MIT Licence
© Alexey A. Shcherbakov, 2024
In this lecture we will consider open optical systems - resonators and waveguides. For brevity, let us rewrite Maxwell's equations in the frequency domain in the operator form \begin{equation}\tag{1} \hat{\mathcal{L}}(\boldsymbol{r},\omega) \boldsymbol{F}(\boldsymbol{r},\omega) = \boldsymbol{J}(\boldsymbol{r},\omega) \end{equation} where $\hat{\mathcal{L}} = \omega\hat{\mathcal{P}} - \hat{\mathcal{D}}$, therfore \begin{equation*} \hat{\mathcal{P}}(\boldsymbol{r},\omega) = \left( \begin{array}{cc} \varepsilon & -i\xi \\ -i\xi^T & \mu \end{array} \right), \;\; \hat{\mathcal{D}} = \left( \begin{array}{cc} 0 & \nabla\times \\ \nabla\times & 0 \end{array} \right) \end{equation*} The field vector comprises the electric and magenetic components $\boldsymbol{F}(\boldsymbol{r},\omega) = (\boldsymbol{E},\;i\boldsymbol{H})^T$ and the joint current vector $\boldsymbol{J}(\boldsymbol{r},\omega) = (-i\boldsymbol{J}_e,\;\boldsymbol{J}_m)^T$. Equation for the Green's function \begin{equation}\tag{2} \hat{\mathcal{L}}(\boldsymbol{r},\omega) \mathcal{G}(\boldsymbol{r},\boldsymbol{r}',\omega) = \hat{\mathbb{I}} \delta(\boldsymbol{r},\boldsymbol{r}') \end{equation} and the integral solution to the MAxwell's equations \begin{equation}\tag{3} \boldsymbol{F}(\boldsymbol{r},\omega) = \int dV' \mathcal{G}(\boldsymbol{r},\boldsymbol{r}',\omega) \boldsymbol{J}(\boldsymbol{r}',\omega) \end{equation}
If an excitation of an open system occurs for a short time arount the moment $t=0$, the temporal response of the system will be \begin{equation*} \boldsymbol{F}(\boldsymbol{r},t) \sim \int d\omega \boldsymbol{F}(\boldsymbol{r},\omega) e^{-i\omega t},\;t>0 \end{equation*} where $\boldsymbol{F}(\boldsymbol{r},\omega)$ is the solution to the equations in the frequency domain for a given spatial distribution of sources at zero time moment. Assuming that the anlytic continuation of the Green's function and, consequently, of the field vector has a countable number of poles $\omega_n = \Omega_n - i\Gamma_n$, $\Gamma_n > 0$, the residue theorem allows us to rewrite the solution as \begin{equation*} \boldsymbol{F}(\boldsymbol{r},t) \sim \sum_n \mathrm{Res}[\boldsymbol{F}(\boldsymbol{r},\omega),\omega_n] e^{-i\omega_n t},\;t>0 \end{equation*} so that the decay of each term is governed by the factor $\exp(-\Gamma_n t)$.
Since the right-hand part of the equation on the Green's function $(2)$ is independent of frequency, the poles of the Green's function must be the eigenvalues of equation \begin{equation}\tag{4} \hat{\mathcal{L}}(\boldsymbol{r},\omega_n) \boldsymbol{F}_n(\boldsymbol{r}) = 0 \end{equation} that is, the eigenvalue equation for modes of open resonator/waveguide. Generally, these values are complex. Outside the resonator/waveguide under consideration, the field is represented in the form of propagating waves, which in the 3D case have a multiplier $\exp(-i\omega_n t + ik_nr)$ where the wavenumber of the free space surrounding the resonator is $k_n=\omega_n\sqrt{\varepsilon\mu}$. Hence it is clear that time-decaying states of open systems are described by exponentially divergent spatial functions, so when decomposing fields into such states one should take this fact into account. Such states are referred to as quasinormal modes or resonant states in the literature.
Equation $(4)$ should be supplemented by appropriate boundary conditions. When solving this equation for complex frequencies, one should keep in mind that any sources in a given configuration of the considered resonators are localized, so that the field outside some finite volume has the form of outgoing waves. Mathematically, we can consider the eigenvalue equation as a limiting case of the general equation $(1)$ with source $\boldsymbol{J}(\boldsymbol{r},\omega) \sim \omega-\omega-\omega_n$ at $\omega\rightarrow\omega_n$. Thus $\boldsymbol{F}(\boldsymbol{r},\omega)\rightarrow \boldsymbol{F}_n(\boldsymbol{r})$.
