## Time response of nonlinear optical processes We now consider the role that time plays in nonlinear optical processes and does not consider the response as instantaneous, and try to describe the nonlinear polarization in time domain. Time domain description is easier for some problems, like pulse lasers, although in principal being the same as the frequency description. >[!Notice]- >Under time domain, we no longer consider electric field $\tilde{E}$ as a superposition of ${E}(\omega)$ in different frequencies, but a single variable, real value function of time $t$. >Or >$\tilde{E}(t)=\frac{1}{2\pi} \int_{-\infty}^{+\infty} {E}(\omega) e^{-i \omega t}\mathrm{d}\omega$ >${E}(\omega)$ does not vary with time, since it's in the Fourier expansion. And back to frequency domain, >$E(\omega) = \int_{-\infty}^{\infty} \tilde{E}(t) e^{i \omega t} \, \mathrm{d}t$ >They are just Fourier transform. Consider the first order response of polarization. Assume we have a liner response function $R$, then the polarization and electric field follows $\tilde{{P}}^{(1)}(t) = \varepsilon_0 \int_0^{\infty} R^{(1)}(\tau) \tilde{{E}}(t - \tau) \, \mathrm{d}\tau$ It's a convolution of the linear response function and electric field. This means the response function gives the contribution to polarization at time $t$ **by** an electric field at earlier time $t-\tau$. The integration "sum" all past contributions together (as convolution suggested). The lower cap is zero instead of $-\infty$, indicate $R^{(1)}(\tau)=0$ for $\tau<0$, the response function obeys the causality condition. Polarization $P(t)$ only depends on past value, not future value. >[!Note] >If $R^{(1)}(\tau)\neq0$ for $\tau<0$, we will have contributions from $\tilde{{E}}(t - \tau)$ and $t - \tau>t$, namely future. This breaks the causality. If we compare expressions from time domain after transformation and frequency domain, we will see that $\tilde{P}^{(1)}(t) = \varepsilon_0 \int_0^{\infty} \mathrm{d}\tau \int_{-\infty}^{\infty} \frac{\mathrm{d}\omega}{2\pi} R^{(1)}(\tau) E(\omega) e^{-i\omega(t-\tau)}= \varepsilon_0 \int_{-\infty}^{\infty} \frac{\mathrm{d}\omega}{2\pi} \colorbox{#a5d8ff66}{\(\int_0^{\infty} \mathrm{d}\tau \, R^{(1)}(\tau) e^{i\omega\tau}\)} E(\omega) e^{-i\omega t}$ Inside the $\colorbox{#a5d8ff66}{blue box}$, we can see it's a Fourier transform like (it's not Fourier transform cause the lower limit is not $-\infty$) of the linear response function, and this quantity times the electric fields in frequency domain, and after another Fourier transform, we get the first order polarization. Cause we know the constitutive relation $P^{(1)}(\omega) = \varepsilon_0 \chi^{(1)}(\omega) E(\omega)$ will become the form we have above in time domain after Fourier transform, namely $\tilde{P}^{(1)}(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} P^{(1)}(\omega) e^{-i \omega t} \mathrm{d}\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} \varepsilon_0 \colorbox{#a5d8ff66}{\(\chi^{(1)}(\omega;\omega)\)}(\omega; \omega) E(\omega) e^{-i \omega t} \mathrm{d}\omega= \varepsilon_0 \int_{-\infty}^{\infty} \frac{\mathrm{d}\omega}{2\pi} \colorbox{#a5d8ff66}{\(\int_0^{\infty} \mathrm{d}\tau \, R^{(1)}(\tau) e^{i\omega\tau}\)} E(\omega) e^{-i\omega t}=\tilde{P}^{(1)}(t)$ So we link the linear susceptibility with the linear response function. $\chi^{(1)}(\omega; \omega) = \int_0^{\infty} \mathrm{d}\tau \, R^{(1)}(\tau) e^{i \omega \tau}$ >[!Notice] >$R^{(1)}$ must be real because it relates two real quantity, $\tilde{E}(t)$ and $\tilde{P}(t)$ by >$\tilde{{P}}^{(1)}(t) = \varepsilon_0 \int_0^{\infty} R^{(1)}(\tau) \tilde{{E}}(t - \tau) \, \mathrm{d}\tau$ >This is the same for higher order cases. Therefore, we can convert the negative frequencies to positive one by taking complex conjugate, namely >$\chi^{(1)}(-\omega)=\chi^{(1)}(\omega)^{*}$ >considering its definition $\chi^{(1)}(\omega; \omega) = \int_0^{\infty} \mathrm{d}\tau \, R^{(1)}(\tau) e^{i \omega \tau}$. This is shown in [[Frequency domain and time domain#^7b97a0]]. Similarly, we can define the second-order response function and polarization, $\tilde{P}^{(2)}(t) = \epsilon_0 \int_0^\infty \mathrm{d}\tau_1 \int_0^\infty \mathrm{d}\tau_2 \, R^{(2)}(\tau_1, \tau_2) E(t - \tau_1) E(t - \tau_2)$The causality condition is $R^{(2)}(\tau_1, \tau_2)$ if $\tau_{1}<0$ or $\tau_{2}<0$. And by the same method, we can write the input fields in frequency domain, $\tilde{P}^{(2)}(t) = \epsilon_0 \int_{-\infty}^{\infty} \frac{\mathrm{d}\omega_1}{2\pi} \int_{-\infty}^{\infty} \frac{\mathrm{d}\omega_2}{2\pi} \int_0^\infty \mathrm{d}\tau_1 \int_0^\infty \mathrm{d}\tau_2 \, R^{(2)}(\tau_1, \tau_2) E(\omega_1) e^{-i\omega_1 (t - \tau_1)} E(\omega_2) e^{-i\omega_2 (t - \tau_2)}= \epsilon_0 \int_{-\infty}^{\infty} \frac{\mathrm{d}\omega_1}{2\pi} \int_{-\infty}^{\infty} \frac{\mathrm{d}\omega_2}{2\pi} \, \chi^{(2)}(\omega_3 ; \omega_1, \omega_2) E(\omega_1) E(\omega_2) e^{-i \omega_3 t}$ to get the expression of $\chi^{(2)}$, here we assume $\omega_{3}=\omega_{1}+\omega_{2}$, $\chi^{(2)}(\omega_3 ; \omega_1, \omega_2) = \int_0^\infty \mathrm{d}\tau_1 \int_0^\infty \mathrm{d}\tau_2 \, R^{(2)}(\tau_1, \tau_2) e^{i(\omega_1 \tau_1 + \omega_2 \tau_2)}$ Also the third order or higher order. Ensuring causality requires $R^{(n)}=0$ if any $\tau$ less than zero. $\tilde{P}^{(3)}(t) = \epsilon_0 \int_{-\infty}^{\infty} \frac{\mathrm{d}\omega_1}{2\pi} \int_{-\infty}^{\infty} \frac{\mathrm{d}\omega_2}{2\pi} \int_{-\infty}^{\infty} \frac{\mathrm{d}\omega_3}{2\pi} \, \chi^{(3)}(\omega_4 ; \omega_1, \omega_2, \omega_3) E(\omega_1) E(\omega_2) E(\omega_3) e^{-i\omega_4 t}$ $\chi^{(3)}(\omega_4 ; \omega_1, \omega_2, \omega_3) = \int_0^\infty \mathrm{d}\tau_1 \int_0^\infty \mathrm{d}\tau_2 \int_0^\infty \mathrm{d}\tau_3 \, R^{(3)}(\tau_1, \tau_2, \tau_3) e^{i(\omega_1 \tau_1 + \omega_2 \tau_2 + \omega_3 \tau_3)}$ >[!Notice] >Such linear response is widely applied. One of our old friend, the Green function also belongs to such linear response (by its own nature). And with adding the self energy using Dyson equation, we have NEGF and may deal with nonlinear, strong correlated problems. Also, [[Kramers–Kronig relations|KK relation]] also works for Green function. > >See https://eduardo.physics.illinois.edu/phys582/LRT.pdf for more information. [[LRT.pdf|Local file]]