Reducibility of quantum harmonic oscillator on $\mathbb {R}^d$ perturbed by a quasi: periodic potential with logarithmic decay

Abstract

We prove the reducibility of quantum harmonic oscillators in ${\mathbb {R}}^d$ perturbed by a quasi-periodic in time potential $V(x,\omega t)$ with $\textit{logarithmic decay}$. By a new estimate built for solving the homological equation we improve the reducibility result by Grébert-Paturel(Annales de la Faculté des sciences de Toulouse : Mathématiques. $\mathbf{28}$, 2019).

Appendix

1.1 Proof of Lemma 2.1

Recall that $\beta >\frac{1}{2}$.

• (i). Since $A\in {\mathcal {M}}_\beta ,B\in {\mathcal {M}}_\beta ^+$, then for any $a,b\in \widehat{{\mathcal {E}}}$

  $$\begin{aligned} \Vert (AB)_{[a]}^{[b]}\Vert&\le \sum _{c\in \widehat{{\mathcal {E}}}}\Vert A_{[a]}^{[c]}B_{[c]}^{[b]}\Vert \le \sum _{c\in \widehat{{\mathcal {E}}}}\Vert A_{[a]}^{[c]}\Vert \cdot \Vert B_{[c]}^{[b]}\Vert \\&\le \sum _{c\in \widehat{{\mathcal {E}}}}\frac{|A|_\beta }{\left( 1+\ln w_a\right) ^\beta \left( 1+\ln w_c\right) ^\beta }\cdot \frac{|B|_{\beta +}}{\left( 1+\ln w_c\right) ^\beta \left( 1+\ln w_b\right) ^\beta (1+|w_b-w_c|)}\\&\le \frac{|A|_\beta |B|_{\beta +}}{\left( 1+\ln w_a\right) ^\beta \left( 1+\ln w_b\right) ^\beta } \sum _{c\in \widehat{{\mathcal {E}}}}\frac{1}{\left( 1+\ln w_c\right) ^\beta (1+|w_b-w_c|)}\\&\le \frac{C|A|_\beta |B|_{\beta +}}{\left( 1+\ln w_a\right) ^\beta \left( 1+\ln w_b\right) ^\beta }\qquad \underline{\text { by Lemma}~5.1\text { and }2\beta >1. } \end{aligned}$$

  If follow that $|AB|_\beta \le C|A|_\beta |B|_{\beta +}$. Similarly, we have $|BA|_\beta \le C|A|_\beta |B|_{\beta +}$.

• (ii). Since $A,B\in {\mathcal {M}}_\beta ^+$, then for any $a,b\in \widehat{{\mathcal {E}}}$

  $$\begin{aligned} \Vert (AB)_{[a]}^{[b]}\Vert&\le \sum _{c\in \widehat{{\mathcal {E}}}}\Vert A_{[a]}^{[c]}B_{[c]}^{[b]}\Vert \le \sum _{c\in \widehat{{\mathcal {E}}}}\Vert A_{[a]}^{[c]}\Vert \cdot \Vert B_{[c]}^{[b]}\Vert \\&\le \sum _{c\in \widehat{{\mathcal {E}}}}\frac{|A|_{\beta +}|B|_{\beta +}}{\left( 1+\ln w_a\right) ^\beta \left( 1+\ln w_b\right) ^\beta \left( 1+\ln w_c\right) ^{2\beta }(1+|w_a-w_c|) (1+|w_b-w_c|)}\\&\le \left( \sum _{\begin{array}{c} c\in \widehat{{\mathcal {E}}}\\ |w_a-w_c|\ge \frac{1}{2}|w_a-w_b| \end{array}}+\sum _{\begin{array}{c} c\in \widehat{{\mathcal {E}}}\\ |w_b-w_c|\ge \frac{1}{2}|w_a-w_b| \end{array}}\right) \\&\le \frac{2|A|_{\beta +}|B|_{\beta +}}{\left( 1+\ln w_a\right) ^\beta \left( 1+\ln w_b\right) ^\beta (1+|w_a-w_b|)}\\&\quad \times \left( \sum _{c\in \widehat{{\mathcal {E}}}}\frac{1}{\left( 1+\ln w_c\right) ^{2\beta }(1+|w_b-w_c|)}+\sum _{c\in \widehat{{\mathcal {E}}}}\frac{1}{\left( 1+\ln w_c\right) ^{2\beta }(1+|w_a-w_c|)}\right) \\&\le \frac{C|A|_{\beta +}|B|_{\beta +}}{\left( 1+\ln w_a\right) ^\beta \left( 1+\ln w_b\right) ^\beta (1+|w_a-w_b|)}\qquad \underline{\text { by Lemma}~5.1\text { and }2\beta >1.} \end{aligned}$$

