Modern Quantum Chemistry

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Exercises: 1

This does not correspond to any chapter in the book, but rather are extra derivations that I found necessary during my coverage of the book, starting with chapter 3.

How are equations (3.219)-(3.221) in the book derived? Let’s start with the STO-1G case and try to maximize the overlap integral

$S=\int dr{\varphi }_{1s}^{SF}\left(\zeta ,r\right){\varphi }_{1s}^{CGF}\left(\zeta ,STO-1G,r\right)$

or, using the explicit forms for ${\varphi }_{1s}^{SF}$ and ${\varphi }_{1s}^{CGF}$

${\varphi }_{1s}^{SF}\left(\zeta ,r\right)=\sqrt{\frac{{\zeta }^{3}}{\pi }}{e}^{-\zeta r}$

${\varphi }_{1s}^{CGF}\left(\alpha ,r\right)={\left(\frac{2\alpha }{\pi }\right)}^{3∕4}{e}^{-\alpha {r}^{2}}$

find $\alpha$ for any $\zeta$ such that

$max=\sqrt{\frac{{\zeta }^{3}}{\pi }}{\left(\frac{2\alpha }{\pi }\right)}^{3∕4}\int {e}^{-\zeta r}{e}^{-\alpha {r}^{2}}dr$

or

$max=4\pi \sqrt{\frac{{\zeta }^{3}}{\pi }}{\left(\frac{2\alpha }{\pi }\right)}^{3∕4}{\int }_{0}^{\infty }{r}^{2}{e}^{-\zeta r}{e}^{-\alpha {r}^{2}}dr$

when we eliminate the $\theta$ and $\varphi$ spatial variables.

Using Mathematica, one can actually solve the integral to get:

Differentiating wrt $\alpha$ we get:

Since Slater exponent $\zeta \ne 0$ and $\alpha \ne \infty$ then we must have that

This nonlinear equation in $\alpha$ can be solved for different values of $\zeta$. A plot of the dependency is given below:

Now let’s consider the STO-2G case. In this case

${\varphi }_{1s}^{CGF}\left({\alpha }_{12},{\alpha }_{22},{d}_{12},{d}_{22},r\right)={d}_{12}{\varphi }_{1s}^{GF}\left({\alpha }_{12},r\right)+{d}_{22}{\varphi }_{1s}^{GF}\left({\alpha }_{22},r\right)$

with

${\varphi }_{1s}^{GF}\left(\alpha ,r\right)={\left(\frac{2\alpha }{\pi }\right)}^{3∕4}{e}^{-\alpha {r}^{2}}$

Therefore, the overlap integral writes:

$S={d}_{12}\int {\varphi }_{1s}^{SF}\left(\zeta ,r\right){\varphi }_{1s}^{GF}\left({\alpha }_{12},r\right)dr+{d}_{22}\int {\varphi }_{1s}^{SF}\left(\zeta ,r\right){\varphi }_{1s}^{GF}\left({\alpha }_{22},r\right)dr$

The overall strategy is as follows. First, write the normalization condition for ${\varphi }_{1s}^{CGF}$ and use the fact that ${\varphi }_{1s}^{GF}\left({\alpha }_{12},r\right)$ and ${\varphi }_{1s}^{GF}\left({\alpha }_{22},r\right)$ are normalized. Thus express ${d}_{22}$ in terms of ${d}_{12}$, ${\alpha }_{12}$, and ${\alpha }_{22}$. Second, replace ${d}_{22}$ expression into the overlap integral. Third, differentiate the overlap integral with respect to ${d}_{12}$, ${\alpha }_{12}$, and ${\alpha }_{22}$ to obtain a system of 3 equations with 3 unknowns which can be solved numerically.

The normalization constraint is:

$\int dr{\left[{d}_{12}{\varphi }_{1s}^{GF}\left({\alpha }_{12},r\right)+{d}_{22}{\varphi }_{1s}^{GF}\left({\alpha }_{22},r\right)\right]}^{2}=1$

or

${d}_{12}^{2}\int dr{\varphi }_{1s}^{G{F}^{2}}\left({\alpha }_{12},r\right)+{d}_{22}^{2}\int dr{\varphi }_{1s}^{G{F}^{2}}\left({\alpha }_{22},r\right)+2{d}_{12}{d}_{22}\int dr{\varphi }_{1s}^{GF}\left({\alpha }_{12},r\right){\varphi }_{1s}^{GF}\left({\alpha }_{22},r\right)=1$

Since ${\varphi }_{1s}^{GF}\left(\alpha ,r\right)$ are normalized, we end up with:

${d}_{12}^{2}+{d}_{22}^{2}+2{d}_{12}{d}_{22}\int dr{\varphi }_{1s}^{GF}\left({\alpha }_{12},r\right){\varphi }_{1s}^{GF}\left({\alpha }_{22},r\right)=1$

Remember that the product of two Gaussians is another Gaussian. In this case the Gaussians are at the same center, hence:

