Reciprocal port normalization gives symmetric scattering

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ and the finite port model at a fixed real $\omega >0$. Let $S(\omega )\in {\mathbb{C}}^{N\times N}$ be its unique scattering matrix. Suppose the continuous device phasor coefficient maps $\epsilon ,\mu :\bar{\mathrm{\Omega }}\to {\mathbb{C}}^{3\times 3}$ obey $\epsilon (x{)}^T=\epsilon (x)$ and $\mu (x{)}^T=\mu (x)$ pointwise, and for every pair of incoming vectors the corresponding source-free scattering solutions lie in $C^1(\bar{\mathrm{\Omega }};{\mathbb{C}}^3)$ and have reciprocal port modes satisfying the exact bilinear pairing stipulated in the model definition. Then

$S(\omega {)}^T=S(\omega )\mathrm{.}$

For $A\in {\mathbb{C}}^{N\times N}$, define its conjugate transpose by $A^{†}:={\bar{A}}^{\mathsf{T}}$, where the overline is componentwise conjugation and a superscript $\mathsf{T}$ is ordinary transpose; $A$ is Hermitian when $A^{†}=A$ and unitary when $A^{†}A=A{A}^{†}=I_N$, with $I_N$ the identity on ${\mathbb{C}}^N$. The transpose relation above uses the unconjugated Lorentz pairing; the unitary relation below uses the separate conjugated Hermitian power pairing. No time-reversal claim follows.

For each incoming vector $a$, define the operational net cycle-averaged material power transferred from the fields in this source-free phasor model by

$P_{\mathrm{b}\mathrm{u}\mathrm{l}\mathrm{k}}(a):=\frac{\omega }{2}{\int }_{\mathrm{\Omega }}\mathrm{Im}({\tilde{E}}^{†}\epsilon \tilde{E}+{\tilde{H}}^{†}\mu \tilde{H})dV\mathrm{.}$

Here $v^{†}={\bar{v}}^T$ for a complex column vector, and this power has units W. It is a signed net transfer, not an assumption of local passivity or a definition of a material realization. Let $P_{\mathrm{\partial }}(a)$ and $P_{\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}}(a)$ be the nonnegative outward wall and unlisted-channel powers of the same finite-port model, whose total outward boundary power is $\mathrm{\parallel }Sa{\mathrm{\parallel }}_2^2-\mathrm{\parallel }a{\mathrm{\parallel }}_2^2+P_{\mathrm{\partial }}(a)+P_{\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}}(a)$. Then

$\mathrm{\parallel }a{\mathrm{\parallel }}_2^2-\mathrm{\parallel }Sa{\mathrm{\parallel }}_2^2=P_{\mathrm{b}\mathrm{u}\mathrm{l}\mathrm{k}}(a)+P_{\mathrm{\partial }}(a)+P_{\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}}(a)\mathrm{.}$

Consequently, if all three powers vanish for every incoming vector $a\in {\mathbb{C}}^N$, then $S^{†}S=S{S}^{†}=I_N$, so $S$ is symmetric and unitary. Pointwise Hermitian $\epsilon$ and $\mu$ suffice to make $P_{\mathrm{b}\mathrm{u}\mathrm{l}\mathrm{k}}(a)=0$ for every input, since their quadratic forms are real; Hermitian symmetry is distinct from the ordinary transpose symmetry required for reciprocity. No constitutive branch beyond the stated phasor model is required for this operational zero-power conclusion.

The following full finite-port contraction hypotheses give additional Lorentz and Drude sufficient cases on this same port map. For every incoming vector, its real field maps $E,H,D,B:\bar{\mathrm{\Omega }}\times \mathbb{R}\to {\mathbb{R}}^3$ and its material-state maps on that domain are jointly $C^{\mathrm{\infty }}$ and $T$-periodic, where $T=2\pi \mathrm{/}\omega$, and satisfy the smooth source-free macroscopic Maxwell equations with ${\rho }_f=J_f=0$.

