# All Science Fair Projects

## Science Fair Project Encyclopedia for Schools!

 Search    Browse    Forum  Coach    Links    Editor    Help    Tell-a-Friend    Encyclopedia    Dictionary

# Science Fair Project Encyclopedia

For information on any area of science that interests you,
enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.).
Or else, you can start by choosing any of the categories below.

# Lamb shift

In physics, the Lamb shift, named after Willis Lamb, is a small difference of energy between two energy levels 2s1 / 2 and 2p1 / 2 of the hydrogen atom in quantum mechanics. According to the Schrödinger equation these two energy levels should only depend on the principal quantum number and should therefore have the same energy.

In 1947 Lamb and Retherford carried out an experiment using microwave techniques to stimulate radio-frequency transitions between 2s1 / 2 and 2p1 / 2 levels. 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 1060MHz of the 2s1 / 2 level above the 2p1 / 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 (QED) the electromagnetic field is quantised and as for the harmonic oscillator in quantum mechanics its lowest state is not zero. So there exist little zero-point oscillations that cause the electron to execute rapid oscillatory motions. The electron is kind of "smeared out" and the radius is changed by r + δr.

The Coulomb potential is therefore perturbed by a small amount and the degeneration of the two energy levels is removed. The new potential can be approximated (using atomar 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

$\Delta E_\mathrm{Lamb}=\alpha^5 m_e c^2 \frac{k(n,0)}{4n^3}\ \mathrm{for}\ \ell=0\,$

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 (< 0.05).

03-10-2013 05:06:04