The Molecular Orbital Approximation  Contents -  Synopsis

The Quantum Mechanical Origin of Orbitals

In 1924, there was much discussion on the wave nature of matter. Davisson and Germer showed that a beam of electrons could be diffracted, in other words that these electrons have a wave nature.

If we represent a particle by a wave

$$\phi=cos(kx)$$ (1)

 or preferably for this discussion

$$\phi=\exp(ikx)$$ (2)

 then De Broglie postulated that the momentum p of the particle is given by

$$p=\hbar k$$ (3)

 where $\hbar$ is Planck's constant. $k=2\pi/\lambda$, where $\lambda$ is the wavelength of the wave. De Broglie's postulate merely says ` the shorter the wavelength, the greater the momentum'.

But classical mechanics conservation of energy says

$$\frac{1}{2}mv^2+V=E$$ (4)

 where $p=mv=\hbar k$. Substituting for k and multiplying by $\phi$ gives

$$\frac{\hbar^2}{2m}k^2\phi+V\phi=E\phi$$ (5)

 Now the big step, the second derivative of the wave gives

$$\frac{d^2\phi}{dx^2}=-k^2\phi$$ (6)

 and thus we have Scrodinger equation for the wave (orbital)

$$(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V)\phi=E\phi$$ (7)

 Finally, because $\phi^{*}\phi=1$ for the wave, and there is one particle, it is argued that $\phi^{*}\phi$ is the probability density. The Schrodinger equation for the H atom is

$$(-\frac{\hbar^2}{2m}\nabla^2-\frac{e^2}{4\pi\epsilon_0r})\phi=E\phi$$ (8)

 The solutions of this equation were obtained in Part IB.

The electron has four coordinates, x,y,z,v, where $-\infty<x,y,z<+\infty$ and $v=\pm\frac{1}{2}$. Therefore v can only be associated with two spin functions, $\alpha(v)$ and $\beta(v)$. Each spatial orbital $\phi$ can be associated with these, $\phi\alpha,\phi\beta$, and thus at most two electrons can be in each spatial orbital. The ground state of H is doubly degenerate, ²S, with spin orbitals 1s$\alpha$,1s$\beta$. 

1998-09-23