Some Novel Thought Experiments Involving Foundations of Quantum Mechanics and Quantum Information
Dissertation
submitted in partial fulfillment of the requirements for the degree of
Doctor of Philosophy in Physics
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\title{\bf Some Novel Thought Experiments Involving Foundations of Quantum Mechanics and Quantum Information}

\author{\large Dissertation\\ submitted in partial fulfillment of the
requirements for the degree of\\ Doctor of Philosophy in
Physics\\[3ex]}
\date{\large{\bf Omid Akhavan}\\ \large{\it Department of Physics,
Sharif University of Technology}\\[3ex] \large Tehran, July
2003\\[40ex]}

\maketitle

\frontmatter




\pagebreak


\vspace*{3.7in}

\begin{center}
\copyright~2003, Omid Akhavan
\end{center}
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\begin{center}
\it To\medskip\\
my mother, who has been with me every day of my life\medskip\\
and to\medskip\\
my wife Eli, who gave me l{$\heartsuit$}ve
\end{center}

\pagebreak

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\vspace*{3in}

\begin{center}
In the memory of\\ our beloved friend and colleague the late Dr.
Majid Abolhasani,\\ who gave me some useful comments on the
foundations of quantum mechanics, and was my initial encourager
for working on quantum information theory.
\end{center}

\pagebreak

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\begin{center}
{\large Acknowledgements}
\end{center}
\addcontentsline{toc}{section}{Acknowledgements}
\bigskip


It is a great pleasure to thank the many people who have
contributed to this dissertation.  My deepest thanks go to Dr.
Mehdi Golshani, my professor, for his moral and financial support
through the years of my PhD, for his unfailing positive attitude
which remounted my morale more than once, for his understanding
and sympathy for my problems with my hands, and for being my guide
through the maze of quantum world. He has been a valued teacher,
and I hope my seven years at Sharif University have given me even
a few of his qualities.

Special thanks go to Ali T. Rezakhani, my close friend,
collaborator and colleague, for his considerable influence on this
dissertation. Much of the view point mentioned here was worked out
in valuable conservations with him.

Warm thanks to Dr. Alireza Z. Moshfegh for introducing me to
experimental physics, for our common works which are not part of
this thesis, and for his moral and financial supports through all
years of my presence at Sharif University.

Thanks also go to Dr. Vahid Karimipour for his stimulating
discussions on quantum information theory, for reading this thesis
and for his illuminating comments.

I am thankful to Drs. Mohammad Akhavan, Mohammadreza Hedayati and
Majid Rahnama for reading this dissertation and for their valuable
comments.

I would like to thank all my teachers, colleagues and friends for
many useful and instructive discussions on physics and life.  I am
grateful also to those who are not mentioned by name in the
following.  In particular let me thank Drs.: Hesam Arfaie, Farhad
Ardalan, Reza Mansouri, Jalal Samimi, and Hamid Salamati as
teachers, and Saman Moghimi, Masoud M. Shafiee, Ahmad Ghodsi, Ali
Talebi, Parviz Kameli, Saeed Parvizi, Husein Sarbolouki, Mohammad
Kazemi, Alireza Noiee, Nima Hamedani, Farhad Shahbazi, Mahdi
Saadat, Hamid Molavian, Mohammad Mardani, Ali A. Shokri, Masoud
Borhani, Javad Hashemifar, Rouhollah Azimirad, Afshin Shafiee, Ali
Shojaie, Fatimah Shojaie, Mohammad M. Khakian, Abolfazl
Ramezanpour, Sohrab Rahvar, Parvaneh Sangpour, Ali Tabeie, Hashem
H. Vafa, Ahmad Mashaie, Akbar Jafari, Alireza Bahraminasab, Akbar
Fahmi, Mohammad R. Mohammadizadeh, Sima Ghasemi, Omid Saremi,
Davoud Pourmohammad, Fredric Faure and Ahmad Mohammadi as
colleagues and friends.

I would also like to thank my teachers at Physics Department of
Uroumieh University who encouraged me to continue physics.
 Particularly, I thank Drs.: Rasoul Sedghi, Mohammadreza Behforouz, Rasoul
 Khodabakhsh, Mostafa Poshtkouhi, Mir Maqsoud Golzan, Jalal
 Pesteh, Shahriar Afshar,
 and Mohammad Talebian.

There are also many people to whom I feel grateful and whom I
would like to thank at this occasion.  Each of the following have
in one way or another affected this dissertation, even if only by
prompting an explanation or turn of phrase.  I thank Drs.: Partha
Ghose, Louis Marchildon, Ward Struyve, Willy De Baere, Marco
Genovese, Adan Cabello, Hrvoje Nikolic, Jean-Francois Van Huele,
Edward R. Floyd, Farhan Saif, Manzoor Ikram, Seth Lloyd,
 Vladimir E. Kravtsov, Antonio Falci,
Ehud Shapiro, Vlatko Vedral, Denis Feinberg, Massimo Palma, Irinel
Chiorescu, Jonathan Friedman, Ignacio Cirac and Paolo Zanardi.





I would like to thank Institute for Studies in Theoretical Physics
and Mathematics (IPM) for financial support of this thesis.

I also appreciate hospitality of the the abdus salam international
centre for theoretical physics (ICTP, Italy) where some part of
this work was completed.

Thanks also go to the following people for a lot of beer: Parisa
Yaqoubi, Edris Bagheri, Khosro Orami, Vaseghinia, Yahyavi,
Beheshti and Nicoletta Ivanissevich.



\hspace{8cm}{\bf Omid Akhavan}\\ {\it Sharif University of
Technology\\July 2003}

\pagebreak
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\begin{center}
{\Large\bf Some Novel Thought Experiments Involving Foundations of
Quantum Mechanics and Quantum Information}
\\ \medskip
{\large by \\ \medskip Omid Akhavan}\\ \medskip \baselineskip=14pt
B.~Sc., Physics, Uroumieh University, Uroumieh, 1996\\
M.~Sc., Physics, Sharif University of Technology, Tehran, 1998\\
PhD, Physics, Sharif University of Technology, Tehran, 2003
\end{center}
\bigskip
\bigskip
\bigskip




\section*{Abstract}
\addcontentsline{toc}{section}{Abstract} In this thesis, we have
proposed some novel thought experiments involving foundations of
quantum mechanics and quantum information theory, using quantum
entanglement property.  Concerning foundations of quantum
mechanics, we have suggested some typical systems including two
correlated particles which can distinguish between the two famous
theories of quantum mechanics, i.e. the standard and Bohmian
quantum mechanics, at the individual level of pair of particles.
Meantime, the two theories present the same predictions at the
ensemble level of particles.  Regarding quantum information
theory, two theoretical quantum communication schemes including
quantum dense coding and quantum teleportation schemes have been
proposed by using entangled spatial states of two EPR particles
shared between two parties. It is shown that the rate of classical
information gain in our dense coding scheme is greater than some
previously proposed multi-qubit protocols by a logarithmic factor
dependent on the dimension of Hilbert space.  The proposed
teleportation scheme can provide a complete wave function
teleportation of an object having other degrees of freedom in our
three-dimensional space, for the first time.  All required unitary
operators which are necessary in our state preparation and Bell
state measurement processes are designed using symmetric
normalized Hadamard matrix, some basic gates and one typical
conditional gate, which are introduced here for the first time.


PACS number(s): 03.65.Ta, 03.65.Ud, 03.67.-a, 03.67.Hk

\pagebreak
\tableofcontents

\pagebreak

\chapter*{Preface} \addcontentsline{toc}{section}{Preface}
The present dissertation consists of two parts which are mainly
based on the following papers and manuscripts:
\begin{itemize}
\item {\em{Bohmian prediction about a two double-slit experiment
and its disagreement with standard quantum mechanics,}} M.
Golshani and O. Akhavan, {\it{J. Phys. A}} {\bf 34}, 5259 (2001);
quant-ph/0103101.

\item {\em{Reply to: Comment on ``Bohmian prediction about a two
double-slit experiment and its disagreement with SQM"}} O. Akhavan
and M. Golshani, quant-ph/0305020.


\item {\em{A two-slit experiment which distinguishes between
standard and Bohmian quantum mechanics,}} M. Golshani and O.
Akhavan, quant-ph/0009040.

\item {\em{Experiment can decide between standard and Bohmian
quantum mechanics,}} M. Golshani and O. Akhavan, quant-ph/0103100.

\item {\em{On the experimental incompatibility between standard
and Bohmian quantum mechanics}}, M. Golshani and O. Akhavan,
quant-ph/0110123.

\item {\em{Quantum dense coding by spatial state entanglement,}}
O. Akhavan, A.T. Rezakhani, and M. Golshani, {\it{Phys. Lett. A}}
{\bf 313}, 261 (2003); quant-ph/0305118.

\item {\em{Comment on ``Dense coding in entangled states",}} O.
Akhavan and A.T. Rezakhani, {\it{Phys. Rev. A}} {\bf 68}, 016302
(2003); quant-ph/0306148.


\item {\em{A scheme for spatial wave function teleportation in
three dimensions}}, O. Akhavan, A.T. Rezakhani, and M. Golshani,
{\it{J. Quant. Inf. Comp.}}, submitted.
\end{itemize}

The first part of this dissertation includes three chapters.  In
chapter 1, an introduction about the foundations of quantum
mechanics, which is mainly concentrated on explanations of; some
problems in the standard quantum mechanics, the quantum theory of
motion, some new insights presented by Bohmian quantum mechanics
and noting some objections that have been advanced against this
theory, has been presented.  In chapter 2, by using position
entanglement property of two particles in a symmetrical two-plane
of double-slit system, we have shown that the standard and Bohmian
quantum mechanics can predict different results at an individual
level of entangled pairs.  However, as expected, the two theories
predict the same interference pattern at an ensemble level of the
particles.  In chapter 3, the predictions of the standard and
Bohmain quantum mechanics have been compared using a double-slit
system including two correlated particles.  It has been shown that
using a selective joint detection of the two particles at special
conditions, the two theories can be distinguished at a statistical
level of the selected particles.  But, by considering all
particles, the predictions of the two theories are still identical
at the ensemble level of particles.


In the second part of the dissertation there are also three
chapters.  In its first chapter, i.e. chapter 4, an introduction
about quantum information theory including quantum dense coding
and teleportation has been presented.  In chapter 5, using a
two-particle source similar to that is applied in chapter 1, a
more efficient quantum dense coding scheme has been proposed.  In
this regard, the suitable encoding and decoding unitary operators
along with its corresponding Bell states have been studied.  The
rate of classical information gain of this scheme has been
obtained and then compared with some other well-known protocols.
Furthermore, possibility of designing of the required position
operators using some basic gates and one conditional position
gates has been investigated.  Next, in chapter 6, using a system
the same as the dense coding scheme, wave function teleportation
of a three dimensional object having some other degrees of freedom
has been studied.  Concerning this, the required operators for
performing Bell state measurement and reconstruction process have
been designed using some position and momentum gates.


In appendix, which consists three sections, some more details on
our considered EPR source, preparing and measuring processes
utilized in some initial cases of the dense coding and
teleportation schemes, and comparison of our dense coding scheme
with some other ones can be found.

\mainmatter

\part{New suggested experiments related to the foundations of quantum mechanics}
\newpage
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\phantom{~~~}

\chapter{Introduction-Foundations of Quantum Mechanics}

\section{Standard quantum mechanics}

The standard view of quantum mechanics (SQM), accepted almost
universally by physicists, is commonly termed the Copenhagen
interpretation.  This interpretation requires complementarity,
e.g. wave-particle duality, inherent indeterminism at the most
fundamental level of quantum phenomena, and the impossibility of
an event-by-event causal representation in a continuous space-time
background {\cite{Cushing}}.  In this regard, some problems
embodied in this interpretation are concisely described in the
following.

\subsection{Some of the major problems of SQM}
\subsubsection{Measurement}

As an example, consider a two-state microsystem whose
eigenfunctions are labelled by $\psi_+$ and $\psi_-$. Furthermore,
there is a macrosystem apparatus with eigenfunctions $\phi_+$ and
$\phi_-$ corresponding to an output for the microsystem having
been in the $\psi_+$ and $\psi_-$ states, respectively.  Since
prior to a measurement we do not know the state of the
microsystem, it is a superposition state given by
\begin{eqnarray}
\psi_0=\alpha\psi_++\beta\psi_-,\hspace{1cm}|\alpha|^2+|\beta|^2=1.
\end{eqnarray}
Now, according to the linearity of Scr\"{o}dinger's equation, the
final state obtained after the interaction of the two systems is
\begin{eqnarray}
\Psi_0=(\alpha\psi_++\beta\psi_-)\phi_0\longrightarrow\Psi_{out}=\alpha\psi_+\phi_++\beta\psi_-\phi_-
\end{eqnarray}
where it is assumed that initially the two systems are far apart
and do not interact.  It is obvious that, the state on the far
right side of the last equation does not correspond to a definite
state for a macrosystem apparatus.  In fact, this result would say
that the macroscopic apparatus is itself in a superposition of
both  plus and minus states.  Nobody has observed such macroscopic
superpositions.  This is the so-called measurement problem, since
the theory predicts results that are in clear conflict with all
observations.  It is at this point that the standard program to
resolve this problem invokes the reduction of wave packet upon
observation, that is,
\begin{eqnarray}
\alpha\psi_+\phi_++\beta\psi_-\phi_-\longrightarrow\left\{%
\begin{array}{ll}
    \psi_+\phi_+, & \hbox{$P_+=|\alpha|^2$;} \\
    \psi_-\phi_-, & \hbox{$P_-=|\beta|^2$.} \\
\end{array}%
\right.
\end{eqnarray}
Various attempts to find reasonable explanation for this reduction
are at the heart of the measurement problem.


\subsubsection{Schr\"{o}dinger's cat}


Concerning the measurement problem, there is a paradox introduced
by Schr\"{o}dinger in 1935.  He suggested the coupling of an
uranium nucleus or atom as a microsystem and a live cat in a box
as a macrosystem. The system is so arranged that, if the nucleus
with a life time $\tau_0$ decays, it triggers a device that kills
the cat. Now the point is to consider a quantum description of the
time evolution of the system.  If $\Psi(t)$, $\phi$ and $\psi$
represent the wave functions of the system, cat and atom,
respectively, then the initial state of the system would be
\begin{eqnarray}
\Psi(0)=\psi_{atom}\phi_{live}.
\end{eqnarray}
This initial state evolves into
\begin{eqnarray}
\Psi(t)=\alpha(t)\psi_{atom}\phi_{live}+\beta(t)\psi_{decay}\phi_{dead}
\end{eqnarray}
and the probabilities of interest are
\begin{eqnarray}
&&P_{live}(t)=|\langle\psi_{atom}\phi_{live}|\Psi(t)\rangle|^2\sim
e^{-t/\tau_0}\\
&&P_{dead}(t)=|\langle\psi_{decay}\phi_{dead}|\Psi(t)\rangle|^2\sim
1-e^{-t/\tau_0}.
\end{eqnarray}
As time goes on, chance looks less for the cat's survival.  Before
one observes the system, $\Psi(t)$ represents a superposition of a
live and a dead cat.  However, after observation the wave function
is reduced to live or dead one.  Now, the main question which was
posed by Schr\"{o}dinger is: what does $\Psi(t)$ represent?
Possible answers are that, it represents (1) our state of
knowledge, and so quantum mechanics is incomplete, and (2) the
actual state of the system which beers a sudden change upon
observation.  If we choose (1) (which is what Schr\"{o}dinger felt
intuitively true), then quantum mechanics is incomplete, i.e.,
there are physically meaningful questions about the system that it
cannot answer-surly the cat was either alive or dead before
observation.  On the other hand, Choice (2) faces us with the
measurement problem, in which the actual collapse of the wave
function must be explained.



\subsubsection{The classical limit}



It is well established that when a theory supersedes an earlier
one, whose domain of validity has been determined, it must reduce
to the old one in a proper limit.  For example, in the special
theory of relativity there is a parameter $\beta=v/c$ such that
when $\beta\ll 1$, the equations of special relativity reduce to
those of classical mechanics. In general relativity, also, the
limit of weak gravitational fields or small space-time curvature
leads to Newtonian gravitational theory.  If quantum mechanics is
to be a candidate for a fundamental physical theory that replaces
classical mechanics, then we would expect that there is a suitable
limit in which the equations of quantum mechanics approach those
of classical mechanics.  It is often claimed that the desired
limit is $\hbar\longrightarrow 0$.  But $\hbar$ is not a
dimensionless constant and it is not possible for us to set it
equal to zero.  A more formal attempt at a classical limit is
Ehrenfest's theorem, according to which expectation values satisfy
Newton's second law as
\begin{eqnarray}
\langle{\bf F}\rangle=m\frac{d^2\langle{\bf r}\rangle}{dt^2}.
\end{eqnarray}
This really only implies that the centroid of the packet follows
the classical trajectory.  However, wave packets spread and the
above equation is just not the same as ${\bf F}=m{\bf a}$.  A
similar formal attempt is the WKB (Wentzel-Kramers-Brillouin)
approximation which is often advertised as a classical limit of
the Schr\"{o}dinger's equation.  Again, there is not a
well-defined limit (in terms of a dimensionless parameter) for
which one can obtain exactly the equations of classical mechanics
for all future of times.  Therefore, if it is not possible to find
a classical description for macroscopic objects in a suitable
limit, then we do not have a complete theory that is applicable to
both the micro and macro domains.



\subsubsection{Concept of the wave function}

The quantum theory that developed in the 1920s is related to its
classical predecessor by the mathematical procedure of
quantization, in which classical dynamics variables are replaced
by operators. Hence, a new entity appeares on which the operators
act, i.e., the wave function.  For a single-body system this is a
complex function, $\psi({\bf x},t)$, and for a field it is a
complex functional, $\psi[\phi,t]$.  In fact, the wave function
introduces a new notion of the state of a physical system.  But,
in prosecuting their quantization procedure, the founding fathers
introduced the new notion of the state not in addition to the
classical variables but instead of them. They could not see, and
finally did not want to see, even when presented with a consistent
example, how to retain in some form the assumption that matter has
substance and form, independently of whether or not it is
observed. The wave function alone was adapted as the most complete
characterizing the state of a system.  Since there was no
deterministic way to describe individual processes using just the
wave function, it seemed natural to claim that these are
indeterminate and unanalyzable in principle. Furthermore, quantum
mechanics appears essentially as a set of working rules for
computing the likely outcomes of certain as yet undefined
processes called measurement.  So, one might well ask what
happened to the original program embodied in the old quantum
theory of explaining the stability of atoms as objective
structures in space-time.  In fact, quantum mechanics leaves the
primitive notion of system undefined; it contains no statement
regarding the objective constitution of matter corresponding to
the conception of particles and fields employed in classical
physics.  There are no electrons or atoms in the sense of distinct
localized entities beyond the act of observation.  These are
simply names attributed to the mathematical symbols $\psi$ to
distinguish one functional form from another one.  So the original
quest to comprehend atomic structure culminated in just a set of
rules governing laboratory practice.  To summarize, according to
the completeness assumption of SQM, the wave function is
associated with an individual physical system.  It provides the
most complete description of the system that is, in principle,
possible.  The nature of the description is statistical, and
concerns the probabilities of the outcomes of all conceivable
measurements that may be performed on the system.  Therefore, in
this view, quantum mechanics does not present a causal and
deterministic theory for the universe.




\section{The quantum theory of motion}

We have seen that, the quantum world is inexplicable in classical
terms.  The predictions concerning the interaction of matter and
light, embodied in Newtonian mechanics and Maxwell's equations,
are inconsistent with the experimental facts at the microscopic
level. An important feature of quantum effects is their apparent
indeterminism, that individual atomic events are unpredictable,
uncontrollable, and literally seem to have no cause.  Regularities
emerge only when one considers a large ensemble of such events.
This indeed is generally considered to constitute the heart of the
conceptual problem posed by quantum phenomena.  A way of resolving
this problem is that the wave function does not correspond to a
single physical system but rather to an ensemble of systems.  In
this view, the wave function is admitted to be an incomplete
representation of actual physical states and plays a role roughly
analogous to the distribution function in classical statistical
mechanics.  Now, to understand experimental results as the outcome
of a causally connected series of individual processes, one can
seek further significance of the wave function (beyond its
probabilistic aspect), and can introduce other concepts (hidden
variables) in addition to the wave function.  It was in this
spirit that Bohm {\cite{Bohm}} in 1952 proposed his theory and
showed how underlying quantum mechanics is a causal theory of the
motion of waves and particles which is consistent with a
probabilistic outlook, but does not require it. In fact, the
additional element that he introduced apart from the wave function
is just a particle, conceived in the classical sense of pursuing a
definite continuous track in space-time.  The basic postulates of
Bohm's quantum
mechanics (BQM) can be summarized as follows:\\
{\it 1.} An individual physical system comprises a wave
propagating in space-time together with a particle which moves
continuously under the guidance of the wave.\\
{\it 2.} The wave is mathematically described by $\psi({\bf x},t)$
which is a solution to the Scr\"{o}dinger's equation:
\begin{eqnarray}
i\hbar\frac{\partial\psi}{\partial
t}=(-\frac{\hbar^{2}}{2m}\nabla^{2}+V)\psi
\end{eqnarray}
{\it 3.} The particle motion is determined by the solution ${\bf
x}(t)$ to the guidance condition
\begin{eqnarray}
\dot{\bf x}=\frac{1}{m}\nabla S({\bf x},t)|_{{\bf x}={\bf x}(t)}
\end{eqnarray}
where $S$ is the phase of $\psi$.\\
These three postulates on their own constitute a consistent theory
of motion.  Since BQM involves physical assumptions that are not
usually made in quantum mechanics, it is preferred to consider it
as a new theory of motion which is appropriately called the
quantum theory of motion {\cite{Holland}}.  In order to ensure the
compatibility of the motions of the ensemble of particles with the
results of quantum mechanics, Bohm added the following further
postulate:\\
{\it 4.} The probability that a particle in the ensemble lies
between the points ${\bf x}$ and ${\bf x}+d{\bf x}$ at time $t$ is
given by
\begin{eqnarray}
R^2({\bf x},t)d^3x
\end{eqnarray}
where $R^2=|\psi|^2$.  This shows that the concept of probability
in BQM only enters as a subsidiary condition on a causal theory of
the motion of individuals, and the statistical meaning of the wave
function is of secondary importance.  Failure to recognize this
has been the source of much confusion in understanding the causal
interpretation.

Now, here, it is proper to compare and contrast Bohm's quantum
theory with the standard one.  It can be seen that, some of the
most perplexing interpretational problems of SQM are simply solved
in BQM.


\subsection{Some new insights by BQM}

\subsubsection{No measurement problem}

One of the most elegant aspects of BQM is its treatment on the
measurement problem, where it becomes a non-problem.  In BQM,
measurement is a dynamical and essentially many-body process.
There is no collapse of the wave function, and so no measurement
problem.  The basic idea is that a particle always has a definite
position before measurement.  So there is no superposition of
properties, and measurement or observation is just an attempt to
discover this position.

To clarify the subject, consider, as an example,  an inhomogeneous
magnetic field which produces a spatial separation among the
various angular momentum components of an incident beam of atoms.
The incident wave packet $g(x)$ moves with a velocity $v_0$ along
the $y$-axis. This function $g(x)$ (e.g., a Gaussian) is fairly
sharply peaked about $x=0$.  The initial quantum state of the atom
is a superposition of angular momentum eigenstates $\psi_{n}(\xi)$
of the atom. Thus, the initial wave function for the system before
the atom has entered the region of the magnetic field can be
written as
\begin{eqnarray}
\Psi_0(x,\xi,t)=g(x-v_0t)\sum_{n}C_n\psi_{n}(\xi).
\end{eqnarray}
The interaction between the inhomogeneous magnetic field and the
magnetic moment of the atom exerts a net force on the atom in the
$z$-direction.  Once the packet emerges from the field, the $n$
components of the packet diverge along separate paths.  After that
sufficient time has elapsed, the $n$ component packets no longer
overlap and have essentially disjoint supports.  Then the wave
function has evolved into
\begin{eqnarray}
\Psi(x,\xi,t)=\sum_{n}C_ng_n[x-x_n(t)]e^{i\varphi_n}\psi_{n}(\xi)
\end{eqnarray}
where $x_n(t)$ show the particle trajectories and the $\varphi_n$
are simply constant phases.  The description given sofar is
similar to an account of a measurement in SQM frame.  So, the next
step would be to apply the projection postulate once an atom has
been observed.  One would simply erase the other packets.  In BQM,
however, the situation is different.  The probability of finding
the atom at some particular position is
\begin{eqnarray}
P(x,\xi,t)=\sum_{n}|C_n|^2|g_n[x-x_n(t)]|^2|\psi_{n}(\xi)|^2.
\end{eqnarray}
There are no interference or cross terms here, because the various
$g_n$ no longer overlap.  After the particle has been found in one
packet, it cannot be in any of the others and has negligible
probability of crossing to other ones (because $P$ effectively
vanishes between the packets).  Now, it is necessary that the
microsystem interact, effectively irreversibly, with a macroscopic
measuring device that has many degrees of freedom to make it
practically impossible (i.e., overwhelmingly improbable) for these
lost wave packets to interfere once again with the one actually
containing the particle.  Thus, the process of measurement is a
two-step one in which (1) the quantum states of the microsystem
are separated into nonoverlapping parts by an, in principle,
reversible interaction and (2) a practically irreversible
interaction with a macroscopic apparatus registers the final
results.


