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In principle, how would we demonstrate the existence of the hydrogen atom in quantum field theory and the standard model?

Has it been done in practice?

Some naive ideas:

  1. Demonstrate that the familiar quantum mechanics model of the hydrogen atom is a limit of the SM in QFT

  2. A non-perturbative calculation numerically on a computer

  3. First simplify the problem by finding a field theory of protons and electrons as a limit of the SM

Qmechanic
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Peter A
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1 Answers1

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You can't have the hydrogen atom as a state in QED with protons and electrons, because it is a bound state and hence it is nonperturbative. QED is inherently a perturbative theory. There are good reasons to believe that QED doesn't even exist nonperturbatively, unless it is associated with a broken phase of a non-abelian gauge theory.

You can, however, deduce the properties of the hydrogen atom from QED by making simplifying assumptions about the interaction:

  1. You want to take the classical limit for the electromagnetic sector.
  2. You want to assume that the Compton wavelength of the electron is much less than the typical scale of the atomic orbitals. This means you can treat the electron and proton as first-quantized particles without worrying about pair production.

You'll end up with the Dirac equation for the electron in Coulomb potential of the proton, which can be solved and leads to the well known result.

Small corrections to this result can also be deduced from perturbative QED. For example, the first-order correction to the photon propagator is responsible for the Lamb shift.

Prof. Legolasov
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    Thank you for such an informative and interesting explanation. Much appreciated – Peter A Nov 01 '20 at 16:57
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    So then does this mean we have no complete quantum relativistic theory of electrodynamics, comparable to Maxwell's equations, where you can just plug in charges and currents and find the EM field? If that's so, then it seems that all this talk of "quantum gravity" and the like is really jumping the gun! Why don't we get electrodynamics down rock solid first, as a single unified cohesive theory that spans both regimes? – The_Sympathizer Jun 21 '22 at 08:17
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    @The_Sympathizer indeed, mathematically sound interacting QFT in 4 dimensions is an open problem. Electrodynamics probably doesn't even exist nonperturbatively because of the Landau pole, but it could be embedded into a compact gauge group of a GUT, which are conjectured to be mathematically well defined (but not proven, see the Yang-Mills existence & mass gap problem). Quantum Gravity is interesting because quantum spacetime is the "ultimate regularizer": the pathologies of QFT are mostly originated at short distances, which is the realm of QG. – Prof. Legolasov Jun 21 '22 at 10:01
  • @Prof. Legolasov: But it seems to me your post suggests a failure of the QFT not just at the level of QG, which is usually expected at a small multiple of the Planck length, but a failure at level as coarse as the Compton wavelength of the electron, about 23 orders of magnitude higher! Wouldn't this have potential impact for suitably high, but not unreasonable, levels of accuracy of measurements of everyday atomic or molecular phenomena? – The_Sympathizer Jul 08 '22 at 15:06
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    @The_Sympathizer a failure of perturbative techniques, not QFT. It would, and it does. Again, there is simply no hydrogen atom in perturbative QFT. To describe atoms/molecules you make simplifying assumptions about how the nonperturbative QFT should work (assumptions about bound states). – Prof. Legolasov Jul 09 '22 at 02:47
  • @Prof. Legolasov: But is there any solid alternative to perturbative QFT that for this scale works as well as ordinary NRQM for which the H atom just "falls out" with the same grace as NRQM (even if not the same mathematical simplicity, of course! I.e. I wouldn't expect whatever comes out to be the simple spherical harmonics b/c of the relativistic corrections and the like)? Or is the "standard model" actually not as complete and cogent as it's made out to be (i.e. supposedly it gives "all physics" not including extreme quantum gravity and other such things)? – The_Sympathizer Jul 09 '22 at 04:43
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    @The_Sympathizer lattice QFT with a cutoff close to the Landau pole scale. It won’t be Lorentz invariant and otherwise nice, but the anomalous terms that violate these symmetries are all small. Perturbative QFT follows from it as an asymptotic expansion, and lattice QFT are known to contain bound states. It wouldn’t be complete though. You’d still need to find the fundamental physics that gives rise to QFT. – Prof. Legolasov Jul 09 '22 at 04:50
  • @Prof. Legolasov: Thanks. – The_Sympathizer Jul 09 '22 at 04:53