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:

enter image description here

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?

  • 1
    $\begingroup$ 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? $\endgroup$ – user1271772 Jan 11 '20 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 ...


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