How are bound states handled in QFT?

Question

QFT seems very well suited to handle scattering amplitudes between particles represented by the fields in the Lagrangian. But what if you want to know something about a bound state without including it as an extra field? For example, suppose we have electron+proton QED (ignoring the proton's structure):

$$\mathcal{L} = -\frac14 (F_{\mu\nu})^2 + \bar{\psi_e} (i\not \partial -m_e)\psi_e + \bar{\psi_p} (i\not \partial -m_p)\psi_p - e \bar{\psi_e} \not A \psi_e + e \bar{\psi_p}\not A \psi_p$$

I can use this with no problem to calculate Rutherford scattering or similar processes. But this Lagrangian should also have the hydrogen atom hidden in it somewhere. For example, I may want to use QFT to calculate the binding energy of hydrogen. Or I might want to calculate the probability of firing an electron at a proton and getting hydrogen plus photons as a result. How can this be done? Obviously this is a broad subject, so I'm just looking for an outline of how it goes.

Related: Bound states in QED(Phys.SE question), "On bound states in quantum field theory"(arXiv link) — ACuriousMind, Nov 09 '15 at 20:32
@ACuriousMind: I've taken a look at that question, it's related but not quite about the same thing. The paper looks interesting though, I'll be sure to look at it. — Javier, Nov 10 '15 at 01:15
In scattering an electron off of a proton, you should see the hydrogen bound states as poles in the S-matrix. To get hydrogen itself as a final state, you might try an effective field theory approach where you add the hydrogen atom to your Lagrangian. — user2309840, Nov 12 '15 at 05:03

Arnold Neumaier · Accepted Answer · 2020-09-18T09:26:39.143

The conventional way to handle bound states in relativistic quantum field theory is the Bethe-Salpeter equation. An old but very informative survey paper on the Bethe-Salpeter equation is

M.M. Broido, Green functions in particle physics, Reports on Progress in Physics 32 (1969), 493-545.

The hydrogen atom is in QFT usually treated in an approximation where the proton is treated as an external Coulomb field (and some recoil effects are handled perturbatively). The basics are given in Weinbergs QFT book Vol. 1 (p.560 for the Bethe-Salpeter equation and Chapter 14 for 1-electron atoms). Weinberg notes on p.560 that

the theory of relativistic effects and radiative corrections in bound states is not yet in entirely satisfactory shape.

This quote from 1995 is still valid today, 20 years later.

On the other hand, quantum chemists use routinely relativistic quantum mechanical calculations for the prediction of properties of heavy atoms. For example, the color of gold or the fluidity of mercury at room temperature can be explained only through relativistic effects. They use the Dirac-Fock approximation of QED.

score 3 · Answer 2 · answered Oct 07 '16 at 00:35

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A bound state such as a hydrogen atom is 'hidden' in the given interaction as a pole in the scattering matrix for the two fields.

answered Oct 07 '16 at 00:35

Ruth

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Hi and welcome to Physics.SE! This is a very tempting answer; could you elaborate a bit? – rob Oct 07 '16 at 00:54
Thanks, you can take a look at this paper for some additional details: https://arxiv.org/pdf/1601.07169v1.pdf -- this is standard in scattering theory – Ruth Oct 09 '16 at 22:28

How are bound states handled in QFT?

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