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.

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    $\begingroup$ Can you solve the Bethe-Salpeter equation exactly? $\endgroup$
    – DanielC
    Commented Aug 12, 2019 at 7:52
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    $\begingroup$ Related: physicsforums.com/threads/stability-of-the-atom-in-qft.576848/… $\endgroup$ Commented Aug 12, 2019 at 9:54
  • $\begingroup$ 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? $\endgroup$
    – paul230_x
    Commented Aug 31, 2021 at 6:54
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    $\begingroup$ @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. $\endgroup$ Commented Aug 31, 2021 at 14:01
  • $\begingroup$ Thank you for the clarification! $\endgroup$
    – paul230_x
    Commented Aug 31, 2021 at 14:49

1 Answer 1


There's a really stupid answer to this, which is that QED is not actually non-perturbatively well-defined. Since QED is not asymptotically free, the Landau pole messes things up at exponentially high (in coupling) energies. Asymptotically free theories are expected to be well-defined, and you can construct such theories that are basically QED at non-ridiculous energies, so this is not too much of an issue. But fundamentally, some level of hand-waving is necessary, since to my knowledge nobody has been able to demonstrate rigorously that QFTs like QCD really do exist (e.g. by showing that lattice QCD has a well-defined continuum limit).

If you are willing to accept enough hand-waving to take that the corrections to the hydrogen spectrum are small, then one can take the non-relativistic solution to the hydrogen wave-function, then use it to create a state in the QFT (by, say, some integral of the proton and electron creation operators along with an appropriate diffused Wilson line connecting the two) and making sure that it plays nice with renormalization. Then, there should be, if constructed right, no divergent contributions to the energy, so if one is willing to accept that corrections are small, then this state should be, at least, lower energy than the un-bound state. But if this is lower energy, then the ground state of the system cannot have the proton and electron arbitrarily far apart, since the correction to the un-bound state would be smaller than the difference between the constructed and un-bound state. And so, the ground state of the one-proton one-electron sector of QED must be some sort of bound state. One presumably should be able to get more information about this bound state by extending this argument some more.

  • $\begingroup$ QED may not have a nonperturbative definition in continuous spacetime, but it does on a lattice, and we can take the lattice spacing to be much finer than any empirically relevant scale. (QED isn't meant to be a Theory of Everything.) I didn't say that in the question, but that's the context I had in mind, because otherwise the question wouldn't be well posed. Maybe it's not well posed anyway, because I'm not sure I really defined "hydrogen atom" correctly, but I tried. :) $\endgroup$ Commented Jun 21, 2022 at 14:39
  • $\begingroup$ If one is willing to take a specific lattice model, then the question is at least definitely well-defined. For a reasonable choice of parameters for the lattice, can one find a temporally localized operator such that the correlator of it with its conjugate decays slower than the unbound energy with increasing separation in euclidean time. As for an answer, I think a person better at searching the literature would know if this is open, but I suspect it is. Doing stuff nonpeturbatively is hard and almost all results are typically done through either peturbation theory or symmetry arguments. $\endgroup$ Commented Jun 21, 2022 at 20:02
  • $\begingroup$ @ChiralAnomaly At the very least, usually the ground state and energy levels are done through peturbation theory, but if you want to demonstrate existence rather than calculate it exactly, one simply has to to show that some state has lower energy than the unbound state. It would be a very weird world if there wasn't some version of the above argument that worked, but I don't know if it is known how to show that this is not the case. $\endgroup$ Commented Jun 21, 2022 at 20:07
  • $\begingroup$ Maybe that would work. QED doesn't have poles at particle or composite-particle masses because those pesky massless photons make the spectrum continuous, but the energy spectrum still has an infimum in each sector -- the single-electron sector, the single "proton" sector, and the 2-particle (electron plus proton) sector. If the infimum in the 2-particle sector is less than the sum of the infima in the two 1-particle sectors, then that would demonstrate the existence of a stable bound state. Seems like perturbation theory should be sufficient for that, like you said. $\endgroup$ Commented Jun 22, 2022 at 22:38

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