# 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.

• 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.