PHY.K02UF Molecular and Solid State Physics

## Molecular orbitals of a conjugated chain

The Roothaan equations for a conjugated chain of $N$ atoms have the form,

$$\left[ \begin{matrix} H_{11} & H_{12} & 0 & \cdots & 0 & 0 \\ H_{12} & H_{11} & H_{12} & 0 & & 0 \\ 0 & H_{12} & H_{11} & H_{12} & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 \\ 0 & & 0 & H_{12} & H_{11} & H_{12} \\ 0 & 0 & \cdots & 0 & H_{12} & H_{11} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ \vdots \\ c_N \end{matrix} \right] = E \left[ \begin{matrix} 1 & S_{12} & 0 & \cdots & 0 & 0 \\ S_{12} & 1 & S_{12} & 0 & & 0 \\ 0 & S_{12} & 1 & S_{12} & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 \\ 0 & & 0 & S_{12} & 1 & S_{12} \\ 0 & 0 & \cdots & 0 & S_{12} & 1 \end{matrix} \right]\left[ \begin{matrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ \vdots \\ c_N \end{matrix} \right].$$

Matrices like this have eigen vectors of the form,

$$\left[ \begin {matrix} \sin\left(\frac{\pi j}{N+1}\right) \\ \sin\left(\frac{2\pi j}{N+1}\right) \\ \sin\left(\frac{3\pi j}{N+1}\right) \\ \vdots \\ \sin\left(\frac{N\pi j}{N+1}\right) \end{matrix} \right]\hspace{1cm}j=1,2,\cdots,N.$$

This can be checked by substituting the eigenvectors into the Roothaan equations,

$$\left[ \begin {matrix} H_{11}\sin\left(\frac{\pi j}{N+1}\right) + H_{12}\sin\left(\frac{2\pi j}{N+1}\right)\\ H_{12}\sin\left(\frac{\pi j}{N+1}\right) +H_{11}\sin\left(\frac{2\pi j}{N+1}\right) +H_{12}\sin\left(\frac{3\pi j}{N+1}\right) \\ H_{12}\sin\left(\frac{2\pi j}{N+1}\right) +H_{11}\sin\left(\frac{3\pi j}{N+1}\right) +H_{12}\sin\left(\frac{4\pi j}{N+1}\right) \\ \vdots \\ H_{12}\sin\left(\frac{(N-1)\pi j}{N+1}\right) + H_{11}\sin\left(\frac{N\pi j}{N+1}\right) \end{matrix} \right]=E\left[ \begin {matrix} \sin\left(\frac{\pi j}{N+1}\right) + S_{12}\sin\left(\frac{2\pi j}{N+1}\right)\\ S_{12}\sin\left(\frac{\pi j}{N+1}\right) +\sin\left(\frac{2\pi j}{N+1}\right) +S_{12}\sin\left(\frac{3\pi j}{N+1}\right) \\ S_{12}\sin\left(\frac{2\pi j}{N+1}\right) +\sin\left(\frac{3\pi j}{N+1}\right) +S_{12}\sin\left(\frac{4\pi j}{N+1}\right) \\ \vdots \\ S_{12}\sin\left(\frac{(N-1)\pi j}{N+1}\right) + \sin\left(\frac{N\pi j}{N+1}\right) \end{matrix} \right].$$

Using the trigonometric relations $\sin a+ \sin b = 2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)$, $\sin a\cos b= \frac{\sin(a+b)}{2} +\frac{\sin(a-b)}{2}$, and $\sin 2a = 2\sin a \cos a$,

$$\left(H_{11}+2H_{12}\cos\left(\frac{\pi j}{N+1}\right)\right)\left[ \begin {matrix} \sin\left(\frac{\pi j}{N+1}\right) \\ \sin\left(\frac{2\pi j}{N+1}\right) \\ \sin\left(\frac{3\pi j}{N+1}\right) \\ \vdots \\ \sin\left(\frac{N\pi j}{N+1}\right) \end{matrix} \right]=E\left(1+2S_{12}\cos\left(\frac{\pi j}{N+1}\right)\right)\left[ \begin {matrix} \sin\left(\frac{\pi j}{N+1}\right) \\ \sin\left(\frac{2\pi j}{N+1}\right) \\ \sin\left(\frac{3\pi j}{N+1}\right) \\ \vdots \\ \sin\left(\frac{N\pi j}{N+1}\right) \end{matrix} \right]\hspace{1cm}j=1,2,\cdots,N.$$

The energies of the molecular orbitals are,

$$$E_{\text{mo},j}=\frac{H_{11}+2H_{12}\cos\left(\frac{\pi j}{N+1}\right)}{1+2S_{12}\cos\left(\frac{\pi j}{N+1}\right)}\hspace{2cm}j=1,2,\cdots,N.$$$

The normalized molecular orbitals are,

$$$\psi_{\text{mo},j}=\sqrt{\frac{2}{N+1}}\sum\limits_{n=1}^N \sin\left(\frac{\pi nj}{N+1}\right)\phi_{pz,n}\hspace{2cm}j=1,2,\cdots,N.$$$

There are valence electrons will occupy the molecular orbitals with the lowest energies. Because $H_{12}<0$, the molecular orbital with the lowest energy is $\psi_{\text{mo},1}$ and the molecular orbital with the highest energy is $\psi_{\text{mo},N}$.

The code below will calculate the energies. The elements of the Hamiltonian matrix and the overlap matrix can be calculated using http://lampz.tugraz.at/~hadley/ss1/molecules/atoms/2pz.php.