Totally symmetric

In general, each nomial mode in a molecule has its own frequency, which is detemiined in the nonnal mode analysis [24]- Flowever, this is subject to the constraints imposed by molecular synmietry [18, 25, 26]. For example, in the methane molecule CFI, four of the nonnal modes can essentially be designated as nonnal stretch modes, i.e. consisting primarily of collective motions built from the four C-FI bond displacements. The molecule has tetrahedral synmietry, and this constrains the stretch nonnal mode frequencies. One mode is the totally symmetric stretch, with its own characteristic frequency. The other tliree stretch nonnal modes are all constrained by synmietry to have the same frequency, and are refened to as being triply-degenerate. 

Each such nonual mode can be assigned a synuuetry in the point group of the molecule. The wavefrmctions for non-degenerate modes have the following simple synuuetry properties the wavefrmctions with an odd vibrational quantum number v. have the same synuuetry as their nonual mode 2the ones with an even v. are totally symmetric. The synuuetry of the total vibrational wavefrmction (Q) is tlien the direct product of the synuuetries of its constituent nonual coordinate frmctions (p, (2,). In particular, the lowest vibrational state. 

Aspects of the Jahn-Teller symmetry argument will be relevant in later sections. Suppose that the electronic states aie n-fold degenerate, with symmetry at some symmetiical nuclear configuration Qq. The fundamental question concerns the symmetry of the nuclear coordinates that can split the degeneracy linearly in Q — Qo, in other words those that appeal linearly in Taylor series for the matrix elements A H B). Since the bras (/1 and kets B) both transform as and H are totally symmetric, it would appear at first sight that the Jahn-Teller active modes must have symmetry Fg = F x F. There... 

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... 

One reason that the symmetric stretch is favored over the asymmetric one might be the overall process, which is electron transfer. This means that most of the END trajectories show a nonvanishing probability for electron transfer and as a result the dominant forces try to open the bond angle during the collision toward a linear structure of HjO. In this way, the totally symmetric bending mode is dynamically promoted, which couples to the symmetric stretch, but not to the asymmetric one. 

If the symmetries of the two adiabatic functions are different at Rq, then only a nuclear coordinate of appropriate symmeti can couple the PES, according to the point group of the nuclear configuration. Thus if Q are, for example, normal coordinates, xt will only span the space of the totally symmetric nuclear coordinates, while X2 will have nonzero elements only for modes of the correct symmetry. 

For states of different symmetry, to first order the terms AW and W[2 are independent. When they both go to zero, there is a conical intersection. To connect this to Section III.C, take Qq to be at the conical intersection. The gradient difference vector in Eq. f75) is then a linear combination of the symmetric modes, while the non-adiabatic coupling vector inEq. (76) is a linear combination of the appropriate nonsymmetric modes. States of the same symmetry may also foiiti a conical intersection. In this case it is, however, not possible to say a priori which modes are responsible for the coupling. All totally symmetric modes may couple on- or off-diagonal, and the magnitudes of the coupling determine the topology. 

Worth and Cederbaum [100], propose to facilitate the search for finding a conical intersection if the two states have different symmetiies If they cross along a totally symmetric nuclear coordinate, then the crossing point is a conical intersection. Even this simplifying criterion leaves open a large number of possibilities in any real system. Therefore, Worth and Cederbaum base their search on large scale nuclear motions that have been identified experimentally to be important in the evolution of the system after photoexcitation. 

In a symmetric top molecule such as NH3, if the transition dipole lies along the molecule s symmetry axis, only k = 0 contributes. Such vibrations preserve the molecule s symmetry relative to this symmetry axis (e.g. the totally symmetric N-H stretching mode in NH3). The additional selection rule AK = 0... 

I>k V k> will be non-zero only if V contains a totally symmetric component (because the... 

The one-dimensional part of the above reducible three-dimensional representation is seen to be the same as the totally symmetric representation we arrived at... 

The totally symmetrical sphere is characterized by a single size parameter its radius. Ellipsoids of revolution are used to approximate the shape of unsymmetrical bodies. Ellipsoids of revolution are characterized by two size parameters. 

We have seen that any two of the C2, ( Jxz), (r Jyz) elements may be regarded as generating elements. There are four possible combinations of + 1 or — 1 characters with respect to these generating elements, + 1 and + 1, + 1 and -1,-1 and +1,-1 and —1, with respect to C2 and (tJxz). These combinations are entered in columns 3 and 4 of the C2 character table in Table A.l 1 in Appendix A. The character with respect to / must always be + 1 and, just as (r Jyz) is generated from C2 and (tJxz), the character with respect to (r Jyz) is the product of characters with respect to C2 and (tJxz). Each of the four rows of characters is called an irreducible representation of the group and, for convenience, each is represented by a symmetry species Aj, A2, or B2. The A] species is said to be totally symmetric since all the characters are + 1 the other three species are non-totally symmetric. 

The dipole moment vector /i must be totally symmetric, and therefore symmetric to all operations of the point group to which the molecule belongs otherwise the direction of the dipole moment could be reversed by carrying out a symmetry operation, and this clearly cannot happen. The vector /i has components fiy and along the cartesian axes of the molecule. In the examples of NH3 and NF3, shown in Figures 4.18(b) and 4.18(e), respectively, if the C3 axis is the z-axis, 7 0 but = 0. Similarly in H2O and cis-... 

A molecule has a permanent dipole moment if any of the translational symmetry species of the point group to which the molecule belongs is totally symmetric. 

In the Cl, C, C and C point groups the totally symmetric symmetry species is A, A, A and Ai (or 2"+), respectively. For example, CHFClBr (Figure 4.7) belongs to the Ci point group therefore /r 7 0 and, since all three translations are totally symmetric, the dipole... 

The BF3 molecule, shown in Figure 4.18(i), is now seen to have /r = 0 because it belongs to the point group for which none of the translational symmetry species is totally symmetric. Alternatively, we can show that /r = 0 by using the concept of bond moments. If the B-F bond moment is /Tgp and we resolve the three bond moments along, say, the direction of one of the B-F bonds we get... 

The molecule tran5 -l,2-difluoroethylene, in Figure 4.18(h), belongs to the C2 , point group in which none of the translational symmetry species is totally symmetric therefore the molecule has no dipole moment. Arguments using bond moments would reach the same conclusion. 

A molecule has a permanent dipole moment if any of the symmetry species of the translations and/or T( and/or 1/ is totally symmetric. Using the appropriate character table apply this principle to each of these molecules and indicate the direction of any non-zero dipole moment. 

There are simple symmetry requirements for the integral of Equation (6.44) to be non-zero and therefore for the transition to be allowed. If both vibrational states are non-degenerate, the requirement is that the symmetry species of the quantity to be integrated is totally symmetric this can be written as... 

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