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Rotating-wave finite-time error

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)). Let H be a complex Hilbert space, let A=A∗∈B(H) and B∈B(H) with ω∈R∖{0}, and put (units ℏ=1) H(t):=A+eiωtB+e−iωtB∗,Havg:=A. Let U(t,s) and Uavg(t,s) be the unitary propagators of Norm Dyson series and uniqueness. Then for all s≤t: ∥U(t,s)−Uavg(t,s)∥≤4∥B∥∣ω∣[1+(t−s)(∥A∥+2∥B∥)]. With ℏ restored, replace A,B by A/ℏ,B/ℏ, i.e. ∥U(t,s)−Uavg(t,s)∥≤4∥B∥ℏ∣ω∣[1+t−sℏ(∥A∥+2∥B∥)].

The bound is linear in t−s; it makes no uniform-in-time claim and no claim about resonance, decay or an unbounded dipole coupling.

Facts & Assumptions

Given: CC; the arbitrary complex Hilbert space H; A=A∗∈B(H), B∈B(H), ω∈R∖{0}; and s≤t. To verify both the normalized and restored formulas at once, take ℏ>0 and let U,Uavg solve the equations with generators −iH(r)/ℏ and −iA/ℏ. Write V(r)=eiωrB+e−iωrB∗, a=∥A∥, b=∥B∥ and τ=t−s.

[F1]

The actual corrected bounded Dyson supplier Norm Dyson series and uniqueness applies on every compact real interval to arbitrary H, norm-continuous bounded self-adjoint families, and any ℏ>0. It gives unitary U and Uavg, their norm-continuous norm derivatives ∂rU(t,r)=(i/ℏ)U(t,r)H(r) and ∂rUavg(r,s)=−(i/ℏ)AUavg(r,s), and identity initial values. Uniqueness identifies their restrictions from overlapping compact intervals. Since H(r)∗=H(r) and ∥H(r)∥≤a+2b for all real r, this supplies the propagators needed for every finite pair s≤t. Its arbitrary-H CC contract carries no general Stone or AC spectral assumption.

[F2]

Operator multiplication is bounded bilinear, ∥CD∥≤∥C∥∥D∥, ∥B∗∥=∥B∥, and the identity has norm at most one (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum; the actual CC bounded-adjoint construction is included in [F1]). Define Z(r)=eiωrB/(iω)−e−iωrB∗/(iω). Direct scalar exponential differentiation gives Z′=V, Z∈C1(R;B(H)) and ∥Z(r)∥≤2b/∣ω∣. The bilinear product rule follows from the norm difference quotient and submultiplicativity.

[F3]

The actual Banach-valued FTC Fundamental theorem of calculus for Banach-valued continuous curves assumes CC, applies with X=B(H), and gives F(t)−F(s)=∫stF′(r)dr for a norm-C1 curve with continuous derivative on the compact interval. These continuous integrands are Bochner integrable and obey ∥∫stf(r)dr∥≤∫st∥f(r)∥dr (Bochner integral norm inequality). The operator Banach-space completeness for arbitrary H is proved in [F1]. The truncation lemma is not an FTC supplier and is not invoked.

Proof

1.1F1F2F3givenalgebra

[Exact Duhamel identity with units restored] Suppose first s<t. Put F(r)=U(t,r)Uavg(r,s) on [s,t]. By [F1]–[F2], F is norm-C1 and F′=(i/ℏ)U(t,r)(H(r)−A)Uavg(r,s)=(i/ℏ)U(t,r)V(r)Uavg(r,s). Its endpoint values are F(t)=Uavg(t,s) and F(s)=U(t,s). The FTC [F3] therefore gives U(t,s)−Uavg(t,s)=−(i/ℏ)∫stU(t,r)V(r)Uavg(r,s)dr. This identity holds as an operator-norm Bochner integral, not merely on a chosen finite-dimensional subspace.

2.1step 1.1F1F2F3givenalgebra

[Integration by parts removes only the middle derivative] Let P(r)=U(t,r), Q(r)=Uavg(r,s) and G(r)=P(r)Z(r)Q(r). All three factors are norm-C1. The exact product rule is G′=P′ZQ+PZ′Q+PZQ′. Solving for the middle term and applying [F3] yields ∫stPZ′Q=G(t)−G(s)−∫st[P′ZQ+PZQ′]dr. Thus the remainder after the boundary term differentiates only the two outer propagators; it does not differentiate the whole product PZQ a second time or retain PZ′Q. By [F1], P′ZQ+PZQ′=(i/ℏ)P(HZ−ZA)Q. Combining this with step 1.1 and Z′=V gives the exact identity U(t,s)−Uavg(t,s)=−(i/ℏ)[Z(t)Uavg(t,s)−U(t,s)Z(s)]−ℏ−2∫stU(t,r)[H(r)Z(r)−Z(r)A]Uavg(r,s)dr. Every order and sign follows from the displayed three-term product rule.

3.1step 2.1F1F2F3givenalgebra

[Full finite-time constant] Unitarity gives norm at most one to both outer propagators. The two boundary summands in step 2.1 have total norm at most (∥Z(t)∥+∥Z(s)∥)/ℏ≤4b/(ℏ∣ω∣). Its remainder integrand has norm at most (∥H(r)∥+∥A∥)∥Z(r)∥≤(2a+2b) 2b/∣ω∣. Applying [F3] and integrating over length τ gives ∥U(t,s)−Uavg(t,s)∥≤4b/(ℏ∣ω∣)+4bτ(a+b)/(ℏ2∣ω∣). Since a+b≤a+2b, this is at most [4b/(ℏ∣ω∣)][1+τ(a+2b)/ℏ], precisely the full restored constant in the Statement. The slightly stronger intermediate estimate is used only to prove that asserted bound.

4.1step 1.1step 2.1step 3.1F1F2F3givenalgebra∎

[Normalized case and all boundary conditions] Taking ℏ=1 in step 3.1 gives the first displayed inequality of the Statement; arbitrary ℏ>0 gives its restored formula without changing ω or introducing an extra factor into Z′=V. The s=t case has U(s,s)=Uavg(s,s)=IH, so the left side is zero and the displayed inequality holds. If B=0, both generators coincide and uniqueness gives equality of propagators, again agreeing with the bound. Negative ω is handled by ∣ω∣ in [F2], and negative absolute times cause no change because [F1] applies on any compact real interval. The zero Hilbert space is included by the norm-at-most-one estimates. CC is used through the actual arbitrary-H Dyson and FTC suppliers; no extra separability, unbounded domain, spectral or AC assumption is introduced. This proves exactly the stated finite-time estimate and retains its explicit absence of a uniform-time claim.

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