1.4b: Bonding and anti-bonding orbitals

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We look, first, at the form of the orbitals that correspond to the energies $\Delta E_{\pm}$, respectively. These can be obtained by solving for the variational coefficients, $C_1$ and $C_2$. These will be given by the matrix equation:

$\begin{displaymath} \left(\matrix{H_{11} & H_{12} \cr H_{21} & H_{22}}\right) \l... ...{12} \cr S_{21} & 1}\right) \left(\matrix{C_1 \cr C_2}\right) \end{displaymath}$

For example, using $E_+=(H_{11}-H_{12})/(1-S)$, the following equations for the coefficients are obtained:

$\displaystyle H_{11} C_1 + H_{12}C_2$	$\textstyle =$	$\displaystyle {H_{11}-H_{12} \over 1-S}(C_1+SC_2)$
$\displaystyle H_{12} C_1 + H_{11}C_2$	$\textstyle =$	$\displaystyle {H_{11}-H_{12} \over 1-S}(SC_1+C_2)$

which are not independent but are satisfied if $C_1=-C_2\equiv C_+$. Similarly, for $E=E_-$, we obtain the two equations:

$\displaystyle (H_{11}C_1 + H_{12}C_2)$	$\textstyle =$	$\displaystyle {H_{11}+H_{12} \over 1+S}(C_1+SC_2)$
$\displaystyle (H_{12}C_1 + H_{11}C_2)$	$\textstyle =$	$\displaystyle {H_{11}+H_{12} \over 1+S}(SC_1+C_2)$

which are satisfied if $C_1=C_2\equiv C_-$. Thus, the two states corresponding to $E_{\pm}$ are

$\displaystyle \vert\psi_-\rangle$	$\textstyle =$	$\displaystyle C_-\left(\vert\psi_1\rangle + \vert\psi_2\rangle \right)$
$\displaystyle \vert\psi_+\rangle$	$\textstyle =$	$\displaystyle C_+\left(\vert\psi_1\rangle - \vert\psi_2\rangle \right)$

The overall constants $C_{\pm}$ are determined by requiring that $\vert\psi_{\pm}\rangle $ both be normalized. For $C_-$, for example, we find

$\displaystyle \langle \psi_-\vert\psi_-\rangle$	$\textstyle =$	$\displaystyle \vert C_-\vert^2 \left(\langle \psi_1\vert + \langle \psi_2\vert\right) \left(\vert\psi_1\rangle + \psi_2\rangle \right)$
	$\textstyle =$	$\displaystyle \vert C_-\vert^2\left(1+1+\langle \psi_1\vert\psi_2\rangle + \langle \psi_2\vert\psi_1\rangle \right)$
	$\textstyle =$	$\displaystyle 2\vert C_-\vert^2(1+S)$
	$\textstyle =$	$\displaystyle 1$

which requires that

\[C_- = {1 \over 2\sqrt{1+S}}\]

Similarly, it can be shown that

\[C_+ = {1 \over 2\sqrt{1-S}}\]

Thus, the two states become

$\displaystyle \vert\psi_-\rangle$	$\textstyle =$	$\displaystyle {1 \over 2\sqrt{1+S}}\left(\vert\psi_1\rangle + \vert\psi_2\rangle \right)$
$\displaystyle \vert\psi_+\rangle$	$\textstyle =$	$\displaystyle {1 \over 2\sqrt{1-S}}\left(\vert\psi_1\rangle - \vert\psi_2\rangle \right)$

Notice that these are orthogonal:

\[\langle \psi_+\vert\psi_-\rangle = 0\]

Projecting onto a coordinate basis, we have

$\displaystyle \psi_-({\bf r})$	$\textstyle =$	$\displaystyle {1 \over 2\sqrt{1+S}}\left(\psi_1({\bf r}) + \psi_2({\bf r})\right)$
$\displaystyle \psi_+({\bf r})$	$\textstyle =$	$\displaystyle {1 \over 2\sqrt{1+S}}\left(\psi_1({\bf r}) - \psi_2({\bf r})\right)$

The state $\psi_-$, which corresponds to the energy $E_-$ admits a chemical bond and is, therefore, called a bonding state. The state $\psi_+$, which corresponds to the energy $E_+$ does not admit a chemical bond and is, therefore, called an anti-bonding state. $\psi_+({\bf r})$ and $\psi_-({\bf r})$ are examples of what are called molecular orbitals. In this case, they are constructed from linear combinations of atomic orbitals.

