I am a beginner in QED and QFT. What is known (or expected to be) about bound states in QED? As far as I understand, in non-relativistic QM electron and positron can form a bound state. Should it be true in QED? Most of the basic text books on QFT I known treat QED with perturbative methods which do not help to study bound states. Is there a literature to read about it?

UPDATE: According to wikipedia, electron and positron form a positronium. It is an unstable particle which can annihilate to two photons. In particular electron and positron cannot form a bound state. But still there is a logical possibility that several electrons and positrons can form a bound state (though probably once one has at least one electron and one positron, they will necessarily annihilate. Is it true?)

On the other hand, if one considers QED with both electrons and muons, then electron and antimuon form muonium. Clearly they cannot annihilate to photons. However in another article in wikipedia it is claimed that muonium is unstable. Is it due to some effects of QED or due to actual presence of other, say weak, interactions?


2 Answers 2


You have to be careful about use of the term bound state in QFT. There is a related notion of resonance, which is a state that looks approximately like a bound state for a while (possibly a long while!) but eventually decays. For example, if you consider the nonrelativistic hydrogen atom, it has bound states corresponding to the electron orbitals (the n,l,m quantum numbers from any first course in QM). However, if you quantize the electric field, the electron is now free to emit a photon and drop down an energy level. As a result, all bound states except the ground state become resonances once you quantize the EM field. This is a general feature when you go from N-particle QM to QFT: because QFT allows particles to spontaneously emit photons (or gluons, Z/W, etc.) with some small but non-zero probability, many states that were bound states in QM become resonances in QFT.

So the articles you are reading that mention positronium or muonium or whatever would be more correct in calling them resonances. They are states that behave approximately like bound states on a short timescale, but because there is a nonzero probability of emitting photons (or other particles) they eventually do, and hence become "unbound".

The book Mathematical Concepts of Quantum Mechanics by Gustafson and Sigal has a very good treatment of resonances and how they come about in QFT. Indeed, most of the second half of the book is devoted to giving a mostly self-contained proof of the aforementioned theorem about bound states of the hydrogen atom. The book is fairly readable and only assumes prior knowledge of basic classical mechanics and EM, as well as some basic properties of the Fourier transform (though these are reviewed).

  • $\begingroup$ Thank you for the answer. It clarifies what is written in the articles. But what about bound states which are not resonances? Do they exist in QED? $\endgroup$
    – MKO
    Jul 21, 2011 at 15:41
  • $\begingroup$ Old post but I think there is a misconception here considering QFT: Not every unstable bound state is a resonance. If the energy of the bound system is below the energy of the particles, it is a bound state. If not, it is a resonance. In that sense, every higher orbital of the electron is still a bound state. And positronium is a bound state. All unstable though. But if you collide an electron and positron and have a contribution of the Z-Boson to the amplitude (at high energies), this is a resonance as the energy of the Z-Boson is above the sum of the e+ e- pair. $\endgroup$
    – Cream
    Sep 18, 2023 at 7:20

Yes, in QED there are bound states. Nobody can forbid us to take the Coulomb potential into account exactly and the rest - by the perturbation theory. In this way they obtain the Lamb shift, for example.

There are different approaches to bound stated in QED: poles of the scattering matrix, Bethe-Salpeter equation, Schwinger approach, Logunov-Tavkhelidze quasi-potential approach, etc.

  • $\begingroup$ Thank you. Just to make sure I understood correctly your answer: for the Lamb shift one needs an exterior field. In my question I did not assume any exterior field. $\endgroup$
    – MKO
    Jul 20, 2011 at 10:25
  • $\begingroup$ No, what you call an exterior field is in fact the Coulomb field. It is simply the equation for the relative variables that looks as an equation for a single particle with a reduced mass in an external field. S-matrix approach has some rigorous results and the it is not obligatory to use S-matrix directly. $\endgroup$ Jul 20, 2011 at 10:34
  • 3
    $\begingroup$ Try looking at hep-ph/9711292, where the authors calculate the Lamb shift using more modern techniques and language. $\endgroup$
    – Simon
    Jul 20, 2011 at 12:06
  • $\begingroup$ @MKO I do not have references at hand, unfortunately, but searching by keywords may help. $\endgroup$ Jul 20, 2011 at 12:59
  • 1
    $\begingroup$ @Simon; interesting paper; I'd assumed that these calculations were done using QED as a correction to Schroedinger's equation. And I managed to get full support to study at Washington State U, perhaps the PhD in 2016. $\endgroup$ Jul 21, 2011 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.