\subsubsection{The classical limit}

By using the guidance condition along with the Schr\"{o}dinger's
equation, the quantum dynamical equation for the motion of a
particle with mass $m$ is given by
\begin{eqnarray}
\frac{dp}{dt}=-\nabla(V+Q)
\end{eqnarray}
where $V$ is the usual classical potential energy and $Q$ is the
so-called quantum potential that is given in terms of the wave
function as
\begin{eqnarray}
Q=-\frac{\hbar^2}{2m}\frac{\nabla^2R}{R}.
\end{eqnarray}
This $Q$ has the classically unexpected feature that its value
depends sensitively on the shape, but not necessarily strongly on
the magnitude of $R$, so that $Q$ need not falloff with distance
as $V$ does.  Now, it is evident that there are no problems in
obtaining the classical equations of motion from BQM, because the
above dynamical equation has the form of Newton's second law.  In
fact, when $(Q/V)\ll 1$ and $(\nabla Q/\nabla V)\ll 1$
(dimensionless parameters) the quantum dynamical equation becomes
just the classical equation of motion.  So the suitable limit is
$Q\longrightarrow 0$ (in the sense of $(Q/V)\longrightarrow 0$ and
also ($\nabla Q/\nabla V)\longrightarrow 0$), rather than anything
like $\hbar\longrightarrow 0$.  It is interesting to know that
there are solutions to the Schr\"{o}dinger's equation with no
classical limit (quantum system with no classical analogue). Thus,
one cannot exclude a priori the possibility that there be a class
of solutions to the classical equations of motion which do not
correspond to the limit of some class of quantum solutions
(classical systems with no quantum analogue).  Therefore, it seems
reasonable to conceive classical mechanics as a special case of
quantum mechanics in the sense that the latter has new elements
($\hbar$ and $Q$) not anticipated in the former.  However, the
possibility that the classical theory admits more general types of
ensemble which cannot be described using the limit of quantum
ensembles, because the latter corresponds to a specific type of
linear wave equation and satisfy special conditions such as being
built from single-valued conserved pure states, suggests that the
two statistical theories can be considered independent while
having a common domain of application.  This domain is
characterized by $Q\longrightarrow 0$ in the quantum theory.  Now,
there is a well-defined conceptual and formal connection between
the classical and quantum domains but, as a new result, they
merely intersect rather than are being contained in the other.


\subsubsection{The uncertainty relations}

One of the basic features of quantum mechanics is the association
of Hermitian operators with physical observables, and the
consequent appearance of noncommutation relations between the
operators.  For example, whatever the interpretation, from SQM or
BQM one can obtain the Heisenberg uncertainty relation
\begin{eqnarray}
\triangle x_i\triangle p_j\geq(\hbar/2)\delta_{ij}
\end{eqnarray}
for operators $x$ and $p$ that satisfy
$[x_i,p_j]=i\hbar\delta_{ij}$ {\cite{Holland}}.  As a result, a
wave function cannot be simultaneously an eigenfunction of $x$ and
$p$.  Since measuring of an observable involves the transformation
of the wave function into an eigenfunction of the associated
operator, it appears that a system cannot simultaneously be in a
state by which its position and momentum are precisely known. How
may one reconcile the uncertainty relation with the assumption
that a particle can be ascribed simultaneously well-defined
position and momentum variables as properties that exist during
all interactions, including measurements?  To answer this, we note
that our knowledge of the state of a system should not be confused
with what the state actually is.  Quantum mechanics is constructed
so that we cannot observe position and momentum simultaneously,
but this fact does not prevent us to think of a particle having a
well-defined track in reality.  Bohm's discussion shows how the
act of measurement, through the influence of the quantum
potential, can disturb the microsystem and thus produce an
uncertainty in the outcome of a measurement \cite{Bohm}.  In other
words, we can interpret the uncertainty relations as an expression
of the different types of motion accessible to a particle when its
wave undergoes the particular types of interaction appropriate to
the measurement.  In fact, the formal derivation of the
uncertainty relations goes through as before, but now we have some
understanding of how the spreads come about physically.  According
to BQM, the particle has a position and momentum prior to, during,
and after the measurement, whether this be of position, momentum
or any other observable.  But in a measurement, we usually cannot
observe the real value that an observable had prior to the
measurement. In fact, as Bohm mentioned \cite{Bohm}, in the
suggested new interpretation, the so-called observables are not
properties belonging to the observed system alone, but instead
potentialities whose precise development depends just as much on
the observing apparatus as on the observed system.


\subsubsection{Concept of the wave function}

As we have seen, to find a connection between the two aspects of
matter, i.e. particle and wave, one can rewrite the complex
Schr\"{o}dinger's equation as a coupled system of equations for
the real fields $R$ and $S$ which are defined by $\psi=Re^{iS}$.
Then, in summary, these fields can play the following several
roles
simultaneously:\\
{\it 1.} They are associated with two physical fields propagating
in space-time and define, together with the particle, an
individual physical system.\\
{\it 2.} They act as actual agents in the particle motion, via the
quantum potential.\\
{\it 3.} They enter into the definition of properties associated
with a particle (momentum, energy and angular momentum).  These
are not arbitrarily specified but are a specific combination of
these fields, and are closely related to the associated quantum
mechanical operators.\\
{\it 4.} They have other meanings which ensure the consistency of
BQM with SQM, and moreover, their connection with the classical
mechanics.\\
Generally in BQM, the wave function plays two conceptually
different roles.  It determines (1) the influence of the
environment on the quantum system and (2) the probability density
by $P=|\psi|^2$.  Now, since the guidance condition along with the
Schr\"{o}dinger's equation uniquely specify the future and past
continuous evolution of the particle and field system, BQM forms
the basis of a causal interpretation of quantum mechanics.


\subsubsection{Wave function of the universe}

By quantizing the Hamiltonian constraint of general relativity in
the standard way one obtains the Wheeler-De Witt's equation, which
is the Schr\"{o}dinger's equation of the gravitational field.  In
this regard, there is an attempt to apply quantum mechanics to the
universe as a whole in the so-called quantum cosmology.  This has
been widely interpreted according to the many-worlds picture of
quantum mechanics.  But there is no need for this, because
acoording to many physicists, quantum cosmology deals with a
single system - our universe. We have seen that, BQM is eminently
suited to a description of systems that are essentially unique,
such as the universe. Therefore, quantum cosmology is independent
of any subsidiary probability interpretation one may like to
attach to the wave function.


\subsubsection{Quantum potential as the origin of mass?}

In BQM it can be shown that the equation of motion of a bosonic
massless quantum field is given by
\begin{eqnarray}
\partial^2\psi({\bf x},t)=-\frac{\delta Q[\psi({\bf x}),t]}{\delta\psi({\bf
x})}|_{\psi({\bf x})=\psi({\bf x},t)}
\end{eqnarray}
which generally implies noncovariant and nonlocal properties of
the field \cite{Holland}.  In fact, these features characterize
the extremes of quantum behavior and, in principle, there exist
states for which the right hand side of the above equation of
motion is a scalar and local function of the space-time
coordinate.  The fact that this term is finite means that although
the wave will be essentially nonclassical but will obey the type
of equation we might postulate for a classical field, in which the
scalar wave equation is equated to some function of the field.
Here, the interesting point is that using quantization of a
massless field it is possible to give mass to the field in the
sense that the quantum wave obeys the classical massive
Klein-Gordon equation
\begin{eqnarray}
(\partial^2+m^2)\psi=0
\end{eqnarray}
as a special case for the equation of motion of a massless quantum
field, where $m$ is a real constant \cite{Holland}.  Therefore,
the quantum potential acts so that the massless quantum field
behaves as if it were a classical field with mass.



\subsection{Some current objections to BQM}

There are some of the typical objections that have been advanced
against BQM.  So, here, these objections are summarized and some
preliminary answers are given to them.


\subsubsection{Predicting nothing new}

It is completely right that BQM was constructed so that its
predictions are exactly the same as SQM's ones at the ensemble
level.  But, BQM permits more detailed predictions to be made
pertaining to the individual processes.  Whether this may be
subjected to an experimental test is an open question, which is
studied here using some examples.


\subsubsection{Nonlocality is the price to be paid}

Nonlocality is an intrinsic and clear feature of BQM.  This
property does not contradict special theory of relativity and the
statistical predictions of relativistic quantum mechanics.  But
sometimes it is considered to be in some way a defect, because
local theories are considered to be preferable.  Yet nonlocality
seems to be a small price to pay if the alternative is to forego
any account of objective processes at all (including local ones).
Furthermore, Aspect's experiment \cite{Aspect} established that,
quantum mechanics is really a nonlocal theory without superluminal
signalling \cite{Bell}.  Therefore, it is not necessary to worry
about this property.


\subsubsection{Existence of trajectories cannot be proved}

BQM reproduces the assertion of SQM that one cannot simultaneously
perform a precise measurement on both position and momentum. But
this cannot be adduced as an evidence against the tenability of
the trajectory concept. Science would not exist if ideas were only
admitted when evidence for them already exists. For example, one
cannot after all empirically prove the completeness postulate. The
argument in favor of trajectory lies elsewhere, in its capacity to
make intelligible a large amount of empirical facts.


\subsubsection{An attempt to return to classical physics}

BQM has been often objected for reintroducing the classical
paradigm.  But, as we mentioned in relation to BQM's classical
limit, BQM is a more complete theory than SQM and classical
mechanics, and includes both of them nearly independent theories
in different domains, and also represents the connection between
them. Therefore, BQM which apply the quantum states to guide the
particle is, in principle, an intelligible quantum theory, not a
classical one.


\subsubsection{No mutual action between the guidance wave and the
particle}

Among the many nonclassical properties exhibited by BQM, one is
that the particle does not react dynamically on the wave that is
guided by.  But, while it may be reasonable to require reciprocity
of actions in classical theory, this cannot be regarded as a
logical requirement of all theories that employ the particle and
field concepts, particularly the ones involving a nonclassical
field.


\subsubsection{More complicated than quantum mechanics}

Mathematically, BQM requires a reformulation of the quantum
formalism, but not an alternation.  The present reason is that SQM
is not the one most appropriate to the physical interpretation.
But, mathematically, the desirable theory, particularly at the
ensemble level, can be considered quantum mechanics, because the
quantum potential is implicit in the Schr\"{o}dinger's equation.





In part I of this dissertation, we have concentrated on the first
objection and studied some thought experiments in which BQM can
predict different results from SQM, at the individual level.


%\newpage
%~~~
%\newpage



%---------------------------------------------------------Ch.1
\chapter{Two double-slit experiment using position entanglement of EPR pair}


\section{Introduction}


According to the standard quantum mechanics (SQM), the complete
description of a system of particles is provided by its wave
function.  The empirical predictions of SQM follow from a
mathematical formalism which makes no use of the assumption that
matter consists of particles pursuing  definite tracks in
space-time.  It follows that the results of the experiments
designed to test the predictions of the theory, do not permit us
to infer any statement regarding the particle--not even its
independent existence.

In the Bohmian quantum mechanics (BQM), however, the additional
element that is introduced apart from the wave function is the
particle position, conceived in the classical sense as pursuing a
definite continuous track in space-time [1-3]. The detailed
predictions made by this causal interpretation explains how the
results of quantum experiments come about, but it is claimed that
they are not tested by them. In fact, when Bohm {\cite{Bohm}}
presented his theory in 1952, experiments could be done with an
almost continuous beam of particles, but not with individual
particles. Thus, Bohm constructed his theory in such a fashion
that it would be impossible to distinguish observable predictions
of his theory from SQM.  This can be seen from Bell's comment
about empirical equivalence of the two theories when he
said:``{\it{It}} [the de Broglie-Bohm version of non-relativistic
quantum mechanics] {\it{is experimentally equivalent to the usual
version insofar as the latter is unambiguous}}"{\cite{Bell}}.  So,
could it be that a certain class of phenomena might correspond to
a well-posed problem in one theory but to none in the other? Or
might definite trajectories of Bohm's theory lead to a prediction
of an observable where SQM would just have no definite prediction
to make?

To draw discrepancy from experiments involving the particle track,
we have to argue in such a way that the observable predictions of
the modified theory are in some way functions of the trajectory
assumption.  The question raised here is whether BQM's laws of
motion can be made relevant to experiment. At first, it seems that
definition of time spent by a particle within a classically
forbidden barrier provides a good evidence for the preference of
BQM.  But, there are difficult technical questions, both
theoretically and experimentally, that are still unsolved about
this tunnelling times {\cite{Cushing}}. Furthermore, a recent work
 indicates that it is not practically feasible to use tunnelling
effect to distinguish between the two theories
{\cite{Abolhasani}}.  In another proposal, Englert {\it{et al.}}
{\cite{Englert}} and Scully {\cite{Scully}} have claimed that in
some cases Bohm's approach gives results that disagree with those
obtained from SQM and, in consequence, with experiment. However,
Dewdney {\it{et al.}} {\cite{Dewdney}} and then Hiley {\it{et
al.}} {\cite{Hiley}} showed that the specific objections raised by
them cannot be sustained. Meanwhile, Hiley believes that no
experiment can decide between the standard and Bohm's
interpretation.  On the other hand, Vigier {\cite{Vigier}}, in his
recent work, has given a brief list of new experiments which
suggests that the U(1) invariant massless photon, with properties
of light within the standard interpretation, are too restrictive
and that the O(3) invariant massive photon causal de Broglie-Bohm
interpretation of quantum mechanics, is now supported by
experiments. In addition, Leggett {\cite{Leggett}} considered some
thought experiments involving macrosystems which can predict
different results for SQM and BQM. \footnote{Leggett
{\cite{Leggett}} assumes that the experimental predictions of SQM
will continue to be realized under the extreme conditions
specified, although a test of this hypothesis is part of the aim
of the macroscopic quantum cohrence program.  In addition, he
considered BQM as another interpretation of the same theory rather
than an alternative theory.}  In other work, Ghose {\it et al.}
\cite{Ghose1,Ghosepla} indicated that although BQM is equivalent
to SQM when averages of dynamical variables are taken over a Gibbs
ensemble of Bohmian trajectories, the equivalence breaks down for
ensembles built over clearly separated short intervals of time in
specially entangled two-bosonic particle systems.  Moreover, Ghose
\cite{Ghose2} showed that BQM is incompatible with SQM unless the
Bohmian system corresponding to an SQM system is ergodic.  Some
other recent work in this regard can be also found in
\cite{Neumaier,Correggi,Alvarenga,Feligioni}.

Here, using an original EPR source \cite{EPR} placed between two
double-slit plane, we have suggested a thought experiment which
can distinguish between the standard and Bohmian quantum mechanics
\cite{0103101,Reply}.  Some details on the considered EPR source
have been examined to clarify the realizability of this
experiment. Finally, an experimental effort for the realization of
this thought experiment has been indicated \cite{Brida}.

\section{Description of the proposed experiment}
\begin{figure}[t]\vspace{15mm}
\includegraphics[width=12cm,height=6cm,angle=0]{fig1.eps}
\caption{A two double-slit experiment configuration. Two identical
particles with zero total momentum are emitted from the source $S$
and then they pass through slits $A$ and $B^{'}$ or $B$ and
$A^{'}$.  Finally, they are detected on $S_{1}$ and $S_{2}$
screens, simultaneously.  It is necessary to note that dotted
lines are not real trajectories.}\label{fig11}
\end{figure}

To distinguish between SQM and BQM we consider the following
scheme. A pair of identical non-relativistic particles with total
momentum zero, labelled by 1 and 2, originate from a source S that
is placed exactly in the middle of a two double-slit screens, as
shown in Fig.~\ref{fig11}.  We assume that the intensity of the
beam is so low that during any individual experiment we have only
a single pair of particles passing through the slits.  In
addition, we assume that the detection screens $S_{1}$ and $S_{2}$
register only those pairs of particles that reach the two screens
simultaneously.  Thus, we are sure that the registration of single
particles is eliminated from final interference pattern. The
detection process at the screens $S_{1}$ and $S_{2}$ may be
nontrivial, but they play no causal role in the basic phenomenon
of the interference of particles waves {\cite{Holland}}.  In the
two-dimensional system of coordinates $(x,y)$ whose origin $S$ is
shown, the center of slits lie at the points $(\pm d,\pm Y)$.
Suppose that before the arrival of the two particles on the slits,
the entangled wave function describing them is given by
\begin{eqnarray}\label{incident}
\psi_{in}(x_1,y_1;x_2,y_2;t)&=&\chi(x_1,x_2)\hbar\int_{-\infty}^{+\infty}exp[ik_y(y_1-y_2)]dk_y e^{-iEt/\hbar}\nonumber\\
&=&2\pi\hbar\chi(x_1,x_2)\delta(y_1-y_2)e^{-iEt/\hbar}
\end{eqnarray}
where $E=E_{1}+E_{2}=\hbar^{2}(k_{x}^{2}+k_{y}^{2})/m$ is the
total energy of the system of the two particles, and
$\chi(x_1,x_2)$ is the $x$-component of the wave function that
could have a form similar to the $y$-component. However, its form
is not important for the present work.  The wave function
~(\ref{incident}) is just the one represented in {\cite{EPR}}, and
it shows that the two particles have vanishing total momentum in
the $y$-direction, and their $y$-component of the center of mass
is exactly located on the $x$-axis.  This is not inconsistent with
Heisenberg's uncertainty principle, because
\begin{eqnarray}{\label{H}}
[p_{y_1}+p_{y_2},y_1-y_2]=0.
\end{eqnarray}
  The plane wave assumption comes from large distance between
source $S$ and double-slit screens.  To avoid the mathematical
complexity of Fresnel diffraction at a sharp-edge slit, we suppose
the slits have soft edges that generate waves having identical
Gaussian profiles in the $y$-direction while the plane wave in the
$x$-direction is unaffected {\cite{Holland}}.  The instant at
which the packets are formed will be taken as our zero of time.
Therefore, the four waves emerging from the slits $A$, $B$,
$A^{'}$ and $B^{'}$ are initially
\begin{eqnarray}\label{psiab}
\psi_{A,B}(x,y)&=&(2\pi\sigma_{0}^{2})^{-1/4}e^{-(\pm
y-Y)^{2}/4\sigma_{0}^{2}}e^{i[k_{x}(x-d)+k_{y}(\pm y-Y)]}\nonumber\\
\psi_{A^{'},B^{'}}(x,y)&=&(2\pi\sigma_{0}^{2})^{-1/4}e^{-(\pm
y+Y)^{2}/4\sigma_{0}^{2}}e^{i[-k_{x}(x+d)+k_{y}(\pm y+Y)]}
\end{eqnarray}
where $\sigma_{0}$ is the half-width of each slit.  At time $t$
the general total wave function at a space point $(x,y)$ of our
considered system for bosonic and fermionic particles is given by
\begin{eqnarray}\label{psi}
\psi(x_{1},y_{1};x_{2},y_{2};t)=N[\psi_{A}(x_{1},y_{1},t)\psi_{B^{'}}(x_{2},y_{2},t)\pm\psi_{A}(x_{2},y_{2},t)\psi_{B^{'}}(x_{1},y_{1},t)\nonumber\\
 +\psi_{B} (x_{1},y_{1},t)\psi_{A^{'}}(x_{2},y_{2},t)\pm\psi_{B} (x_{2},y_{2},t)\psi_{A^{'}}
(x_{1},y_{1},t)]\nonumber\\
\end{eqnarray}
where $N$ is a reparametrization constant that its value is
unimportant in this work and
\begin{eqnarray}\label{psiabt}
\psi_{A,B}(x,y,t)&=&(2\pi \sigma_{t}^{2})^{-1/4}e^{-(\pm
y-Y-\hbar k_yt/m)^{2}/4\sigma_{0}\sigma_{t}}\nonumber\\
&&\times e^{i[k_{x}(x-d)+k_{y}(\pm
y-Y)-Et/\hbar]}\nonumber\\
\psi_{A^{'},B^{'}}(x,y,t)&=&(2\pi \sigma_{t}^{2})^{-1/4}e^{-(\pm
y-Y-\hbar k_yt/m)^{2}/4\sigma_{0}\sigma_{t}}\nonumber\\
&&\times e^{i[-k_{x}(x+d)+k_{y}(\pm y-Y)-Et/\hbar]}
\end{eqnarray}
with
\begin{eqnarray}
\sigma_{t}=\sigma_{0}(1+\frac{i\hbar t}{2m\sigma_{0}^{2}}).
\end{eqnarray}
In addition, the upper and lower sings in the total wave function
(\ref{psi}) refer to symmetric and anti-symmetric wave function
under exchange of particle 1 to particle 2, corresponding to
bosonic and fermionic property, while in Eq.~(\ref{psiabt}) they
refer to upper and lower slits, respectively.  In the next
section, we shall use BQM to derive some of the predictions of
this proposed experiment.


\section{Bohmian quantum mechanics prediction}
In BQM, the complete description of a system is given by
specifying the position of the particles in addition to their wave
function which has the role of guiding the particles according to
following guidance condition for $n$ particles, with masses $m_1,
m_2,..., m_n$
\begin{equation}\label{guidance}
\dot{\bf x}_{i}({\bf x},t)=\frac{1}{m_{i}} {\nabla_{i}}S({\bf
x},t)=\frac{\hbar}{m_{i}}Im\left( \frac{\nabla_{i}\psi({\bf
x},t)}{\psi({\bf x},t)}\right)
\end{equation}
where ${\bf x}=({\bf x}_1 ,{\bf x}_2,\ldots, {\bf x}_n)$ and
\begin{equation}
\psi({\bf x},t)= R({\bf x},t) e^{iS({\bf x},t)/\hbar}
\end{equation}
is a solution of Schr$\ddot{o}$dinger's wave equation.  Thus,
instead of SQM with indistinguishable particles, in BQM the path
of particles or their individual histories distinguishes them and
each one of them can be studied separately {\cite{Holland}}.  In
addition, Belousek {\cite{Belousek}} in his recent work, concluded
that the problem of Bohmian mechanical particles being
statistically (in)distinguishable is a matter of theory choice
underdetermined by logic and experiment, and that such particles
are in any case physically distinguishable.  For our proposed
experiment, the speed of the particles 1 and 2 in the
$y$-direction is given , respectively, by
\begin{eqnarray}
&&\dot{y}_{1}(x_{1},y_{1};x_{2},y_{2};t)=\frac{\hbar}{m}Im(\frac{\partial_{y_{1}}\psi(x_{1},y_{1};x_{2},y_{2};t)
}{\psi(x_{1},y_{1};x_{2},y_{2};t)})\nonumber\\
&&\dot{y}_{2}(x_{1},y_{1};x_{2},y_{2};t)=\frac{\hbar}{m}Im(\frac{\partial_{y_{2}}\psi(x_{1},y_{1};x_{2},y_{2};t)
}{\psi(x_{1},y_{1};x_{2},y_{2};t)}).
\end{eqnarray}
With the replacement of $\psi(x_{1},y_{1};x_{2},y_{2};t)$ from
Eqs.~(\ref{psi}) and (\ref{psiabt}), we have
\begin{eqnarray}\label{v1v2}
\dot{y}_{1}=&N\frac{\hbar}{m}&Im\{\frac{1}{\psi}[[-2(y_{1}-Y-\hbar
k_yt/m)/4\sigma_{0}\sigma_{t}+ik_{y}]\psi_{A_{1}}\psi_{B^{'}_{2}}\nonumber\\
           &\pm&[-2(y_{1}+Y+\hbar
           k_yt/m)/4\sigma_{0}\sigma_{t}-ik_{y}]\psi_{A_{2}}\psi_{B^{'}_{1}}\nonumber\\
           &+&[-2(y_{1}+Y+\hbar
           k_yt/m)/4\sigma_{0}\sigma_{t}-ik_{y}]\psi_{B_{1}}\psi_{A^{'}_{2}}\nonumber\\
           &\pm&[-2(y_{1}-Y-\hbar
           k_yt/m)/4\sigma_{0}\sigma_{t}+ik_{y}]\psi_{B_{2}}\psi_{A^{'}_{1}}]\}\nonumber\\
\dot{y}_{2}=&N\frac{\hbar}{m}&Im\{\frac{1}{\psi}[[-2(y_{2}+Y+\hbar
k_yt/m)/4\sigma_{0}\sigma_{t}-ik_{y}]\psi_{A_{1}}\psi_{B^{'}_{2}}\nonumber\\
           &\pm&[-2(y_{2}-Y-\hbar
           k_yt/m)/4\sigma_{0}\sigma_{t}+ik_{y}]\psi_{A_{2}}\psi_{B^{'}_{1}}\nonumber\\
           &+&[-2(y_{2}-Y-\hbar
           k_yt/m)/4\sigma_{0}\sigma_{t}+ik_{y}]\psi_{B_{1}}\psi_{A^{'}_{2}}\nonumber\\
           &\pm&[-2(y_{2}+Y+\hbar k_yt/m)/4\sigma_{0}\sigma_{t}-ik_{y}]\psi_{B_{2}}\psi_{A^{'}_{1}}]\}
\end{eqnarray}
where, for example, the short notation
$\psi_{A}(x_{1},y_{1},t)=\psi_{A_{1}}$ is used. Furthermore, from
Eq.~(\ref{psiabt}) it is clear that
\begin{eqnarray}\label{symmetry}
&&\psi_{A}(x_{1},y_{1},t)=\psi_{B}(x_{1},-y_{1},t)\cr
&&\psi_{A}(x_{2},y_{2},t)=\psi_{B}(x_{2},-y_{2},t)\cr
&&\psi_{B^{'}}(x_{1},y_{1},t)=\psi_{A^{'}}(x_{1},-y_{1},t)\cr
&&\psi_{B^{'}}(x_{2},y_{2},t)=\psi_{A^{'}}(x_{2},-y_{2},t)
\end{eqnarray}
 which indicates the reflection symmetry  of $\psi(x_{1},y_{1};x_{2},y_{2};t)$ with
respect to the $x$-axis.  Utilizing this symmetry in
Eq.~(\ref{v1v2}), we can see that
\begin{eqnarray}\label{-v}
&&\dot{y}_{1}(x_{1},y_{1};x_{2},y_{2};t)=-\dot{y}_{1}(x_{1},-y_{1};x_{2},-y_{2};t)\cr&&
\dot{y}_{2}(x_{1},y_{1};x_{2},y_{2};t)=-\dot{y}_{2}(x_{1},-y_{1};x_{2},-y_{2};t)
\end{eqnarray}
which are valid  for both bosonic and fermionic particles.
Relations (\ref{-v}) show that if $y_{1}(t)=y_{2}(t)=0$, then the
speed of each particles in the $y$-direction is zero. This means
that none of the particles can cross the $x$-axis nor are they
tangent to it, provided both of them are simultaneously on this
axis.  There is the same symmetry of the velocity about the
$x$-axis as for an ordinary double-slit experiment
{\cite{Holland}}.

