This heuristic derivation of the electrodynamic level shift follows
Theodore A. Welton's approach. The fluctuations in the electric and magnetic fields associated with the
QED vacuum perturbs the
electric potential due to the
atomic nucleus. This
perturbation causes a fluctuation in the position of the
electron, which explains the energy shift. The difference of
potential energy is given by :\Delta V = V(\vec{r}+\delta \vec{r})-V(\vec{r})=\delta \vec{r} \cdot \nabla V (\vec{r}) + \frac{1}{2} (\delta \vec{r} \cdot \nabla)^2V(\vec{r})+\cdots Since the fluctuations are
isotropic, :\langle \delta \vec{r} \rangle _{\rm vac} =0, :\langle (\delta \vec{r} \cdot \nabla )^2 \rangle _{\rm vac} = \frac{1}{3} \langle (\delta \vec{r})^2\rangle _{\rm vac} \nabla ^2. So one can obtain :\langle \Delta V\rangle =\frac{1}{6} \langle (\delta \vec{r})^2\rangle _{\rm vac}\left\langle \nabla ^2\left(\frac{-e^2}{4\pi \epsilon _0r}\right)\right\rangle _{\rm at}. The classical
equation of motion for the electron displacement (
δr) induced by a single mode of the field of
wave vector and
frequency ν is :m\frac{d^2}{dt^2} (\delta r)_{\vec{k}}=-eE_{\vec{k}}, and this is valid only when the
frequency ν is greater than
ν0 in the Bohr orbit, \nu > \pi c/a_0. The electron is unable to respond to the fluctuating field if the fluctuations are smaller than the natural orbital frequency in the atom. For the field oscillating at
ν, :\delta r(t)\cong \delta r(0)(e^{-i\nu t}+e^{i\nu t}), therefore :(\delta r)_{\vec{k}} \cong \frac{e}{mc^2k^2} E_{\vec{k}}=\frac{e}{mc^2k^2} \mathcal{E} _{\vec{k}} \left (a_{\vec{k}}e^{-i\nu t+i\vec{k}\cdot \vec{r}}+h.c. \right) \qquad \text{with} \qquad \mathcal{E} _{\vec{k}}=\left(\frac{\hbar ck/2}{\epsilon _0 \Omega}\right)^{1/2}, where \Omega is some large normalization volume (the volume of the hypothetical "box" containing the hydrogen atom), and h.c. denotes the hermitian conjugate of the preceding term. By the summation over all \vec{k}, :\begin{align} \langle (\delta \vec{r} )^2\rangle _{\rm vac} &=\sum_{\vec{k}} \left(\frac{e}{mc^2k^2} \right)^2 \left\langle 0\left |(E_{\vec{k}})^2 \right |0 \right \rangle \\ &=\sum_{\vec{k}} \left(\frac{e}{mc^2k^2} \right)^2\left(\frac{\hbar ck}{2\epsilon _0 \Omega} \right) \\ &=2\frac{\Omega}{(2\pi )^3}4\pi \int dkk^2\left(\frac{e}{mc^2k^2} \right)^2\left(\frac{\hbar ck}{2\epsilon_0 \Omega}\right) && \text{since continuity of } \vec{k} \text{ implies } \sum_{\vec{k}} \to 2 \frac{\Omega}{(2\pi)^3} \int d^3 k \\ &=\frac{1}{2\epsilon_0\pi^2}\left(\frac{e^2}{\hbar c}\right)\left(\frac{\hbar}{mc}\right)^2\int \frac{dk}{k} \end{align} This integral diverges as the wave number approaches zero or infinity. As mentioned above, this method is expected to be valid only when \nu > \pi c/a_0, or equivalently k > \pi/a_0. It is also valid only for wavelengths longer than the
Compton wavelength, or equivalently k . Therefore, one can choose the upper and lower limit of the integral and these limits make the result converge. :\langle(\delta\vec{r})^2\rangle_{\rm vac}\cong\frac{1}{2\epsilon_0\pi^2}\left(\frac{e^2}{\hbar c}\right)\left(\frac{\hbar}{mc}\right)^2\ln\frac{4\epsilon_0\hbar c}{e^2}. For the
atomic orbital and the
Coulomb potential, :\left\langle\nabla^2\left(\frac{-e^2}{4\pi\epsilon_0r}\right)\right\rangle_{\rm at}=\frac{-e^2}{4\pi\epsilon_0}\int d\vec{r}\psi^*(\vec{r})\nabla^2\left(\frac{1}{r}\right)\psi(\vec{r})=\frac{e^2}{\epsilon_0}|\psi(0)|^2, since it is known that :\nabla^2\left(\frac{1}{r}\right)=-4\pi\delta(\vec{r}). For
p orbitals, the nonrelativistic
wave function vanishes at the origin (at the nucleus), so there is no energy shift. But for
s orbitals there is some finite value at the origin, :\psi_{2S}(0)=\frac{1}{(8\pi a_0^3)^{1/2}}, where the
Bohr radius is :a_0=\frac{4\pi\epsilon_0\hbar^2}{me^2}. Therefore, :\left\langle\nabla^2\left(\frac{-e^2}{4\pi\epsilon_0r}\right)\right\rangle_{\rm at}=\frac{e^2}{\epsilon_0}|\psi_{2S}(0)|^2=\frac{e^2}{8\pi\epsilon_0a_0^3}. Finally, the difference of the potential energy becomes: :\langle\Delta V\rangle=\frac{4}{3}\frac{e^2}{4\pi\epsilon_0}\frac{e^2}{4\pi\epsilon_0\hbar c}\left(\frac{\hbar}{mc}\right)^2\frac{1}{8\pi a_0^3}\ln\frac{4\epsilon_0\hbar c}{e^2} = \alpha^5 mc^2 \frac{1}{6\pi} \ln\frac{1}{\pi\alpha}, where \alpha is the
fine-structure constant. This shift is about 500 MHz, within an order of magnitude of the observed shift of 1057 MHz. This is equal to an energy of only . Welton's heuristic derivation of the Lamb shift is similar to, but distinct from, the calculation of the
Darwin term using
Zitterbewegung, a contribution to the
fine structure that is of lower order in \alpha than the Lamb shift. == Lamb–Retherford experiment ==