χ$_{i}$ ( t < 0) = 0. The Fourier transform of Eq. 3 for a purely ac field E ( t ) = E cos( ωt ) gives: where the polarization P and the susceptibilities χ$_{i}$ are now taken in the frequency domain and the dots indicate again infinite sums involving higher order terms. The response P ( t ) to E ( t ) = E cos( ωt ) can thus be written To obtain Eq. (5), we have used the fact that because χ$_{1}$ and χ$_{3}$ are real in the time domain, their Fourier transform verify χ ∗ $_{1}$( ω ) = χ$_{1}$ ( − ω ) and χ ∗ $_{3}$( ω$_{1}$, ω$_{2}$, ω$_{3}$ ) = χ$_{3}$ ( − ω$_{1}$, − ω$_{2}$, − ω$_{3}$ ) (the star denotes the complex conjugate), and the invariance of χ$_{3}$ by permutation of its argu- ments. For simplicity, we write χ$_{3}$ ( ω ) = χ$_{3}$ ( ω, ω, ω ) and χ ¯ $_{3}$( ω ) = χ$_{3}$ ( − ω, ω, ω ). Eq. (5) can be written where the susceptibilities χ$_{i}$ are given as a function of their real and imaginary parts χ ′ i and χ ′′ $_{i}$. For practical applications, the modulii and arguments | χ$_{i}$ | and δ$_{i}$ are rather used: We see in the rhs in Eqs 5-7 that the nonlinear susceptibility of interest, namely χ$_{3}$ 13 P ( t ) /ϵ$_{0}$ = E ( χ ′ $_{1}$cos ωt + χ ′′ 1 sin ωt ) + 3 / 4 E $^{3}$( χ ′ ¯ $_{3}$cos ωt + χ ′′ ¯ 3 sin ωt ) + ... + 1 / 4 E $^{3}$( χ ′ $_{3}$cos 3 ωt + χ ′′ 3 sin 3 ωt ) + ..., (6) P ( t ) /ϵ$_{0}$ = E | χ$_{1}$ | cos( ωt − δ$_{1}$ ) + 3 / 4 E 3 | χ ¯ $_{3}$| cos( ωt − δ ¯ $_{3}$)+ + ... + 1 / 4 E 3 | χ$_{3}$ | cos(3 ωt − δ$_{3}$ ) + ... (7) P ( ω ′ ) ϵ$_{0}$ = E 2 [ χ$_{1}$ ( ω ) + 3 E 2 4 χ$_{3}$ ( − ω, ω, ω ) + ... ] δ ( ω ′ − ω ) + E 2 [ χ$_{1}$ ( − ω ) + 3 E 2 4 χ$_{3}$ ( ω, − ω, − ω ) + ... ] δ ( ω ′ + ω ) + E 3 8 χ$_{3}$ ( ω, ω, ω ) δ ( ω ′ − 3 ω ) + E 3 8 χ$_{3}$ ( − ω, − ω, − ω ) δ ( ω ′ + 3 ω ) + ..., (4) P ( t ) /ϵ$_{0}$ = Re [ ( Eχ$_{1}$ ( ω ) + 3 / 4 E $^{3}$χ ¯ $_{3}$( ω ) + ... ) e − $^{iωt}$] + Re [ 1 / 4 E $^{3}$χ$_{3}$ ( ω ) e − i 3 ωt + ... ] + .... (5)