## Nonlinear susceptibility Now we formally introduce the concept of nonlinear susceptibility, as a complex property with absorption accounted. For electric field, $\tilde{\mathbf{E}}(\mathbf{r}, t) = \sum_n' \tilde{E}_n(\mathbf{r}, t)$ where $\tilde{\mathbf{E}}_n(\mathbf{r}, t) = \mathbf{E}_n(\mathbf{r}) e^{-i \omega_n t} + \text{c.c.}$ The prime $ denotes for the restricted summation, namely there are limitations on $n$. Here, the summation is to be taken over the positive frequencies only (cause negative frequencies included in c.c.). Here we use a **Fourier type decomposition** to separate individual frequencies, and have time removed from $\mathbf{E}_{n}(\mathbf{r})$, therefore, the tilde sign is removed. And by applying [[Slowly varying envelope approximation|spatially slowly varying envelope approximation]], we can write, $\mathbf{E}_n(\mathbf{r}) = \mathbf{A}_n e^{i \mathbf{k}_n \cdot \mathbf{r}}$ This envelope $\mathbf{A}$ is not a function of time or space, only frequency ($\omega_{n}$, or say $n$) dependent. So we have the full expression of the electric field as $\tilde{\mathbf{E}}(\mathbf{r}, t) = \sum_n' \mathbf{A}_n e^{i (\mathbf{k}_n \cdot \mathbf{r} - \omega_n t)} + \text{c.c.}$ >[!Notice] >Sometimes people write $\mathbf{E}_{n}$ as $\mathbf{E}(\omega_{n})$ and $\mathbf{A}_{n}=\mathbf{A}(\omega_{n})$ to emphasize the frequencies. Because of the complex conjugate, negative frequencies would be $\mathbf{E}(-\omega_n) = \mathbf{E}(\omega_n)^*$ and $\mathbf{A}(-\omega_n) = \mathbf{A}(\omega_n)^*$. >$\tilde{\mathbf{E}}(\mathbf{r}, t) = \sum_n \mathbf{E}(\omega_n) e^{-i \omega_n t}= \sum_n \mathbf{A}(\omega_n) e^{i (\mathbf{k}_n \cdot \mathbf{r} - \omega_n t)}$ >>[!Important] >>It could be confusing, cause the notation $\tilde{\mathbf{E}}_n(\mathbf{r}, t)$ is applied for both single frequency case $\omega$ and as a decomposition of frequency $\omega_{n}$, especially under this new notations. This is because the frequency is not written explicitly before, here we will write them as parameters. >>In the beginning, we have >>$\tilde{\mathbf{E}}(\mathbf{r}, t;\omega_{1}, \omega_{2},\cdots,\omega_{n}) = \sum_n' \tilde{E}_n(\mathbf{r}, t;\omega_{n})$ >>Similarly, we have >>$\tilde{\mathbf{E}}_n(\mathbf{r}, t;\omega_{n}) = \mathbf{E}_n(\mathbf{r};\omega_{n}) e^{-i \omega_n t} + \text{c.c.}$ >>and >>$\mathbf{E}_n(\mathbf{r};\omega_{n}) = \mathbf{A}_n(;\omega_{n}) e^{i \mathbf{k}_n \cdot \mathbf{r}}$ >>These are the old notation, $\omega_{n}$ is always a parameter. >>And in the new notation, we neglected the subscript $n$, and with no parameters explicitly written, one could not see it is a summation or a single frequency. Like the expression in >>$\tilde{\mathbf{E}}(\mathbf{r}, t) = \mathcal{E} \cos(\mathbf{k} \cdot \mathbf{r} - \omega t)$ >>This actual is >>$\tilde{\mathbf{E}}(\mathbf{r}, t;\omega) = \mathcal{E}(;\omega) \cos(\mathbf{k} \cdot \mathbf{r} - \omega t)$ >>instead of the $\tilde{\mathbf{E}}(\mathbf{r}, t;\omega_{1}, \omega_{2},\cdots,\omega_{n})$ we had above. Similarly, the complex amplitudes >>$\mathbf{E}(\mathbf{r};\omega) = \frac{1}{2} \mathcal{E} e^{i \mathbf{k} \cdot \mathbf{r}}, \quad \mathbf{E}(\mathbf{r};-\omega) = \frac{1}{2} \mathcal{E} e^{-i \mathbf{k} \cdot \mathbf{r}}$ >>Here the term "complex amplitude" is not