Unbounded Trace Orbits of Thue–Morse Hamiltonian

Abstract

It is well known that, an energy is in the spectrum of Fibonacci Hamiltonian if and only if the corresponding trace orbit is bounded. However, it is not known whether the same result holds for the Thue–Morse Hamiltonian. In this paper, we give a negative answer to this question. More precisely, we construct two subsets $\Sigma _{II}$ and $\Sigma _{III}$ of the spectrum of the Thue–Morse Hamiltonian, both of which are dense and uncountable, such that each energy in $\Sigma _{II}\cup \Sigma _{III}$ corresponds to an unbounded trace orbit. Exact estimates on the norm of the transfer matrices are also obtained for these energies: for $E\in \Sigma _{II}\cup \Sigma _{III},$ the norms of the transfer matrices behave like

$$\begin{aligned} e^{c_1\gamma \sqrt{n}}\le \Vert T_{ n}(E)\Vert \le e^{c_2\gamma \sqrt{n}}. \end{aligned}$$

However, two types of energies are quite different in the sense that each energy in $\Sigma _{II}$ is associated with a two-sided pseudo-localized state, while each energy in $\Sigma _{III}$ is associated with a one-sided pseudo-localized state. The difference is also reflected by the local dimensions of the spectral measure: the local dimension is 0 for energies in $\Sigma _{II}$ and is larger than 1 for energies in $\Sigma _{III}.$ As a comparison, we mention another known countable dense subset $\Sigma _I$. Each energy in $\Sigma _I$ corresponds to an eventually constant trace map and the associated eigenvector is an extended state. In summary, the Thue–Morse Hamiltonian exhibits “mixed spectral nature”.

Appendix

In this appendix, we give another proof for the fact that the Thue–Morse Hamiltonian has no point spectrum.

Theorem 8.1

([19, 29]) $H_\lambda$ has no point spectrum.

We begin with the following observation:

Lemma 8.2

If $E\in \mathbb R$ is such that $H_\lambda \phi =E\phi$ has a solution $\phi$ with $\phi _{\pm n}\rightarrow 0$ when $n\rightarrow \infty ,$ then E is a type-II energy.

Proof

Since the two-sided Thue–Morse sequence is reflection-symmetric about 1 / 2, there are only two possibilities: either $\vec {\phi }_0\parallel v_{\pi /4}$ or $\vec {\phi }_0\parallel v_{-\pi /4}$, as argued in [19]. Without loss of generality, we assume $\vec {\phi }_0=v_{\pi /4}=(1,-1)^t/\sqrt{2}$. Then we have

$$\begin{aligned} A_{2n}\vec {\phi }_0= (\phi _{2^{2n}+1},\phi _{2^{2n}})^t\rightarrow \vec {0} \end{aligned}$$

when $n\rightarrow \infty .$ On the other hand, by Remark 2.3, $A_{2n}$ has the following form( see also [19]):

$$\begin{aligned} A_{2n}= \begin{pmatrix} a_n&{}\quad b_n\\ -b_n&{}\quad c_n \end{pmatrix}. \end{aligned}$$

Thus $a_n-b_n=\sqrt{2}\phi _{2^{2n}+1}$ and $-b_n-c_n=\sqrt{2}\phi _{2^{2n}}$. Consequently

$$\begin{aligned} t_{2n}(E)=a_n+c_n=\sqrt{2}(\phi _{2^{2n}+1}-\phi _{2^{2n}})\rightarrow 0. \end{aligned}$$

By the recurrence relation of $t_n$, we get

$$\begin{aligned} t_{2n+1}(E)=2-\frac{2-t_{2n+2}(E)}{t_{2n}^2(E)}\rightarrow -\infty . \end{aligned}$$

By Lemma 4.6, E is a type-II energy. $\square$

Proof of Theorem 8.1

Assume on the contrary that E is an eigenvalue of $H_\lambda$. Then there exists nonzero $\phi \in \ell ^2(\mathbb Z)$ such that $H_\lambda \phi =E\phi .$ Then by the above lemma, E is a type-II enenrgy. However this will contradict with Theorem 1.5, since by that theorem, no $\ell ^2$ solution is possible for $E\in \Sigma _{II}.$ $\square$