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\section*{Quantum dynamics for a Cantor spectrum, tight-binding model - log-periodic behaviour as a consequence of time-averaging}

This note is about the signature of the fractal structure of the spectrum in time-averaged dynamical quantities. The model studied is the same as in the notes of June 1 (tight-binding model and Cantor spectrum).

\medskip

As showed in the notes of 1 June, the wave function is given by :
\begin{eqnarray*}
\lefteqn{\psi(k,t) = \int_{K}\phi_{\epsilon}(1)^{*}\phi_{\epsilon}(k)e^{-i\epsilon t}d\mu}\\
& = & \int_K |\phi_1(\epsilon)|^2p_{k}(\epsilon)e^{-i\epsilon t}d\mu \\
\end{eqnarray*}

We define $|\phi_1(\epsilon)|^2 d\mu$ as the local measure associated to $\psi$ ; for simplicity we write now $ |\phi_1(\epsilon)|^2 d\mu = d\mu$. Using : $p_k(x) = \sum_{0\leq j_k \leq k} c_{j_k}x^{j_k}$, we have :

$$\psi(k,t) = \sum_{j_k\leq k}c_{j_k}\int_{K}\epsilon^{j_k}e^{-i\epsilon t}d\mu$$


Since the RMS (and many other dynamical quantities) can be expressed in terms of the amplitude of $\psi$ at the sites of the lattice, namely $|\psi(k,t)|^2$, the dynamics of the system depends essentially on the cumulants :
$$ h_k(t) = \int_K e^{-i\epsilon t}\epsilon^k d\mu(\epsilon)$$

The $h_k(t)$ are not scaling functions, and therefore neither is the RMS. However, we showed that there is a structure and some invariant properties, with dynamical implications which were discussed  in the notes of June 1.

\medskip

We discuss here the influence of time-averaging : instead of studying instantaneous dynamical quantities, we are interested in their time average, given by the transformation $\frac{1}{t}\int_0^t$. Thus, we study $\frac{1}{t} |\psi(k,t')|^2 dt'$, which narrows down to the study of the quantities :
$$\frac{1}{t}\int_{0}^t h_k^{*}(t') h_l(t') dt'$$

Since $|\psi(k,t)|^2$ is real, we need only to consider the real part of $\frac{1}{t}\int_{0}^t h_k^{*}(t') h_l(t') dt'$, which we note $f_{k,l}(t)$. A standard calculation leads to : 
\begin{eqnarray*}
\lefteqn{f_{k,l}(t) = \Re\left(\frac{1}{t}\int_{0}^t h_k^{*}(t') h_l(t') dt'\right)}\\
& = & \iint_{K\times K} x^k x'^l sinc((x'-x)t)d\mu(x)d\mu(x')
\end{eqnarray*}

Note that $f_{0,0}(t)$ is the time-averaged return probability $C(t) = \frac{1}{t}\int_0^t |\langle\psi(0)|\psi(t')\rangle|^{2} dt'$, which we calculated exactly and which we know to have the following asymptotic expression :
\[C(t) \sim_{t \to \infty} \frac{t^{-d_K}}{\ln(3)}\sum_{n}\gamma_{n}t^{-\frac{2i\pi n}{\ln(3)}}\sin\left(\frac{\pi}{2}(d_{K,n}-1)\right)\Gamma(d_{K,n}-1)\]

with $d_K = \frac{\ln 2}{\ln 3}$.
The asymptotic behaviour of $C(t)$ can be intuited by the following observation : using the scaling property of the measure, we find that : 
$$C(t) = \frac{1}{2}C(t/3) + \frac{1}{2}\iint_{K\times K} sinc(\frac{x-x'+2}{3}t)d\mu(x)d\mu(x')$$ 
When $t \to \infty$, the second term of the sum on the right tends to zero, since for all $x,x'$ in $K$, $1<x-x'+2<3$. Thus, $C(t)$ is asymptotically a scaling function : 
$$C(t) \sim_{t \to \infty} \frac{1}{2}C(t/3)$$ 
This is due to the $sinc$ in the expression of $C(t)$ : $C(t)$ is a wavelet transform of the spectrum. For the same reason, we expect $\iint x^k x'^l sinc((x'-x)t)d\mu(x)d\mu(x')$ to be, asymptotically, a scaling function.