denoting amplitude, but "[phasor](https://en.wikipedia.org/wiki/Phasor)", an amplitude with phase included. These $\mathbf{E}$ only depend on frequency (and real space vector $\mathbf{r}$), and it's purely algebraic transformation. Since >>$\cos(\mathbf{k} \cdot \mathbf{r} - \omega t) = \frac{1}{2} \left( e^{i (\mathbf{k} \cdot \mathbf{r} - \omega t)} + e^{-i (\mathbf{k} \cdot \mathbf{r} - \omega t)} \right)$ >>Only keep the dependence of space, we have the expression of $\mathbf{E}(\omega)$. Or we can remove the spatial dependence to have >>$\mathbf{A}(\omega) = \frac{1}{2} \mathcal{E}, \quad \mathbf{A}(-\omega) = \frac{1}{2} \mathcal{E}$ > >>[!Note] >>Although it is common to have $\sum$ as $\int$ in physics, and here we do have Fourier type decomposition, but I do not think this is strict Fourier transform. Since we are still mainly dealing them in a discrete manner, and although we are talking $\tilde{\mathbf{E}}(\mathbf{r}, t;\omega)$ and $\mathcal{E}$ in frequency domain, but this is only for these components, not the entire electric field. >>And the mathematical expression does not strictly follows the Fourier transform. Though thinking in this manner could facilitate the understandings. Apply the new notation, and here we no longer have the $ to restrict the summation to have only positive frequencies. Therefore, the field for a single frequency $\omega$ is $\tilde{\mathbf{E}}(\mathbf{r}, t) = \mathcal{E} \cos(\mathbf{k} \cdot \mathbf{r} - \omega t)$ And we have separate positive and negative parts, we write the complex amplitudes separately, $\mathbf{E}(\omega) = \frac{1}{2} \mathcal{E} e^{i \mathbf{k} \cdot \mathbf{r}}, \quad \mathbf{E}(-\omega) = \frac{1}{2} \mathcal{E} e^{-i \mathbf{k} \cdot \mathbf{r}},$ and the envelope amplitudes, $\mathbf{A}(\omega) = \frac{1}{2} \mathcal{E}, \quad \mathbf{A}(-\omega) = \frac{1}{2} \mathcal{E}$ As stated in the callout, the factor $\frac{1}{2}$ is by separating positive and negative frequencies. With the same notation, the polarization can be written as, $\tilde{\mathbf{P}}(\mathbf{r}, t) = \sum_n \mathbf{P}(\omega_n) e^{-i \omega_n t}$ with positive and negative frequencies separated. With the frame established, the second order polarization could be written as $P_i(\omega_n + \omega_m) = \varepsilon_0 \sum_{jk} \sum_{(nm)} \chi_{ijk}^{(2)}(\omega_n + \omega_m, \omega_n, \omega_m) E_j(\omega_n) E_k(\omega_m)$ This $P_{i}$ is not written in `\mathbf` form because it is the $i$ component. The indices $i, j, k$ are Cartesian components of the field, or direction. The notation ($nm$) denotes the **summation** of frequencies with these notation will be held unchanged, although individual frequency could vary (like change the order of $\omega_{n}$ and $\omega_{m}$). Now let's look at the 2nd susceptibility itself. The $\chi_{ijk}^{(2)}(\omega_n + \omega_m, \omega_n, \omega_m)$ contains two inputs, $\omega_{m}$ and $\omega_{n}$, and an output, $\omega_{m}+\omega_{n}$. The convention is write the first frequency as the summation of the latter ones. But something people write $\chi^{(2)}(\omega_{3};\omega_{1}, \omega_{2})$ to emphasize the generate (output) frequency and separate from the input values. The ${i,j,k}$ corresponds to the direction of inputting electric fields and that of polarization. The