  • $\begingroup$ Related: physics.stackexchange.com/q/65636/2451 , physics.stackexchange.com/q/282574/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jan 14, 2023 at 16:55
  • $\begingroup$ @Qmechanic I am afraid that I'm unable to recognize/extract an answer to my question from the links, although I can see that they discuss a similar(or is it the same?) problem. $\endgroup$
    – Hans Wurst
    Commented Jan 14, 2023 at 19:20
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    $\begingroup$ This answer from the second question Qmechanic linked answers your question: The spectrum of bound states is - at least by the definition of "bound state" in that answer - always the pure point spectrum of an observable, i.e. there are no continuous bound states. $\endgroup$
    – ACuriousMind
    Commented Jan 17, 2023 at 15:32

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Please have a look at this reference: Thierry Jecko, On the mathematical treatment of the Born-Oppenheimer approximation, J. Math. Phys. 55, 053504 (2014) https://arxiv.org/abs/1303.5833 Some discussion of the continuous spectrum is at p. 14. In general the commonly-given description of the Born-Huang expansion is very "handwavey". Sutcliffe and Wolley have discussed this many times. ASAIK in practice the Born-Huang expansion is used only with two (maybe three) surfaces to address near-degeneracies and intersections, never for very accurate work, so that the question of the completeness of the expansion never comes up. Accurate non-adiabatic results (mostly limited to high-accuracy studies of H2+ and H2) use other approaches (eg by Wolniewicz J Chem Phys 103, 1792 (1995) or Pachucki-Komasa J Chem Phys 130, 164113 (2009) --- these are two-step approaches relatively similar to the standard one. Otherwise fully nonadiabatic approaches are possible, eg pursued by Adamowicz or Matyus).

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