Resonant state expansion of the Green's function is based on the Mittag-Leffler theorem: if the function $f(z)$ is analytic everywhere except for a countable number of poles $z_n$ with residues $R_n$, and $\lim_{z\rightarrow\infty}f(z)/z^p=0$, then this function can be represented as \begin{equation*} f(z) = f_p(z) + \sum_n\frac{R_n}{z-z_n} \end{equation*} where $f_p$ - is a polynomial of the power $p-1$. Let us apply this theorem to write the Green's function, taking into account that $\lim_{\omega\rightarrow\infty}\mathcal{G}=0$: \begin{equation}\tag{5} \mathcal{G}(\boldsymbol{r},\boldsymbol{r}',\omega) = \sum_n \frac{\mathcal{Q}_n(\boldsymbol{r},\boldsymbol{r}')}{\omega-\omega_n} \end{equation} Let us substitute this expression into the equation on the Green's function $(2)$, multiply by some field $\boldsymbol{F}_b(\boldsymbol{r})$, localized in some finite volume, and integrate both parts of the equation, which gives \begin{equation*} \sum_n \frac{\hat{\mathcal{L}}(\boldsymbol{r},\omega)\boldsymbol{A}_n(\boldsymbol{r})}{\omega-\omega_n} = \boldsymbol{F}_b(\boldsymbol{r}) \end{equation*} where \begin{equation*} \boldsymbol{A}_n(\boldsymbol{r}) = \intop_{V'} dV' \mathcal{Q}_n(\boldsymbol{r},\boldsymbol{r}') \boldsymbol{F}_b(\boldsymbol{r}') \end{equation*} In the limit $\omega\rightarrow\omega_m$ we obtain a system of independent equations $\hat{mathcal{L}}(\boldsymbol{r},\omega_n)\boldsymbol{A}_n(\boldsymbol{r}) = 0$, whence it follows that the resulting vector amplitudes are proportional to the resonant states defined by the equation $(4)$. And, since the choice of the field $\boldsymbol{F}_b(\boldsymbol{r})$ is arbitrary, it is necessary to require that the application of the Green's operator makes a projection onto the $n$-th resonance state: \begin{equation*} \mathcal{Q}_n(\boldsymbol{r},\boldsymbol{r}') \sim \boldsymbol{F}_n(\boldsymbol{r}) \otimes \boldsymbol{X}(\boldsymbol{r'}) \end{equation*} To determine the vector $\boldsymbol{X}$, invoke the reciprocity relation $\mathcal{G}(\boldsymbol{r}, \boldsymbol{r}',\omega) = \mathcal{G}^T (\boldsymbol{r}',\boldsymbol{r},\omega)$. Hence it immediately follows that $\boldsymbol{X}\sim\boldsymbol{F}_n(\boldsymbol{r}')$. Thus, choosing an appropriate normalization of the resonant states, one can write \begin{equation}\tag{6} \mathcal{G}(\boldsymbol{r},\boldsymbol{r}',\omega) = \sum_n \frac{\boldsymbol{F}_n(\boldsymbol{r}) \otimes \boldsymbol{F}_n(\boldsymbol{r'})}{\omega-\omega_n} \end{equation}
In somle problems the representation $(6)$ is insufficient, so that it is necessary to take into account cuts in the complex plane and the choice of suitable Riemann surfaces, for example, due to the occurrence of a square root in the dispersion equation for plane waves. In a more general form, the Green's tensor expansion is written as \begin{equation}\tag{6a} \mathcal{G}(\boldsymbol{r},\boldsymbol{r}',\omega) = \sum_n \frac{\boldsymbol{F}_n(\boldsymbol{r}) \otimes \boldsymbol{F}_n(\boldsymbol{r'})}{\omega-\omega_n} + \sum_p \mathcal{G}_p(\boldsymbol{r},\boldsymbol{r}',\omega) \end{equation} where the index $p$ enumerates contribution of individual cuts: \begin{equation*} \mathcal{G}_p(\boldsymbol{r},\boldsymbol{r}',\omega) = \frac{1}{2\pi i} \int_{C_p} d\omega' \frac{\Delta \mathcal{G}(\boldsymbol{r},\boldsymbol{r}',\omega')}{\omega-\omega'} \end{equation*} Here $C_m$ is the path along a cut, and $\Delta \mathcal{G}$ is the difference of the Green's function on different Riemannian surfaces. It should be noted that in numerical solutions the contribution of cuts can also be written in the form $(6)$.