fields, written as amplitudes only, include $E_{j}(\omega_{n})$ and $E_{k}(\omega_{m})$. $E_{j}(\omega_{n})$ in the field expression is related to a phase factor $e^{-i\omega_{n}t}$, while $E_{k}(\omega_{m})$ is related to $e^{-i\omega_{m}t}$. So the combined form would have a factor with $e^{-i(\omega_{m}+\omega_{n})t}$, the summation of frequency. >[!Notice] >Here we say related does not mean the amplitude contains time, and it does not contains time. Now let's check how such an definition will work. Consider SFG, define input field $\omega_{1}$, $\omega_{2}$ and output field $\omega_{3}=\omega_{1}+\omega_{2}$. The polarization at direction $i$ would be $P_i(\omega_3) = \varepsilon_0 \sum_{jk} \sum_{(nm)} \chi^{(2)}_{ijk}(\omega_3, \omega_n, \omega_m) E_j(\omega_n) E_k(\omega_m)$First expand $\sum_{(nm)}$, $P_i(\omega_3) = \varepsilon_0 \sum_{jk} \left[ \chi^{(2)}_{ijk}(\omega_3, \omega_1, \omega_2) E_j(\omega_1) E_k(\omega_2) + \chi^{(2)}_{ijk}(\omega_3, \omega_2, \omega_1) E_j(\omega_2) E_k(\omega_1) \right]$ One could already see that $j$ and $k$ are actually dummy indices and interchangeable, cause they are assigned to the same Cartesian components ($x, y, z$). So one can write the second part $\chi^{(2)}_{ikj}(\omega_3, \omega_2, \omega_1) E_k(\omega_2) E_j(\omega_1)$. And apply **intrinsic permutation symmetry** $\chi^{(2)}_{ikj}(\omega_m + \omega_n, \omega_m, \omega_n) = \chi^{(2)}_{ijk}(\omega_m + \omega_n, \omega_m, \omega_n)$ One gets, $P_i(\omega_3) = 2 \varepsilon_0 \sum_{jk} \chi^{(2)}_{ijk}(\omega_3, \omega_1, \omega_2) E_j(\omega_1) E_k(\omega_2)$ Alternative, to clearly see all the terms, one could expand $jk$, so one have $\begin{align*} P_i(\omega_3) = \varepsilon_0 \Big( & \chi^{(2)}_{ixx}(\omega_3, \omega_1, \omega_2) E_x(\omega_1) E_x(\omega_2) + \chi^{(2)}_{ixy}(\omega_3, \omega_1, \omega_2) E_x(\omega_1) E_y(\omega_2) \\ & + \chi^{(2)}_{ixz}(\omega_3, \omega_1, \omega_2) E_x(\omega_1) E_z(\omega_2) + \chi^{(2)}_{iyx}(\omega_3, \omega_1, \omega_2) E_y(\omega_1) E_x(\omega_2) \\ & + \chi^{(2)}_{iyy}(\omega_3, \omega_1, \omega_2) E_y(\omega_1) E_y(\omega_2) + \chi^{(2)}_{iyz}(\omega_3, \omega_1, \omega_2) E_y(\omega_1) E_z(\omega_2) \\ & + \chi^{(2)}_{izx}(\omega_3, \omega_1, \omega_2) E_z(\omega_1) E_x(\omega_2) + \chi^{(2)}_{izy}(\omega_3, \omega_1, \omega_2) E_z(\omega_1) E_y(\omega_2) \\ & + \chi^{(2)}_{izz}(\omega_3, \omega_1, \omega_2) E_z(\omega_1) E_z(\omega_2) \Big) \\ + \varepsilon_0 \Big( & \chi^{(2)}_{ixx}(\omega_3, \omega_2, \omega_1) E_x(\omega_2) E_x(\omega_1) + \chi^{(2)}_{ixy}(\omega_3, \omega_2, \omega_1) E_x(\omega_2) E_y(\omega_1) \\ & + \chi^{(2)}_{ixz}(\omega_3, \omega_2, \omega_1) E_x(\omega_2) E_z(\omega_1) + \chi^{(2)}_{iyx}(\omega_3, \omega_2, \omega_1) E_y(\omega_2) E_x(\omega_1) \\ & + \chi^{(2)}_{iyy}(\omega_3, \omega_2, \omega_1) E_y(\omega_2) E_y(\omega_1) + \chi^{(2)}_{iyz}(\omega_3, \omega_2, \omega_1) E_y(\omega_2) E_z(\omega_1) \\ & + \chi^{(2)}_{izx}(\omega_3, \omega_2, \omega_1) E_z(\omega_2) E_x(\omega_1) + \chi^{(2)}_{izy}(\omega_3, \omega_2, \omega_1) E_z(\omega_2) E_y(\omega_1) \\ & + \chi^{(2)}_{izz}(\omega_3, \omega_2, \omega_1) E_z(\omega_2) E_z(\omega_1) \Big) \end{align*} $ In this expression, for example, $\chi^{(2)}_{ixy}(\omega_3, \omega_1, \omega_2) E_x(\omega_1) E_y(\omega_2)$ are exactly the same as $\chi^{(2)}_{iyx}(\omega_3, \omega_2, \omega_1) E_y(\omega_2) E_x(\omega_1)$, according to intrinsic permutation symmetry. This will be discussed in [[General symmetries and simplifications]]. But we will have the same result, $P_i(\omega_3) = 2 \varepsilon_0 \sum_{jk} \chi^{(2)}_{ijk}(\omega_3, \omega_1, \omega_2) E_j(\omega_1) E_k(\omega_2)$ >[!Notice] >Till here, we still did **not** assign the orientation of the crystal or the material. The Cartesian system can be seen as an absolute spatial coordinate for electromagnetic waves. So **DO NOT** consider this $\chi^{(2)}$ as an intrinsic material property, but something after tensor operator. If both fields are only polarized at $x$ direction, then we have $P_i(\omega_3) = 2 \varepsilon_0 \chi^{(2)}_{i xx}(\omega_3, \omega_1, \omega_2) E_x(\omega_1) E_x(\omega_2)$ Now consider SHG, this time we have only one input frequency, $\omega_{1}$, and the output is $\omega_{3}=2\omega_{1}$. Since we no longer have difference $m$ and $n$, we have no $\sum_{(nm)}$, so $P_i(\omega_3) = \varepsilon_0 \sum_{jk} \chi^{(2)}_{ijk}(\omega_3, \omega_1, \omega_1) E_j(\omega_1) E_k(\omega_1)$ Again if the input field polarize at direction $x$ only, then $P_i(\omega_3) = \varepsilon_0 \chi^{(2)}_{i xx}(\omega_3, \omega_1, \omega_1) E_x(\omega_1)^2$ >[!Notice] >One can see that for SHG, the factor 2 is gone, which is kind of wired at the first look, for SHG seems to be SFG with $\omega_{2}=\omega_{1}$. But it is easy to understand that for SFG, we have two input fields, while for SHG, there is only one. The polarization caused by SFG is for sure stronger if the field amplitudes are the same. >One can consider this problem in such a way. Keep the total intensity the same. For intensity $I$, we have >$I\propto E^{2}$ >For SFG, let $E_{x,SFH}(\omega_{1})=E_{x,SFG}(\omega_{2})$, then the total intensity would be >$I_{SFG}\propto 2E_{x,SFG}^{2}(\omega_{1})$ >To have the same intensity, for SHG, the $E_{x,SHG}(\omega_{1})$ should then be >$E_{x,SHG}(\omega_{1})=\sqrt{2}E_{x,SFH}(\omega_{1})$ >If we put this $E_{x,SHG}(\omega_{1})$ into $P_i(\omega_3) = \varepsilon_0 \chi^{(2)}_{i xx}(\omega_3, \omega_1, \omega_1) E_x(\omega_1)^2$, we get the same amount of polarization. People write the permutation of frequencies explicitly from the summation as a degeneracy factor $D$, namely $P_i(\omega_n + \omega_m) = \varepsilon_0 D \sum_{jk} \chi^{(2)}_{ijk}(\omega_n + \omega_m, \omega_n, \omega_m) E_j(\omega_n) E_k(\omega_m)$ and this $D$ is the number of distinct permutations of the applied frequencies. For SFG, we have seen that this $D=2$, from intrinsic permutation. For SHG, $D=1$. The higher order susceptibility follows the same manner. For example, for 3rd order susceptibility, we have $P_i(\omega_o + \omega_n + \omega_m) = \varepsilon_0 \sum_{jkl} \sum_{(mno)} \chi^{(3)}_{ijkl}(\omega_o + \omega_n + \omega_m, \omega_o, \omega_n, \omega_m) E_j(\omega_o) E_k(\omega_n) E_l(\omega_m)$ and if replace permutations of frequencies with degeneracy factor, $P_i(\omega_o + \omega_n + \omega_m) = \varepsilon_0 D \sum_{jkl} \chi^{(3)}_{ijkl}(\omega_o + \omega_n + \omega_m, \omega_o, \omega_n, \omega_m) E_j(\omega_o) E_k(\omega_n) E_l(\omega_m)$ This $D$ is the number of distinct permutations of the frequencies.