# Exact solution to electron-electron scattering?

One of the first results seen in elementary quantum mechanics is the closed form solution to the bound states of the hydrogen atom. In the usual approach, scattering theory is placed on the opposite side of spectrum; exclusively though of as a perturbative process. I am interested in knowing about the scattering states of the hydrogen atom, but more specifically in terms of electron-electron scattering.

My question is regarding the scattering states of electron-eletron scattering, which has the (non-relativistic) Hamiltonian below.

$$H=\left(\frac{\hbar^2}{2m_e}\nabla^2_1+\frac{\hbar^2}{2m_e}\nabla^2_2\right)+\frac{e^2}{|\mathbf{r}_1-\mathbf{r}_2|}$$

Because the problem is so similar to that of the electron-proton system, I would expect a closed form solution for dealing with the scattering states non-perturbatively. However, this problem (Rutherford scattering) is usually treated in textbooks using the Born approximation, with divergences at low deflection angles where the approach breaks down. I am especially interested in this regime.

So my questions are the following:

1. In the non-relativistic limit, is there a non-perturbative solution to electron-electron scattering? By solution I (naively) am thinking of something akin to an expression for the scattering cross-section.

2. When relativity is introduced (either with the Dirac-equation, or the full machinery of QED), at what point is a non-perturbative solution out of reach? Are there any special cases (e.g. total energies are below that needed for pair-production) that are significantly easier to handle?

• I'm guessing the answer is "no" as I've never heard of such a solution mentioned before. – Virgo Nov 22 '16 at 6:30

The problem is isomorphic to hydrogen problem. It's the same two-body problem, but with same-sign charges. It's separable into standard spherical harmonic equation and radial equation. The radial equation is solvable in terms of Tricomi confluent hypergeometric function $U$ and Kummer confluent hypergeometric function $_1F_1$. The second one is regular and by itself can represent standing waves, while the first one is singular at origin and can be combined with the first one to form incoming/outgoing waves.
• @KFGauss motion in Coulomb potential is discussed e.g. in Laundau&Lifshitz Quantum Mechanics $\S36$. It includes discussion of continuous spectrum including the case of repulsive potential. – Ruslan Nov 27 '16 at 6:23
To solve this problem, it is better to use the relativistic equation M2. $$\Delta \Psi -\frac{1}{\hbar^{2}}\left [ \frac{m^{4}c^{6}}{\left ( E-U\left ( \overrightarrow{r}\right ) \right )^{2}}-m^{2}c^{2} \right ]\Psi =0$$ The problem is analogous to the hydrogen atom. It is necessary in the equation for the hydrogen atom to replace the mass of the electron m by the reduced mass $\frac{m}{2}$. And instead of the Coulomb potential of attraction$U\left ( \overrightarrow{r} \right )=-\frac{Ze^{2}}{4\pi \varepsilon _{0}r}$, substitute the Coulomb repulsion potential $U\left ( \overrightarrow{r} \right )=+\frac{Ze^{2}}{4\pi \varepsilon _{0}r}$. As a result, we obtain the equation: $$\Delta \Psi -\frac{1}{\hbar^{2}}\left [ \frac{\left ( \frac{m}{2} \right )^{4}c^{6}}{\left ( E-\frac{Ze^{2}}{4\pi \varepsilon _{0}r}\right )^{2}}-\left ( \frac{m}{2} \right )^{2}c^{2} \right ]\Psi =0$$ $\left ( Z=1 \right )$ If energy is emitted during the electron-electron interaction, this means the formation of a potential well. In this case, the solutions show the possibility of forming a composite particle of two electrons.