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Static Bohr-frequency evaluation of the jump integral

Statement

Assume Countable Choice. Work on Cd, d≥1, with its standard complex inner product, conjugate-transpose adjoint †, and induced operator norm. Let HS=HS† have an orthonormal eigenbasis ∣m⟩, 1≤m≤d, with real eigenvalues Em. Let Xα=Xα† and ∥Xα∥≤1, 1≤α≤N. Use the normalized N×N frequency and time kernels h and the positive parameter γ of the normalization lemma. Assume the entrywise Fourier identities ∫Rh(u)eiωu du=2πh(ω) at every Bohr frequency ω=En−Em. For real s,t, define U(t,s)=e−iHS(t−s) by the finite-dimensional matrix-exponential series. For 1≤λ≤N, the Schrödinger-picture jump Lλ(t)=γ∑α=1N∫Rhλα(t−s)U(t,s)XαU(t,s)† ds exists absolutely, is independent of t, and equals 2πγ∑m,n=1d∑α=1Nhλα(En−Em)(Xα)mn∣m⟩⟨n∣, where (Xα)mn denotes the matrix entry in the given eigenbasis.

Facts & Assumptions

[F1]

The normalized time kernel h has finite zeroth and first mixed-norm moments, and γ>0. ULE kernel adjoint symmetry and rate identities

[F2]

Assuming Countable Choice, on a finite-dimensional system, measurable Hermitian couplings of operator norm at most one give absolutely convergent jump integrals for this kernel. Convergent jump integrals and Hermitian Lamb operators

[F3]

For matrices on Cd with the standard complex inner product, the adjoint is the conjugate transpose, the operator norm is the unit-ball supremum, and ezA=∑k=0∞zkAk/k! converges absolutely in operator norm for every complex z. Finite-dimensional matrix exponential

[F4]

If A=A† on Cd and r∈R, then e−irA is unitary and depends differentiably, hence continuously, on r in operator norm. This is the supplier with ℏ=1. Unitarity and differentiation of the finite Hermitian matrix exponential

[A1]

Assume Countable Choice for the convergence supplier [F2] and the change-of-variables theorem [F7]. The Axiom of Countable Choice (AC_ω)

[F5]

For a measurable real function p, its positive and negative parts are p+=max⁡{p,0} and p−=max⁡{−p,0}. For integrable p, both parts have finite integral and ∫p=∫p+−∫p−. For a complex integrable f=p+iq, the integral is ∫f=∫p+i∫q, and integrability means finite integral of ∣f∣. Integrable real and complex functions, and their integrals

[F6]

The space L1(μ) is a complex vector space and its Lebesgue integral is complex-linear. Thus finite sums and constant complex factors commute with the integral of integrable functions. The Lebesgue integral is linear on L¹(μ)

[F7]

Assume Countable Choice. For open U,V⊆Rn, a C1 diffeomorphism T:U→V, and nonnegative Lebesgue measurable a, the change-of-variables formula is ∫Va(y) dy=∫Ua(T(x))∣det⁡DT(x)∣ dx. We use U=V=R and Tt(u)=t−u, for which Tt−1=Tt and ∣Tt′∣=1. A C¹ diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions

Proof

Given: Countable Choice, the finite-dimensional complex system, Hermitian HS, real times, bounded Hermitian couplings, normalized kernel, and pointwise inversion identities in the Statement. Goal: Prove absolute convergence and evaluate Lλ(t).

1.1givenF3F4

Put u=t−s. Since HSk∣m⟩=Emk∣m⟩, apply the norm-convergent series in [F3] to each eigenvector to obtain U(t,s)∣m⟩=e−iEmu∣m⟩. As Em,u are real, its adjoint is diagonal with entries eiEmu. By [F4], U(t,s) is unitary for every real s,t.

1.2A1F5F7algebra

Fix real t and put Tt(u)=t−u. For any complex f=p+iq∈L1(R), apply [F7] to ∣f∣ to obtain ∫∣f∘Tt∣=∫∣f∣<∞. Apply [F7] separately to the four nonnegative measurable functions p+,p−,q+,q− in [F5]. Their integrals are finite, and composition with Tt commutes with taking positive and negative parts. Subtract the two real-part identities and the two imaginary-part identities, then combine them by [F5]. This proves ∫Rf(s) ds=∫Rf(t−u) du for complex integrable f. In particular, reflection introduces no sign in a Lebesgue integral because the Jacobian is used in absolute value.

2.1givenA1F1F2F3F4step 1.1

For each fixed real t, put Yα(t)(s)=U(t,s)XαU(t,s)†. By [F4] its matrix entries are continuous in s, hence measurable. Conjugate-transpose multiplication gives (Yα(t)(s))†=Yα(t)(s). Unitarity gives ∥Uv∥=∥v∥=∥U†v∥ for every vector v; consequently ∥UXαU†v∥≤∥Xα∥∥v∥, so the unit-ball supremum in [F3] gives ∥Yα(t)(s)∥≤1. Together with [A1], these meet the hypotheses needed to apply [F2], with the kernel from [F1], proving absolute convergence of each Lλ(t).

3.1F1F5F6step 1.1step 2.1step 1.2algebra

Step 1.1 gives (Yα(t)(s))mn=(Xα)mnei(En−Em)(t−s). The entrywise integrands are integrable by step 2.1. Apply step 1.2 to each such entry and use [F6] to extract the constant (Xα)mn and commute the finite matrix and coupling sums with their entrywise integrals. The functions u↦hλα(u)ei(En−Em)u are integrable: their modulus is ∣hλα(u)∣, bounded by the mixed norm whose integral is finite in [F1]. Thus the coefficient multiplying ∣m⟩⟨n∣ is γ∑α=1N(Xα)mn∫Rhλα(u)ei(En−Em)u du.

4.1givenstep 3.1∎

Apply the stated entrywise Fourier identity at each real En−Em to the coefficient in step 3.1. This gives exactly the finite sum in the Statement, containing no t, and proves time independence as well as the claimed formula. The Fourier identities remain explicit hypotheses; no pointwise inversion conclusion is inferred from integrability alone.

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