Content-Type: multipart/mixed; boundary="-------------0107310922640" This is a multi-part message in MIME format. ---------------0107310922640 Content-Type: text/plain; name="01-293.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="01-293.keywords" fractals, box-counting dimension, quantum mechanics ---------------0107310922640 Content-Type: application/x-tex; name="proof.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="proof.tex" \documentclass[cmp,final]{svjour} % Remove option referee for final version \usepackage{graphics} \usepackage{epic,eepic} \usepackage{amsfonts,amssymb} \newcommand{\const}{\mbox{\rm const}} \newcommand{\supp}{\mbox{\rm supp}} \newcommand{\osc}{\mbox{\rm osc}} \newcommand{\var}{\mbox{\rm Var}} \newcommand{\vol}{\mbox{\rm vol}} \newcommand{\mod}{\mbox{\rm mod}} \newcommand{\mb}{\mathbf} \newcommand{\ep}{{\ \hfill $\Box$}} \begin{document} \title{Fractality of certain quantum states} \author{Daniel W\'ojcik\inst{1}\thanks{\emph{On leave from:} Center for Theoretical Physics, Polish Academy of Sciences, Al. Lotnik\'ow 32/46, 26-680 Warszawa, Poland}\and Karol \.Zyczkowski\inst{2}% } \institute{Institute for Physical Science and Technology, University of Maryland, College Park, MD, 20742, USA \and Center for Theoretical Physics, Polish Academy of Sciences, Al. Lotnik\'ow 32/46, 26-680 Warszawa, Poland } \date{Received: 1 August 2001 / Accepted: date} \communicated{name} \maketitle \begin{abstract} We prove the theorem announced in Phys. Rev. Lett. {\bf 85}:5022, 2001 concerning the existence and properties of fractal states for the Schr\"odinger equation in the infinite one-dimensional well. \end{abstract} \section{Introduction} \label{sec:intro} Fractals are sets and measures of non-integer dimension~\cite{Mandelbrot82da}. They are good models of phenomena and objects in various areas of science. They are often connected with non-equilibrium problems of growth~\cite{Meakin98da} and transport~\cite{Gaspard98sa,Dorfman99s}. In this direction their importance can be expected to grow in view of the recent results connecting transport coefficients with fractal properties of hydrodynamic modes~\cite{Gilbert00s,Gaspard01s}. Their ubiquity in dynamical systems theory as attractors, repellers, and attractor boundaries is well-known~\cite{Ott93da,Falconer90da}. Fractals have also been found in quantum mechanics. For instance, quantum models related to the problem of chaotic scattering often reveal the fractal structure \cite{Eckhardt88da,Ketzmerick96da,Casati00da}. Quantum field theories in fractal spacetimes have recently been reviewed in \cite{Kroeger00da}. Recently fractals have been shown to satisfy Schr\"odinger equations for simplest non-chaotic potentials~\cite{Berry96da,wojcik01db}. In this paper we prove some of the results announced in~\cite{wojcik01db}. This work has two objectives. One is to present the first (as far as we know) rigorous proof of fractality of any quantum states. The other is to illustrate a convenient method of calculating dimensions of graphs of continuous functions introduced by Claude Tricot~\cite{Tricot95da}. One of the oldest fractals is Weierstrass function \cite{Weierstrass72da,Edgar93da}: \begin{equation} W(x)=\sum_{n=0}^\infty a^n \cos (b^n x\pi). \label{Weier} \end{equation} introduced as an example of everywhere continuous nowhere differentiable