What is the sign of the ``horizontal'' $\beta$?

We need to establish a value for $\beta^h$. Technically, $\beta^h$ corresponds to evaluating an integral (see C. W. David, J. Chem. Ed., 59, 288 (1982) and C. W. David, J. Chem. Ed., 68, 129 (1991)) such as

$$\beta^h = \int \left ((z-R/2)e^{-\alpha r_1}\right )H \left ( (z+R/2)e^{-\alpha r_4}\right ) d \tau$$ (18)

where r₁ and r₄ are the distances from nucleus 1 and 4 respectively to the electron, and R is the distance between the carbon atoms one and four. In a consistent coordinate system (x,y,z),

$$\beta^h = \int \int \int \left ((z-R/2)e^{-\alpha \sqrt{x^2+y... ...( (z+R/2)e^{-\alpha \sqrt{x^2+y^2+(z+R/2)^2}}\right ) dx dy dz$$

using the coordinate system previously defined (the relevant carbon atoms are located at (0,0,R/2) and (0,0,-R/2)). This is the reason why we chose to place Carbons 1 and 4 on the z-axis. Otherwise, we would have to have written things like

$$\left ( (z+R/2)e^{-\alpha \sqrt{(x-\delta_x)^2+(y-\delta_y)^2+(z+R/2)^2}}\right )$$

where $\delta_x$ and $\delta_y$ were offset coordinates for the x- and y- directions, similar to the R/2 offset term for the z- coordinates. The one-electron Hamiltonian Operator (in the same coordinate system) is

$$- \frac{\hbar^2}{2m_e}\nabla^2 - \frac{Z_1e^2}{\sqrt{x^2+y^2+(z-R/2)^2}} -\frac{Z_4e^2}{\sqrt{x^2+y^2+(K)^2}}$$