Finite JRS adiabatic projector bound

Put $D(s)=[P^{\mathrm{\prime }}(s),P(s)]$. Since $P,P^{\mathrm{\prime }}$ are self-adjoint, $D^{\ast }=-D$, so $K_A(s)=TH(s)+iD(s)$ is self-adjoint. Let $U_A$ solve $i{U}_A^{\mathrm{\prime }}=K_A{U}_A$, $U_A(0)=I$; [F4] gives existence and uniqueness, and differentiating $U_A^{\ast }{U}_A$ gives unitarity. Differentiating $P^2=P$ yields $P{P}^{\mathrm{\prime }}P=0$ and $P^{\mathrm{\prime }}P+P{P}^{\mathrm{\prime }}=P^{\mathrm{\prime }}$, hence $[D,P]=P^{\mathrm{\prime }}$. Therefore $P$ solves $i{P}^{\mathrm{\prime }}=[K_A,P]$, and so does $F(s)=U_A(s)P(0)U_A(s{)}^{\ast }$. Both have initial value $P(0)$; uniqueness for this finite linear matrix ODE gives $F(s)=P(s)$, or equivalently $U_A(s)P(0)U_A(s{)}^{\ast }=P(s)$.

Define the unitary wave operator $\mathrm{\Omega }(s)=U_A(s{)}^{\ast }{U}_T(s)$. Differentiating with the two evolution equations gives ${\mathrm{\Omega }}^{\mathrm{\prime }}(s)=-U_A(s{)}^{\ast }D(s)U_T(s)$ and therefore $\mathrm{\Omega }(s)=I-{\int }_0^s{U}_A(r{)}^{\ast }D(r)U_A(r)\mathrm{\Omega }(r)dr$. The intertwining identity in step 1.1 implies that the conjugated kernel is off-diagonal relative to $P(0),Q(0)$ and that ${\displaystyle Q(0)\mathrm{\Omega }(s)P(0)=-{\int }_0^{s}Q(0)P^{\mathrm{\prime }}(r)[r]P(0)\mathrm{\Omega }(r)P(0)dr,}$ where $X[r]=U_A(r{)}^{\ast }X(r)U_A(r)$.

For any differentiable $X$, the commutator identity from Finite spectral-band derivative estimates for adiabatic paths and step 1.1 gives ${\displaystyle \frac{d}{dr}(Q(0)\tilde{X}[r]P(0))=-iTQ(0)X[r]P(0)+Q(0)\dot{\tilde{X}}[r]P(0)\mathrm{.}}$ Indeed, $Q[H,\tilde{X}]P=-QXP$, and the extra commutator with $iD\mathrm{/}T$ has zero $QP$ block because $P[D,\tilde{X}]Q=Q[D,\tilde{X}]P=0$ when $D$ and $\tilde{X}$ are off-diagonal. Integrating this identity by parts in step 2.1 with $X=P^{\mathrm{\prime }}$ and $Y(r)=P(0)\mathrm{\Omega }(r)P(0)$ yields ${\displaystyle Q(0)\mathrm{\Omega }(s)P(0)=-\frac{i}{T}[Q(0)\widetilde{P^{\mathrm{\prime }}}[r]P(0)Y(r){\mid }_0^s-{\int }_0^{s}Q(0)\dot{\widetilde{P^{\mathrm{\prime }}}}[r]P(0)Y(r)dr-{\int }_0^{s}Q(0)\widetilde{P^{\mathrm{\prime }}}[r]P(0)P(0){\mathrm{\Omega }}^{\mathrm{\prime }}(r)P(0)dr]\mathrm{.}}$ Here $\mathrm{\parallel }Y(r)\mathrm{\parallel }\le 1$ and $P(0){\mathrm{\Omega }}^{\mathrm{\prime }}(r)P(0)=P(0)P^{\mathrm{\prime }}[r]Q(0)\mathrm{\Omega }(r)P(0)$, because the $PQ$ block of $[P^{\mathrm{\prime }},P]$ is $-P{P}^{\mathrm{\prime }}Q$. The last integrand is therefore bounded by $\mathrm{\parallel }Q\widetilde{P^{\mathrm{\prime }}}P\mathrm{\parallel }\mathrm{\parallel }{P}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }Q(0)\mathrm{\Omega }P(0)\mathrm{\parallel }$. Repeating the same calculation with $P(0)$ and $Q(0)$ interchanged bounds $\mathrm{\parallel }P(0)\mathrm{\Omega }Q(0)\mathrm{\parallel }$ by the same expression.