function by Karl Weierstrass around 1872. Maximum range of parameters for which the above series has required properties was found by Godfrey Harold Hardy in 1916~\cite{Hardy16da}, who also showed that \begin{equation} \label{weierosc} \sup\{ |f({x})-f({y})| : |x-y|\leq \delta\} \sim \delta^H, \end{equation} where \[ H=\frac{\ln (1/a)}{\ln b}. \] From this it easily follows (see below) that the box-counting dimension of the graph of the Weierstrass function $W(x)$ is \begin{equation} D_W= 2 + H = 2+\frac{\ln a}{\ln b} = 2-\left|\frac{\ln a}{\ln b}\right|. \end{equation} Functions whose graphs have non-integer box-counting dimension will be called {\em fractal functions}. Even though the box-counting dimension of the Weierstrass function is easy to calculate~\cite{Tricot95da}, the proof that its Hausdorff dimension has the same value is still lacking. Lower bounds on the Hausdorff dimension of the graph were found by Mauldin~\cite{Mauldin86da,Mauldin86db}. Graphs of random Weierstrass functions were shown to have the same Hausdorff and box-counting dimensions for almost every distribution of phases~\cite{Hunt98da}. The above construction can easily be realized in quantum mechanics. Consider solutions of the Schr{\"o}dinger equation \begin{equation} i\partial_t \Psi(x,t) = -\nabla^2 \Psi(x,t) \label{Schrowell} \end{equation} for a particle in one-dimensional infinite potential well. The general solutions satisfying the boundary conditions $\Psi(0,t)=0=\Psi(\pi,t)$ have the form \begin{equation} \Psi(x,t)=\sum_{n=1}^{\infty} a_n \sin (nx) e^{-i n^2 t}, \label{well1} \end{equation} where \begin{equation} a_n=\frac2\pi \int_0^{\pi} \!\! dx\, \sin (nx) \Psi(x,0). \label{wellcoeff} \end{equation} Weierstrass quantum fractals are wave functions of the form \begin{equation} \Psi_M(x,t)=N_M \sum_{n=0}^M q^{n(s-2)} \sin(q^n x) e^{-iq^{2n} t}, \label{weierwell} \end{equation} where $q=2,3,\dots$, $s\in (0,2)$. In the physically interesting case of finite $M$ the wave function $\Psi_M$ is a solution of the Schr\"odinger equation. The limiting case \begin{equation} \label{wellinf} \Psi(x,t):=\lim_{M\rightarrow\infty} \Psi_M(x,t)= N \sum_{n=0}^\infty q^{n(s-2)} \sin(q^n x) e^{-iq^{2n} t}, \end{equation} with the normalization constant $N=\sqrt{\frac2\pi(1-q^{2(s-2)})}$, is continuous but nowhere differentiable. It is a weak solution of the Schr\"odinger equation. Note that~(\ref{wellinf}) converges for \( |q^{s-2}|<1 \quad \equiv \quad s<2. \) Later on we will show that the probability density of~(\ref{wellinf}) shows fractal features for $s>0$. Thus the interesting range of $s$ is $(0,2)$. We show that not only the real part of the wave function $\Psi(x,t)$, but also the physically important probability density $P(x,t) : = |\Psi(x,t)|^2$ exhibit fractal nature. This is not obvious, because $|\Psi(x,t)|^2$ is the sum of squares of real and imaginary part having usually equal dimensions. One can easily show that the dimension of the graph of a sum of functions whose graph have the same dimensions $D$ can be anything\footnote{Let $f_1$ and $f_2$ be functions with graphs having dimensions, respectively, $1\leq D_1M : \quad n < (q^\varepsilon)^n. \] This leads to the following estimate of the oscillation of $R_n$: \[ \osc_\delta(x)R_n \leq d_2(s,q) \delta^{2-s-\varepsilon}, \] where \[ d_2(s,q)=\frac{4q^{s-2}}{(1-q^{s-2})^2}. \] Thus