Finite spectral-band derivative estimates for adiabatic paths

Fix $s$ and write $P={\sum }_{\lambda \in {\mathrm{\Lambda }}_P}{P}_{\lambda }$. The displayed definition makes $\tilde{X}$ off-diagonal: $P\tilde{X}P=Q\tilde{X}Q=0$. On each band-to-complement block, $[H,\tilde{X}]$ multiplies $-P_{\lambda }X{\widehat{R}}_{\lambda }$ by $\lambda -H{\mathrm{\mid }}_Q$, giving $P_{\lambda }XQ$; on the reverse block it gives $-QX{P}_{\lambda }$. Summing yields $[H,\tilde{X}]=PX-XP$.

Choose an orthonormal basis $e_1,\dots ,e_m$ of $\mathrm{ran}P$ adapted to the mutually orthogonal $P_{\lambda }$. For each basis vector $e_j$ in $\mathrm{ran}{P}_{\lambda }$, $Q\tilde{X}{e}_j=-{\widehat{R}}_{\lambda }QX{e}_j$, so [F6] gives ${\sum }_j\mathrm{\parallel }Q\tilde{X}{e}_j{\mathrm{\parallel }}^2\le g^{-2}{\sum }_j\mathrm{\parallel }QX{e}_j{\mathrm{\parallel }}^2\le m\mathrm{\parallel }QXP{\mathrm{\parallel }}^2\mathrm{/}{g}^2$. For a unit vector $u={\sum }_j{u}_j{e}_j$, Cauchy–Schwarz [F4] yields $\mathrm{\parallel }Q\tilde{X}Pu\mathrm{\parallel }\le {\sum }_j\mathrm{\mid }{u}_j\mathrm{\mid }\mathrm{\parallel }Q\tilde{X}{e}_j\mathrm{\parallel }\le ({\sum }_j\mathrm{\mid }{u}_j{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}({\sum }_j\mathrm{\parallel }Q\tilde{X}{e}_j{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}$, proving $\mathrm{\parallel }Q\tilde{X}P\mathrm{\parallel }\le \sqrt{m}\mathrm{\parallel }QXP\mathrm{\parallel }\mathrm{/}g$. Since each ${\widehat{R}}_{\lambda }$ and $P_{\lambda }$ is self-adjoint, taking adjoints in the defining sum gives $\widetilde{X^{\ast }}={\tilde{X}}^{\ast }$. Applying the column estimate to $X^{\ast }$ and using [F5] gives $\mathrm{\parallel }P\tilde{X}Q\mathrm{\parallel }\le \sqrt{m}\mathrm{\parallel }PXQ\mathrm{\parallel }\mathrm{/}g$. Since $\tilde{X}$ is off-diagonal, for $u=p+q$ one has $\mathrm{\parallel }\tilde{X}u{\mathrm{\parallel }}^2=\mathrm{\parallel }P\tilde{X}Qq{\mathrm{\parallel }}^2+\mathrm{\parallel }Q\tilde{X}Pp{\mathrm{\parallel }}^2$; hence its full norm is the maximum of the two block norms. This proves $\mathrm{\parallel }\tilde{X}\mathrm{\parallel }\le \sqrt{m}\mathrm{\parallel }X\mathrm{\parallel }\mathrm{/}g$.

On a local fixed Riesz contour put $R_z=(H-zI{)}^{-1}$ and ${\widehat{R}}_z=Q(H{\mathrm{\mid }}_Q-zI{)}^{-1}Q$. Then $\tilde{X}=(2\pi i{)}^{-1}\oint R_{z}X{R}_{z}dz$. For $G(A,B)=(2\pi i{)}^{-1}\oint R_{z}A{R}_{z}B{R}_{z}dz$, using $[H-z,R_z]=0$ gives ${\displaystyle [H,G(A,B)]=\frac{1}{2\pi i}\oint (A{R}_{z}B{R}_z-R_{z}A{R}_{z}B)dz=A\tilde{B}-\tilde{A}B\mathrm{.}}$ Indeed, $R_z[H-z,A]R_z=A{R}_z-R_{z}A$ and $R_z[H-z,B]R_z=B{R}_z-R_{z}B$; adding the two commutator contributions cancels the middle terms. Write $C=A\tilde{B}-\tilde{A}B$. On each band eigenspace $P_{\lambda }$, the Sylvester equations from this commutator give ${\displaystyle PGQ=P\tilde{C}Q,QGP=-Q\tilde{C}P,}$ because $P_{\lambda }\tilde{C}Q=-P_{\lambda }C{\widehat{R}}_{\lambda }$ and $Q\tilde{C}{P}_{\lambda }=-{\widehat{R}}_{\lambda }C{P}_{\lambda }$.

Differentiate $HP=PH$ and multiply on the left by $Q$ and on the right by each $P_{\lambda }$. The resulting Sylvester equation is $(H{\mathrm{\mid }}_Q-\lambda I)Q\dot{P}{P}_{\lambda }=-Q\dot{H}{P}_{\lambda }$, so $Q\dot{P}{P}_{\lambda }=-{\widehat{R}}_{\lambda }\dot{H}{P}_{\lambda }$. Differentiating $P^2=P$ gives $P\dot{P}P=Q\dot{P}Q=0$; self-adjointness gives the other off-diagonal block by adjoint. These identities equal the reduced-resolvent formula for $\tilde{\dot{H}}$, proving $\dot{P}=\tilde{\dot{H}}$. Step 2.1 then gives $\mathrm{\parallel }\dot{P}\mathrm{\parallel }\le \sqrt{m}\mathrm{\parallel }\dot{H}\mathrm{\parallel }\mathrm{/}g$.

