.

$$p_{h_4} = (z+R/2)e^{-\alpha r_4} = \frac{R}{2}(\lambda \mu + 1) e^{-\alpha (\lambda - \mu)R/2}$$

We now have, substituting appropriately into Equation 19,

$$\beta^h = -\frac{\hbar^2}{2 m_e} \int_0^{2\pi} d\phi \int_{-1}^{+... ...}{8}(\lambda^2-\mu^2) \frac{R}{2} (\lambda \mu - 1)e^{-\alpha (\lambda+\mu)R/2}$$

$${4\over {R^2 \left ( \lambda^2 - \mu^2 \right )}} \left \{ \left(... ... }\right) \right \} \frac{R}{2}( \lambda \mu + 1)e^{-\alpha (\lambda-\mu)R/2} +$$

$$\int_0^{2\pi} d\phi \int_{-1}^{+1}d\mu \int_1^\infty d\lambda \fr... ...}{8}(\lambda^2-\mu^2) \frac{R}{2}( \lambda \mu - 1)e^{-\alpha (\lambda+\mu)R/2}$$

$$\left (-\frac{Z_1 e^2 }{\frac{R(\lambda+\mu)}{2}} -\frac{Z_4 e^2 ... ...da-\mu)}{2}} \right ) \frac{R}{2} (\lambda \mu + 1)e^{-\alpha (\lambda-\mu)R/2}$$ (20)

where we have dropped the partial derivative with respect to $\phi$ term since we are dealing with $\phi$ independent $\sigma$ states ($\pi$ states and higher need to include this term!) resulting in a function for $\beta^h(R,\alpha)$. We would then have to assume an R value (based on the C-C bond length), and an $\alpha$ value, presumably obtained through variational calculations. Standard practice instead is to assign a value from thermochemical considerations.

$\beta^h$ is sometimes thought of as being proportional to the overlap, so that twisting the bond involved results in decrease in both the resonance integral and the overlap.

At the termini of the molecule, we would have

$$S_{1-4} = \int ( p_z sin \omega + p_y cos \omega)_1 ( p_z sin \omega + p_y cos \omega)_4 d \tau$$

Let us calculate the overlap between p_h1 and p_h4 and ask if the appropriate Hamiltonian matrix element can be proportional to this overlap. We have (in analogy with the development of H_1-4 (above) that the integral we need to evaluate is

$$S_{1-4} = \int ( p_z sin \omega )_1 ( p_z sin \omega )_4 d \tau$$

for controtatory ring closure.

$$\frac{S_{1-4}(R,\alpha) }{\sin^2 \omega} = \int_0^{2\pi} d\p... ...R/2} \frac{R}{2} (\lambda \mu + 1)e^{-\alpha (\lambda-\mu)R/2}$$

where we have explicitly shown which variables will remain after integration. We rewrite this equation as

$$\frac{S_{1-4}(R,\alpha) }{\sin^2 \omega} = 2\pi \frac{R^5}{3... ...^2) (\lambda \mu - 1) (\lambda \mu + 1) e^{-\alpha \lambda R}$$

which becomes

$$\pi \frac{R^5}{16} \int_{-1}^{+1}d\mu \int_1^\infty d\lambda ... ...2-\mu^2) \left( \lambda^2 K^2-1 \right) e^{-\alpha \lambda R}$$