for all $x$ and $\delta=q^{-n}$, where $(\ln n)/n < \varepsilon\ln q$, we have \begin{eqnarray*} \osc_\delta(x)P & \leq & \osc_\delta(x)P_n + \osc_\delta(x)R_n\\ & \leq & (2d_1+d_2) \delta^{2-s-\varepsilon}. \end{eqnarray*} From proposition~\ref{propo:estdim} it follows that \[ \dim_B {\rm graph}\,P_t(x) \leq 2- (2-s-\varepsilon)=s+\varepsilon \mathop{\longrightarrow}_{\varepsilon\rightarrow0} s. \] \item Fix $t$. Let $f(x)=P(x,t)$. We want to show that \[ W:=\int_a^b |f(x+\delta)-f(x-\delta)| dx \geq c \delta^{2-s}. \] Take $a=0$, $b=\pi$. Notice that (we skip the normalization constant) \begin{eqnarray} W & = & \int_0^\pi |f(x+\delta)-f(x-\delta)| dx \label{def.w}\\ & = & \int_0^\pi \left| \sum_{k=0}^\infty q^{k(s-2)} \sum_{l=0}^k \{\sin q^l(x+\delta) \sin q^{k-l}(x+\delta) +\right.\nonumber\\ & & \left. - \sin q^l(x-\delta) \sin q^{k-l}(x-\delta) \} \,a_{kl} \right| dx \nonumber\\ & = & \int_0^\pi \left| \sum_{k=0}^\infty q^{k(s-2)} \sum_{l=0}^k \{ \cos q^lx \sin q^{k-l}x \sin q^l \delta \cos q^{k-l}\delta \} \,a_{kl}\right| \,dx , \nonumber \end{eqnarray} where \[ a_{kl} = \cos (q^{2k}-q^{2l}) t. \] Take $|h(x)|\leq 1$. Observe, that \begin{eqnarray} \int_a^b \left|\sum_i f_i(x)\right| dx & \geq & \int_a^b \left|h(x)\right|\left|\sum_i f_i(x)\right| dx \nonumber\\ \label{h(x)} & \geq & \left|\int_a^b \sum_i h(x)f_i(x) dx \right| \\ & \geq & \left|\int_a^b h(x)f_k(x) dx \right| - \sum_{i\neq k} \left|\int_a^b h(x) f_i(x)dx \right|. \nonumber \end{eqnarray} One can interchange the order of summation and integration because $f(x)$ is absolutely convergent. Let us take $\delta=q^{-N}$, $h(x) = \sin q^m x \cos q^n x$. After substitution in (\ref{def.w}) using (\ref{h(x)}) we obtain \begin{eqnarray*} W & \geq &\left| \sum_{k=0}^\infty q^{k(s-2)} \sum_{l=0}^k \right. \sin q^{l-N} \sin q^{k-l-N} \cdot\\ &&\cdot \left. \int_0^\pi \!\! dx\,\sin q^lx \cos q^{k-l}x \cos q^m x \cos q^n x \,a_{kl} \right| \\ & = & \frac{\pi}{4} q^{(m+n)(s-2)} \left| \cos q^{m-N} \sin q^{n-N} \cos (q^{2(m+n)}-q^{2m}) t \right| \\ & =: & \frac{\pi}{4}\widetilde{W} \end{eqnarray*} In what follows we will show three consecutive proofs that \[ \exists c: \quad \widetilde{W} = q^{(m+n)(s-2)} \left| \cos q^{m-N} \sin q^{n-N} \cos (q^{2(m+n)}-q^{2m}) t \right| \geq c q^{N(s-2)}, \] for $t=\frac{k\pi}{q^l}$, $t/\pi\in {\mathbb Q}$, and a general proof for arbitrary real $t$. We will take advantage of the fact that $q$ is integer. Let $N=m+n$. Then \[ \widetilde{W} = q^{N(s-2)} \left| \cos q^{m-N} \sin q^{-m} \cos (q^{2N}-q^{2m}) t \right|. \] It is enough to consider $t\in[0,\pi[$. %%% nie jest takie jasne po co az TRZY dowody. proponuje jasniej %%% rozroznic kolejne przypadki %% \begin{enumerate} \item Let $t=\frac{k\pi}{q^l}$. Take $m$ such that \( 2m\geq l \), for instance $m=l$. Then \[ (q^{2N}-q^{2m})\frac{k\pi}{q^l} = (q^{2N-l}-q^l)k\pi, \] therefore \[ \left| \cos\left[ (q^{2N}-q^{2l}) \frac{k\pi}{q^l}\right]\right| = 1. \] We also have $\sin q^{-l}=\const$ and \[ \cos 1 = \cos q^0 \leq \cos q^{l-N} \leq \cos q^{-\infty} = \cos 0. \] This means that for $t=\frac{k\pi}{q^l}$, for sufficiently large $N$ \[ \widetilde{W} \geq q^{N(s-2)} \cos 1 \cdot \sin q^{-l} \cdot \cos 0 = \const\ q^{N(s-2)}. \] \item %%% Let $t/\pi\in {\mathbb Q}$. For the next two proofs let us write $t/\pi$ in $q$ basis \begin{equation} \label{eq:expansion} \frac{t}{\pi} = \frac{a_1}{q} +\frac{a_2}{q^2} +\frac{a_3}{q^3} +\dots = \sum_{k=1}^\infty \frac{a_k}{q^k}, \end{equation} where $a_k\in\{0,1,\dots,q-1\}$. Therefore \begin{eqnarray} \lefteqn{\cos[(q^{2N}-q^{2m}) t]} \nonumber\\ & = & \cos [\pi ( q^{2N-1}a_1 + q^{2N-2}a_2 + \dots + a_{2N} + q^{-1}a_{2N+1} + \dots + \nonumber\\ && +q^{2m-1}a_1 + q^{2m-2}a_2 + \dots + a_{2m} + q^{-1}a_{2m+1} + \dots)] \nonumber\\ & = & \cos \left[\pi \left (\frac{a_{2N+1}-a_{2m+1}}{q} + \frac{a_{2N+2}-a_{2m+2}}{q^2} + \dots \right)\right]. \label{eq:cosine} \end{eqnarray} If we could only choose $m$ so that the first two terms in this series cancel out, we would have a lower estimate on the cosine, because in this case \[ \left|\frac{a_{2N+3}-a_{2m+3}}{q^3} + \dots \right| \leq (q-1)\left(\frac{1}{q^3}+ \frac{1}{q^4}+ \dots \right) = \frac{1}{q^2}. \] Thus \[ \cos[(q^{2N}-q^{2m}) t] \geq \cos(\pi/q^2) \geq \cos \pi/4. \] Take a rational $t$. In any basis $q$ the expansion~(\ref{eq:expansion}) of $t/\pi$ is finite or periodic. Finite expansions have been treated in the previous point, thus we assume here the expansion of $t/\pi$ is periodic with period $T$. Thus for sufficiently large $n$ we have \[ a_{n+T}=a_n \] so that $t/\pi$ can be written as \[ t/\pi = 0.a_1 a_2 \dots a_K (a_{K+1} \dots a_{K+T}), \] where $(a_{K+1} \dots a_{K+T})$ means ``repeat $a_{K+1} \dots a_{K+T}$ periodically ad infinitum''. We shall now estimate $\widetilde{W}$ for this $t$. Every $N > K$ can be written as $N=K+pT+r$, where $p$ is natural or 0 and $r\in\{1,2,\dots,T\}$. If we now take $m=K+r$ then not only the first two but all the terms in~(\ref{eq:cosine}) cancel out, therefore \[ \cos[(q^{2N}-q^{2m}) t] =1. \] Obviously, \begin{eqnarray*} \sin q^{-m} & \geq & \sin q^{-(K+T)},\\ \cos q^{m-N} & = & \cos q^{-pT} \geq \cos 1, \end{eqnarray*} Therefore \[ \widetilde{W} \geq q^{N(s-2)}\, \sin q^{-(K+T)} \cos 1 = \const\ q^{N(s-2)}. \] \item %%% General case of an arbitrary $t/\pi\in [0,1]$. Its expansion in $q$ basis is given by~(\ref{eq:expansion}). Let $A$ be the set of all the two element sequences with elements from the set $\{0,1,\dots,q-1\}$. Thus \[ A=\{\{0,0\}, \{0,1\}, \dots, \{0,q-1\}, \{1,0\}, \dots, \{q-1,q-1\} \} , \] we write $A_{k,l}:=\{k,l\},\, k,l \in \{0,1,\dots,q-1\}$. Consider all the pairs of consecutive $q$-digits of $t/\pi$ of the form \begin{equation} \{a_{2m+1}, a_{2m+2} \}, \label{eq:pairs} \end{equation} i.e.\ $\{a_1, a_2 \}, \{a_3, a_4 \},$ etc. Every such pair is equal to some $A_{k,l}$. Let $N_{k,l}$ be the first such $m$ for which \[ A_{k,l}=\{a_{2m+1}, a_{2m+2} \}. \] If $A_{k,l}$ for given $k,l$ doesn't appear in the sequence of all the pairs~(\ref{eq:pairs}) we set $N_{k,l}=0$. Let \[ M = \sup_{k,l} N_{k,l}. \] Thus if $n>M$ the sequence $\{a_{2n+1}, a_{2n+2}\}$ has appeared at least once among the pairs $\{a_1, a_2 \}, \{a_3, a_4 \},\dots, \{a_{2M+1}, a_{2M+2} \}$. Therefore, for every $N>M$ we can find such an $m\in{1,2,\dots,M}$ that \[ \left|\cos\left[(q^{2N}-q^{2m})\frac{t}{\pi} \pi\right] \right| \geq \cos\frac{\pi}{q^2} \geq \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}. \] Also \begin{eqnarray*} \sin q^{-m} & \geq & \sin q^{-M},\\ \cos q^{m-N} & \geq & \cos q^{M-N} \geq \cos 1, \end{eqnarray*} which leads to \[ \widetilde{W} \geq q^{N(s-2)} \frac{\sqrt{2}}{2}\sin q^{-M} \cos 1 = \const\ q^{N(s-2)}. \] \end{enumerate} We