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For a system has both continuous and discrete spectrum, is it possible that a physical states is something like:

$$\psi(x)=\sum_{n=n_1}^{n=n2}c_n\psi_n(x)+\int_{E_0-\Delta_E}^{E_0+\Delta_E}\psi_E(x)\mathrm{d}E$$

I am asking this because of some comments under my answer to this Phys.SE question: How is a bound state defined in quantum mechanics?

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    $\begingroup$ Yes, it is perfectly possible. The superposition principle wouldn't make much sense without this possibility. $\endgroup$ – Void Oct 8 '14 at 8:41
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Yes, this is perfectly possible. The simplest example is the hydrogen atom, which has an infinite sequence of discrete eigenstates $|nlm\rangle$ with negative energy, and an energy continuum at $E>0$, which are known as Coulomb waves. It is perfectly possible for a state to be in a superposition of those:

  • First of all, it is required by the superposition principle. If $|\psi\rangle$ and $|\phi\rangle$ are states, the superposition $|\psi\rangle+|\phi\rangle$ must also be a state. This is the essential leap in the postulate that the state space of quantum mechanics is a Hilbert space.

  • Secondly, these superpositions get used all the time. If you ionize an atom, you will typically only remove the electron with a finite probability $p<1$, which means that you will leave population in both the bound and continuum subspaces of state space.

    I should stress that in such an ionization event you do (sometimes) need to account for the state of the system as a coherent superposition. This is particularly the case if the ionized portion of the wavefunction returns to interact with the core, as is the case in High Harmonic Generation (see e.g. this review or the original paper for bound + continuum superpositions) or Above-Threshold Ionization (as in this paper).

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  • $\begingroup$ In the paragraph Secondly..., I think you are talking about an ensemble rather than a genuine electron state. $\endgroup$ – an offer can't refuse Oct 22 '14 at 9:44
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    $\begingroup$ No, that is not the case. The experiment is indeed done on macroscopic samples, but all the atoms that contribute have the exact same quantum state. By 'population' here I mean $|\psi|^2$, i.e. a fraction of the wavefunction of a single atom. If you simply assume only a fraction of the atoms has been ionized, you will fail to get all the physics. (In particular, that decohered sample would not display HHG or ATI rings.) Do read the papers - if they are unclear I can probably suggest better references but I need to know what you're having trouble with. $\endgroup$ – Emilio Pisanty Oct 22 '14 at 10:35
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Yes, this is possible. Note, also, that a continuous spectrum is often an approximation, since materials are of finite extent and thus have always have a discrete spectrum but the continuous spectrum is a good approximation.

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  • $\begingroup$ This type of reasoning is good. $\endgroup$ – an offer can't refuse Nov 27 '17 at 2:14

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