If we consider $y=(y_{1}+y_{2})/2$ to be the vertical coordinate
of the center of mass of the two particles \footnote{Here, one may
argue that this center of mass definition seems inconsistent with
the previous definition introduced in the incident wave function
(\ref{incident}).  But, in appendix A, we have shown that these
two  definitions of the center of mass coordinate are two
consistent representations.}, then we can write
\begin{eqnarray}\label{eqmotion}
\dot{y}&=&(\dot{y}_{1}+\dot{y}_{2})/2\cr
       &=&-N\frac{\hbar}{2m}Im\{{\frac{1}{\psi}(\frac{y_{1}+y_{2}}{2\sigma_{0}\sigma_{t}})
       (\psi_{A_{1}}\psi_{B^{'}_{1}}\pm\psi_{A_{2}}\psi_{B^{'}_{1}}+\psi_{B_{1}}\psi_{A^{'}_{2}}\pm\psi_{B_{2}}\psi_{A^{'}_{1}})}\}\cr
       &=&\frac{(\hbar/2m0\sigma_{0}^{2})^{2}}{1+(\hbar/2m\sigma_{0}^{2})^{2}t^{2}}yt
       .
\end{eqnarray}
Solving the equation of motion (\ref{eqmotion}), we obtain the
path of the $y$-coordinate of the center of mass
\begin{equation}\label{y(t)}
y(t)=y(0)\sqrt{1+(\hbar/2m\sigma_{0}^{2})^{2}t^{2}}.
\end{equation}
If it is assumed that, at $t=0$ the center of mass of the two
particles is exactly on the $x$-axis, that is $y(0)=0$, then the
center of mass of the particles will always remain on the
$x$-axis. Thus, according to BQM, the two particles will be
detected at points symmetric with respect to the $x$-axis, as
shown in Fig.~\ref{fig11}.

It seems that calculation of quantum potential can give us another
perspective of this experiment.  As we know, to see the connection
between the wave and particle, the Schr$\ddot{o}$dinger equation
can be rewritten in the form of a generalized Hamilton-jacobi
equation that has the form of the classical equation, apart from
the extra term
\begin{equation}\label{Q}
Q({\bf x},t)=-\frac{\hbar^{2}}{2m}\frac{\nabla^{2}R({\bf
x},t)}{R({\bf x},t)}
\end{equation}
where the function Q has been called quantum potential. However,
it can be seen that the calculation and analysis of Q, by using
our total wave function (\ref{psi}), is not very simple.  On the
other hand, we can use the form of Newton's second law, in which
the particle is subject to a quantum force $(-\nabla Q)$, in
addition to the classical force $(-\nabla V)$, namely
\begin{equation}\label{F}
{\bf F}=-{\bf\nabla} (Q+V).
\end{equation}
Now, if we utilize the equation of motion of the center of mass
$y$-coordinate (\ref{y(t)}) and Eq.~(\ref{F}), it is possible to
obtain the quantum potential for the center of mass motion
$(Q_{cm})$. Thus, we can write
\begin{equation}\label{Qx}
-\frac{\partial Q}{\partial x}=m\ddot{x}=0
\end{equation}
\begin{equation}
-\frac{\partial Q}{\partial
y}=m\ddot{y}=\frac{my(0)(\hbar/2m\sigma_{0}^{2})^{2}}{(1+(\hbar
t/2m\sigma_{0}^{2})^{2})^{3/2}}=\frac{my^{4}(0)}{y^{3}}(\frac{\hbar}{2m\sigma_{0}^{2}})^{2}
\end{equation}
where the result of Eq.~(\ref{Qx}) is clearly due to motion of
plane wave in the $x$-direction.  In addition, we assume that
$\nabla V=0$ in our experiment.  Thus, our effective quantum
potential is only a function of the $y$-variable and it has the
form
\begin{equation}
Q=\frac{my^{4}(0)}{2y^{2}}(\frac{\hbar}{2m\sigma_{0}^{2}})^{2}=\frac{1}{2}my^{2}(0)\frac{(\hbar/2m\sigma_{0}^{2})^{2}}{1+(\hbar
t/2m\sigma_{0}^{2})^{2}}.
\end{equation}
If it is assumed that $y(0)=0$, the quantum potential for the
center of mass of the two particles is zero at all times and it
remains on the $x$-axis. However, if $y(0)\neq 0$, then the center
of mass can never touch or cross the $x$-axis.  These conclusions
are consistent with our earlier results (Eq.~(\ref{y(t)})).


\section{Predictions of standard quantum mechanics}


So far, we have been studying the results obtained from BQM at the
individual level.  Now it is well known from SQM that the
probability of simultaneous detection of two particles at $y_{M}$
and $y_{N}$, at the screens $S_{1}$ and $S_{2}$, is equal to
\begin{equation}
P_{12}(y_{M},y_{N})=\int_{y_{M}}^{y_{M}+\triangle}dy_{1}\int_{y_{N}}^{y_{N}+\triangle}dy_{2}|\psi(x_{1},y_{1};x_{2},y_{2};t)|^{2}.
\end{equation}
The parameter $\Delta$, which is taken to be small, is a measure
of the size of the detectors.  It is clear that the probabilistic
prediction of SQM is in disagreement with the symmetrical
prediction of BQM for the $y(0)=0$ condition, because SQM predicts
that probability of asymmetrical detection, at the individual
level of a pair of particles, can be different from zero, in
opposition to BQM's symmetrical predictions.  In the other words,
based on SQM's prediction, the probability of finding the two
particles on one side of the $x$-axis can be nonzero while we
showed that BQM's prediction forbids such events in our scheme,
and its probability is exactly zero.  Thus, if necessary
arrangements to perform this experiment are provided, one can
choose one of the two theories as a more complete description of
the quantum universe.


\section{Statistical distribution of the center of mass coordinate around the $x$-axis}


We have assumed that the two particles are entangled so that in
spite of a position distribution for each particle, $y(0)$ can be
always considered to be on the $x$-axis. However, one may argue
that, it is necessary to consider a position distribution for
$y(0)$, that is, $\triangle y(0)\neq0$ while $\langle
y(0)\rangle=0$. Therefore, it may seem that, not only symmetrical
detection of the two particles is violated, but also they can be
found on one side of the $x$-axis on the screens, because the
majority of the pairs can not be simultaneously on the $x$-axis
{\cite{Marchildon}}.  In this regard, Ghose {\cite{Ghose3}}
believes that the two entangled bosonic particles cannot cross the
symmetry axis even if we have the situation $(y_1+y_2)_{t=0}\neq
0$. However, even by accepting Marchildon's argument about this
situation {\cite{Marchildon}}, this problem can be solved if we
adjust $\triangle y(0)$ to be very small. We assume that, to keep
symmetrical detection about the $x$-axis with reasonable
approximation, the center of mass dispersion in the $y$-direction
must be smaller than the distance between any two neighboring
maxima on the screens, that is,
\begin{eqnarray}\label{dely}
\triangle y(t)\ll\frac{\lambda D}{2Y}\simeq\frac{\pi\hbar t}{Ym}
\end{eqnarray}
where $\lambda$ is the de Broglie wavelength.  By using
Eq.~(\ref{y(t)}), we can write
\begin{eqnarray}
\triangle
y(0)\ll\frac{2a\pi}{\sqrt{1+a^2}}\frac{\sigma_{0}^{2}}{Y}
\end{eqnarray}
where $a=\hbar t/2m\sigma_{0}^{2}$.  Since considering $a\geq 1$
condition is a reasonable assumption in interferences experiments,
we have
\begin{eqnarray}
\triangle y(0)\ll 2\pi\frac{\sigma_{0}^{2}}{Y}.
\end{eqnarray}
This relation shows that $Y\ll\sigma_0$ condition can be
considered as a suitable choice in our scheme.  For usual
condition of $Y\sim\sigma_0$, we have
\begin{eqnarray}
\triangle y(0)\ll 2\pi\sigma_0.
\end{eqnarray}
Therefore, to prevent joint detection of the two particles on the
one side of the $x$-axis, and also, to obtain an acceptable
symmetrical joint detection around this axis, it is enough to
adjust the $y$ dispersion of the center of mass position of the
two particles very smaller than the width of slits.  In this case,
we only lose our information about the initial trajectory of
bosonic particles.  It is evident that, if one considers
$\triangle y(0)\sim\sigma_{0}$, as was done in \cite{Marchildon},
the incompatibility between the two theories will disappear. But
because of the entanglement of the two particles in the
$y$-direction, it is possible to adjust $y(0)$ independent of
$\sigma_{0}$, so that
\begin{eqnarray}\label{interval}
0\leq y(0)=\frac{1}{2}(y_{1}+y_{2})_{t=0}\ll\sigma_{0}.
\end{eqnarray}
Although it is obvious that $(\triangle y_{1})_{t=0}=(\triangle
y_{2})_{t=0}\sim\sigma_{0}$, but the position entanglement of the
two particles at the source $S$ in the $y$-direction makes them
always satisfy Eq.~(\ref{interval}), which is not feasible in the
one-particle double-slit devices with $\triangle
y(0)\sim\sigma_0$.


\section{Comparison between SQM and BQM at the ensemble level}


Now, one can compare the results of SQM and BQM at the ensemble
level of the particles. To do this, we consider an ensemble of
pairs of particles that have arrived at the detection screens at
different times $t_{i}$.  It is well known that, in order to
ensure compatibility between SQM and BQM for ensemble of
particles, Bohm added a further postulate to his three basic and
consistent postulates [1-3].  Based on this further postulate, the
probability density that a particle in the ensemble lies between
${\bf x}$ and ${\bf x}+d{\bf x}$, at time $t$, is given by
\begin{eqnarray}
P({\bf x},t)=R^{2}({\bf x},t).
\end{eqnarray}
Thus, the joint probability of simultaneous detection for all
pairs of particles of the ensemble, arriving on the two screens at
different time $t_i$, with $y(0)=0$, is
\begin{eqnarray}
P_{12}=\lim_{N\rightarrow\infty}\sum_{i=1}^{N}R^{2}(y_{1}(t_{i}),-y_{1}(t_{i}),t_{i})\equiv\int^{+\infty}_{-\infty}dy_{1}\int^{+\infty}_{-\infty}dy_{2}|\psi(y_{1},y_{2},t)|^{2}=1\nonumber\\
\end{eqnarray}
where every term in the sum shows only one pair arriving on the
screens at the symmetrical points about the $x$-axis at time
$t_{i}$, with the intensity of $R^{2}$.  If all times $t_{i}$ in
the sum are taken to be $t$, the summation on $i$ can be converted
to an integral over all paths that cross the screens at that time.
Now, one can consider that the joint detection of two points on
the two screens at time $t$ is not symmetrical around the
$x$-axis, but we know that they are not detected simultaneously.
So, it is possible to consider the joint probability of detecting
two particles at two arbitrary points $y_{M}$ and $y_{N}$ as
follows
\begin{equation}
P_{12}(y_{M},y_{N},t)=\int_{y_{M}}^{y_{M}+\Delta}dy_{1}\int_{y_{N}}^{y_{N}+\Delta}dy_{2}|\psi(y_{1},y_{2},t)|^{2},
\end{equation}
which is similar to the prediction of SQM, but obtained in a
Bohmian way.  Thus, it appears that under such conditions, the
possibility of distinguishing the two theories at the statistical
level is impossible, as expected.

Here, to show equivalence of the two theories, we have assumed for
simplicity that $y(0)=0$.  If one consider $y(0)\neq 0$ or
$\triangle y(0)\neq 0$, the equivalence of the two theories is
maintained, as it is argued by Marchildon {\cite{Marchildon}}.
But, using this special case, we show that assumption of $y(0)=0$
is consistent with statistical results of SQM, and in consequence,
finding such a source may not be impossible.



\section{Quantum equilibrium hypothesis and our proposed experiment}

In some of recent comments \cite{Str,Struyve,Guay,Nikolic}, the
quantum equilibrium hypothesis (QEH) is utilized in order to show
that our proposed experiment cannot distinguish between SQM and
BQM. In this section, we have presented some explanations to show
that their argument may not be right and our basic conclusions
about this scheme are still intact.


 We have
seen that, when the entangled particles pass through the slits,
the transformation
\begin{eqnarray}{\label{6}}
\psi_{in}(x_1,y_1;x_2,y_2)\longrightarrow \psi(x_1,y_1;x_2,y_2)
\end{eqnarray}
occurs to the wave function describing the system.  Now, it is
interesting to know what can happen to the entanglement when the
two particles emerge from the slits producing the Gaussian wave
functions represented by Eq.~(\ref{psiab}).  In the other words,
there is a question as to whether the position entanglement
property of the two particles is kept after this transformation.
To answer this question, one can first examine the effect of the
total momentum operator on the wave function of the system,
$\psi(x_1,y_1;x_2,y_2;t)$, which yields
\begin{eqnarray}{\label{ppsi}}
(p_{y_1}+p_{y_2})\psi(x_1,y_1;x_2,y_2;t)&=&-i\hbar
(\frac{\partial}{\partial y_1}+\frac{\partial}{\partial
y_2})\psi(x_1,y_1;x_2,y_2;t)\nonumber\\
&=&i\hbar
(\frac{y_1(t)+y_2(t)}{2\sigma_0\sigma_t})\psi(x_1,y_1;x_2,y_2;t)\nonumber\\
\end{eqnarray}
where one can see that the wave function is an eigenfunction of
the total momentum operator.  Now, if we can assume that the total
momentum of the particles remains zero at all times (an assumption
about which we shall elaborate later on), then it can be concluded
that the center of mass of the two particles in the $y$-direction
is always located on the $x$-axis.  In other words, a momentum
entanglement in the form $p_1+p_2=0$ leads to the position
entanglement in this experiment. However, Born's probability
principle, i.e. $P=|\psi|^2$, which is a basic rule in SQM, shows
that the probability of asymmetrical joint detection of the two
particles can be non-zero on the screens. Thus, there is no
position entanglement and consequently no momentum entanglement
between the two particles.  This compels us to believe that,
according to SQM, the momentum entanglement of the two particles
must be erased during their passage through the slits, and the
center of mass position has to be distributed according to
$|\psi|^2$.



In BQM, however, Born's probability principle is not so important
as a primary rule and all particles follow well-defined tracks
determined by the wave function  $\psi({\bf{x}},t)$, using the
guidance condition (\ref{guidance}) with the unitary time
development governed by Schr$\ddot{o}$dinger's equation.  However,
to ensure the consistency of statistical results of BQM with SQM,
Bohm \cite{Bohm} added QEH, i.e. $P=|\psi|^2$, to his
self-consistent theory just as an additional assumption. Now, let
us review the previous details, but this time in BQM frame. Based
on our supposed EPR source, there are momentum and position
entanglements between the two particles before they were arrived
on the slits. Then, the wave function of the emerging particles
from the slits suffers a transformation represented by
Eq.~({\ref{6}}).  It is not necessary to know in details how this
transformation acts, but what is important is that the two
double-slit screens are considered to be completely identical.
Thus, we expect that the two particles in the slits undergo the
same transformation(s), and so the momentum entanglement, i.e.
$p_1+p_2=0$, must still be valid in BQM which is a deterministic
theory, contrary to SQM. Then, according to Eq. ~({\ref{ppsi}}),
the validity of the momentum entanglement immediately leads to the
position entanglement
\begin{eqnarray}{\label{9}}
y(t)=\frac{1}{2}(y_1(t)+y_2(t))=0.
\end{eqnarray}
We would like to point out that this entanglement is obtained by
using the quantum wave function of the system.  Therefore, this
claim that the supposed position entanglement can not be
understood by using the assumed wave function for the system is
not correct.  By the way, if we accept that the momentum
entanglement is not kept and consequently $y(0)$ obeys QEH, then
deterministic property of BQM, which is a main property of this
theory, must be withdrawn.  However, it is well known that Bohm
{\cite{Bohm}} put QEH only as a subsidiary constraint to ensure
the consistency of the motion of an ensemble of particles with
SQM's results.  Thus, although in this experiment, the center of
mass position of the two entangled particles turns out to be a
constant in BQM frame, the position of each particle is consistent
with QEH so that the final interference pattern is identical to
what is predicted by SQM. Therefore, Bohm's aim concerning QEH is
still satisfied and the deterministic property of BQM is left
intact. In addition, superluminal signals resulting from nonlocal
conditions between our distant entangled particles are precisely
masked by considering QEH for the distribution of each entangled
particle.


So far, we have shown that in BQM frame, the center of mass
position of the two entangled particles can be considered to be a
constant, without any distribution.  So, this property provides a
way to make a discrepancy between SQM and BQM, even for an
ensemble of entangled particles.  For instance, suppose that we
only consider those pairs one of which arrives at the upper half
of the right screen.  Thus, BQM predicts that only detectors
located on the lower half of the left screen become ON and the
other ones are always OFF.  In fact, we obtain two identical
interference patterns at the upper half of the right screen and
the lower half of the left screen.  Instead, SQM is either silent
or predicts a diluted interference pattern at the left screen.
Therefore, concerning the validity of the initial constraint
$y(0)=0$ in BQM, selecting of some pairs to obtain a desired
pattern, which is called selective detection in {\cite{0103101}},
can be applied to evaluate the two theories, at the ensemble level
of pairs.





\section{Approaching realization of the experiment}


Based on a recent work on Bohmian trajectories for photons
{\cite{Ghosepla}}, the first effort for realization of a typical
two-particle experiment was performed very recently by Brida {\it
et al.} {\cite{Brida,BridaPRA}}, using correlated photons produced
in type I parametric down conversion (PDC). In this realization a
beam of a 351 nm pump laser of 0.4 W power with 1 mm in diameter
is directed into a lithium iodate crystal, where correlated pairs
of photons are generated by type I PDC \cite{Brida1}. The two
photons are emitted at the same time (within femtoseconds, whilst
correlation time is some orders of magnitude larger) in a
well-defined direction for a specific frequency. By means of an
optical condenser the produced photons, within two correlated
directions corresponding to 702 nm emission (the degenerate
emission for a 351 nm pump laser), are sent on a double slit
(obtained by a metal deposition on a thin glass by a
photolithographic process) placed just before the focus of the
lens system. The two slits are separated by  100 $\mu$m and have a
width of 10 $\mu$m. They lay in a plane orthogonal to the incident
laser beam and are orthogonal to the table plane. Two single
photon detectors are placed at a 1.21 and a 1.5 m distance after
the slits. They are preceded by an interferential filter at 702 nm
of 4 nm full width at half height and by a lens of 6 mm diameter
and 25.4 mm focal length. The output signals from the detectors
are routed  to a two channel counter, in order to have the number
of events on a single channel, and to a  time to amplitude
converter (TAC) circuit, followed by a single channel analyzer,
for selecting and counting the coincidence events. Figure
\ref{fig12} illustrates this experimental set-up.
\begin{figure}[t]
\begin{center}
\includegraphics[width=7cm,height=12cm,angle=-90]{fig12.eps}
\caption{The experimental apparatus. A pump laser at 351 nm
generates parametric down conversion of type I in a lithium-iodate
crystal. Conjugated photons at 702 nm are sent to a double-slit by
a system of two piano-convex lenses in a way that each photon of
the pair crosses a well defined slit. The first photodetector is
placed at  1.21 m and the second one at 1.5 m from the slit. Both
the single photon detectors (D) are preceded by an interferential
filter at 702 nm (IF) and a lens (L) of 6 mm diameter and 25.4 mm
focal length. Signals from detectors are sent to a time amplitude
converter and then to the acquisition system (multi-channel
analyzer and counters)\cite{Brida}.}\label{fig12}
\end{center}
\end{figure}



By scanning the diffraction pattern using the first detector and
leaving the second fixed at 55 mm from the symmetry axis, it is
found that the coincidences pattern perfectly followed SQM's
predictions, as Fig.~\ref{fig13} shows. The last ones are given by
\begin{eqnarray}
C(\theta_1,\theta_2)&=&g(\theta _1, \theta _i^A )^2 g(\theta _2,
\theta _i^B)^2 + g(\theta _2, \theta _i^A )^2 g(\theta _1, \theta
_i^B)^2\nonumber\\
&&+2 g(\theta _1, \theta _i^A ) g(\theta _2, \theta _i^B) g(\theta
_2, \theta _i^A ) g(\theta _1, \theta _i^B) cos [2kY(sin \theta _1
- sin \theta _2)]\nonumber\\
\end{eqnarray}
in which
\begin{eqnarray}
g( \theta , \theta _i^l)  = { sin ( k\sigma_0 ( sin (\theta) - sin
(\theta _i^l)) \over k\sigma_0 ( sin (\theta) - sin (\theta
_i^l))}
\end{eqnarray}
where $\theta _{1}$ ($\theta _{2}$) is the diffraction angle of
the photon observed by detector 1 (2), and $\theta_i^l$ is the
incidence angle of the photon on the slit $l$ (A or B).
\begin{figure}[t]
\begin{center}
\includegraphics[width=7.5cm,height=11cm,angle=-90]{fig13.eps}
\caption{Coincidences data in the region of interest compared with
SQM's predictions. The second detector is kept fixed at -55 mm
from the $x$-axis. The x errors bars represent the width of the
lens before the detector \cite{Brida}.}\label{fig13}
\end{center}
\end{figure}



The next result of this experiment is that  a  coincidence peak is
observed (see Fig.~\ref{fig14}) also when the first detector is
placed inside enough the same semiplane of the second one.  The
coincidences acquisition with a temporal window of 2.6 ns is
considered, and the background is evaluated shifting the delay
between start and stop of TAC of 16 ns and acquiring data for the
same time of the undelayed acquisition. When the  center of the
lens of the first detector is placed 17 mm after the median
symmetry axis  of the two slits and the second detector is kept at
55 mm, with 35 acquisitions of 30 minutes, it is obtained 78 $\pm$
10 coincidences per 30 minutes after background subtraction,
whilst in this situation BQM's prediction for coincidences is
strictly zero. Furthermore, even when the two detectors were
placed in the same semiplane, the first at 44.4 mm and the second
at 117 mm from the symmetry axis, in correspondence of the second
diffraction peak, a clear coincidence signal was still observed
(albeit less evident than in the former case): in fact, after
background subtraction, an average of 41 $\pm$ 14 coincidences per
hour with 17 acquisitions of one hour (and a clear peak appeared
on the multichannel analyzer) is obtained.
\begin{figure}[t]
\begin{center}
\includegraphics[width=8cm,height=4cm,angle=0]{fig14.eps}
\caption{Observed coincidence peak (output of the multi-channel
analyzer) when the center of the lens of the first photodetector
is placed 17 mm after the median symmetry axis of the double slit
in the same semiplane of the other photodetector, which is kept at
55 mm after the median symmetry axis. Acquisition time lasts  17
hours. No background subtraction is done. A coincidence peak is
clearly visible for a delay between start (first photodetector)
and stop (second photodetector) of 9 ns (the delay inserted on the
second line signal)\cite{Brida}.}\label{fig14}
\end{center}
\end{figure}


Although performing of this experiment using photons, by Brida
{\it et al.} \cite{Brida}, shows that it is feasible to realize
the proposed thought experiment, however, their work does not
satisfy all of our necessary conditions to enter into the region
in which BQM's prediction is different from SQM's.  For instance,
in sec. 1.5, we have shown that the constraint $\triangle y\ll
\lambda D/2Y$ is necessary to keep the symmetrical detection of
the two particles in BQM frame of nonrelativistic domain.  So, by
considering $\hbar t/2m\sigma_{0}^{2}\geq 1$, one obtains
$\triangle y(0)\ll 2\pi\sigma_{0}^{2}/Y$.  This roughly means that
in the considered experimental set-up, with $\sigma_0=5 \mu$m and
$Y=50 \mu$m, we also should adjust $\triangle y(0)\ll 1 \mu$m to
observe a clear difference between the standard and Bohmian
predictions. But in \cite{Brida}, the applied $\triangle y(0)$ in
the laser beam is as much as 1 mm \cite{Genovese}.  Therefore, it
seems that we still need more elaborate efforts to complete the
realization of this thought experiment.


