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\begin{document}
\heading{Counting Statistics of Many-Particle Quantum Walks} 
\auname{Avry Shirakov}


\date{July, 2012}

\maketitle
\sect{Introduction}

Ever since the discovery of quantum mechanics people have been puzzled
by the counter-intuitive character of the laws of nature. Over time
we have learned to accept the more and more effects that unimaginable
in a classical world. Recent years research has uncovered many new
effects that are strikingly different from their classical counterparts.
In this work we will talk about one of the many surprising aspects
of quantum world, it is quantum random walk. 

To illustrate quantum random walks and give an intuition, we start
with an example. We define the model in one dimension, on a line or
a circle. Next step, we take the Hilbert space spanned by the positions
of the particle. For a circle of size $M$ we have
\begin{equation}
\mathcal{H}_{x}=\left\{ |x\rangle:x=0,...,M-1\right\} 
\end{equation}
 The position Hilbert space is extended by a 'coin'-space, which is
spanned by two basis states
\begin{equation}
\mathcal{H}_{c}=\left\{ |\uparrow\rangle,|\downarrow\rangle\right\} 
\end{equation}
States of the total system are in the space $\mathcal{H}=\mathcal{H}_{c}\otimes\mathcal{H}_{x}$.
The conditional translation of the system can be described by the
following unitary operator 
\begin{equation}
S=|\uparrow\rangle\langle\uparrow|\otimes\sum_{x}|x+1\rangle\langle x|+|\downarrow\rangle\langle\downarrow|\otimes\sum_{x}|x-1\rangle\langle x|
\end{equation}
where the index $x$ runs over $0\le x\le M-1$. $S$ transforms the
basis state $|\uparrow\rangle\otimes|x\rangle$ to $|\uparrow\rangle\otimes|x+1\rangle$
and $|\downarrow\rangle\otimes|x\rangle$ to $|\downarrow\rangle\otimes|x-1\rangle$.
From now on we will use the standard ordering $\mathcal{H}=\mathcal{H}_{x}\otimes\mathcal{H}_{c}$
which is represented by $|x\,\sigma\rangle$ vectors.

The first step of a random walk is a rotation in the 'coin'-space
$\mathcal{H}_{c}$. The unitary transformation $\mathcal{C}$ is very
arbitrary and therefore a rich family of walks with different behavior
can be defined by modifying $\mathcal{C}$. A frequently used balanced
unitary 'coin' is so called Hadamard coin $H$, 
\begin{equation}
H^{(1)}=\frac{1}{\sqrt{2}}\left(\begin{matrix}1 & 1\\
1 & -1
\end{matrix}\right)
\end{equation}
and for $M$ sites it looks like $H^{\left(M\right)}=H\otimes H\otimes\cdots\otimes H$
($2M\times2M$ matrix). 

It is easy to see that the Hadamard coin is balanced:
\begin{equation}
H\left(|0\rangle\otimes|\uparrow\rangle\right)=\frac{1}{\sqrt{2}}\left(|0\rangle\otimes\left(|\uparrow\rangle+|\downarrow\rangle\right)\right)
\end{equation}
\begin{equation}
S\left(\frac{1}{\sqrt{2}}\left(|0\rangle\otimes\left(|\uparrow\rangle+|\downarrow\rangle\right)\right)\right)=\frac{1}{\sqrt{2}}\left(|1\rangle\otimes|\uparrow\rangle+|-1\rangle\otimes|\downarrow\rangle\right)
\end{equation}
Measuring such coin state in the standard basis gives each of $\left\{ |1\rangle\otimes|\uparrow\rangle,|-1\rangle\otimes|\downarrow\rangle\right\} $
with probability $\frac{1}{2}$. After this measurement there is no
correlation between the positions left. If we continue such a measurement
at each iteration we obtain the plain classical random walk on the
circle. Its limiting distribution on the line (for a large number
of iterations $t$) approaches a Gaussian distribution with mean zero
and variance $\sigma^{2}=t$.

In the quantum random walk we will not measure the coin register during
intermediate iterations, but keep the quantum correlations between
different positions and let them interfere in subsequent steps. This
interference will cause a radically different behavior of the quantum
walk.

