TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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Wigner–Smith matrix for differentiable finite channels

Statement

Let $I \subset R$ be an open energy interval with fixed finite channel number $r \geq 1$ , let $S : I \to U (r)$ be $C^{1}$ in the finite-channel convention of Wave, scattering and channel operators, and let $ℏ > 0$ . Define

$Q (E) : = - i ℏ S (E)^{*} S^{'} (E), E \in I .$

Then $Q (E)$ is Hermitian and

$tr Q (E) = - i ℏ \frac{(\det S)^{'} (E)}{\det S (E)} = - i ℏ \frac{d}{d E} \log \det S (E),$

where the last expression means the logarithmic derivative $(\det S)^{'} / \det S$ , independent of a choice of logarithm branch. If the incoming and outgoing bases are changed by energy-independent unitary matrices $U_{i n}$ and $U_{o u t}$ , so that $S^{♯} = U_{o u t}^{*} S U_{i n}$ , then

$Q^{♯} = U_{i n}^{*} Q U_{i n} .$

If $E$ is measured in joules and $ℏ$ in joule-seconds, $Q$ has units of seconds. This is an algebraic scattering-matrix result; it does not by itself establish an operational arrival-time or dwell-time interpretation.

Statement references

Wave, scattering and channel operators

Facts & Assumptions

Given: The open interval $I$ , fixed $r$ , $C^{1}$ unitary matrix field $S (E)$ , and $ℏ > 0$ .

[F1]

A finite-channel scattering matrix is a matrix field $S (E) \in U (r)$ ; $S^{*}$ is conjugate transpose, while trace and determinant are the usual finite-dimensional trace and alternating determinant (Wave, scattering and channel operators).

[F2]

The finite-dimensional derivative, product and trace operations are entrywise operations on $C^{1}$ scalar functions; determinant is the alternating multilinear polynomial in the matrix entries (Wave, scattering and channel operators).

Proof

Given: [F1], [F2] and the data in the Statement.

1.1givenF1F2algebra

Differentiating $S (E)^{*} S (E) = I_{r}$ entrywise gives $S^{'} (E)^{*} S (E) + S (E)^{*} S^{'} (E) = 0$ . Thus $A : = S^{*} S^{'}$ satisfies $A^{*} = - A$ , and $Q^{*} = (- i ℏ A)^{*} = i ℏ A^{*} = - i ℏ A = Q$ .

2.1step 1.1F1F2algebra

Since $S$ is unitary, $S^{- 1} = S^{*}$ and $\det S \neq 0$ . Multilinearity of the alternating determinant gives $\det (I_{r} + h A + o (h)) = 1 + h tr (A) + o (h)$ for $A = S (E)^{- 1} S^{'} (E)$ : terms linear in $h$ are exactly the $r$ choices of one diagonal entry, while every nonidentity permutation or two off-diagonal factors contributes $o (h)$ . Applying this to $S (E + h) = S (E) (I_{r} + h S (E)^{- 1} S^{'} (E) + o (h))$ yields $(\det S)^{'} (E) = \det S (E) tr (S (E)^{- 1} S^{'} (E)) = \det S (E) tr (S (E)^{*} S^{'} (E))$ . Multiplication by $- i ℏ / \det S (E)$ proves the trace formula.

3.1step 1.1givenalgebra∎

For constant unitary $U_{i n}, U_{o u t}$ , differentiation gives $(S^{♯})^{'} = U_{o u t}^{*} S^{'} U_{i n}$ , while $(S^{♯})^{*} = U_{i n}^{*} S^{*} U_{o u t}$ . Therefore $(S^{♯})^{*} (S^{♯})^{'} = U_{i n}^{*} S^{*} S^{'} U_{i n}$ , proving the stated conjugation rule. Energy-dependent basis changes introduce derivatives of the basis matrices, so they are not covered by this claim.

Depends on

Wave, scattering and channel operators