The Molecular Hamiltonian and the avoidance of Overcounting

Question

Whenever I see the total non-relativistic molecular Hamiltonian,

$\hat{H}_{molecular} = \hat{T}_{e} + \hat{T}_{n} + \hat{V}_{ee} + \hat{V}_{nn} + \hat{V}_{en}$

I always notice that the sums describing the electron-electron and the nucleus-nucleus interactions respectively are corrected for overcounting (i.e. the second index is greater than the first),

$\hat{V}_{ee} = \sum_{i}\sum_{j>i}\frac{1}{|\mathbf{r_{i} - r_{j}}|}$

$\hat{V}_{nn} = \sum_{I}\sum_{J>I}\frac{Z_{I}Z_{J}}{|\mathbf{R_{I} - R_{J}}|}$

whereas the sum describing electron-nucleus interactions is not corrected for overcounting:

$\hat{V}_{en} = \sum_{I}\sum_{j}\frac{Z_{I}}{\mathbf{|R_{I} - r_{j}}|}$

Could anyone possibly tell me why this is?

Thank-you!

Answer 1

Let's call the two types or particle $n$ and $e$.

When summing over the interactions between particles of type $n$, say, you have a sum like this

$\sum_i\sum_{j>i} H_{ij}$

where $i$ and $j$ are both sums over the $n$s. You need the $j>i$ because without it the pair $i=2, j=3$ (say) means the second particle of type $n$ and the third particle of type $n$. But this very same pair also reoccurs when $j=3, i=2$. You don't want to count both.

But when summing over pairs of different types of particles then in the sum

$\sum_i\sum_j H_{ij}$

the $i$ is over particles of type $n$ and the $j$ is over type $e$ (say). You can't possibly be double counting because no particle of type $n$ is a particle of type $e$. So the pair given by $i=2, j=3$, say, isn't ever the same as the pair given by $j=2, i=3$.