## 2 Results
### Expressions of Polarizability in the Rayleigh Limit
\begin{pmatrix}g_{NN-}^{A_{ED}}&g_{NN-}^{A_{EB}}&g_{NN-}^{A_{ED}} &g_{NN0-}^{A_{EB}}&g_{NN0-}^{A_{EB}}&g_{NN-}^{A_{ED}}&g_{NN-}^{A_{EB}}\\ g_{NN-}^{A_{HD}}&g_{NN-}^{A_{HB}}&g_{NN0-}^{A_{HB}}&g_{NN0-}^{A_{HB}}&g_{NN-}^{A _{HD}}&g_{NN-}^{A_{HD}}\\ g_{NN-}^{A_{ED}}&g_{NN-}^{A_{ED}}&g_{NN-}^{A_{ED}}&g_{NN0-}^{A_{ED}}&g_{NN0-}^{A _{ED}}&g_{NN-}^{A_{ED}}\\ g_{NN-}^{A_{ED}}&g_{NN-}^{A_{ED}}&g_{NN-}^{A_{ED}}&g_{NN0-}^{A_{ED}}&g_{NN0-}^{ A_{ED}}&g_{NN0-}^{A_{ED}}&g_{NN0-}^{A_{ED}}\\ g_{NN0-}^{A_{ED}}&g_{NN-}^{A_{ED}}&g_{NN0-}^{A_{ED}}&g_{NN00}^{A_{ED}}&g_{NN00}^{ A_{HD}}&g_{NN0-}^{A_{ED}}&g_{NN0-}^{A_{ED}}\\ g_{NN0-}^{A_{ED}}&g_{NN-}^{A_{ED}}&g_{NN0-}^{A_{ED}}&g_{NN00}^{A_{ED}}&g_{NN00}^{ A_{ED}}&g_{NN0-}^{A_{ED}}&g_{NN0-}^{A_{ED}}\\ g_{NN0-}^{A_{ED}}&g_{NN-}^{A_{ED}}&g_{NN-}^{A_{ED}}&g_{NN0+}^{A_{ED}}&g_{NN0+}^{ A_{ED}}&g_{NN+}^{A_{ED}}&g_{NN+}^{A_{ED}}\\ g_{NN+}^{A_{ED}}&g_{NN-}^{A_{ED}}&g_{NN0+}^{A_{ED}}&g_{NN0+}^{A_{ED}}&g_{NN+}^{ A_{ED}}&g_{NN+}^{A_{ED}}\\ \end{pmatrix} \]  
\[\text{while }\hat{U}= \begin{pmatrix}-1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&-1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&-1&0\\ 0&0&0&0&0&1\end{pmatrix} \hat{V}= \begin{pmatrix}0&1&0&0&0&0\\ -1&0&0&0&0&0\\ 0&0&0&1&0&0\\ 0&0&-1&0&0&0\\ 0&0&0&0&1\\ \end{pmatrix} .\]  
Excluding \(F_{Mq}=i(k_{q}/k_{0})\hat{V}\hat{G}_{N}F_{Nq}\) from the system Equations (29) and (30), we obtained an eigenproblem for \(F_{Nq}\), which corresponded to six bianisotropic orbitals in the Rayleigh limit  
\[\left(\hat{U}\hat{G}_{M}\hat{V}\hat{G}_{N}\right)F_{Nq}= \begin{pmatrix}k_{0}/k_{ q}\end{pmatrix} ^{2}F_{Nq} \tag{31}\]  
The boundary conditions of Equations (17)-(20) turn into  
\[j_{1}(k_{0}R)\left(pq\right)+h_{1}(k_{0}R)(ab)=\sum_{q}A_{q}j_{1}\big{(}k_{q}R )diag\{i,1,i,1,i,1\}\hat{G}_{M}F_{Mq} \tag{32}\]  
\[ \begin{array}{c}\frac{\partial}{\partial r}\left[r\,j_{1}(kr)\right]_{r=R} \left(pq\right)+\frac{\partial}{\partial r}\left[r\,h_{1}(kr)\right]_{r=R}(ab) \\ =\sum_{q}A_{q}\Big{(}\frac{k_{0}}{k_{q}}\Big{)}diag\{1,i,1,i,1,i\}\times\Big{(} \frac{\partial}{\partial r}\left[r\,j_{1}(k_{q}r)\right]_{r=R}\hat{G}_{N}F_{Nq}+j_ {1}\big{(}k_{q}R\big{)}\hat{G}_{L}F_{Nq}\Big{)}\end{array}  \tag{33}\]  
