Lithium Atom

The spin factor affects primarily the degeneracy of the energy levels associated with the hydrogen and helium atoms. To a good approximation, the spin factors do not affect the energy levels of such atoms.

The lithium atom, however, has three electrons. An antisymmetric spin wave function of three electrons could in principle be written as the Slater determinant,

$$\chi = \frac{1}{\sqrt{6}} \left \vert \begin{matrix}\alpha(1) & \... ...) & \alpha(2) \\ \alpha(3) & \beta(3) & \alpha(3) \\ \end{matrix} \right \vert.$$ (47)

Such Slater determinant, however, is equal to zero because two of the columns are equal to each other. This fact rules out the possibility of having a zero order wave function that is the Fock product of three hydrogenlike functions:

$$\psi^{(0)}= 1S(1) \ 1S(2) \ 1S(3)$$ (48)

Only if the construction of an antisymmetric spin wave function was possible, we could proceed in analogy to the Helium atom and compute the perturbation due to repulsive coupling terms as follows,

$$E^{(1)} = <\psi\vert\frac{e^2}{r_{12}}\vert\psi> +<\psi\vert\frac{e^2}{r_{23}}\vert\psi> +<\psi\vert\frac{e^2}{r_{13}}\vert\psi>$$

where $\psi$ is the product of hydrogenlike functions of Eq. (48).

Having ruled out such possibility, we construct the zeroth order ground-state wave function for lithium in terms of a determinant similar to Eq. (47), but where each element is a spin-orbital (i.e., a product of a one electron spatial orbital and one-electron spin function),

$$\psi^{(0)} = \frac{1}{\sqrt{6}} \left \vert \begin{matrix}1S(1)\a... ...\ 1S(3)\alpha(3) & 1S(3)\beta(3) & 2S(3)\alpha(3) \\ \end{matrix} \right \vert,$$ (49)

where the third column includes the spatial orbital $2S$, instead of the orbital $1S$, because the Pauli exclusion principle rules out the possibility of having two electrons in the same spin-orbital. It is important to note that Eq. (49) is not simply a product of spatial and spin parts as for the $H$ and $He$ atoms. In contrast, the wave function of $Li$ involves a linear combination of terms which are products of non-factorizable spatial and spin wavefunctions.

Exercise 44: Show that for the lithium atom, treating the electron-electron repulsion interaction $\hat{H}_{rep}$ as a perturbation,

$$E^{(0)}=E_{1S}^{(0)} +E_{1S}^{(0)} +E_{2S}^{(0)},$$

and,

$$E^{(1)} = 2 <1S(1)2S(2)\vert\frac{e^2}{r_{12}}\vert 1S(1) 2S(2)> +<1S(1)1S(2)\vert\frac{e^2}{r_{12}}\vert 1S(1) 1S(2)>$$

$$-<1S(1)2S(2)\vert\frac{e^2}{r_{12}}\vert 2S(1) 1S(2)>.$$