The quantum random walk of $t$ steps is defined as the transformation
$U^{t}$ which can be written as
\begin{equation}
U^{t}=U\left(t\right)=UUUU\cdot\cdot\cdot U
\end{equation}
where $U$ is acting on $\mathcal{H}=\mathcal{H}_{x}\otimes\mathcal{H}_{c}$,
and is given by
\begin{equation}
U=SH^{\left(M\right)}
\end{equation}
$U$ is represented as a multiplication the following matrices, 
\begin{equation}
S=\left(\begin{matrix}\ddots & \boldsymbol{R} & \boldsymbol{0} & \boldsymbol{0}\\
\boldsymbol{R^{\dagger}} & \boldsymbol{0} & \boldsymbol{R} & \boldsymbol{0}\\
\boldsymbol{0} & \boldsymbol{R^{\dagger}} & \boldsymbol{0} & \boldsymbol{R}\\
\boldsymbol{0} & \boldsymbol{0} & \boldsymbol{R^{\dagger}} & \ddots
\end{matrix}\right)\,\,\,\,\,\,\,\,\,\, H^{\left(M\right)}=\left(\begin{matrix}\ddots & \boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0}\\
\boldsymbol{0} & \boldsymbol{H} & \boldsymbol{0} & \boldsymbol{0}\\
\boldsymbol{0} & \boldsymbol{0} & \boldsymbol{H} & \boldsymbol{0}\\
\boldsymbol{0} & \boldsymbol{0} & \boldsymbol{0} & \ddots
\end{matrix}\right)
\end{equation}
while $\boldsymbol{R},\,\boldsymbol{H},\,\boldsymbol{0}$ are defined
as the following $2\times2$ matrices 
\begin{equation}
\boldsymbol{0}=\left(\begin{matrix}0 & 0\\
0 & 0
\end{matrix}\right)\,,\,\boldsymbol{R}=\left(\begin{matrix}0 & 0\\
0 & 1
\end{matrix}\right)\,,\,\boldsymbol{R^{\dagger}}=\left(\begin{matrix}1 & 0\\
0 & 0
\end{matrix}\right)\,,\,\boldsymbol{H}=\frac{1}{\sqrt{2}}\left(\begin{matrix}1 & 1\\
1 & -1
\end{matrix}\right).
\end{equation}
To illustrate the difference between the quantum random walk and its
classical counterpart let us evolve the walk for some steps starting
in the initial state $|\downarrow\rangle\otimes|0\rangle$. The results
can be found in Fig(a) and Fig(b) respectively. Below is an example
of an algebra used in such calculations.
\begin{equation}
|\downarrow\rangle\otimes|0\rangle\rightarrow U\left(|\downarrow\rangle\otimes|0\rangle\right)=\frac{1}{\sqrt{2}}\left(|\uparrow\rangle\otimes|1\rangle-|\downarrow\rangle\otimes|-1\rangle\right)
\end{equation}
\begin{figure}
\subfloat[The probability of being at position $x$ after $t$ steps of the
classical random walk on the line starting at $0$. ]{%
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline 
t\textbackslash{}x  & -3  & -2  & -1  & 0  & 1  & 2  & 3\tabularnewline
\hline 
\hline 
0  &  &  &  & 1  &  &  & \tabularnewline
\hline 
1  &  &  & 1/2  &  & 1/2  &  & \tabularnewline
\hline 
2  &  & 1/4  &  & 1/2  &  & 1/4  & \tabularnewline
\hline 
3  & 1/8  &  & 3/8  &  & 3/8  &  & 1/8\tabularnewline
\hline 
\end{tabular}

}~~~~~~~~~~~~~~~~\subfloat[The probability of being at position $x$ after $t$ steps of the
quantum random walk on the line with the initial state $|\Phi_{in}\rangle=|\downarrow\rangle\otimes|0\rangle$. ]{%
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline 
t\textbackslash{}x  & -3  & -2  & -1  & 0  & 1  & 2  & 3\tabularnewline
\hline 
\hline 
0  &  &  &  & 1  &  &  & \tabularnewline
\hline 
1  &  &  & 1/2  &  & 1/2  &  & \tabularnewline
\hline 
2  &  & 1/4  &  & 1/2  &  & 1/4  & \tabularnewline
\hline 
3  & 1/8  &  & 5/8  &  & 1/8  &  & 1/8\tabularnewline
\hline 
\end{tabular}

