Consider the hamiltonian for a helium atom, $$ H=\frac12\mathbf p_1^2+\frac12\mathbf p_2^2 - \frac{2}{r_1}-\frac{2}{r_2} + a \, \delta(\mathbf r_1-\mathbf r_2), $$ where I have taken out the electrostatic interactions between the two electrons and changed it to a contact interaction of strength $a$ to try and simplify things.

I would like to know to what extent this model is solvable, and if possible to what extent it has been explored in the literature. It is a natural thing to try, but it is also likely to be buried under the (more relevant) physics of actual electrostatic interactions between the electrons, so who knows what's out there.

I ask because, as I mentioned in this answer, this is a very simple model that is still likely to show autoionization for all doubly excited states, as normal helium does, and it's a good test case to see how that comes about and what the bare-bones features of that mechanism actually are.

As such, I'm interested in

  • what one can say about the ground and singly excited states in a Hartree-Fock perspective,
  • to what extent one can fully solve the two-body Schrödinger equation for bound states with this interaction,
  • how the different Hartree-Fock configurations look like for the doubly excited states as well as the corresponding singly ionized continua at those energies,
  • how the Fano mechanism uses the entangling contact interaction to couple those two sectors, and what the resulting Fano resonances look like, and
  • whether the simplified interaction lets us say more about those autoionizing resonances beyond what one can say via the Fano theory.

As mentioned in the comments, this localized contact interaction is unlikely to impose autoionization on double excitations that have an exchange-symmetric spin state (since then the spatial part is antisymmetric and the electrons never coincide), so the main thrust of the question is on the (antisymmetric) spin singlet states, where the nontrivial physics should happen.

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    $\begingroup$ At least half of the states is as simple as in the case of $a=0$: the ones antisymmetric with respect to exchange of electrons. So at least those doubly-excited states having this property will not be auto-ionizing. $\endgroup$ – Ruslan Feb 22 '17 at 12:45
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    $\begingroup$ @Ruslan You mean those with a symmetric spin part, but yes, you're correct. $\endgroup$ – Emilio Pisanty Feb 22 '17 at 12:46
  • $\begingroup$ @EmilioPisanty See physics.stackexchange.com/a/24826/84967 In short: $V(\vec r)\propto \delta(\vec r-\vec a)$ is an ill-defined potential. $\endgroup$ – AccidentalFourierTransform Feb 22 '17 at 15:29
  • $\begingroup$ @AccidentalFourierTransform Note that this is a repulsive, not an attractive, delta-function potential, which makes the arguments in that answer moot. I can't rule out more fundamental problems, though - can you comment on those? $\endgroup$ – Emilio Pisanty Feb 22 '17 at 17:00
  • $\begingroup$ @AccidentalFourierTransform you're mistaking $\delta(|\vec r_1-\vec r_2|)$ for 6D delta with point support $\delta(|\vec r-\vec a|)$ where $\vec a$ is a point. The delta-potential here has a hypersurface as singular support: all the points in 6D space where $\vec r_1=\vec r_2$. For a simpler example, compare $\delta(y)$ in 2D space vs $\delta(\sqrt{x^2+y^2})$: the former is more like the potential here, while the latter is more like the potential discussed in your link. $\endgroup$ – Ruslan Feb 22 '17 at 19:09

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