Schrödinger Equation for Two Electrons?

Question

I'm having trouble finding a simple answer to this question (maybe because there isn't one), but I'm just confused about how the Schrödinger Equation would look for two electrons. I understand that it would exist in 6 dimensional configuration space, but how does the potential look? It's confusing that the potential would be different for each electron depending on where the other one is.

Answer 1

Let's have 2 electrons and nothing else. Then the Schrödinger wave function $\Psi$ depends on the positions $\vec{r}_1$ and $\vec{r}_2$ of the two electrons (i.e. on 6 coordinates) and on time $t$: $$\Psi(\vec{r}_1,\vec{r}_2,t)$$

The potential energy is just the Coulomb repulsion energy between the 2 electrons. It depends on the distance $|\vec{r}_1-\vec{r}_2|$: $$V(\vec{r}_1,\vec{r}_2)=\frac{e^2}{4\pi\epsilon_0|\vec{r}_1-\vec{r}_2|}$$

So you get the time-dependent Schrödinger equation in a straight-forward way: $$ -\frac{\hbar^2}{2m}\Delta_1\Psi(\vec{r}_1,\vec{r}_2,t) -\frac{\hbar^2}{2m}\Delta_2\Psi(\vec{r}_1,\vec{r}_2,t) +\frac{e^2}{4\pi\epsilon_0|\vec{r}_1-\vec{r}_2|}\Psi(\vec{r}_1,\vec{r}_2,t) =i\hbar\frac{\partial\Psi(\vec{r}_1,\vec{r}_2,t)}{\partial t} $$ where $\Delta_1$ and $\Delta_2$ are the Laplace operators with respect to $\vec{r}_1$ and $\vec{r}_2$.

Answer 2

The six coordinates can be combined into 3 $\vec R_{CM}$, the center of mass position, and 3 $\vec r$ using the usual reduced mass:

$$ \frac 1 m = \frac 1 {m_1} + \frac 1 {m_2} = \frac 1 {m_e} + \frac 1 {m_e} = \frac 2 {m_e} $$

Then:

$$ [\frac{\hat p^2}{2m} + V(\vec r)]\psi(\vec r) = 0 $$

The potential is just the Coulomb potential:

$$ V(\vec r) = V(r) = +\frac 1 {4\pi\epsilon_0}\frac{e^2} r$$

where I explicitly included the $+$ sign to indicated that it is repulsive.

Putting the for of the momentum operator in position space:

$$ [\frac{-\hbar^2}{2m}\nabla^2 + \frac 1 {4\pi\epsilon_0}\frac{e^2} r]\psi(\vec r) = 0 $$

At this point you can factor out the angular dependence so that:

$$\psi(\vec r) = R(r)Y^,_l(\theta, \phi) $$

and rewrite the different equation only in terms of $r$.

A full treatment would require a magnetic dipole-dipole term:

$$ H_m(\vec r) = -\frac{\mu_0\gamma^2\hbar^2}{4\pi r^3}[3(\vec S_1\cdot\hat r)(\vec S_2 \cdot \hat r)-\vec S_1 \cdot \vec S_2] $$

where

$$\gamma = \frac e {2m_e}(2 + \frac{\alpha}{2\pi} + \cdots) = 1.76085963023\times 10^{-11}\,{\mathrm s^{-1}T^{-1}} $$

is the electron gyromagnetic ratio and $\vec S_i$ are their spin operators.