Excitation of Multiple Surface Plasmon-Polaritons by a Metal Layer Inserted in an Equichiral Sculptured Thin Film

Abstract

Excitation of multiple surface plasmon-polaritons (SPPs) by an equichiral sculptured thin film with a metal layer defect was studied theoretically in the Sarid configuration, using the transfer matrix method. Multiple SPP modes were distinguished from waveguide modes in optical absorption for p-polarized plane wave. The degree of localization of multiple SPP waves was investigated by calculation of the time-averaged Poynting vector. The results showed that the long-range and short-range SPP waves can simultaneously be excited at both interfaces of metal core in this proposed structure which may be used in a broad range of sensing applications.

Appendix A: The Transfer Matrix Method

The transfer matrix of the bottom, top of ECSTFs and metallic thin film can be obtained by solving the source-free Maxwell curl postulates in that regions:

$$ {\displaystyle \begin{array}{l}\underline{\nabla}\times \underline{E}\left(\underline{r}\right)= i\omega \underline{B}\left(\underline{r}\right)\\ {}\underline{\nabla}\times \underline{H}\left(\underline{r}\right)=- i\omega \underline{D}\left(\underline{r}\right)\end{array}} $$

(5)

where D and B are the displacement electric and the magnetic induction fields, respectively. The constitutive relations for D and B are as

$$ {\displaystyle \begin{array}{l}\underline{D}\left(\underline{r}\right)={\varepsilon}_0\;{\underline {\underline{\varepsilon}}}_{ECSTF}.\underline{E}\left(\underline{r}\right)\\ {}\underline{B}\left(\underline{r}\right)={\mu}_0\underline{H}\left(\underline{r}\right)\end{array}} $$

(6)

in lth arm of the ECSTF and

$$ {\displaystyle \begin{array}{l}\underline{D}\left(\underline{r}\right)={\varepsilon}_0\;{\varepsilon}_{met}\underline{E}\left(\underline{r}\right)\\ {}\underline{B}\left(\underline{r}\right)={\mu}_0\underline{H}\left(\underline{r}\right)\end{array}} $$

(7)

in metal mediums. The ε_met is the metal homogenous dielectric permittivity, and the nonhomogeneous dielectric permittivity ${\underline {\underline{\varepsilon}}}_{ECSTF}$ for lth arm of the ECSTF is defined as [12]

$$ {\underline {\underline{\varepsilon}}}_{ECSTF}={\underline {\underline{S}}}_z^T\left(\zeta \right).{\underline {\underline{S}}}_y\left(\chi \right).{\underline {\underline{\varepsilon}}}_{ref}.{\underline {\underline{S}}}_y^T\left(\chi \right).{\underline {\underline{S}}}_z\left(\zeta \right) $$

(8)

where the superscript T indicates the transpose of a dyadic, ζ = h(l − 1) (2π/Q) − γ, χ is tilt angle of the nanocolumns of the ECSTF,h =  + 1(or − 1) is the right-handedness (or left-handedness) of the ECSTF, and γ is the offset angle(the starting point for growth) of the nanocolumns of the ECSTF relative to the x axis in x-y plane. In our work, we fixed ψ_inc = 0^° and the structure rotates in clockwise direction so that it is considered as −γ in the ζ angle (Fig. 1).

The local relative permittivity, rotation, and tilt dyadics are, respectively, as [35]

$$ {\displaystyle \begin{array}{l}{\underline {\underline{\varepsilon}}}_{ref}={\varepsilon}_a\;{\underline{u}}_z{\underline{u}}_z+{\varepsilon}_b\;{\underline{u}}_x{\underline{u}}_x+{\varepsilon}_c\;{\underline{u}}_y{\underline{u}}_y\\ {}{\underline {\underline{S}}}_z=\left(\;{\underline{u}}_x{\underline{u}}_x+{\underline{u}}_y{\underline{u}}_y\right)\;\cos \zeta +\left(\;{\underline{u}}_y{\underline{u}}_x-{\underline{u}}_x{\underline{u}}_y\right)\;\sin \zeta +{\underline{u}}_z{\underline{u}}_z\\ {}{\underline {\underline{S}}}_y=\left(\;{\underline{u}}_x{\underline{u}}_x+{\underline{u}}_z{\underline{u}}_z\right)\;\cos \chi +\left(\;{\underline{u}}_z{\underline{u}}_x-{\underline{u}}_x{\underline{u}}_z\right)\;\sin \chi +{\underline{u}}_y{\underline{u}}_y\end{array}} $$

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where ε_{a, b, c} are the relative permittivity scalars.

With considering electromagnetic fields in dielectric and metal regions are as follows:

$$ {\displaystyle \begin{array}{l}\underline{E}\left(\underline{r}\right)=\underline{e}(z)\;{e}^{ikx}\\ {}\underline{H}\left(\underline{r}\right)=\underline{h}(z)\;{e}^{ikx}\end{array}} $$

(10)

and using Eqs. 5–7, one can obtain four ordinary differential equations and two algebraic equations. The two algebraic equations for e_z and h_z in ECSTF (for lth arm) and metal mediums are, respectively, as follows:

$$ {\displaystyle \begin{array}{l}{e}_z(z)=\frac{\frac{\left({\varepsilon}_a-{\varepsilon}_b\right)}{2}\sin 2\chi\;A-\frac{k}{\varepsilon_0\omega }{h}_y(z)}{B}\\ {}{h}_z(z)=\frac{k}{\mu_0\omega }{e}_y(z)\end{array}} $$

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where A = (e_x(z) cos ζ − e_y(z) sin ζ) and B = (ε_acos²χ + ε_bsin²χ).