\medskip

For the calculation of $C(t)$, we used the change of variable $l = |x-x'|$ and reduced the problem to calculating the area $dS(l,\mu)$ (see fig. 1). The log-periodic behaviour comes from the expression of $dS(l,\mu)$.

\begin{figure} %1
\centering
\includegraphics[scale=0.6]{cant-carpet3.PNG}
\caption{2 dimensional Cantor dust (represented here at the second step of construction) ; the infinitesimal surface $dS(l,\mu)$ corresponds to the surface of the Cantor dust, in the sense of $d\mu$, contained in the two red strips.}
\end{figure}

If we make the approximation that : $sinc(xt) = 1$ for $|xt|<\pi$ and $sinc(xt) = 0$ elsewhere, we get the following approximate expression for $C(t)$ :
$$C(t) = \int_{l=0}^{\pi/t} dS(l,\mu) = t^{-\ln 2/\ln 3}\sum_{n} d_n t^{-2i\pi n/\ln 3}$$

which accounts for the typical asymptotic behaviour, although the coefficients $d_n$ are a little bit different from the coefficients in the exact expression.\\

If we apply the same reasoning for  $f_{k,l}(t) = \iint x^k x'^l sinc((x'-x)t)d\mu(x)d\mu(x')$, we find : 
$$f_{k,l}(t) \approx \iint_{|x-x'|<\frac{\pi}{t}} x^k x'^l d\mu(x)d\mu(x')$$
Using a similar method as for $C(t)$, we can show that (see detail in the next section) :
$$\iint_{|x-x'|<\frac{\pi}{t}} x^k x'^l d\mu(x)d\mu(x') = t^{-\ln 2/\ln 3}\left[G(\ln t) + \mathcal{O}\left(\frac{1}{t}\right)\right]$$
See a graph of $f_{1,2}(t)$ fig. 2.

Therefore, $f_{k,l}(t)$ is a scaling function in first approximation, and verifies asymptotically the relation $f_{k,l}(t) = \frac{1}{2}f_{k,l}(t/3)$.

\begin{figure} %2
\centering
\includegraphics[scale=0.8]{f12-talpha-loglog.pdf}
\caption{Loglog scale graphs of $f_{1,2}(t)$ (green) and $t^{-\ln 2/\ln 3}$ (blue)}
\end{figure}

\medskip

Note that this implies that the wave function has the following asymptotic behaviour : $\frac{1}{t}\int_0^t|\psi(k,t')|^2 dt'\sim t^{-\ln 2/\ln 3}G_{per}(\ln t)$, with $G_{per}(\ln t)$ periodic.

\medskip

Now, since 
$$f_{k,l}(t) =  \iint x^k x'^l sinc((x'-x)t)d\mu(x)d\mu(x') \sim_{t \to \infty}t^{-\ln2/\ln 3}G(\ln t/\ln 3)$$ 
we deduce, adapting the argument of Guarneri in [1], the following lower-bound for the time-averaged RMS : 
$$ \frac{1}{t}\int_0^t \Delta x^2(t') dt' \geq t^{\frac{2\ln 2}{3\ln 3}}$$

which is consistent with the numerics (fig. 3).

\begin{figure} %3
\centering
\includegraphics[scale=0.8]{rmsmoycantor.pdf}
\caption{Loglog scale graphs of $\frac{1}{t}\int_0^t \Delta x(t') dt'$ (blue) and $t^{\frac{\ln 2}{2\ln 3}}$ (green)}
\end{figure}

Let us now give the detail of the computation.

