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Consider a gedanken(=thought) experiment where I am tracking the motion of the electron in a hydrogen atom with a time resolution of (say) $\Delta t = 10^{-20}$ seconds. Further assume (for simplicity) that the proton is stationary. In quantum field theory we talk about forces carried by virtual particles. Therefore, the Coulomb interaction between the proton and electron in this hydrogen atom is a result of the exchange of virtual photons. However, we do not need to pull out our QED machinery because the electron on average is not energetic (or fast) enough. Also, the number of virtual photons exchange between a single pair of proton-electron that the quantum-ness of EM field is washed out and we can treat it classically. In other words, in this limit we do not need to think in terms of virtual photons.

Now, in this thought experiment the proton vanishes at $t = 0$. The absence of the proton will not be felt by the electron until $t = a_0/c \,(\approx 17\Delta t$); $a_0$ and $c$ are the Bohr radius and speed of light. Assuming causality I would expect the electron to fly off (because it is no longer bound) at this point in time ($17\Delta t$). But virtual photons can exceed $c$. Then won't the electron know that proton is disappeared before $17\Delta t$?

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    $\begingroup$ A small detail: It's easy to get unphysical results when thinking experiments where charge can 'vanish'. In particular, you've just created monopole radiation. $\endgroup$
    – emarti
    Commented Dec 18, 2012 at 10:10
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    $\begingroup$ It seems to me you are asking why the EM force is transmitted at $c$ when virtual photons can move faster than $c$. Is this a fair simplification of your question? $\endgroup$
    – John Rennie
    Commented Dec 18, 2012 at 10:58
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    $\begingroup$ Yes, John Rennie. The question can effectively be reduced to that. $\endgroup$
    – PhHEP
    Commented Dec 18, 2012 at 11:05
  • $\begingroup$ @PhHEP - John Baez discusses this in math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html but even he doesn't manage a clear pop-science explanation. I'm not sure there is a easy way to understand it. $\endgroup$
    – John Rennie
    Commented Dec 18, 2012 at 16:07

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I will address the premise that the electron is in an orbit around the hydrogen atom. This is a classical picture overlayed on the basic quantum mechanical one. The electron around the hydrogen atom is in a "spherical" probability cloud about the proton.

enter image description here

The above is the n=1,l=0,m=0 probability distribution, which is the lowest energy state.

A single measurement, would get one point in this cloud . Your gedanken experiment is a single measurement ( though of unphysical dimensions as far as we know, protons do not fall down perpendicular dimensions).

Now, in this thought experiment the proton vanishes at t=0. The absence of the proton will not be felt by the electron until t=a0/c(≈17Δt); a0 and c are the Bohr radius and speed of light.

That is not accurate in the quantum mechanical picture ( let alone QFT). In the S state above, the electron has a high probability of being on top of the proton or it could be at the outer edge of the probability envelope. One cannot know the distance from r=0. It is a draw of the luck.

Assuming causality I would expect the electron to fly off (because it is no longer bound) at this point in time (17Δt). But virtual photons can exceed c. Then won't the electron know that proton is disappeared before 17Δt?

This 17Δt is not correct. The Bohr radius is semiclassical and nothing to do with virtual photons or the real QM solution of the potential problem.

In the simple quantum mechanical problem the electron will be free with energy E_n=1, since the potential that gave the solution to the hydrogen atom V has become 0, and it will conserve angular momentum and momentum presumably, depending on how the proton disappeared. Virtual photons have nothing to do with it.

Now if we go to a QFT picture with virtual photons being exchanged between the proton and the electron we find a cloud of virtual photons too building up the potential V. One would need a different calculation than the simple one you are using to see whether the outlier, faster than c, virtual photons can transmit the information of the disappearance to some electrons. It will be a probabilistic answer anyway with expectation values.

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Your seemingly unrealistic gedanken experiment is in fact a quite realistic. First, one can kick out the proton with help of a fast neutron. Next, to increase your "delay" time, you can consider a Rydberg atom with a high enough $n$, so the electron velocity is rather small with respect to the light (and the maximum proton) velocity.

What happens to the atom in this case? Briefly speaking, the atom gets ionized with a sudden perturbation. Sudden perturbations are considered within QM as not changing the state (the initial cloud in your case), but as changing Hamiltonians. The electron in QM feels the Coulomb potential changes instantly, but reacts depending on the work done by the force. Suddenly new Hamiltonian acts somehow on the original state and makes it evolve during some (electron reaction) characteristic times.

EDIT: In QFT and in a classical relativistic field theory all potentials are "retarded", to tell the truth.

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