2.9. MO-LCAO for two atomic orbitals

Here, we derive expressions for the molecular orbitals (MOs) and their energies resulting from the overlap of two atomic orbitals (AOs), based on the variational method.

We start with a molecular orbital of the form

(2.152)\[\psi = c_1\phi_1 + c_2\phi_2\]

where \(\phi_1\) and \(\phi_2\) are AOs that are not necessarily normalized. \(c_1\) and \(c_2\) are the expansion coefficients we would like to determine.

The expectation value of the energy for \(\psi\) is

(2.153)\[E = \frac{\langle\psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle} = \cdots = \frac{c_1^2H_{11}+2c_1c_2H_{12}+c_2^2H_{22}}{c_1^2S_{11}+2c_1c_2S_{12}+c_2^2S_{22}}\]

where \(H_{ij} = \langle\phi_i|\hat{H}|\phi_j\rangle\) and \(S_{ij} = \langle\phi_i|\phi_j\rangle\).

The variational principle states that at the optimum values of \(c_1\) and \(c_2\) (i.e. that give a \(\psi\) that satisfies the time-independent Schrödinger equation as best as possible), the energy \(E\) should be minimal and closest to the ground-state energy. To calculate this minimum, we set the two partial derivatives of the energy with respect to \(c_1\) and \(c_2\) equal to zero.

(2.154)\[\frac{\partial E}{\partial c_1} = 0 \qquad \text{and} \qquad \frac{\partial E}{\partial c_2} = 0\]

After applying the quotient rule for differentiation, and some additional algebra, this gives the following homogeneous system of linear equations (the so-called secular equations):

(2.155)\[\begin{split}\begin{eqnarray} (H_{11}-ES_{11})c_1+(H_{12}-ES_{12})c_2 &=& 0 \\ (H_{12}-ES_{12})c_1+(H_{22}-ES_{22})c_2 &=& 0 \end{eqnarray}\end{split}\]

In matrix notation, this is

(2.156)\[\begin{split}\begin{pmatrix} H_{11}-ES_{11} & H_{12}-ES_{12} \\ H_{21}-ES_{21} & H_{22}-ES_{22} \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\end{split}\]

Using the trivial solution \(c_1=c_2=0\) would clearly satisfy this system of equations, but that would result in \(\psi=0\), which is nonphysical. Non-trivial solutions are only possible if the secular determinant is zero:

(2.157)\[\begin{split}\begin{vmatrix} H_{11}-E S_{11} & H_{12}-E S_{12} \\ H_{12}-E S_{12} & H_{22}-E S_{22} \end{vmatrix} = 0\end{split}\]

Expanding the determinant gives

(2.158)\[(H_{11}-E S_{11})(H_{22}-E S_{22}) - (H_{12}-E S_{12})^2 = 0\]

Rearranging this to collect all terms of equal power in \(E\) gives the following quadratic equation for the energy:

(2.159)\[E^2(S_{11}S_{22}-S_{12}^2) +E(-H_{11}S_{22}-H_{22}S_{11}+2H_{12}S_{12}) + H_{11}H_{22}-H_{12}^2 = 0\]

The solutions \(E\) to this quadratic equation gives the orbital energies \(E\). Only for these values, non-trivial solutions for \(c_1\) and \(c_2\) are possible such that the MO is non-vanishing. In this case, \(c_2\) can be obtained from \(c_1\) via

(2.160)\[c_2 = -\frac{H_{11}-ES_{11}}{H_{12}-ES_{12}}c_1\]

which we obtained by rearranging the first equation of the system of equations. To fix \(c_1\), we additionally need to invoke the normalization requirement for \(\psi\).

For a homodiatomic situation (\(H_{11}=H_{22}\) and \(S_{11}=S_{22}\)), it is possible to get a simple solution. In this case, the quadratic equation simplifies to

(2.161)\[(H_{11}-E S_{11})^2 = (H_{12}-E S_{12})^2 \quad\rightarrow\quad H_{11}-E S_{11} = \mp(H_{12}-E S_{12})\]

This rearranges to

(2.162)\[H_{11}\pm H_{12} = E S_{11} \pm E S_{12}\]

and finally

(2.163)\[E = \frac{H_{11}\pm H_{12}}{S_{11}\pm S_{12}}\]

(If the AOs are normalized, then \(S_{11}=1\).) For the coefficients, we get \(c_2=\pm c_1\). After normalization, this yields

(2.164)\[\psi = \frac{1}{\sqrt{2}}\frac{\phi_1\pm\phi_2}{\sqrt{1\pm S_{12}}}\]