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The Variational Principle by Hand — Linear variation — the secular equation

Exercises

Quantum Chemistry I › Unit 2 · The variational game › The Variational Principle by Hand › all problems

Practice Problem 6 of 8

Linear variation — the secular equation

Problem

Two non-orthogonal basis functions on

H_{2}^{+}

have

H_{11} = H_{22} = - 0.5

H_{12} = - 0.5

, overlap

S_{11} = S_{22} = 1

S_{12} = 0.5

(hartree). Solve the

2 \times 2

secular problem

det (H - E S) = 0

for the lower (bonding) root.

Solution

Predict before reading on. Earlier problems set a derivative to zero. With more than one basis function, minimizing gives this determinant instead. What replaced $d E / d ζ = 0$ ?

By symmetry the roots are $E_{\pm} = \frac{H _{11} \pm H _{12}}{1 \pm S _{12}}$ . The bonding (lower) root:

E_{+} = \frac{- 0.5 + ( - 0.5 )}{1 + 0.5} = \frac{- 1.0}{1.5} = - 0.667 hartree

Setting the gradient with respect to the linear coefficients to zero turns minimization into the generalized eigenvalue problem $Hc = E Sc$ — exactly Roothaan–Hall, two units ahead. The derivative became a determinant.