  This leads to that $|AB|_{\beta +}\le C|A|_{\beta +}|B|_{\beta +}$.

• (iii). Use assertion (ii) of Lemma 2.1.

• (iv). Let $A\in {\mathcal {M}}_\beta$ and $s\ge 1$. Then for any $\xi \in \ell _s^2$, we have

  $$\begin{aligned} \Vert A\xi \Vert _{-s}^2&=\sum _{a\in \widehat{{\mathcal {E}}}}w_a^{-s}\Vert \sum _{b\in \widehat{{\mathcal {E}}}}A_{[a]}^{[b]}\xi _{[b]}\Vert ^2 \le \sum _{a\in \widehat{{\mathcal {E}}}} w_a^{-s} \left( \sum _{b\in \widehat{{\mathcal {E}}}}\Vert A_{[a]}^{[b]}\Vert \cdot \Vert \xi _{[b]}\Vert \right) ^2\\&\le |A|_\beta ^2\sum _{a\in \widehat{{\mathcal {E}}}}w_a^{-s}\left( \sum _{b\in \widehat{{\mathcal {E}}}}\frac{w_b^{s/2}\Vert \xi _{[b]}\Vert }{(1+\ln w_a)^\beta (1+\ln w_b)^\beta w_b^{s/2}}\right) ^2\\&\le |A|_\beta ^2\sum _{a\in \widehat{{\mathcal {E}}}}\frac{1}{(1+\ln w_a)^{2\beta }w_a^s}\left( \sum _{b\in \widehat{{\mathcal {E}}}}\frac{1}{(1+\ln w_b)^{2\beta }w_b^s}\right) \left( \sum _{b\in \widehat{{\mathcal {E}}}}w_b^s\Vert \xi _{[b]}\Vert ^2\right) \\&\le |A|_\beta ^2\sum _{a\in \widehat{{\mathcal {E}}}}\frac{1}{(1+\ln w_a)^{2\beta }w_a}\left( \sum _{b\in \widehat{{\mathcal {E}}}}\frac{1}{(1+\ln w_b)^{2\beta }w_b}\right) \Vert \xi \Vert _s^2\\&\le C^2|A|_\beta ^2\Vert \xi \Vert _s^2. \end{aligned}$$

• (v). Case 1: $s\in [0,1]$. In this case we first prove

  $$\begin{aligned} (I):=\sum _{b\in \widehat{{\mathcal {E}}}}\frac{(w_a/w_b)^s}{(1+\ln w_b)^{2\beta }(1+|w_a-w_b|)}\le C, \quad \forall \,s\in [0,1]\text { and }\beta >\frac{1}{2}. \end{aligned}$$

  (5.1)

  We split the series above into two parts as follows:

  $$\begin{aligned} (I)=\left( \sum _{w_b>\frac{w_a}{2}}+\sum _{w_b\le \frac{wa}{2}}\right) \frac{(w_a/w_b)^s}{(1+\ln w_b)^{2\beta }(1+|w_a-w_b|)}:=(I_1)+(I_2). \end{aligned}$$

  For the former, $(w_a/w_b)^s\le 2^s\le 2$. Thus

  $$\begin{aligned} (I_1)\le \sum _{b\in \widehat{{\mathcal {E}}}}\frac{2}{(1+\ln w_b)^{2\beta }(1+|w_a-w_b|)}\le C. \end{aligned}$$

  (5.2)

