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\begin{document}

\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 $\epsilon$ : 
\begin{equation}
\phi_{\epsilon}(k) = p_k(\epsilon)
\end{equation}
We are interested of the evolution of a wave packet, described by the normalized wave function $\psi(t)$, evolving according to Schrodinger's equation :
$$i\frac{d\psi(t)}{dt} = H\psi(t)$$
The wave function is initially localized at the site 1 : $\langle k |\psi(t=0)\rangle = \delta(k-1)$. At time $t$, the system is in the state : $|\psi (t)\rangle = e^{-iHt}|\psi(0)\rangle$, and the value of the wave function at site $k$ is given by : 
$$|\psi(k,t)| = \int_K \phi_{\epsilon}(1)^{*}\phi_{\epsilon}(k)e^{-i\epsilon t}d\mu$$  
Using (2), we obtain :
\begin{eqnarray}
\lefteqn{\psi(k,t) = \int_{K}\phi_{\epsilon}(1)^{*}\phi_{\epsilon}(k)e^{-i\epsilon t}d\mu}\\
& = & \int_K p_1(\epsilon)p_{k}(\epsilon)e^{-i\epsilon t}d\mu \\
& = & \sum_{j_k\leq k}c_{j_k}\int_{K}\epsilon^{j_k}e^{-i\epsilon t}d\mu
\end{eqnarray}
with $p_k(x) = \sum_{0\leq j_k \leq k} c_{j_k}x^{j_k}$
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 maxima 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\geq 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 :
$$Z_0 = \{t_{m} = \frac{3\pi}{2}(2m+1), m\in \mathbb{Z}\}$$

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 : 
$$h_0(v_{n,m}) = \prod_{k\geq n+1} \frac{1+e^{-i2t/3^{\overbrace{k-n}^{k'}}}}{2} = \prod_{k'\geq 1} \frac{1+e^{-i2t/3^{k'}}}{2} = h_0(v_{0,m})$$
Furthermore, the product $\prod_{k'\geq 1} \frac{1+e^{-i2t/3^{k'}}}{2}$ 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 : 
$$\{t_{m,n} = \frac{\pi}{2}(2m+1)3^{k+1}, m\in \mathbb{Z}\}$$

First, we deduce that, if $t_{n,m} = \frac{\pi}{2}(2m+1)3^2$, $m\in \mathbb{Z}\}$, 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 : $\{t_{m,n} = \frac{\pi}{2}(2m+1)3^2, m\in \mathbb{Z} \} \subset Z_1$. 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 
$$Z_k = \{t_{m,n} = \frac{\pi}{2}(2m+1)3^{k+1}, m\in \mathbb{Z}\}$$
but it is not rigorously proven ; the numerics however seem to support this hypothesis (see fig. 1).

\begin{figure} %1
\centering
\includegraphics[scale=0.7]{h0h1h31.pdf}
\caption{Numerical curves of $|h_0(t)|$ (blue), $|h_1(t)|$ (red) and $|h_{31}(t)|$ (yellow).}
\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. The maxima of $|h_{0}(t)|$ are reached at the points : $t_{m,n} = 3^n m \pi$. 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 (see fig. 2).

\begin{figure} %2
\centering
\includegraphics[scale=0.7]{9emeh0tiersh1h2.pdf}
\caption{Numerical curves of $|h_0(t/9)|$ (blue), $|h_1(t/3)|$ (red) and $|h_{2}(t)|$ (yellow).}
\end{figure}


\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 into :
$$h_k(t) \propto h_{k-1}(t/3)$$
Note however that it is not a rigorous proportionality relation.       

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

In order to study how the fractal feature of the spectrum affects the dynamics of the system, we will now consider the RMS :
$$\Delta x (t)^2 = \sum_{k = 1}^{\infty} k^2 |\psi(k,t)|^2$$

and the participation ratio :
$$P_{\psi} = \frac{1}{\int_{espace} |\langle x|\psi\rangle|^{4}dx}$$

Injecting (3) in the definition of the RMS, we obtain :
\begin{eqnarray*}
\lefteqn{\Delta x(t) = \sum_{k = 1}^{\infty} k^2 |\psi(k,t)|^2}\\
& = & \sum_{k = 1}^{\infty} k^2 |\sum_{j_k\leq k}c_{j_k}\int_{K}\epsilon^{j_k}e^{-i\epsilon t}d\mu|^2 \\
& = & \sum_{k = 1}^{\infty} k^2 |\sum_{j_k\leq k}c_{j_k}h_{j_k}(t)|^2
\end{eqnarray*}

Note that, even if the $h_k(t)$ have periodically spaced  zeroes, this does not imply that $\Delta x(t)$ cancels for some $t$ : in fact, since the zeroes of $h_k$ are odd multiples of $3^{k+1}$, the infinite sum in the expression of $\Delta x$ insures that, for any time $t$, we are always before the first zero of infinitely many  $h_k$. 

