Symmetric product representations

The matrices (27) provide one representation of SU(2). Other representations can be constructed by taking symmetric product representations with itself. The transformations of the symmetric products u2,uv,v2(= x, x2,x2) according to (27) are... 

The electronic state of corannulene anion with C5v symmetry is 2El. The symmetric product representation of Ex is decomposed as follows ... 

Suppose now that A) and B) belong to an electronic representation I ,. Since H is totally symmetric, Eq. (6) implies that the matrix elements (A II TB) belong to the representation of symmetrized or anti-symmetrized products of the bras (A with the kets 7 A). However, the set TA) is, however, simply a reordering of the set ( A). Hence, the symmetry of the matrix elements in the even- and odd-electron cases is given, respectively, by the symmetrized [Ye x Te] and antisymmetrized Ff x I parts of the direct product of I , with itself. A final consideration is that coordinates belonging to the totally symmetric representation, To, cannot break any symmetry determined degeneracy. The symmetries of the Jahn-Teller active modes are therefore given by... 

In order to apply the direct product representation to the derivation of selection rules, recognize that a matrix element of the form ipi, O lpj) will be equal to zero for symmetry reasons if there is even one symmetry operation that takes the integrand into its negative. The argument follows exactly the course of that of section 10.2. Thus the matrix element will vanish unless the direct product representation is totally symmetric (Ai), or contains A upon reduction. 

Simplification of secular equations. Because the Hamiltonian is totally symmetric - that is, for a molecule of C2v symmetry such as H2O, of symmetry species Ai - the matrix elements Hij = ipi, Ti. ipj) as well as the overlap integrals Sij = (tpi, ipj) will be equal to zero unless the direct product representation r. contains Ai. This is the basis for the assertion that states of different symmetry do not mix. ... 

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. 

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. 

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... 

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. 

Let us illustrate this conclusion by the example of spherical-top molecules. In the degenerate states E, T , T2, and G3/2 there is the E representation in the decomposition of the appropriate symmetric product, and hence in these cases the matrix elements are nonzero for anisotropic components of the tensors of polarizability and quadrupole moment, since E is first met in the decomposition of the spherical representation D2. Moreover, in tetrahedral systems in states of the type T and G3/2 the matrix elements of the dipole moment are nonzero, since [T2] and G3/2 contain the representation T2. 

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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