\section{Conclusions}


In conclusion, we have suggested a two-particle system which can
be  adjusted to yield only symmetrical detections for the two
entangled particles in BQM frame whilst according to SQM the
probability for asymmetrical detections is not zero.  The main
reason for the existence of the mentioned differences between SQM
and BQM in this thought experiment is that, in BQM as a
deterministic theory, the position and momentum entanglements are
kept at the slits, while in SQM, due to its probabilistic
interpretation, we must inevitably accept that the entanglements
of the two particles are erased when the two particles pass
through the slits. Incidentally, the saved position entanglement
in BQM, i.e. $y(0)=0$, which is a result of the deterministic
property of the thoery is not inconsistent with QEH, because we
are still able to reproduce SQM's prediction for an ensemble of
such particles, just as QEH requires. Therefore, our proposed
experiment is a suitable candidate to distinguish between the
standard and Bohmian quantum mechanics.

%---------------------------------------------------------Ch.2


\chapter{Study on double-slit device with two correlated particles}


\section{Introduction}


The statistical interpretation of the wave function of the
standard quantum mechanics (SQM) is consistent with all performed
experiments.  An interference pattern on a screen is built up by a
series of apparently random events, and the wave function
correctly predicts where the particle is most likely to land in an
ensemble of trials.  One may, however, take the view that the
characteristic distribution of spots on a screen which builds up
an interference pattern is an evidence for the fact that the wave
function has a more potent physical role.  If one attempts to
understand the experimental results as the outcome of a causally
connected series of individual process, then one is free to
inquire about further significance of the wave function and to
introduce other concepts in addition to it.  Bohm {\cite{Bohm}},
in 1952, showed that an individual physical system comprises a
wave propagating in space-time together with a point particle
which moves continuously under the guidance of the wave [1-3]. He
applied his theory to a range of examples drawn from
non-relativistic quantum mechanics and speculated on the possible
alternations in the particle and field laws of motion such that
the predictions of the modified theory continue to agree with
those of SQM where this is tested, but it could disagree in as yet
unexplored domains {\cite{Holland}}. For instance, when Bohm
presented his theory in 1952, experiments could be done with an
almost continuous beam of particles.  Thus, it was impossible to
discriminate between the standard and the Bohmian quantum
mechanics (BQM) at the individual levels.  In fact, the two
theories can be discriminated at this level, because SQM is a
probabilistic theory while BQM is a precisely defined and
deterministic theory.


In this chapter, we have studied entangled and disentangled wave
functions that can be imputed to a two-particle interference
device, using a Gaussian wave function as a real representation.
Then, SQM and BQM predictions are compared at both the individual
and the statistical levels \footnote{The individual level refers
to our experiment with pairs of particles which are emitted in
clearly separated short intervals of time, and by statistical
level we mean our final interference pattern.}[29-30].



\section{Description of the two-particle experiment}


Consider the famous double-slit experiment.  In the
two-dimensional coordinate system, the centers of the two slits
are located at $(0,\pm Y)$.  Instead of the usual one-particle
emitting source, consider a special source $S_{1}$, so that a pair
of identical non-relativistic particles originate simultaneously
from it.  We assume that, the intensity of the beam is so low that
at a time we have only a single pair of particles passing through
the slits and the detection screen $S_{2}$ registers only those
pairs of particles that reach it simultaneously, and so the
interference effects of single particles will be eliminated. For
mathematical simplicity, we avoid slits with sharp edges which
produce the mathematical complexity of Fresnel diffraction, i.e.,
we assume that the slits have soft edges, so that the Gaussian
wave packets are produced along the $y$-direction, and that the
plane wave along the $x$-axis remain unchanged {\cite{Holland}}.
We take the time of the formation of the Gaussian wave to be
$t=0$. Then, the emerging wave packets from the slits $A$ and $B$
are respectively
\begin{eqnarray}
&&\psi_{A}(x,y)=(2\pi\sigma_{0}^{2})^{-1/4}e^{-(y-Y)^{2}/4\sigma_{0}^{2}}e^{i[k_{x}x+k_{y}(y-Y)]}\nonumber\\
&&\psi_{B}(x,y)=(2\pi\sigma_{0}^{2})^{-1/4}e^{-(y+Y)^{2}/4\sigma_{0}^{2}}e^{i[k_{x}x-k_{y}(y+Y)]}
\end{eqnarray}
where $\sigma_{0}$ is the half-width of each slit.  Moreover at
time $t$ we can write
\begin{eqnarray}\label{psiabt2}
&&\psi_{A}(x,y,t)=(2\pi
\sigma_{t}^{2})^{-1/4}e^{-(y-Y-\hbar k_yt/m)^{2}/4\sigma_{0}\sigma_{t}}e^{i[k_{x}x+k_{y}(y-Y)-Et/\hbar]}\nonumber\\
&&\psi_{B}(x,y,t)=(2\pi \sigma_{t}^{2})^{-1/4}e^{-(y+Y+\hbar
k_yt/m)^{2}/4\sigma_{0}\sigma_{t}}e^{i[k_{x}x-k_{y}(y+Y)-Et/\hbar]}\nonumber\\
\end{eqnarray}
where
\begin{equation}
\sigma_{t}=\sigma_{0}(1+\frac{i\hbar t}{2m\sigma_{0}^{2}}).
\end{equation}


Concerning the two-particle source, in general, we can have two alternatives:\\
{\it 1.} The wave function describing the system (the two emitted
particles+double slit screen) is so entangled that if one particle
passes through the upper (lower) slit, the other particle must go
through lower (upper) slit.  In
 other words, the total momentum and the center of mass of the
two particles in the
$y$-direction is considered zero at the source.\\
{\it 2.} The wave function describing the system is not entangled.
In other words, the emission of each particle is done freely and
the two particles can be
considered independently.  However, as a key point, the two correlated particles are still emitted simultaneously.\\


In the following, we shall study each one of the two alternatives,
separately, and explain SQM's predictions.  Then, Bohmian
predictions have been compared with those of SQM.


\section{Entangled wave function}



We take the wave incident on the double-slit screen to be a plane
wave of the form
\begin{eqnarray}{\label{incident2}}
\psi_{in}(x_1,y_1;x_2,y_2;t)&=&\chi(x_1,x_2)\hbar\int_{-\infty}^{+\infty}exp[ik_y(y_1-y_2)]dk_y e^{-iEt/\hbar}\nonumber\\
&=&2\pi\hbar\chi(x_1,x_2)\delta(y_1-y_2)e^{-iEt/\hbar}
\end{eqnarray}
where $\chi(x_1,x_2)$ is the $x$-component of the wave function
and $E=E_{1}+E_{2}=\hbar^{2}(k_{x}^{2}+k_{y}^{2})/m$ is the total
energy of the system of the two identical particles. The parameter
$m$ is the mass of each particle and $k_{i}$ is the wave number of
particle in $i$-direction.

For this two-particle system, the total wave function after
passing the two particles through the slits can be written as
\begin{eqnarray}\label{psientangled}
\psi(x_{1},y_{1};x_{2},y_{2};t) =
N[\psi_{A}(x_{1},y_{1},t)\psi_{B}(x_{2},y_{2},t)\pm\psi_{A}(x_{2},y_{2},t)\psi_{B}(x_{1},y_{1},t)]\nonumber\\
\end{eqnarray}
where $N$ is a normalization constant which its value is not
important here.  Moreover, we consider that $\psi_A$ and $\psi_B$
are the Gaussian wave function introduced in Eq.~(\ref{psiabt2}).
 Also note that, the upper and lower signs in the total entangled
wave function (\ref{psientangled}) are due to symmetric and
anti-symmetric wave function under the exchange of particles 1 and
2, corresponding to bosonic and fermionic property, respectively.


\section{Disentangled wave function}


In this case, the incident plane wave can be considered to be
\begin{equation}
\tilde{\psi}_{in}(x_{1},y_{1};x_{2},y_{2};t)=ae^{i[k_{x}(x_{1}+x_{2})+k_{y}(y_{1}+y_{2})]}e^{-iEt/\hbar}
\end{equation}
where $a$ is a constant. Now, for such a two-particle system, the
total wave function after passing the two particles through the
slits at time $t$ can be written as
\begin{eqnarray}\label{psifactor}
\lefteqn{\tilde{\psi}(x_{1},y_{1};x_{2},y_{2};t) =}\cr&&
\widetilde{N}[\psi_{A}(x_{1},y_{1},t)\psi_{B}(x_{2},y_{2},t)+\psi_{A}(x_{2},y_{2},t)\psi_{B}(x_{1},y_{1},t)\cr&&
 +\psi_{A} (x_{1},y_{1},t)\psi_{A}(x_{2},y_{2},t)+\psi_{B}
(x_{1},y_{1},t)\psi_{B} (x_{2},y_{2},t)]\cr&&
=\widetilde{N}[\psi_{A}(x_{1},y_{1},t)+\psi_{B}(x_{1},y_{1},t)][\psi_{A}(x_{2},y_{2},t)+\psi_{B}(x_{2},y_{2},t)]
\end{eqnarray}
where $\widetilde{N}$ is another normalization constant and
$\psi_A$ as well as $\psi_B$ can be considered the Gaussian wave
function represented in Eq.~(\ref{psiabt2}).


\section{Standard quantum mechanics predictions}

Based on SQM, the wave function can be associated with an
individual physical system.  It provides the most complete
description of the system that is, in principle, possible.  The
nature of description is statistical, and concerns the
probabilities of the outcomes of all conceivable measurements that
may be performed on the system.  It is well known from SQM that,
the probability of simultaneous detection of the particles at
$y_{M}$ and $y_{N}$, on the screen $S_{2}$, located at
$x_{1}=x_{2}=D$ and $t=Dm/\hbar k_x$, is equal to
\begin{equation}\label{p12}
P_{12}(y_{M},y_{N},t)=\int_{y_{M}}^{y_{M}+\triangle}dy_{1}\int_{y_{N}}^{y_{N}+\triangle}dy_{2}|\psi(x_{1},y_{1};x_{2},y_{2};t)|^{2}.
\end{equation}
The parameter $\Delta$, which is taken to be small,  is a measure
of the size of the detectors.  We shall compare this prediction of
SQM with that of BQM.




\section{Bohmian predictions for the entangled case}


Consider the entangled wave function (\ref{psientangled}).  By
substituting it in the guidance condition (\ref{guidance}), one
can obtain
\begin{eqnarray}\label{v1v21}
\dot{y}_{1}=&N\frac{\hbar}{m}&Im\{
\frac{1}{\psi}[[-2(y_{1}-Y-\hbar
k_yt/m)/4\sigma_{0}\sigma_{t}+ik_{y}]\psi_{A_{1}}\psi_{B_{2}}\nonumber\\
           &\pm&[-2(y_{1}+Y+\hbar
           k_yt/m)/4\sigma_{0}\sigma_{t}-ik_{y}]\psi_{A_{2}}\psi_{B_{1}}]\}\nonumber\\
\dot{y}_{2}=&N\frac{\hbar}{m}&Im\{\frac{1}{\psi}[[-2(y_{2}+Y+\hbar
k_yt/m)/4\sigma_{0}\sigma_{t}-ik_{y}]\psi_{A_{1}}\psi_{B_{2}}\nonumber\\
           &\pm&[-2(y_{2}-Y-\hbar
           k_yt/m)/4\sigma_{0}\sigma_{t}+ik_{y}]\psi_{A_{2}}\psi_{B_{1}}]\}.
\end{eqnarray}
On the other hand, from Eq.~(\ref{psiabt2}) one can see that,
\begin{eqnarray}\label{psiapsib2}
&&\psi_{A}(x_{1},y_{1},t)=\psi_{B}(x_{1},-y_{1},t)\cr
&&\psi_{A}(x_{2},y_{2},t)=\psi_{B}(x_{2},-y_{2},t)
\end{eqnarray}
 which indicate the reflection symmetry  of $\psi(x_{1},y_{1};x_{2},y_{2};t)$ with
respect to the $x$-axis. Using this symmetry in Eq.~(\ref{v1v21}),
we have
\begin{eqnarray}\label{v1pmv1}
&&\dot{y}_{1}(x_{1},y_{1};x_{2},y_{2};t)=\mp\dot{y}_{1}(x_{1},-y_{1};x_{2},-y_{2};t)\cr&&
\dot{y}_{2}(x_{1},y_{1};x_{2},y_{2};t)=\mp\dot{y}_{2}(x_{1},-y_{1};x_{2},-y_{2};t).
\end{eqnarray}
These relations show that if $y_{1}(t)=y_{2}(t)=0$, i.e., two
particles are on the $x$-axis, simultaneously, then the speed of
each bosonic particles in the $y$-direction is zero along the
symmetry axis $x$, but we have no such  constraint on fermionic
particles, as was also mentioned by Ghose {\cite{Ghose1}}.
However, in the previous chapter, we have shown that, there is
such a constraint on both bosonic and fermionic particles, using
the two entangled particles in a two double-slit device.
\begin{figure}[t]
\begin{center}\vspace{20mm}
\includegraphics[width=12cm,height=5cm,angle=0]{Figure1.EPS}
\caption{A two-slit scheme in which two identical entangled
particles are emitted from the source $S_{1}$.  Then, they pass
through the slits $A$ and $B$, and finally they are detected on
the screen $S_{2}$, simultaneously. We have assumed that,
$y(0)=0$, or $\langle y(0)\rangle=0$ under $\triangle
y(0)\ll\sigma_{0}$ and $\hbar t/2m\sigma_{0}^{2}\sim 1$
conditions. It is clear that dashed lines are not real
trajectories.}\label{fig21}
\end{center}
\end{figure}



If we consider $y=(y_{1}+y_{2})/2$ to be the vertical coordinate
of the center of mass of the two particles, then we can write
\begin{eqnarray}
\dot{y}&=&(\dot{y}_{1}+\dot{y}_{2})/2\cr
       &=&N\frac{\hbar}{2m}Im\{\frac{1}{\psi}(-\frac{y_{1}+y_{2}}{2\sigma_{0}\sigma_{t}})(\psi_{A_{1}}\psi_{B_{2}}\pm\psi_{A_{2}}\psi_{B_{1}})\}\cr
       &=&\frac{(\hbar/2m\sigma_{0}^{2})^{2}}{1+(\hbar/2m\sigma_{0}^{2})^{2}t^{2}}yt.
\end{eqnarray}
Solving this differential equation, we get the path of the
$y$-coordinate of the center of mass
\begin{equation}\label{y(t)21}
y(t)=y(0)\sqrt{1+(\hbar/2m\sigma_{0}^{2})^{2}t^{2}}.
\end{equation}
Using Eq.~(\ref{y(t)21}) and doing the same as what was done in
chapter 1, one obtains the quantum potential for the center of
mass motion
\begin{equation}
Q_{cm}=\frac{my^{4}(0)}{2y^{2}}(\frac{\hbar}{2m\sigma_{0}^{2}})^{2}=\frac{1}{2}my^{2}(0)\frac{(\hbar/2m\sigma_{0}^{2})^{2}}{1+(\hbar
t/2m\sigma_{0}^{2})^{2}}.
\end{equation}
If the center of mass of the two particles is exactly on the
$x$-axis at $t=0$ , then $y(0)=0$, and the center of mass of the
particles will always remain on the $x$-axis.  In addition, the
quantum potential for the center of mass of the two particles is
zero at all times. Thus, we have $y_{1}(t)=-y_{2}(t)$ and the two
particles, in both the bosonic and fermionic case,  will be
detected at points symmetric with respect to the $x$-axis, as is
shown in Fig.~\ref{fig21}. This differs from the prediction of
SQM, as the probability relation (\ref{p12}) shows. SQM predicts
that the probability of asymmetrical detection of the pair of
particles can be different from zero in contrast to BQM's
symmetrical prediction. Furthermore, according to SQM's
prediction, the probability of finding two particles at one side
of the $x$-axis can be non-zero while it is shown that BQM forbids
such events, provided that $y(0)=0$. Figure \ref{fig21} shows one
of the typical inconsistencies which can be considered at the
individual level. Based on BQM, bosonic and fermionic particles
have the same results, but, we know that if one bosonic particle
passes through the upper (lower) slit, it must detected on the
upper (lower) side on the $S_{2}$ screen, due to relations
(\ref{v1pmv1}). Instead, there is no such  constraint on fermionic
particles.  If $y(0)\neq 0$ and we have a distributed source with
$\langle y(0)\rangle=0$, then for the conditions $\hbar
t/2m\sigma_{0}^{2}\sim 1$ and $Y\sim\sigma_0$, it is easy to show
that the constraint $\triangle y(0)\ll\sigma_0$ still yields
reasonable symmetrical detection around the $x$-axis on the
screen.  But, now, in addition to the fermionic particles, the
bosonic ones can cross the symmetry axis.




\section{Bohmian predictions for the disentangled case}


By considering the disentangled wave function (\ref{psifactor})
and the guidance condition (\ref{guidance}), Bohmian velocities of
particle 1 and 2 can be obtained as
\begin{eqnarray}\label{v1v22}
\dot{y}_{1}=&\widetilde{N}&\frac{\hbar}{m}Im\{\frac{1}{(\psi_{A_{1}}+\psi_{B_{1}})}([\frac{-2(y_{1}-Y-\hbar
k_yt/m)}{4\sigma_{0}\sigma_{t}}+ik_{y}]\psi_{A_{1}}\nonumber\\
           &+&[\frac{-2(y_{1}+Y+\hbar
           k_yt/m)}{4\sigma_{0}\sigma_{t}}-ik_{y}]\psi_{B_{1}}\}\nonumber\\
           \dot{y}_{2}=&\widetilde{N}&\frac{\hbar}{m}Im\{\frac{1}{(\psi_{A_{2}}+\psi_{B_{2}})}([\frac{-2(y_{2}-Y-\hbar
k_yt/m)}{4\sigma_{0}\sigma_{t}}+ik_{y}]\psi_{A_{2}}\nonumber\\
           &+&[\frac{-2(y_{2}+Y+\hbar
           k_yt/m)}{4\sigma_{0}\sigma_{t}}-ik_{y}]\psi_{B_{2}}\}.
\end{eqnarray}
Thus, as expected, the speed of each particle is independent of
the other.  Using relations (\ref{v1v22}) as well as
Eq.~(\ref{psiapsib2}), we obtain
\begin{eqnarray}\label{v1=v22}
&&\dot{y}_{1}(x_{1},y_{1},t)=-\dot{y}_{1}(x_{1},-y_{1},t)\cr&&
\dot{y}_{2}(x_{2},y_{2},t)=-\dot{y}_{2}(x_{2},-y_{2},t).
\end{eqnarray}
This implies that the $y$-component of the velocity of each
particle would vanish on the $x$-axis.  Although these relations
are similar to the relations that were obtained for the entangled
wave function, but here we have an advantage: none of the
particles can cross the $x$-axis nor are tangent to it,
independent of the other particle's position.  This property can
be used to show that BQM's predictions are incompatible with
SQM's.

\begin{figure}[t]
\begin{center}\vspace{20mm}
\includegraphics[width=11cm,height=7.5cm,angle=0]{Figure2.EPS}
\caption{Schematic drawing of a two-slit device in which two
identical disentangled particles are simultaneously emitted by the
source $S_1$. The symmetrical detection is not predicted at the
central maximum, using both SQM and BQM.  But, using BQM, we can
have symmetrical detection at the other maxima (for example, at
$y_1$ as the first acceptable maximum) under the condition $Y\ll
2\pi\sigma_0$.}\label{fig22}
\end{center}
\end{figure}



  To see this
incompatibility, we use a special detection process on the screen
$S_{2}$ that we call it selective detection.  In this selective
detection, we register only those pair of particles which are
detected on the two sides of the $x$-axis, simultaneously.  That
is, we eliminate the cases of detecting only one particle or
detecting both particles of the pair on the upper or lower part of
the $x$-axis on the screen.  Again, it is useful to obtain the
equation of motion of the center of mass in the $y$-direction.
Using Eq.~(\ref{v1v22}), one can show that,
\begin{eqnarray}\label{ydot2}
\dot{y}=\frac{(\hbar/2m\sigma_{0}^{2})^{2}yt}{1+(\hbar/2m\sigma_{0}^{2})^{2}t^{2}}
+\widetilde{N}\frac{\hbar}{2m}Im\{\frac{1}{\psi}(\frac{Y+\hbar k_yt/m}{\sigma_{0}\sigma_{t}}+2ik_{y})(\psi_{A_{1}}\psi_{A_{2}}-\psi_{B_{1}}\psi_{B_{2}})\}.\nonumber\\
\end{eqnarray}
We can assume that the distance between the source and the
two-slit screen is so large that we have $k_y\simeq 0$.  Then,
using the special case $Y\ll\sigma_0$, the second term in
Eq.~(\ref{ydot2}) becomes negligible and the equation of motion
for the $y$-coordinate of the center of mass is reduced to
\begin{equation}
\dot{y}\simeq\frac{(\hbar/2m\sigma_{0}^{2})^{2}}{1+(\hbar/2m\sigma_{0}^{2})^{2}t^{2}}yt
\end{equation}
and similar to the entangled case, we have
\begin{equation}\label{y(t)22}
y(t)\simeq y(0)\sqrt{1+(\hbar/2m\sigma_{0}^{2})^{2}t^{2}}.
\end{equation}
Since for this special source there was not any entanglement
between the two particles, we must have $\triangle
y(0)\sim\sigma_0$, according to quantum equilibrium hypothesis
(QEH). Now, consider the case in which $\langle y(0)\rangle=0$. To
obtain symmetrical detection with reasonable approximation it is
enough to assume that the center of mass variation is smaller than
the distance between any two neighboring maxima on the screen.
Then, according to Eq.~(\ref{dely}), one can write
\begin{equation}\label{dely0}
\triangle y(0)\ll\frac{\pi\hbar t}{Ym}
\end{equation}
which yields
\begin{equation}
Y\ll 2\pi\sigma_0
\end{equation}
where in Eq.~(\ref{dely0}) we consider $\hbar
t/2m\sigma_{0}^{2}\sim 1$ so that $y(t)\sim y(0)$. Therefore,
under these conditions, BQM's symmetrical prediction is
incompatible with SQM's asymmetrical one. Figure \ref{fig22} shows
a schematic drawing of BQM's symmetrical detection occurred at a
maximum for the conditions $Y\ll 2\pi\sigma_0$ and $\hbar
t/2m\sigma_0^2\sim 1$.  It should be noted that, under these
conditions, the two wave packets are overlapped on the screen in
an interval of the order of $\sigma_0$.  In this interval neither
BQM nor SQM predict symmetrical detection around the $x$-axis.  In
fact, the symmetrical detection predicted by BQM happens far
(relative to $\sigma_0$) from the $x$-axis on the screen.  In
other words, save the central peak, which does not show symmetry
with respect to the $x$-axis, other less prominent maxima of the
diffraction pattern appear at the locations
\begin{eqnarray}
&&y_{n_{+}}\simeq n_+\frac{\pi\hbar t}{2m\sigma_0}\pm\triangle
y\cr&& y_{n_{-}}\simeq -n_-\frac{\pi\hbar
t}{2m\sigma_0}\pm\triangle y
\end{eqnarray}
where $y_{n_{\pm}}$ refer to $y$-component of the maxima above or
below the $x$-axis on the screen, respectively.  In addition,
$n_{\pm}$ represent positive integers.  BQM's symmetrical
prediction puts the following constraint:
\begin{equation}
n_+=n_-
\end{equation}
for great $n$'s. However, SQM's probabilistic prediction does not
require this constraint.

  Now, consider conditions in which $\langle y(0)\rangle\neq 0$, $\triangle
  y(0)\sim\sigma_0$, and $\hbar t/2m\sigma_{0}^{2}\gg 1$.  Then, the $x$-axis will not
be an axis of symmetry and we have a new region on the $S_{2}$
screen around which all pairs of particles will be detected
symmetrically.  Thus, using Eqs.~(\ref{v1=v22}) and (\ref{y(t)22})
as well as selective detection for the two particles, which
requires registration of those two particles that are detected at
the two sides of the $x$-axis simultaneously and omission of the
others, BQM can predict a rather empty interval with low intensity
of particles that has a length
\begin{equation}
L\simeq 2\langle y\rangle\simeq\frac{\hbar
t}{m\sigma_{0}^{2}}\langle y(0)\rangle
\end{equation}
if the constraint $\triangle y\ll L$ is satisfied.  The last
constraint at $\hbar t/2m\sigma_{0}^{2}\gg 1$ condition,
corresponds to $\triangle y(0)\ll\langle y(0)\rangle$.  Figure
\ref{fig23} shows that according to BQM and under the conditions
\begin{eqnarray}
&&\frac{\hbar t}{2m\sigma_{0}^{2}}\gg 1 \cr&&
Y\ll\sigma_0\ll\langle y(0)\rangle
\end{eqnarray}
a considerable position change in the $y$-coordinate of the source
produces a region with very low intensity on the screen which is
not predicted by SQM.
\begin{figure}[t]
\begin{center}\vspace{20mm}
\includegraphics[width=11cm,height=7.5cm,angle=0]{Figure3.EPS}
\caption{Schematic drawing of a two-disentangled particle applied
in a two-slit device in which the conditions
$Y\ll\sigma_0\ll\langle y(0)\rangle$ and $\hbar
t/2m\sigma_{0}^{2}\gg 1$ along with a special selective detection
are considered. The length $L$ shows the low intensity interval in
the final interference pattern.}\label{fig23}
\end{center}
\end{figure}
In fact, based on SQM, we have two alternatives:\\
{\it 1.} The joint probability relation (\ref{p12}) is still valid
and there is only
a reduction in the intensity throughout the screen $S_{2}$, due to the selective detection.\\
{\it 2.} SQM is silent about our selective detection.\\
In the first case, there is a disagreement between the predictions
of SQM and BQM and in the second case, BQM has a better predictive
power than SQM, even at the statistical level.


