First-order amplitude tail

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Adopt the hypotheses and notation of Strong bounded interaction picture: $H$ is a complex Hilbert space with inner product linear in the first variable, $H_0$ is self-adjoint with domain $D(H_0)$, $\mathrm{\hslash }>0$, $V:[a,b]\to \mathcal{B}(H)$ is strongly continuous with self-adjoint values and $v={\sup }_t\mathrm{\parallel }V(t)\mathrm{\parallel }<\mathrm{\infty }$, $K=v\mathrm{/}\mathrm{\hslash }$, and $U(t,s)$ is the mild propagator of that item. Suppose in addition that $({\phi }_n{)}_{n\in \mathbb{N}}$ is a complete orthonormal family of $H$ consisting of eigenvectors of $H_0$, with $H_0{\phi }_n=E_n{\phi }_n$ and $E_n\in \mathbb{R}$ (Orthonormal families, complete orthonormal systems and Hilbert bases), and fix indices $a_0,b_0$ and $s\le t$ in $[a,b]$, $\tau :=t-s$. Define the transition amplitude $A_{b_0{a}_0}(t,s):=e^{i{E}_{b_0}(t-s)\mathrm{/}\mathrm{\hslash }}⟨U(t,s){\phi }_{a_0},{\phi }_{b_0}⟩\mathrm{.}$ Then $A_{b_0{a}_0}(t,s)={\delta }_{a_0{b}_0}-\frac{i}{\mathrm{\hslash }}{\int }_s^t{e}^{i(E_{b_0}-E_{a_0})(r-s)\mathrm{/}\mathrm{\hslash }}⟨V(r){\phi }_{a_0},{\phi }_{b_0}⟩dr+R_{b_0{a}_0}(t,s),$ where the remainder satisfies the explicit bound $\mathrm{\mid }{R}_{b_0{a}_0}(t,s)\mathrm{\mid }\le e^{K\tau }\frac{(K\tau {)}^2}{2}\mathrm{.}$

Consequently, if $\eta :[a,b]\to {\mathbb{R}}_{\ge 0}$ is integrable with $\mathrm{\parallel }V(t)-V_0(t)\mathrm{\parallel }\le \eta (t)$ for a family $V_0$ of self-adjoint bounded operators, assume the selected scalar coefficient $c_0(r)=⟨V_0(r){\phi }_{a_0},{\phi }_{b_0}⟩$ is Lebesgue measurable (equivalently here, require it to be integrable). Then the first-order amplitude computed with $V_0$ differs from the exact one by at most $\frac{1}{\mathrm{\hslash }}{\int }_s^t\eta (r)dr+e^{K\tau }(K\tau {)}^2\mathrm{/}2$. No physical transition probability, no measurement postulate and no long-time limit are asserted.

The former backward pairing fails already when $V=0$, $H_0=I$ on ${\mathrm{\ell }}^2(\mathbb{N})$, $\mathrm{\hslash }=1$, $\tau =\pi \mathrm{/}2$ and $a_0=b_0$. It gives $e^{i\tau }⟨{\phi }_a,e^{-i\tau }{\phi }_a⟩=e^{2i\tau }=-1$, whereas the claimed zero-remainder free amplitude would be $1$. The forward pairing is $1$. The matrix element also needs the same ordering: on two coordinate vectors let $H_0=0$ and $V=\left(\begin{array}{cc}0& i\\ -i& 0\end{array}\right)$, extended by zero on the remaining coordinates. The forward amplitude from the first to the second coordinate is $-\sin \tau$, with derivative $-1$; $-i⟨V{\phi }_a,{\phi }_b⟩=-1$, while the former backward matrix element gives $+1$.

A norm bound alone does not imply measurability of an arbitrary selected coefficient. In a setting admitting a non-Lebesgue-measurable $A\subseteq [a,b]$, $V=0$ and $V_0(r)={\mathbf{1}}_A(r)I$ obey $\mathrm{\parallel }V-V_0\mathrm{\parallel }\le 1$ but have the nonmeasurable diagonal coefficient ${\mathbf{1}}_A$. This conditional example asserts no existence of such a set from CC alone; the comparison hypothesis now states the exact scalar measurability it uses. The retained remainder constant also has the direct check $K{\int }_0^{\tau }(e^{Kx}-1)dx\le K{\int }_0^{\tau }Kx{e}^{Kx}dx\le (K\tau {)}^2{e}^{K\tau }\mathrm{/}2$, including $K=0$ without division. This removes the extra factor of $K$ in the former intermediate equality.