A Raman-delayed nonlinearity for elliptically polarized ultrashort optical pulses

Abstract

We address the role of the delayed Kerr nonlinearity associated with stimulated rotational Raman scattering on the propagation of intense ultrashort two-color filaments with elliptical polarization in gases and examine its impact on the generation of optical frequencies as well as terahertz frequencies. We find that in air the nonlinearity associated with stimulated Raman scattering strongly impacts the dynamics of the laser filaments and modifies the overall pulse spectral content. In particular, it alters the THz spectra upon propagation in the filamentation regime. Differences between linearly and circularly polarized two-color pump pulses are discussed.

Appendix

1.1 The microscopic Raman response

Here we derive the nonlinear polarization vector due to Raman scattering in air molecules, assumed to be only due to the rotations of non-interacting dinitrogen molecules. We shall suppose that $N_2$ molecules have only three energy levels $W_1$, $W_2$ and $W_3$. The first two levels are rotational states attainable by Stockes scattering, and the third is a virtual level of energy much higher than the other two, such that it cannot be directly excited by the laser, i.e., $W_3 \gg W_2 - W_1$. We note $\Omega _1$, $\Omega _2$ and $\Omega _3$ the respective natural frequencies of these states and $\Omega _{nm}=\Omega _n-\Omega _m$. In view of the above hypotheses, one has $\Omega _{21} \ll \omega _0 \ll \Omega _{32}$.

Let us define $\psi (t,{\overrightarrow {r}})$ the molecular wave function of dinitrogen. We also note $H_0$ the unperturbed Hamiltonian and $V(t,{\overrightarrow { r}}) = -{\overrightarrow {\mu }}\cdot {\overrightarrow {E}} = -\mu _xE_x - \mu _yE_y = V_x + V_y$ the perturbation operator of the Hamiltonian, with ${\overrightarrow {\mu }} = -e{\overrightarrow {r}}$ the dipole moment vector of $N_2$ molecules. The Schrödinger equation is therefore written:

$$\begin{aligned} (H_0 + V_x + V_y) \psi = i\hbar \frac{\partial \psi }{\partial t}. \end{aligned}$$

(15)

The molecular wave function $\psi$ is decomposed into the sum of eigenstates of $H_0$, $u_n({\overrightarrow {r}})$, where n is an integer between 1 and 3, $\hbar \Omega _n$ the associated eigenvalues, and $\langle u_m |u_n \rangle = \delta _{mn}$, with $\delta _{mn}$ being the Kronecker symbol. Following [23], we substitute

$$\begin{aligned} \psi (t,{\overrightarrow {r}}) = \sum _{n=1}^3 c_n(t)e^{-i\Omega _nt} u_n({\overrightarrow {r}}) \end{aligned}$$

(16)

into Eq. (15) and treat separately each of its contributions defining $({\overrightarrow {\mu} } \cdot {\overrightarrow {E}})_{mn} \equiv \mu _{mn}^x E_x + \mu _{mn}^y E_y$ with $\mu _{mn}^{x,y} = \langle u_m |\mu _{x,y} |u_n\rangle$. The time dependent coefficients $c_n$ must then satisfy

$$\begin{aligned} \frac{\partial c_m(t)}{\partial t} = \frac{i}{\hbar }\sum _{n=1}^3 c_n(t) ({\overrightarrow {\mu }} \cdot {\overrightarrow {E}})_{mn} e^{-i\Omega _{nm}t}\,. \end{aligned}$$

(17)

The microscopic polarization vector ${\overrightarrow {p}}$ is defined as

$$\begin{aligned} \begin{aligned} {\overrightarrow {p}}(t)&= \langle \psi |{\overrightarrow {\mu }} |\psi \rangle \\&= \sum _{m=1}^3 \sum _{n=1}^3 \rho _{nm} e^{i\Omega _{mn}t}\left( {\begin{array}{c}\mu ^x_{mn}\\ \mu ^y_{mn}\end{array}}\right) \,, \end{aligned} \end{aligned}$$

(18)

with $\rho _{mn} = c_m c_n^*$. By noticing that $\rho _{nm}^* = \rho _{mn}$, $\mu _{mn}^{x,y} = (\mu _{nm}^{x,y})^*$ and $({\overrightarrow {\mu }} \cdot {\overrightarrow {E}})_{mn} = ({\overrightarrow {\mu }} \cdot {\overrightarrow {E}})_{nm}^*$, one gets

$$\begin{aligned} \begin{aligned} \frac{\partial \rho _{nm}}{\partial t} = \frac{i}{\hbar } \sum _{j=1}^3&\left[ \rho _{jm}({\overrightarrow {\mu }} \cdot {\overrightarrow {E}})_{nj}e^{-i\Omega _{jn}t} \right. \\&\left. - \rho _{nj}({\overrightarrow {\mu }} \cdot {\overrightarrow {E}})_{jm}e^{-i\Omega _{mj}t}\right] \,. \end{aligned} \end{aligned}$$

(19)

Similar assumptions as in [23], among which $\rho _{33} = \mu ^{x,y}_{12} = \mu ^{x,y}_{ii} = 0$ ($1 \le i \le 3$), imply that the non-zero $\mu ^x_{mn}$ (resp. $\mu ^y_{mn}$) are all equal and assimilate all the related coefficients $({\overrightarrow {\mu }} \cdot {\overrightarrow {E}})_{mn}$ to $({\overrightarrow {\mu }} \cdot {\overrightarrow {E}})$.

