An efficient and unified method for band structure calculations of 2D anisotropic photonic-crystal fibers

Abstract

In this article, band structure calculations of two dimensional (2D) anisotropic photonic-crystal fibers (PhCFs) are considered. In 2D PhCFs, Maxwell’s equations for the transversal electric and magnetic mode become decoupled, but the difficulty, arising from the anisotropic permittivity ${{\varvec{\varepsilon }}}$ and/or permeability ${{\varvec{\mu }}},$ plaguing the frequency-domain finite difference method, especially the original Yee’s scheme, is our top concern. To resolve this difficulty, we re-establish the connection between the lowest order finite element method with the quasi-periodic condition and Yee’s scheme using 2D non-orthogonal mesh, whereby the decoupled Maxwell’s equations in 2D anisotropic PhCFs are readily discretized into a generalized eigenvalue problem (GEP). Moreover, we spell out the nullspace of the resulting GEP, if it exists, and explicitly construct the Moore–Penrose pseudoinverse of the singular coefficient matrix, whose smallest positive eigenvalues can be solved by the inverse Lanczos method. Extensive band structures of 2D PhCFs are calculated and benchmarked against reliable results to demonstrate the accuracy and efficiency of our method.

Appendices

Appendix A: Derivation of (7)

Using vector calculus identities, we have

$$\begin{aligned} \nabla \times {\varvec{\mu }}^{-1}\cdot \nabla \times (E_z\vec {{\textbf {e}}}_z)= \nabla \times {\varvec{\mu }}^{-1}\cdot (-\vec {{\textbf {e}}}_z\times \nabla E_z). \end{aligned}$$

For convenience, denote $\vec {{{\textbf {v}}}}:={\varvec{\mu }}^{-1}\cdot (\vec {{\textbf {e}}}_z\times \nabla E_z).$ After projecting both sides of (3a) onto $\vec {{\textbf {e}}}_z,$ we obtain

$$\begin{aligned}&\omega ^2\varepsilon _{z} E_{z}\vec {{\textbf {e}}}_{z}\cdot \vec {{\textbf {e}}}_{z} = \omega ^2\varepsilon _{z} E_{z}, \nonumber \\&\vec {{\textbf {e}}}_{z} \cdot \nabla \times (-\vec {{{{\textbf {v}}}}}) = -\nabla \cdot ({\vec {{{\textbf {v}}}}}\times \vec {{\textbf {e}}}_{z})= \nabla \cdot (\vec {{\textbf {e}}}_{z}\times \vec {{{\textbf {v}}}}), \end{aligned}$$

for the RHS and LHS of (3a), respectively.

Remember that $E_z$ is a function of ${{\textbf {r}}}_{\perp },$ hence $\nabla E_z$ should be understood as $[\nabla _{\perp } E_z, 0],$ then $\vec {{\textbf {e}}}_z\times \nabla E_z$ is just $\nabla _{\perp } E_z$ rotated counterclockwise by $\pi /2$ within the plane perpendicular to $\vec {{\textbf {e}}}_z$ and represented by $[\vec {{\textbf {e}}}_z\times \nabla _{\perp } E_z,0].$ Furthermore, by (4), $\vec {{{\textbf {v}}}}$ should be understood as $[{\varvec{\mu }}_{\perp }^{-1}\cdot (\vec {{\textbf {e}}}_z\times \nabla _{\perp } E_z),0]$ with $\vec {{\textbf {e}}}_z\times \vec {{{\textbf {v}}}}$ being understood similarly to $\vec {{\textbf {e}}}_z\times \nabla E_z,$ therefore $\nabla \cdot (\vec {{\textbf {e}}}_z\times \vec {{{\textbf {v}}}})$ should be seen as $\nabla _{\perp }\cdot (\vec {{\textbf {e}}}_z\times \vec {{{\textbf {v}}}})+0.$ Thus, after using the definition of $\vec {{{\textbf {v}}}}$ above, we obtain exactly the LHS of (7a).

Appendix B: Derivation of (28c)

With $\varvec{a}_1$ and $\varvec{a}_2$ as basis vectors of ${\varvec{\Omega }}$ shown in (15a), we can define the reciprocal vectors $\varvec{a}^1$ and $\varvec{a}^2$ such that $\varvec{a}^\ell \cdot \varvec{a}_{\ell '} = \delta _{\ell ',\ell }, \ell ',\ell =1,2,$ or $[\varvec{a}^1,\varvec{a}^2][\varvec{a}_1, \varvec{a}_2]^{\top } = I_2$ in terms of matrix language. Then, given the basis function $\phi _{ij}, i\in {\mathbb {N}}_1,j\in {\mathbb {N}}_2,$ it follows from [29] that

$$\begin{aligned} \nabla _{\perp }\phi _{ij} = \varvec{a}^1\partial _\xi \phi _{ij} + \varvec{a}^2\partial _\eta \phi _{ij}. \end{aligned}$$

(B1)

Furthermore, it follows from (15b) and [29] that

$$\begin{aligned} \varvec{a}_1 = (\varvec{a}^2\times \vec {{\textbf {e}}}_z)\Vert \varvec{a}_1\times \varvec{a}_2\Vert _2,\quad \varvec{a}_2 =(\vec {{\textbf {e}}}_z \times \varvec{a}^1)\Vert \varvec{a}_1\times \varvec{a}_2\Vert _2. \end{aligned}$$

(B2)

Immediately, we see that (B1) and (B2) lead to (28c).