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The cause of Lamb shift is interaction between vacuum fluctuations and electron in the atom. Will electron (or photon) in free space interact with vacuum fluctuations? How momentum will transfer vacuum fluctuations to the electron?

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Here are the higher order diagrams contributing to the Lamb shift

enter image description here

(Here the dot on the right is the external vertex for the background electromagnetic field.)

The electron on the left is a "real" particle, i.e. it has an energy momentum vector , a four vector of special relativity. The photon leaving the central vertex is virtual, a mathematical construct under an integral for the final calculations, that has the quantum numbers of the photon but not the mass.

From the law of momentum conservation, ( which has not been falsified and is always validated in experiments) the dot on the right has to be a vertex with another particle (or field that can eventually transfer momentum to another real particle).

So the electron on the left in absolutely free space, running up the y axis, will not radiate unless there is an interaction with a field or particle. Vacuum fluctuations (f type diagram) are no exception, momentum has to be conserved and a virtual loop by itself cannot conserve momentum , because by construction it has to be under an integral describing an interaction which has inputs and outputs on real four vectors

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  • $\begingroup$ Yet, will electron (photon) momentum in free space fluctuate, being conserved in average? $\endgroup$
    – user214453
    Commented Dec 3, 2018 at 6:09
  • $\begingroup$ It cannot happen by momentum conservation, even for small delta(t). Violations of momentum conservation have never been observed. $\endgroup$
    – anna v
    Commented Dec 3, 2018 at 6:55
  • $\begingroup$ At small delta(t) energy conservation violate (e.g. W boson), so momentum conservation also can be violated $\endgroup$
    – user214453
    Commented Dec 3, 2018 at 10:20
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    $\begingroup$ no energy conservation is not violated either, in the feynman diagram calculations, and that is all we have. $\endgroup$
    – anna v
    Commented Dec 3, 2018 at 12:44

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