## Nonlinear wave equation ### Helmholtz equation for electromagnetic waves As shown [[Maxwell equation recap#^79091d|previously]], simply by Maxwell equations, we may obtain the wave equation for electric and magnetic field. In vacuum, we have $\nabla^2 \mathbf{E} = \mu_0\varepsilon_0\frac{\partial^2 \mathbf{E}}{\partial t^2}$ $\nabla^2 \mathbf{H} = \mu_0\varepsilon_0\frac{\partial^2 \mathbf{H}}{\partial t^2}$ Since it's vacuum, we do not have material dependence and free charges. If in material without free charges, we would only have to add the material dependence term $\epsilon_{r}$ ($\mu_{r}$ is a constant and equals to 1), this gives $\nabla^2 \mathbf{E} = \mu_0\varepsilon_0 \varepsilon_{r}\frac{\partial^2 \mathbf{E}}{\partial t^2}$ $\nabla^2 \mathbf{H} = \mu_0\varepsilon_0\varepsilon_{r}\frac{\partial^2 \mathbf{H}}{\partial t^2}$ Here material properties are included inside $\varepsilon_{r}$ term, so we do not explicitly write the polarization. If consider the nonlinearity, we may derived the expression from the original form of Maxwell equation in medium. With the assumption: - Materials are non-magnetic - No free charges inside the material, i,e., $\nabla \cdot D = \rho_{ext} = 0$. (In the note, [[NLO_Chapter 1.pdf#page=4&selection=151,1,172,4|NLO_Chapter 1, page 4]], this is defined as no free carriers, but if no free carrier, the conductivity should goes to zero, which does not make sense in our case since we do have some conductivity in the wave eq. But it is worth noting that there are indeed many cases free carriers are zero, like undoped semiconductors and isolators) - The medium is homogeneous, namely $\nabla \varepsilon =0$. Under these assumptions, we may write the divergence of $\mathbf{D}$, $\nabla \cdot (\varepsilon_{0}\mathbf{E}) = \nabla \cdot \mathbf{D} -\nabla \cdot \mathbf{P}$ $\nabla \cdot \mathbf{E} = -\nabla \cdot \frac{\mathbf{P}}{\varepsilon_{0}} $ >[!Note] >In linear material, we have a stronger result, $\nabla \cdot \mathbf{E} =0$. >If the material is linear, i.e., $\mathbf{P}=\mathbf{P_{L}}=\varepsilon \chi \mathbf{E}$, we have $\mathbf{D} = \varepsilon \mathbf{E} = \varepsilon_{0} \mathbf{E} + \mathbf{P_{L}}$, the divergence of $\mathbf{D}$ would be >$0=\nabla \cdot \mathbf{D} = \nabla \varepsilon \cdot \mathbf{E} + \varepsilon \cdot \nabla \cdot \mathbf{E}$ >$\nabla \cdot \mathbf{E} = -\frac{1}{ \varepsilon } \cdot \nabla \cdot \mathbf{E}$ >Since the material is homogeneous, $\nabla \varepsilon = 0$, so $\nabla \cdot \mathbf{E} =0$. And the nonlinear wave equation is derived using exactly the same trick we did for the vacuum case before, just replace $\mathbf{B}$ with $\mathbf{H}$ and $\mu_{0}\mu_{r}$. $\mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{E}) = - \frac{\partial }{\partial t} (\mathbf{\nabla} \times \mathbf{B})$ $\begin{align} \nabla^2 \mathbf{E} - \nabla(\nabla \cdot \mathbf{E}) &= \frac{\partial }{\partial t} (\mathbf{\nabla} \times \mathbf{B}) \\ \nabla^2 \mathbf{E} - \nabla(\nabla \cdot \mathbf{E}) &= \mu_0\mu_{r}\frac{\partial }{\partial t} (\mathbf{\nabla} \times \mathbf{H})\\ \nabla^2 \mathbf{E} - \nabla(\nabla \cdot \mathbf{E}) &= \mu_0\mu_{r}\frac{\partial }{\partial t} \left( \mathbf{J_{ext}} +\frac{\partial \mathbf{D}}{ \partial t } \right) \end{align} $ >[!Notice] >**In the following derivation** we will assume there is no net free charges inside the material, so we have $\nabla(\nabla \cdot \mathbf{E}) = 0$. >However, even if we have free charges, we can still write >$\nabla (\nabla \cdot \mathbf{E}) = \nabla \left( \frac{1}{\varepsilon} (\rho_{ext}-\nabla \varepsilon \cdot \mathbf{E} ) \right) = \nabla \left( \frac{1}{ \varepsilon} \rho_{ext} \right)$ >In some