3.27: A Symbolic Huckel MO Calculation Using Mathcad

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Enter and solve the butadiene Huckel determinant for energy eigenvalues.

\[ \begin{array}{c|c} \left| \begin{pmatrix} \alpha - E & \beta & 0 & 0 \\ \beta & \alpha - E & \beta & 0 \\ 0 & \beta & \alpha - E & \beta \\ 0 & 0 & \beta & \alpha - E \end{pmatrix} \right| = 0 & _{ \text{float, 4}}^{ \text{solve, E}} \rightarrow \begin{pmatrix} \alpha + .6180 \beta \\ \alpha - 1.618 \beta \\ \alpha + 1.618 \beta \\ \alpha - .6180 \beta \end{pmatrix} \end{array} \nonumber \]

Calculate the eigenvectors:

\[ \begin{array}{c|c} \text{eigenvecs} \left( \begin{pmatrix} \alpha - E & \beta & 0 & 0 \\ \beta & \alpha - E & \beta & 0 \\ 0 & \beta & \alpha - E & \beta \\ 0 & 0 & \beta & \alpha - E \end{pmatrix} \right) = 0 & _{ \text{float, 4}}^{ \text{simplify}} \rightarrow \begin{pmatrix} .317 & .6013 & -.6013 & -.3717 \\ .6014 & -.3716 & .3716 & .6014 \\ .6014 & .3716 & .3716 & -.6014 \\ .3717 & .6013 & .6013 & .3717 \end{pmatrix} \end{array} \nonumber \]

Construct an energy level diagram and show the occupied levels.

\[ \begin{matrix} \text{Energy} & \text{Occupancy} & \text{Wave function coefficients} \\ \alpha - 0.618 \beta & - & \begin{pmatrix} -.3717 & .6014 & -.6014 & .3717 \end{pmatrix} \\ \alpha - 0.618 \beta & - & \begin{pmatrix} -.6014 & .3717 & .3717 & -.6014\end{pmatrix} \\ \alpha + 0.618 \beta & _{-}xo _{-} & \begin{pmatrix} -.6013 & -.3716 & .3716 & .6013 \end{pmatrix} \\ \alpha + 1.618 \beta & _{-}xo_{-} & \begin{pmatrix} .3717 & .6013 & .6013 & .3717 \end{pmatrix} \end{matrix} \nonumber \]

Calculate the π-electron energy:

\[ E_ \pi = \left[ 2 \left( \alpha + 1.618 \beta \right) + 2 \left( \alpha + 0.618 \beta \right) \right] \rightarrow E_ \pi = 4 \alpha + 4.472 \beta \nonumber \]

Calculate the delocalization energy:

\[ E_d = \left[ 4 \alpha + 4.472 \beta - 2 \left( 2 \alpha + 2 \beta \right) \right] \rightarrow E_d = .472 \beta \nonumber \]

Calculate the wavelength of the photon required for the HOMO-LUMO transition.

\[ \begin{array}{c|c} \frac{hc}{ \lambda} = \left( \alpha - 0.618 \beta \right) - \left( \alpha + 0.618 \beta \right) & _{ \text{float, 3}} ^{ \text{solve, } \lambda} \rightarrow -.809 h \frac{c}{ \beta} \end{array} \nonumber \]