Mathematical Methods for Physics and Engineering - Matematica.NET

29.11 PHYSICAL APPLICATIONS OF GROUP THEORYCase (i). The manganese atomic orbital φ 1 =(3z 2 − r 2 )f(r), lying at the centre of themolecule, is not affected by any of the symmetry operations since z and r are unchangedby them. It clearly transforms according to the identity irrep A 1 . We therefore need toknow which combination of the iodine orbitals Ψ x (N) andΨ y (N), if any, also transformsaccording to A 1 .We use the projection operator (29.24). If we choose Ψ x (1) as the arbitrary onedimensionalstarting vector, we unfortunately obtain zero (as the reader may wish toverify), but Ψ y (1) is found to generate a new non-zero one-dimensional vector transformingaccording to A 1 . The results of acting on Ψ y (1) with the various symmetry elements Xcan be written down by inspection (see the discussion in section 29.2). So, for example, theΨ y (1) orbital centred on iodine atom 1 and aligned along the positive y-axisischangedby the anticlockwise rotation of π/2 produced by R ′ into an orbital centred on atom 4and aligned along the negative x-axis; thus R ′ Ψ y (1) = −Ψ x (4). The complete set of groupactions on Ψ y (1) is:I, Ψ y (1); Q, −Ψ y (3); R, Ψ x (2); R ′ , −Ψ x (4);m x , Ψ y (1); m y , −Ψ y (3); m d , Ψ x (2); m d ′, −Ψ x (4).Now χ (A1) (X) =1forallX, so (29.24) states that the sum of the above results for XΨ y (1),all with weight 1, gives a vector (here, since the irrep is one-dimensional, just a wavefunction)that transforms according to A 1 and is therefore capable of forming a chemicalbond with the manganese wavefunction φ 1 .ItisΨ (A1) =2[Ψ y (1) − Ψ y (3) + Ψ x (2) − Ψ x (4)],though, of course, the factor 2 is irrelevant. This is precisely the ring orbital Ψ 1 given inthe problem, but here it is generated rather than guessed beforehand.Case (ii). The atomic orbital φ 2 =(x 2 − y 2 )f(r) behaves as follows under the action oftypical conjugacy class members:I, φ 2 ; Q, φ 2 ; R, (y 2 − x 2 )f(r) =−φ 2 ; m x ,φ 2 ; m d , −φ 2 .From this we see that φ 2 transforms as a one-dimensional irrep, but, from table 29.4, thatirrep is B 1 not A 1 (the irrep according to which Ψ 1 transforms, as already shown). Thusφ 2 and Ψ 1 cannot form a bond. ◭The original question did not ask for the the ring orbital to which φ 2 maybond, but it can be generated easily by using the values of XΨ y (1) calculated incase (i) and now weighting them according to the characters of B 1 :Ψ (B1) =Ψ y (1) − Ψ y (3) + (−1)Ψ x (2) − (−1)Ψ x (4)+Ψ y (1) − Ψ y (3) + (−1)Ψ x (2) − (−1)Ψ x (4)=2[Ψ y (1) − Ψ x (2) − Ψ y (3) + Ψ x (4)].Now we will find the other irreps of 4mm present in the space spanned bythe basis functions Ψ x (N) andΨ y (N); at the same time this will illustrate theimportant point that since we are working with characters we are only interestedin the diagonal elements of the representative matrices. This means (section 29.2)that if we work in the natural representation D nat we need consider only thosefunctions that transform, wholly or partially, into themselves. Since we have noneed to write out the matrices explicitly, their size (8 × 8) is no drawback. All theirreps spanned by the basis functions Ψ x (N) andΨ y (N) can be determined byconsidering the actions of the group elements upon them, as follows.1107