People always says that the infinity of the coulomb potential $V(r)=\frac{k}{r}$ as $r$ approach to zero is resolved in quantum field theory. I would like to know how this is done.
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I think it depends on what one means by "resolving the infinity". One could argue that this infinity is already resolved in quantum mechanics, since one can solve the Schrödinger equation in Coulomb potential without encountering any divergence. The difference from the classical case is the quantum uncertainty, which means that the electron can never "fall" into the infinite hole, where its position and momentum would be both zero.
The Coulomb potential does pose problems in a one-dimensional setting - a seemingly abstract problem (first treated in a journal for teachers), which did become a serious physical problem in the context of calculating exciton energies in carbon nanotubes. More refined calculations however solve this problem, since, after all, the nanotubes have finite diameter, which imposes a cutoff on the potential.
Roger V.
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what do you mean by the electron can never "fall" into the infinite hole, where its position and momentum would be both zero. Why can we not have a particle whose position is zero but momentum undefined? – amilton moreira Dec 09 '20 at 17:45
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Then this particle won't stay in the potential well. You have probably seen the beginner's QM problem: determine the minimum depth of a square well, when it can have a bound state, using only the uncertainty principle. Same argument can be done for the Coulomb potential - it will have a bound state, but its energy is finite. – Roger V. Dec 09 '20 at 17:48
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Quantum field theory is based on the solutions of quantum mechanical equations , the creation and annihilation operators operate on the solutions of the appropiate equations without a potential term( Klein Gordon, or Dirac, or quantized Maxwell for photons).
For bound states there is no need for QED, because at the quantum mechanical level the potential defines the energy levels allowed in the bound problem. The orbitals of the energy level solutions allow the electrons to overlap the nucleons because there is no classical "attraction" , orbitals are probability loci and the classical model does not hold. See the possible orbitals for the electron in the hydrogen atom here..
Quantum field theory is used for calculating crossections and decays of elementary particles in scattering experiments, and there the Coulomb potential of the particles is transmuted to exchanged virtual particles in Feynman diagrams. In electron-electron scattering one get a Feynman diagram:
and the potential between the two electrons is the exchanged virtual photon. This is a first order diagram, one would have to sum higher orders to get an accurate result, but again: at the quantum level the Coulomb potential has a different representation.
In the case of opposite charges, $e^+ e^-$ the Heisenberg uncertainty (HUP) is in built in the QED theory, and there is a probability of the two incoming leptons to annihilate to two gamma with the following diagram
In this case the Coulomb potential role is taken by the virtual electron , and annihilation together with the HUP make sure that the (0,0,0) is just another probability locus. For higher energies a plethora of particles come out, as studied at $e^+e^-$ colliders.
For electron-proton scattering analogous diagrams would exist with virtual photon exchange being the effect/carrier of the Coulomb potential for the scattering case.
anna v
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