Home Forums Hyperfine course a nucleus with a general shape ? Perturbation theory without multipole expansion

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    rkolesni
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    If we take QxV as a perturbation Hamiltonian, it is not necessarily going to be a small perturbation, since it still contains the monopole term. Furthermore, choosing how many terms to retain in multipole expansion allows us to control the amount of detail we want to resolve and reduce the computational cost by discarding higher order terms.

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