Is the helium atom with only a contact interaction between the electrons solvable?

Question

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.

Answer 1

Note that for the less relevant case of $a \lt 0$ the problem may be ill-posed, depending on he size of $a$. See @EmilioPisanty's comment. Otherwise this model is solvable for triplets and perhaps also for singlets.

For triplets, just solve the hydrogenlike equation with Z=2 and occupy two different orbitals. This works as there is no interaction at all between the electrons.

For singlets and $a=0$ the result is the same as for triplets, except that you can also put both electrons in the same orbital or, generally, in partially overlapping orbitals. If $a \gt 0$ a repulsive $a|\psi|^4$ interaction occurs. I don't know if an exact solution is known or possible. The Kummer equation unifies the Schrödinger, Dirac and Klein Gordon radial hydrogenlike equations. You could look for extensions of it with third order terms. However, this seems a disproportionate effort for what is just a toy model, when atomic MCHF solutions are widely available.

Note that very accurate solutions are achievable with Hylleraas functionals.

As a further note, while it is possible to write down the solutions of the Schrödinger equation, this itself is only an approximation. For accuracy one should use the also exactly soluble Dirac equation, augmented with QED corrections. Also there are hyperfine interactions with nuclear moments and the effect of finite nuclear size to consider.

Answer 2

There is a fuzzy wavefunction solution.

1. When electrons are at the same distance from the nucleus. Then the coordinates of both electrons are linked, as soon as Kronecker delta is 1.

2. When energy of both electrons are independent eachother, because both distances to nucleus are not equal. Then delta Kronecker is zero.

3. Special case when repulsive Coulomb interaction between electrons is minimal. This is two blade electrons $$\vec{r_1} = -\vec{r_2}.$$

First case fullfills next Schrödinger Equation. Two times the Hamiltonian of hydrogen-like with Z=2 atom, considering constant $$a_n =\frac{e}{4\pi\epsilon_0}\left\langle\psi_1\left|\frac{1}{r_{12}}\right|\psi_1\right\rangle=\frac{e}{4\pi\epsilon_0}\left\langle R_n(r)\left|\frac{1}{r}\right|R_n(r)\right\rangle \hspace{0.1in} for \hspace{0.1in} l=0$$ where in the denominator stays the distance between He electrons. $$2H_{Z=2}\psi_1=(E-a_n) \psi_1 = p^2-\frac{4}{r}$$ where the energy levels only depends on the radius, for these reason is considered twice the kinetic energy, then $$E-a_n = \frac{-108.8}{n^2}$$ and the result in eV is $$E = \frac{-108.8}{n^2}+a_n.$$

Next case, is the energies of two electrons with independent coordinates atracted by the nucleus with Z=2.

$$E =\frac{-54.4}{n_1^2} - \frac{54.4}{n_2^2} \hspace{0.1in}(eV)$$ and none electron repulsion is considered, because of delta Kronecker is zero. And wave function is the product of the electron wave functions $$\psi_2=\psi_{21}\hspace{0.1in}\psi_{22}$$

And finally, a singularity from the first case, with analytical resolution with electron repulsion, is when $$a \hspace{0.1in}\delta(r_1-r_2) =\frac{1}{2r}$$, that is two blade electrons. It could be obtained next energy levels. Twice the level energies of hydrogen-like atom with $$Z=\frac{7}{4}=2-\frac{1}{4} \\ E = -\frac{83.3}{n^2} \hspace{0.1 in}(eV)$$

because $$H\psi_3=\left(2\frac{p^2}{2}-2\frac{2}{r}+\frac{1}{2r}\right)\psi_3$$ That solves the Schrödinger equation raised above. No Hartree-Fock statements are applied. Perhaps this Fuzzy Wave Function and related energy levels could explain Fano resonance of He.