# Plane-wave molecular integrals

1. Aug 26, 2010

### Morberticus

Hi,

Just wondering about the state of the literature regarding plane-wave molecular integrals. The coulomb potential seems to be causing problems for me. I assumed I would need to consider a general multipole expansion to integrate over a potential |r-r'|^-1 but I have had a look around and some people seem to use a Fourier transform to avoid difficulties. Most of the literature I have come across concerns Gaussian wavefunctions (presumably due to their usefulness in quantum chemistry) and I am wondering if there is an equivalent review of plane-wave basis functions.

Thanks

2. Aug 26, 2010

### alxm

Plane waves are lousy basis functions for molecules; I don't know why you'd want to use them. That said, it's been done and tried if you look in the literature.

The area where they have real applications (in this context) is solid-state systems, where they're known as Bloch functions. (or rather, Bloch functions describe the expansion) Any SS book should tell you all about it.

3. Aug 27, 2010

### wsttiger

The coulomb integral is diagonal in fourier space. It's just 1 / k^2 where k is the wavevector.

4. Sep 1, 2010

### Morberticus

I think the reason the fourier transform is used is possibly to exploit

$$\int e^{ik(r-r')}dk = 2\pi\delta(r-r')$$

or to approximate 1/|r-r'| by considering a finite interval for the fourier integral, but I have not figured out how yet.

I am effectively looking for a valid approximation of |r-r|^-1. I have tried a general multiple explansion but it seems to break galilean invariance.

5. Sep 2, 2010

### wsttiger

I'm having trouble understanding what you are asking.

Fourier transforms are used to transform a real-space function (like an orbital) into momentum space (i.e., change the basis into a plane-wave basis). Once this is done, the |r-r'|^-1 convolution can be performed by dividing by k^2 (where k is the wavevector). Then the result can be Fourier transformed back into real space.

Is this not what you were asking?

BTW, like the above poster said .... plane waves make lousy basis functions for molecules; even for solid state crystals, plane wave codes need to use pseudopotentials to represent the core electrons.

6. Sep 3, 2010

### Morberticus

I know they're normally bad, but their nodal structure is useful for what I'm trying to do. It's just that I'm having trouble formulating the two-electron integrals <ij|g|kl> in a computer-friendly way.

7. Sep 3, 2010

### alxm

You wouldn't mind explaining what you're trying to do, instead? It's easier to give advice that way.
I ask because I recall answering another of your questions, where it turned out that what you were ultimately trying to do wasn't going to work.

Two-electron integral evaluation, under any circumstance, is not a cakewalk. After all, we're talking about what John Pople got the Nobel prize for, and spent much of his career working on.

8. Sep 5, 2010

### Morberticus

Do you meant he multipole expansion? I am still trying to get that to work. The trouble is the epansion breaks galilean invariance for me, though perhaps you're right, and that is unavoidable.

I'm not trying to do anything too exotic, just trying to see if there's an easy way to evaluate the integral:

$$\int e^{iAr}e^{iBr'}\frac{1}{|r-r'|}e^{-iCr}e^{-iDr'}drdr'$$

Where the middle term |r-r'|^-1 seems to be giving me all the hassle.

9. Sep 6, 2010

### DrDu

That does not look too difficult: introduce A'=A-C and B'=B-D so that you only have two exponentials. Then introducing X=(A'+B')/2, Y=(A'-B')/2 and Q=(r+r')/2 and q=(r-r')/2
the integral becomes 1/2 int exp(i (2XQ+2Yq)) /|q| dq dQ. The integral over Q becomes a delta function for X and the integral over q is the Fourier transform of the coulomb potential which is proportional to 1/Y^2.

10. Sep 6, 2010

### Morberticus

Thanks. I think the fourier transform is causing the the greatest difficulty for me.
An older thread of mine: https://www.physicsforums.com/showthread.php?t=425274

If it is proportional to 1/Y^2 does that mean there is a simple expression for it?

11. Sep 6, 2010

### DrDu

Yes, but the derivation of the fourier transform of the coulomb potential can be found in any book on quantum chemistry or solid state physics.

12. Sep 6, 2010

### DrDu

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