where \(\hat{G}_{L}\) is composed of the coefficients \(g_{NL}\), similar to the matrix \(\hat{G}_{N}\) provided above, \((pq)=(p_{1-1},q_{1-1},p_{10},q_{10},p_{11},q_{11})^{T}\), and \((ab)=(a_{1-1},b_{1-1},a_{10},b_{10},a_{11},b_{11})^{T}\). Taking the limit of \(k_{q}R,k_{0}R\to 0\) in the spherical Bessel functions, and excluding the scattered amplitudes \((ab)\), and substituting Equation (29) for \(\hat{G}_{M}F_{Mq}\), we obtained  
\[3(pq)=diag\{1,i,1,i,1,i\}\big{(}\hat{1}+2\hat{G}_{N}+\hat{G}_{L}\big{)}\hat{F}_{N} \overset{\rightarrow}{A} \tag{34}\]Equation (34) can be re-written using \( \begin{pmatrix}\mathbf{D}\\ \mathbf{B}\end{pmatrix} =\frac{1}{\sqrt{3\pi}}\hat{T}^{-1}\hat{F}_{N}\overset{\to}{A}, \begin{pmatrix}\mathbf{v}\\ \mathbf{h}\end{pmatrix} =\sqrt{3\pi}\hat{T} \begin{pmatrix}\mathbf{\epsilon}\\ \mathbf{h}\end{pmatrix} \), and the matrix \(\hat{T}\) given by \(\hat{T}= \begin{pmatrix}-i&+1&0&0&0&0\\ 0&0&0&1&+i&0\\ 0&0&-i\sqrt{2}&0&0&0\\ 0&0&0&0&0&\sqrt{2}\\ i&+1&0&0&0&0\\ 0&0&0&-1&+i&0\end{pmatrix} \). In accordance with Equation (34), \(\left(\hat{1}+2\hat{G}_{N}+\hat{G}_{L}\right)=\hat{T}\left(\hat{1}+2\hat{M}^{-1 }\right)\hat{T}^{-1}\), or \(3\hat{T}^{-1}\)\(\left(\hat{1}+2\hat{G}_{N}+\hat{G}_{L}\right)^{-1}\hat{T}=3\hat{M}\left(\hat{M}+2 \hat{1}\right)^{-1}\). This resulted in  
\[ \begin{pmatrix}\mathbf{D}\\ \mathbf{B}\end{pmatrix} =3\hat{M}\left(\hat{M}+2\hat{1}\right)^{-1} \begin{pmatrix} \mathbf{\epsilon}\\ \mathbf{h}\end{pmatrix}  \tag{35}\]  
\[ \begin{pmatrix}\mathbf{D}\\ \mathbf{B}\end{pmatrix} = \begin{pmatrix}\mathbf{E}\\ \mathbf{H}\end{pmatrix} +4\pi \begin{pmatrix}\mathbf{P}\\ \mathbf{M}\end{pmatrix} =\hat{M}^{-1} \begin{pmatrix}\mathbf{D}\\ \mathbf{B}\end{pmatrix} +4\pi \begin{pmatrix}\mathbf{P}\\ \mathbf{M}\end{pmatrix} \text{, we found } \begin{pmatrix}\mathbf{D}\\ \mathbf{B}\end{pmatrix} =4\pi\left(\hat{1}-\hat{M}^{-1}\right)^{-1}