Determination of the Polarizability Tensors of the Magnetic Substates ofP23Metastable Argon

Abstract

The atomic beam $E\ensuremath{-}H$ gradient-balance method has been employed to measure the $z$-component ${\ensuremath{\alpha}}_{z}({m}_{J})$ of the electric polarizability tensors of the ${m}_{J}=+2$ and ${m}_{J}=+1$ magnetic substates of $^{3}P_{2}$ metastable argon. The measurements were normalized to the polarizability of $^{3}S_{1}$ metastable helium, which is calculable to within an accuracy of 1%. In this method, a metastable beam is passed through a region containing transverse, congruent, inhomogeneous electric and magnetic fields. The condition that atoms in a particular magnetic substate experience no deflection in such a field region is ${\ensuremath{\alpha}}_{z}({m}_{J})E(\frac{\ensuremath{\delta}E}{\ensuremath{\delta}z})={\ensuremath{\mu}}_{\mathrm{eff}}(\frac{\ensuremath{\delta}H}{\ensuremath{\delta}z})$, where ${\ensuremath{\mu}}_{\mathrm{eff}}$ is the magnetic moment in the field $H$, and $\frac{\ensuremath{\delta}E}{\ensuremath{\delta}z}$, $\frac{\ensuremath{\delta}H}{\ensuremath{\delta}z}$ are the $z$ components of the gradients of the magnitudes of the electric and magnetic fields, respectively. Because of the congruency of the $E$ and $H$ fields $\frac{(\frac{\ensuremath{\delta}E}{\ensuremath{\delta}z})}{E}=\frac{(\frac{\ensuremath{\delta}H}{\ensuremath{\delta}z})}{H}$, and therefore ${\ensuremath{\alpha}}_{z}({m}_{J})=\frac{{\ensuremath{\mu}}_{\mathrm{eff}}H}{{E}^{2}}$. Because of the normalization, it was unnecessary to find either $E$ or $H$. The results, in units of ${10}^{\ensuremath{-}24}$ ${\mathrm{cm}}^{3}$, are ${\ensuremath{\alpha}}_{z}(\ifmmode\pm\else\textpm\fi{}1)=50.4\ifmmode\pm\else\textpm\fi{}3.5$ and ${\ensuremath{\alpha}}_{z}(\ifmmode\pm\else\textpm\fi{}2)=44.5\ifmmode\pm\else\textpm\fi{}3.1$. The measured values of ${\ensuremath{\alpha}}_{z}(+1)$ and ${\ensuremath{\alpha}}_{z}(\ifmmode\pm\else\textpm\fi{}2)$ can be used to determine the polarizability tensors of each of the magnetic substates. The other components of the diagonalized tensors, with ${\ensuremath{\alpha}}_{x}({m}_{J})={\ensuremath{\alpha}}_{y}({m}_{J})$, in units of ${10}^{\ensuremath{-}24}$${\mathrm{cm}}^{3}$ are ${\ensuremath{\alpha}}_{x}(0)=46.4\ifmmode\pm\else\textpm\fi{}2.4$, ${\ensuremath{\alpha}}_{x}(\ifmmode\pm\else\textpm\fi{}1)=47.2\ifmmode\pm\else\textpm\fi{}2.3$, ${\ensuremath{\alpha}}_{x}(\ifmmode\pm\else\textpm\fi{}2)=50.4\ifmmode\pm\else\textpm\fi{}3.5$, and ${\ensuremath{\alpha}}_{z}(0)=52.4\ifmmode\pm\else\textpm\fi{}4.8$.