Using $h_k(t) \propto h_{k-1}(t/3)$, we obtain in first approximation : 
$$ \Delta x (t) \propto \Delta x (t/3)$$
which leads to : 
$$ \Delta x (t) = t^{\alpha} G(\ln t)$$
with $G$ periodic.\\

We can make a more precise calculation by using directly the recurrence relation $h_k(t) = \frac{3^{-k}}{2} h_k(t/3) + \frac{3^{-k}}{2}e^{-i2 t/3}\sum_{j = 0}^{k} \binom {k} {j} h_j(t/3) 2^{k-j}$ : 
\begin{eqnarray*}
\lefteqn{\Delta x(t) = \sum_{k = 1}^{\infty} k^2 |\sum_{j_k\leq k}c_{j_k}h_{j_k}(t)|^2}\\
& = & \sum_{k = 1}^{\infty} k^2 |\sum_{j_k\leq k}c_{j_k}\frac{3^{-j_k}}{2} h_{j_k}(t/3) + \frac{3^{-j_k}}{2}e^{-i2 t/3}\sum_{l = 0}^{j_k} \binom {j_k} {l} h_l(t/3) 2^{j_k-l}(t)|^2\\
& \approx & \frac{1}{b} \Delta x (t/3) + R(t)
\end{eqnarray*}

where $b$ can be approximated numerically (there does not seem to be a simple theoretical value for $b$) and $R(t)$ is a sum of terms of the form $k^2 h_j(t)^{*}h{j'}(t)$. 

\medskip

A similar reasoning seems to indicate that the participation ratio also has a log-periodic feature, however the calculation is more complicated and I have not yet finished it.

\section*{Annexe}

\subsection*{Calculation of the $h_k(t)$}

We derive the expression of the functions $h_k(t) =  \int_K e^{-ix t}\epsilon^k d\mu(-)$, $k\in \mathbb{N}$ using Mellin's transform. 

\medskip

The Mellin transform of $h_k$ is :
\begin{eqnarray*}
\lefteqn{m_k (s) = \int_0^{\infty} t^{s-1}h_k(t) dt}\\
& = & \int_K d\mu(x)x^k \int_0^{\infty} t^{s-1}e^{-ix t}dt\\
& = & \Gamma(s)e^{-i\pi s/2}\underbrace{\int_K d\mu(x)x^{k-s}}_{l_k(s)}
\end{eqnarray*}

where $\Gamma$ is Euler's gamma function. The integral converges on the strip $0<\Re(s)<1/2$ but the final expression can be analytically extended on all $\mathbb{C}$, except on the countable set of poles we shall now determine. Applying (1) :


\begin{eqnarray*}
\lefteqn{l_k(s) = \int_K d\mu(x) x^{k-s}}\\
& = & \frac{1}{2}\int_K d\mu(x)\frac{x^{k-s}}{3^{k-s}}+\frac{1}{2}\int_K d\mu(x)\frac{(x+2)^{k-s}}{3^{k-s}}\\
& = & \frac{3^{s-k}}{2}l_k(s) + r_k(s)
\end{eqnarray*}

which leads to :
\begin{eqnarray*}
\lefteqn{l_k(s) = \frac{r_k(s)}{1-3^{s-k}/2}}
\end{eqnarray*}

$r_k$ has no poles because $x+2$ is never null for $x \in K$. $l_k$ has poles at the points : $s_{m,k} = k + \frac{\ln 2}{\ln 3} +\frac{2i\pi m}{\ln 3}$. Since $Gamma(z)$ has poles at the negative integers $-n$, $n \in \mathbb{N}$, with residue $Res(\Gamma)_{z = -n} = \frac{(-1)^n}{n!}$, we conclude that $m_k (s)$ has poles at the points : 
$$\{s_{m,k} = k + \frac{\ln 2}{\ln 3} +\frac{2i\pi m}{\ln 3}, m \in \mathbb{Z} \} \cup \mathbb{Z}^{-}$$

We now apply Mellin's inverse formula :
$$h_k(t) = \frac{1}{2i\pi} \int_{c-i\infty}^{c+\infty} m_k(s)t^{-s} ds$$
with $0<c<1/2$ to insure the convergence of the integral. We apply the residue formula on a contour defined as the limit for $r \to \infty$ of the semicircle of figure 3. The asymptotic behaviour of the $\Gamma$ function insures that the integral on the circular part goes to zero as $r \to \infty$.

Therefore, the zeroes of the $\Gamma$ function alone are taken into account in the residue formula, which yields :

$$ h_k(t) = \sum_{n\neq 0} \frac{(-it)^n}{n!}\mu_{n+k}$$
with $\mu_n = \int_K x^n d\mu$

\begin{figure} %3
\centering
\includegraphics[scale=0.7]{contour2.JPG}
\caption{Contour used in the residue formula.}
\end{figure}



\section*{References}

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

\end{document}
