Totally symmetric irreducible representation

The right-hand side of Eq. (96) is of course the weighted direct sum of the irreducible representations. By convention the totally symmetric irreducible representation corresponds to t = 1. Thus, if n(1> = 0, the integral in Eq. (95) vanishes. The transitions m -> nandm n are then forbidden by the symmetry selection rules. Thte principle can be illustrated by the following example. 

The Pauli antisymmetry principle tells us that the wave function (including spin degrees of freedom), and thus the basis functions, for a system of identical particles must transform like the totally antisymmetric irreducible representation in the case of fermions, or spin (for odd k) particles, and like the totally symmetric irreducible representation in the case of bosons, or spin k particles (where k may take on only integer values). 

Various schemes exist to try to reduce the number of CSFs in the expansion in a rational way. Symmetry can reduce the scope of the problem enormously. In die TMM problem, many of die CSFs having partially occupied orbitals correspond to an electronic state symmetiy other than that of the totally symmetric irreducible representation, and dius make no contribution to the closed-shell singlet wave function (if symmetry is not used before the fact, die calculation itself will determine the coefdcients of non-contributing CSFs to be zero, but no advantage in efdciency will have been gained). Since this application of group dieoiy involves no approximations, it is one of the best ways to speed up a CAS calculation. 

Since the matrices 1 1, 1 [,. .. form a one-dimensional totally symmetric irreducible representation of any point group, it is customary to put the corresponding characters in the first row and so 1(C<) — 1. 

In the second place, the Hamiltonian operators which occur and commute with all Or belong to the totally symmetric irreducible representation T1 (see Appendix A. 10-3) and integrals over them dT vanish unless T = T (see eqn (8-4.5)). Thus, in carrying out an approximate solution of the electronic Schrodinger equation, changing to a set of basis functions which belong to the irreducible representations will allow us, by inspection, to put many of the integrals which occur equal to zero. There will also, because of this, be an... 

If we consider in the direct product representation rH P then since Hitf belong to P, rH P = P and therefore TH — P. Hence, any operator which commutes with all 0M of a point group can be said to belong to the totally symmetric irreducible representation P. 

The three classical Kekule structures (already alluded to in section III.E) of naphthalene are shown in Scheme 36a. Two of them are designated as Ki and K2 and represent the annulenic resonance along the perimeter of the naphthalene, while the third one, Kc, has a double bond in the center and transforms as the totally symmetric irreducible representation, Ag of the Dzh group. The Ki and K2 structures are mutually interchangeable by the i, C2, and ov symmetry operations of the point group, much as in the case of benzene. An in-phase combination transforms, therefore, as Ag, whereas an out-of-phase one transforms as B2u. These symmetry adapted wave func-... 

As usual here we only assume that U belongs to the totally symmetric irreducible representation. 

The Slater-Roothaan method uses solid-harmonic Gaussians [18] to fit the molecular orbitals and five other nonnegative quantities, namely the total density, the cube root of the partitioned density for both spins, and the 2/3 power of the partitioned density for both spins. They are treated as five additional orbitals of the totally symmetric irreducible representation. Changing them slightly does not affect the robust energy at all. 

Another useful relationship As the totally symmetric irreducible representation Fys in every group is always associated with one or more binary products ofx, y, and z and it follows that totally symmetric vibrational modes are always Raman active. 

It is always possible to find a behavior that remains unchanged under any of the symmetry operations of the given point group. Thus, there is always an irreducible representation which has only +1 characters. This is the totally symmetric irreducible representation, and it is always the first one in any character table. 

Looking at the C2v character table, we can say that v, and v2 belong to the totally symmetric irreducible representation A, and v3 belongs to52. 

The vibrational wave function of the ground state belongs to the totally symmetric irreducible representation of the point group of the molecule. The wave function of the first excited state will belong to the irreducible representation to which the normal mode undergoing the particular transition belongs. 

First, however, consider the symmetry properties of the central atom orbitals. Take the C v point group as an example. Its character table is presented in Table 6-1. The pz and dz2 atomic orbitals of the central atom belong to the totally symmetric irreducible representation Ah the d- y2 orbital belongs to B and dAJ to B2. The symmetry properties of the (px,Py) and (dxz, dyz) orbitals present a good opportunity... 

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... 

This expression contains important information regarding the symmetry of the excited states as well. Only those excited states can participate in the reaction whose symmetry matches the symmetry of both the ground state and the reaction coordinate. We already know that Qr belongs to the totally symmetric irreducible representation except at maxima and minima. This implies that only those excited states can participate in the reaction whose symmetry is the same as that of the ground state. This information is instrumental in the construction of correlation diagrams, as will be seen later. 

The expression for the intermolecular potential must satisfy two symmetry requirements. First, it must be invariant if we rotate the molecular frame of either of the two molecules through specific Euler angles wg that correspond with a symmetry element of the molecule in question. This means that our basis must be invariant under rotations of the outer direct product group Gp GP-, where GP is the symmetry group of molecule P and GP- that of molecule F Acting with the projection operator of the totally symmetric irreducible representation of the group GP (of order GP),... 

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