On the Interaction of the Electron with the Vacuum Fluctuations of Electromagnetic Field

A new approach is proposed for evaluating the interaction energy between the electron and the vacuum ﬂuctuations of electromagnetic ﬁeld. It is applied to two cases: when the electron is free and when it is in a potential-a hydrogen atom. The results are consistent with previous relevant treatments of people.


On the Interaction of the Electron with the Vacuum Fluctuations of Electromagnetic Field
Xiaowen Tong * Abstract A new approach is proposed for evaluating the interaction energy between the electron and the vacuum fluctuations of electromagnetic field. It is applied to two cases: when the electron is free and when it is in a potential-a hydrogen atom. The results are consistent with previous relevant treatments of people.
The Hamiltonian of the radiation field iŝ We always drop the infinite constant which is called the zero-point energy since it has no material effect in most cases. Still, the zero-point oscillations of the electromagnetic field exist and give rise to some observable results. Their effect hide in the action of the creation operatorâ † kσ on |0 . Is there a direct way to evaluate the interaction bettwen the electron and these oscillations, for example, the interacting energy?
As we have seen, unlike the states in which real photons present every mode of the zero-point energy posseesses a half of one light quanta. Therefore, we guess that the interaction energy of the electron and the zero-point oscillations is the half of the one between the electron and a state of radiation field in which every mode is occupied by one real photon (one photon for one polarization) and which we call the full one-photon state. The latter can be readily calculated as follows: analogous to the self-action [1], the modification caused by the interaction to the wave function of the free electron is to multiply it a phase factor exp(−i∆ET ) [1]. The first-order of this modification corresponds to the process where only one photon is absorbed or emitted. We know that the S-matrix element of the Compton scattering in which the electron is scattered into the same eigenstate as its initial state is Since there is one photon in every mode, we need to sum all the modes to get the modification of the wave function. Thus the interaction energy ∆E between the electron and the full one-photon state of radiation field is given by In momentum space, where in the last step we have used a direct cut-off(Wick rotation) [1]. Hence the energy shift of the electron caused by the fluctuations is This result is in accordance with previous treatments[2] * Next we consider the Lamb shift which, as thought, is completely caused by the vacuum fluctuations of the radiation field. For simiplicity, we will derive it in nonrelativistic theory. The quantized transverse wave of electromagnetic field isÂ

The interaction Hamiltonian iŝ
2 (x, t) * Our result is a half of Dalibard's. This is because the square of our cut-off momentum is the double of his. Weisskopf had also obtained this energy which, however, is incorrect [3]. He evaluated the mean square of the field fluctuations wrongly(Eq.(25)) which shall be one half of his value(for example, refer to [4]). He also calculated the integral incorrectly in Eq.(18) in his paper.
Before calculating the Lamb shift, let us first evaluate the interaction energy for the free electron using our approach in nonrelativistic theory. We designate the state of the electron with momentum p and the full one-photon state as |p and |1 f ull rad , repectively. According to our method, the second-order interaction energy is then ∆E (2) p,vf = 1 2 where ∆E vac is the energy of the electron interacting with the vacuum which is given by There is no this subtraction in the previous relativistic calculation for it has been implemented via the time-ordered product. We then have [5] ∆E (2) p,vf = 1 2 Being accurate enough for our case, the approximation v/c 1, ω/mc 2 1 is applied to ∆E (2) p,vf . We obtain ∆E (2) p,vf = 0 (4) Now we are ready to evaluate the Lamb shift: ∆Ẽ (2) n,vf = 1 2 e 2 m 2 c 2 I rad 1 f ull | n|p ·Â |I I|p ·Â |n |1 f ull rad E n − E I − ∆E vac −∆E (2) n,vf (mass) Here |n is the eigenstate we considered of the hydrogen atom. The reason of the appearance of ∆E vac is the same with ∆E vac . ∆E (2) n,vf (mass) is the correction to the energy corresponding to the variation of the electron mass. It equals to 0 in virtue of (4). Thus ∆Ẽ (2) n,vf = 1 3π e 2 m 2 c 3 n | n |p |n | 2 (E n − E n ) log mc 2 |E n − E n |