Let us now calculate the elements participating in the macroscopic dipole moment. Using the relation Eq. (18), we obtain

$$\begin{aligned} {\overrightarrow {p}}(t) = (\rho _{13}e^{i\Omega _{31}t} + \rho _{32}e^{-i\Omega _{32}t}){\overrightarrow {\mu }} + c.c. \end{aligned}$$

(20)

where $\rho _{13}$ and $\rho _{32}$ are governed by Eq. (19).

To integrate the two equations for $\rho _{13}$ and $\rho _{32}$, we neglect the time variations of $\rho _{11}$, $\rho _{12}$, $\rho _{21}$ and $\rho _{22}$ compared to those of $\rho _{13}$ and $\rho _{32}$, because $\Omega _{31} \gg \Omega _{21}$ and $\Omega _{31} \sim \Omega _{32}$. With initially $\rho _{13}(t=0) = \rho _{32}(t=0) = 0$, this integration results in

$$\begin{aligned} \rho _{13}(t)&= \frac{({\overrightarrow {\mu }} \cdot {\overrightarrow {E}})}{\hbar \Omega _{31}}\left( \rho _{11}e^{-i\Omega _{31}t} + \rho _{12}e^{-i\Omega _{32}t}\right) \,, \end{aligned}$$

(21)

$$\begin{aligned} \rho _{32}(t)&= \frac{({\overrightarrow {\mu }} \cdot {\overrightarrow {E}})}{\hbar \Omega _{31}}\left( \rho _{12}e^{i\Omega _{31}t} + \rho _{22}e^{i\Omega _{32}t}\right) \,. \end{aligned}$$

(22)

Since $\rho _{11} + \rho _{22} \simeq 1$ and state 2 remains sparsely populated compared to state 1, we can also assume $\rho _{11} - \rho _{22} \simeq 1$. $\rho _{12}$ is thus given by Eq. (19) as

$$\begin{aligned} \frac{\partial \rho _{12}}{\partial t} = -\frac{i ({\overrightarrow {\mu }} \cdot {\overrightarrow {E}})^2}{\hbar ^2\Omega _{31}}e^{-i\Omega _{21}t}\,. \end{aligned}$$

(23)

Following [23], we add an heuristic population decrease of $\rho _{12}$, related to a decay term with lifetime $\tau _2$. Introducing $\tau _1 = \Omega _{21}^{-1}$, we get

$$\begin{aligned} \frac{\partial \rho _{12}}{\partial t} = -\frac{\rho _{12}}{\tau _2} - \frac{i ({\overrightarrow {\mu }} \cdot {\overrightarrow {E}})^2}{\hbar ^2 \Omega _{31}}e^{-i\frac{t}{\tau _1}}\,. \end{aligned}$$

(24)

Using these expressions, we can simplify the nonlinear part of the polarization vector oriented along ${\hat{\mu }}={\overrightarrow {\mu }}/\mu$ to

$$\begin{aligned} {\overrightarrow {p}}_R(t) = \frac{2 \mu ^2}{\hbar \Omega _{31}} ({\hat{\mu }} \cdot {\overrightarrow {E}})\rho _{12}e^{i\Omega _{21}t}{\hat{\mu }} + c.c.\,, \end{aligned}$$

(25)

and finally obtain after solving Eq. (24)

$$\begin{aligned}&{\overrightarrow {p}}_R(t) = \frac{1}{\hbar } \left( \frac{2 \mu ^2 }{\hbar \Omega _{31}}\right) ^2 [{\hat{\mu }} \cdot {\overrightarrow {E}}(t)] \nonumber \\&\quad \times \int ^t_{-\infty } [{\hat{\mu }} \cdot {\overrightarrow {E}}(t')]^2 \sin \!\left( \frac{t-t'}{\tau _1}\right) e^{\frac{t'-t}{\tau _2}} \ dt' {\hat{\mu }}\,. \end{aligned}$$

(26)

Finally let $\xi$ be the angle between ${\overrightarrow {\mu }}$ and ${\overrightarrow { e}}_x$. To obtain a macroscopic expression of the Raman polarization, it is necessary to average the above relation over all angles $\xi$ and multiply the expression by the molecular density of the medium $N_a$. For isotropic air, the angle $\xi$ is distributed uniformly between 0 and $2\pi$. The macroscopic Raman polarization is thus given by

$$\begin{aligned} {\overrightarrow {P}}_R(t) = \frac{N_a}{2\pi } \int _0^{2\pi } {\overrightarrow {p}}_R(t, \xi ) d\xi \,. \end{aligned}$$

(27)

After integration, we obtain the final form of the Raman polarization Eq. (5) for a vector laser field. Note that if we set $E_y = 0$, we recover the standard one-dimensional expression of [23] for a linearly polarized laser pulse.