}
\end{figure}


This example shows that the probability distribution induced by the
quantum walk (fig(b)) differs from the classical one (fig(a)), and
that it is asymmetrically distributed on the positions (it is drifting
to the left). The drift to the left arises from different treatment
of $|\uparrow\rangle$ and $|\downarrow\rangle$ by the Hadamard.
Intuitively, this induces more cancellations for paths going right
(destructive interference), whereas particles moving to the left interfere
constructively. To obtain symmetric distribution of $|\uparrow\rangle$
and $|\downarrow\rangle$we can use either: define initial state $\frac{1}{\sqrt{2}}\left(|0\rangle\otimes\left(|\uparrow\rangle+i|\downarrow\rangle\right)\right)$
or use a different balanced 'coin' - $\frac{1}{\sqrt{2}}\left(\begin{matrix}1 & i\\
i & 1
\end{matrix}\right)$.

\begin{figure}
\includegraphics[scale=0.6]{QRW}

\caption{Quantum random walk vs. classical random walk. The quantum spreading
is proportional to $t$ while in classical case it is proportional
to $\sqrt{t}$. Taken from \cite{QRDWolf}. }
\end{figure}


The classical symmetric random walk on the line after $t$ steps has
a variance $\sigma^{2}=t$, so the expected distance from the origin
is of order $\sigma=\sqrt{t}$. By contrast the quantum random walk
has variance that scales with $\sigma^{2}\sim t^{2}$, which implies
that the expected distance from the origin is of order $\sigma\sim t$.
This result can be understood by thinking of Bloch waves in a periodic
lattice, in this case the motion of the waves in a lattice is ballistic,
(when a $\sigma\sim t$ the motion is called ballistic).

A simulation that presents both walks can be found here \href{http://demonstrations.wolfram.com/QuantumRandomWalk/}{Quntum Random Walk}.

Quantum versions of random walks have diverse applications that are
motivating experimental implementations, however the main driving
behind this interest is the use in quantum algorithms, which have
always employed the quantum walks form of a program running on a quantum
computer. Moreover, quantum walks used to model transport phenomena
in spin chains and bio-molecules broaden their scope beyond algorithms.

%%%%%%%%%%
\sect{Occupation Statistics:}

%%%%%%%%%%
\textbf{The Classical Problem:}

We want to solve the following combinatorial problem. Find the distribution
of $N$ identical particles on $M$ sites. In a classical manner we
have to count on all the possible arrangements of the system. In the
following the arrangement of $N$ particles in the $M$ modes will
be denoted by a vector $\boldsymbol{n}_{k}=\left(n_{1},n_{2},..,n_{M}\right)$,
with $n_{k}$ the number of orbitals in the ``output'' mode $k$,
and $\sum_{i}n_{i}=N$. For distinguishable, non interacting particles
the probability for a certain arrangement $\boldsymbol{n}_{k}$ is:
\begin{equation}
P_{cl}\left(\boldsymbol{n}_{k}\right)=\frac{N!}{M^{N}}\prod_{i=0}^{M-1}\frac{1}{n_{i}!}
\end{equation}
This situation is called classical since there are no interferences
between the particles (probabilities are summed instead of amplitudes).
Example: (here we take $N=M$) Coincident events $\boldsymbol{n}_{c}=(1,1,...,1)$
are realized with probability $\frac{N!}{N^{N}}$, and bunching events
(all particles are in one specific mode) $\boldsymbol{n}_{b}=(0,0,..,N,..,0)$
are realized with probability $\frac{1}{N^{N}}$. Both events are
unlikely for large $N$ .

%%%%%%%%%%
\textbf{The Quantum problem:}

The analogous problem with identical quantum particles is best formulated
in second quantization with the following creation and annihilation
rules:
\begin{equation}
\,\left[\hat{a}_{i},\hat{a}_{j}^{\dagger}\right]\,=\delta_{i,j}\,\,;\,\,\,\,\left[\hat{a}_{i},\hat{a}_{j}\right]\,=\,\left[\hat{a}_{i}^{\dagger},\hat{a}_{j}^{\dagger}\right]\,=0\,\,\, bosons
\end{equation}
\begin{equation}
\left\{ \hat{a}_{i},\hat{a}_{j}^{\dagger}\right\} =\delta_{i,j}\,\,;\,\,\left\{ \hat{a}_{i},\hat{a}_{j}\right\} =\left\{ \hat{a}_{i}^{\dagger},\hat{a}_{j}^{\dagger}\right\} =0\,\, fermions
\end{equation}
Since applications are possible with today's optical technologies
we focus on bosons.

