homepage › Forums › a nucleus with a general shape ? › Pertrubation theory with no multipole expansion
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Carlos M Fajardo ZParticipant
The first multipole expansion that is used in the formulation of the nucleuselectron system is in the interaction made between a free nucleus and the free electron cloud. Since the monopole term that results from this expansion is part of the usual electronic problem (Hydrogen atom problem), the unperturbed hamiltonian will be, for this case, the sum of the free nuclei and free electronic hamiltonian. Leaving the interaction between the two as the perturbed hamiltonian. This by itself means that the perturbation theory is only valid if the interaction between the two induces a small perturbation in the energy values of the two systems (which clearly wouldn’t be the case). Ignoring that we are already under brittle ground, the next step is to give a full expression that describes this interaction. Since the different electrons will feel a different potential from the nucleus due to the screening of their partners closer to the nucleus, an effective Coulomb interaction for each electron would have to be defined. Thus, the energy of Coulomb interaction between an electron and a nucleus would be the classical energy described in the interaction between only one electron and a nucleus as a point charge times a coefficient of effective Coulomb interaction of charged particles in the atom. This coefficient will depend on the relative position of the different electrons with the nucleus and with the other electrons. From this point on, only computational solutions for manybody interactions can be applied to solve the problem, e.g., the HartreeFock method. Making it impossible to arrive at an analytical solution to the problem.
Part of my reasoning was developed using the paper:
Shuvalov, V. A. (2007). Electron and nuclear dynamics in manyelectron atoms, molecules and chlorophyllâ€“protein complexes: A review. Biochimica et Biophysica Acta (BBA)Bioenergetics, 1767(6), 422433. 
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