have thus shown that for every $t$, for natural $q$ and for $\delta=q^{-N}$ \[ W \geq \const\ \delta^{2-s}, \] therefore (proposition~\ref{propo:estdim}) \[ \dim_B {\rm graph}\,P_t(x) \geq 2- (2-s)=s. \] \end{enumerate} \item[3.] {\em For almost every $x$, $D_t(x)=\dim_B \mbox{\rm graph} P_x(t)=D_t:=1+s/2$}. \\ We will use the form~(\ref{eq:probdens2}) of the probability density. It is enough to analyze the dimension of %$P_{xt}(x,t)$: \[ \widetilde{P}(t) := \sum_{c=1}^\infty q^{2c(s-2)} \sin q^c x \sum_{d=1}^c q^{-d(s-2)}\sin q^{c-d}x \cos [(q^{2c}-q^{2(c-d)})t]. \] \begin{enumerate} \item Let \[ P_n(t):= \sum_{c=1}^n q^{2c(s-2)} \sin q^c x \sum_{d=1}^c q^{-d(s-2)}\sin q^{c-d}x \cos[ (q^{2c}-q^{2(c-d)})t]. \] Then \begin{eqnarray*} |P_n'(t)| & = & \left| \sum_{c=1}^n q^{2c(s-2)} \sin q^c x \sum_{d=1}^c q^{-d(s-2)} \sin q^{c-d}x \cdot\right.\\ & & \left. \cdot (q^{2c}-q^{2(c-d)}) \sin [(q^{2c}-q^{2(c-d)})t] \right| \\ & \leq & \sum_{c=1}^n q^{2c(s-2)} \sum_{d=1}^c q^{-d(s-2)} (q^{2c}-q^{2(c-d)})\\ & = & \sum_{c=1}^n q^{2c(s-2+1)} \sum_{d=1}^c q^{-d(s-2)} (1-q^{-2d})\\ & = & \frac{q^{2-s}}{q^{2-s}-1} \left[ q^s \frac{q^{ns}-1}{q^s -1} - q^{2(s-1)}\frac{q^{2n(s-1)}-1}{q^{2(s-1)}-1} \right] + \\ & & - \frac{q^{-s}}{q^{-s}-1} \left[ q^{s-2} \frac{q^{n(s-2)}-1}{q^{s-2}-1} - q^{2(s-1)} \frac{q^{2n(s-1)}-1}{q^{2(s-1)}-1} \right]. \end{eqnarray*} Therefore, for $n$ large enough \[ |P_n'(t)| \leq c_1 q^{n \max\{s,2(s-1), s-2\}} = c q^{ns}. \] Let $\delta=q^{-\alpha n}$. Then $q^n=\delta^{-1/\alpha}$ and \[ \osc_\delta(t) P_n \leq c_1 2 \delta q^{ns} = 2c_1 \delta^{1-s/\alpha}. \] Let \[ R_n(t):=\widetilde{P}(t)-P_n(t). \] Then \begin{eqnarray*} \osc_\delta(t) R_n & \leq & 2 \sum_{c=n+1}^\infty q^{2c(s-2)} \sum_{d=1}^c q^{-d(s-2)}\\ & \leq & \frac{2q^{2-s}}{q^{2-s}-1} \sum_{c=n+1}^\infty q^{2c(s-2)+c(2-s)}\\ & = & c_2 \delta^{(s-2)/\alpha}. \end{eqnarray*} To obtain consistent estimate we must set \[ 1-\frac{s}{\alpha} =\frac{2}{\alpha} -\frac{s}{\alpha}, \] which gives $\alpha=2$. Thus \[ \osc_\delta(t)\widetilde{P} \leq (2c_1+c_2) \delta^{1-s/2}. \] \item Now we want to show that \[ W = \int_a^b \!\! dt\, |\widetilde{P}(t+\delta) - \widetilde{P}(t-\delta)| \geq c \delta^{1-s/2}. \] Set $a=0, b=2\pi$ for convenience. Then \begin{eqnarray*} W & = & \int_0^{2\pi} \!\! dt \,\left| \sum_{c=1}^\infty q^{2c(s-2)} \sin q^c x \sum_{d=1}^c q^{-d(s-2)}\sin q^{c-d}x \right.\cdot\\ & & \left.\cdot\left\{\cos [(q^{2c}-q^{2(c-d)})(t+\delta)]- \cos [(q^{2c}-q^{2(c-d)})(t-\delta)]\right\}\right| \\ & = & \int_0^{2\pi} \!\! dt \, \left|\sum_{c=1}^\infty q^{2c(s-2)} \sin q^c x \sum_{d=1}^c q^{-d(s-2)}\sin q^{c-d}x \right. \cdot\\ & & \cdot\left.\left\{ -2 \sin [(q^{2c}-q^{2(c-d)})t] \sin [(q^{2c}-q^{2(c-d)})\delta]\right\}\right| . \end{eqnarray*} Using our standard arguments we multiply the integrand by a suitable function smaller or equal to 1: \begin{eqnarray*} W & \geq & \int_0^{2\pi} \!\!dt \,|h(t)||\widetilde{P}(t+\delta) - \widetilde{P}(t-\delta)| \\ & \geq & \left| \int_0^{2\pi} \!\!dt \,h(t)[\widetilde{P}(t+\delta) - \widetilde{P}(t-\delta)]\right|. \end{eqnarray*} We choose $h(t) = \sin[(q^{2c}-q^{2(c-d)})t]$ and set $\delta=q^{-2N}$. It follows that \[ W \geq 2\pi q^{(2c-d)(s-2)} \left|\sin (q^c x) \sin (q^{c-d}x) \sin [(q^{2c}-q^{2(c-d)})q^{-2N}]\right|. \] We now want to show that for almost all $x$ \[ W \geq c_3 \delta^{1-s/2} = c_3 q^{N(s-2).