Going from the QED Lagrangian to low energy Hydrogen scattering

Question

Many introductory texts and lecture notes on effective field theory (such as this and this) use the example of Rayleigh Scattering to demonstrate power counting and dimensional analysis. They use an "atom creation operator" $\phi$ with mass dimension $\frac{3}{2}$ and couple it to $F_{\mu\nu}$, and from there it follows using dimensional analysis that at very low energy the scattering of light off the atom goes as $\sigma \sim a_0^6 E_\gamma^4$.

My question is: All the information needed to describe this physics should be contained somewhere in the QED Lagrangian for an electron and proton:

$$ \mathcal{L} = \bar{\psi}_e(i\gamma^\mu D_\mu -m_e)\psi_e + \bar{\psi}_p(i\gamma^\mu D_\mu -m_p)\psi_p - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} $$

How would one construct the "hydrogen creation operator" $\phi$ with mass dimension $\frac{3}{2}$, as above, from this? It seems to me that such an operator must exist and can be used to describe extremely low energy physics as an effective field theory (as it is used in the Rayleigh scattering example), but it's not clear to me exactly how one would do this or what limits or assumptions (other than $E_\gamma \ll E_\text{binding}$) one would have to make in order to do it. How would one even know, a priori, to use a single composite degree of freedom $\phi$ to describe electron-proton systems at these low energies?

Answer 1

It seems to me that such an operator must exist...

Yes, but that doesn't necessarily mean we know how to express it in terms of the basic fields in QED. In fact, this is a very difficult problem.

Why it's so difficult

To get some appreciation for just how difficult it is, spend a few minutes thinking about how you would answer this seemingly-easier question: Which state in QED has the lowest energy?$^\dagger$ As far as I know, the answer is known only when the coupling constant $e$ is zero. For $e\neq 0$, we can use perturbation theory to calculate correlation functions, but we cannot use perturbation theory to calculate states. When we try, we discover that the large (formally infinite) number of degrees of freedom in quantum field theory overwhelms the smallness of the coupling constant: the slightest change in the coupling constant makes all of the perturbed states orthogonal to all of the original states.

$^\dagger$ ^{Actually, this isn't quite the right way to ask the question. When doing quantum field theory in a space of infinite volume, we can always choose a Hilbert-space representation in which there isn't any state of lowest energy. But one of the general principles of QFT instructs us to use a representation that does have a state of lowest energy. The right way to ask the question is: How do we construct such a represetation?}

Here's why that matters: Particles are defined as excitations above the lowest-energy (vacuum) state, which has zero particles by definition. Since we don't know which state has zero particles, we also don't know which combinations of field operators would create single-particle states — whether the single particle in question is an electron, a photon, or a hydrogen atom.

Where perturbation theory does work well

Many introductions to quantum field theory are almost exclusively about scattering, because scattering amplitudes are one important thing that we can calculate with perturbation theory. Even though we don't know which state represents the vacuum state or which operators represent single-particle creation operators, we do know how to relate particles to poles in correlation functions.$^{\dagger\dagger}$ When the coupling is small enough, we can use perturbation theory to find single-particle poles in simple correlation functions like $\langle 0|\psi(x)\psi^\dagger(x)|0\rangle$. This is the idea behind interpolating fields: even though the field operators may not create single-particle states, they still create states that include a single-particle term in the superposition. Using the connection between particles and poles, we can isolate those terms in scattering amplitudes. Regarding what happens at strong coupling, see this answer by Accidental Fourier Transform.

$^{\dagger\dagger}$ ^{Actually, the situation in QED is more complicated, because photons are massless. As a result, the basic particles in QED don't produce poles in the strict sense.}

Effective field theory

One nice thing about effective field theory is that we can use it even if we don't know how to express the effective theory's fields in terms of the original theory's fields. It's basically a guess-and-check method, using things like symmetries to help motivate a plausible guess, and using things like scattering amplitudes (which can be calculated in perturbation theory) to choose the values of the parameters that aren't fixed by those symmetries.

How would one even know, a priori, to use a single composite degree of freedom $\phi$ to describe electron-proton systems at these low energies?

If we have some reason (either empirical or theoretical) to think that something like a hydrogen atom does exist in QED, then we can try constructing an effective field theory for it and see how it works out. The guess-and-check method will tell us whether or not it works, but first we have to make a guess so that we have something to check. We have plenty of good reasons to think that QED does have a hydrogen atom (we can start with the nonrelativistic version of QED to see this more directly), but I'm still calling it a "guess" because I don't know of any watertight end-to-end mathematical proof.$^{\dagger\dagger\dagger}$ That's why I posted this question: How can we deduce that a hydrogen atom is stable in relativistic QED?

$^{\dagger\dagger\dagger}$ ^{Whenever the word "proof" is mentioned in the neighborhood of QED, people are sometimes quick to say "but QED doesn't even exist, so why expect a proof?" My response is: QED does exist on a lattice, and that's good enough for all practical purposes. A proof using lattice QED might be too messy for human consumption, but that's just a practical problem, not a problem in principle.}