Let us substitute the derived Green's function expansion into the volume integral equation for the field vector: \begin{equation}\tag{7} \boldsymbol{F}(\boldsymbol{r},\omega) = \sum_n \frac{\boldsymbol{F}_n(\boldsymbol{r})}{\omega-\omega_n} \int dV' \boldsymbol{F}_n(\boldsymbol{r'}) \cdot \boldsymbol{J}(\boldsymbol{r}',\omega) \end{equation} As discussed above $\boldsymbol{F}(\boldsymbol{r}, \omega) \underset{\omega\rightarrow\omega_{n}}{\rightarrow}\boldsymbol{F}_{n}(\boldsymbol{r})$ for sources of the form $\boldsymbol{J}(\boldsymbol{r}, \omega) = (\omega-\omega_n) \tilde{\boldsymbol{J}}(\boldsymbol{r})$. Then taking the limit $\omega\rightarrow\omega_n$ in equation $(7)$, one obtains \begin{equation}\tag{8} \int dV' \boldsymbol{F}_n(\boldsymbol{r'}) \cdot \tilde{\boldsymbol{J}}(\boldsymbol{r}') = 1 \end{equation} Now let us take equations $(1)$ and $(4)$ for the field of some localized sources and resonance states, multiply the former by the state $\boldsymbol{F}_{n}(\boldsymbol{r})$, and the latter - by the solution for the sources $\boldsymbol{F}(\omega,\boldsymbol{r})$, taking into account the explicit representation of the operator, and add the equations: \begin{align*} \omega \boldsymbol{F}_{n}(\boldsymbol{r}) \cdot \hat{\mathcal{P}}(\boldsymbol{r},\omega) \boldsymbol{F}(\boldsymbol{r},\omega) - \omega_n \boldsymbol{F}(\boldsymbol{r},\omega) \cdot \hat{\mathcal{P}}(\boldsymbol{r},\omega_n) \boldsymbol{F}_{n}(\boldsymbol{r}) + \\ + \boldsymbol{F}(\boldsymbol{r},\omega) \cdot \hat{\mathcal{D}}(\boldsymbol{r},\omega_n) \boldsymbol{F}_{n}(\boldsymbol{r}) - \boldsymbol{F}_{n}(\boldsymbol{r}) \cdot \hat{\mathcal{D}}(\boldsymbol{r},\omega_n) \boldsymbol{F}(\boldsymbol{r},\omega) = \\ = (\omega-\omega_n) \boldsymbol{F}_{n}(\boldsymbol{r}) \cdot \tilde{\boldsymbol{J}}(\boldsymbol{r}) \end{align*} Substituting the explicit form of the field vectors and the operator $\hat{\mathcal{D}}$ into the second line, and using the rule for decomposition of a divergence of a vector product, we obtain, \begin{equation*} \boldsymbol{F}(\boldsymbol{r},\omega) \cdot \hat{\mathcal{D}}(\boldsymbol{r},\omega_n) \boldsymbol{F}_{n}(\boldsymbol{r}) - \boldsymbol{F}_{n}(\boldsymbol{r}) \cdot \hat{\mathcal{D}}(\boldsymbol{r},\omega_n) \boldsymbol{F}(\boldsymbol{r},\omega) = i\nabla \cdot (\boldsymbol{E}_{n} \times \boldsymbol{H} - \boldsymbol{E} \times \boldsymbol{H}_n) \end{equation*} We integrate the whole equality over some finite volume including localized sources and take the limit $\omega\rightarrow\omega_n$. In this case, we use the decomposition of the analytic continuation of field vectors and the product $\omega \hat{\mathcal{P}}(\boldsymbol{r},\omega)$ near the point $\omega=\omega_n$: \begin{equation*} \omega \hat{\mathcal{P}}(\boldsymbol{r},\omega) \approx \omega_n \hat{\mathcal{P}}(\boldsymbol{r},\omega_n) + (\omega-\omega_n) {[\omega \hat{\mathcal{P}}(\boldsymbol{r},\omega)]'}_{\omega=\omega_n} \end{equation*}
