ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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A visible subblock need not be unitary

Physical scope

A finite-dimensional matrix example showing the distinction between the full two-channel output and a one-channel observation. The swap matrix is stipulated as a unitary scattering-data example; no particular Maxwell device realization is asserted. The zero observed output is not material absorption.

Example

Assume ${A C}_{ω}$ .

For $z, w \in C^{2}$ , use the Hermitian inner product $⟨ z, w ⟩ = {\overline{z}}_{1} w_{1} + {\overline{z}}_{2} w_{2}$ and norm $∥ z ∥_{2}^{2} = ⟨ z, z ⟩$ . For $A \in C^{2 \times 2}$ , let $A^{†} = {\overline{A}}^{T}$ , with componentwise conjugation and ordinary transpose; $A$ is unitary when $A^{†} A = A A^{†} = I_{2}$ , where $I_{2}$ is the identity on $C^{2}$ .

Consider the algebraic two-channel scattering matrix

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

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 $P = diag (1, 0)$ , which satisfies $P^{2} = P = P^{†}$ and is the orthogonal projection onto $span {e_{1}}$ . Identify its range $ran P = span {e_{1}}$ with $C$ using the basis vector $e_{1}$ . For incident vector $a = A e_{1}$ , where $∣ A ∣^{2} = 1 W$ , the full output is $S a = A e_{2}$ , while the observed output is $P S a = 0$ . The observed operator $S_{o b s} : = P S ∣_{ran P} : ran P \to ran P$ 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.

Statement references

The Axiom of Countable Choice ( ${A C}_{ω}$ )

Finite propagating-port scattering map

Passivity gives a contractive scattering matrix

Facts & Assumptions

Given: ${A C}_{ω}$ , the displayed two-channel matrix, and the projection retaining channel 1.

[A1]

${A C}_{ω}$ is carried by the finite scattering theorem cited for the lossless full-channel interpretation. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

For a physical finite port map satisfying the contraction theorem's exact hypotheses—a bounded $C^{1}$ port domain, ${A C}_{ω}$ , fixed $ω > 0$ , a unique solution map, smooth $T$ -periodic real Maxwell fields with $ρ_{f} = J_{f} = 0$ , one stated homogeneous Lorentz or Drude branch, and the stated wall and outgoing-power bookkeeping—vanishing material, wall, and unlisted-output losses imply $S^{†} S = I_{N}$ . The port model assigns power to the full amplitude vector. Finite propagating-port scattering map Passivity gives a contractive scattering matrix

Verification

1.1givenA1F1algebra

The matrix is real and symmetric, so $S^{†} = S$ . Direct multiplication gives $S^{†} S = S^{2} = I_{2}$ ; for $a = A e_{1}$ we have $S a = A e_{2}$ and $∥ A e_{2} ∥_{2}^{2} = ∣ A ∣^{2} = 1 W = ∥ A e_{1} ∥_{2}^{2}$ . Thus this stipulated full-channel matrix satisfies the algebraic unitary condition that [F1] gives for a physical lossless model under its separate smooth Lorentz/Drude hypotheses; no such physical realization is asserted.

2.1step 1.1algebra

Since $P e_{1} = e_{1}$ and $P e_{2} = 0$ , the restriction $S_{o b s} = P S ∣_{ran P}$ sends the basis vector $e_{1}$ to $P S e_{1} = 0$ ; its one-dimensional matrix is $[0]$ . Hence $S_{o b s}^{†} S_{o b s} = [0] \neq I_{1}$ , so the observed one-channel map is not unitary.

3.1step 1.1step 2.1algebra∎

For the chosen input, the omitted component is $A e_{2}$ and has squared norm $∣ A ∣^{2} = 1 W$ , equal to the incident power. The retained component vanishes, but the full lossless map sends all power to channel 2; therefore the vanishing observed output is not evidence of material absorption.

A visible subblock need not be unitary

Example

Statement references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources