Mathematical Methods for Physics and Engineering : A Comprehensive Guide

29.11 PHYSICAL APPLICATIONS OF GROUP THEORY

Case(i). The manganese atomic orbitalφ 1 =(3z^2 −r^2 )f(r), lying at the centre of the
molecule, is not affected by any of the symmetry operations sincezandrare unchanged
by them. It clearly transforms according to the identity irrep A 1. We therefore need to
know which combination of the iodine orbitals Ψx(N)andΨy(N), if any, also transforms
according to A 1.
We use the projection operator (29.24). If we choose Ψx(1) as the arbitrary one-
dimensional starting vector, we unfortunately obtain zero (as the reader may wish to
verify), but Ψy(1) is found to generate a new non-zero one-dimensional vector transforming
according to A 1. The results of acting on Ψy(1) with the various symmetry elementsX
can be written down by inspection (see the discussion in section 29.2). So, for example, the
Ψy(1) orbital centred on iodine atom 1 and aligned along the positivey-axisischanged
by the anticlockwise rotation ofπ/2 produced byR′into an orbital centred on atom 4
and aligned along the negativex-axis; thusR′Ψy(1) =−Ψx(4). The complete set of group
actions on Ψy(1) is:

I,Ψy(1); Q,−Ψy(3); R,Ψx(2); R′,−Ψx(4);

mx,Ψy(1); my,−Ψy(3); md,Ψx(2); md′,−Ψx(4).

Nowχ(A^1 )(X)=1forallX, so (29.24) states that the sum of the above results forXΨy(1),
all with weight 1, gives a vector (here, since the irrep is one-dimensional, just a wave-
function) that transforms according to A 1 and is therefore capable of forming a chemical
bond with the manganese wavefunctionφ 1 .Itis

Ψ(A^1 )=2[Ψy(1)−Ψy(3) + Ψx(2)−Ψx(4)],

though, of course, the factor 2 is irrelevant. This is precisely the ring orbital Ψ 1 given in
the problem, but here it is generated rather than guessed beforehand.
Case(ii). The atomic orbitalφ 2 =(x^2 −y^2 )f(r) behaves as follows under the action of
typical conjugacy class members:

I, φ 2 ; Q, φ 2 ; R,(y^2 −x^2 )f(r)=−φ 2 ; mx,φ 2 ; md,−φ 2.

From this we see thatφ 2 transforms as a one-dimensional irrep, but, from table 29.4, that
irrep is B 1 not A 1 (the irrep according to which Ψ 1 transforms, as already shown). Thus
φ 2 and Ψ 1 cannot form a bond.

The original question did not ask for the the ring orbital to whichφ 2 may

bond, but it can be generated easily by using the values ofXΨy(1) calculated in

case (i) and now weighting them according to the characters ofB 1 :

Ψ(B^1 )=Ψy(1)−Ψy(3) + (−1)Ψx(2)−(−1)Ψx(4)

+Ψy(1)−Ψy(3) + (−1)Ψx(2)−(−1)Ψx(4)

=2[Ψy(1)−Ψx(2)−Ψy(3) + Ψx(4)].

Now we will find the other irreps of 4mmpresent in the space spanned by

the basis functions Ψx(N)andΨy(N); at the same time this will illustrate the

important point that since we are working with characters we are only interested

in the diagonal elements of the representative matrices. This means (section 29.2)

that if we work in the natural representationDnatwe need consider only those

functions that transform, wholly or partially, into themselves. Since we have no

need to write out the matrices explicitly, their size (8×8) is no drawback. All the

irreps spanned by the basis functions Ψx(N)andΨy(N) can be determined by

considering the actions of the group elements upon them, as follows.