Direct product degenerate

Whenever a fiinction can be written as a product of two or more fiinctions, each of which belongs to one of the synnnetry classes, the symmetry of the product fiinction is the direct product of the syimnetries of its constituents. This direct product is obtained in non-degenerate cases by taking the product of the characters for each symmetry operation. For example, the fiinction xy will have a symmetry given by the direct product of the syimnetries of v and ofy this direct product is obtained by taking the product of the characters for each synnnetry operation. In this example it may be seen that, for each operation, the product of the characters for Bj and B2 irreducible representations gives the character of the representation, so xy transfonns as A2. 

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. 

Except for the multiplication of by we follow the rules for forming direct products used in non-degenerate point groups the characters under the various symmetry operations are obtained by multiplying the characters of the species being multiplied, giving... 

It was explained in Section 4.3.2 that the direct product of two identical degenerate symmetry species contains a symmetric part and an antisymmetric part. The antisymmetric part is an A (or 2") species and, where possible, not the totally symmetric species. Therefore, in the product in Equation (7.79), Ig is the antisymmetric and Ig + Ag the symmetric part. 

When spin-orbit coupling is introduced the symmetry states in the double group CJ are found from the direct products of the orbital and spin components. Linear combinations of the C"V eigenfunctions are then taken which transform correctly in C when spin is explicitly included, and the space-spin combinations are formed according to Ballhausen (39) so as to be diagonal under the rotation operation Cf. For an odd-electron system the Kramers doublets transform as e ( /2)a, n =1, 3, 5,... whilst for even electron systems the degenerate levels transform as e na, n = 1, 2, 3,. For d1 systems the first term in H naturally vanishes and the orbital functions are at once invested with spin to construct the C functions. 

Distribution of the two additional electrons to 8 required for dianion formation among the degenerate LUMO orbitals of E, symmetry gives rise to four new states, since, within the 5v symmetry group, the direct product Ej Ej may be reduced to a sum of Aj, and 2 irreducible representations. The A2 state represents a triplet, while Aj and 2 are singlet states. 

Problem 10-5. In a homonuclear diatomic molecule, taking the molecular axis as z, the pair of LCAO-MO s tpi = 2p A + PxB and tp2 = 2 PyA + 2 PyB forms a basis for a degenerate irreducible representation of D h, as does the pair 3 = 2pxA PxB and 4 = PxA — PxB Identify the symmetry species of these wave functions. Write down the four-by-four matrices for the direct product representation by examining the effect of the group elements on the products 0i 03, 0i 04, V 2 03) and 02 04- Verify that the characters of the direct product representation are the products of the characters of the individual representations. 

The techniques used earlier for linear molecules extend easily to non-linear molecules. One begins with those states that can be straightforwardly identified as unique entries within the box diagram. For polyatomic molecules with no degenerate representations, the spatial symmetry of each box entry is identical and is given as the direct product of the open-shell orbitals. For the formaldehyde example considered earlier, the spatial symmetries of the n7t and %% states were A2 and Aj, respectively. 

The Jahn-Teller theorem was proved by showing that for all symmetry groups except and there was at least one normal mode of vibration which belonged to a non symmetric representation f,- such that the direct product of F/ with the representation Fy of the degenerate electronic state contained the representation Fy. 

The direct product enables one to find the symmetry of a wave function when the symmetries of its factors are known. For example, consider In the harmonic-oscillator approximation, the vibrational wave function is the product of 3N—6 harmonic-oscillator functions, one for each normal mode. To find the symmetry of we first examine the symmetries of its factors. Let the distinct vibrational frequencies of the molecule be vx>v2,..., vk,...,vn, and let vk be <4-fold degenerate let the harmonic-oscillator... 

This can easily be demonstrated, using direct product representations for orbitals alone, when only nondegenerate orbitals are involved. The same is true when degenerate orbitals are involved, but more sophisticated methods of proof are required. 

R. C. Haddon. I think there is something special. We do need more fullerenes with triply degenerate orbitals unfortunately none have been isolated beside C60. It comes about as a result of the direct product of the t-type orbitals. 

In other words, the irreducible representation / of the vibrational mode has to be contained in the direct product of those (i"j and / ) of the electronic states [26], For the intra-state, or JT, couplings, the selection rule (1) involves the symmetrised direct product (J3)2 for the degenerate electronic state and leads to the well-known result ... 

The total symmetry of the molecule is D2h, so we cannot have point group degenerate states. The accidentally degenerate states represented above have C2v symmetry. The direct product C2v Cs recovers properly the full symmetry of the system. Depending on the particular excitation and final spin state four states can be generated. In Table 4 we compare the results of Hartree-Fock (HF), configuration interaction with singles and doubles excitations (CI-SD) and the two structure GMS calculations, with the available experimental data. 

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... 

The symmetry of the normal mode of vibration that can take the molecule out of the degenerate electronic state will have to be such as to satisfy Eq. (6-7). The direct product of E with itself (see Table 6-11) reduces to A + A 2 + E. The molecule has three normal modes of vibration [(3 x 3) - 6 = 3], and their symmetry species are A + E. A totally symmetric normal mode, A, does not reduce the molecular symmetry (this is the symmetric stretching mode), and thus the only possibility is a vibration of E symmetry. This matches one of the irreducible representations of the direct product E E therefore, this normal mode of vibration is capable of reducing th eZ)3/, symmetry of the H3 molecule. These types of vibrations are called Jahn-Teller active vibrations. 

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