${\varphi }_{1s}^{GF}\left({\alpha }_{12},r\right){\varphi }_{1s}^{GF}\left({\alpha }_{22},r\right)={\left[\frac{2{\alpha }_{12}{\alpha }_{22}}{\left({\alpha }_{12}+{\alpha }_{22}\right)\pi }\right]}^{3∕4}{\left[\frac{2\left({\alpha }_{12}+{\alpha }_{22}\right)}{\pi }\right]}^{3∕4}{e}^{-\left({\alpha }_{12}+{\alpha }_{22}\right){r}^{2}}={\left(\frac{4{\alpha }_{12}{\alpha }_{22}}{{\pi }^{2}}\right)}^{3∕4}{e}^{-\left({\alpha }_{12}+{\alpha }_{22}\right){r}^{2}}$

Replacing into the normalization condition we get:

${d}_{12}^{2}+{d}_{22}^{2}+2{d}_{12}{d}_{22}{\left(\frac{4{\alpha }_{12}{\alpha }_{22}}{{\pi }^{2}}\right)}^{3∕4}\cdot 4\pi \cdot {\int }_{0}^{\infty }dr{r}^{2}{e}^{-\left({\alpha }_{12}+{\alpha }_{22}\right){r}^{2}}=1$

After evaluating the integral, we are left with:

${d}_{12}^{2}+{d}_{22}^{2}+2{d}_{12}{d}_{22}{\left[\frac{4{\alpha }_{12}{\alpha }_{22}}{{\left({\alpha }_{12}+{\alpha }_{22}\right)}^{2}}\right]}^{3∕4}=1$

We can formulate this as a 2nd degree equation and choose the solution (positive value for ${d}_{22}$):

${d}_{22}=-{d}_{12}{\left[\frac{4{\alpha }_{12}{\alpha }_{22}}{{\left({\alpha }_{12}+{\alpha }_{22}\right)}^{2}}\right]}^{3∕4}+\sqrt{{d}_{12}^{2}{\left[\frac{4{\alpha }_{12}{\alpha }_{22}}{{\left({\alpha }_{12}+{\alpha }_{22}\right)}^{2}}\right]}^{3∕2}-\left({d}_{12}^{2}-1\right)}$

Now, let’s replace this value into the overlap integral $S$:

$\begin{array}{rcll}S& =& {d}_{12}\int {\varphi }_{1s}^{SF}\left(\zeta ,r\right){\varphi }_{1s}^{GF}\left({\alpha }_{12},r\right)dr+& \text{}\\ & & +\left\{\sqrt{{d}_{12}^{2}{\left[\frac{4{\alpha }_{12}{\alpha }_{22}}{{\left({\alpha }_{12}+{\alpha }_{22}\right)}^{2}}\right]}^{3∕2}-\left({d}_{12}^{2}-1\right)}-{d}_{12}{\left[\frac{4{\alpha }_{12}{\alpha }_{22}}{{\left({\alpha }_{12}+{\alpha }_{22}\right)}^{2}}\right]}^{3∕4}\right\}\int {\varphi }_{1s}^{SF}\left(\zeta ,r\right){\varphi }_{1s}^{GF}\left({\alpha }_{22},r\right)dr& \text{}\end{array}$

Proceeding as above, the integrals evaluate to:

The overlap integral $S$ is then:

or, after pulling out constant factors:

Differentiating wrt ${d}_{12}$, ${\alpha }_{12}$ and ${\alpha }_{22}$ we get:

Now let’s consider the STO-3G case. The normalization condition writes:

$\int dr\left[{d}_{13}{\varphi }_{1s}^{GF}\left({\alpha }_{13},r\right)+{d}_{23}{\varphi }_{1s}^{GF}\left({\alpha }_{23},r\right)+{d}_{33}{\varphi }_{1s}^{GF}\left({\alpha }_{33},r\right)+\right]=1$

or

$\begin{array}{rcll}{d}_{13}^{2}\int dr{\varphi }_{1s}^{G{F}^{2}}\left({\alpha }_{13},r\right)+{d}_{23}^{2}\int dr{\varphi }_{1s}^{G{F}^{2}}\left({\alpha }_{23},r\right)+{d}_{33}^{2}\int dr{\varphi }_{1s}^{G{F}^{2}}\left({\alpha }_{33},r\right)+& & & \text{}\\ 2{d}_{13}{d}_{23}\int dr{\varphi }_{1s}^{GF}\left({\alpha }_{13},r\right){\varphi }_{1s}^{GF}\left({\alpha }_{23},r\right)+2{d}_{13}{d}_{33}\int dr{\varphi }_{1s}^{GF}\left({\alpha }_{13},r\right){\varphi }_{1s}^{GF}\left({\alpha }_{33},r\right)+& & & \text{}\\ +2{d}_{23}{d}_{33}\int dr{\varphi }_{1s}^{GF}\left({\alpha }_{23},r\right){\varphi }_{1s}^{GF}\left({\alpha }_{33},r\right)=1& & & \text{}\end{array}$