$$ {\displaystyle \begin{array}{l}{e}_z(z)=-\frac{k}{\varepsilon_0{\omega \varepsilon}_{met}}{h}_y(z)\\ {}{h}_z(z)=\frac{k}{\mu_0\omega }{e}_y(z)\end{array}} $$

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The four ordinary differential equations can be sorted as a matrix equation:

$$ \frac{d}{dz}\left[\underline{f}(z)\right]=i\left[\underline {\underline{P}}(z)\right]\left[\underline{f}(z)\right] $$

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where $\underline{\Big[f}(z)\Big]={\left[{e}_x(z)\kern0.36em {e}_y(z)\kern0.36em {h}_x(z)\kern0.48em {h}_y(z)\;\right]}^T$ is a column vector. The $\left[\underline {\underline{P}}(z)\right]$ is a 4 × 4 matrix, and the elements of it forlth arm of the ECSTF are

$$ {P}_{11}=\frac{k\left({\varepsilon}_a-{\varepsilon}_b\right)\cos \zeta \sin 2\chi }{2B},{P}_{12}=-{P}_{11}\tan \zeta, {P}_{14}={\mu}_0\omega -\frac{k^2}{\varepsilon_0\omega\;B},{P}_{23}=-{\mu}_0\omega, {P}_{31}={\varepsilon}_0\omega \frac{\sin 2\zeta }{2}\left(\frac{\varepsilon_a{\varepsilon}_b}{B}-{\varepsilon}_c\right),{P}_{32}=\frac{k^2}{\mu_0\omega }-{\varepsilon}_0\omega\;\left(\frac{\varepsilon_a{\varepsilon}_b{\sin}^2\zeta }{B}+{\varepsilon}_c{\cos}^2\zeta \right),{p}_{34}=-{P}_{12},{P}_{41}={\varepsilon}_0\omega \left(\frac{\varepsilon_a{\varepsilon}_b{\cos}^2\zeta }{B}+{\varepsilon}_c{\sin}^2\zeta \right),{P}_{42}=-{P}_{31},{P}_{44}={P}_{11} $$

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and the other elements are zero. The elements of $\left[\underline {\underline{P}}(z)\right]$ in metal region are

$$ {\displaystyle \begin{array}{l}{P}_{14}={\mu}_0\omega -\frac{k^2}{\varepsilon_0{\omega \varepsilon}_{met}},{P}_{23}=-{\mu}_0\omega, {P}_{32}=\frac{k^2}{\mu_0\omega }-{\varepsilon}_0{\omega \varepsilon}_{met},{P}_{41}={\varepsilon}_0{\omega \varepsilon}_{met}\\ {}\end{array}} $$

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and the else elements are zero.

We now consider an ECSTF with thickness of d = N t, where t is the thickness of an arm of ECSTF and N is the number of arms. The transfer matrix of a columnar thin film with thickness of t is ${e}^{i\;\left[\underline {\underline{P}}\right]\;t}$. Therefore, the transfer matrix of an ECSTF is [33, 34]

$$ \left[\underline {\underline{M}}\right]={\left[\underline {\underline{M}}\right]}_N\;{\left[\underline {\underline{M}}\right]}_{N-1}\dots {\left[\underline {\underline{M}}\right]}_3{\left[\underline {\underline{M}}\right]}_2{\left[\underline {\underline{M}}\right]}_1 $$

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where ${\left[\underline {\underline{M}}\right]}_l={e}^{i\;\left[{\underline {\underline{P}}}_l\right]\;t},l=1,2,\dots, N$. Then, this method can be used to obtain the transfer matrix of the bottom and top of ECSTFs in Eq. 3 ($\left[{\underline {\underline{M}}}_b\right]$ and $\left[{\underline {\underline{M}}}_t\right]$). Also, easily, the transfer matrix of metal is just ${\left[\underline {\underline{M}}\right]}_m={e}^{i\;\left[\underline {\underline{P}}\right]\;{d}_{met}}$.

By rewriting Eq. 1 as $\left[\underline{f}\right(z=-{\left({d}_b+\frac{d_{met}}{2}\right)}_{-}\Big]=\left[\underline {\underline{K}}\left({\theta}_{inc}\right)\right]{\left[{a}_S\;{a}_P\;{r}_S\kern0.36em {r}_P\right]}^T$ in incident medium (including incident and reflected electric fields) and$\left[\underline{f}\right(z={\left({d}_t+\frac{d_{met}}{2}\right)}_{+}\Big]=\left[\underline {\underline{K}}\left({\theta}_{tr}\right)\right]{\left[{t}_S\;{t}_P\;0\kern0.36em 0\right]}^T$in transmitted medium, the nonzero elements of $\left[\underline {\underline{K}}\left({\theta}_{inc}\right)\right]$ and $\left[\underline {\underline{K}}\left({\theta}_{tr}\right)\right]$, respectively, are

$$ {K}_{12}=-{K}_{14}=-\cos {\theta}_{inc},{K}_{21}={K}_{23}=1,{K}_{31}=-{K}_{33}=\frac{n_1}{\eta {}_0}{K}_{12},{K}_{42}={K}_{44}=\frac{-{n}_1}{\eta {}_0} $$

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and

$$ {K}_{12}=-{K}_{14}=-\cos {\theta}_{tr},{K}_{21}={K}_{23}=1,\kern0.36em {K}_{31}=-{K}_{33}=\frac{n_2}{\eta {}_0}{K}_{12},{K}_{42}={K}_{44}=\frac{-{n}_2}{\eta {}_0} $$

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