Representation, degenerate totally symmetric

The symmetry argument actually goes beyond the above deterniination of the symmetries of Jahn-Teller active modes, the coefficients of the matrix element expansions in different coordinates are also symmetry determined. Consider, for simplicity, an electronic state of symmetiy in an even-electron molecule with a single threefold axis of symmetry, and choose a representation in which two complex electronic components, e ) = 1/v ( ca) i cb)), and two degenerate complex nuclear coordinate combinations Q = re " each have character T under the C3 operation, where x — The bras e have character x. Since the Hamiltonian operator is totally symmetric, the diagonal matrix elements e H e ) are totally symmetric, while the characters of the off-diagonal elements ezf H e ) are x. Since x = 1, it follows that an expansion of the complex Hamiltonian matrix to quadratic terms in Q. takes the form... 

An example of the application of Eq. (47) is provided by the group < 3v whose symmetry operations are defined by Eqs. (18). If the same arbitrary function,

symmetry operation can be worked out, as shown in the last column of Table 13. With the use of the projection operator defined by Eq. (47) and the character table (Table 6), it is found (problem 16) that the coordinate z is totally symmetric (representation Ai). However, it is the sum xy + zx that is preserved in the doubly degenerate representation, E. It should not be surprising that the functions xy and zx are projected as the sum, because it was the sum of the diagonal elements (the trace) of the irreducible representation that was employed in each case in the... 

From these symmetry coordinates, Q is associated with the totally symmetric representation Aj, while the pair (Q2,Q3) is associated with the doubly degenerate representation , The geometric meaning of these coordinates is illustrated in Fig. 5. From the coordinates (Q2, Q3) we may define the polar coordinates... 

Figure 5. Displacements associated with the Q coordinates which transform as irreducible representations of the group S3 (which is isomorphic with the symmetry point group C3r, itself a subgroup of D J. Q, represents the totally symmetric breathing mode of the equilateral triangle, and Q2 represents the antisymmetric distortion coordinate, which is degenerate with the Q3 bending mode for Dih geometries.

Since MA is degenerate, its direct product with itself will always contain the totally symmetric irreducible representation and, at least, one other irreducible representation. For the integral to be nonzero, q must belong either to the totally symmetric irreducible representation or to one of the other irreducible representations contained in the direct product of vJ/ 0 with itself. A vibration belonging to the totally symmetric representation, however, does not decrease the symmetry... 

The symmetry of the normal mode of vibration that can take the molecule out of the degenerate electronic state will have to be such as to satisfy Eq. (6-7). The direct product of E with itself (see Table 6-11) reduces to A + A 2 + E. The molecule has three normal modes of vibration [(3 x 3) - 6 = 3], and their symmetry species are A + E. A totally symmetric normal mode, A, does not reduce the molecular symmetry (this is the symmetric stretching mode), and thus the only possibility is a vibration of E symmetry. This matches one of the irreducible representations of the direct product E E therefore, this normal mode of vibration is capable of reducing th eZ)3/, symmetry of the H3 molecule. These types of vibrations are called Jahn-Teller active vibrations. 

In this sequence, the totally symmetric stretching vibrations are compared, since they are the only ones in this irreducible representation in which the symmetry coordinate is identical with the normal coordinate. However, the same trend is often observed for the degenerate stretching modes v p2), although in this case interaction with the deformation mode of the same symmetry va F2) can in principle lead to frequency changes ... 

Since V(r, Q) is totally symmetric, the derivative 9V/9Qy must transform in the same way as Qy. Therefore it is easily seen by elementary group theory that the relationship (9) is valid if the irreducible representation T is contained in the symmetric square [T jsyn, where F is the representation corresponding to the electronic level (j) . For non-degenerate levels [F ] = Ai holds and no lowering of symmetry occurs. The situation is different for degenerate levels, however, because the system will slip in equivalent minima with a lower symmetry, which is determined by the JT active vibrations. ... 

For the diagonal element, the vibronic coupling constant, wF is nonzero if and only if the symmetric product of F, [F ] contains F. For a non-degenerate state, F x r = Ai, where, 4i is a totally symmetric representation. Therefore, the distortions are totally symmetric in a non-degenerate electronic state. As for a degenerate state, the symmetric product contains some non-totally symmetric representations that cause Jahn-Teller distortions. 

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