  Then turn to the latter. Since $w_b\le \frac{w_a}{2}$, then $1+|w_a-w_b|\ge w_a-w_b\ge \frac{w_a}{2}\ge w_b$. Thus $1+|w_a-w_b|=(1+|w_a-w_b|)^s(1+|w_a-w_b|)^{1-s}\ge (w_a/2)^sw_b^{1-s}\ge \frac{1}{2}w_a^sw_b^{1-s}.$ Therefore, one obtains $(I_2)\le \sum _{b\in \widehat{{\mathcal {E}}}}\frac{2}{(1+\ln w_b)^{2\beta }w_b}\le C.$ Collecting the last estimate and (5.2) leads to the results (5.1).

  Now, we prepare to prove the assertion of (v) when $s\in [0,1]$. Since $A\in {\mathcal {M}}_\beta ^+$, then for any $\xi \in \ell _s^2$,

  $$\begin{aligned} \Vert A\xi \Vert _s^2&\le \sum _{a\in \widehat{{\mathcal {E}}}}w_a^s\left( \sum _{b\in \widehat{{\mathcal {E}}}}\Vert A_{[a]}^{[b]}\Vert \cdot \Vert \xi _{[b]}\Vert \right) ^2\\&\le \sum _{a\in \widehat{{\mathcal {E}}}}\frac{|A|_{\beta +}^2}{(1+\ln w_a)^{2\beta }}\left( \sum _{b\in \widehat{{\mathcal {E}}}}\frac{(w_a/w_b)^{s/2}}{(1+\ln w_b)^\beta (1+|w_a-w_b|)^{1/2}}\frac{w_b^{s/2}\Vert \xi _{[b]}\Vert }{(1+|w_a-w_b|)^{1/2}}\right) ^2\\&\le \sum _{a\in \widehat{{\mathcal {E}}}}\frac{|A|_{\beta +}^2}{(1+\ln w_a)^{2\beta }}\left( \sum _{b\in \widehat{{\mathcal {E}}}}\frac{(w_a/w_b)^s}{(1+\ln w_b)^{2\beta }(1+|w_a-w_b|)}\right) \left( \sum _{b\in \widehat{{\mathcal {E}}}}\frac{w_b^s\Vert \xi _{[b]}\Vert ^2}{1+|w_a-w_b|}\right) \\&\le C|A|_{\beta +}^2\sum _{b\in \widehat{{\mathcal {E}}}}w_b^s\Vert \xi _{[b]}\Vert ^2\sum _{a\in \widehat{{\mathcal {E}}}}\frac{1}{(1+\ln w_a)^{2\beta }(1+|w_a-w_b|)}\quad \underline{~\text {by }(5.1)~}\\&\le C^2|A|_{\beta +}^2\Vert \xi \Vert _{s}^2. \quad \underline{\text { by Lemma}~5.1 ~} \end{aligned}$$

  Next turn to the other case: $s\in [-1,0)$. Repeating similar procedures as the first case and noting that

  $$\begin{aligned} \sum _{a\in \widehat{{\mathcal {E}}}}\frac{(w_a/w_b)^s}{(1+\ln w_a)^{2\beta }(1+|w_a-w_b|)}\le C,\quad \forall \,s\in [-1,0)\text { and }\beta >\frac{1}{2}, \end{aligned}$$

  we complete the proof. $\square$

1.2 Some auxiliary lemmas

Lemma 5.1

(see Lemma A1 in [44]) For $j\ge 1$ and $\delta >1$, there exists a positive constant $C\equiv C(\delta )$ independent of j such that $\sum _{l\ge 1}\frac{1}{(1+\ln l)^\delta (1+|l-j|)}\le C.$

The following lemma is classical.

Lemma 5.2

Let $f:[0,1]\mapsto {\mathbb {R}}$ be a ${\mathcal {C}}^1$ map satisfying $|f'(x)|\ge \delta$ for all $x\in [0,1]$ and let $\kappa >0$ then $\text {Meas}\big (\{x\in [0,1]:|f(x)|\le \kappa \}\big )\le \frac{2\kappa }{\delta }.$