  of energy levels in hydrogen - relativistic corrections to the Bohr model]]  In physics, the '''Lamb shift''', named after Willis Lamb (1913–2008), is a small difference in energy between two energy levels ^2S_{1/2} and ^2P_{1/2} (in term symbol notation) of the hydrogen atom in quantum electrodynamics (QED). According to Dirac, the ^2S_{1/2} and ^2P_{1/2} orbitals should have the same energies. However, the interaction between the electron and the vacuum causes a tiny energy shift on ^2S_{1/2}. Lamb and Robert Retherford measured this shift in 1947, and this measurement provided the stimulus for renormalization theory to handle the divergences. It was the harbinger of modern QED as developed by Julian Schwinger, Richard Feynman, and Shinichiro Tomonaga. Lamb won the Nobel Prize in Physics in 1955 for his discoveries related to the Lamb shift.  == Derivation == This heuristic derivation of the electrodynamic level shift following Welton is from ''Quantum Optics''.  The fluctuation in the electric and magnetic fields associated with the vacuum perturbs the Coulomb 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 + \frac{1}{2} (\delta \vec{r} \cdot \nabla)^2V(\vec{r})+... Since the fluctuations are isotropic, :\langle \delta \vec{r} \rangle _{vac} =0 :\langle (\delta \vec{r} \cdot \nabla )^2 \rangle _{vac} = \frac{1}{3} \langle (\delta \vec{r})^2\rangle _{vac} \nabla ^2. So we can obtain :\langle \Delta V\rangle =\frac{1}{6} \langle (\delta \vec{r})^2\rangle _{vac}\langle \nabla ^2\left(\frac{-e^2}{4\pi \epsilon _0r}\right)\rangle _{at}.  The classical equation of motion for the electron displacement (\delta r)_{\vec{k}} induced by a single mode of the field of wave vector \vec{k} and frequency \nu is :m\frac{d^2}{dt^2} (\delta r)_{\vec{k}}=-eE_{\vec{k}}, and this is valid only when the frequency \nu is greater than \nu _0 in the Bohr orbit, \nu &gt;\frac{\pi c}{a_0}.  For the field oscillating at \nu, :\delta r(t)\cong \delta r(0)e^{-i\nu t}+c.c., therefore :(\delta r)_{\vec{k}} \cong \frac{e}{mc^2k^2} E_{\vec{k}}=\frac{e}{mc^2k^2} \mathcal{E} _{\vec{k}}(a_{\vec{k}}e^{-i\nu t+i\vec{k}\cdot \vec{r}}+h.c.).  By the summation over all \vec{k}, :\langle (\delta \vec{r} )^2\rangle _{vac}=\sum_{\vec{k}} \left(\frac{e}{mc^2k^2} \right)^2\langle 0|(E_{\vec{k}})^2|0\rangle =\sum_{\vec{k}} \left(\frac{e}{mc^2k^2} \right)^2\left(\frac{\hbar ck}{2\epsilon _0V} \right)), where :\mathcal{E} _{\vec{k}}=(\hbar ck/2\epsilon _0V)^{1/2}.  The summation is changed into the integral because of the continuity of \vec{k}, so :\langle (\delta \vec{r} )^2\rangle _{vac}=2\frac{V}{(2\pi )^3}4\pi \int dkk^2\left(\frac{e}{mc^2k^2} \right)^2\left(\frac{\hbar ck}{2\epsilon_0V}\right)=\frac{1}{2\epsilon_0\pi^2}\left(\frac{e^2}{\hbar c}\right)\left(\frac{\hbar}{mc}\right)^2\int \frac{dk}{k}.  This result diverges when there is no limit about the integral. But this method is valid only when \nu &gt;\frac{\pi c}{a_0}, or equivalently k&gt;\frac{\pi}{a_0}. It is also valid only for wavelengths longer than the Compton wavelength, or equivalently k. Therefore we can choose the upper and lower limit of the integral and these limits make the result converge. :\langle(\delta\vec{r})^2\rangle_{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, :\langle\nabla^2\left(\frac{-e^2}{4\pi\epsilon_0r}\right)\rangle_{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 we know that :\nabla^2(\frac{1}{r})=-4\pi\delta(\vec{r}).  For p orbitals, the nonrelativistic wave function vanishes at the origin, 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 :\langle\nabla^2\left(\frac{-e^2}{4\pi\epsilon_0r}\right)\rangle_{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}. This shift is about 1&amp;nbsp;GHz, very similar with the observed energy shift.  == Experimental work == In 1947 Willis Lamb and Robert Retherford carried out an experiment using microwave techniques to stimulate radio-frequency transitions between ^2S_{1/2} and ^2P_{1/2} levels of hydrogen. By using lower frequencies than for optical transitions the Doppler broadening could be neglected (Doppler broadening is proportional to the frequency). The energy difference Lamb and Retherford found was a rise of about 1000&amp;nbsp;MHz of the ^2S_{1/2} level above the ^2P_{1/2} level.  This particular difference is a one-loop effect of quantum electrodynamics, and can be interpreted as the influence of virtual photons that have been emitted and re-absorbed by the atom. In quantum electrodynamics the electromagnetic field is quantized and, like the harmonic oscillator in quantum mechanics, its lowest state is not zero. Thus, there exist small zero-point oscillations that cause the electron to execute rapid oscillatory motions. The electron is &quot;smeared out&quot; and the radius is changed from r to r+\delta r.  The Coulomb potential is therefore perturbed by a small amount and the degeneracy of the two energy levels is removed. The new potential can be approximated (using atomic units) as follows:  :\langle E_\mathrm{pot} \rangle=-\frac{Ze^2}{4\pi\epsilon_0}\left\langle\frac{1}{r+\delta r}\right\rangle.  The Lamb shift itself is given by  :\Delta E_\mathrm{Lamb}=\alpha^5 m_e c^2 \frac{k(n,0)}{4n^3}\ \mathrm{for}\ \ell=0\,   with k(n,0) around 13 varying slightly with n, and  :\Delta E_\mathrm{Lamb}=\alpha^5 m_e c^2 \frac{1}{4n^3}\left[k(n,\ell)\pm \frac{1}{\pi(j+\frac{1}{2})(\ell+\frac{1}{2})}\right]\ \mathrm{for}\ \ell\ne 0\ \mathrm{and}\ j=\ell\pm\frac{1}{2},  with k(n,\ell) a small number (    