“Know how to solve every problem that has been solved.” “What I cannot create, I do not understand.” — Richard Feynman

Basis Sets by Hand — The Slater orbital, for contrast

Exercises

Quantum Chemistry I › Unit 3 · Basis sets › Basis Sets by Hand › all problems

Practice Problem 2 of 8

The Slater orbital, for contrast

Problem

Now normalize a 1s Slater-type orbital

ϕ (r) = N e^{- ζ r}

— the

N

making

\int ∣ ϕ ∣^{2} d^{3} r = 1

— and note how its radial integral differs from the Gaussian case.

Solution

The exponential does not factorize in Cartesians; use spherical coordinates, $d^{3} r = 4 π r^{2} d r$ for an s-function:

\int ∣ ϕ ∣^{2} d^{3} r = N^{2} 4 π \int_{0}^{\infty} r^{2} e^{- 2 ζ r} d r = N^{2} 4 π \frac{2}{( 2 ζ ) ^{3}} = N^{2} \frac{π}{ζ ^{3}} = 1

N = (\frac{ζ ^{3}}{π})^{1/2}

The STO uses a factorial radial integral $\int_{0}^{\infty} r^{2} e^{- 2 ζ r} d r$ ; the Gaussian used a Cartesian product of error integrals. The exponential decays more slowly and has a cusp at the origin — the physically correct shape the Gaussian lacks.