# exchange integral at large distances

I want to ask a question about the exchange integral $$\ce{H_{AB}}$$ -

If I consider an exchange integral $$\ce{H_{AB}}$$ the following exchange integral can be evaluated: so $$\ce{S_{AB}}$$ is $$0$$ at $$\ce{R = \infty}$$

but in this presentation I was shown today, at large values of R, $$\ce{H_{AB}}$$ is preportional to the overap integral $$\ce{S_{AB}}$$.

Surely at large distances and towards infinity, as $$\ce{S_{AB}}$$ goes to zero, $$\ce{H_{AB}}$$ also goes to zero too.

Where does the preportionality

$$H_{AB} \propto S_{AB}$$

come from then using these approimations when R is large?

• Great question @vik1245. Would you be interested in committing to a brand new stack exchange dedicated towards computational modeling? area51.stackexchange.com/proposals/122958/…. I wonder if you might be interested in committing to it? – user1271772 Jan 11 at 3:01

The author of the presentation is talking about large, but not yet very large/infinite distance here. Note the annotations: $$\frac{e^2}{4 \pi \epsilon R} = k/R$$ when $$R \rightarrow \infty$$: slowly to zero
$$S_{AB}$$ when $$R \rightarrow \infty$$: to zero
$$\left< 1s_A\left| k/r_A \right| 1s_B\right>$$
when $$R \rightarrow \infty$$: rapidly to zero
So there is some regime somewhere, in which the change of the entire expression is dominated by the change in $$S_{AB}$$. The verification of the correctness of the different notations is apparently left to the reader ...