Does the Uehling potential have any observable effect? Uehling potential is the correction made to the coulomb potential due to vacuum polarization (electron-positron pair creation and annihilation). How big is this correction compared to the coulomb potential (0.01% for example), and can the change it makes to the coulomb potential have a significant observable effect? If so, where?
 A: Yes. The Uehling potential due to vacuum polarization by virtual electron-positron pairs is the dominant contribution — 205.0073 meV — to the Lamb shift between the $2P_{1/2}$ and $2S_{1/2}$ states of muonic hydrogen. (See equation (15) here.)
Muonic hydrogen is a bound state of a proton and a muon. Since a muon is more massive than an electron, the muon orbitals lie closer to the proton than electron orbitals do. The Uehling potential is significant only at short distances where the electrostatic field of the proton is intense, so tightly bound muonic hydrogen is better able to probe the region close to the proton.
The Lamb shift is a QED effect due to virtual particles; the $2S_{1/2}$ and $2P_{1/2}$ states have exactly the same energy if you simply calculate energy levels using the Dirac equation. The Lamb shift in electronic hydrogen is mainly due to the self-energy of the electron (i.e., its interaction with virtual photons). By contrast, the Lamb shift in muonic hydrogen is mainly due to vacuum polarization (i.e., the muon interacting indirectly via virtual photons with virtual electrons and positrons).
The Lamb shift in muonic hydrogen was measured in 2010 by Pohl et. al. There are other contributions coming from the finite size of the proton, so this experiment constituted a measurement of the proton radius. However, it got a result that differed from other techniques for measuring the proton size, and this has led to the proton radius puzzle.
When two charges are separated by the reduced Compton wavelength of an electron, their potential energy is greater than what Coulomb's Law predicts by a factor of 1.0000556435. When their separation is only one tenth of this, the correction factor grows to 1.00172446. You can obtain these numbers by numerically evaluating the integral in the Coulomb-plus-Uehling-correction potential of a charge $q$,
$$V(r)=\frac{q}{4\pi\epsilon_0 r}\left(1+\frac{2\alpha}{3\pi}\int_1^\infty dx\,e^{-2x r/\lambda_{e,r}}\frac{2x^2+1}{2x^4}\sqrt{x^2-1}\right)$$
Here $\alpha$ is the fine structure constrant and $\lambda_{e,r}=\hbar/m_e c$ is the reduced Compton wavelength of the electron.
In electronic hydrogen, the radius of the ground-state orbital (the Bohr radius) is about 137 times larger than the reduced Compton wavelength of the electron. (This is the reciprocal of the fine structure constant.) At this distance, the Uehling correction is negligible because of the exponential factor in the integral. But in muonic hydrogen, this radius is reduced by a factor of about 207, the ratio of the muon mass to the electron mass. This makes the muonic ground-state orbital radius just 0.663 of the reduced Compton wavelength of the electron, and makes the Uehling correction measurable.
