Product Rules

To determine the total electronic configuration, the term symbols and to characterize the transition dipole moment, the product rules for symmetry species are beneficial. They are based on the fact that (with the only constraint that the symmetry species is not degenerated) a direct multiplication of characters is possible. This operation yields a row of characters identical to the characters for some other irreducible representation of the group. This is shown for a combination of symmetry species within point group C2v.
 C2v, 2mm E C 2 σv(xz) σ'v(yz) h = 4 A1 1 1 1 1 z, z2, x2, y2 A2 1 1 -1 -1 xy Rz B1 1 -1 1 -1 x, xz Ry B2 1 -1 -1 1 y, yz Rx

To get the product of the symmetry species of A2 and B1 we just multiply the characters of the two rows:
 C2v E C2 σv(xz) σ'v(yz) A2 1 1 -1 -1 B1 1 -1 1 -1 A2 x B1 1 -1 -1 1

The characters recieved for the product A2 x B1 are identical to the entries for symmetry species B2 listed in the original character table of group C2v. Such products have been tabulated in literature, e.g. by Herzberg, Molecular Spectroscopy, Volume III. Below, we present the multiplication table for point group C2v.
 C2v A1 A2 B1 B2 A1 A1 A2 B1 B2 A2 A2 A1 B2 B1 B1 B1 B2 A1 A2 B2 B2 B1 A2 A1

Note that
 A1 x A1  =  A1 B1 x B1  =  A1...., i.e. the square of any not degenerated symmetry species is totally symmetric. B1 x A1  =  B1 B2 x A1  =  B2...., i.e. if a totally symmetric species is one factor, the other factor and the resulting product are identical.

The last table summarizes the rules for not degenerated symmetry species. Using Mulliken's notation with A,' , g, +, B, ", u and −,  there is an obvious similarity with the sign rules known from algebraic multiplication.
 i: g x g  =  g u x u  =  g g x u  =  u x g  =  u σh: (') x (')  =  (') ('') x ('')  =  (') (') x ('')  =  ('') Cp: A x A  =  A B x B  =  A A x B  =  B x A  =  B C2: 1 x 1  =  1 2 x 2  =  1 1 x 2  =  2 x 1  =  2

Outlook: In another symmetry group (e.g. D2h), the product rules may remind of a cyclic substitution.
 D2h: 1 x 2  =  3 2 x 3  =  1 3 x 1  =  2

Thus B2g x B3u = B1u

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