CamCASP/Programming/5

CamCASP => Programming => Auxiliary basis sets

Spherical versus Cartesian GTOs

• What are the advantages of the two basis types: Spherical and Cartesian?
• Which one results in more accurate energies and why?

Spherical GTOs:

1. Smaller in size
2. More well-defined set of linear equations during DF
3. Lower accuracy.
4. Seemingly erratic behaviour in aug-cc-pVnZ series.
5. Rotations made easy as we have the Wigner rotation matrices.

Cartesian GTOs:

1. Larger in size
2. DF equations seem OK.
3. Higher accuracy of energies
4. More consistent behaviour of aug-cc-pVnZ series.
5. Don't have rotation matrices coded, but these have been derived.

Simple example: 1

Here's an example that demonstrates the inadequacy of the spherical GTOs. Consider a 1p1s (yes, in this order) main basis and a single occupied orbital. This would be the px orbital. The space the auxiliary basis needs to span is 1p1s x 1p1s = 1s1p1d. So let us construct this auxiliary basis and perform the DF.

Here's the question: Can the density be exactly fitted using this auxiliary basis?

The answer depends on which basis type is used for the auxiliary basis.

• Cartesian: The auxiliary basis consists of the following combinations:

<math>

 s \ \  x \ \  y \ \  z \ \  x^2  \ \  xy  \ \ xz  \ \  y^2  \ \  yz  \ \  z^2

</math>

So the density, <math>x^2</math>, can be exactly described.

• Spherical: Now the auxiliary basis consists of the functions:

<math>

 s \ \   x  \ \   y  \ \  z \ \ \sqrt{3}xy \ \ \sqrt{3}yz \ \ z^2-1/2(x^2+y^2) \ \ \sqrt{3}xz \ \ \sqrt{3}/2(x^2-y^2)    

</math>

Amazingly, the density <math>x^2</math> cannot be exactly written as a linear combination of these functions.

Here are the input files for a simple CamCASP calculation that demonstrates this. The example calculates the nuclear integral of the occupied orbital with itself. The exact integral is <math>2.60588</math>, and what we get is:

• Cartesian: 2.60588
• Spherical: 2.53362

I.e., a 2.8% error in the spherical case. This error is reflected in the normalization constraints. These are satisfied exactly with the Cartesian auxiliary basis, but the spherical basis results in a normalization constraint of 1.1227 for the occupied x occupied orbital product. This should be exactly one. --alston 17:08, 12 May 2009 (BST)