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:

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

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

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):

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

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

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

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:

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

References

K Veera Reddy (1998). Symmetry And Spectroscopy Of Molecules. New Age International. ISBN 8122411428.

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[Sym-1] K Veera Reddy (1998). Symmetry And Spectroscopy Of Molecules. New Age International. ISBN 8122411428.

Hole formalism

Example

See also

References