# How can we deduce that a hydrogen atom is stable in relativistic QED?

Consider relativistic quantum electrodynamics (QED) with three quantum fields: the electromagnetic field $$A_\mu$$, one fermion field $$\psi$$ for electrons/positrons, and one fermion field $$\psi'$$ for protons/antiprotons. The protons/antiprotons are treated as elementary in this model, just like electron/positrons but with opposite charge and larger mass. The lagrangian is \begin{align*} L &\sim \overline\psi\gamma^\mu(i\partial_\mu-A_\mu)\psi +m\overline\psi\psi \\ &+ \overline\psi'\gamma^\mu(i\partial_\mu-A_\mu)\psi' +m'\overline\psi'\psi' \\ &-\frac{1}{4q^2}F_{\mu\nu}F^{\mu\nu} \end{align*} where $$q$$ is the elementary charge. In this model, the electron/positron electric charge observable $$Q$$ and the proton/antiproton electric charge observable $$Q'$$ are separately conserved. The justification for this claim is given below. The lowest-energy state in the $$(Q,Q')=(-q,+q)$$ subspace is presumably the state with a single hydrogen atom having zero total momentum. If that presumption is true, it would immediately prove that the hydrogen atom is stable (in this model), because there is no state with lower energy into which it could decay without violating the $$Q$$ or $$Q'$$ conservation laws.

Question: Is that presumption true? Is the lowest-energy state in the $$(Q,Q')=(-q,+q)$$ subspace really a zero-momentum hydrogen atom?

I'm sure it is true, but my confidence is based on intuition from non-relativistic models, along with some reputable but hand-waving effective-field-theory arguments about how those non-relativistic models are related to relativistic QED. It's the hand-waving part that bugs me. Such arguments have stood the test of time, and I've often used them myself without flinching, but mathematically it's a weak link. I'm looking for a more mathematical argument for the stability of hydrogen that uses relativistic QED itself, specifically using the model described above.

• There are other questions on Physics SE about the stability of hydrogen, but the ones I found all either use a non-relativistic approximation (sometimes with relativistic perturbations) or address peripheral issues using perturbation theory. I'm asking a non-perturbative question and seeking a non-perturbative answer.

• Relativistic QED can be constructed non-perturbatively with no mathematical ambiguity if we replace continuous space with a finite lattice. (We can make the lattice so fine and so large that it might as well be continuous and infinite.) So I think the question is well-posed, except that I haven't defined what I mean by "hydrogen atom" ...

• By "hydrogen atom," I mean a state in which the correlation function between the charge densities $$j_0(x)$$ and $$j_0'(y)$$ falls off exponentially with increasing spacelike distance $$|x-y|$$, but falls off more slowly (if at all) when $$|x-y|$$ is less than a characteristic scale that can be identified as the size of the atom. The charge densities $$j_0$$ and $$j'_0$$ correspond to the charges $$Q=\int d^3x\ j_0(x) \hskip2cm Q'=\int d^3x\ j'_0(x).$$ I'm not quite happy with this definition of "hydrogen atom," but hopefully the intent is clear.

• I claimed that $$Q$$ and $$Q'$$ are separately conserved in this model. Here's the basis for that claim. The model has two independent global $$U(1)$$ symmetries, namely $$\psi(x)\to\exp(i\theta)\psi(x) \hskip2cm \psi'(x)\to\exp(i\theta')\psi'(x)$$ The transformation parameters $$\theta$$ and $$\theta'$$ are independent of each other because the model doesn't have any terms in which these two fields are multiplied together. They are coupled only via the electromagnetic field, which isn't affected by these global transformations. Using Noether's theorem, we get two separately conserved currents associated with these two independent symmetries. We can also verify these conservation laws directly, by using the Heisenberg equations of motion to calculate $$\partial_\mu j^\mu$$ and $$\partial_\mu (j')^\mu$$, with $$j^\mu\sim q\overline\psi\gamma^\mu\psi \hskip2cm (j')^\mu\sim q\overline\psi'\gamma^\mu\psi'.$$

• Of course, $$Q$$ and $$Q'$$ are not separately conserved in the real world (a neutron can decay into a proton, electron, and neutrino), but that's because there is more to the real world than QED. I'm asking this question in the context of QED. A similar comment applies to the possibility of proton instability, which may occur in the real world but not in the model described above.

• Can you solve the Bethe-Salpeter equation exactly? Commented Aug 12, 2019 at 7:52
• Commented Aug 12, 2019 at 9:54
• What is the source for electromagnetic field $A_{\mu}$? Classically, each electron and proton will create its own EM field: $A_{\mu}$ for electron, and $A'_{\mu}$ for proton So I guess the Lagrangian should be modified. It appears that your present lagrangian L models a system of proton and electron which are far separated but existing on a common EM background field. Any progress made using this model so far?
– KP99
Commented Aug 31, 2021 at 6:54
• @KP99 No, that's not correct. This is standard QED, and it's valid for arbitrary separation (close or far). There is only one EM field in QED, just like there is only one EM field in classical electrodynamics. You might be confusing the concept of different configurations of the field with the concept of different fields. Different arrangements of electrons and protons produce different configurations of the EM field, but there is only one EM field. It's a dynamic field: its configuration is determined by (well, influenced by) the configuration of electrons and protons. Commented Aug 31, 2021 at 14:01
• Thank you for the clarification!
– KP99
Commented Aug 31, 2021 at 14:49