\begin{equation*} \boldsymbol{E}(\boldsymbol{r},\omega) \approx \boldsymbol{E}_n(\boldsymbol{r}) + (\omega-\omega_n) {[\boldsymbol{E}(\boldsymbol{r},\omega)]'}_{\omega=\omega_n} = \boldsymbol{E}_n(\boldsymbol{r}) + (\omega-\omega_n) \frac{1}{\omega_n} (\boldsymbol{r}\nabla) \boldsymbol{E}_n(\boldsymbol{r}) \end{equation*}Here the prime denotes the frequency derivative, and the last equality is obtained by accounting to the fact that at the boundary of the specified volume the field is represented as a decomposition into propagating waves, and hence we can write $\boldsymbol{E}(\boldsymbol{r},\omega) = \boldsymbol{E}(\omega\boldsymbol{r})$. As a result, taking into account the relation $(8)$ we obtain \begin{equation}\tag{9} V_n + S_n = 1 \end{equation} where \begin{equation}\tag{10} V_n = \intop_V dV' \boldsymbol{F}_n(\boldsymbol{r}') \cdot {[\omega \hat{\mathcal{P}}(\boldsymbol{r}',\omega)]'}_{\omega=\omega_n} \boldsymbol{F}_n(\boldsymbol{r}') \end{equation}
\begin{equation}\tag{11} S_n = i \oint_{\partial V} d\boldsymbol{\Sigma} (\boldsymbol{E}_{n} \times \boldsymbol{H}'_n - \boldsymbol{E}'_n \times \boldsymbol{H}_n) \end{equation}In calculations, the frequency derivative of the field in the resonant state can be calculated as follows. Let us return to equation $(7)$ and consider it in the neighbourhood of the state $m$, substituting the corresponding source representation: \begin{equation*} \boldsymbol{F}(\boldsymbol{r},\omega) = \sum_n \frac{\omega-\omega_m}{\omega-\omega_n} \boldsymbol{F}_n(\boldsymbol{r}) \int dV' \boldsymbol{F}_n(\boldsymbol{r'}) \cdot \tilde{\boldsymbol{J}}_m(\boldsymbol{r}',\omega) = \sum_n \frac{\omega-\omega_m}{\omega-\omega_n} \boldsymbol{F}_n(\boldsymbol{r}) I_{nm} \end{equation*} Here $I_{mm} = 1$ as derived above. The field outside the resonant system is decomposed into outgoing vector waves. Let us write such a decomposition for each resonant state $\boldsymbol{F}_n(\boldsymbol{r}) = \sum_L \alpha_{Ln} \boldsymbol{O}(\boldsymbol{r},\omega)$ and differentiate by frequency at the point $\omega=\omega_m$: \begin{equation}\tag{12} \boldsymbol{F}_m'(\boldsymbol{r}) = \sum_L [\alpha_{Lm} \boldsymbol{O}'(\boldsymbol{r},\omega_m) + \alpha_{Lm}' \boldsymbol{O}(\boldsymbol{r},\omega_m)] \end{equation} When substituted into the orthogonality relation, contribution of the second summand is zero due to the following reason. Consider the fields $\boldsymbol{F}_{1,2}(\boldsymbol{r},\omega)$ excited by sources $\boldsymbol{J}_{1,2}(\boldsymbol{r},\omega)$ localized in the resonator region. Then for the Lorentz reciprocal medium \begin{equation*} \boldsymbol{F}_1 \cdot \hat{\mathcal{D}} \boldsymbol{F}_2 - \boldsymbol{F}_2 \cdot \hat{\mathcal{D}} \boldsymbol{F}_1 = \boldsymbol{F}_2 \cdot \boldsymbol{J}_1 - \boldsymbol{F}_1 \cdot \boldsymbol{J}_2 \end{equation*} Integration over a volume comprising resonators and local sources, and invocation of the Gauss theorem give \begin{equation*} i \oint_{\partial V} d\boldsymbol{\Sigma} (\boldsymbol{E}_2\times\boldsymbol{H}_1-\boldsymbol{E}_1\times\boldsymbol{H}_2) = \int_V dV (\boldsymbol{F}_2 \cdot \boldsymbol{J}_1 - \boldsymbol{F}_1 \cdot \boldsymbol{J}_2) \end{equation*} Furhter substituting into the right-hand side of the solutions for the field via the Green's function and applying the reciprocity theorem gives \begin{equation}\tag{13} i \oint_{\partial V} d\boldsymbol{\Sigma} (\boldsymbol{E}_2\times\boldsymbol{H}_1-\boldsymbol{E}_1\times\boldsymbol{H}_2) = 0 \end{equation} This equation shows that by substituting $(12)$ into $(9)$ the second summand goes to zero, and it is sufficient to calculate the frequency derivatives of the outgoing vector waves, which in the case of plane, spherical, or cylindrical waves can be done analytically.
Doing the same derivation as was used for the equality $(13)$ for two resonant states $\boldsymbol{F}_{m,n}(\boldsymbol{r})$, we obtain the second equation expressing the orthogonality: \begin{equation}\tag{14} \int_V dV \boldsymbol{F}_m \cdot [\omega_n \hat{\mathcal{P}}(\omega_n) - \omega_m \hat{\mathcal{P}}(\omega_m)] \boldsymbol{F}_n + i \oint_{\partial V} d\boldsymbol{\Sigma} (\boldsymbol{E}_{m} \times \boldsymbol{H}_n - \boldsymbol{E}_n \times \boldsymbol{H}_m) = 0 \end{equation}
Turning to the question of completeness of the system of resonant states, we substitute the decomposition $(6)$ into the equation for the Green's function $(2)$ and use the explicit operator form and the equation on the states $(4)$. After simple transformations we obtain \begin{equation}\tag{15} \sum_n \frac{\omega \hat{\mathcal{P}}(\boldsymbol{r},\omega) - \omega_n \hat{\mathcal{P}}(\boldsymbol{r},\omega_n)}{\omega-\omega_n} \boldsymbol{F}_n(\boldsymbol{r}) \otimes \boldsymbol{F}_n(\boldsymbol{r}') = \hat{\mathbb{I}} \delta(\boldsymbol{r}-\boldsymbol{r}') \end{equation} The obtained equality expresses the closurer relation for system of resonance states and is a necessary condition for completeness. In the case of absence of dispersion $\hat{\mathcal{P}}(\boldsymbol{r},\omega)=\hat{\mathcal{P}}(\boldsymbol{r})$ for materials, this relation simplifies \begin{equation}\tag{15a} \hat{\mathcal{P}}(\boldsymbol{r}) \sum_n \boldsymbol{F}_n(\boldsymbol{r}) \otimes \boldsymbol{F}_n(\boldsymbol{r}') = \hat{\mathbb{I}} \delta(\boldsymbol{r}-\boldsymbol{r}') \end{equation} In the more realistic case of Drude-Lorentz dispersion \begin{equation*} \hat{\mathcal{P}}(\boldsymbol{r},\omega) = \hat{\mathcal{P}}_{\infty} (\boldsymbol{r}) + \sum_{i} \frac{\hat{\mathcal{P}}_{i}(\boldsymbol{r})}{\omega-\Omega_{i}} \end{equation*} and the linear independence of lorentzians leads to the following closure expression together with additional sum rules: \begin{equation}\tag{15b} \hat{\mathcal{P}}_{\infty}(\boldsymbol{r}) \sum_n \boldsymbol{F}_n(\boldsymbol{r}) \otimes \boldsymbol{F}_n(\boldsymbol{r}') = \hat{\mathbb{I}} \delta(\boldsymbol{r}-\boldsymbol{r}') \end{equation}
\begin{equation*} \hat{\mathcal{P}}_{i}(\boldsymbol{r}) \sum_n \frac{\boldsymbol{F}_n(\boldsymbol{r}) \otimes \boldsymbol{F}_n(\boldsymbol{r}')}{\omega_n-\Omega_i} = \hat{\mathbb{I}} \delta(\boldsymbol{r}-\boldsymbol{r}') \end{equation*}The presence of such a set of relations may indicate that this basis is overcomplete, nevertheless, in practice, when the decomposition is performed over a small number of states, it is possible to attain a sufficient accuracy of calculations. The question of completeness also remains open in the presence of exceptional points in the system, in which both eigenfrequencies and eigenvectors degenerate simultaneously.
