Advanced Chemistry - Mono Mole

December 7, 2022April 1, 2026

Slater-type orbitals

Slater-type orbitals (STO) are mathematical functions that resemble hydrogenic wavefunctions. Introduced by John Slater in 1930, they are used as trial wavefunctions in computational chemistry to approximate the energies of molecular systems. The general form of an STO is:

$\psi(r,\theta,\phi)=Nr^{n-1}e^{-\zeta r}Y_{l,ml}(\theta,\phi)\;\;\;\;\;\;\;\;(26)$

where

N is a normalisation constant

n, $l$ and $m_l$ are quantum numbers

$r$ is the distance between an electron and the nucleus of an atom.

$\zeta$ is the effective charge of the nucleus

$Y_{l,ml}(\theta,\phi)$ represents the spherical harmonics

STOs with different n but the same $l$ and $m_l$ are not orthogonal to one another. When working with such STOs, the Gram-Schmidt process is used to convert the non-orthogonal orbitals to orthogonal ones.

Question

Show that the normalisation constant of a 1s orbital is $\sqrt\frac{\zeta^3}{\pi}$ .

Answer

$N^2\int_0^{2\pi}d\phi\int_0^{\pi}sin\theta d\theta\int_0^{\infty}e^{-2\zeta r}r^2dr=1$
Using the identity $\int_0^{\infty}x^ne^{-\alpha x}dx=\frac{n!}{\alpha^{n+1}}$ (see this article for proof), we have $\sqrt\frac{\zeta^3}{\pi}$ .

Question

Using the identity $\int_0^{\infty}x^ne^{-\alpha x}dx=\frac{n!}{\alpha^{n+1}}$ , show that the 1s orbitals are orthogonal to each other but not orthogonal to the 2s orbital.

Answer

$\frac{\zeta_1^3}{\pi}\int e^{-\zeta_1r}e^{-\zeta_1r}d\tau=1$
$\sqrt\frac{\zeta_1^3\zeta_3^5}{3\pi^2}\int e^{-\zeta_1r}re^{-\zeta_3r}d\tau=\frac{\sqrt{192\zeta_1^3\zeta_3^5}}{(\zeta_1+\zeta_3)^4}$

For the orbitals to be orthogonal to each other, either $\zeta_1$ or $\zeta_3$ needs to be zero, which is forbidden because the effective nuclear charge is not zero. Hence, the orbitals are not orthogonal to each other.

Next article: Numerical solution of the Hartree self-consistent method for He

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December 7, 2022November 26, 2025

Simpson’s rule

Simpson’s rule is a numerical method that uses quadratic functions to approximate definite integrals.

With reference to the diagram above, $\int_{-h}^hf(x)dx$ is the area under the cubic curve demarcated by two equal intervals, each of magnitude h. The area is estimated using 3 points on the quadratic function $P(x)=Ax^2+Bx+C$ , where

$\int_{-h}^hf(x)dx\approx \int_{-h}^hP(x)dx=\frac{h}{3}(2Ah^2+6C)\;\;\;\;\;\;\;\;(23)$

Substituting $P(0)=y_1$ , $P(-h)=y_0$ and $P(h)=y_2$ into $P(x)$ gives $C=y_1$ , $Ah^2-Bh+C=y_0$ and $Ah^2+Bh+C=y_2$ . Combining the last two equations gives $2Ah^2+2C=y_0+y_2$ . Substituting this equation and $C=y_1$ into eq23 yields

$\int_{-h}^hf(x)dx\approx\frac{h}{3}(y_0+4y_1+y_2)$

Similarly, if we consider the domain from h to h‘ (see above diagram),

$\int_{h}^{h'}f(x)dx\approx\frac{h}{3}(y_2+4y_3+y_4)$

Therefore, using 5 points on $f(x)$ results in:

$\int_{-h}^{h'}f(x)dx\approx\frac{h}{3}(y_0+4y_1+2y_2+4y_3+y_4)$

Extending the logic for n points on $f(x)$ ,

$\int_{a}^{b}f(x)dx\approx\frac{h}{3}\{f(a)+4[f(x_1)+f(x_3)+\cdots+f(x_{n-1})]$
$+2[f(x_2)+f(x_4)+\cdots+f(x_{n-2})]+f(b)\}\;\;\;\;\;\;\;\;(24)$

where $a=x_0$ and $b=x_n$ .