cases this gradient can also be considered as 0 (like even distributed charges). With the assumptions considered, we have $\nabla^2 \mathbf{E} = \mu_0\mu_{r}\frac{\partial }{\partial t} \left( \mathbf{J_{ext}} +\frac{\partial \mathbf{D}}{ \partial t } \right)$ write the electric displacement as $\mathbf{P}$ and $\mathbf{E}$, we have $\nabla^2 \mathbf{E} - \mu_0\mu_{r}\frac{\partial }{\partial t} \mathbf{J_{ext}} - \mu_0\mu_{r} \varepsilon_{0}\frac{\partial^{2} }{\partial t^{2}} \mathbf{E}=\mu_0\mu_{r}\frac{\partial^{2} }{\partial t^{2}} \mathbf{P} $ And if replace the external current term with conductivity, we have $\nabla^2 \mathbf{E} - \mu_0\mu_{r}\sigma\frac{\partial }{\partial t} \mathbf{E} - \mu_0\mu_{r} \varepsilon_{0}\frac{\partial^{2} }{\partial t^{2}} \mathbf{E}=\mu_0\mu_{r}\frac{\partial^{2} }{\partial t^{2}} \mathbf{P} $ >[!Notice] >Here we assume the conductivity is frequency independent. This is valid at lower frequency, but at high frequency (>THz), one have to write it as a frequency dependent term $\sigma(\omega)$, or change the form of wave equation. Now we have the general form of nonlinear wave equation. In most of the cases, our system is perfect dielectric and no external current is involved. In this case, we have $\nabla^2 \mathbf{E} - \mu_0\mu_{r} \epsilon_{0}\frac{\partial^{2} }{\partial t^{2}} \mathbf{E}=\mu_0\mu_{r}\frac{\partial^{2} }{\partial t^{2}} \mathbf{P} $ ### Nonlinear wave equation in perfect dielectrics Here we adapt the conventions applied before, namely tilde sign $\tilde{}$ stands for rapidly changing fields. And ignore the current and magnetic effects, which is the most common case in crystals. $\nabla^2 \tilde{\mathbf{E}} - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \tilde{\mathbf{E}} = \frac{1}{\varepsilon_0 c^2} \frac{\partial^2}{\partial t^2} \tilde{\mathbf{P}}$ Here $c=\frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}$, the material response is implicitly given in the polarization. One can write this with electric displacement considering $\tilde{\mathbf{D}} = \varepsilon_0 \tilde{\mathbf{E}} + \tilde{\mathbf{P}}$. $\nabla^2 \tilde{\mathbf{E}} - \frac{1}{\varepsilon_0 c^2} \frac{\partial^2}{\partial t^2} \tilde{\mathbf{D}} = 0$ In many cases, we write the linear part and nonlinear part separately. For polarization, we have $\tilde{\mathbf{P}} = \tilde{\mathbf{P}}^{(1)} + \tilde{\mathbf{P}}_{\mathrm{NL}}$, and for electric displacement, we have $\tilde{\mathbf{D}} = \tilde{\mathbf{D}}^{(1)} + \tilde{\mathbf{P}}_{\mathrm{NL}}$, where $\tilde{\mathbf{D}}^{(1)}=\varepsilon_0 \tilde{\mathbf{E}} + \tilde{\mathbf{P}}^{(1)}$. This helps separating them in the wave equation, $\nabla^2 \tilde{\mathbf{E}} - \frac{1}{\varepsilon_0 c^2} \frac{\partial^2}{\partial t^2} \tilde{\mathbf{D}}^{(1)} = \frac{1}{\varepsilon_0 c^2} \frac{\partial^2}{\partial t^2} \tilde{\mathbf{P}}_{\mathrm{NL}}$ #### Lossless, dispersionless medium First consider the simple case, the linear electric displacement $\tilde{\mathbf{D}}^{(1)}$ can be written as the dot product of dielectric constant and electric field, namely $\tilde{\mathbf{D}}^{(1)} = \varepsilon_0 \boldsymbol{\varepsilon}^{(1)} \cdot \tilde{\mathbf{E}}$. This is due to the tensor nature of the susceptibility. If consider the isotropic material, then we have $\tilde{\mathbf{D}}^{(1)} = \varepsilon_0 {\varepsilon}^{(1)} \tilde{\mathbf{E}}$ and the wave equation is $\nabla^2 \tilde{\mathbf{E}} - \frac{{\varepsilon}^{(1)}}{c^2} \frac{\partial^2}{\partial t^2} \tilde{\mathbf{E}} = \frac{1}{\varepsilon_0 c^2} \frac{\partial^2}{\partial t^2} \tilde{\mathbf{P}}_{\mathrm{NL}}$ This is a form of "driven" equation. The nonlinear polarization at right hand