} \] Set $2c-d=N$. Then \begin{equation} \label{ineq:w} W \geq 2\pi q^{N(s-2)} \left|\sin (q^c x) \sin (q^{N-c}x) \sin [q^{2(c-N)}-q^{-2c}]\right|. \end{equation} Thus it is enough to bound \begin{equation} \left| \sin (q^c x) \sin (q^{N-c}x) \sin [q^{2(c-N)}-q^{-2c}] \right| \label{eq:tobound} \end{equation} from below. Choose rational $x/\pi$. All the rational numbers in a given basis $q$ have one of the two possible forms: finite or periodic. In the first case ($x/\pi=k/q^l$) we cannot find the lower bound on~(\ref{eq:tobound}). We cannot succeed, because at these points the function $P_x(t)$ is smooth (cf.\ the next point of the proof). The other case means that $x/\pi$ can be written as \[ x/\pi = 0.a_1 a_2\dots a_K (a_{K+1}\dots a_{K+T}), \] where again $(a_{K+1}\dots a_{K+T})$ denotes the periodic part. Therefore, for every $N>K$, $q^n x \,\mod\, \pi$ can take only one of $T$ values: $q^{K+1} x \,\mod\, \pi, \dots, q^{K+T} x \,\mod\,\pi$. Let us take $c=1$, $N>K$. Then $|\sin (q^c x)|= |\sin (qx)|>0$ and is a constant. $|\sin (q^{N-1} x)|$ takes one of $T$ values, none of which is $0$, therefore it is always bounded from below by \[ \inf_{l=1,2,\dots,T} |\sin (q^{K+l} x)| > 0. \] Also the last term can be bounded: \[ |\sin(q^{-2(N-c)}-q^{-2c})| \geq \sin (q^{-2} - q^{-2(N-1)}) \geq \sin q^{-3} \] for $N\geq3$. Thus for rational $x/\pi$ with periodic expansion in $q$ \[ W \geq c_3 q^{N(s-2)} , \] where $c_3=2\pi |\sin (qx)\sin (q^{-3} x)|\inf_{l=1,2,\dots,T} |\sin (q^{K+l} x)| $. Consider now irrational $x/\pi$. Inequality (\ref{ineq:w}) for $c=N$ takes form \begin{eqnarray} W & \geq & 2\pi q^{N(s-2)} \left|\sin (q^N x) \sin x \sin [1-q^{-2N}]\right| \nonumber\\ & \geq & c q^{N(s-2)} \left|\sin (q^N x)\right|, \label{ineq:w1} \end{eqnarray} for $N\geq 2$. Instead of showing it can be bounded from below we will use it to prove that for almost every $x$ \begin{equation} \dim_B \mbox{\rm graph} P_x(t) \geq 1+s/2. \end{equation} Let \begin{equation} \label{eq:Renyi} x_n := q^n (x/\pi) \, \mod \, 1. \end{equation} Let \[ F_N^\alpha := \left\{ x \, : \, \exists \,n\geq N \quad \left(x_n\leq \frac{1}{q^{N\alpha}}\right) \vee \left(1-x_n\leq \frac{1}{q^{N\alpha}}\right)\right\}, \] $\alpha\in [0,1]$. Let \[ F_\infty^\alpha := \bigcap_{N=1}^\infty F_N^\alpha \] Clearly, \[ F_N^\alpha \supset F_{N+1}^\alpha \supset F_{N+2}^\alpha \dots \] Since Renyi map (\ref{eq:Renyi}) preserves the Lebesgue measure we have \begin{equation} \mu(F_N^\alpha) \leq 2\left(\frac{1}{q^N\alpha} + \frac{1}{q^{N+1}\alpha} + \frac{1}{q^{N+2}\alpha} + \dots\right) = \frac{2q}{q-1} \frac{1}{q^{N\alpha}}. \end{equation} Therefore \begin{equation} 0 \leq \mu(F_\infty^\alpha) \leq \inf_N \mu(F_N^\alpha) = 0 \end{equation} It follows that for almost every $\{x_n\}$ \begin{equation} \lim_{n\rightarrow\infty} \frac{\ln |\sin x_{n}|}{n} \geq \lim_{n\rightarrow\infty} \frac{\ln q^{-n\alpha}}{n} \geq \lim_{n\rightarrow\infty} \frac{q^{-n\alpha}}{2n} \geq -\alpha \ln q. \end{equation} Thus for every $\alpha>0$ and for almost every $x/\pi\in[0,1]$ we have \begin{eqnarray} \dim_B \mbox{\rm graph} P_x(t) & = & \lim\left(2 - \frac{\ln \var_\delta P_x(t)}{\ln q^{-2N}}\right) \\ & \geq & 2+\lim\left(\frac{\ln \var_\delta P_x(t)}{2N\ln q}\right) \end{eqnarray} But \begin{equation} \var_\delta P_x(t) \geq W \end{equation} therefore from (\ref{ineq:w1}) \begin{eqnarray} \dim_B \mbox{\rm graph} P_x(t) & \geq & 2+\lim\left(\frac{\ln c + N(s-2)\ln q + \ln |\sin (x_n \pi)|}{2N\ln q}\right) \\ & = & 1+s/2 + \lim\left(\frac{\ln |\sin (x_n \pi)|}{2N\ln q}\right) \\ & \geq & 1+s/2 - \alpha. \end{eqnarray} But $\alpha$ is arbitrary, thus \begin{equation} \dim_B \mbox{\rm graph} P_x(t) \geq 1+s/2. \end{equation} \end{enumerate} \item[4.] {\em For a discrete, dense set of points $x_d$, $D_t(x_d) = \dim_B \mbox{\rm graph} P_{x_d}(t)=1$}.