# 22.2.5: v. Exercise Solution

1. Two Slater type orbitals, i and j, centered on the same point results in the following overlap integrals:
$S_{ij} = \int\limits^{2\pi}_{0}\int\limits^{\pi}_0\int\limits^{\infty}_0 \left( \dfrac{2\xi_i}{a_0} \right)^{n_i+\dfrac{1}{2}}\sqrt{\dfrac{1}{(2n_i)!}} r^{(n_i-1)}e^{\left( \dfrac{-\xi_i r}{a_0} \right)} Y_{l_i,m_i}(\theta,\phi ) \left( \dfrac{2\xi_j}{a_0} \right)^{n_j+\dfrac{1}{2}} \sqrt{ \dfrac{1}{(2n_j)!}} r^{(n_j-1)}e^{\left( \dfrac{-\xi_j r}{a_0} \right)} Y_{l_j,m_j}(\theta,\phi ) r^2sin \theta drd\theta d\phi .$

For these s orbitals l = m = 0 and $$Y_{0,0}(\theta ,\phi ) = \dfrac{1}{\sqrt{4\pi}}.$$ Performing the integrations over $$\theta$$ and $$\phi$$ yields $$4\pi$$ which then cancels with these Y terms. The integral then recuces to:
$S_{ij} = \left( \dfrac{2\xi_i}{a_0} \right)^{n_i+\dfrac{1}{2}} \sqrt{\dfrac{1}{(2n_i)!}} \left( \dfrac{2\xi_j}{a_0} \right)^{n_j + \dfrac{1}{2}} \sqrt{\dfrac{1}{(2n_j)!}} \int\limits^{\infty}_{0} r^{(n_i-1+n_j-1)}e^{\left( \dfrac{-(\xi_i+\xi_j)r}{a_0} \right)}r^2dr$
$= \left( \dfrac{2\xi_i}{a_0} \right)^{n_i+\dfrac{1}{2}} \sqrt{\dfrac{1}{(2n_i)!}} \left( \dfrac{2\xi_j}{a_0} \right)^{n_j + \dfrac{1}{2}}\sqrt{\dfrac{1}{(2n_j)!}}\int\limits^{\infty}_{0} r^{(n_i-1+n_j-1)}e^{\left( \dfrac{-(\xi_i+\xi_j)r}{a_0} \right)}r^2dr$

Using integral equation (4) the integral then reduces to:

$S_{ij} = \left( \dfrac{2\xi_i}{a_0} \right)^{n_i + \dfrac{1}{2}} \sqrt{\dfrac{1}{(2n_i)!}} \left( \right)^{n_j+\dfrac{1}{2}} \sqrt{ \dfrac{1}{(2n_j)!}}(n_i+n_j)! \left( \dfrac{a_0}{\xi_i+\xi_j} \right)^{n_i+n_j+1}.$

We then substitute in the values for each of these constants:
$\text{for i=1; n=1, l=m=0, and } \xi = 2.6906$
$\text{for i=2; n=2, l=m=0, and } \xi = 0.6396$
$\text{for i=3; n=3, l=m=0, and } \xi = 0.1503.$

Evaluating each of these matrix elements we obtain:

$S_{11} = (12.482992)(0.707107)(12.482992)(0.707107)(2.00)(0.006417) = 1.000000$
$S_{21}=S_{12} = (1.850743)(0.204124)(12.482992)(0.707107)(6.00)(0.008131) = 0.162673$
$S_{22} = (1.850743)(0.204124)(1.850743)(0.204124)(24.00)(0.291950) = 1.00$
$S_{31} = S_{13} = (0.0144892)(0.037268)(12.482992)(0.707107)(24.00)(0.005404) = 0.000635$
$S_{32} = S_{23} = (0.014892)(0.037268)(1.850743)(0.204124)(120.00)(4.116872) = 0.103582$
$S_{33} = (0.014892)(0.037268)(0.014892)(0.037268)(720.00)(4508.968136) = 1.00$

