Answer complication perturbation without expansion.

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    mmvdwiel
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    a) A reason for the multipole expansion of the interaction hamiltonian before the perturbation, is that you can expand it into a sum of interactions (nuclear, atomic and hyperfine) with different physical origins, and therefor vastly different energy levels. This first order interaction can be found analytically, so its solution can be considered to be known. The free nucleus hamiltonian summed with this interaction Hamiltonian will now be used as the unperturbed hamiltonian, while further, smaller, not necessarily analytically calculable interaction contributions (caused by any non-sphericality of the nucleus) will be considered the perturbation.
    Without this multipole expansion, the first order interaction would have been inseperatable from the rest of the interaction, so the first order interaction (the largest nucleus-electron) would be a part of the perturbation. This would make the perturbation far larger than with a multipole expansion, so perturbation theory would have been less applicable.

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