Hole formalism

Hole formalism in quantum chemistry states that for many electronic properties, one may consider systems with e or (n-e), the number of unoccupied sites or “holes”, to be equivalent.^[1] The number of microstates (N) of a system corresponds to the total number of distinct arrangements for a number “e” of electrons to be placed in a number “n” of possible orbital positions:

${\displaystyle N={\binom {n}{e}}={\frac {n!}{e!(n-e)!}}}$

In hole formalism, the commutative property of multiplication applies, meaning that in the above equation, ${\displaystyle e!\times (n-e)!=(n-e)!\times e!}$

Example[]

For a set of p orbitals, n = 6 since there are two vacant positions in each of the three orbitals (p_x, p_y, p_z). Therefore, for P2 (e = 2 and n = 6):

${\displaystyle N={\binom {6}{2}}={\frac {6!}{2!(6-2)!}}={\frac {6\times 5\times 4\times 3\times 2\times 1}{(2\times 1)\times (4\times 3\times 2\times 1)}}=15}$

Based on the hole formalism, microstates of P4 (e = 4 and n = 6) are:

${\displaystyle N={\binom {6}{4}}={\frac {6!}{4!(6-4)!}}=15}$

Here it is seen that P4 (2 holes) and P2 (4 holes) give equivalent values of N. The same is true for all p, d, f, … systems such as d1/d9 or f2/f12.

This is characterised by the symmetry of k-combinations in general, that is:

${\displaystyle {\binom {n}{e}}={\binom {n}{n-e}}}$ For all positive integers n, e where n ≥ e.

See also[]

• Term symbol
• Electron hole

References[]