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I consistently had this question of how could the light-matter interaction be described in terms of the fundamental language of QED. To be more specific, is there a way to 'derive' the interaction Hamiltonian such as Jaynnes-Cummings Hamiltonian from the fundamental interaction Lagrangian such as

$$\mathcal{L}=i\overline{\psi}\gamma^{\mu}(\partial_{\mu}-ieA_{\mu})\psi-m\overline{\psi}\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}.$$

used nearly exclusively in the particle physics context to describe electron photon interaction?

I found a similar question raised in

What is the difference between QED and quantum optics?

but the answer to the question only stated that full quantum field theoretic description is rarely used and semiclassical description suffices and is convenient in most cases.

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I'm not an expert in this field but I comment briefly: The answer to your question is yes. The first term in your expression essentially couples the electron momentum $p$ to the field amplitude $A$ so that you have $H = A \cdot p$. From this you can go through a few steps (including the Power Zienau transform and dipole approximation) to eventually get the $E\cdot d$ Hamiltonian from which usual "AMO" type interactions are derived.

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  • $\begingroup$ Can I do the same thing to nucleus part of the hydrogen atom as well? Or do I need to write a whole new Lagrangian to describe the situation of coupling to composite particle? I wonder if I can build the Lagrangian in a step by step manner $\endgroup$
    – 류민석
    Commented Jun 26, 2023 at 2:21
  • $\begingroup$ @류민석 The procedures I'm familiar with usually involve treating the nucleus as fixed in place. This is usually justified because (1) we are talking about QED, so it is an electric or magnetic field interacting with the atom and (2) the mass of the nucleus is ~2000x the mass of the electron. So the electron moves much more in response the external fields than the atom. Also, there's no QED lagrangian for the nucleus because it is a composite of many particles. so you would be outside of the scope of QED if you wanted to include a microscopic model for the nucleus as well. $\endgroup$
    – Jagerber48
    Commented Jun 26, 2023 at 2:31
  • $\begingroup$ I see what you mean! So we can treat the nucleus as static potential source which exchanges virtual photons with the electron and then we may model the rest using photon electron interaction in QED. Thanks for clarification. $\endgroup$
    – 류민석
    Commented Jun 26, 2023 at 3:35
  • $\begingroup$ You can look up the papers computing the precision, QED tests of H atom. The nucleus does not need to be fixed for the H atom, and H atom alone, because someone found the relativistic version of the reduced mass scheme. For the other atoms and molecules, things are much more complicated, and we know that some systems need Beyond Born-Oppenheimer approximation to get precise results, i.e. it is not always the case that fixed nucleus is sufficient. Needless to say, these things are so difficult that almost nobody uses them. $\endgroup$ Commented Jun 26, 2023 at 3:54

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