However, since the two particles in this scheme are emitted in a
disentangled state, the results obtained may seem unbelievable. In
this regard, Struyve {\cite{Stru}} based on our aforementioned
factorizable wave function (\ref{psifactor}), believes that the
two independent particles of this experiment cannot produce
different predictions for SQM and BQM.  He argues that
{\cite{Stru}}, the results of the experiment will not be altered
if we emit the two particles simultaneously or emit only one
particle at a time, because the two particles are totally
independent.   But, it should be noted that, we can use the
selective detection only when the source emits two identical
particles at the same time, and if the source emits just one
particle at each time, then it is meaningless to utilize our
selective detection.  In the following, once again,  we
substantiate our previous arguments about this thought experiment,
 in some details.

At first, we examine the applied condition $Y\ll\sigma_0$, in a
double-slit device with two disentangled particles.  One may argue
that this condition is meaningless, because based on the
specifications of the set-up, Y represents the distance between
the center of each slit to the $x$-axis, and therefore, the
minimum value of $Y$ approaches $\frac{1}{2}\epsilon +\sigma_0$,
where $\epsilon$ is considered to be very small and represents the
length of the plane that separates the two slits.  But, this
objection can be answered by considering the overlapping of each
particle's two Gaussian wave functions which are generated at the
two near slits. The overlap causes the peak of each Gaussian wave
to approach  more and more to the $x$-axis.  In addition, under
this condition, the Gaussian wave functions lose their symmetrical
form at each slit.  Our argument becomes clearer when we consider
$\epsilon=0$ as a limiting case, i.e, we have only one slit.  In
this limiting case, it is clear that $Y=0$.  Therefore, when the
two slits are very near together, the peak of Gaussian wave
functions, i.e. $Y$, come very near to the $x$-axis and the
condition $Y\ll\sigma_0$ is completely satisfied.

Another problem can arise when one thinks of the two independent
particles. To handle this problem, let us reconsider the second
term in Eq.~(\ref{ydot2}), particularly the coefficient
$(\psi_{A_1}\psi_{A_2}-\psi_{B_1}\psi_{B_2})$. Using the Gaussian
wave (\ref{psiabt2}) and the condition $k_y\simeq 0$, the latter
coefficient can be written in the form
\begin{eqnarray}
\psi_{A_1}\psi_{A_2}-\psi_{B_1}\psi_{B_2}&=&(2\pi\sigma_t)^{-1/4}e^{2i(k_xx-Et/\hbar)}e^{-(y_{1}^{2}+y_{2}^{2})/4\sigma_0\sigma_t}e^{-(Y+u_yt)^2/2\sigma_0\sigma_t}\nonumber\\
&&\times[e^{(y_1+y_2)(Y+u_yt)/2\sigma_0\sigma_t}-e^{-(y_1+y_2)(Y+u_yt)/2\sigma_0\sigma_t}]\nonumber\\
\end{eqnarray}
where $u_y=\hbar k_y/m$. If we require that $y_1(t)+y_2(t)=0$,
i.e., the $y$-component of the two particles are entangled, then
we would obtain the equation of motion (\ref{y(t)21}), as
expected. But our two particles in this scheme are initially
disentangled, and it is not necessary to have $y_1(t)+y_2(t)=0$.
Instead, we can have another choice on the geometry of the
two-slit set-up. In fact, if we apply the condition
$Y\ll\sigma_0$, again the behavior of the equation of motion of
the two particles in the $y$-direction is similar to the motion of
the two entangled particles, while the two particles were
disentangled.  Hence, we can state that the classical interaction
of the wave function of the two disentangled particles with the
two-slit plane barrier for the condition $Y\ll\sigma_0$, results
in a wave function which now guides the $y$-component of the
center of mass of the two apparently disentangled particles in the
same way as the case of two entangled particles with the initial
condition $-\sigma_0\leq(y_1+y_2)_{t=0}\leq\sigma_0$, for those
pairs of particles that pass through the two slits.  Thus, we have
shown that the results obtained in the two-slit device, using two
synchronized identical particles along with the selective
detection, are completely different from the ones obtained in a
single-particle double-slit experiment. In fact, it seems that the
motion of either particle is now dependent on its own location and
the location of the other particle, although the apparent form of
the wave function of the system can be efficiently represented by
the use of the disentangled form in (\ref{psifactor}).  Based on
QEH, the factorizable wave function (\ref{psifactor}) results that
both SQM and BQM must yield the same interference pattern for the
whole particle arrived on the screen. But, one can see that this
is not true for our specified conditions in which we have used a
selective detection. If we study the interference pattern without
using selective detection, we must obtain the same results for the
two theories. However, using selective detection along with the
guidance condition, it is clear that not only the two theories do
not have the same statistical predictions, but also BQM clarifies
and illuminates SQM, as D\"{u}rr {\it et al.} \cite{Durr}
said:{\em ``by selectively forgetting results we can dramatically
alter the statistics of those that we have not forgotten.  This is
a striking illustration of the way in which Bohmian mechanics does
not merely agree with the quantum formalism, but, eliminating
ambiguities, clarifies, and sharpens it."}.  In our selective
detection, we have forgotten detected single-particle and the
two-particle contributions on the one side of the $x$-axis (of the
screen $S_{2}$).



\section{Conclusions}


In this chapter, we have studied two nearly similar thought
experiments which can give different results for the standard and
Bohmian quantum mechanics in some particular conditions. The
suggested experiments consist of a two-slit interferometer with a
special source which emits two identical non-relativistic
particles, simultaneously.  We have shown that, according to the
characteristic of the source, our two-particle system can be
described by two kinds of wave functions: the entangled and the
disentangled wave functions. For the entangled case, we have
obtained some disagreement between SQM and BQM at the individual
level while the two theories predict the same statistical results,
as expected. For the disentangled case, the predictions of the two
theories can be different at the individual level, too. Again, the
results of the two theories were the same at the ensemble level.
However, the use of selective detection can dramatically alter the
interference pattern, so that not only the statistical results of
BQM do not agree with those of SQM, but BQM may also increase our
predictive power. Therefore, our suggested thought experiments
provide examples for which the standard and Bohmian quantum
mechanics may yield different predictions.






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%___________________________________________________________________PartII__________________________________________________