Use [F2] on the endpoint and derivative terms of step 3.1, and $\mathrm{\parallel }Q\widetilde{P^{\mathrm{\prime }}}P\mathrm{\parallel }\le \sqrt{m}\mathrm{\parallel }{P}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{/}g$ on its product term. Since $\mathrm{\parallel }Q(0)\mathrm{\Omega }P(0)\mathrm{\parallel }\le 1$, this gives ${\displaystyle \mathrm{\parallel }Q(0)\mathrm{\Omega }P(0)\mathrm{\parallel }\le \frac{1}{T}[\frac{\sqrt{m}\mathrm{\parallel }{P}^{\mathrm{\prime }}(0)\mathrm{\parallel }}{g(0)}+\frac{\sqrt{m}\mathrm{\parallel }{P}^{\mathrm{\prime }}(s)\mathrm{\parallel }}{g(s)}+{\int }_0^s(\frac{\sqrt{m}\mathrm{\parallel }Q{P}^{\mathrm{\prime }\mathrm{\prime }}P\mathrm{\parallel }}{g}+\frac{\sqrt{m}\mathrm{\parallel }{P}^{\mathrm{\prime }}{\mathrm{\parallel }}^2}{g}+\frac{2m\mathrm{\parallel }{H}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }{P}^{\mathrm{\prime }}\mathrm{\parallel }}{g^2})dr]\mathrm{.}}$ Now $\mathrm{\parallel }{P}^{\mathrm{\prime }}\mathrm{\parallel }\le \sqrt{m}\mathrm{\parallel }{H}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{/}g$. The first two integral terms are respectively at most $m\mathrm{\parallel }{H}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\parallel }\mathrm{/}{g}^2+4m\sqrt{m}\mathrm{\parallel }{H}^{\mathrm{\prime }}{\mathrm{\parallel }}^2\mathrm{/}{g}^3$ and $m\sqrt{m}\mathrm{\parallel }{H}^{\mathrm{\prime }}{\mathrm{\parallel }}^2\mathrm{/}{g}^3$; the last is at most $2m\sqrt{m}\mathrm{\parallel }{H}^{\mathrm{\prime }}{\mathrm{\parallel }}^2\mathrm{/}{g}^3$. Their sum gives exactly the integrand in $A_m(s)$, and the endpoint terms give its two endpoint summands. The same bound holds for $\mathrm{\parallel }P(0)\mathrm{\Omega }Q(0)\mathrm{\parallel }$.

Unitary conjugation by $U_A$ and step 1.1 give ${\displaystyle \mathrm{\parallel }{U}_{T}P(0)U_T^{\ast }-P(s)\mathrm{\parallel }=\mathrm{\parallel }\mathrm{\Omega }P(0){\mathrm{\Omega }}^{\ast }-P(0)\mathrm{\parallel }=\mathrm{\parallel }Q(0)\mathrm{\Omega }P(0)-P(0)\mathrm{\Omega }Q(0)\mathrm{\parallel }\mathrm{.}}$ The last operator is off-diagonal with the two blocks bounded in step 4.1; its norm is the maximum of their norms, so it is at most $A_m(s)$. For a unit $\psi \in \mathrm{ran}P(0)$, $Q(s)U_T(s)\psi =U_A(s)Q(0)\mathrm{\Omega }(s)P(0)\psi$ by step 1.1. Its squared norm is the transition probability and is at most $\mathrm{\parallel }Q(0)\mathrm{\Omega }(s)P(0){\mathrm{\parallel }}^2\le A_m(s{)}^2$.