The material closure is one of these two homogeneous, constant-coefficient branches. In the finite Lorentz branch, choose finite $M\ge 1$ and maps $P_{\mathrm{\ell }}:\bar{\mathrm{\Omega }}\times \mathbb{R}\to {\mathbb{R}}^3$ for $1\le \mathrm{\ell }\le M$. They satisfy

$D={\epsilon }_{\mathrm{\infty }}E+{\sum }_{\mathrm{\ell }=1}^M{P}_{\mathrm{\ell }},B=\mu H,{\mathrm{\partial }}_t^2{P}_{\mathrm{\ell }}+{\gamma }_{\mathrm{\ell }}{\mathrm{\partial }}_t{P}_{\mathrm{\ell }}+{\omega }_{\mathrm{\ell }}^2{P}_{\mathrm{\ell }}=a_{\mathrm{\ell }}E,$

with constant ${\epsilon }_{\mathrm{\infty }},\mu ,a_{\mathrm{\ell }},{\omega }_{\mathrm{\ell }}>0$ and ${\gamma }_{\mathrm{\ell }}\ge 0$. In the Drude branch, take maps $P,V:\bar{\mathrm{\Omega }}\times \mathbb{R}\to {\mathbb{R}}^3$. They satisfy

$D={\epsilon }_{\mathrm{\infty }}E+P,B=\mu H,{\mathrm{\partial }}_{t}P=V,{\mathrm{\partial }}_{t}V+\gamma V=aE,$

with constant ${\epsilon }_{\mathrm{\infty }},\mu ,a>0$ and $\gamma \ge 0$. Their SI units are those specified in the finite-port contraction model.

For each input, require the real fields and material states to have only this frequency: let $\tilde{E},\tilde{H},\tilde{D},\tilde{B}:\bar{\mathrm{\Omega }}\to {\mathbb{C}}^3$ be their phasor amplitudes with $E(x,t)=\mathrm{Re}(\tilde{E}(x)e^{-i\omega t})$, and analogously for $H,D,B$. In the Lorentz branch also write $P_{\mathrm{\ell }}(x,t)=\mathrm{Re}({\tilde{P}}_{\mathrm{\ell }}(x)e^{-i\omega t})$ with ${\tilde{P}}_{\mathrm{\ell }}:\bar{\mathrm{\Omega }}\to {\mathbb{C}}^3$; in the Drude branch write $P,V$ in the same form with amplitudes $\tilde{P},\tilde{V}:\bar{\mathrm{\Omega }}\to {\mathbb{C}}^3$. Substitution into the closure equations gives

${\epsilon }_{\mathrm{L}}(\omega )=({\epsilon }_{\mathrm{\infty }}+{\sum }_{\mathrm{\ell }=1}^M\frac{a_{\mathrm{\ell }}}{{\omega }_{\mathrm{\ell }}^2-{\omega }^2-i{\gamma }_{\mathrm{\ell }}\omega })I_3,{\epsilon }_{\mathrm{D}}(\omega )=({\epsilon }_{\mathrm{\infty }}-\frac{a}{{\omega }^2+i\gamma \omega })I_3,$

where every Lorentz denominator is required nonzero, $I_3$ is the $3\times 3$ identity, and the magnetic coefficient is $\mu I_3$ in either branch. Require these harmonic coefficient maps to be the continuous coefficients of this same reciprocal-phasor port model. The matrices are symmetric under ordinary transpose. The branch loss density is $\mathrm{\ell }={\sum }_{j=1}^M({\gamma }_j\mathrm{/}{a}_j)\mathrm{\mid }{\mathrm{\partial }}_t{P}_j{\mathrm{\mid }}^2$ in the Lorentz branch and $\mathrm{\ell }=(\gamma \mathrm{/}a)\mathrm{\mid }V{\mathrm{\mid }}^2$ in the Drude branch, using the real vector norm $\mathrm{\mid }v{\mathrm{\mid }}^2=v\cdot v$. For the solution with incoming vector $a$, define $P_{\mathrm{m}\mathrm{a}\mathrm{t}}(a):=T^{-1}{\int }_0^T{\int }_{\mathrm{\Omega }}\mathrm{\ell }dVdt$, and let $P_{\mathrm{\partial }}(a)$ and $P_{\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}}(a)$ be its nonnegative passive-wall and unlisted-outgoing powers under the finite-port model's bookkeeping. The field, material-state, and parameter units are exactly those specified in that model.

In either harmonic branch, $P_{\mathrm{b}\mathrm{u}\mathrm{l}\mathrm{k}}(a)=P_{\mathrm{m}\mathrm{a}\mathrm{t}}(a)$. Thus $P_{\mathrm{m}\mathrm{a}\mathrm{t}}(a)=P_{\mathrm{\partial }}(a)=P_{\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}}(a)=0$ for every incoming vector $a$ is a sufficient case of the operational zero-power criterion, also supplied by the contraction result. These branch hypotheses are sufficient cases, not restrictions on the general phasor power identity above.

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)

Finite propagating-port scattering map

Reciprocity in reciprocal linear media

Passivity gives a contractive scattering matrix

The divergence and curl of a cross product

Divergence on a bounded C1 Euclidean domain