\\ Let $x_{k,m}=\frac{m\pi}{q^k}$, where $k\in{\mathbb N}$, $m=0,1,\dots,q^k-1$. The set $\{x_{k,m}\}$ is dense in $[0,1]$. At these points $\Psi(x_{k,m},t)$ is a sum of a finite number of terms: \[ \Psi\left(\frac{m\pi}{q^k},t\right)= \sqrt{\frac2\pi\left(1-\frac{1}{q^{2(2-s)}}\right)} \sum_{n=0}^{k-1} q^{(s-2)n} \sin (q^{n-k} m\pi )\,e^{-iq^{2n}t}. \] Therefore, \[ \dim_B \mbox{\rm graph} \left|\Psi\left(\frac{m\pi}{q^k},t\right)\right|^2 = 1. \] \item[5.] {\em For even $q$ the average velocity $\frac{d\langle x\rangle}{dt}(t)$ is fractal with the dimension of its graph equal to $D_v=\max\{(1+s)/2,1\}$}.\\ Heuristically, this is rather obvious because \[ \frac{d\langle x\rangle}{dt}(t) \approx \sum_{k=1}^\infty \frac{q^{k(s-1)}}{q^{2k}} \sin q^{2k} t = \sum_{k=1}^\infty q^{2k(s-3)/2} \sin q^{2k} t. \] Thus the average velocity is essentially a Weierstrass-like function and the dimension of its graph should be \[ 2-\frac{3-s}{2} =\frac{1+s}{2}. \] It is enough to consider \[ W(t) := \sum_{k=1}^\infty \frac{q^{k(s-1)}}{q^{2k}-1} \sin (q^{2k}-1)t. \] \begin{enumerate} \item Let \[ W_n(t) := \sum_{k=1}^n \frac{q^{k(s-1)}}{q^{2k}-1} \sin (q^{2k}-1)t. \] Set $\delta=q^{-\alpha n}$. Then \[ |W_n'(t)| = \left|\sum_{k=1}^n q^{k(s-1)} \cos (q^{2k}-1)t\right| \leq \sum_{k=1}^n q^{k(s-1)} \leq c_1 \delta^{(1-s)/\alpha}, \] where \[ c_1 = \frac{q^{s-1}}{q^{s-1}-1}. \] Therefore \[ \osc_\delta(t)W_n \leq 2 c_1 \delta^{(1-s)/\alpha}\delta = 2 c_1 \delta^{1+(1-s)/\alpha}. \] Now, for \[ P_n(t):=W(t)-W_n(t) \] we have \begin{eqnarray*} |P_n(t)| & = & \left|\sum_{k=n+1}^\infty \frac{q^{k(s-1)}}{q^{2k}-1} \sin (q^{2k}-1)t\right| \leq \sum_{k=n+1}^\infty \frac{q^{k(s-1)}}{q^{2k}-1}\\ & \leq & 2 \sum_{k=n+1}^\infty \frac{q^{k(s-1)}}{q^{2k}} = c_2 \delta^{-\frac{s-3}{\alpha}}, \end{eqnarray*} where \[ c_2=\frac{2q^{s-3}}{1-q^{s-3}}. \] Thus \[ \osc_\delta(t)P_n\leq 2 c_2 \delta^{\frac{3-s}{\alpha}}. \] To obtain consistent estimate for both $P_n$ and $W_n$ we must set \[ 1+\frac{1-s}{\alpha} = \frac{3-s}{\alpha}, \] thus $\alpha=2$ and $\delta=q^{-2N}$. Therefore \[ \osc_\delta(t)W \leq \osc_\delta(t)W_n+\osc_\delta(t)P_n \leq 2 (c_1+c_2) \delta^{2-\frac{s+1}{2}}. \] \item Consider \begin{eqnarray*} \lefteqn{ \int_a^b |W(t+\delta)-W(t-\delta)| dt}\\ & = & \int_a^b \!\!dt \, \left|\sum_{k=1}^\infty \frac{q^{k(s-1)}}{q^{2k}-1} \cos(q^{2k}-1)t \,\sin (q^{2k}-1)\delta\right| \\ & \geq & \left| \int_a^b\!\!dt \, h(t) f_N(t) \right| - \sum_{k\neq N} \left| \int_a^b \!\!dt \,h(t) f_k(t)\right| , \end{eqnarray*} where \[ f_k(t) = \frac{q^{k(s-1)}}{q^{2k}-1}\cos(q^{2k}-1)t \, \sin (q^{2k}-1)\delta. \] Let $h(t)=\cos (q^{2N}-1)t$, $\delta=q^{-2N}$. Then \begin{eqnarray*} \lefteqn{\left| \int_a^b h(t) f_N(t) dt\right|}\\ & = & \frac{q^{N(s-1)}}{q^{2N}-1} \sin(1-q^{-2N}) \int_a^b \!\!\cos^2(q^{2N}-1)t\,dt\\ & \geq &\frac{q^{N(s-1)}}{q^{2N}} \sin \frac{\pi}{6} \int_a^b \!