Lately, I've been studying Hückel theory where I learned about the approximation of neglecting all overlaps and making overlap matrix an identity matrix that is $S_{AB}= 0\;_; \;S_{AA}= 1$ or $S_{ij}= \delta _{ij}\;.$ He then replaced the off-diagonal Hamiltonian elements with parameter viz. $H_{AB}= \beta_{AB}\;_,$ where $\beta_{AB}$ is a negative quantity.

Question then arose if the overlap is zero as is evident from $S_{AB}$ being zero, how can there be a non-zero value of $H_{AB}$. It was known to me that $H_{AB}$ is the

contribution to the energy due to the accumulation of electron density where the two atomic orbitals overlap, including, for instance, the Coulombic attraction between the extra accumulation of electron density and both nuclei.

So, it is contradictory to simultaneously have the condition $H_{AB}\ne 0$ with $S_{AB}= 0.$

I asked this querying for the same; there I got a nice answer from Brian where he showed that even if $S_{AB}=0$; it need not be necessary to have $H_{AB}= 0$; the graphs he provided in his answer prove this otherwise a counterintuitive fact.

Present thinking & associated query:

Brian's answer confirms the fact that $S_{AB}= 0$ and $H_{AB}\ne 0$ can go with each other. But doesn't that mean there is no need to have a net overlapping of AOs between the internuclear region in order to have accumulation of electron density in the internuclear region?

For example, while using Simple Hückel Theory for ethene, Atkins put all $S= 0$ as usual per the approximation and then wrote:

[...] Finally, as explained in the derivation, he also supposed that the terms $H_{AB}$ and $H_{BA} ,$ which represent the energy of the interaction of the two nuclei with the accumulation of electron density in the internuclear region .......

Really, if he writes this after considering $S_{AB}= 0\;_,$ doesn't that mean there can be accumulation of electron in the internuclear region even if the the participating AOs are orthogonal and there is no net overlapping? He is considering the case of ethene $\pi$-network which means there is electron density that leads to $\pi$-bonding even if $S=0$.

So, is it right to conclude that there can be accumulation of electron density in the internuclear region even if there is no net overlap in the atomic orbitals?

  • 2
    $\begingroup$ Yep, definitely better as a new question. :-) $\endgroup$
    – hBy2Py
    Dec 31, 2015 at 16:22
  • $\begingroup$ @Brian: Also, I did ponder a while & wondered how Hückel actually made this approximation of $S= 0$; making this would make the basis sets orthogonal. But orthogonality of AOs depends on the geometry of the molecule, isn't it? I googled a bit & didn't found anything regarding this. Nevertheless I do have a book Computational Chemistry by Errol Lewars which briefly mentioned it citing 'a twisted allyl species'. So, definitely, Hückel Theory is really eerie: he didn't consider those overlaps & didn't pay heed to the geometry of the molecule:/ $\endgroup$
    – user5764
    Jan 1, 2016 at 4:46
  • 1
    $\begingroup$ Oh, certainly Hückel's bases had $S_{AB}\neq 0$ in actuality -- this was higgsss's point in the comments on my answer at the other question, I believe. In order to make for a tractable calculation, though, I think Hückel just made his approximation by editing the matrices to coerce $S_{AB}=0$. Absolutely, this makes the mathematical problem diverge from the physical system, but in many cases it still gives qualitatively reasonable results. $\endgroup$
    – hBy2Py
    Jan 1, 2016 at 12:01
  • 1
    $\begingroup$ Using approximations to simplify calculations and having the given parameters physically zero are two completely independent problems. If neglecting S makes no big difference in the results, or simple parametrisation can mostly compensate this simplification, it is justified irrespective if the actual S is zero or not. $\endgroup$
    – Greg
    Jan 1, 2016 at 12:11

1 Answer 1


If you define overlap as given by the overlap integral,

$ S_{ij} = \langle \psi_{i} | \psi_{j} \rangle $

than you can have an accumulation of electron density in a specific part in space even though $S_{ij} = 0$.

Of course, if the complete wave functions of the two AOs are completely non-overlapping in all space, for instance because the centers of the two AOs are seperated by a long distance (say more than 5 A), then there can be no accumulation of electron density anywhere. This is of course logical. There is hardly any accumulation of electron density between two atoms that are quite far apart.

An interesting scenario exists where two wave functions have opposite sign and an overlap of these wave functions actually gives a decrease of electron density. This is termed deconstructive interference opposite to constructive interference when there is an accumulation of electron density.

  • $\begingroup$ Thanks for answering the question; so that means even if the net overlap is zero, if there is, at some place, non-zero $\psi_i\psi_j$, it can have a net Hamiltonian $H_{ij},$ is it so? Also, in ethene, the $p$ orbitals do overlap to give $\pi$ bonding; although Hückel put $S=0$. $\endgroup$
    – user5764
    Jan 1, 2016 at 4:14
  • $\begingroup$ Another thing I want to ask you is: if the the sum(integral) of $\psi_i\psi_j$ is zero, even though there may be places where $\psi_i\psi_j$ is non-zero; does that mean even $\psi_{i} \hat H\psi_j$ is non-zero at some places, but the net sum is zero since the net overlap is zero? Brian in his answer has shown that this doesn't necessarily happen. So, can I conclude that $H_{ij}$ gives the average energy of those $\psi_{i}\psi_{j}$ which are non-zero? $\endgroup$
    – user5764
    Jan 1, 2016 at 4:21
  • $\begingroup$ There are two things to keep in mind here; you have the product of two wave functions $\psi_{i} \psi_{j} (\vec{r})$ and you have $S_{ij} = \langle \psi_{i} | \psi_{j} \rangle$ which is the integral over all space of this product. Even if this $S_{ij} = 0$, the former can have a non-zero value at a specific $\vec{r}$. Obviously, if the value of $\psi_{i} \psi_{j} (\vec{r}) = 0$ at all $\vec{r}$, then the integral will be zero as a consequence of that. $\endgroup$
    – Ivo Filot
    Jan 1, 2016 at 12:07
  • $\begingroup$ Furthermore, if $\psi_{i} \psi_{j} (\vec{r}) = 0$ at some $\vec{r}$, then $\psi_{i} \hat{H} \psi_{j} (\vec{r}) = 0$ as well. Because if there is no overlap density at some point in space, then there can be no energy density at that point. This is a consequence of the following: If $a \cdot b = 0$, then either $a=0$, $b=0$, or $a = 0$ and $b=0$. $\endgroup$
    – Ivo Filot
    Jan 1, 2016 at 12:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.