# Minimal basis CI of H2 dissociation

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.

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.

• $K_{ij}$ 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. Aug 1 '17 at 16:21
• if the orbitals are real then it's all the same. Anyway, happy to hear the answer was of use to you Aug 1 '17 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).

• 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. Jul 31 '17 at 17:08
• 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. Jul 31 '17 at 17:13
• I would have thought so as well, but they use this statement to claim that the exchange integral K_12=1/2(11|11) Jul 31 '17 at 17:17
• 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. Jul 31 '17 at 17:32
• 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. Aug 2 '17 at 11:33