I came across something called overlap integrals. I could understand that it is something about the overlapping of orbitals. What is it actually?

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    $\begingroup$ It basically gives you a quantitative measure for the overlap between orbitals. It should appear in any physical chemistry textbook. I don't feel confident giving a detailed explanation so I'll leave that to someone else. $\endgroup$ – Vlad Oct 9 '16 at 17:35

I guess you are talking about the overlap integral in as used chemical bonding. If we consider the simplest case of $\ce{H_2^+}$. The $1s$ orbitals are $\psi_1(r)$ on one atom and $\psi_2(r)$ on the other, and are functions of the electron's position r in the molecule. The overlap integral (conventionally called S) is defined as $$ S= \int \psi_1(r)\psi_2(r) d^3r$$ (The $d^3r$ is shorthand for the fact that this is a triple integral as the orbitals are 3 dimensional and we have to work out the distance of the electron from each nuclei using special set of coordinates, which is important, but is a maths detail).
As the atomic orbitals are normalised the maximum overlap is 1 and minimum zero. The overlap only contains contributions from regions of space where where the atomic orbitals are not zero. Thus as the nuclei become closer to one another the overlap increases and vice versa, thus the electrons are only ever partially 'overlap' in a molecule. Note, however, if the atomic orbitals are orthogonal to one another the overlap integral is zero at all distances.

On the same theme you will probably have have met the the Coulomb and exchange integrals. The Coulomb integral is $$ J= \int \psi_1(r)\frac{e^2}{r_2}\psi_1(r) d^3r$$ which is the electrostatic interaction between the charge distribution associated with the electron when in orbital 1 (associated with nucleus 1), with the proton in the nucleus of atom 2 (and similarly for orbital 2 and nucleus 1). The Exchange or Resonance integral is defined as $$ K= \int \psi_1(r)\frac{e^2}{r_1}\psi_2(r) d^3r$$ and is difficult to explain in classical terms because it origin is quantum in nature.

  • $\begingroup$ Minor comment: You forgot to complex conjugate $\psi_1$ in all your integrals. The overlap integral vanishes as the distance between the atoms $R\to \infty$, and there is no bonding, I guess. @porphyrin $\endgroup$ – SRS Nov 23 '17 at 15:49
  • $\begingroup$ yes technically that is true, but the $\ce{H2} orbitals are real so it would make no difference in this case, and yes the overlap decreases rapidly as r increases because the wave function decreases exponentially with increase in r. $\endgroup$ – porphyrin Nov 24 '17 at 16:23

Reduce the problem to one dimension first. The overlap integral between two functions, $f = f(x)$ and $g = g(x)$, is defined by

$$S_{fg} = \int_{-\infty}^\infty [f(x)]^*g(x)\,\mathrm{d}x$$

where $^*$ denotes complex conjugation. $S$ is the common symbol for the overlap integral; the subscript $fg$ indicates that it is the overlap between $f$ and $g$.

So, for example, let's try a simple case: let

$$f(x) = g(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \exp\left(-\frac{m\omega x^2}{2\hbar}\right)$$

(the ground state of the quantum harmonic oscillator). This function is real, so the complex conjugation doesn't do anything. We have

$$\begin{align} S_{fg} &= \int_{-\infty}^\infty \left(\frac{m\omega}{\pi \hbar}\right)^{1/2} \exp\left(-\frac{m\omega x^2}{\hbar}\right)\,\mathrm{d}x \\ &= \sqrt{\frac{m\omega}{\pi\hbar}} \int_{-\infty}^\infty \exp\left(-\frac{m\omega x^2}{\hbar}\right)\,\mathrm{d}x \\ &= \sqrt{\frac{m\omega}{\pi\hbar}} \sqrt{\frac{\pi\hbar}{m\omega}} \\ &= 1 \end{align}$$

(I used a standard integral, $\int_{-\infty}^\infty \exp{(-ax^2)}\,\mathrm{d}x = \sqrt{\pi/a}$). So this tells you that the overlap of this function with itself is equal to 1. That's not very interesting, because if you substitute $g(x) = f(x)$ into the original definition, you'll find that that is simply the normalisation condition for a wavefunction. I just wanted to use it as an illustration.

In general, the overlap between two normalised states ranges from 0 (when they are orthogonal) to 1 (when they are identical). This applies to three dimensions as well.

Of course, the notion of overlap is usually talked about in a three-dimensional context. So, all you have to do is to generalise the definition to three dimensions. Let's say we have two functions $\psi_1(x,y,z)$ and $\psi_2(x,y,z)$. Then we simply have

$$S_{12} = \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty \psi_1^*\psi_2 \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$

or in more concise notation

$$S_{12} = \int \psi_1^*\psi_2 \,\mathrm{d}\tau$$

where $\mathrm{d}\tau$ is the volume element (note that Wikipedia denotes it as $\mathrm{d}V$) and the limits of integration are assumed to be over all space.

This won't get you anywhere with atomic orbitals, though. Orbitals are frequently expressed in terms of spherical coordinates. So, you need to know what the volume element in spherical coordinates is, and that is:

$$\mathrm{d}\tau = r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$$

(note again that Wikipedia can be confusing sometimes, because the symbols used for spherical coordinates are different: $\theta$ and $\phi$ in chemistry have exactly the opposite meaning from $\theta$ and $\phi$ in mathematics.)

The limits of the integral are still "over all space", but in spherical coordinates this is expressed differently. $r$ takes values from zero to infinity, $\theta$ from $0$ to $\pi$, and $\phi$ from $0$ to $2\pi$. So:

$$S_{12} = \int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^\infty \psi_1^*\psi_2 r^2 \sin\theta \,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$$

All you need to do now, is to find the form of the function, plug it into the integral, and evaluate the integral. Warning: That's best done by a computer.

  • $\begingroup$ I have a question - if you have $f(x) = g(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \exp\left(-\frac{m\omega x^2}{2\hbar}\right)$ and then you're doing $f(x) \times g(x)$ surely the integral would evaluate $\exp\left(-\frac{m\omega x^2}{2\hbar}\right)^2$ right since the exponential part of the function is squared? I can see you doubled the power that you raised the normalisation constant, but surely the same applies to the exponential part too? $\endgroup$ – vik1245 Oct 22 '20 at 13:35
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    $\begingroup$ @vik1245, $f(x) = g(x)$ has a factor of 2 in the denominator of the exponential. This was removed inside the integral as $\exp(a/2) \cdot \exp(a/2) = \exp(a)$. Tough to spot, I'll admit! $\endgroup$ – orthocresol Oct 22 '20 at 14:20

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