Measurement of the electric polarizability of lithium by atom interferometry

We have built an atom interferometer and, by applying an electric field on one of the two interfering beams, we have measured the static electric polarizability of lithium $\alpha =(24.33\pm 0.16)\times {10}^{-30}{\mathrm{m}}^3$ with a 0.66% uncertainty. Our experiment is similar to an experiment done on sodium in 1995 by Pritchard and co-workers, with several improvements: the electric field can be calculated analytically and the interference signals have a large intensity and a high visibility, resulting in accurate phase measurements. This experiment illustrates the extreme sensitivity of atom interferometry: when the atom enters the electric field, its velocity increases and the fractional change, equal to $4\times {10}^{-9}$ for our largest field, is measured with a ${10}^{-3}$ accuracy.

Figure 1

Schematic drawing of our Mach-Zehnder atom interferometer: a collimated atomic beam, coming from the left, is diffracted by three laser standing waves and the output beam 1 selected by a slit is detected by a hot-wire detector $D$. The capacitor with a septum is placed just before the second laser standing wave. The $x$, $y$, and $z$ axis are defined.

Figure 2

Schematic drawing of the capacitor. The septum is parallel to the $z$ axis and the electrodes are located at $x=\pm h\approx 2\mathrm{mm}$. The high voltage electrodes at the potential $V_0$ extends from $z=-a$ to $z=+a$, while the guard electrodes extend outside with $\mid z\mid >a$, with $a\approx 25\mathrm{mm}$. The septum and the guard electrodes are at $V=0$.

Figure 3

Experimental signals and their fits (full curves) corresponding to $V=0$ (black dots) and $V_0\approx 260\mathrm{V}$ (gray dots): the phase shift is close to $3\pi$ with a reduced visibility.

Figure 4

Phase shift $⟨\varphi \left(V_0\right)⟩$ as a function of the applied voltage $V_0$: the best fit using Eqs. (4, 5) is represented by the full curve and the residuals are plotted in the lower graph.

Figure 5

Relative fringe visibility $⟨\mathcal{V}⟩∕{\mathcal{V}}_0$ (with ${\mathcal{V}}_0=62\%$) as a function of the applied voltage $V_0$ and the best fit using Eqs. (4, 5) (full curve).