Excitation boson operator

The kinetic energy of the mode Q being invariant, we may write for the molecular hamiltonian, using the excitation boson operators B, B ... 

Passing to the Boson operators by aid of Table II, and after neglecting the zero-point-energy of the fast mode, we obtain a quantum representation we shall name I, in which the effective Hamiltonians of the slow mode corresponding respectively to the ground and first excited states of the fast mode are... 

Here a and are the usual oscillator creation and annihilation operators with bosonic commutation relations (73), and 0i,..., 1 ,..., 0Af) denotes a harmonic-oscillator eigenstate with a single quantum excitation in the mode n. According to Eq. (80a), the bosonic representation of the Hamiltonian (79) is given by... 

Linear terms are absent because of the Brillouin theorem. The coefficients Ap p. and Bap p, can be calculated by equating the nonzero matrix elements of the RPA Hamiltonian [Eq. (122)], in the basis of Eq. (121), to the corresponding matrix elements of the exact Hamiltonian [Eq. (23)] in the same basis. From the translational symmetry of the mean field states it follows that the A and B coefficients do not depend on the complete labels P = n, i, K and P = n, /, K1, but only on the sublattice labels /, AT and /, K. The second ingredient of the RPA formalism is that we assume boson commutation relations for the excitation and de-excitation operators (Raich and Etters, 1968 Dunmore, 1972). 

Thus far we have only considered one (boson) vector field, namely, the direct product field R Xn of creation and annihilation operators. The coefficients of the creation and annihilation operator pairs in fact also constitute vector fields this can be shown rigorously by construction, but the result can also be inferred. Consider that the Hamiltonian and the cluster operators are index free or scalar operators then the excitation operators, which form part of the said operators, must be contracted, in the sense of tensors, by the coefficients. But then we have the result that the coefficients themselves behave like tensors. This conclusion is not of immediate use, but will be important in the manipulation of the final equations (i.e., after the diagrams have contracted the excitation operators). Also, the sense of the words rank and irreducible rank as they have been used to describe components of the Hamiltonian is now clear they refer to the excitation operator (or, equivalently, the coefficient) part of the operator. 

Energy excitations in 1-d Fermi systems are effectively Bose excitations with zero mass. A suitable representation of Fermion field operators in terms of Bosons has been given by... 

The Hamiltonian equations 15.1-15.4 is applicable to various processes characteristic of molecular systems, including the dynamics at conical intersections (Coin s) [1-4] and excitation energy transfer (EET) processes [5,6,8]. Its simplest realization corresponds to a single system operator, in which case the classical spin-boson Hamiltonian [12,18] is obtained, where the bath coordinates couple to the energy gap operator = a) (a — J3) (J31 of a two-level system (TLS). 

Here, 1 > and 2 > are the diabatic electronic states (exciton and CT states, respectively), J their electronic coupling parameter, uj the vibrational frequency, 6 and h the boson creation and annihilation operators, and g and g2 the equilibrium position shifts in the excited states 1 > and 2 >, and AE the zeroth order splitting between the two electronic states. The zero energy is set to ( AE - Eg (jS) where ft = 1. and Eg is the ground state electronic energy of a Born-Oppenheimer model. 

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