This implies that there must be an even number of equal intervals for the rule to work, i.e. n must be odd.

Question

Evaluate
$DI=\int_0^\infty\phi_i(x_i)\left(\int_0^\infty f(x_i,x_j)dx_j\right)\phi_i(x_i)dx_i$

Answer

Substituting eq24 into the above equation yields:

$DI=\int_0^{\infty}\phi_i^2(x_i)\frac{h}{3}\{f(x_i,x_0)+4[f(x_i,x_1)+f(x_i,x_3)+\cdots+f(x_i,x_{\infty-1})]$ $+2[f(x_i,x_2)+f(x_i,x_4)+\cdots+f(x_i,x_{\infty-2})]+f(x_i,x_{\infty})\}dx_i$

So,

$DI=\frac{h}{3}[A+B+C+\cdots+D]\;\;\;\;\;\;\;\;(25)$

where (form 1)

$A=\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_0)dx_i=\frac{h}{3}[\phi_0^2(x_0)f(x_0,x_0)+4\phi_1^2(x_1)f(x_1,x_0)$ $+2\phi_2^2(x_2)f(x_2,x_0)+\cdots+\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_0)]$

$B=4\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_1)dx_i=\frac{4h}{3}[\phi_0^2(x_0)f(x_0,x_1)+4\phi_1^2(x_1)f(x_1,x_1)$ $+2\phi_2^2(x_2)f(x_2,x_1)+\cdots+\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_1)]$

$C=2\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_2)dx_i=\frac{2h}{3}[\phi_0^2(x_0)f(x_0,x_2)+4\phi_1^2(x_1)f(x_1,x_2)$ $+2\phi_2^2(x_2)f(x_2,x_2)+\cdots+\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_2)]$

$D=\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_{\infty})dx_i=\frac{h}{3}[\phi_0^2(x_0)f(x_0,x_{\infty})+4\phi_1^2(x_1)f(x_1,x_{\infty})$ $+2\phi_2^2(x_2)f(x_2,x_{\infty})+\cdots+\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_{\infty})]$

or equivalently (form 2),

$A=\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_0)dx_i=\frac{h}{3}[\phi_0^2(x_0)f(x_0,x_0)+4\phi_0^2(x_0)f(x_0,x_1)$ $+2\phi_0^2(x_0)f(x_0,x_2)+\cdots+\phi_0^2(x_0)f(x_0,x_{\infty})]$

$B=4\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_1)dx_i=\frac{4h}{3}[\phi_1^2(x_1)f(x_1,x_0)+4\phi_1^2(x_1)f(x_1,x_1)$ $+2\phi_1^2(x_1)f(x_1,x_2)+\cdots+\phi_1^2(x_1)f(x_1,x_{\infty}]$

$C=2\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_2)dx_i=\frac{2h}{3}[\phi_2^2(x_2)f(x_2,x_0)+4\phi_2^2(x_2)f(x_2,x_1)$ $+2\phi_2^2(x_2)f(x_2,x_2)+\cdots+\phi_2^2(x_2)f(x_2,x_{\infty})]$

$D=\int_0^{\infty}\phi_i^2(x_i)f(x_i,x_{\infty})dx_i=\frac{h}{3}[\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_0)+4\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_1)$ $+2\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_2)+\cdots+\phi_{\infty}^2(x_{\infty})f(x_{\infty},x_{\infty})]$

These integrals are useful for computing the numerical solution of the Hartree self-consistent field method for helium.

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