\part{New proposed experiments involving quantum information theory}
\newpage
~~~ \thispagestyle{empty}
%\newpage
\phantom{~~~}

\chapter{Introduction-Quantum Information Theory}
\section{Introduction}
In 1982 Feynman \cite{Feynman} observed  that certain quantum
mechanical effects cannot be simulated efficiently on a classical
computer. This observation led to speculation that perhaps
computation in general can be done more efficiently if it uses
these quantum effects. But building quantum computers proved
tricky, and as no one was sure how to use the quantum effects to
speed up computation, the field developed slowly. It was not until
1994, when Shor \cite{Shor94,Shor97} surprised the world by
describing a polynomial time quantum algorithm for factoring
integers, that the field of quantum computing came into its own.
This discovery prompted a flurry of activity, both among
experimentalists trying to build quantum computers and
theoreticians trying to find other quantum algorithms. Additional
interest in the subject has been created by the invention of
quantum dense coding \cite{BW92}, quantum teleportation
\cite{BB93} and quantum key distribution \cite{IEEE} as well as,
more recently, popular press \cite{BEZ,Galindo} accounts of
experimental successes in quantum teleportation and the
demonstration of a three-bit quantum computer.

The concept of entanglement is the distinctive and responsible
feature that allows quantum information to overcome some of the
limitations posed by classical information, as exemplified by the
new phenomena of teleportation, dense coding and key distribution
which are explained in the following sections.  In fact,
entanglement leads to profound experimental consequences like
non-local correlations: when two distantly apart parties Alice and
Bob share, say, an EPR pair, the measurement by Alice on her state
simultaneously determines the state on the Bob side. Apparently,
this implies instant information transmission, in sharp contrast
to Einstein's relativity. However, to reconcile both facts we must
notice that the only way Bob has to know about his state (without
measuring it) is by receiving a classical communication from
Alice, which does propagate no faster than the speed of light.



\section{Quantum dense coding}

 Classical information can also be sent through
quantum channels: to transmit the word 10011, it is enough that
Alice prepares 5 qubits in the states $|1\rangle$, $|0\rangle$,
$|0\rangle$, $|1\rangle$, $|1\rangle$, sends them to Bob through
the quantum channel, and Bob measures each of them in the basis
{$|0\rangle$, $|1\rangle$}. Each qubit carries a cbit, and this is
the most it can do in isolation. But, as theoretically proposed by
Bennett and Wiesner \cite{BW92}, if Alice and Bob share beforehand
an entangled state, then 2 cbits of information can be sent from
Alice to Bob with a single qubit.  As a matter of fact,
entanglement is a computing resource that allows more efficient
ways of coding information. Assume, for instance, an entangled
state of two photons. One of the photons goes to Alice, the other
one to Bob. She performs one of the following operations on the
polarization of her arriving photon: identity, flipping, change of
$\pi$ in the relative phase, and the product of the last two. Once
this is done, she sends back the photon to Bob, who measures in
which of the four Bell states the photon pair is. Then, in this
fashion we have been able to send 2 bits of information over one
single particle with only 2 states, that is, by means of a qubit.
It doubles what can be accomplished classically which results the
name of quantum dense coding or super dense coding. Moreover, if
Eve, as an eavesdropper, intercepts the qubit, she cannot get from
it alone any information.  Because, all the information lies in
the entangled state, and Bob possesses half of the pair. Actually,
Alice has sent Bob 2 qubits, but the first one long ago, as part
of the initial entangled state. This fact has allowed them to
communicate more efficiently, resorting to the entangled state
they shared. Dense coding is kind of the inverse process to
teleportation. In the latter the communication of two cbits allows
us to reproduce a qubit state, while in the former the
communication of a qubit carries along two cbits of information.


The following is a review of the dense coding protocol which is
explained in \cite{Polak}.  Consider an EPR source which supplies
Alice and Bob with a two-particle state like
\begin{eqnarray}
|\psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)
\end{eqnarray}
one of whose particles goes to Alice and the other one to Bob, who
keep them. Alice is supplied with 2 cbits, which represent the
numbers 0, 1, 2, 3 as 00, 01, 10, 11.\\
{\it Step 1.} Coding: According to the value of that number, Alice
effects on her EPR half the unitary operation I,X,Z,Y, which
brings the EPR state to 00+11, 10+01, 00-11, 10-01. Once this is
done, she sends her half to Bob.\\
{\it Step 2.} Decoding: Upon reception, Bob effects on the EPR
pair first a CNOT operation, such that the state becomes 00+10,
11+01, 00-10, 11-01. He then measures the second qubit; if Bob
finds 0, he already knows that the message was 0 or 2, and if he
finds 1, the message was 1 or 3. That is, he has gotten the second
bit of the two-bit message. In order to know the first one, Bob
next applies a Hadamard transformation on the first qubit, thereby
the state becomes 00, 01, 10, -11, and after measuring the first
bit, if he finds 0, he knows that the message was 0 or 1, and if
he finds 1, the message was 2 or 3, that is, he has just gotten
the second bit of the message.\\ An experiment of this nature has
been performed in Innsbruck \cite{Mattle}, by using the
polarized-entangled photons of type II parametric down conversion
that a non-linear crystal of $\beta$-barium borate produces: UV
photons get disintegrated (though with low probability) in a pair
of softer photons, with polarizations which in a certain geometric
configuration are entangled. In that experiment, they managed to
send ${\rm log}_2 3 = 1.58$ cbits per qubit.  In a recent
experiment \cite{Fang}, in which the qubits are the spins of
$^{1}$H and $^{13}$C in a clorophorm molecule $^{13}$CHCl$_{3}$
marked with $^{13}$C, and RMN techniques are employed to
initialize, manipulate and read out the spins, the authors claim
to have reached the 2 cbits per qubit.

 The initial preparation of
the entangled pair and the posterior transmission of the
information qubit may have opposite senses; for example, Bob sends
to Alice one half of the entangled state, keeping the other half
for himself, and then Alice uses her qubit to send to Bob the
desired information. This may be of interest if the cost in the
transmission in one way is higher than in the reverse way.
 On the other hand, intercepting the message from Alice to
Bob does not provide any information to Eve, because the message
is entangled with the part of the EPR pair possessed by Bob.
Therefore it is automatically a secure emission of information
(except if Eve intercepts both the original pair and the message
and she replaces them).


\section{Quantum teleportation}



Copying classical states  has never posed unsurmountable
difficulties to experts. It suffices to thoroughfully observe the
original as much as it may be required, avoiding to damage it, to
retrieve the information needed to make a copy of it.  This
careful observation does not alter, in a noticeable way, its
state. But if the original to be reproduced is a quantum system in
an unknown state $\psi$, then any measurement (incompatible with
$P_\psi$) made on the system to get information on $\psi$ will
disturb, uncontrollably, the state destroying the original.
Moreover, even in the case of having an unlimited number of copies
of that state, infinitely many measurements will be necessary to
determine that unknown state.

For example, assume that Alice has one spin-$\frac{1}{2}$)
particle as a qubit in a pure state. Bob needs it, but Alice does
not have any quantum channel to transmit it to him.  If Alice
knows the precise state of her qubit (for example, if she knows
that her spin-$\frac{1}{2}$ is oriented in the direction ${\bf
n}$), it is enough for her to give Bob in a letter (classical
channel) that information (the components of ${\bf n}$) to enable
him preparing a qubit exactly equal to Alice's.  But if she
happens not to know the state, she may choose to confess it to
Bob, who would then be inevitably driven to prepare his qubit in a
random way, obtaining a 50\% fidelity on average.  But Alice can
also try to be more cooperative, making for example a measurement
on her qubit of ${\bf n}'\cdot\bf \sigma$, with ${\bf n}'$
arbitrarily chosen, and then transmitting to Bob both the
components of ${\bf n}'$ and the result $\epsilon=\pm 1$ thus
obtained.  Armed with this information, Bob can prepare his qubit
in the state
\begin{eqnarray}
\frac{1}{2}(I+\epsilon{\bf n}'\cdot\bf \sigma).
\end{eqnarray}
 The average fidelity so obtained is larger than before: 2/3.
However, it is not still enough.


If Alice and Bob share an EPR pair, there exists a protocol known
as quantum teleportation, devised by Bennett {\it et al.}
\cite{BB93} in 1993, which resorts to the quantum entanglement of
states, and the non-locality of quantum mechanics allows Bob to
reproduce Alice's unknown quantum state with the assistance of
only 2 cbits of information sent by Alice to Bob through a
classical channel. This procedure necessarily destroys Alice's
state (otherwise it would violate the quantum no-cloning theorem).
 The following is a description of the teleportation scheme as
 explained in \cite{Polak}.

Assume that Alice likes to teleport a qubit with the following
quantum state
\begin{eqnarray}
\ket{\phi}=\alpha\ket{0}+\beta\ket{1}
\end{eqnarray}
in which
\begin{eqnarray}
\alpha=\cos\frac{1}{2}\theta,
\beta=e^{i\phi}{\sin}\frac{1}{2}\theta.
\end{eqnarray}
 In addition, consider
\begin{eqnarray}
\ket{\psi}=\frac{1}{\sqrt{2}}(\ket{00}+\ket{11})
\end{eqnarray}
 be the EPR state
shared between Alice and Bob, with Alice having the first of its
qubits, and Bob the second.  The initial state is thus
$\ket{\phi}\otimes\ket{\psi}$, of which Alice can locally
manipulate its two first bits and Bob the third one.

\noindent {\em Step 1.} Alice applies to the initial state the
unitary operator
\begin{eqnarray}
U=(({\rm H}\otimes I){\rm CNOT})\otimes I
\end{eqnarray}
acting with the CNOT gate on the first two qubits and next with
the Hadamard unitary gate H on the first one.  The resulting state
is
\begin{eqnarray}
\frac{1}{2}(\ket{00}\otimes\ket{\phi}+\ket{01}\otimes X\ket{\phi}+
\ket{10}\otimes Z\ket{\phi}+\ket{11}\otimes Y\ket{\phi}).
\label{qi10}
\end{eqnarray}


\noindent {\em Step 2.} Alice then measures the first two qubits,
obtaining $\ket{00}, \ket{01}, \ket{10}$,  or $\ket{11}$
equiprobably.\footnote{Steps 1+2 amount to performing a Bell
measurement on the initial state, thus correlating the Bell states
$00\pm 11,01\pm 10$ of Alice's two qubits with the states of Bob's
qubit.  It suffices to note that
\begin{eqnarray*}
&\ket{\phi}\ket{\psi}=\frac{1}{\sqrt{2}}
\ket{\phi}(\ket{00}+\ket{11})=\frac{1}{2\sqrt{2}}
((\ket{00}+\ket{11})\ket{\phi}+\\ &(\ket{01}+\ket{10})X\ket{\phi}+
(\ket{00}-\ket{11})Z\ket{\phi}+ (\ket{01}-\ket{10})Y\ket{\phi}).
\end{eqnarray*}}

Alice lets Bob know the result thus obtained, sending him two
cbits: the pair of binary digits 00, 01, 10, 11 that characterizes
it.  As a byproduct of Alice's measurement, the first bit ceases
to be in the original state $\ket{\phi}$, while the third qubit
gets projected onto
$\ket{\phi},X\ket{\phi},Z\ket{\phi},Y\ket{\phi}$, respectively.


\noindent {\em Step 3.} Once Bob receives the classical
information sent by Alice, he just needs to apply on his qubit the
corresponding gate I, $X, Z, Y$, in order to drive it to the
desired state $\ket{\phi}$.

Notice that this teleportation sends an unknown quantum state from
one place (where it vanishes) to another place (where it shows up)
without really traversing the intermediate space.  It does not
violates causality, though.  In the first part of the process,
quantum correlations get established between the Bell states
obtained by Alice and the associated states of Bob's qubit.  In
the remaining part to conclude the teleportation, information is
transmitted by classical means, in the standard non-superluminal
fashion.  Notice also that in this ``noncorporeal'' process, it is
the information about the quantum state, the qubit, and not the
physical state itself, what gets passed from Alice to Bob.  There
has been no transportation whatsoever of matter, energy or
information at a speed larger than the speed of light.

It is nevertheless surprising in the quantum teleportation that
all the information needed to reproduce the state
$\ket{\phi}=(\cos\frac{1}{2}\theta)\ket{0}+e^{i\phi}
({\sin}\frac{1}{2}\theta)\ket{1}$ (information that is infinite
for it requires to fix a point $(\theta,\phi)$ on the Bloch sphere
with infinite precision, thus requiring infinitely many qubits),
can be accomplished with only 2 cbits, provided an EPR state is
shared. This state, by itself, only generates potentially an
infinite number of random and correlated bit pairs.




Quantum teleportation was realized experimentally with photons for
the first time in two laboratories \cite{Bouwmeester97,Boschi}.
This is at least what these authors claim, although several
criticisms have been raised [48-50] (see however
\cite{Bouwmeester98,Bouwmeester99}). In the experiment by the Roma
group \cite{Boschi}, the initial state to be teleported from Alice
to Bob was a photon polarization, but not an arbitrary one,
because it coincided with that of the Alice's photon in the shared
EPR photon pair.  In the experiment by the Innsbruck group
\cite{Bouwmeester97}, however, the teleported state was arbitrary.
Teleportation was reached with a high fidelity of $0.80\pm
0.05$,\footnote{This fidelity overcomes the value $\frac{2}{3}$
corresponding to the case in which Alice measures her qubit and
communicates the result to Bob classically.} but with a reduced
efficiency (a 25\% of cases).  Teleportation has also been
realized for states which are parts of entangled states
\cite{Pan}.


It is also worthwhile to mention that quantum teleportation of
states of infinite dimensional systems, namely, the teleportation
of coherent optical states leaning on pairs of EPR squeezed
states, which is theoretically proposed by Braunstein ana Kimble
\cite{BraunsteinKimble}, is realized experimentally by Furasawa
{\it et al.} \cite{Furasawa}. In this experiment, whose fidelity
is $0.58\pm 0.02$ (higher than the maximum $\frac{1}{2}$ expected
without resorting to entanglement), a third party, the {\em
verifier} Victor, supplies Alice with one state that is known to
him, but not to her.  After teleporting that state from Alice to
Bob, Victor verifies on the output if Bob's state is similar to
the one he provided to Alice. In this sense, this experiment is
different from all the others, and led the authors to claim
priority in the realization of teleportation.


It does not seem to be easy to implement the theoretical protocol
with a 100\% effectiveness.  The Bell operator (which
distinguishes among the four Bell states of 2 qubits) cannot be
measured unless both qubits interact appreciably with each other
(as it occurs with the CNOT gate used in the protocol explained
above), something which is very hard to achieve with photons.
However, with atoms in EM cavities the hopes are high.


Perhaps the most realistic application of quantum teleportation
outside of pure physics research is in the field of quantum
computation.  In fact, quantum teleportation, which doubtlessly
will be extended to entangled states from different kinds of
systems (photons and atoms, ions and phonons, etc.), might have,
in the future, remarkable applications for quantum computers and
in computer networks (for example, combined with prior
distillation of good EPR pairs), as well as in the production of
quantum memory records by means of teleportation of information on
systems such as photons to other systems as trapped and
well-isolated ions in cavities \cite{Bouwmeester97,Bennett95}.  A
quantum computer can work on superposition of many different
inputs at once.  For instance, it can run at algorithm
simultaneously on one million inputs, using only as many qubits as
a conventional computer would need bits to run the algorithm once
on a single input.  Theorists have proved that some algorithms
running on quantum computers can solve certain problems faster
(i.e., in fewer computational steps) than any known algorithm
running on a classical computer \cite{Shor94,Shor97,Grover}.  The
problems include, for example, factoring large numbers, which is
of great interest for breaking secret codes, and searching for
items in a database.  So far only the most basic elements of
quantum computers have been built: logic gates that can process
one or two qubits \cite{BarencoD,Monroe}. The realization of even
a small scale quantum computer is still far away.  A key problem
is transferring quantum data reliably between different logic
gates or processors, whether within a single quantum computer or
across quantum networks.  Quantum teleportation is one solution.
In fact, Gottesman and Chuang \cite{Gottesman} recently proved
that a general quantum computer can be built out of three basic
components: entangled particles, quantum teleporters, and gates
that operate on a single qubit at a time.  This result provides a
systematic way to construct two-qubit gates.  The trick of
building a two-qubit gate from a teleporter is to teleport two
qubits from the gate's input to its output, using carefully
modified entangled pairs.  The entangled pairs are modified in
just such a way that the gate's output receives the appropriately
processed qubits.  Performing quantum logic on two unknown qubits
is thus reduced to the tasks of preparing specific predefined
entangled states and teleporting. Meantime, the complete Bell
state measurement needed to teleport 100\% efficiency is itself a
type of two-qubit processing.


Teleportation of complicate objects can be also considered as an
interesting and dreamy subject which one can think about
\cite{Zeilinger}.  Can we really hope to teleport complicated
objects?  It seems that, concerning this, there are many severe
obstacles.  For instance, the object has to be in a pure quantum
state, and such states are very fragile.  Photons do not interact
with environment much, so our experiments can be done in the open
space, but experiments with atoms and larger objects must be done
in a isolated vacuum to avoid interactions with gas molecules and
environments.  In addition, the larger an object becomes, the
easier it is to disturb its quantum state-even by thermal
radiation from the walls of the apparatus.  This is why we do not
routinely see quantum effects in our everyday world.  Quantum
interference, an easier effect to produce than entanglement or
teleportation, has been demonstrated with buckyballs, sphere made
of 60 carbon atoms.  Such work will proceed to larger objects,
perhaps even small viruses, but we might not expect to continue
this for much larger objects.  Other problems is due to finding
proper procedures for teleporting all desired kinds of objects.
Foe example, it has not yet been proposed a realizable scheme to
make possible teleportation of the spatial wave function of an
object-even a particle.  Also, the Bell state measurement is an
obstacle.  In addition to some experimental difficulties relating
to this kind of measurement, what would it mean to do a Bell state
measurement of, e.g., a virus consisting of about $10^7$ atoms?
Clearly it is a serious problem to extract the $10^8$ bits of
information that such a measurement would generate.  Meantime, for
an object of just a few grams the numbers become apparently
impossible: $10^{24}$ bits of data.  Teleportation of more
complicated objects particularly a person with $10^{32}$ bits of
information can be considered as an exciting but perhaps a
fictitious subject.  Being in the same quantum state does not seem
necessary for being the same person.  It seems that we change our
quantum state all the time and remain the same people. Conversely,
identical twins or biological clones are not the same people,
because they have different memories.  So may be Heisenberg
uncertainty as well as quantum no-cloning theorem do not prohibit
us from replicating a person precisely enough for him to think he
was the  same as the original.  Therefore, it seems that the
teleportation of a person does not require quantum accuracy, and
provides a surprising procedure for long travel, while the time
for the person is frozen.


In the part two of this dissertation, we have studied quantum
dense coding in a spatial scheme resulting in more efficiency than
some other well-known protocols, and also we have proposed a
theoretical scheme which can provide position state teleportation
of an object.  To do this, once again we have considered an
entangled source between two parties which works very similar to
that is used in our two double-slit scheme.


%\newpage


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%----------------------------------------------------------Ch.4-----------------------------------------------------------
\chapter{Quantum dense coding by spatial state entanglement}
\section{Introduction}

The quantum entanglement property is providing new methods of
information transfer, in some cases much more powerful than their
classical counterparts.  In quantum information theory,
entanglement, as a key concept is used for a wide range of
applications, such as quantum dense coding {\cite{BW92}},
teleportation {\cite{BB93}}, secret sharing {\cite{HBB99}} and key
distribution {\cite{IEEE}}. To find more about the mentioned
topics and the efforts done on them, both theoretically and
experimentally, one can refer to {\cite{BEZ,Cabello}} and
references therein.

Quantum dense coding protocol was proposed originally by Bennett
and Wiesner {\cite{BW92}} in 1992.  The protocol describes a way
to transmit two bits of classical information through manipulation
of only one of the entangled pair of spin-$\frac{1}{2}$ particles,
while each of the pair individually could carry only one bit of
classical information.  The first experimental realization of
dense coding has been reported by Mattle {\it et al.}
{\cite{Mattle}} in 1996.

There have been attempts to generalize dense coding protocol to
achieve higher channel capacity. The first proposition in this
regard was due to the original reference of dense coding
{\cite{BW92}} by using a pair of $n$-state particles prepared in a
completely entangled state (instead of an EPR spin pair in a
singlet state) to encode $n^{2}$ values.  In practice, there might
be some limitations for finding $n$-state particles with high
$n$'s and controlling them.  Very recently, some progress has been
made in this direction {\cite{Molina}}, but it is still worthy,
both theoretically and experimentally, to examine other
alternatives for achieving this goal. For example, Bose {\it{et
al.}} {\cite{Bose}} studied $\textsf{N}$ pairwise entangled states
in which each party gets one particle except Bob with
${\textsf{N}}$ qubits. Also in a more efficient scheme, they
considered ${\textsf{N}}+1$ parties sharing maximally entangled
qubits in a way that each party, including Bob, possesses one
qubit. Moreover, some recent attempts on generalization of quantum
dense coding can be found in [66-69]. In an elegant alternative,
Vaidman {\cite{Vaidman94}} proposed a method for utilizing
canonical continuous variables $x$ (position variable) and $p$
(linear momentum variable) to perform a quantum communication. In
this regard, Braunstein and Kimble {\cite{Braunstein2000}}
presented a typical realization for continuous variable dense
coding using the quadrature amplitudes of the electromagnetic
fields in which the mean photon number in each channel should be
considered very large.  Recently, in another way, some effort has
been also done on experimental realization of continuous variables
dense coding {\cite{Li}}.

Here, we have presented a theoretical extended version of dense
coding protocol using entangled position state of two particles
shared between two parties.  In this regard, all necessary Bell
states and their corresponding unitary operators are presented to
encode and decode information {\cite{Dense}}.  This version, at
large $N$'s, can be considered as a conceivable scheme for
Vaidman's idea {\cite{Vaidman94}} except that we have considered
just the position variable (not both canonical variables $x$ and
$p$) for communication.  Finally, the efficiency of our scheme is
compared with some other known ones.


\section{Description of the dense coding set-up}

Consider an original EPR source \cite{EPR} which emits
 a pair of identical (fermionic or bosonic) particles
with vanishing total linear momentum, isotropically.  Many
examples similar to this process can be found in different
branches of physics {\cite{Selleri}}. For instance, the molecule
NO can decay from an excited state to a state of two free atoms N
and O which propagate in opposite directions, and in another
example, the $\Lambda$ hyperon decays into a proton and a
negatively charged $\pi$ meson.
 In our scheme, the source $S$ is placed exactly in the middle of the two parties,
``Alice" and ``Bob", where each one has an array of receivers
aligned on a vertical line. The receivers just {\em{receive}} the
particles and do not perform any destructive measurement. Total
number of the receivers of each party is considered to be $2N$.
Since Alice and Bob are assumed to be very far from each other and
the source is considered isotropic, there is an equal probability
for every receiver to obtain one of the emitted particles.  We
label the receivers placed at the upper (lower) part of the
$x$-axis with positive (negative) integers; 1, 2, \ldots, $N$ (-1,
-2, \ldots, $-N$). Figure~{\ref{fig41}} shows an illustration of
this scheme. Now, the position state of the {\em system} (the
arrays + the source) can be written in the form
\begin{eqnarray}{\label{state12}}
|\psi_{1,2}\rangle=\frac{1}{\sqrt{2N}}\sum_{n=1}^{N}[|n,-n\rangle\pm|-n,n\rangle]
\end{eqnarray}
where the subscripts $1$ and $2$ are related to $\pm$ signs,
respectively, and $n$ refers to label of the receivers.  In
addition, the order in writing the state is according to the
direct product of Alice's state and Bob's.  The signs $\pm$ in
$|\psi_{1,2}\rangle$ indicate the symmetry and anti-symmetry
property of the position state with respect to the exchange of
particles.  In the following, without loss of generality, we
assume bosonic property for our system, that is, $|\psi_1\rangle$
represents the initial state of the system.

\begin{figure}[t]
\begin{center}\vspace{15mm}
\includegraphics[width=4.5cm,height=9cm,angle=-90]{fig41}
\caption{Schematic drawing of the proposed dense coding scheme. An
original EPR source is located between two array of receivers. The
$2N$ receivers of each party are connected to its corresponding
lab by quantum channels. The main quantum channels can be
connected together in each lab using side quantum channels.
Alice's lab is equipped with $O_{(k^{\pm},j)}$ encoder unitary
operators, and Bob's lab contains a decoder unitary operator like
${\cal H}$. The Bell state measurement is completed by just a
