Boson irreps

Odd number of electrons —> Fermion irreps of the double group Even number of electrons —> Boson irreps of the double group... 

For two-particle states the wave function must transform according to a boson irrep. The simplest case is that of a two-particle state made up of a spinor and its Kramers partner. In the absence of other degeneracies this corresponds to a closed shell, and we would expect this product to transform as the totally symmetric irrep. 

Otherwise, the product can span two irreps. In this case, we can take a linear combination of the product and its time-reversed partner to obtain a function that transforms as an irrep. For example, in C2v the functions that transform as boson irreps are... 

We have now shown that the symmetry of a spinor overlap density depends on the symmetry of /q and r(R ). These two symmetries are the same in the groups C , C h, and S2n, but not in the other groups. If the product of the ket irrep and the conjugate of the bra irrep lies in one of the degenerate boson irreps, the density could have a well-defined symmetry. In analogy to the two-electron wave functions, we can establish the symmetry by subtracting the mj values for the irreps and performing the modulo division to arrive at an me value for a boson irrep. 

However, if the product spans nondegenerate boson irreps, the symmetry is not well-defined, and it is the real and imaginary parts of the density that have a well-defined symmetry. This is the case even for p = and r =, for the groups in which r(R ) /q. We can of course make the classification of the integrals into the four classes regardless of whether the densities belong to a particular boson irrep, but if we wish to exploit boson symmetry we have to address this issue. 

For N even, the overall symmetry of the wave function must be that of a boson irrep. The application of time reversal maps each set of determinants onto itself. The two sets form bases for states of different boson symmetry, and the set with Mk even contains the basis for the totally symmetric irrep. For the complex binary groups C2 and there are only two boson irreps, and so the Hamiltonian matrix is fully blocked by the use of Kramers pairs. Thus selection of the parity of Mk determines the state symmetry. The last of the complex groups, C j, has inversion symmetry, which gives rise to two more boson irreps, but again inversion may be treated by nonrelativistic methods. 

If the Kramers partners are in different rows of the same, doubly degenerate irrep, as in the group then the matrix elements are real and the gradient is zero. Likewise, the Hamiltonian matrix element between the two wave functions is zero, because the two configurations span two boson irreps and the -I- and - linear combinations have specific boson symmetries. This situation is somewhat similar to the nonrelativistic case of and I states, which are linear combinations of two determinants, or to the open-shell singlet case. 

As an example, in the group the configuration (lei/2)H2ei/2) has the symmetry El, and the time-reversed configuration (le i/2) (2e i/2) has the symmetry of the other component of the E boson irrep, A-i. The configuration (161/2) 1 3/2) however, spans the Bi and B2 representations along with (le i/2) l6-3/2) and the two configurations must be combined to form symmetry-adapted wave functions. 

The same considerations also apply to the case of two open shells where the product of the fermion irreps for the open shells belongs to a doubly degenerate boson irrep. In this case the reference is a single determinant, related to its partner by the time-reversal operator. Because there is no symmetry between the open shells, we cannot derive relations between the amplitudes for Kramers partners. 

Table D3 Basis functions for fermion irreps in terms of boson irreps for groups that have only one axis of order greater than 2...

The molecular spinors are expanded in terms of the four quaternion units (1, i, j, k). Two-dimensional contour maps of the large components are created in this work for the molecular spinors so to illustrate the nodal structure, and we review the relation between the quaternion representation and the normal four-component complex representation. [32] Each quaternion unit belongs to one of the boson irreducible representations (boson irreps) of C2v provided that the small components are neglected. 

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