$S = \begin{bmatrix} & 1.000000 & & & \\ & 0.162673 & 1.000000 & & \\ & 0.000635 & 0.103582 & 1.000000 & \end{bmatrix}$

We now solve the matrix eigenvalue problem S U = $$\lambda$$ U.
The eigenvalues, $$\lambda$$, of this overlap matrix are:
[ 0.807436 0.999424 1.193139 ],
and the corresponding eigenvectors, U, are:

\begin{bmatrix} & 0.596540 & -0.537104 & -0.596372 & \\ & -0.707634 & -0.001394 & 0.706578 & \\ & 0.378675 & 0.843515 & -0.380905 & \end{bmatrix}

The $$\lambda^{-\dfrac{1}{2}}$$ matrix becomes:

$\lambda^{-\dfrac{1}{2}} = \begin{bmatrix} & 1.112874 & 0.000000 & 0.000000 & \\ & 0.000000 & 1.000288 & 0.000000 & \\ & 0.000000 & 0.000000 & 0.915492 & \end{bmatrix}.$

Back transforming into the original eigenbasis gives $$S^{-\dfrac{1}{2}}$$, e.g.
$S^{-\dfrac{1}{2}} = U\lambda^{-\dfrac{1}{2}} U^T$
$S^{-\dfrac{1}{2}} = \begin{bmatrix} & 1.010194 & & & \\ & -0.083258 & 1.014330 & & \\ & 0.006170 & -0.052991 & 1.004129 & \end{bmatrix}$

The old ao matrix can be written as:

$C = \begin{bmatrix} & 1.000000 & 0.000000 & 0.000000 & \\ & 0.000000 & 1.000000 & 0.000000 & \\ & 0.000000 & 0.000000 & 1.000000 & \end{bmatrix}$

The new ao matrix (which now gives each ao as a linear combination of the original aos) then becomes:

$C^{\prime} = S^{-\dfrac{1}{2}}C = \begin{bmatrix} & 1.010194 & -0.083258 & 0.006170 & \\ & -0.083258 & 1.014330 & -0.052991 & \\ & 0.006170 & -0.052991 & 1.004129 & \end{bmatrix}$

These new aos have been constructed to meet the orthonormalization requirement $$C^{\prime T}SC^{\prime} = 1$$ since:

$\left( S^{-\dfrac{1}{2}}C \right)^T SS^{-\dfrac{1}{2}} C = C^TS^{-\dfrac{1}{2}} SS^{-\dfrac{1}{2}} C=C^TC = 1.$

But, it is always good to check our result and indeed:

$C^{\prime T}SC^{\prime} = \begin{bmatrix} & 1.000000 & 0.000000 & 0.000000 & \\ & 0.000000 & 1.000000 & 0.000000 & \\ & 0.000000 & 0.000000 & 1.000000 & \end{bmatrix}$
2. The least time consuming route here is to evaluate each of the needed integrals first. These are evaluated analogous to exercise 1, letting $$\chi_i$$ denote each of the individual Slater Type Orbitals.

\begin{align} \int\limits^{\infty}_{0} \chi_i r \chi_j r^2 dr & = & \langle r \rangle_{ij} \\ & = & \left( \dfrac{2\xi_i}{a_0} \right)^{n_i + \dfrac{1}{2}} \sqrt{\dfrac{1}{(2n_i)!}} \left( \dfrac{2\xi_j}{a_0} \right)^{n_j + \dfrac{1}{2}} \sqrt{\dfrac{1}{(2n_j)!}}\int\limits_{0}^{\infty}r^{(n_i+n_j+1)}e^{\left( -\dfrac{(\xi_i+\xi_j)r}{a_0} \right)}dr \end{align}

Once again using integral equation (4) the integral recduces to:
$= \left( \dfrac{2\xi_i}{a_0} \right)^{n_i + \dfrac{1}{2}} \sqrt{\dfrac{1}{(2n_i)!}} \left( \dfrac{2\xi_j}{a_0} \right)^{n_j+\dfrac{1}{2}} \sqrt{\dfrac{1}{(2n_j)!}}\int\limits_0^{\infty}r^{(n_i+n_j+1)}e^{\left(-\dfrac{(\xi_i+\xi_j )r}{a_0}\right)}dr$