\! \cos^2(q^{2N}-1)t\,dt\\ & \geq & \frac12 q^{N(s-3)} \left[\frac{b-a}{2} + \frac{\sin(2b(q^{2N}-1))-\sin(2a(q^{2N}-1))}{4(q^{2N}-1)}\right]\\ & \geq & \frac12\delta^{\frac{3-s}{2}} \left[\frac{b-a}{2} - \frac{2\cdot2}{4\cdot q^{2N}}\right]\\ &= &\frac12\delta^{\frac{3-s}{2}} \left[\frac{b-a}{2} - \delta\right]\\ & \geq &\frac18\delta^{\frac{3-s}{2}}(b-a). \end{eqnarray*} On the other hand \begin{eqnarray*} \lefteqn{\left| \int_a^b h(t) f_k(t) dt\right|}\\ & = &\frac{q^{k(s-1)}}{q^{2k}-1} \sin\left[(q^{2k}-1)q^{-2N} \right] \int_a^b \!\! \cos(q^{2k}-1)t \cos(q^{2N}-1)t\,dt\\ & \leq & \frac{q^{k(s-1)}}{q^{2k}-1} \left| \frac{\sin(b(q^{2N}-q^{2k})) - \sin(a(q^{2N}-q^{2k}))}{2(q^{2N}-q^{2k})} + \right.\\ &&\left.\frac{\sin(b(q^{2N}+q^{2k}))-\sin(a(q^{2N}+q^{2k}))}{2(q^{2N}+q^{2k})} \right| \\ & \leq & 2 q^{k(s-3)}\left[\frac{1}{|q^{2N}-q^{2k}|} + \frac{1}{q^{2N}+q^{2k}}\right]\\ & \leq & 2 q^{k(s-3)}\left[\frac{1}{q^{2N}-q^{2(N-1)}} + \frac{1}{q^{2N}} \right] \\ & \leq & 5 \cdot q^{k(s-3)} \delta. \end{eqnarray*} Therefore \begin{eqnarray*} W & \geq & \frac18\delta^{(3-s)/2}(b-a) - \sum_k 5 q^{k(s-3)} \delta\\ & \geq & \frac18\delta^{(3-s)/2}(b-a) - 5 \frac{q^{s-3}}{q^{s-3}-1}\delta. \end{eqnarray*} But $\frac{3-s}{2}<1$, thus for large enough $N$ (small enough $\delta$) the first term dominates the other, therefore \[ W \geq c \delta^{2-(1+s)/2}, \] with $c=\frac{b-a}{16}$, for example. \end{enumerate} From theorem~\ref{twierdze:tricot} it follows that \[ D_v=\frac{1+s}{2}. \] \item[6.] {\em The surface $P(x,t)$ has dimension $D_{xy}=2+s/2$}. \\ Setting $x$ or $t$ constant we have shown that oscillations are bounded by $c\delta^H$ where exponent $H$ is one of $1, s, s/2$. We also showed the lower bound of variation is always $c\delta^H$ again with $H$ being one of $1, s, s/2$. What's more, there is a dense set of points $x$ for which $\var_\delta P_x(t) \geq c\delta^{s/2}$. One can take for instance all rational $x/\pi$ with periodic $q$-expansion. Thus from theorem~\ref{twie:multidim} we have \begin{equation} D_{xt}=1+\max\{D_x,D_t\}=2+\frac{s}{2}. \end{equation} \ep \end{enumerate} %%% Tutaj az prosilby sie jeden (JEDEN) akapit %%% podkreslajacy co i po co zostalo udowodnione... %%% Niestety nie dam rady napisac go dzisiaj.. \appendix \section{Appendix: Auxiliary calculations} \label{app:aux} \subsection{Probability density} Take the fractal wave function~(\ref{wellinf}) \[ \Psi(x,t) = \sqrt{\frac2\pi\left(1-q^{2(s-2)}\right)} \sum_{n=0}^\infty q^{n(s-2)} \sin(q^n x) e^{-iq^{2n} t}. \] Let us calculate two useful forms of the probability density $P(x,t)$ \begin{eqnarray*} P(x,t) & = & |\Psi(x,t)|^2\\ & = & \frac2\pi\left(1-q^{2(s-2)}\right) \sum_{m,n=0}^\infty q^{(m+n)(s-2)} \sin q^n x \sin q^m x\, e^{-i(q^{2n} - q^{2m})t}. \end{eqnarray*} Taking $k=m+n,l=n$ we obtain \begin{eqnarray} P(x,t)& = & \frac2\pi\left(1-q^{2(s-2)}\right) \sum_{k=0}^\infty q^{k(s-2)} \sum_{l=0}^k \sin q^l x \sin q^{k-l}x \,e^{-i(q^{2l} - q^{2(k-l)})t},\nonumber\\ & = & \frac2\pi\left(1-q^{2(s-2)}\right) \sum_{k=0}^\infty q^{k(s-2)} \sum_{l=0}^k \sin q^l x \sin q^{k-l}x \cos(q^{2l} - q^{2(k-l)})t,\qquad\quad. \label{eq:probdens} \end{eqnarray} Substitute $c=m,\,d=m-n$ \begin{eqnarray} P(x,t) & = & \frac2\pi\left(1-q^{2(s-2)}\right) \sum_{m=0}^\infty \left\{ q^{2m(s-2)} \sin^2 q^m x + \right.\nonumber\\ && \left.+ 2\sum_{n