Plasma Luminescence from Femtosecond Filaments in Air: Evidence for Impact Excitation with Circularly Polarized Light Pulses

Filaments produced in air by intense femtosecond laser pulses emit UV luminescence from excited ${\mathrm{N}}_2$ and ${\mathrm{N}}_2^{+}$ molecules. We report on a strong dependence at high intensities ($I\ge 1.4\times {10}^{14}\mathrm{W}/{\mathrm{cm}}^2$) of this luminescence with the polarization state of the incident laser pulses. We attribute this effect to the onset of new impact excitation channels from energetic electrons produced with circularly polarized laser pulses above a threshold laser intensity.

Figure 1

Luminescence spectrum of the air plasma filaments for linearly and circularly polarized laser pulses of $250\mu \mathrm{J}$ (a) and 8.3 mJ (b). In (a), the signal obtained with circular laser polarization is multiplied by a factor of 2 for visibility. The vibration quantum numbers related to the second positive system of the ${\mathrm{N}}_2$ and the first negative system of ${\mathrm{N}}_2^{+}$ are denoted.

Figure 2

Luminescence at 337 nm [(a)–(d)] and 391 nm [(a′)–(d′)] as a function of the rotation angle of the quarter-wave plate. The incident pulse energy was $250\mu \mathrm{J}$, $600\mu \mathrm{J}$, 2.7 mJ, and 10 mJ for (a) and (a′), (b) and (b′), (c) and (c′), (d) and (d′), respectively. All the measurements were performed around the center of the filament. Angle 0° corresponds to linearly polarized light.

Figure 3

Calculated electron energy distribution in the case of linearly (a), elliptically with $ϵ=1/2$ (b), and circularly (c) polarized laser pulses. The laser intensity used in the calculation was $1.4\times {10}^{14}\mathrm{W}/{\mathrm{cm}}^2$. The red lines represent analytic results and the bars numerical simulation.

Figure 4

Luminescence at 337 nm as a function of the rotation angle of the quarter-wave plate for air pressure of 10 mbar. The incident pulse energy was $350\mu \mathrm{J}$ and the calculated laser intensity is $1.45\times {10}^{14}\mathrm{W}/{\mathrm{cm}}^2$.