From $\frac{1}{t}\int_0^t|\psi(k,t')|^2 dt'\sim t^{-\ln 2/\ln 3}G_{per}(\ln t)$, we deduce that there exists a constant $c$ such that : $\frac{1}{t}\int_0^t|\psi(k,t')|^2 dt'\leq ct^{-\ln 2/\ln 3}$. Let $y\in\mathbb{R}_{+}$. Now : 
\begin{eqnarray*}
\lefteqn{ \frac{1}{t}\int_0^t \Delta x^2(t') dt' = \sum_{k\geq 0} k^2 \langle |\psi(k,t)|^2\rangle }\\
& \geq & \sum_{k\geq sqrt{y}}k^2 \langle |\psi(k,t)|^2\rangle \\
& \geq & y(1-\sum_{k=0}^{sqrt{y}} k^2 \langle |\psi(k,t)|^2\rangle \\
& \geq & y(1-ct^{-\ln 2/\ln 3}\frac{x^{\frac{3}{2}}}{3})
\end{eqnarray*}

The maximum value of the function $f(y) = y(1-ct^{-\ln 2/\ln 3}\frac{x^{\frac{3}{2}}}{3})$ is reached for $y = \left(\frac{6}{5}\right)^{\frac{2}{3}}c^{-\frac{2}{3}}t^{\frac{2\ln 2}{3\ln3}}$ ; thus :
$$\frac{1}{t}\int_0^t \Delta x^2(t') dt'\geq bt^{\frac{2\ln 2}{3\ln3}}$$
with $b$ a real constant. This completes the proof.

\section*{Calculation of $f_{k,l}(t)$}

We give the analytical expression of the approximated function for $f_{k,l}(t)$, which we call 
$$r_{k,l}(t) =  \iint_{|x-y|<\frac{\pi}{t}} x^k y^l d\mu(x)d\mu(y)$$
We show that :
$$ r_{k,l}(t) = t^{-\ln 2/\ln 3}\left[G(\ln t) + \mathcal{O}\left(\frac{1}{t}\right)\right]$$

We follow a similar method as for $C(t)$.

First, we calculate the infinitesimal area $dS_{k,l}(u,\mu)$, which is the integral of $x^ky^l$, for the measure $d\mu$, on the two strips $y = x+u$ and $y = x-u$ (red strips on fig. 1) : 
$$dS_{k,l}(u,\mu) = \int_K (x^k(x+u)^l\theta(1-u-x) + x^k(x-u)^l\theta(x-u))d\mu(x)$$
To derive $dS_{k,l}(u,\mu)$, we study the poles of its Mellin transform and apply the residue theorem. The Mellin transform is : 
$$m_{k,l}(s) = \int_0^1 u^{s-1}dS_{k,l}(u,\mu) = \iint_{K\times K} x^ky^l|x-y|^{s-1}d\mu d\mu$$
We now show by recurrence that the poles of $m_{k,l}(s)$ is the set :
$$S_{k,l} = \{ 1-\frac{\ln 2}{\ln 3} - j +\frac{2i\pi n}{\ln 3} ; j\in [|0,k+l|], n\in\mathbb{Z}\}$$ 

Initialization : \\
$m_{0,0}(s)  = \iint_{K\times K}|x-y|^{s-1}$ has been calculated for $C(t)$ and has poles at 
$$S_{0,0} = \{ 1-\frac{\ln 2}{\ln 3} +\frac{2i\pi n}{\ln 3} ;  n\in\mathbb{Z}\}$$

Inductive step : \\
using the symmetry of the measure, we find :