The functional form of the two molecular orbitals for H$_2^+$ within the current approximation scheme is

$\displaystyle \langle {\bf r}\vert\psi_-\rangle$	$\textstyle =$	$\displaystyle \psi_-({\bf r}) = {1 \over 2\sqrt{1+S}}\left({1 \over \pi a_0^3}\... ...rt/a_0}+e^{-\left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert/a_0}\right]$
$\displaystyle \langle {\bf r}\vert\psi_+\rangle$	$\textstyle =$	$\displaystyle \psi_+({\bf r}) = {1 \over 2\sqrt{1-S}}\left({1 \over \pi a_0^3}\... ...rt/a_0}-e^{-\left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert/a_0}\right]$

The contours of these functions are sketched below (the top plot shows the two individual two atomic orbitals, while the middle and bottom show the linear combinations $\psi_-({\bf r})$ and $\psi_+({\bf r})$, respectively):

$\psi_1({\bf r})$ and $\psi_2({\bf r})$ have the same value at the origin and very similarly values near the origin, it is clear that the electron probability density in this region will intensify for $\psi_-({\bf r})$ which is the sum of $\psi_1({\bf r})$ and $\psi_2({\bf r})$. This clearly corresponds to a chemical bonding situation. In contrast, for $\psi_+({\bf r})$, which is the difference between $\psi_1({\bf r})$ and $\psi_2({\bf r})$, there will be a deficit of electron density in the region between the two protons, which is indicative of a non-bonding situation.

\(\displaystyle H_{11} C_1 + H_{12}C_2\)	\(\textstyle =\)	\(\displaystyle {H_{11}-H_{12} \over 1-S}(C_1+SC_2)\)
\(\displaystyle H_{12} C_1 + H_{11}C_2\)	\(\textstyle =\)	\(\displaystyle {H_{11}-H_{12} \over 1-S}(SC_1+C_2)\)

\(\displaystyle (H_{11}C_1 + H_{12}C_2)\)	\(\textstyle =\)	\(\displaystyle {H_{11}+H_{12} \over 1+S}(C_1+SC_2)\)
\(\displaystyle (H_{12}C_1 + H_{11}C_2)\)	\(\textstyle =\)	\(\displaystyle {H_{11}+H_{12} \over 1+S}(SC_1+C_2)\)

\(\displaystyle \vert\psi_-\rangle\)	\(\textstyle =\)	\(\displaystyle C_-\left(\vert\psi_1\rangle + \vert\psi_2\rangle \right)\)
\(\displaystyle \vert\psi_+\rangle\)	\(\textstyle =\)	\(\displaystyle C_+\left(\vert\psi_1\rangle - \vert\psi_2\rangle \right)\)

\(\displaystyle \langle \psi_-\vert\psi_-\rangle\)	\(\textstyle =\)	$\displaystyle \vert C_-\vert^2 \left(\langle \psi_1\vert + \langle \psi_2\vert\right) \left(\vert\psi_1\rangle + \psi_2\rangle \right)$
	\(\textstyle =\)	\(\displaystyle \vert C_-\vert^2\left(1+1+\langle \psi_1\vert\psi_2\rangle + \langle \psi_2\vert\psi_1\rangle \right)\)
	\(\textstyle =\)	\(\displaystyle 2\vert C_-\vert^2(1+S)\)
	\(\textstyle =\)	\(\displaystyle 1\)

\(\displaystyle \vert\psi_-\rangle\)	\(\textstyle =\)	\(\displaystyle {1 \over 2\sqrt{1+S}}\left(\vert\psi_1\rangle + \vert\psi_2\rangle \right)\)
\(\displaystyle \vert\psi_+\rangle\)	\(\textstyle =\)	\(\displaystyle {1 \over 2\sqrt{1-S}}\left(\vert\psi_1\rangle - \vert\psi_2\rangle \right)\)

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\(\displaystyle \psi_-({\bf r})\)	\(\textstyle =\)	\(\displaystyle {1 \over 2\sqrt{1+S}}\left(\psi_1({\bf r}) + \psi_2({\bf r})\right)\)
\(\displaystyle \psi_+({\bf r})\)	\(\textstyle =\)	\(\displaystyle {1 \over 2\sqrt{1+S}}\left(\psi_1({\bf r}) - \psi_2({\bf r})\right)\)

\(\displaystyle \langle {\bf r}\vert\psi_-\rangle\)	\(\textstyle =\)	$\displaystyle \psi_-({\bf r}) = {1 \over 2\sqrt{1+S}}\left({1 \over \pi a_0^3}\... ...rt/a_0}+e^{-\left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert/a_0}\right]$
\(\displaystyle \langle {\bf r}\vert\psi_+\rangle\)	\(\textstyle =\)	$\displaystyle \psi_+({\bf r}) = {1 \over 2\sqrt{1-S}}\left({1 \over \pi a_0^3}\... ...rt/a_0}-e^{-\left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert/a_0}\right]$