simple position measurement on Bob's outgoing channels.}
\label{fig41}
\end{center}
\end{figure}


\section{A representation for Bell states}

In general, the entangled state ({\ref{state12}}) can be
considered as a member of a larger family, that is,
\begin{eqnarray}\label{firstfamily}
&&|\psi_{(1,j)}\rangle={\frac{1}{\sqrt{2N}}} \sum_{n=1}^{N}[h_{j,2n-1}|n,-n\rangle+h_{j,2n}|-n,n\rangle]\nonumber\\
&&\hspace{30mm}1\leq j\leq 2N
\end{eqnarray}
in which
\begin{eqnarray}
h_{i,j}=\sqrt{2N}[{\bf{H}}]_{i,j}
\end{eqnarray}
where $\bf{H}$ is a $2N$-dimensional {\em{normalized symmetric}}
Hadamard matrix which satisfies the property ${\bf H}^{2}=I$.
Furthermore, the above entangled states can be generalized to a
more complete set of orthonormal and maximally entangled states,
which can be defined as
\begin{eqnarray}\label{Bell}
&&|\psi_{(k^{\pm},j)}\rangle={\frac{1}{\sqrt{2N}}}\sum_{n=1}^{N}[h_{j,2n-1}|n,f_{k^{\pm}}(n)\rangle+h_{j,2n}|-n,-f_{k^{\pm}}(n)\rangle]\nonumber\\
&&\hspace{30mm}1\leq k\leq N, 1\leq j\leq 2N
\end{eqnarray}
where $k^{\pm}$ and $j$ are called family and member indices,
respectively, and
\begin{eqnarray}
f_{k^{\pm}}(n)=\pm(n+k-1)_{{\rm mod}(N)}.
\end{eqnarray}
If one uses the convention
\begin{eqnarray}
(k^{\pm},j)\longrightarrow 2N(k^{\pm}-1)+j
\end{eqnarray}
in which
\begin{eqnarray}
&k^{+}\equiv 2k\nonumber\\
&k^{-}\equiv 2k-1
\end{eqnarray}
then the mentioned subscript $(k^{\pm},j)$ will find a natural
form, as we have shown in appendix B.

Now, the states ({\ref{Bell}}) form {\em{Bell bases}} for our
dense coding scheme. In fact, it is straightforward to check that
these states have the orthonormal and maximally entanglement
properties, as we have shown in the following
\begin{eqnarray}{\label{orthonormality}}
\langle\psi_{(k^{\pm},j)}|\psi_{(k'^{\pm},j')}\rangle
&=&\frac{1}{2N}\sum_{n,m=1}^{N}[h_{j,2n-1}h_{j',2m-1}\langle n,f_{k\pm}(n)|m,f_{k'\pm}(m)\rangle\nonumber\\
&&+h_{j,2n-1}h_{j',2m}\langle n,f_{k\pm}(n)|-m,-f_{k'\pm}(m)\rangle\nonumber\\
&&+h_{j,2n}h_{j',2m-1}\langle -n,-f_{k\pm}(n)|m,f_{k'\pm}(m)\rangle\nonumber\\
&&+h_{j,2n}h_{j',2m}\langle -n,-f_{k\pm}(n)|-m,-f_{k'\pm}(m)\rangle]\nonumber\\
&=&\frac{1}{2N}\sum_{n=1}^{N}(h_{j,2n-1}h_{j',2n-1}+h_{j,2n}h_{j',2n})\delta_{k,k'}\nonumber\\
&=&\frac{1}{2N}\sum_{n=1}^{N}(h_{j,n}h_{j',n})\delta_{k,k'}\nonumber\\
&=&\delta_{j,j'}\delta_{k,k'}
\end{eqnarray}
which shows the orthonormal property of our Bell bases.  Meantime,
for the maximally entanglement property we can write
\begin{eqnarray}
&&{\rm{tr}}_2(|\psi_{(k^{\pm},j)}\rangle\langle\psi_{(k^{\pm},j)}|)\nonumber\\&=&
{\rm tr}_2(\sum_{n,m=1}^{N}\frac{1}{2N}[h_{j,2n-1}h_{j,2m-1}|n,f_{k^{\pm}}(n)\rangle\langle m,f_{k^{\pm}}(m)|\nonumber\\
&&+h_{j,2n-1}h_{j,2m}|n,f_{k^{\pm}}(n)\rangle\langle -m,-f_{k^{\pm}}(m)|\nonumber\\
&&+h_{j,2n}h_{j,2m-1}|-n,-f_{k^{\pm}}(n)\rangle\langle m,f_{k^{\pm}}(m)|\nonumber\\
&&+h_{j,2n}h_{j,2m}|-n,-f_{k^{\pm}}(n)\rangle\langle -m,-f_{k^{\pm}}(m)|])\nonumber\\
&=&\frac{1}{2N}\sum_{n,m=1}^{N}[h_{j,2n-1}h_{j,2m-1}\delta_{n,m}|n\rangle\langle
m|+h_{j,2n-1}h_{j,2m}\delta_{n,-m-2k+2}|n\rangle\langle -m|\nonumber\\
&&+h_{j,2n}h_{j,2m-1}\delta_{n,-m-2k+2}|-n\rangle\langle m|+h_{j,2n}h_{j,2m}\delta_{n,m}|-n\rangle\langle -m|]\nonumber\\
&=&\frac{1}{2N}\sum_{n,m=1}^{N}[h_{j,2n-1}h_{j,2m-1}|n\rangle\langle n|+h_{j,2n}h_{j,2n}|-n\rangle\langle -n|]\nonumber\\
&=&\frac{1}{2N}\sum_{n=-N(n\neq 0)}^{N}|n\rangle\langle
n|=\frac{1}{2N}I_2.
\end{eqnarray}

Here it should be noted that, the order of a typical real Hadamard
matrix can be 1, 2 and $4k$ where $k$ is a positive integer
{\cite{Wallis,Agaian}}.  Physically, it means that in other cases,
i.e. for odd and half odd $N$ cases, one cannot make enough
necessary entangled orthonormal states to perform an efficient
dense coding. However, it is not a serious limitation in our
proposed scheme.


\section{Alice's encoding process}


For the allowed $N$'s, Alice can find unitary encoding operators
which transform the state describing the system in
Eq.~(\ref{state12}), $|\psi_1\rangle$, as a Bell state with $j=1$
in the first family into the others given in Eq.~(\ref{Bell}).
Then, the representation of $4N^2$ suitable unitary operators for
this task can be considered, for example, as
\begin{eqnarray}{\label{O}}
&&O_{(k^{\pm},j)}=\sum_{n=1}^{N}[h_{j,2n-1}|n\rangle\langle
f_{k^{\pm}}(n)|+h_{j,2n}|-n\rangle\langle -f_{k^{\pm}}(n)|]
\end{eqnarray}
which operate on Bell bases in Alice's lab as follows
\begin{eqnarray}\label{opsi}
O_{(k^{\pm},j)}|\psi_{(k'^{\pm},j')}\rangle=|\psi_{(k''^{\pm},j'')}\rangle
\end{eqnarray}
in which
\begin{eqnarray}
h_{j'',i}=h_{j,i}h_{j',i}{\label{h}}\\
\end{eqnarray}
when $j'=1$, and
\begin{eqnarray}
k''^{\pm}=(k+k'-1)^{ss'}_{{\rm mod}(N)} {\label{k}}
\end{eqnarray}
where $s$ and $s'$ are signs of superscripts of $k$ and $k'$,
respectively.  After the completion of Alice's encoding process,
she sends her particle to Bob in a parallel line to the $x$-axis.

It is useful to see the act of $O_{(k^{\pm},j)}$ operator on the
Bell states in details.  Thus, we have
\begin{eqnarray}
O_{(k^{\pm},j)}|\psi_{(k'^{\pm},j')}\rangle &=&
\frac{1}{\sqrt{2N}}\sum_{n=1}^{N}\sum_{m=1}^{N}[h_{j,2n-1}h_{j',2m-1}\langle
f_{k^{\pm}}(n)|m\rangle|n,f_{k'^{\pm}}(m)\rangle\nonumber\\
&&+h_{j,2n-1}h_{j',2m}\langle f_{k^{\pm}}(n)|-m\rangle|n,-f_{k'^{\pm}}(m)\rangle\nonumber\\
&&+h_{j,2n}h_{j',2m-1}\langle -f_{k^{\pm}}(n)|m\rangle|-n,f_{k'^{\pm}}(m)\rangle\nonumber\\
&&+h_{j,2n}h_{j',2m}\langle
-f_{k^{\pm}}(n)|-m\rangle|-n,-f_{k'^{\pm}}(m)\rangle]
\end{eqnarray}
which by using orthonormality condition of the states leads to
\begin{eqnarray}
O_{(k^{\pm},j)}|\psi_{(k'^{\pm},j')}\rangle &=&
\frac{1}{\sqrt{2N}}\sum_{n=1}^{N}\sum_{m=1}^{N}[h_{j,2n-1}h_{j',2m-1}\delta_{\pm(n+k-1),m}|n,f_{k'^{\pm}}(m)\rangle\nonumber\\
&&+h_{j,2n-1}h_{j',2m}\delta_{\pm(n+k-1),-m}|n,-f_{k'^{\pm}}(m)\rangle\nonumber\\
&&+h_{j,2n}h_{j',2m-1}\delta_{\mp(n+k-1),m}|-n,f_{k'^{\pm}}(m)\rangle\nonumber\\
&&+h_{j,2n}h_{j',2m}\delta_{\mp(n+k-1),-m}|-n,-f_{k'^{\pm}}(m)\rangle].
\end{eqnarray}
Now, there are two $k^+$ and $k^-$ cases which should be
considered.  For the $k^+$ case, we obtain
\begin{eqnarray}
&&O_{(k^{+},j)}|\psi_{(k'^{\pm},j')}\rangle=\nonumber\\
&&h_{j,2n-1}h_{j',2m-1}|n,f_{k''^{\pm}}(n)\rangle+h_{j,2n}h_{j',2m}|-n,-f_{k''^{\pm}}(n)\rangle
\end{eqnarray}
in which $m=n+k-1$ and $k''\equiv k+k'-1$.  In a similar way, for
the $k^-$ case we have
\begin{eqnarray}
&&O_{(k^{-},j)}|\psi_{(k'^{\pm},j')}\rangle=\nonumber\\
&&h_{j,2n-1}h_{j',2m}|n,f_{k''^{\mp}}(n)\rangle+h_{j,2n}h_{j',2m-1}|-n,-f_{k''^{\mp}}(n)\rangle
\end{eqnarray}
where again $m=n+k-1$ and Eq.~(\ref{k}) is satisfied.  Since
Alice's encoding process is done on the initial state of the
system with $j'=1$, and also in a normalized symmetric Hadamard
matrix $h_{1,i}=+1$, therefore, the act of the $O_{(k^{\pm},j)}$
on the $|\psi_{(k'^{\pm},j')}\rangle$ results in the Bell states
introduced in Eq.~(\ref{Bell}) with $h_{j'',i}=h_{j,i}$ as well as
Eq.~(\ref{k}).


\section{Introducing basic gates and their realizability}


It is a relevant question to ask about how to implement
$O_{(k^{\pm},j)}$ operators in practice.  So here, we propose one
way of performing this task by using some basic and conceivable
operators. For example, similar to Pauli's operators in the
spin-$\frac{1}{2}$ space, suitable basic unitary operators in the
position space can be considered to be
\begin{eqnarray}{\label{Nn}}
&&N_{n}|\pm n\rangle=\pm|\pm n\rangle \\
&&P_{n}|\pm n\rangle=|\mp n\rangle
\end{eqnarray}
where the subscript $n$ means that the operator acts as a local
gate on the $\pm n$-th channels.  The latter operator, $P_n$, is
defined to relate the two groups of receivers in the upper and
lower halves of the $x$-axis. We also need to define a ladder
operator $L_{+}$ to relate the receivers located on each half of
the $x$-axis, so that
\begin{eqnarray}{\label{L+}}
&&L_{+}|\pm n\rangle=|\pm(n+1)\rangle_{{\rm mod}(N)}.
\end{eqnarray}

Now, it is worthy to consider theoretical realization of the above
mentioned operators in this protocol.  The operation of $N_n$ gate
can be easily conceived, for instance, by a kind of phase shifter,
whose action can be represented as
\begin{eqnarray}\label{N_n}
N_n=e^{i\theta(\mp n)\pi}
\end{eqnarray}
where $\theta(n)$ is the conventional unit step function.  To
conceive this operator, one can consider a typical switch gate
acting on the $-n$-th channel and its side channel, as is shown in
Fig.~{\ref{fig42}}.  If $N_n$ turned ON (OFF), then the direct
(side) channel is closed and the side (direct) channel is opened.
It is clear that this set-up is required only for the channels
having negative label.
\begin{figure}[ht]
\begin{center}\vspace{36mm}
\includegraphics[width=6.5cm,height=1.9cm,angle=0]{fig42}
\caption{Quantum circuit diagram of $N_n$ gate in an OFF state.}
\label{fig42}
\end{center}
\end{figure}

On the other hand, $P_n$ operator can be considered as a kind of
swap gate between the channels $n$ and $-n$.  So, to design this
operation, one can use two local switch gates acting on these two
channels. If $P_n$ turned ON (OFF) it means that direct (side)
channels are closed and side (direct) channels are open (closed).
Figure~{\ref{fig43}} schematically shows a $P_n$ gate which is
constructed with two quantum switch gates and side channels.
\begin{figure}[ht]
\begin{center}\vspace{25mm}
\includegraphics[width=5cm,height=3cm,angle=0]{fig43.EPS}
\caption{Circuit diagram of $P_n$ gates in an OFF state.}
\label{fig43}
\end{center}
\end{figure}

In a similar manner, the action of the $L_+$ operations can be
conceived using $2N(N-1)$ numbers of the switch gates and side
quantum channels, as shown in Fig.~{\ref{fig44}}.
\begin{figure}[ht]
\begin{center}\vspace{15mm}
\includegraphics[width=9.5cm,height=5.5cm,angle=00]{fig44.EPS}
\caption{Schematic application of $L_+$ operators, by using
$2N(N-1)$ switch gates and side quantum channels, for $n>0$. If
$L^{k}_{+}$ is required, then the all $S_{k}$ gates must be turned
ON. } \label{fig44}
\end{center}
\end{figure}
Therefore, our basic operators, i.e. $N_{n}$, $P_{n}$ and $L_{+}$,
can be constructed using some quantum switch gates and side
channels.

As a proposal, if our introduced channels in the scheme are
considered, for instance, superconductor wires containing an
entangled current according to (\ref{state12}), then these basic
operators can be realized by using switching process in the
superconductor circuits. In fact, when a basic gate is OFF, its
main channel(s) is (are) superconducting and the side channel(s)
is (are) not superconducting, and conversely, if the gate is ON
then the main channel(s) is (are) not superconducting and the
suitable and corresponding side channel(s) is (are)
superconducting. However, full realization of this proposition is
left as an experimental challenge.

Now using the above basic gates, it is possible to show that one
can construct operators which transform any member of a family to
any other member of the other ones, that is, $O_{(k^{\pm},j)}$.
Here, our construction mechanism is to find a set of operators,
for instance $O_j$, to transform the $j$ label of the Bell states
of any given family. Next, we should introduce another set of
operators, $F_{k^{\pm}}$, to transform $k$ label of the Bell
states of any given member. Thus, our desired total operators have
the form
\begin{eqnarray}{\label{totalO}}
O_{(k^{\pm},j)}=F_{k^{\pm}}O_{j}.
\end{eqnarray}
The explicit form of the $O_j$ operators based on the basic gates
can be constructed using a normalized symmetric Hadamard matrix as
follows
\begin{eqnarray}\label{O_j}
&&O_j=\prod_{i=1}^{N}(P_{i}N_{i}P_{i})^{(h_{a,2i-1}-h_{j,2i})/2}N_{i}^{(h_{a,2i}-h_{j,2i})/2}\nonumber\\
&&\hspace{28mm}1 \leq j,a\leq 2N
\end{eqnarray}
where $a$ is an arbitrary fixed positive integer. One can easily
check that, the operator $O_j$ has the property
\begin{eqnarray}
O_{j}|\psi_{(k^{\pm},j')}\rangle=|\psi_{(k^{\pm},j'')}\rangle
\end{eqnarray}
where $h_{j^{''},i}$ obeys the rule in Eq.~(\ref{h}) for every
$j'$.  In fact, in a detailed form we can write
\begin{eqnarray}
&&O_{j}|\psi_{(k^{\pm},j')}\rangle=\frac{1}{\sqrt{2N}}\sum_{n=1}^{N}[h_{j',2n-1}(-1)^{(h_{a,2n-1}-h_{j,2n-1})/2}|n,f_{k^{\pm}}(n)\rangle\nonumber\\
&&\hskip3.5cm +h_{j',2n}(-1)^{(h_{a,2n}-h_{j,2n})/2}|-n,-f_{k^{\pm}}(n)\rangle]\nonumber\\
&&=\frac{1}{\sqrt{2N}}\sum_{n=1}^{N}[h_{j',2n-1}h_{j,2n-1}|n,f_{k^{\pm}}(n)\rangle+h_{j',2n}h_{j,2n}|-n,-f_{k^{\pm}}(n)\rangle]\nonumber\\
&&=\frac{1}{\sqrt{2N}}\sum_{n=1}^{N}[h_{j'',2n-1}|n,f_{k^{\pm}}(n)\rangle+h_{j'',2n}|-n,-f_{k^{\pm}}(n)\rangle].
\end{eqnarray}

Furthermore, the explicit form for $F_{k^{\pm}}$ can be considered
as
\begin{eqnarray}\label{F_k}
F_{k^{\pm}}=L_{+}^{(N-k+1)} \prod_{i=1}^{N}P_{i}^{(2k-k^{\pm}-1)}
\end{eqnarray}
in which $k^{+}\equiv 2k$ and $k^{-}\equiv 2k-1$. It can be easily
seen that they transform families to each other according to
\begin{eqnarray}
F_{k^{\pm}}|\psi_{(k'^{\pm},j')}\rangle=|\psi_{(k''^{\pm},j')}\rangle
\end{eqnarray}
where $k^{''\pm}$ satisfies the rule in Eq.~(\ref{k}).  Therefore,
the obtained explicit form for $O_{(k^{\pm},j)}$ operators is
equivalent to their representation in Eq.~(\ref{O}) and
(\ref{opsi}).  In appendix B, one can find examples concerning
explicit forms of $O_{(k^{\pm},j)}$ operators for some initial
cases.



\section{Bob's decoding process}

It was seen that after completing the encoding process, Alice
sends her particle to Bob.  Thus, to each of Bob's receivers, two
outgoing quantum channels should be considered for which we use
the labels ``control" and ``target" channel (in brief; C and
T-channel, respectively).  Now we assume that, at first Bob opens
all T-channels and closes all C-channels.  Thus, he receives his
own particle, which was devoted to him by the source, at one of
the T-channels.  Then, he sends it to a quantum storage media and
closes all T-channels as well as opens all C-channels and waits to
receive the processed particle which is sent to him by Alice.
After synchronizing the two particles, Bob is ready to perform a
Bell state measurement (BSM) on them. To do this, Bob needs to
apply a {\it grand} unitary and Hermitian operator on all the
channels in his lab which should be followed by a position state
measurement on his outgoing channels.
 The representation of the grand operator can be considered as
\begin{eqnarray}\label{grand}
{\cal{H}}=\sum_{j=1}^{2N}\sum_{k=1}^{N}|j,f'_{k^{\pm}}(j)\rangle\langle\psi
'_{(k^{\pm},j)}|
\end{eqnarray}
where
\begin{eqnarray}
f'_{k^{\pm}}(j)=\pm(j+k-1)_{mod(2N)}
\end{eqnarray}
and
\begin{eqnarray}
|\psi'_{(k^{\pm},j)}\rangle=\frac{1}{\sqrt{2N}}\sum_{n=1}^{2N}h_{j,n}|j,f'_{k^{\pm}}(j)\rangle
\end{eqnarray}
which is a compact form of Eq.~(\ref{Bell}), and can be
transformed to it by applying some local unitary operator.  It is
easy to check that the properties
\begin{eqnarray}
{\cal{H}}|\psi '_{(k^{\pm},j)}\rangle=|j,f'_{k^{\pm}}(j)\rangle
\end{eqnarray}
 and also ${\cal{H}}^2=I$ are satisfied.  Here, to make the former property
consistent with our channel labelling it suffices, just as an
example, to use the following convention: if in a ket like
$|i\rangle$ we have $1\leq |i|\leq N$, then we should change $i$
to $-i$, and in the case of $N+1\leq |i|\leq 2N$, to $|i|-N$.
Moreover, in the following, we have shown that the grand unitary
operator is also a reversible one.
\begin{eqnarray}
{\cal{H}}^{2}&&=\sum_{j,j'=1}^{2N}\sum_{k,k'=1}^{N}|j,f'_{k^{\pm}}(j)\rangle\langle\psi'_{(k^{\pm},j)}|j',f'_{k'^{\pm}}(j')\rangle\langle\psi'_{(k'^{\pm},j')}|\nonumber\\
&&=\sum_{j,j'=1}^{2N}\sum_{k,k'=1}^{N}\sum_{n,n'=1}^{2N}h_{j,n}h_{j',n'}|j,f'_{k^{\pm}}(j)\rangle\langle
n',f'_{k'^{\pm}}(n')|\delta_{j',n}\delta_{k,k'}\nonumber\\
&&=\sum_{j,k,n,n'}h_{j,n}h_{n,n'}|j,f'_{k^{\pm}}(j)\rangle\langle
n',f'_{k^{\pm}}(n')|\nonumber\\
&&=\sum_{j,k}|j,f'_{k^{\pm}}(j)\rangle\langle
j,f'_{k^{\pm}}(j)|=1.
\end{eqnarray}




It can be argued that any general quantum physical process is
required to operate by ``finite means", i.e., it is equipped only
with the possibility of applying any operation of some finite
fixed set of basic unitary operations {\cite{Deutsch}}. Also, it
is shown that various quite small collections of unitary operators
(so called ``universal sets" of operations) suffice to approximate
any unitary operation on any number of qubits to arbitrary
accuracy {\cite{Barenco, DeutschB}}.   Furthermore, in a recent
work, Bremner {\it et al.} {\cite{Bremner}} showed that any
unitary operation can be constructed using finite numbers of an
arbitrary conditional operator.  Therefore, the grand unitary
operator $\cal H$ is constructible using some finite basic
operators.  In the following, as an example, a realizable
theoretical scheme for the BSM will be presented, using our
introduced basic gates.


\section{A conceivable scheme for Bell state measurement}


In this section, we want to find an explicit form for $\cal{H}$
operator based on the introduced basic operators.  At first, Bob
needs one kind of logic gates to entangle two particles. The first
presentation and demonstration of these kinds of gates, called
CNOT, was performed in 1995 by Barenco {\it et al.}
{\cite{BarencoD}} and Monroe {\it et al.} {\cite{Monroe}},
respectively.  Here, we have considered another similar gate as a
non-local operator which acts conditionally on the $|l,m\rangle$
state as
\begin{eqnarray}\label{PCS}
{\rm PCS}|l,m\rangle=\theta(-l)(I\otimes
P_m)|l,m\rangle+\theta(l)|l,m\rangle
\end{eqnarray}
where PCS stands for position controlled swap operator and $\theta
(l)$ is the conventional unit step function.

To understand the action of this operator there is a way using
four typical spin-$\frac{1}{2}$ CNOT gates. Figure \ref{fig45}
shows a theoretical sketch of this construction. The
$\overline{\rm CNOT}$ is somehow a complement of the usual CNOT in
the sense that
\begin{eqnarray}
&{\rm CNOT}|x,y\rangle=|x,x+y}\rangle_{\rm mod(2)\nonumber\\
&\overline{\rm CNOT}|x,y\rangle=|x,\overline{x}+y}\rangle_{\rm
mod(2).
\end{eqnarray}
 If a spin-$\frac{1}{2}$ particle passes through the control channel
then the two first CNOT and $\overline{\rm CNOT}$ definitely
change spin of an intermediate spin-$\frac{1}{2}$ particle, named
spin target, which is preset to a fixed state , e.g., $|0\rangle$.
It is assumed that any change in the state describing the spin
target means switching the swap gate for the channels $m$ and
$-m$. The two latter CNOT and $\overline{\rm CNOT}$ are just to
reset the spin target.  To show this PCS gate in our quantum
circuits, we adopt a simplified representation for
 it, as depicted in Fig.~{\ref{fig46}}.  Our PCS gate can be realized using, for
example, superconductor circuits.  It is obvious that any change
in the spin target results into a change in the magnetic flux
passing through a superconductor circuit.  For switching the swap
gate the induced current can be amplified by a superconducting LC
circuit in order to produce at least a critical magnetic field to
set the direct (side) channel not superconducting
(superconducting).
\begin{figure}[tp]
\begin{center}\vspace{28mm}
\includegraphics[width=5.8cm,height=3.3cm]{fig45}\caption{Schematic diagram of a position controlled-swap gate.}
\label{fig45}
\end{center}
\end{figure}

\begin{figure}[ht]
\begin{center}\vspace{22mm}
\includegraphics[width=4cm,height=1.5cm,angle=00]{fig46.EPS}
\caption{Schematic representation of a PCS gate. For simplicity,
we omit dotted cross lines in our quantum circuits.} \label{fig46}
\end{center}
\end{figure}


On the other hand, the same as the spin-$\frac{1}{2}$ case, Bob
can use Hadamard operators in the position space with the form
\begin{eqnarray}
H_{x_{n}}=\frac{1}{\sqrt{2}}(P_{n}+N_{n}) \label{Hxn}
\end{eqnarray}
where $H_{x_n}$ is a local operator acting on the $\pm n$-th
channels.  Although the position Hadamard operator has not a
multiplicative form, but its design is relatively straightforward
in our scheme.  Figure \ref{fig47} shows a schematic circuit for
the Hadamard gate acting on the $n$-th channel in which we have
assumed that the probability of passing a particle through each
basic gate is the same. In fact, in our scheme as an advantage, it
is not necessary to find a multiplicative form to design such
operators using the basic ones.
\begin{figure}[ht]
\begin{center}\vspace{30mm}
\includegraphics[width=6cm,height=1.8cm]{fig47}\caption{Quantum circuit diagram of $H_{x_n}$ gate.}
\label{fig47}
\end{center}
\end{figure}

For $N\geq 2$ cases, however, applying just PCS and $H_{x_n}$
operators does not produce pure position states in a disentangled
and measurable form.  So, we should introduce a non-local unitary
operator for this mean which acts like
\begin{eqnarray}\label{U}
U_{(N)}|\pm l,\pm
m\rangle={\frac{1}{\sqrt{N}}}\sum_{n=1}^{N}h_{m,m+n-1}|f_{l^{\pm}}(n),f_{m^{\pm}}(n)\rangle_{{\rm
mod}(N)}
\end{eqnarray}
where $l$ and $m$ are positive integers.  Moreover, here,
$h_{m,n}$ is an element of an $N$-dimensional normalized symmetric
Hadamard matrix.  It is straightforward to check that
$U_{(N)}^2=I$, as we have shown in the following
\begin{eqnarray}
&&U_{(N)}U_{(N)}|\pm l,\pm
m\rangle=\nonumber\\
&&{\frac{1}{N}}\sum_{n=1}^{N}\sum_{k=1}^{N}h_{m,m+n-1}h_{m+n-1,m+n+k-2}|f_{l^{\pm}}(n+k-1),f_{m^{\pm}}(n+k-1)\rangle_{{\rm mod}(N)}\nonumber\\
&&=\frac{1}{N}\sum_{a}\sum_{k}h_{m,a-k+1}h_{a-k+1,a}|\pm(l+a-m),\pm
a\rangle\nonumber\\
&&=\sum_{a}\delta_{m,a}|\pm(l+a-m),\pm a\rangle=|\pm l,\pm
m\rangle.
\end{eqnarray}

 The same as $O_{(k^{\pm},j)}$ operator, it is also
possible to find an explicit form for $U_{(N)}$ operator based on
the basic gates.  Here, for example, we have considered it as
\begin{eqnarray}\label{ghoul}
U_{(N)}={\frac{1}{\sqrt{N}}} \sum_{n=0}^{N/2-1}\{L_{+}^{2n}\otimes
[\prod_{i=1}^{N}P_{i}^{(h_{b,i}-h_{2n+i,i})/2}N_{i}P_{i}^{(h_{b,i}-h_{N-2n+i,i})/2}]L_{+}^{2n}\nonumber\\
+L_{+}^{2n+1}\otimes[\prod_{i=1}^{N}(P_{i}N_{i}P_{i}N_{i})^{(h_{b,i}-h_{N-2n-1+i,i})/2}]L_{+}^{2n+1}
\}
\end{eqnarray}
where $h_{i,j}$ is an element of an $N$-dimensional real symmetric
Hadamard matrix.  In addition, the subscript $b$ is an arbitrary
constant positive integer.  This relatively complex form results
from the generalization procedure introduced in appendix B. If we
represent $U_{(N)}$ operator in the form
\begin{eqnarray}
U_{(N)}=\frac{1}{\sqrt{N}}\sum_{n=0}^{N-1}(L_{+}^{n}\otimes
A_{(n+1)}L_{+}^{n})
\end{eqnarray}
then, in a similar manner used to design $H_{x_n}$ operator, it
will be possible to conceive $U_{(N)}$ operator using the local
operator $A_{(n+1)}$ which can be designed by the basic gates, and
a nonlocal operator working like PCS gate, as is shown in
Fig.~\ref{fig48}.  Here, we have assumed that each of Bob's
channels is divided into $N$ identical branches, so that the
probability of entering the particle into each branch is the same.
Finally, each branch is joined to its correspondence channels in
Bob's lab.  It should be noted that, the mechanism of the required
conditional gate in this quantum circuit is the same as PCS gate,
but it should turn on the $S_k$ gates (instead of $P_n$ gate) to
operate $L_{+}^{k}$ on Alice's particle.
\begin{figure}[t]\label{fig48}
\begin{center}\vspace{30mm}
\includegraphics[width=10cm,height=4.8cm]{fig48}\caption{Quantum circuit diagram of $U_{(N)}$ operator for the $n$-th T-channel.  Here, each branch
of the T-channel acts as a control channel for the $S_k$ gates
applied on the C-channels.}
\end{center}
\end{figure}

Now, using the above mentioned operators, Bob can perform his BSM
in this way
\begin{eqnarray}\label{Hexplicit}
U_{(N)}(H_{x_1}H_{x_2}\ldots H_{x_n}\otimes I){\rm
PCS}|\psi_{(k^{\pm},j)}\rangle=|m,n\rangle
\end{eqnarray}
which should be followed by a position measurement on the outgoing
channels.  Figure \ref{fig49} shows a schematic quantum circuit of
Bob's lab.  In Eq.~(\ref{Hexplicit}), $m$ and $n$ are some unique
functions of $k^{\pm}$ and $j$. In addition, action of the set of
operators in this equation is equivalent to the action of the
grand operator (\ref{grand}).


\begin{figure}[ht]
\begin{center}\vspace{10mm}
\includegraphics[width=7cm,height=8.7cm,angle=0]{fig49}
\vspace{5mm}\caption{Some details of Bob's laboratory. Bold lines,
between gray boxes and PC show classical channels (i.e., ordinary
wires), and the others are quantum ones.  Gray boxes represent
quantum memories which our programmed PC controls their whole
performance, including opening and closing C and T-channels on
time and the synchronization of the two particles.  The connection
points between quantum channels are shown by the bold points.}
\label{fig49}
\end{center}
\end{figure}


\section{The rate of classical information gain}


We have seen that, there are $4N^2$ Bell basis and the same number
of different unitary operators $O_{(k^{\pm},j)}$ for encoding in
our dense coding scheme. This obviously corresponds to encoding
$4N^2$ different messages by Alice.  Thus, she can send
$2~\log_{2}(2N)$ bits of classical information per particle to
Bob. Now, he needs $N$ PCS, $2N$ Hadamard and one $U_{(N)}$ gates
to read out the sent classical information during the BSM. Since
Bob performs one BSM on just two particles, it is possible to
consider that all PCS and then all Hadamard gates operate in a
parallel form, i.e. {\it{concurrently}}. If the operation times
for the PCS, Hadamard and $U_{(N)}$ gates are $t_p$, $t_h$ and
$t_u$, respectively, the rate of classical information gain R,
defined as sent classical bits of information per unit time and
sent particle, is
\begin{eqnarray}\label{logbitnumber}
R_x=\frac{2~\log_{2}(2N)}{t_p+t_h+t_u}.
\end{eqnarray}
  In a similar way, one can calculate R for a dense coding
protocol which works using ${\textsf{N}}$ pairwise entangled
qubits and/or $\textsf{N}$ maximally entangled qubits shared
between two parties {\cite{Bose,Comment}}. In the pairwise
entangled case, $2{\textsf{N}}$ classical bits of information are
transferred from Alice to Bob. Meantime, in this case,
$\textsf{N}$ separate CNOT and Hadamard gates are required by Bob
to decode the Bell states. Thus, its R in terms of bits per unit
time per sent particle is
\begin{eqnarray}\label{logbitnumber}
R_p=\frac{2{\textsf{N}}}{{\textsf{N}}^2(t_c+t_h)}
\end{eqnarray}
where $t_c$ is the operation time of a CNOT gate.  Recently,
however, Lee {\it et al.} {\cite{Lee}} assert that the rate of
information gain for this case is as high as $2^{N}/N(t_c+t_h)$
bits per unit time.  Concerning this, we have shown in appendix C
that their claims are unfounded.  On the other hand, for the
maximally entangled case, the number of sent classical bits is
$\textsf{N}$. In this case, Bob needs $({\textsf{N}}-1)$
successive CNOT and one Hadamard gates so that he gains
\begin{eqnarray}\label{logbitnumber}
R_m=\frac{{\textsf{N}}}{({\textsf{N}}-1)[({\textsf{N}}-1)t_c+t_h]}
\end{eqnarray}
bits per time and particle.  Now, if we assume that both $N$ and
${\textsf{N}}$ are very large and all basic gates operate in an
equal time interval, i.e. $t_c\sim t_h\sim t_p/4\sim t_u/N\equiv
t$, then $R_x=2\log_{2}(2N)/Nt$ and $R_p=R_m=1/{\textsf{N}}t$.
Therefore, it can be easily seen that at large $N$'s, if
$N={\textsf{N}}$ is considered, i.e. dimensions of Hilbert spaces
are identical, then our protocol is more efficient than both the
pairwise and the maximally entangled cases with a {\em
logarithmic} factor.

Furthermore, other degrees of freedom such as spin, polarization
and squeezation can be added to our protocol in order to obtain a
more powerful quantum dense coding.  Here, for instance, we
consider that each particle can also have spin $S$.  To introduce
