Power-normalized scattering and passivity — Examples

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let $\mathcal{I}\subset (0,\mathrm{\infty })$ be a set of admissible angular frequencies at which a closed lossless one-channel finite port model is defined, with no unlisted outgoing power. At each $\omega \in \mathcal{I}$, assume the solution satisfies all hypotheses of Passivity gives a contractive scattering matrix: the source-free macroscopic Maxwell equations with ${\rho }_f=J_f=0$, a smooth $T$-periodic real solution and material states on one of the stated homogeneous finite Lorentz or Drude branches, and the stated passive-wall and outgoing-power bookkeeping. Here $T=2\pi \mathrm{/}\omega$; material, wall, and unlisted-output losses vanish for every input. Write the scalar scattering matrix at each frequency as $S(\omega )=[r(\omega )]$. Then $\mathrm{\mid }r(\omega )\mathrm{\mid }=1$ for each such frequency. This fixes reflected power but leaves the phase unconstrained by power conservation: any unit-modulus scalar has the same power factor. For any real-valued function $\varphi :\mathcal{I}\to \mathbb{R}$, $r_{\varphi }(\omega )=e^{i\varphi (\omega )}$ has unit modulus; when this circle-valued function is nonconstant, its phase varies. If $\mathcal{I}$ contains two distinct frequencies, such a varying family exists: set $\varphi$ equal to $0$ and $\pi$ at those two frequencies, respectively, and set $\varphi =0$ elsewhere. This is an algebraic freedom allowed by the power constraint, not a claim that every such phase law is realized by a Maxwell device.

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

Finite propagating-port scattering map

Passivity gives a contractive scattering matrix

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$.

For $z,w\in {\mathbb{C}}^2$, use the Hermitian inner product $⟨z,w⟩={\bar{z}}_1{w}_1+{\bar{z}}_2{w}_2$ and norm $\mathrm{\parallel }z{\mathrm{\parallel }}_2^2=⟨z,z⟩$. For $A\in {\mathbb{C}}^{2\times 2}$, let $A^{†}={\bar{A}}^{\mathsf{T}}$, with componentwise conjugation and ordinary transpose; $A$ is unitary when $A^{†}A=A{A}^{†}=I_2$, where $I_2$ is the identity on ${\mathbb{C}}^2$.

Consider the algebraic two-channel scattering matrix

$S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\mathrm{.}$

The displayed matrix is unitary. Let $e_1=(1,0{)}^T$ and $e_2=(0,1{)}^T$. A one-channel observer retains only channel 1, represented by $P=\mathrm{diag}(1,0)$, which satisfies $P^2=P=P^{†}$ and is the orthogonal projection onto $\mathrm{span}\{e_1\}$. Identify its range $\mathrm{ran}P=\mathrm{span}\{e_1\}$ with $\mathbb{C}$ using the basis vector $e_1$. For incident vector $a=A{e}_1$, where $\mathrm{\mid }A{\mathrm{\mid }}^2=1\mathrm{W}$, the full output is $Sa=A{e}_2$, while the observed output is $PSa=0$. The observed operator $S_{\mathrm{o}\mathrm{b}\mathrm{s}}:=PS{\mathrm{\mid }}_{\mathrm{ran}P}:\mathrm{ran}P\to \mathrm{ran}P$ is represented by the one-by-one matrix $[0]$, so it is not unitary even though the full two-channel map conserves power. The omitted second output accounts for the power; there is no material absorption in this example. The matrix is a consistency example, not a claim that a specified waveguide device realizes it.

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

Finite propagating-port scattering map

Passivity gives a contractive scattering matrix