Renner-Teller degeneracy

Clearly f will go to zero when E2 = Et, independently of the magnitude of. Note, however, that the gradient will also go to zero if Et is different from E2 but the two surfaces are parallel (i.e., Xj, the gradient difference vector, has zero length). In this case the method would fail. This situation will occur for a Renner-Teller-like degeneracy, for example. Of course, in this case, the geometry can be found by normal unconstrained geometry optimization. 

Walsh s rules predict that the excited states may be bent and, in particular, Renner-Teller interaction will lift the degeneracy of the state. The... 

Simple examples of these effects were discussed in Section IV as limiting cases of the pseudo-Jahn Teller and pseudo-Renner-Teller effects. There it was shown that instead of using the adiabatic approximation, we should, in the interest of convenience, adopt a diabatic approximation in which the electronic degeneracy is maintained in zeroth order. The splitting can then simply be ascribed to adiabatic coupling proportional to Q, (or Qf, etc.) or to Qf (or Qf, etc.) between the two electronic components that remain degenerate at Q, = 0. The selection rule for pseudo-Jahn Teller coupling is thus <0° 0 or, iff denotes representation, T , x F,- x F 3-4,. 

It follows from Table 2 that nondegenerate modes can be Jahn-Teller but not Renner-Teller active. Their Jahn-Teller activity lifts the electronic but not the vibronic degeneracy since these modes do not add degeneracies of their own (see also Section IV,B)- Degenerate modes can be Jahn-Teller active, Renner-Teller active, or both. This activity generally leads to splitting of vibronic energy levels. 

It appears that Renner-Teller coupling due to c-type modes is even weaker, since no overtones of these modes have as yet been reported. As shown in Section VII,D, this coupling lifts the degeneracy and would therefore be expected to give rise to first overtones under resonance with doubly... 

---