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\section*{Summary : Quantum dynamics for a fractal spectrum : Thouless coefficient - Thouless formula - quantum dynamics for a Fibonacci potential and for a Cantor spectrum in the tight-binding model}

We note $K$ the triadic Cantor set, embedded in $[0,1]$, and $d\mu$ the associated measure. $d\mu$ verifies :

\begin{equation}
\int_k f(x) d\mu(x) = \frac{1}{2}\int_k f\left(\frac{x}{3}\right) d\mu(x)+\frac{1}{2}\int_k f\left(\frac{x+2}{3}\right) d\mu(x)
\end{equation}

\subsection*{Return probability}

cf. correlation-fun-cantor.pdf

\subsection*{Thouless coefficient}

\subsection*{Thouless formula}

\subsection*{Fibonacci potential in the tight-binding model}

\subsection*{Quantum dynamics for a triadic Cantor set spectrum in the tight-binding model}

%We consider an infinite 1D lattice, with sites $k$ labelled from $0$ to $\infty$. We define a potential $V$ on this system such that the discrete Schrodinger operator : $H = -\Delta+V$, which acts on $l_2(\mathbb{Z})$, has a triadic Cantor set spectrum. As in the previous paragraph, the eigenfunctions $\{\phi_{\epsilon}\}_{\epsilon \in K}$, evaluated on the sites $k$, can be seen as polynomials in $\epsilon$ : 
%$$\phi_{\epsilon}(k) = p_k(\epsilon)$$

Therefore, the wave function $\psi(k,t)$, describing the wave packet, is given by :
%\begin{eqnarray*}
%\lefteqn{\psi(k,t) = \int_{K}g(\epsilon)\phi_{\epsilon}(k)e^{-i\epsilon t}d\mu}\\
%& = & \int_K g(\epsilon)p_{k}(\epsilon)e^{-i\epsilon t}d\mu \\
%& = & \sum_{j_k\leq k}c_{j_k}\int_{K}g(\epsilon)\epsilon^{j_k}e^{-i\epsilon t}d\mu
%\end{eqnarray*}

Thus, the dynamical behaviour of the system depends essentially on the functions :
%$$ h_k(t) = \int_K e^{-i\epsilon t}\epsilon^k d\mu(\epsilon)$$
aka the Fourier transforms on $K$ of the moments $x^k$.

\subsubsection*{Study of $h_k(t)$}

Using the Mellin transform (see annexe), we may express $h_k(t)$ as a series : 
%$$ h_k(t) = \sum_{n\neq 0} \frac{(-it)^n}{n!}\mu_{n+k}$$
%with $\mu_n = \int_K x^n d\mu$. (Note that we obtain the same expression by writing $e^(-i\epsilon t) = \sum_{n \leq 0}\frac{(-i\epsilon t)^n}{n!} $ and reversing the signs sum and integral). 

It is not clear, from this expression, whether or not $h_k(t)$ has a log-periodic feature, which is expected as a fingerprint of the scaling symmetry of the spectrum expressed in (1). We shall now show that there is, however, such a behaviour. We will do so by calculating the zeroes of $h_k(t)$ and the maximas of $|h_k(t)|$.

\medskip

First note that $h_0(t)$ verifies the following functional equation, derived by direct application of (1) :
%\begin{eqnarray*}
%\lefteqn{h_0(t) = \int_K e^{-i\epsilon t}d\mu}\\
%& = & \frac{1}{2}\int_K e^{-i\epsilon t/3}d\mu+\frac{e^{-i 2t/3}}{2}\int_K e^{-i\epsilon t/3}d\mu\\
%& = & \frac{1+e^{-i2t/3}{2} h_0(t/3) 
%\end{eqnarray*}

By repeating this process, and using $h_0(0) = \int_K 1d\mu = 1$, we find the exact expression of $h_0(t)$ :

%\begin{equation}
%h_0(t) = \prod_{k\neq 1} \frac{1+e^{-i2t/3^k}{2}
%\end{equation}

This is consistent with the expression derived in [1]. Furthermore, there is a perfect fit between the graph of the theoretical expression (2) and the one obtained from the direct calculation of $h_0(t)$ using : 
%$$ h_0(t) = \lim_{n \to \infty} \left(\frac{3}{2} \right)^{n} \sum_{a_{j_n}\in K_{n}} \int_{a_{j_n}}^{a_{j_n}+3^{-n}} e^{-ixt}dx$$
with $K_n$ the set of the left edges of the remaining segments after $n$ iterations in the construction of the Cantor set from the segment $[0,1]$.\\

From (2), we easily deduce the zeroes of $h_0(t)$ : it is the set :
%$$Z_0 = \{t_{m,n} = \frac{\pi}{2}(2m+1)3^n, m\in \mathbb{Z}, n \in \mathbb{N}^{*}\}$$