Consider a volume whose surface covers the resonator under consideration. On this surface any field can be decomposed by the complete set of incoming and outgoing vector waves, $\boldsymbol{I}_{L}(\boldsymbol{r},\omega)$ and $\boldsymbol{O}_{L}(\boldsymbol{r},\omega)$: \begin{equation}\tag{16} \boldsymbol{F}(\boldsymbol{r},\omega) = \sum_L [\alpha^{inc}_L(\omega) \boldsymbol{I}_{L}(\boldsymbol{r},\omega) + \alpha^{out}_L(\omega) \boldsymbol{O}_{L}(\boldsymbol{r},\omega)] \end{equation} The set of waves can be chosen to satisfy the relations \begin{equation*} \oint_{\partial V} d\boldsymbol{\Sigma} \cdot (\boldsymbol{E}^{inc}_{L}\times\boldsymbol{H}^{out}_{L'} - \boldsymbol{E}^{out}_{L'}\times\boldsymbol{H}^{inc}_{L}) = \delta_{LL'} \end{equation*}
\begin{equation*} \oint_{\partial V} d\boldsymbol{\Sigma} \cdot (\boldsymbol{E}^{\sigma}_{L}\times\boldsymbol{H}^{\sigma}_{L'} - \boldsymbol{E}^{\sigma}_{L'}\times\boldsymbol{H}^{\sigma}_{L}) = 0,\;\sigma=inc,out \end{equation*}Then, the coefficients \begin{equation*} \alpha^{inc}_L = -i \oint_{\partial V} d\boldsymbol{\Sigma} \cdot (\boldsymbol{E}^{out}_{L}\times\boldsymbol{H} - \boldsymbol{E}\times\boldsymbol{H}^{out}_{L}) \end{equation*}
\begin{equation*} \alpha^{sca}_L = i \oint_{\partial V} d\boldsymbol{\Sigma} \cdot (\boldsymbol{E}^{inc}_{L}\times\boldsymbol{H} - \boldsymbol{E}\times\boldsymbol{H}^{inc}_{L}) \end{equation*}The interaction of the resonator with the incident field can be described in terms of the scattering matrix: \begin{equation}\tag{17} \alpha^{out}_L= \sum_{L'} S_{LL'} \alpha^{inc}_{L'} \end{equation}
Let us express explicitly the components of the scattering matrix by identifying in the physical system some background operator $\hat{\mathcal{P}}_b$ and a perturbation (not neccessary small) $\Delta\hat{\mathcal{P}} = \hat{\mathcal{P}} - \hat{\mathcal{P}}_b$, describing the local measurement of material parameters due to the presence of the resonator. Let us write the total field in the system as the sum of the background and scattered field $\boldsymbol{F}_{tot} = \boldsymbol{F}_b + \boldsymbol{F}_{sca}$, where the background field consists of incoming and outgoing waves and satisfies the equation $\hat{\mathcal{L}}_b \boldsymbol{F}_b = 0$. Hence we can write the equation for the scattered field as \begin{equation}\tag{18} \hat{\mathcal{L}}(\boldsymbol{r},\omega) \boldsymbol{F}_{sca}(\boldsymbol{r},\omega) = -\omega \Delta\hat{\mathcal{P}}(\boldsymbol{r},\omega) \boldsymbol{F}_b \end{equation} The solution of this equation through the volume integral equation with Green's function decomposed by resonance states leads to the expression \begin{equation}\tag{19} \boldsymbol{F}_{tot}(\boldsymbol{r},\omega) = \boldsymbol{F}_b(\boldsymbol{r},\omega) - \sum_n \frac{I_n(\omega)}{\omega-\omega_n}\boldsymbol{F}_n(\boldsymbol{r}) \end{equation} with the overlat integral \begin{equation}\tag{20} I_n(\omega) = \omega \int dV \boldsymbol{F}_n(\boldsymbol{r}) \cdot \Delta\hat{\mathcal{P}}(\boldsymbol{r},\omega) \boldsymbol{F}_b(\boldsymbol{r},\omega) \end{equation} This representation is used for the resonance decomposition of the field in the near zone. To represent the field in the far field, we pass to the vector wave decomposition: $\boldsymbol{F}_{tot,L} = \boldsymbol{F}_{b,L} + \boldsymbol{F}_{sca,L}$. The scattering matrix can be represented accordingly: \begin{equation}\tag{21} S_{LL'}(\omega) = S^b_{LL'}(\omega) + S^{sca}_{LL'}(\omega) \end{equation} Using expressions for the coefficients of the incoming and outgoing waves, we can obtain \begin{equation}\tag{22} S^b_{LL'}(\omega) = i \oint_{\partial V} d\boldsymbol{\Sigma} \cdot (\boldsymbol{E}^{inc}_{L}\times\boldsymbol{H}_{b,L'} - \boldsymbol{E}_{b,L'}\times\boldsymbol{H}^{inc}_{L}) \end{equation}
\begin{equation}\tag{23} S^{sca}_{LL'}(\omega) = i \oint_{\partial V} d\boldsymbol{\Sigma} \cdot (\boldsymbol{E}^{inc}_{L}\times\boldsymbol{H}_{sca,L'} - \boldsymbol{E}_{sca,L'}\times\boldsymbol{H}^{inc}_{L}) \end{equation}Substituting the expression through the difference of material operators, we arrive at the expression \begin{equation}\tag{24} S_{LL'}(\omega) = S^b_{LL'}(\omega) - \sum_n \frac{\alpha_{L}^{(n)}(\omega)I^b_{n,L'}(\omega)}{\omega-\omega'} \end{equation} where \begin{equation*} \alpha^{(n)}_L = -i \oint_{\partial V} d\boldsymbol{\Sigma} \cdot (\boldsymbol{E}^{inc}_{L}\times\boldsymbol{H}_n - \boldsymbol{E}_n\times\boldsymbol{H}^{inc}_{L}) \end{equation*}
\begin{equation*} I^b_{n,L}(\omega) = \omega \int dV \boldsymbol{F}_n(\boldsymbol{r}) \cdot \Delta\hat{\mathcal{P}}(\boldsymbol{r},\omega) \boldsymbol{F}_{b,L}(\boldsymbol{r},\omega) \end{equation*}One way to find the decomposition by resonance states of complex systems is to use known or easily computable parameters of resonance states in simpler systems. By analogy with the decomposition of the problem into a background and a resonance problem, suppose that the operator $\hat{\mathcal{L}}_0$ (and its material part $\hat{\mathcal{P}}_0$) corresponds to the basic problem with known decomposition, and its difference from the problem under consideration is given by the operator $\delta\hat{\mathcal{P}}_0$. Then the eigenstate problem is rewritten as \begin{equation*} \hat{\mathcal{L}}_0(\boldsymbol{r},\omega_{\nu}) \boldsymbol{F}_{\nu}(\boldsymbol{r}) = -\omega_{\nu} \delta\hat{\mathcal{P}}_0(\boldsymbol{r},\omega_{\nu}) \boldsymbol{F}_{\nu}(\boldsymbol{r}) \end{equation*} Here the modes of the complex system are numbered by the index $\nu$ to distinguish them from the modes of the basic simple system. The solution of the equation is \begin{equation*} \boldsymbol{F}_{\nu}(\boldsymbol{r}) = -\sum_n \frac{\omega_{\nu}\boldsymbol{F}_{n}(\boldsymbol{r})}{\omega_{\nu}-\omega_n} \int_V dV' \boldsymbol{F}_{n}(\boldsymbol{r}') \delta\hat{\mathcal{P}}_0(\boldsymbol{r}',\omega_{\nu}) \boldsymbol{F}_{\nu}(\boldsymbol{r}') \end{equation*} We will look for the unknown states in the form of the expansion $\boldsymbol{F}_{\nu} = \sum_n c_n \boldsymbol{F}_{n}$. Substituting it into both parts of the last equation, we arrive at a system of algebraic equations expressing the eigenvalue problem: \begin{equation}\tag{25} (\omega_{\nu}-\omega_n) c_n = -\omega_{\nu} \sum_m V_{nm}(\omega_{\nu}) c_m \end{equation} with \begin{equation*} V_{nm} = \int_V dV' \boldsymbol{F}_{n}(\boldsymbol{r}') \delta\hat{\mathcal{P}}_0(\boldsymbol{r}',\omega_{\nu}) \boldsymbol{F}_{m}(\boldsymbol{r}') \end{equation*}
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