The initial state is defined as
\begin{equation}
|\left(1,1,..,1\right)_{x}\rangle=\prod_{x=0}^{M-1}\hat{a}_{x}^{\dagger}|0\rangle
\end{equation}
 The input creation operators $\hat{a}_{x}^{\dagger}$ are mapped
to output creation operators $\hat{b}_{k}^{\dagger}$ through unitary
matrix $U$, such that 
\begin{equation}
\hat{b}_{k}^{\dagger}=\sum_{x=0}^{M-1}U_{xk}\hat{a}_{x}^{\dagger}
\end{equation}
 Formally, the unbiased Bell multiport beam splitter under consideration
corresponds to the unitary operation given by the Fourier matrix,
defined for any dimension $M$ by
\begin{equation}
U_{xk}=\frac{e^{\frac{2\pi i}{M}\cdot x\cdot k}}{\sqrt{M}}
\end{equation}


The possible states with fixed particle number per port after the
scattering process read
\begin{equation}
|\left(n_{0},n_{1},...\right)_{k}\rangle=\left(\prod_{k=0}^{M-1}\frac{1}{\sqrt{n_{k}!}}\left(\hat{b}_{k}^{\dagger}\right)^{n_{k}}\right)|0\rangle
\end{equation}
 In order to describe the event probability of a given arrangement
$\boldsymbol{n}_{k}$, we define a vector $\boldsymbol{X}$ of length
$N$ with entries that specify each particle's output port. It is
constructed by concatenating $n_{k}$ times the port number $\left(k\right)$,
($\boldsymbol{X}\left(\boldsymbol{n}_{k},k\right)$ maps particle
$k$ according to $\boldsymbol{n}_{k}$ into output ports).
\begin{equation}
\boldsymbol{X}=\left(\underbrace{1,1,..,1,}_{n_{1}}\underbrace{2,2,..,2,}_{n_{2}}...,\underbrace{n,n,..,n}_{n_{n}}\right)
\end{equation}
or in a short way it can be written as $\boldsymbol{X}=T\boldsymbol{n}_{k}$,
where $T$ denotes a mapping matrix between two vectors. More comprehensive
description of $T$ is that it maps vectors from Fock space to vectors
into Hilbert space. One can see that for the vector $\boldsymbol{n}_{k}=\left(3,2,3,0,2\right)$
in Fock space, the vector that is accepted in Hilbert space is $\boldsymbol{X}\left(\boldsymbol{n}_{k}\right)=\left(1,1,1,2,2,3,3,3,5,5\right)$.
So we can write the transition probability to a specific output arrangement
$\boldsymbol{n}_{k}$ as:
\begin{equation}
P_{qm}\left(\boldsymbol{n_{k}}\right)=\left|\langle\left(n_{0},n_{1},...\right)_{k}|\left(1,1,...,1\right)_{x}\rangle\right|^{2}=\prod_{j=0}^{M-1}\frac{1}{\sqrt{n_{j}!}}\left|\sum_{K\in P_{n}}\prod_{a=1}^{N}U_{X_{a}\left(\boldsymbol{n_{k}}\right),K\left(a\right)}\right|^{2}
\end{equation}
 where $P_{n}$ denotes the set of all permutations of $\left\{ 1,...,n\right\} $.
$K\left(a\right)$ is a permutation function that maps the particles
into the orbitals $K\left(a\right)=k$. For example: a given permutation
$K$ that is defined as:
\begin{equation}
K=\left(\begin{matrix}1 & 2 & 3\\
0 & 2 & 1
\end{matrix}\right)
\end{equation}
As an example of use of eq $\left(11\right)$: we calculate $P_{qm}\left(\boldsymbol{n}_{k}\right)$
for different $\boldsymbol{n}_{k}$ while the initial state of the
system is $|\left(1,1\right)_{x}\rangle$ and $N=2$. We calculate
explicitly just for a special case, for the output $\boldsymbol{n}_{k}=\left(1,1\right)\rightarrow X=\left(1,2\right)$.
The algebra shown below can be applied for calculating the other cases,
the results are shown in Fig (2).
\begin{equation}
P_{qm}\left(1,1\right)=\left|\langle\left(1,1\right)_{k}|\left(1,1\right)_{x}\rangle\right|^{2}=\prod_{j=0}^{M-1}\frac{1}{\sqrt{n_{j}!}}\left|\sum_{K\in P_{n}}\prod_{a=1}^{N}U_{X_{a}\left(1,1\right),K\left(a\right)}\right|^{2}
\end{equation}
\begin{equation}
=\frac{1}{1!1!}\left|U_{X_{1}\left(1,1\right),K_{1}\left(1\right)}U_{X_{2}\left(1,1\right),K_{1}\left(2\right)}+U_{X_{1}\left(1,1\right),K_{2}\left(1\right)}U_{X_{2}\left(1,1\right),K_{2}\left(2\right)}\right|^{2}
\end{equation}
\begin{equation}
=\frac{1}{1!1!}\left|U_{1,0}U_{2,1}+U_{1,1}U_{2,0}\right|^{2}
\end{equation}
\begin{equation}
=\left|\frac{e^{\frac{2\pi i}{2}\cdot0}}{\sqrt{2}}\frac{e^{\frac{2\pi i}{2}\cdot1}}{\sqrt{2}}+\frac{e^{\frac{2\pi i}{2}\cdot0}}{\sqrt{2}}\frac{e^{\frac{2\pi i}{2}\cdot0}}{\sqrt{2}}\right|^{2}=\left|\frac{1}{\sqrt{2}}\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\right|^{2}=0
\end{equation}


