Electron-helium scattering in a 1.17 eV laser field: The effect of polarization direction

We report measurements of one-photon emission during the elastic scattering of electrons by He atoms through ${90}^{\circ }$ in the presence of 1.17 eV photons from a Nd:YAG laser. The incident energy of the electrons was in the range 30–200 eV and the linear polarization direction of the laser was varied over ${180}^{\circ }$ in the plane, perpendicular to the scattering plane, that contains the momentum transfer direction. Our results are perfectly consistent with the Kroll-Watson approximation. In particular, we see no evidence of free-free transitions when the polarization is perpendicular to the momentum transfer direction.

Figure 1

The form of the free-free signal for one-photon emission ($n=1$) as a function of the angle $\theta$ between the laser polarization and the momentum transfer direction. The dashed line is a KWA calculation given by Eq. (2) with $x=cos\theta$. The solid line is the ${cos}^2\theta$ approximation given by Eq. (6).

Figure 2

The experimental layout for the present experiments. A: Nd:YAG laser, ${\mathrm{B}}_1:\lambda /2$ plate, ${\mathrm{B}}_2$: polarizing beam splitter cube, C: $\lambda /2$ plate and computer controlled rotatable mount, D: polarizing beam splitter cube, E: lens, F: He nozzle, G: scattered-electron detector, H: electron gun, J: vacuum chamber, K: beam dump, L: power meter (for setup). The direction of the momentum transfer $\widehat{Q}$ is indicated.

Figure 3

The free-free signal as a function of the angle $\theta$ between the laser polarization and the momentum transfer direction for incident-electron energies 30, 60, 120, and 200 eV, and the sum of all the results. The data have been scaled to give the best fits to ${cos}^2\theta$ (solid lines).

Figure 4

Schematic of an electron-atom collision in an oscillating electric field. ${\mathit{k}}_1$ is the electron momentum at the beginning of an oscillatory cycle. ${\mathit{k}}_2$ and ${\mathit{k}}_3$ are the momenta just before and after the collision at time $t_c$. ${\mathit{k}}_4$ is the momentum at the end of the cycle. See text for details.

Figure 5

Scattering kinematics showing the scattering plane ($xy$ plane) and the plane perpendicular to the momentum transfer $\mathit{Q}$ ($xz$ plane) that bisects the scattering angle formed by ${\mathit{k}}_1$ and ${\mathit{k}}_4$. For any polarization vector that lies in this plane ${\widehat{\mathbf{\epsilon }}·\widehat{\mathbf{k}}}_1={\widehat{\mathbf{\epsilon }}·\widehat{\mathbf{k}}}_4$.