side acts as a **source**. >[!Notice] >As for the expression **source**. This could be understood as that external fields generate the nonlinear polarization $\tilde{\mathbf{P}}_{\mathrm{NL}}$, and this nonlinear polarization acts as a source, generates new components of electric fields. >(Because if we are in vacuum or in linear material, the expression would be like >$\nabla^2 \tilde{\mathbf{E}} - \frac{{\varepsilon}^{(1)}}{c^2} \frac{\partial^2}{\partial t^2} \tilde{\mathbf{E}} = 0$ >and we don't have the source!) >In this process, both input and output fields are governed by this wave equation. Also don't remember that we have the constitutive equation (in [[Nonlinear susceptibility]]) connecting polarization and electric fields. >If we have conductive materials, charges, or magnetic effects cannot be ignored, the source term would be more complicated. #### Dispersive medium For dispersive medium things are more complicated but we can play the same trick. Use Fourier-transform-like method to decompose different frequency components. $\tilde{\mathbf{E}}(\mathbf{r}, t) = \sum_n' \tilde{\mathbf{E}}_n (\mathbf{r}, t)$ $\tilde{\mathbf{D}}^{(1)}(\mathbf{r}, t) = \sum_n' \tilde{\mathbf{D}}^{(1)}_n (\mathbf{r}, t)$ $\tilde{\mathbf{P}}_{\mathrm{NL}}(\mathbf{r}, t) = \sum_n' \tilde{\mathbf{P}}_{\mathrm{NL}, n} (\mathbf{r}, t)$ Like what we had in [[Nonlinear susceptibility]], the prime stands for the summation over positive frequencies only. And each frequency component looks like, $\tilde{\mathbf{E}}_n (\mathbf{r}, t) = \mathbf{E}_n (\mathbf{r}) e^{-i \omega_n t} + \mathrm{c.c.}$ $\tilde{\mathbf{D}}^{(1)}_n (\mathbf{r}, t) = \mathbf{D}_n^{(1)} (\mathbf{r}) e^{-i \omega_n t} + \mathrm{c.c.}$ $\tilde{\mathbf{P}}_{\mathrm{NL}, n} (\mathbf{r}, t) = \mathbf{P}_{\mathrm{NL}, n} (\mathbf{r}) e^{-i \omega_n t} + \mathrm{c.c.}$ Write the linear electric displacement as $\mathbf{D}_n^{(1)}(\mathbf{r}) = \varepsilon_0 \boldsymbol{\varepsilon}^{(1)}(\omega_n) \cdot \mathbf{E}_n(\mathbf{r})$. Then insert into the expression in time domain for each component, we have $\nabla^2 \mathbf{E}_n(\mathbf{r}) + \frac{\omega_n^2}{c^2} \boldsymbol{\varepsilon}^{(1)}(\omega_n) \cdot \mathbf{E}_n(\mathbf{r}) = -\frac{\omega_n^2}{\varepsilon_0 c^2} \mathbf{P}_{\mathrm{NL},n}(\mathbf{r})$ This $\boldsymbol{\varepsilon}^{(1)}(\omega_n)$ is a complex quantity, and we removed the time dependence. This is often called **nonlinear Helmholtz equation**. If in time domain and dissipation can be neglected, we have $\nabla^2 \tilde{\mathbf{E}}_n - \frac{\boldsymbol{\varepsilon}^{(1)}(\omega_n)}{c^2} \cdot \frac{\partial^2 \tilde{\mathbf{E}}_n}{\partial t^2} = \frac{1}{\varepsilon_0 c^2} \frac{\partial^2 \tilde{\mathbf{P}}_{\mathrm{NL},n}}{\partial t^2}$ >[!Note] >If we still write in time domain, and use $\tilde{\mathbf{D}}_n^{(1)}(\mathbf{r}, t) = \varepsilon_0 \boldsymbol{\varepsilon}^{(1)}(\omega_n) \cdot \tilde{\mathbf{E}}_n(\mathbf{r}, t)$, then we have >$\nabla^2 \tilde{\mathbf{E}}_n - \frac{\boldsymbol{\varepsilon}^{(1)}(\omega_n)}{c^2} \cdot \frac{\partial^2 \tilde{\mathbf{E}}_n}{\partial t^2} = \frac{1}{\varepsilon_0 c^2} \frac{\partial^2 \tilde{\mathbf{P}}_{\mathrm{NL},n}}{\partial t^2}$ >In the book [[Nonlinear Optics]], this is under the condition that dissipation could be ignored. This should also be true if we have instantaneous response of dissipation, although in most of the time (and in theory) this will not happen and one cannot write the susceptibility simply as a time independent complex tensor. >[!Info] >Helmholtz equation: see https://en.wikipedia.org/wiki/Helmholtz_equation >Dispersion relation: see https://en.wikipedia.org/wiki/Dispersion_relation