The diagonal blocks follow by expanding $R_z={\sum }_{\lambda \in {\mathrm{\Lambda }}_P}{P}_{\lambda }\mathrm{/}(\lambda -z)+{\widehat{R}}_z$. In $QGQ$, the only non-holomorphic terms have the middle resolvent in $P_{\lambda }$, and their integrals are $-{\widehat{R}}_{\lambda }A{P}_{\lambda }B{\widehat{R}}_{\lambda }$; terms with all three resolvents in $Q$ are holomorphic inside the contour. Hence $QGQ=-Q\tilde{A}\tilde{B}Q$. In $PGP$, terms with three band resolvents have zero integral because their rational integrands decay as $z^{-3}$ and all their poles lie inside the contour. The remaining terms have two band poles $\lambda ,\mu$ and one $Q$ resolvent. Their scalar contour coefficient equals ${\widehat{R}}_{\lambda }^2$ if $\lambda =\mu$ by Cauchy's derivative formula, and equals $({\widehat{R}}_{\lambda }-{\widehat{R}}_{\mu })\mathrm{/}(\lambda -\mu )={\widehat{R}}_{\lambda }{\widehat{R}}_{\mu }$ if $\lambda \mathrm{\ne }\mu$ by the resolvent identity. Thus $PGP=P\tilde{A}\tilde{B}P$. Together with step 2.2, these blocks prove ${\displaystyle G(A,B)=(P-Q)(\tilde{A}\tilde{B}+\widetilde{A\stackrel{~}{B}}-\widetilde{\stackrel{~}{A}B})\mathrm{.}}$ All contours enclose the whole band; coincident band eigenvalues are grouped into one $P_{\lambda }$, and the complementary resolvent is holomorphic throughout the enclosed region.

Differentiating the contour formula and using ${\dot{R}}_z=-R_z\dot{H}{R}_z$ gives ${\displaystyle \dot{\tilde{X}}=\tilde{\dot{X}}-G(\dot{H},X)-G(X,\dot{H})\mathrm{.}}$ Apply step 3.2 to both $G$ terms and use $\dot{P}=\tilde{\dot{H}}$ from step 3.1. Their sum is ${\displaystyle (P-Q)(\dot{P}\tilde{X}+\tilde{X}\dot{P}+\widetilde{[\dot{H},\stackrel{~}{X}]}-\widetilde{[\dot{P},X]}),}$ because the remaining terms are $\widetilde{\dot{H}\stackrel{~}{X}+X\dot{P}-\dot{P}X-\stackrel{~}{X}\dot{H}}=\widetilde{[\dot{H},\stackrel{~}{X}]}-\widetilde{[\dot{P},X]}$. Negating this sum in the derivative formula yields ${\displaystyle \dot{\tilde{X}}=\tilde{\dot{X}}+(Q-P)(\dot{P}\tilde{X}+\tilde{X}\dot{P}+\widetilde{[\dot{H},\stackrel{~}{X}]}-\widetilde{[\dot{P},X]})\mathrm{.}}$ The contour formula is local, so it remains valid at internal band eigenvalue coincidences; no selected eigenvalue is crossed by the contour.

Apply step 4.1 to $X=\dot{H}$. Since $\tilde{\dot{H}}=\dot{P}$ is off-diagonal, ${\dot{P}}^2$ is block diagonal and has zero $QP$ block. Also $[\dot{P},\dot{H}]=-[\dot{H},\dot{P}]$. Therefore ${\displaystyle Q\ddot{P}P=Q\tilde{\ddot{H}}P+2Q\widetilde{[\dot{H},\dot{P}]}P\mathrm{.}}$ The block estimate in step 2.1 bounds the first term by $\sqrt{m}\mathrm{\parallel }\ddot{H}\mathrm{\parallel }\mathrm{/}g$ and the second by $2(\sqrt{m}\mathrm{/}g)\mathrm{\parallel }[\dot{H},\dot{P}]\mathrm{\parallel }\le 4\sqrt{m}\mathrm{\parallel }\dot{H}\mathrm{\parallel }\mathrm{\parallel }\dot{P}\mathrm{\parallel }\mathrm{/}g\le 4m\mathrm{\parallel }\dot{H}{\mathrm{\parallel }}^2\mathrm{/}{g}^2$ using submultiplicativity [F3] and step 3.1. This is the claimed bound.

Apply step 4.1 to $X=\dot{P}$. The products $\dot{P}\tilde{\dot{P}}$ and $\tilde{\dot{P}}\dot{P}$ are block diagonal, and $[\dot{P},\dot{P}]=0$. Hence ${\displaystyle Q\dot{\tilde{\dot{P}}}P=Q\tilde{\ddot{P}}P+Q\widetilde{[\dot{H},\stackrel{~}{\dot{P}}]}P\mathrm{.}}$ The block bound from step 2.1 makes the first term at most $\sqrt{m}\mathrm{\parallel }Q\ddot{P}P\mathrm{\parallel }\mathrm{/}g$. The second is at most $(\sqrt{m}\mathrm{/}g)\mathrm{\parallel }[\dot{H},\tilde{\dot{P}}]\mathrm{\parallel }\le (2\sqrt{m}\mathrm{/}g)\mathrm{\parallel }\dot{H}\mathrm{\parallel }\mathrm{\parallel }\tilde{\dot{P}}\mathrm{\parallel }\le 2m\mathrm{\parallel }\dot{H}\mathrm{\parallel }\mathrm{\parallel }\dot{P}\mathrm{\parallel }\mathrm{/}{g}^2$, using submultiplicativity [F3] and steps 2.1 and 3.1.