We also get the local maxima of $|h_0(t)|$ : 
%$$M_0 = \{v_{n,m} = \pi m 3^n, m\in \mathbb{Z}, n \in \mathbb{N}^{*}\}$$
with the property : 
%$$h_0(v_{n,m}) = \prod_{k\neq n+1} \frac{1+e^{-i2t/3^\overbrace{k-n}^{k'}}{2} = \prod_{k'\neq 1} \frac{1+e^{-i2t/3^{k'}}{2} = h_0(v_{0,m})$$
%Furthermore, the product $\prod_{k'\neq 1} \frac{1+e^{-i2t/3^{k'}}{2}$ is convergent and consists of nonzero terms, therefore it is nonzero. Taking $n\to \infty$, we can therefore deduce that $h_0(t)$ does not converge to zero when $t \to \infty$ (as also found in [1]). 

\medskip

Now, using (1) again, we derive the following recurrence formula for $h_k(t)$ : 
\begin{eqnarray*}
\lefteqn{h_k(t) = \int_K e^{-i\epsilon t}\epsilon^k d\mu}\\
& = & \frac{3^{-k}}{2}\int_K e^{-i\epsilon t/3}\epsilon^k d\mu+\frac{3^{-k}e^{-i 2t/3}}{2}\int_K e^{-i\epsilon t/3}(\epsilon+2)^k d\mu\\
& = & \frac{3^{-k}}{2} h_k(t/3) + \frac{3^{-k}}{2}e^{-i2 t/3}\sum_{j = 0}^{k} \binom {k} {j} h_j(t/3) 2^{k-j}
\end{eqnarray*}

We will use this relation to show, by recurrence, that the set $Z_k$ of zeroes of $h_k$, $k \neq 0$ contains the set : 
%$$\{t_{m,n} = \frac{\pi}{2}(2m+1)3^k, m\in \mathbb{Z}\}$$

%First, we deduce that, if $t_{n,m} = \frac{\pi}{2}(2m+1)3^n$, $m\in \mathbb{Z}$, $n \in \mathbb{N}\ \{0,1\}$, then :
%$$h_1(t_{n,m}) =\frac{3^{-1}}{2} h_1(t_{n,m}/3) - \frac{3^{-1}}{2}(h_1(t_{n,m}/3)+\underbrace{h_0(t_{n,m}/3)}_{0}$$
%since $t_{n,m}/3 \in Z_0$.
%Thus : $\{t_{m,n} = \frac{\pi}{2}(2m+1)3^2, m\in \mathbb{Z} \} \in Z_1$. The generalization to higher $k$ is trivially derived using the recurrence relation above. \\

It seems that there are no other zeroes for $h_k(t)$, e.g. that 
%$$Z_k = \{t_{m,n} = \frac{\pi}{2}(2m+1)3^k, m\in \mathbb{Z}\}$$
but it is not rigorously proven ; the numerics however seem to support this hypothesis (see fig. 2).

\begin{figure} %2
\centering
\includegraphics[scale=0.7]{loglogrmsfibothzoom2.png}
\caption{Numerical curves of $h_0(t)$, $h_1(t)$ and $h_{31}(t)$.}
\end{figure}

  
We therefore deduce that $h_{k}(t)$ has the same zeroes than $h_{k-1}(t/3)$.

\medskip

Using a similar reasoning, we can study the points at which the module $|h_{k}(t)|$ has a local maximum. One can show (see annexe) that these maxima are reached, for $|h_{0}(t)|$, at points of the form : $t_{m,n} = 3^n m \pi$. It is also 
easy to show that, for all $n,m$, one has : $h_0(t_{m,n})=h_0(t_{m,0})$. Using the recurrence relation between the $h_k$,
one can see that $h_{k}(t_{m,n}) \approx 3^{-k}h_0(t_{m,n-1})$.\\

Thus, we may infer, with a good approximation, that : $$h_k(t) \propto h_{k-1}(t/3)$$
This is confirmed by the numerics.

\medskip

To summarize, the functions $h_k(t)$ themselves are not log-periodic, and have periodically spaced zeroes of the form 
$t_{m,n} = \frac{\pi}{2}(2m+1)3^n$. The fingerprint of the fractal spectrum lies in the recurrence relation :
$$h_k(t) = \frac{3^{-k}}{2} h_k(t/3) + \frac{3^{-k}}{2}e^{-i2 t/3}\sum_{j = 0}^{k} \binom {k} {j} h_j(t/3) 2^{k-j}$$

which, up to a good approximation, can be simplified in the following renormalization relation :
$$h_k(t) \propto h_{k-1}(t/3)$$
in good approximation (e.g, it is not a rigorous proportionality relation, since the $h_k$        

\subsubsection*{Application to the RMS and the participation ratio}



\section*{References}

[1] E. HILLE and J. D. TAMARKIN, The American Mathematical Monthly, Vol. 36, No. 5 (May, 1929), pp. 255-264

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