Analytic perturbation remainder

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) and the hypotheses of Simple-level reduced-resolvent coefficients on an arbitrary complex Hilbert space $H$, with inner product linear in the first variable: $H_0$ is self-adjoint on its exact dense domain, $E$ is an isolated simple eigenvalue with gap at least $g>0$, ${\psi }_0$ is a normalized eigenvector, and $V=V^{\ast }$ is bounded with $v=\mathrm{\parallel }V\mathrm{\parallel }$. Let $P_{\lambda },E_{\lambda }$ be the contour projection and eigenvalue, $Q=I-P_0$, $S={\iota }_Q(H_0{\mathrm{\mid }}_{D(H_0)\cap QH}-E{)}^{-1}Q$ the actual reduced inverse, and $\psi (\lambda )$ the gauge-fixed eigenvector with $⟨\psi (\lambda ),{\psi }_0⟩=1$.

For $v>0$ put $R=g\mathrm{/}(8v)$. On the full complex disk $\mathrm{\mid }\lambda \mathrm{\mid }<R$:

(a) $q=2\mathrm{\mid }\lambda \mathrm{\mid }v\mathrm{/}g\le 1\mathrm{/}4$, $\mathrm{\parallel }{P}_{\lambda }-P_0\mathrm{\parallel }\le 1\mathrm{/}3$, $\mathrm{\parallel }\psi (\lambda )\mathrm{\parallel }\le 2$, and the holomorphic scalar energy identity is $E_{\lambda }=⟨(H_0+\lambda V)\psi (\lambda ),{\psi }_0⟩=E+\lambda ⟨V\psi (\lambda ),{\psi }_0⟩\mathrm{.}$

(b) The exact coefficients are $E^{\mathrm{\prime }}(0)=⟨V{\psi }_0,{\psi }_0⟩,{\psi }^{\mathrm{\prime }}(0)=-SV{\psi }_0,E^{\mathrm{\prime }\mathrm{\prime }}(0)=-2⟨VSV{\psi }_0,{\psi }_0⟩,$ with bounds $\mathrm{\mid }{E}^{\mathrm{\prime }}(0)\mathrm{\mid }\le v$, $\mathrm{\parallel }{\psi }^{\mathrm{\prime }}(0)\mathrm{\parallel }\le v\mathrm{/}g$ and $\mathrm{\mid }{E}^{\mathrm{\prime }\mathrm{\prime }}(0)\mathrm{\mid }\le 2v^2\mathrm{/}g$. The whole complementary spectral subspace, including its continuum, contributes through $S$. The cubic remainder satisfies $\mid E_{\lambda }-E-\lambda E^{\mathrm{\prime }}(0)-\frac{{\lambda }^2}{2}{E}^{\mathrm{\prime }\mathrm{\prime }}(0)\mid \le \frac{Rv(\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{/}R{)}^3}{1-\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{/}R}\mathrm{.}$ At $\mathrm{\mid }\lambda \mathrm{\mid }=R\mathrm{/}2$ the displayed bound is $Rv\mathrm{/}4=g\mathrm{/}32$. For $\mathrm{\mid }\lambda \mathrm{\mid }\le R\mathrm{/}2$ it is at most $128v^3\mathrm{\mid }\lambda {\mathrm{\mid }}^3\mathrm{/}{g}^2$.

(c) If $v=0$, then $V=0$ and $P_{\lambda }=P_0$, $E_{\lambda }=E$, $\psi (\lambda )={\psi }_0$ for every complex $\lambda$; no radius involving division by $v$ is used in this branch.

These conclusions are conditional on the strict positive gap and bounded perturbation, and do not assert an unbounded or degenerate perturbation theorem.

The former moving-vector energy identity fails even for real coupling in the advertised small disk. Take $H_0=\mathrm{diag}(1,3)$, $E=1$, $g=2$, $V=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and $\lambda =20\mathrm{/}99<R=1\mathrm{/}4$. The eigenvalue is $97\mathrm{/}99$ and the gauge vector is $(1,-1\mathrm{/}10)$, of squared norm $101\mathrm{/}100$. Thus $⟨H_{\lambda }\psi ,\psi ⟩=9797\mathrm{/}9900\mathrm{\ne }97\mathrm{/}99$, whereas pairing with the fixed ${\psi }_0=(1,0)$ gives the correct energy.

The former interior resolvent exclusion also fails: for $H_0=\mathrm{diag}(0,1)$, $V=\mathrm{diag}(1,0)$, $E=0$, $g=v=1$, and $\lambda =z=1\mathrm{/}16$, $z$ is an eigenvalue although $\mathrm{\mid }z-E\mathrm{\mid }<g\mathrm{/}2-\mathrm{\mid }\lambda \mathrm{\mid }v=7\mathrm{/}16$. The valid Neumann condition is distance from all of $\sigma (H_0)$ greater than $\mathrm{\mid }\lambda \mathrm{\mid }v$, and leads to one-sided spectral inclusion as proved in step 1.1. The former evaluated remainder value $g\mathrm{/}64$ was an arithmetic error; the displayed general bound evaluates to $g\mathrm{/}32$, without any assertion that it is attained.