In Modern Quantum Chemistry by Szabo and Ostlund, they give an example of a minimal basis, full CI calculation of the dissociation of $\ce{H2}$ (on p.$241$ if you have the book). In explaining why this correctly describes the dissociation, they say:

To see this, recall that as $\mathrm{R}\rightarrow\infty$, ... that all molecular orbital two-electron integrals tend to $\frac{1}{2}(\phi_{1}\phi_{1}|\phi_{1}\phi_{1})$, where $\phi_{1}$ is a hydrogenic orbital.

I could not find where they justified this statement. The closest I could find was in a previous section describing the dissociation using RHF (p.$166$) they state:

[Besides $H^{\text{core}}_{11}$] All other integrals go to zero as $\mathrm{R}\rightarrow\infty$, except the one-center electron-electron repulsion integral $(\phi_{1}\phi_{1}|\phi_{1}\phi_{1})$.

What justifies this first statement? Is this unique to this case of a diatomic at infinite separation? These two comments seem to contradict rather than support each other.


2 Answers 2


Using Szabo and Ostlund's notation (page 85):

\begin{align} J_{ij} &= (\psi_i\psi_i|\psi_j\psi_j)& \text{and}&& K_{ij} &= (\psi_i\psi_j|\psi_i\psi_j)\tag1\label{integrals} \end{align}

For minimal basis set $\ce{H2}$ (page 162):

\begin{align} \psi_1 &= [2(1+S_{12})]^{-1/2}(\phi_1+\phi_2)& \text{and}&& \psi_2 &= [2(1-S_{12})]^{-1/2}(\phi_1-\phi_2)\tag2\label{minimal} \end{align}

Substitute equation $\eqref{minimal}$ into $\eqref{integrals}$, set $S_{12}$ and any integral involving both $\phi_1$ and $\phi_2$ to zero (large $R$) and you see that

$$J_{11} = K_{12} = \frac{1}{2}(\phi_1\phi_1|\phi_1\phi_1)$$

Remember that $(\phi_1\phi_1|\phi_1\phi_1)=(\phi_2\phi_2|\phi_2\phi_2)$ since you are dealing with a homonuclear diatomic.

  • $\begingroup$ $K_{ij}$ should be $(ij|ji)$, but I think it still works out the same. Thank you, this helped a lot! I guess part of my confusion was I was inadvertently equating R and r12. $\endgroup$
    – Tyberius
    Aug 1, 2017 at 16:21
  • $\begingroup$ if the orbitals are real then it's all the same. Anyway, happy to hear the answer was of use to you $\endgroup$
    – Jan Jensen
    Aug 1, 2017 at 17:20

Here is my intuition. For two atomic orbitals $\phi_1$ and $\phi_2$ located on distinct atomic centers 1 and 2, respectively, there are two ways of justifying this.

  1. The bra or the ket can be taken alone as an overlap integral over two Gaussians. At $r_{12} = \left|\mathbf{r}_1 - \mathbf{r}_2\right|$, $(\phi_1|\phi_1)$ and $(\phi_2|\phi_2)$ need not be zero because there is no "distance" to speak of; it's an integral over a squared Gaussian somewhere in space. However, because $\lim_{r_{12} \to \infty} r_{12} = \infty$, the overlap $(\phi_1|\phi_2)$ will approach (but not exactly be) zero. It is a matter of the overlap of two Gaussians at infinite separation being effectively zero.

  2. Consider the full two-electron four-center integral, where the bra is on one center and the ket on the other: $$(\phi_{1}\phi_{1}|\phi_{2}\phi_{2}) = \int\!\int d\mathbf{r}_{1} d\mathbf{r}_{2} \, \phi_{1}^{*}(\mathbf{r}_{1})\phi_{1}(\mathbf{r}_{1})\frac{1}{r_{12}}\phi_{2}^{*}(\mathbf{r}_{2})\phi_{2}(\mathbf{r}_{2}).$$ Following the logic from above, the bra and ket taken separately as overlaps or charge distributions may be non-zero and large, however $\lim_{r_{12} \to \infty} \frac{1}{r_{12}} = 0$.

Now, if $\phi_1$ and $\phi_2$ are molecular orbitals that are delocalized, this will no longer necessarily hold; in that case, there is also the presence of the density $P_{\mu\nu}$. I am assuming that their hydrogenic orbitals are just like atomic orbital (Gaussian basis functions).

  • $\begingroup$ I might have been unclear with my question. The 2nd statement doesn't bother me, I think they explain it the same way in the book as you just did. My issue is with the first statement, which seems to contradict the 2nd. $\endgroup$
    – Tyberius
    Jul 31, 2017 at 17:08
  • $\begingroup$ I misread, sorry; I see why it's bothersome. Unless I'm missing something, it's worded poorly. I think it means that they all slowly disappear and you're only left with things that look like (11|11), so you "approach" only having (11|11). It isn't possible (by convention, at least) for the specific integral (11|22) to become (11|11). I don't think it's referring to the antibonding H2 molecular orbital, which would also go to 0 at infinite separation. $\endgroup$ Jul 31, 2017 at 17:13
  • $\begingroup$ I would have thought so as well, but they use this statement to claim that the exchange integral K_12=1/2(11|11) $\endgroup$
    – Tyberius
    Jul 31, 2017 at 17:17
  • $\begingroup$ I'll have to find my copy of the book and remind myself of their operator notation then, because if J_12=(11|22), I would expect K_12=(12|12), which would go 0 under point 1. $\endgroup$ Jul 31, 2017 at 17:32
  • 1
    $\begingroup$ I was also inadvertently equating interatomic distance with interelectronic distance. This is true; all one can say is that the instantaneous interelectronic distance may be larger more often once $R_{AB}$ increases, but the $g_{12} = \frac{1}{r_{12}}$ operator is clearly not the reason for this going to zero. The wording for point 1 is correct but the mathematical reasoning is wrong there as well. $\endgroup$ Aug 2, 2017 at 11:33

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