\begin{figure}
\begin{tabular}{|c|c|c|c|}
\hline 
$|\Phi_{k}\rangle$  & $|\boldsymbol{n}_{k}\rangle$  & $P_{cl}$  & $P_{qm}$\tabularnewline
\hline 
\hline 
$\begin{matrix}k=1\\
k=0\\
\\
\end{matrix}$\includegraphics[scale=0.05]{twozero}  & ~~$\left(2,0\right)_{k}$~~  & ~~$1/4$~~  & ~~$1/2$~~\tabularnewline
\hline 
$\begin{matrix}k=1\\
k=0\\
\\
\end{matrix}$\includegraphics[scale=0.05]{oneone}  & ~~$\left(1,1\right)_{k}$~~  & ~~$1/2$~~  & ~~$0\,\,$\tabularnewline
\hline 
$\begin{matrix}k=1\\
k=0\\
\\
\end{matrix}$\includegraphics[scale=0.05]{zerotwo}  & ~~$\left(0,2\right)_{k}$~~  & ~~$1/4$~~  & ~~$1/2$~~\tabularnewline
\hline 
\end{tabular}\caption{The probability to find two bosons in s state $|\Phi_{k}\rangle$
while beginning at the state $|\left(1,1\right)_{x}\rangle$. The
probabilities of quantum and classical cases are shown.}
\end{figure}


It is possible to recover a general bosonic behavior by grouping many
final arrangements in larger classes which are characterized by the
number of occupied ports $k$ or by the number of particles in one
port $m$. The event probability for such a class is given by the
sum of the probabilities of the single events that pertain to the
class. Very generally, for bosons, quantum states with large occupation
numbers are favored. This behavior is reflected by the formalism.
According to eq$\left(20\right)$, the probabilities $P_{qm}\left(\boldsymbol{n}_{k}\right)$
are given in terms of sum over permutations of scattering amplitudes
(over complex numbers of equal modulus - products of matrix elements
of $U_{x,k}$). Since these numbers typically have different phases,
they tend to add up destructively. However, all $n_{j}!$ permutations
$K$ that interchange the $n_{j}$ particles that exit in port $j$
leave the scattering amplitude invariant, so that $n_{j}!$ terms
in the sum have equal phases and add up constructively. This motivates
the following approximation for the translation probability:
\begin{equation}
P_{approx}\left(\boldsymbol{n_{k}}\right)=\frac{\left(\prod_{j}n_{j}!\right)P_{cl}\left(\boldsymbol{n_{k}}\right)}{\sum_{\boldsymbol{r}}\left(\prod_{j}r_{j}!\right)P_{cl}\left(\boldsymbol{r}\right)}
\end{equation}
We present probability distribution for the number of occupied ports,
for the classical calculation eq$\left(12\right)$, for the bosonic
quantum case eq$\left(20\right)$ and for the approximation eq$\left(26\right)$,
for $k=14$ in Fig(3).

\begin{figure}[h]
\includegraphics[scale=0.25]{fig1}

\caption{Events probability for a given number of occupied ports, for $k=14$.
(red) Rectangles indicate classical combinations, (blue) triangles
the quantum mechanical probability distribution, and (black) circles
(bosonic) approximation for the quantum result. \cite{tichy2010zero}}


\end{figure}


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