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When considering interactions, the free propagator $S_0(p)$ of fermions for example gets "dressed" due to the self-energy of the fermion. The complete propagator then becomes $$ S(p)=\frac{/\!\!\!p}{p^2-m_0^2-\Sigma(p)} $$ where $m_0$ is the bare mass, and $\Sigma(p)$ is the self-energy of the fermion. Independently of that, when one tries to do one loop calculations, the self-energy loop usually gets inserted on an internal fermion line. See the example below:

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Question: When calculating the Feynman amplitude of such a diagram, do we have to use the dressed propagator? Or is dressed propagator never used? If it is used instead of the bare propagator, do we also have to use the dressed propagator in the leading order diagrams? I am asking this, since there are the following possibilities for the diagram above:

  1. Only use bare propagators, i.e. $\bar{v}(p_1)\ldots S_0(p_1-p_3)\Sigma(p_1-p_3)S_0(p_1-p_3)\ldots u(p_2)$.

  2. Only use dressed propagators, i.e. $\bar{v}(p_1)\ldots S(p_1-p_3)\Sigma(p_1-p_3)S(p_1-p_3)\ldots u(p_2)$.

  3. Since the self-energy is absorbed inside the dressed propagator, completely ignore the self-energy loop in this graph and write the following instead: $\bar{v}(p_1)\ldots S(p_1-p_3)\ldots u(p_2)$.

Which one of these possibilities would be applicable? Does anything change in the application if one prefers the MS-scheme over the on-shell scheme?

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