Again, upon substituting in the values for each of these constants, evaluation of these expectation values yields:

$\langle r \rangle_{11} = (12.482992)(0.707107)(12.482992)(0.707107)(6.00)(0.001193) = 0.557496$
$\langle r \rangle_{21} = \langle r \rangle_{12} (1.850743)(0.204124)(12.482992)(0.707107)(24.00)(0.002441) = 0.195391$
$\langle r \rangle_{22} = (1.850743)(0.204124)(1.850743)(0.204124)(120.00)(0.228228) = 3.908693$
$\langle r \rangle_{31} = \langle r \rangle_{13} = (0.014892)(0.0337268)(12.482292)(0.707107)(120.00)(0.001902) = 0.001118$
$\langle r \rangle_{32} = \langle r \rangle_{23} = (0.014892)(0.037268)(1.850743)(0.204124)(720.00)(5.211889) = 0.786798$
$\langle r \rangle_{33} = (0.014892)(0.037268)(0.014892)(0.037268)(5040.00)(14999.893999) = 23.286760$

$\int\limits_{0}^{\infty}\chi_i r\chi_j r^2 dr = \langle r \rangle_{ij} \begin{bmatrix} & 0.557496 & & & \\ & 0.195391 & 3.908693 & & \\ & 0.001118 & 0.786798 & 23.286760 & \end{bmatrix}$

Using these integrals one then proceeds to evaluate the expectation values of each of the orthogonalized aos, $$\chi^{\prime}_n$$, as:
$\int\limits^{\infty}_{0} \chi^{\prime}_n r \chi^{\prime}_n r^2 dr = \sum\limits_{i=1}^3 \sum\limits_{j=1}^3 C^{\prime}_{ni}C_{nj}^{\prime} \langle r \rangle_{ij}.$
This results in the following expectation values (in atomic units):

\begin{align} \int\limits_{0}^{\infty} \chi^{\prime}_{1s} r \chi^{\prime}_{1s} r^2 dr & = & 0.563240 \text{ bohr} \\ \int\limits_{0}^{\infty} \chi^{\prime}_{2s} r \chi^{\prime}_{2s} r^2 dr & = & 3.973199 \text{ bohr} \\ \int\limits_{0}^{\infty} \chi^{\prime}_{3s} r \chi^{\prime}_{3s} r^2 dr & = & 23.406622 \text{ bohr} \end{align}
3. The radial density for each orthogonalized orbital, $$\chi^{\prime}_n$$, assuming integrations over θ and $$\phi$$ have already been performed can be written as:

$\int\limits_{0}^{\infty} \chi^{\prime}_n\chi^{\prime}_n r^2 dr = \sum\limits_{i=1}^3\sum\limits_{j=1}^3 C^{\prime}_{ni}C^{\prime}_{nj} \int\limits_{0}^{\infty} R_iR_j r^2 dr \text{ , where } R_i \text{ and } R_j \text{ are the radial portions of the individual Slater Type Orbitals, e.g.,}$

$R_iR_jr^2 = \left( \dfrac{2\xi_i}{a_0} \right)^{n_i + \dfrac{1}{2}}\sqrt{\dfrac{1}{(2n_i)!}}\left( \dfrac{2\xi_j}{a_0} \right)^{n_j+\dfrac{1}{2}}\sqrt{\dfrac{1}{(2n_j)!}}r^{(n_i+n_j)}e^{\left(-\dfrac{(\xi_i+\xi_j)r}{a_0} \right)}$

Therefore a plot of the radial probability for a given orthogonalized atomic orbital, n, will be: $$\sum\limits_{i=1}^3\sum\limits_{j=1}^3 C^{\prime}_{ni}C^{\prime}_{nj}R_iR_jr^2 \text{ vs. r.}$$

Plot the orthogonalized 1s orbital probability density vs r; note there are no nodes.