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In doing second order time-independent perturbation theory in non-relativistic quantum mechanics one has to calculate the overlap between states

$$E^{(2)}_n ~=~ \sum_{m \neq n}\frac{|\langle m | H' | n \rangle|^2}{E_n^{(0)}-E_m^{(0)}}$$

(where $E^{(k)}_n$ represents the kth order correction to the nth energy level).

For the Hydrogenic atom the spectrum of the Hamiltonian consists of a discrete set corresponding to "bound states" (the negative energy states) and a continuum of "scattering states" (the positive energy states).

Is there an example where the overlap between a bound state and the scattering states makes a measurable contribution to the energy in the perturbative regime? Should the scattering states be included in the perturbative calculations? Are there experimental results that backs this up? (For example, perhaps the electron-electron interaction in highly excited states of Helium).

As an aside, what "is" the Hilbert space of the Hydrogen atom in the position representation? I've often read the basis of eigenstates of the Hydrogen atom Hamiltonian isn't complete without the scattering states, but I've not seen any convincing argument of this. I've read that the radial bound states are dense in $L^2((0, \infty))$ (e.g. here), so including the scattering states the Hilbert space must strictly contain this.

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2 Answers 2

Your main question was "Is there an example where the overlap between a bound state and the scattering states makes a measurable contribution to the energy in the perturbative regime?"

Actually, I disagree with the statement of the other answer that the scattering states must be included in the perturbative calculations only if the result is to be highly accurate. In fact, as for the hydrogen atom, if you take, as a simple example, the perturbative potential $\epsilon/r$, than the continuum in second order accounts for TWO THIRDS of the total contribution! If you take $\epsilon/r^2$, it is even 75%!

For other examples, see the following articles:

  1. Here you can see, that "the continuum contribution to the sum is sometimes quite large, accounting, in extreme cases, for all but a few percent of the total."

  2. For a textbook example, you can have a look at Schiff (p.263-265), where the Stark effect of hydrogen atom is worked out with the continuum contribution.

If you are interested, I can provide you with more material.

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The scattering states must be included in the perturbative calculations if the result is to be highly accurate. In particular, it is not justified to ignore the continuous spectrum at energies close to the dissociation threshold.

The Hilbert space in the position representation is the space of square integrable functions on $R^3\setminus\{0\}$ with respect to the inner product $$\langle\phi|\psi\rangle:=\int \frac{dx}{|x|}\phi(x)^*\psi(x).$$ The bound states alone are not dense in this space.

For thorough treatments of the hydrogen spectrum see the books
G R. Gilmore, Lie Groups, Lie Algebras & Some of Their Applications, Wiley 1974, Dover, 2002. pp. 427-430
and
A.O. Barut and R. Raczka, Theory of group representations and applications, 2nd. ed., Warzawa 1980. Chapter 12.2.

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1  
Thanks, that makes sense to me. The bound states are complete on the subspace of square integrable functions on $R^3 \setminus \{0\}$ with respect to the inner product $\langle\phi|\psi\rangle:=\int dx\phi(x)^*\psi(x)$ and the scattering states are the orthogonal complement of this subspace in the Hilbert space you mentioned. Though I'm surprised it's $\frac{dx}{|x|}$ and not $\frac{dx}{|x|^2}$. –  Edward Ross Nov 22 '12 at 22:49
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The power of the denominator is really 1 and not 2. This singular inner product is relevant to get a unitary representation of the dynamical symmetry group $SO(4,2)$ of hydrogen. –  Arnold Neumaier Nov 23 '12 at 10:57
    
This is a very strange inner product. –  Jiang-min Zhang Mar 9 at 10:57
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@Jiang-minZhang: You may understand it by reverting to the usual inner product by a transformation of the wave functions to $\bar\psi(x):=|x|^{-1/2}\psi(x)$. Then you get the standard inner product on the transformed wave functions. Instead you have now to worry about the right boundary conditions at $x=0$. –  Arnold Neumaier Mar 10 at 13:26

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