\begin{eqnarray*}
\lefteqn{m_{k,l}(s) = \iint_{K\times K} x^ky^l|x-y|^{s-1}d\mu d\mu}\\
& = & \frac{3^{1-s-(k+l)}}{4} \iint_{K\times K} x^ky^l|x-y|^{s-1}d\mu d\mu \\
& + & \frac{3^{1-s-(k+l)}}{4} \iint_{K\times K} (x+2)^k(y+2)^l|x-y|^{s-1}d\mu d\mu + R_{k,l}(s)
\end{eqnarray*} 
with 
$$R_{k,l}(s) = \frac{3^{1-s-(k+l)}}{4} \iint_{K\times K} (x+2)^ky^l|x-y+2|^{s-1}d\mu d\mu + \frac{3^{1-s-(k+l)}}{4} \iint_{K\times K} x^k(y+2)^l|x-y-2|^{s-1}d\mu d\mu $$ 
has no poles. Expanding the second term :
\begin{eqnarray*}
\lefteqn{m_{k,l}(s) =  \frac{3^{1-s-(k+l)}}{4} \iint_{K\times K} x^ky^l|x-y|^{s-1}d\mu d\mu}\\
& + & \frac{3^{1-s-(k+l)}}{4} \sum_{i=0}^k \sum_{j = 0}^l \binom {k} {i}\binom {l} {j}2^{k+l-i-j}\iint_{K\times K} x^iy^j|x-y|^{s-1}d\mu d\mu + R_{k,l}(s)\\
& = & \frac{3^{1-s-(k+l)}}{2}m_{k,l}(s) + \frac{3^{1-s-(k+l)}}{4} \mathop{\sum_{i=1}^{k}\sum_{j=1}^{l}}_{(i,j)\neq (k,l)}  \binom {k} {i}\binom {l} {j}2^{k+l-i-j}m_{i,j}(s) +R_{k,l}(s)
\end{eqnarray*} 

Thus, the poles of $m_{k,l}(s)$ is the set of the poles of all the $m_{i,j}(s)$ with $i\leq k$ and $j\leq l$, plus the set of complex points $ \{ 1-\frac{\ln 2}{\ln 3} - (k+l) +\frac{2i\pi n}{\ln 3} ; n\in\mathbb{Z}\}$. Using the induction hypothesis, we conclude that the poles of $m_{k,l}(s)$ is the set :
$$S_{k,l} = \{ 1-\frac{\ln 2}{\ln 3} - j +\frac{2i\pi n}{\ln 3} ; j\in [|0,k+l|], n\in\mathbb{Z}\}$$ 
which completes the proof.

Now, we apply the inverse Mellin transform and use the residue formula to find $dS_{k,l}(u,\mu)$ : 
\begin{eqnarray*}
\lefteqn{dS_{k,l}(u,\mu) = \frac{1}{2i\pi}\int_{\gamma -i\infty}^{\gamma+i\infty}u^{-s}m_{k,l}(s)ds}\\
& = & \sum_{j=0}^{k+l}\sum_{n \in \mathbb{Z}}c_{j,n}u^{j+\alpha-1-2i\pi/\ln 3}
\end{eqnarray*} 
with $\alpha = \frac{\ln 2}{\ln 3}$ and 
$$c_{j,n} = R_{k,l}\left(1-\frac{\ln 2}{\ln 3} - j +\frac{2i\pi n}{\ln 3}\left)+\mathop{\sum_{i=1}^{k}\sum_{i'=1}^{l}}_{i+i'\neq j}  \binom {k} {i}\binom {l} {i'}2^{k+l-i-j}m_{i,j}\left(1-\frac{\ln 2}{\ln 3} - j +\frac{2i\pi n}{\ln 3}\right)$$ 
By integrating, we obtain $r_{k,l}(t)$ :
\begin{eqnarray*}
\lefteqn{r_{k,l}(t) = \int_0^{1/t}dS(u,\mu)}\\
& = & \sum_{j=0}^{k+l}\sum_{n \in \mathbb{Z}}\int_0^{1/t}c_{j,n}u^{j+\alpha-1-2i\pi/\ln 3}\\
& = & t^{-\alpha}\sum_{j=0}^{k+l}\sum_{n \in \mathbb{Z}}c_{j,n}\frac{t^{-j+2i\pi n/\ln 3}}{\alpha+j-2i\pi n/\ln 3}
\end{eqnarray*} 

Thus : 
$$r_{k,l}(t) =  t^{-\alpha}\left[G(\ln t) + \mathcal{O}\left(\frac{1}{t}\right)\right]$$
with $G(\ln t) = \sum_{n\in \mathbb{Z}}\frac{t^{2i\pi n/\ln 3}}{\alpha -2i\pi n/\ln 3}c_{0,n}$.


\section*{References}

[1] Guarneri, "Spectral Properties of Quantum Diffusion
on Discrete Lattices", Eurquhys. Lett., 10 (2), pp. 95-100 (1989)

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