Totally symmetric representations

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... 

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... 

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. 

We will find an excitation which goes from a totally symmetric representation into a different one as a shortcut for determining the symmetry of each excited state. For benzene s point group, this totally symmetric representation is Ajg. We ll use the wavefunction coefficients section of the excited state output, along with the listing of the molecular orbitals from the population analysis ... 

If the system contains symmetry, there are additional Cl matrix elements which become zero. The symmetry of a determinant is given as the direct product of the symmetries of the MOs. The Hamilton operator always belongs to the totally symmetric representation, thus if two determinants belong to different irreducible representations, the Cl matrix element is zero. This is again fairly obvious if the interest is in a state of a specific symmetry, only those determinants which have the correct symmetry can contribute. 

As described above, the ground state vibrational wavefunction is totally symmetric for most common molecules. Therefore, the product, -(1)0 must at least contain a totally symmetric component. The direct product of two irreducible representations contains the totally symmetric representation only if the two irreducible representations are identical. Therefore, transitions can occur from a symmetrical initial state only to those states that have the same symmetry properties as the transition operator, 0. 

Figure 6. Representation of the three totally symmetric alg vibrations of (EuBr Mgi) that is responsible of the electron phonon coupling in the 4f6Sd1—4fi transitions of Eu2+ in CsMgBr3. Color code Eu2+ in violet, Br in red, and Mg2 in yellow.

Although the resulting direct product may not be reduced, it can be made so by application of the magic formula, or often by inspection. The nonvanishing of the integral is then determined by the existence of the totally symmetric representation in the resulting direct sum. This procedure will be illustrated by the development of spectroscopic selection rules in Section 12.3.3. 

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... 

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. 

E (for the identity) in Table 6 are accounted for. Furthermore, the totally symmetric representation is r(1) e A the latter notation is dial usually used by speetroscopists The construction of the remainder of the character table is accomplished by application of the orthogonality property of the characters [see Eq. (30) and problem 131. Standard character tables have been derived in this way for the more common groups, as given in Appendix VQI. 

The internal coordinates for the water molecule are chosen as changes in the structural parameters defined in Fig. 3. The effect of each symmetry operation of the symmetry group ( 2 on these internal coordinates is specified in Table 2. Clearly, the internal coordinate Ace is totally symmetric, as the characters xy(Aa) correspond to those given for the irreducible representation (IR) Ai. On die other hand, the characters x/(Ar), as shown, can not be identified with a specific IR. By inspection of Table 2, however, it is apparent that the direct sum Ai B2 corresponds to the correct symmetry of these coordinates. In more complicated cases the magic formula can always be employed to achieve the correct reduction of the representation in question. 

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... 

The A and B labels in Table 1 follow the convention that A s have characters of +1 for the rotation axis of highest order (C2 in the present case) while B s have character -1. A, by convention, is the totally symmetric i.r., since all operations of the group turn something of A symmetry into itself. Every group has a totally symmetric i.r. I.r. s with suffix 1 are symmetric (character +1) under av, whereas those with suffix 2 are antisymmetric (character -1). Table 1 is an example of a character table. Two-dimensional representations are denoted by symbols E. 

For the case of chiral ligands (with which this section is exclusively concerned), it is easy to convince oneself that the representation (n)u of <5n, whose diagram is shown in Fig. 16, is chiral for every skeleton. For, the representation is totally symmetric under all pure permutations, in particular under those belonging to , and antisymmetric under all group elements involving to, in particular under those in the coset of D in . The representations of [Pg.63]

If there is a molecular symmetry group whose elements leave the hamiltonian 36 invariant, then the closed-shell wavefunction belongs to the totally symmetric representation of both the spin and symmetry groups.8 It is further true that under these symmetry operations the molecular orbitals transform among each other by means of an orthogonal transformation, such as mentioned in Eq. (5) 9) and, therefore, span a representation of the molecular symmetry group. In general, this representation is reducible. 

Since the Hartree-Fock wavefunction 0 belongs to the totally symmetric representation of the symmetry group of the molecule, it is readily seen that the density matrix of Eq. (10) is invariant under all symmetry operations of that group, and the same holds, therefore, for the Hartree-Fock operator 7. 

The representations of U(2) are, in general, labeled by two quantum numbers. However, since we are considering only boson realizations, these representations must be totally symmetric and can thus be characterized by only one quantum number, that is, the first entry in the Young tableau... 

Here N is the vibron number (related to the number of bound states, as shown in the sections to follow). The U(3) representations are characterized in general by three quantum numbers. However, in the reduction of totally symmetric states of U(4), only totally symmetric states of U(3) appear, characterized by a single quantum number, nK, which can take the values... 

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). 

Thus this corresponds to a one-dimensional representation. The totally symmetric representation is created in a similar fashion, but with all of the boxes arranged vertically ... 

For the homonuclear (HON) species, the permutation-symmetry operator had the following form Y = 83) <8) Ye S2), where 83) is a Young operator for the third-order symmetric group which permutes the nuclear coordinates and 82) is a Young operator for the second-order symmetric group which permutes the electronic coordinates. For the fermionic nuclei (H and T, spin = 1/2) the Young operators corresponded to doublet-type representations, while for the bosonic D nuclei we use operators that correspond to the totally symmetric representation. In all cases the electronic operator corresponded to a singlet representation. 

For C2v, the representation Ai is symmetric with respect to every operation. Such a one-dimensional representation, in which every element is assigned the number 1, occurs for every group, and is called the totally symmetric representation. Two representations are symmetric with respect to C2 each is denoted A. Two are symmetric with respect to symmetry species a unique symbol, and " need not be used in this case. 

The four numbers on the right-hand sides of these equations, (1,1,1,1), are the character vector for the totally-symmetric representation Ai. Accordingly, we say that Az transforms as Ai in C2v... 

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