spin, it is sufficient to assume that the source emits entangled
pair of particles not only with vanishing total momentum but also
with zero total spin. Therefore, the quantum state of the system
would be
\begin{eqnarray}
|\psi_{1}\rangle_{xs}&=&\frac{1}{\sqrt{2N(2S+1)}}\sum_{n=1}^{N}[|n,-n\rangle+|-n,n\rangle]
\nonumber\\&&\times\sum_{s=0}^{2S}(\pm 1)^{s}|(S-s),-(S-s)\rangle
\end{eqnarray}
which is simply a tensor product of position and spin states of
the system.  Therefore, now Alice is capable of sending
$2~\log_{2}[2N(2S+1)]$ bits per particle.


\section{Conclusions}


We have proposed a theoretical extended version of dense coding
protocol by using entangled spatial states of the two particles
shared between two parties.  Our construction is based on using
the well known real symmetric Hadamard matrix for representing
orthogonal states, required encoding and decoding operators, and
hence is subject to its intrinsic characteristics. Furthermore, we
have given a typical proposition for initial realization of the
scheme. In this regard, some basic gates as well as a position
conditional gate have been introduced and it is shown that whole
of the scheme can be established based upon them. By comparing our
scheme with some of the previously proposed multi-qubit protocols,
it is shown that the rate of classical information gain in our
case is better than them by a logarithmic factor. Also we have
shown that capability of considering internal degrees of freedom,
like spin, strengthens the scheme in a straightforward manner.




%----------------------------------------------------------Ch.5
\chapter{A scheme towards complete state teleportation}
\section{Introduction}
Teleportation scheme proposed by Bennett {\em{et al.}}
{\cite{BB93}} is a protocol for disembodied transmission of an
{\em{unknown}} quantum state of a spin-$\frac{1}{2}$ particle by a
sender, Alice, to a receiver, Bob, by conveying two classical bits
of information.  In addition, the successful experimental
realizations of this protocol, particularly for polarization
states of photons {\cite{Bouwmeester97,Boschi}}, stimulated
studying of teleportation protocol for more complex systems.  For
instance, extensions to $N$-dimensional Hilbert spaces recently
have attracted much attentions.  In [83-86] an $N$-level source
and in {\cite{Ikram,Zhou}} some two level EPR sources have been
utilized to accomplish $N$-level state teleportation.  Using the
above entangled versions, it is theoretically possible to teleport
any system with discretizable Hilbert space.

But, what about teleportation of position states of a quantum
{\em{object}}, $\psi({\bf{x}})$? The theoretical answer to this
question was first provided by Vaidman
{\cite{Vaidman94,VaidmanYoran}}, utilizing perfect entanglement
between position and momentum which results in a scheme for
teleportation of continuous variables, or the wave function
$\psi(\mathbf{x})$. Interestingly, reliable teleportation of
continuous variables is shown to be implementable in experiments.
In fact, Braunstein and Kimble {\cite{BraunsteinKimble}} made a
{\em{realistic}} proposal for teleportation of quantum state of a
single mode of the electromagnetic field. In other words, their
scheme was a typical implementation of Vaidman's method. Then,
experimental realization of Braunstein-Kimble method was performed
by Furasawa {\em{et al.}} {\cite{Furasawa}}. But, it is believed
that this type of experiment cannot be expected to improve, due to
difficulties in establishing highly squeezed light fields
{\cite{Vaidman98}}, and even in ideal experimental conditions,
details of Bob's reconstruction process results in an
{\em{almost}} perfect replica of light fields {\cite{BEZ}}. Also,
as Vaidman pointed out, Braunstein-Kimble method is not applicable
directly for teleporting $\psi({\textbf{x}})$, where
${\textbf{x}}$ is the spatial position of a quantum object
{\cite{VaidmanYoran}}.  As a solution, Vaidman proposed a way to
overcome this disability by introducing a quantum-quantum
interaction which can convert the continuous variable of a real
position to the electromagnetic field amplitude variable. However,
other methods or alternatives might be possible to solve this
problem.

In this chapter, we have presented an alternate theoretical scheme
to achieve realization of teleporting wave function
$\psi_{S}(\mathbf{x})$ of an object having spin. To do this, we
have offered a scheme, including a special EPR source and some
arrays of receivers to teleport a multi-level {\em{entangled
position-spin}} state in our three-dimensional space
\cite{Teleport}.  Moreover, some theoretical details on Bell state
measurement (BSM) and the reconstruction process have been
illustrated in our novel spatial scheme.



\section{Description of the teleportation set-up}
Consider a theoretical set-up including two parties, Alice and
Bob, who have labs which contain 2$N$-receiver arrays, as shown in
Fig.~\ref{fig1}.  An EPR source is placed in the center of the
symmetry axis ($z$-axis) of the two receiver lines, which emits,
{\em{isotropically}}, two identical {\em{maximally entangled}}
particles with vanishing total momentum.  The receivers are
labelled by $\pm 1$, $\pm 2$,\ldots, $\pm N$. We assume that they
do not perform any destructive measurements on the received
particles. All lines that connect each receiver to the related lab
are representatives of quantum channels.  Each quantum object with
an unknown state, as an initial import to Alice's lab, is sent,
after quantum scanning, to her lab to be combined with her own
particle. Then, she performs a BSM on her two particles. Based
upon the result of this measurement, she communicates with Bob
through a classical channel, sending related binary codes. Bob,
after receipt of Alice's message, reconstructs the initial quantum
state using his lab.




\section{A representation for Bell bases}
After this illustration, it is possible to explain more details
about the scheme for teleportation of quantum position state of an
object. At the first step, we study the teleportation of
one-dimensional position state of a spinless particle.  In this
case, the total state describing the {\em{system}} (EPR source+the
arrays) is
\begin{eqnarray}{\label{position}}
|\psi_{1(2)}\rangle_{x}=\frac{1}{\sqrt{2N}}\sum_{n=1}^{N}[|n,-n\rangle\pm|-n,n\rangle]
\end{eqnarray}
where $n$ refers to the label of the receivers.  Additionally, we
adopt the convention that in the tensor products the left (right)
states belong to Alice (Bob).  The signs $\pm$ in
$|\psi_{1(2)}\rangle_{x}$, in which 1(2) indicates +(-), stand for
symmetry and antisymmetry of the state with respect to the
two-party exchange which simply means bosonic or fermionic
property of our system. Here, without any loss of generality, we
admit that our system is bosonic. In general, the entangled state
({\ref{position}}) can be considered as a  member of a larger
family, that is,
\begin{eqnarray}\label{firstfamily}
&&|\psi_{(1,j)}\rangle_x={\frac{1}{\sqrt{2N}}}
\sum_{n=1}^{N}[h_{j,2n-1}|n,-n\rangle+h_{j,2n}|-n,n\rangle]\nonumber\\
&&\hspace{3.2cm}1\leq j\leq 2N
\end{eqnarray}
in which $h_{i,j}=\sqrt{2N}[{\bf H}]_{i,j}$, and ${\bf H}$ is a
{\em{normalized symmetric}} $2N\times 2N$ Hadamard matrix which
satisfies the property ${\bf H}^{2}=I$. Furthermore, the above
states can be generalized to a more complete set of orthonormal
and maximally entangled states, which are defined as
\begin{eqnarray}\label{totPsi}
&&|\psi_{(k^{\pm},j)}\rangle_x={\frac{1}{\sqrt{2N}}}\sum_{n=1}^{N}[h_{j,2n-1}|n,f_{k^{\pm}}(n)\rangle+h_{j,2n}|-n,-f_{k^{\pm}}(n)\rangle]\nonumber\\
&&\hspace{3cm}1\leq k\leq N, 1\leq j\leq 2N
\end{eqnarray}
where $j$ and $k^{\pm}$ are member and family indices,
respectively, and we have also adopted the convention
$f_{k^{\pm}}(n)=\pm(n+k-1)_{{\rm mod}(N)}$.  By the way, the
states ({\ref{totPsi}}) form {\em{Bell bases}} for our
teleportation scheme. In fact, it can be simply checked that these
states have the properties
\begin{eqnarray}
&\langle\psi_{(k^{\pm},j)}|\psi_{(k'^{\pm},j')}\rangle=\delta_{j,j'}\delta_{k,k'}\nonumber\\
&{\rm{tr}}_{1(2)}(|\psi_{(k^{\pm},j)}\rangle\langle\psi_{(k^{\pm},j)}|)=\frac{1}{2N}I_{1(2)}.
\end{eqnarray}
\begin{figure}[tp]
\begin{center}\vspace{5mm}
\includegraphics[width=4.7cm,height=9cm,angle=-90]{prlfig1}\caption{A
primary scheme for teleportation of a position state.}
\label{fig1}
\end{center}
\end{figure}
In addition, as known, the order of a real Hadamard matrix is 1, 2
and $4k$ with $k$ as a positive integer {\cite{Wallis,Agaian}}.
Thus, in other conditions, there does not exist any related
teleportation scheme.  In other words, we cannot make all
necessary entangled orthonormal states to perform teleportation
for odd and half odd $N$ cases.


\section{Unitary transformation of Bell bases}
For the permitted $N$'s, one can find unitary operators which
transform the initial state of the system into the other Bell
states given in Eq.~(\ref{totPsi}). The representation of suitable
operator for this mean is
\begin{eqnarray}\label{Ot}
O'_{(k^{\pm},j)}=\sum_{n=1}^{N}[h_{j,2n-1}|n\rangle\langle
-f_{k^{\pm}}(n)|+h_{j,2n}|-n\rangle\langle f_{k^{\pm}}(n)|]
\end{eqnarray}
which acts on the Bell bases as
\begin{eqnarray}
O'_{(k^{\pm},j)}|\psi_{(k'^{\pm},j')}\rangle_x=|\psi_{(k''^{\pm},j'')}\rangle_x
\end{eqnarray}
where
\begin{eqnarray}\label{hj}
h_{j'',i}=h_{j,i}h_{j',i}
\end{eqnarray}
at $j'=1$, and
\begin{eqnarray}\label{k''}
k''^{\pm}=(k+k'-1)^{-ss'}_{{\rm mod}(N)}
\end{eqnarray}
in which $s$ and $s'$ are signs of superscripts of $k$ and $k'$,
respectively.  In the following, it will be seen that the operator
$O'_{(k^{\pm},j)}$ are applied by Bob to reconstruct the unknown
wave functions which are teleported to him by Alice. Moreover, an
example for realization of these operators can be considered the
same as $O_{(k^{\pm},j)}$ mentioned for the dense coding scheme in
the previous chapter.


Here, it is useful to have a review on the details of action of
this unitary operator on the Bell bases.  Thus, using
orthonormality of the position bases, we have
\begin{eqnarray}
O'_{(k^{\pm},j)}|\psi_{(k'^{\pm},j')}\rangle_x&=&\frac{1}{\sqrt{2N}}\sum_{n,m=1}^{N}[h_{j,2n-1}h_{j',2m-1}\delta_{\mp(n+k-1),m}|n,f_{k'^{\pm}}(m)\rangle\nonumber\\
&&+h_{j,2n-1}h_{j',2m}\delta_{\mp(n+k-1),-m}|n,-f_{k'^{\pm}}(m)\rangle\nonumber\\
&&+h_{j,2n}h_{j',2m-1}\delta_{\pm(n+k-1),m}|-n,f_{k'^{\pm}}(m)\rangle\nonumber\\
&&+h_{j,2n}h_{j',2m}\delta_{\pm(n+k-1),-m}|-n,-f_{k'^{\pm}}(m)\rangle].
\end{eqnarray}
Now, there are two $k^+$ and $k^-$ cases which can be examined
separately.  For the $k^+$ case, we have
\begin{eqnarray}
O'_{(k^{+},j)}|\psi_{(k'^{\pm},j')}\rangle_x&=&\frac{1}{\sqrt{2N}}\sum_{n=1}^{N}[h_{j,2n-1}h_{j',2(n+k-1)}|n,f_{k''^{\mp}}(n)\rangle\nonumber\\
&&+h_{j,2n}h_{j',2(n+k-1)-1}|-n,-f_{k''^{\mp}}(n)\rangle]
\end{eqnarray}
where $k''=k+k'-1$.  Similarly, for the $k^-$ case we can write
\begin{eqnarray}
O'_{(k^{-},j)}|\psi_{(k'^{\pm},j')}\rangle_x&=&\frac{1}{\sqrt{2N}}\sum_{n=1}^{N}[h_{j,2n-1}h_{j',2(n+k-1)-1}|n,f_{k''^{\pm}}(n)\rangle\nonumber\\
&&+h_{j,2n}h_{j',2(n+k-1)}|-n,-f_{k''^{\pm}}(n)\rangle]
\end{eqnarray}
where once again $k''=k+k'-1$.  Therefore, it is clear now that
for $j'=1$ Eqs.~(\ref{hj}) and (\ref{k''}) are satisfied.






\section{General procedure for teleporting an object}

In the one-dimensional scheme, the most general position states
which Alice is able to teleport is
\begin{eqnarray}{\label{PHIX}}
|\phi\rangle_x=\sum_{n=1}^{N}[a_{n}|n\rangle+a_{-n}|-n\rangle].
\end{eqnarray}
Concisely, the mapping of the object's wave function to this state
can be conceived by a typical scanning, for instance, using two
approaching arrays of receivers sweeping the object.  After this
scanning, Alice has to combine her own particle with the given
one. By rewriting the total state of the three particles based on
our proposed Bell bases, she must perform a BSM on her two
possessed particles and then send the classical obtained results,
which encode the retrieving process, to Bob as the following
%\begin{widetext}
\begin{eqnarray}\label{rewrite}
|\phi\rangle_{x}|\psi_{1}\rangle_{x}&&=\frac{1}{\sqrt{2N}}\sum_{
   n=-N(n\neq0)}^{N}~ \sum_{m=1}^{N}a_{n}[|n,m,-m\rangle+|n,-m,m\rangle]\nonumber\\
&&=\frac{1}{2N}\sum_{k=1}^{N}\sum_{j=1}^{2N}|\psi_{(k^{\pm},j)}\rangle_{x}O'^{\dag}_{(k^{\pm},j)}|\phi\rangle_{x}\nonumber\\
&&{\begin{array}{c}
  {\tiny{\rm{Alice's}}}\vspace{-2mm}\\
 \longrightarrow \vspace{-1mm}\\
  {\tiny{\rm{BSM}}}
\end{array}}{{|\psi_{(k^{\pm},j)}}}\rangle_{x}O'^{\dag}_{(k^{\pm},j)}|\phi\rangle_{x}
\end{eqnarray}
where
\begin{eqnarray}
O'^{\dag}_{(k^{\pm},j)}=\sum_{n=1}^{N}[h_{j,2n-1}|-f_{k^{\pm}}(n)\rangle\langle
n|+h_{j,2n}|f_{k^{\pm}}(n)\rangle\langle -n|].
\end{eqnarray}


Now, it suffices for Bob to know the classical information
identifying the corresponding operator, i.e. $(k^{\pm},j)$, to
reconstruct the initial state $|\phi\rangle_x$ using a suitable
$O'_{(k^{\pm},j)}$ operator.  Because this operator is a unitary
operator, as we can see in the following
\begin{eqnarray}
O'_{(k^{\pm},j)}O'^{\dag}_{(k^{\pm},j)}&=&\sum_{n,m=1}^{N}[h_{j,2n-1}h_{j,2m-1}\delta_{n,m}|n\rangle\langle m|\nonumber\\
&&+h_{j,2n-1}h_{j,2m}\delta_{n,-m-2k+2}|n\rangle\langle -m|\nonumber\\
&&+h_{j,2n}h_{j,2m-1}\delta_{n,-m-2k+2}|-n\rangle\langle m|\nonumber\\
&&+h_{j,2n}h_{j,2m}\delta_{n,m}|-n\rangle\langle -m|\nonumber]\\
&=&\sum_{n=1}^{N}|n\rangle\langle n|+|-n\rangle\langle -n|=I.
\end{eqnarray}
Here, it should be pointed out that in the original scheme for the
continuous variables teleportation
{\cite{Vaidman94,VaidmanYoran}}, Alice needs to apply two separate
kinds of measurements on the position and momentum to accomplish a
BSM for teleportation of $\psi(x)$.  But, in our proposed scheme,
Alice needs to perform just one kind of BSM on the position
variable.

It is straightforward to check the rewriting done in the second
line of Eq.~(\ref{rewrite}).  To do this,  we first consider
\begin{eqnarray}
O'^{\dag}_{(k^{\pm},j)}|\phi\rangle_{x}&=&\sum_{n,m=1}^{N}[h_{j,2n-1}a_{m}\delta_{n,m}|-f_{k^{\pm}}(n)\rangle+h_{j,2n-1}a_{-m}\delta_{n,-m}|-f_{k^{\pm}}(n)\rangle\nonumber\\
&&+h_{j,2n}a_{m}\delta_{-n,m}|f_{k^{\pm}}(n)\rangle+h_{j,2n}a_{-m}\delta_{n,m}|f_{k^{\pm}}(n)\rangle]\nonumber\\
&=&\sum_{m=1}^{N}[h_{j,2m-1}a_{m}|-f_{k^{\pm}}(m)\rangle+h_{j,2m}a_{-m}|f_{k^{\pm}}(m)\rangle].
\end{eqnarray}
Then, by substituting this result in the second line of
Eq.~(\ref{rewrite}), we obtain
\begin{eqnarray}
&&\frac{1}{2N}\sum_{k=1}^{N}\sum_{j=1}^{2N}|\psi_{(k^{\pm},j)}\rangle_{x}O'^{\dag}_{(k^{\pm},j)}|\phi\rangle_{x}\nonumber\\
&&=\frac{1}{2N}\sum_{k=1}^{N}\sum_{j=1}^{2N}({\frac{1}{\sqrt{2N}}}\sum_{l=1}^{N}[h_{j,2l-1}|l,f_{k^{\pm}}(l)\rangle+h_{j,2l}|-l,-f_{k^{\pm}}(l)\rangle])\nonumber\\
&&\hskip17mm\times(\sum_{m=1}^{N}[h_{j,2m-1}a_{m}|-f_{k^{\pm}}(m)\rangle+h_{j,2m}a_{-m}|f_{k^{\pm}}(m)\rangle])\nonumber\\
&&=\frac{1}{2N\sqrt{2N}}\sum_{k=1}^{N}\sum_{j=1}^{2N}\sum_{l,m=1}^{N}[h_{j,2l-1}h_{j,2m-1}a_{m}|l,f_{k^{\pm}}(l),-f_{k^{\pm}}(m)\rangle\nonumber\\
&&\hskip35mm+h_{j,2l-1}h_{j,2m}a_{-m}|l,f_{k^{\pm}}(l),f_{k^{\pm}}(m)\rangle\nonumber\\
&&\hskip35mm+h_{j,2l}h_{j,2m-1}a_{m}|-l,-f_{k^{\pm}}(l),-f_{k^{\pm}}(m)\rangle\nonumber\\
&&\hskip35mm+h_{j,2l}h_{j,2m}a_{-m}|-l,-f_{k^{\pm}}(l),f_{k^{\pm}}(m)\rangle]\nonumber\\
&&=\frac{1}{\sqrt{2N}}\sum_{k,m=1}^{N}[a_{m}|m,f_{k^{\pm}}(m),-f_{k^{\pm}}(m)\rangle+a_{-m}|-m,-f_{k^{\pm}}(m),f_{k^{\pm}}(m)\rangle]\nonumber\\
&&=\frac{1}{\sqrt{2N}}\sum_{n=-N,(n\neq
0)}^{N}\sum_{m=1}^{N}[a_{m}|m,n,-n\rangle+a_{-m}|-m,-n,n\rangle]\nonumber\\
&&=|\phi\rangle_{x}|\psi_1\rangle_{x}.
\end{eqnarray}
In the last three lines of this proof, we have used the orthogonal
property of rows or columns of the Hadamard matrix, that is,
\begin{eqnarray}
\sum_{j}h_{j,m}h_{j,n}=2N\delta_{m,n}.
\end{eqnarray}




\section{Alice's Bell state measurement}

To do a BSM, Alice needs to apply a {\it grand} unitary and
Hermitian operator on all the channels followed by a position
state measurement on Alice's outgoing receivers.  Here, for
instance, we can consider the representation of the grand operator
as
\begin{eqnarray}\label{grandx}
{\cal{H}}_x=\sum_{j=1}^{2N}\sum_{k=1}^{N}|j,f'_{k^{\pm}}(j)\rangle\langle\psi
'_{(k^{\pm},j)}|
\end{eqnarray}
\begin{figure}[tp]
\begin{center}\vspace{25mm}
\includegraphics[width=6cm,height=3cm]{fig52}\caption{Schematic diagram of a position controlled-gate.}
\label{fig2}
\end{center}
\end{figure}
where $f'_{k^{\pm}}(j)=\pm(j+k-1)_{{\rm mod}(2N)}$ and
$|\psi'_{(k^{\pm},j)}\rangle_x=\frac{1}{\sqrt{2N}}\sum_{n=1}^{2N}h_{j,n}|j,f'_{k^{\pm}}(j)\rangle_x$.
It is easy to check that ${\cal{H}}_x|\psi
'_{(k^{\pm},j)}\rangle_x=|j,f'_{k^{\pm}}(j)\rangle_x$, and
${\cal{H}}_{x}^{2}=I$.  In addition, to make the former property
consistent with our channel labelling, it suffices to apply a
proper unitary operator. It is seen that the required grand
operator can be considered as a conditional gate which acts in a
2$N$-dimensional (channel) Hilbert space. Furthermore, based on
the works of Deutsch {\cite{Deutsch}} and Barenco {\em et al.}
{\cite{Barenco}} as well as Bremner {\em et al.} \cite{Bremner},
we can construct the grand operator using, for example, a set of
controlled gates. Thus, we introduce our position controlled gate
(PCG) which acts like
\begin{eqnarray}\label{PCS}
{\rm{PCG}}|n,m\rangle=\theta(-n)(I\otimes
G)|n,m\rangle+\theta(n)|n,m\rangle
\end{eqnarray}
where $\theta (n)$ is the unit step function, and $G$ is a general
form for $N_m$, $P_m$ and $L_+$ basic gates. To understand the
action of this operator there is a way using four well-known
spin-$\frac{1}{2}$ CNOT gates. Figure \ref{fig2} shows a sketch of
this theoretical construction. The $\overline{\rm{CNOT}}$ is
somehow a complement to the usual CNOT in the sense that
${\rm{CNOT}}|x,y\rangle=|x,x+y\rangle_{{\rm mod}(2)}$ and
$\overline{\rm{CNOT}}|x,y\rangle=|x,\overline{x}+y\rangle_{{\rm
mod}(2)}$. If, for instance, a spin-$\frac{1}{2}$ particle passes
through the control channel, then the two first CNOT and
$\overline{\rm{CNOT}}$ definitely change spin of an intermediate
spin-$\frac{1}{2}$ particle, named spin target, which is preset to
a fixed state , e.g. $|0\rangle$. It is assumed that any change in
the state describing the spin target means switching the gate for
acting on the channel $m$. The two latter CNOT and
$\overline{\rm{CNOT}}$ are just to reset the spin target.  This
conditional gate can be realized using, for example,
superconductor circuits. It is obvious that any change in the spin
target results into a change in the magnetic flux passing through
a superconductor circuit.  Thus, for switching the gate, the
induced current can be amplified by a superconducting LC circuit
in order to produce at least a critical magnetic field to set the
direct (side) channel not superconducting (superconducting).

\section{Teleportation of an object having spin}

At this stage, one can think of an entangled position-spin state
teleportation of a particle having spin $S$.  To accomplish this,
it is only sufficient to assume that the source emits entangled
pair of particles not only with vanishing total momentum but also
with zero total spin. Therefore, the quantum state of the system
would be
\begin{eqnarray}
|\psi_{1}\rangle_{xs}=\frac{1}{\sqrt{2N(2S+1)}}\sum_{n=1}^{N}\sum_{s=0}^{2S}(\pm
1)^{s}[|n,-n\rangle+|-n,n\rangle] |(S-s),-(S-s)\rangle\nonumber\\
\end{eqnarray}
which is simply a tensor product of position and spin states of
the source.  So, as an advantage, it is not necessary to have any
entanglement between these two degrees of freedom. By this system,
Alice would be able to send unknown quantum states which have the
following general form
\begin{eqnarray}
|\phi\rangle_{xs}=\sum_{n=-N(n\neq
0)}^{N}\sum_{s=0}^{2S}a_{ns}|n,S-s\rangle.
\end{eqnarray}
As is seen, to teleport spin state along with position state, it
does not need to change our formalism drastically, but it simply
needs the map $N\rightarrow N(2S+1)$ in our mathematical stages.
So, just by adapting our protocol to work for $4N^{2}(2S+1)^{2}$
dimensions of Hilbert space, we can simultaneously accomplish spin
state as well as position state teleportation for the entangled
state in the form $|\phi\rangle_{xs}$. This procedure can be
generalized to teleportation of any other degrees of freedom of an
object, provided that its necessary entanglement is supplied in
the source.


\section{Teleportation of a 2-dimensional object using a planar quantum scanner}

To extend the scheme to the teleportation of a two-dimensional
position state of an object, we assume that our arrays are planar
so that the state of our system would be
\begin{eqnarray}
\label{positionxy}
&&{|\psi_{1(2)}\rangle}_{xy}={\frac{1}{\sqrt{4N_{x}N_{y}}}}\sum_{n_{x}=1}^{N_{x}}\sum_{n_{y}=1}^{N_{y}}
\nonumber\\ &&[|(n_{x},n_{y}),(-n_{x},-n_{y})\rangle\pm|(-n_{x},-n_{y}),(n_{x},n_{y})\rangle]+\nonumber\\
&&[|(n_{x},-n_{y}),(-n_{x},n_{y})\rangle\pm|(-n_{x},n_{y}),(n_{x},-n_{y})\rangle].\hspace{6mm}
\end{eqnarray}
 By this simple assumption, Alice can now teleport the states in
general form
\begin{eqnarray}
|\phi\rangle_{xy}=\sum_{n_{x}=-N_x(n_x\neq
0)}^{N_x}\sum_{n_{y}=-N_y(n_y\neq 0)}^{N_y}
a_{n_{x}n_{y}}|(n_{x},n_{y})\rangle\hspace{3mm}
\end{eqnarray}
to Bob, which are representatives of states that result in
two-dimensional wave functions $\psi(x,y)$ at large $N$'s.  Now,
by insight obtained from one-dimensional case, it is clear that
two-dimensional position state teleportation is mathematically
equivalent to one-dimensional case just with a larger (channel)
Hilbert space dimension, that is, $N=N_xN_y$.


\section{Teleportation of the 3rd dimension using momentum basis}
There is an intricacy in extension of the protocol for
teleportation of three-dimensional wave functions $\psi(x,y,z)$.
It does not seem trivial how to achieve teleportation of
three-dimensional objects, utilizing our {\em{planar}} receivers.
To overcome this problem, we use the fact that any quantum state
$|\phi\rangle_z$ can be expanded in terms of the momentum basis as
well as the position ones, that is
\begin{eqnarray}
|\phi\rangle_{z}=\sum_{z}c_{z}|z\rangle=\sum_{p}b_{p}|p\rangle.
\end{eqnarray}
Thus, to teleport a state represented in the momentum basis we add
another property to our source, without any change in the form of
our planar receivers.  Suppose that the emitted entangled
particles from the source, can also get a definite momentum whose
value is $p_1,p_2,\ldots,p_M$ with the same probability in the
$z$-direction.  Fulfilling these properties for the source might
be an experimental challenge (for example, as a primary proposal
in this regard, see \cite{Farhan}). Anyway, the state of the
system in this direction can be written as
\begin{eqnarray}
|\varphi_{1}\rangle_{z}=\frac{1}{\sqrt{2M}}\sum^{M}_{m=1}[|p_{m},-p_{m}\rangle+|-p_{m},p_{m}\rangle].
\end{eqnarray}
Utilizing this source, Alice can send the most general states in
the form
\begin{eqnarray}
|\phi\rangle_{z}=\sum_{m=1}^{M}[b_{m}|p_{m}\rangle+b_{-m}|-p_{m}\rangle]\equiv\sum_{p}b_{p}|p\rangle.
\end{eqnarray}
All  teleportation processes of $|\phi\rangle_{z}$ once again can
be performed very similar to the one-dimensional position state
teleportation.  In fact, it is sufficient to consider the
conversion $|\pm n\rangle\longleftrightarrow|\pm p_{n}\rangle$ in
all relations which we have obtained in the one-dimensional case.
For example, similar to Eq.~(\ref{totPsi}), the momentum Bell
bases can be represented as
\begin{eqnarray}
&&|\varphi_{(q^{\pm},r)}\rangle_z={\frac{1}{\sqrt{2M}}}\sum_{m=1}^{M}[h_{r,2m-1}|p_m,\pm p_{f_{q}(m)}\rangle+h_{r,2m}|-p_m,\mp p_{f_{q}(m)}\rangle]\nonumber\\
&&\hspace{3.2cm}1\leq q\leq M, 1\leq r\leq 2M
\end{eqnarray}
where
\begin{eqnarray}
f_q(m)=(m+q-1)_{{\rm mod}(M)}.
\end{eqnarray}


\section{Towards complete teleportation of a 3-dimensional object}

\begin{figure}[t]
\begin{center}\vspace{20mm}
\includegraphics[width=12cm,height=7cm]{telset}
\caption{Schematic drawing of the proposed theoretical set-up for
teleporting the quantum state of an object in our
three-dimensional space.  The initial state is scanned by two
approaching arrays of receivers which are properly connected to
Alice's lab.} \label{telset}
\end{center}
\end{figure}
Equipped with this novel kind of source, Alice is now able to
teleport three-dimensional states.  Regarding the facts of the
previous sections, it is sufficient to consider $x$ (as a
representative of $xy$) and $z$ components to explain function of
the protocol for three-dimensional case. Therefore, we consider
general states like
\begin{eqnarray}
|\phi\rangle_{xz}&=&\sum_{x,z}c_{xz}|\phi\rangle_{x}|\phi\rangle_{z}\nonumber\\
&=&\sum_{x,z}\sum_{n,m=-N,-M(n,m\neq 0)}^{N,M}\hskip -5mm
c_{xz}a_{nx}b_{mz}|n\rangle_{x}|p_{m}\rangle_{z}\hspace{5mm}
\end{eqnarray}
which Alice wishes to teleport.  Here, the proposed teleportation
set-up in Fig.~\ref{fig1} should be changed a little so that the
system (arrays+source) shown in Fig.~\ref{telset} can now describe
the three-dimensional quantum state
$|\psi_1\rangle_x|\varphi_1\rangle_z$. Again, Alice combines her
own particle with the scanned particle corresponding to the state
$|\phi\rangle_{xz}$.  This combination can be represented as
\begin{eqnarray}
|\phi\rangle_{xz}|\psi_{1}\rangle_x|\varphi_{1}\rangle_z&&=\frac{1}{4NM}\sum_{x,z}\sum_{n=-N(n\neq
0)}^{N}\sum_{m=-M(m\neq
0)}^{M}c_{xz}a_{nx}b_{mz}\nonumber\\
&&\times\sum_{k,j=1}^{N,2N}|\psi_{(k^{\pm},j)}\rangle_x
O'^{\dag}_{(k^{\pm},j)}|n\rangle_{x}\sum_{q,r=1}^{M,2M}|\varphi_{(q^{\pm},r)}\rangle_z
T^{\dag}_{(q^{\pm},r)}|p_{m}\rangle_{z}.\nonumber\\
\end{eqnarray}
in which $|\varphi_{(q^{\pm},r)}\rangle_z$ states are momentum
analog of Bell bases $|\psi_{(k^{\pm},j)}\rangle_x$, and also
$T_{(q^{\pm},r)}$ operators work like momentum analog of
$O'_{(k^{\pm},j)}$ operators generated by the replacement of $|\pm
n\rangle\longleftrightarrow|\pm p_{n}\rangle$ in Eq.~(\ref{Ot}).
After performing two separate BSM's on the position and momentum
bases of the two particles in the $x$ and $z$ directions,
respectively, the protocol is completed and one can obtain
\begin{eqnarray}
&&|\psi_{(k^{\pm},j)}\rangle_x|\varphi_{(q^{\pm},r)}\rangle_z\hskip -3mm\sum_{x,z,n,m}\hskip -3mm c_{xz}a_{nx}b_{mz}O'^{\dag}_{(k^{\pm},j)}T^{\dag}_{(q^{\pm},r)}|n\rangle_{x}|p_{m}\rangle_{z}\nonumber\\
&&=|\psi_{(k^{\pm},j)}\rangle_x|\varphi_{(q^{\pm},r)}\rangle_z
O'^{\dag}_{(k^{\pm},j)}T^{\dag}_{(q^{\pm},r)}|\phi\rangle_{xz}.
\end{eqnarray}
Now Bob needs to know the classical information $(k^{\pm},j)$ and
$(q^{\pm},r)$ via a classical channel in order to reconstruct the
initial state $|\phi\rangle_{xz}$.


\section{Examination on the realizability of the momentum gates}

\subsection{Momentum basic gates}

Similar to the position basic gates introduced in the previous
chapter, here, it is also possible to consider some basic gates
for the momentum space. Concerning this, we have introduced the
following operators
\begin{eqnarray}\label{PRD}
&&{\cal P}_m|\pm p_m\rangle=\pm|\pm p_m\rangle\nonumber\\
&&{\cal R}_m|\pm p_m\rangle=|\mp p_m\rangle\nonumber\\
&&D_+|\pm p_m\rangle=|\pm p_{(m+1)_{{\rm mod}(M)}}\rangle\nonumber\\
\end{eqnarray}
in the momentum space which are called normalized momentum,
reversal and drift operators corresponding to $N_n$, $P_m$ and
$L_+$ operators in the position space, respectively.  This
correspondence helps one to use all the relations obtained up to
now for the position case, just by the following replacements:
\begin{eqnarray}
&&|\pm n\rangle\longrightarrow|\pm p_m\rangle\nonumber\\
&&N_n\longrightarrow {\cal P}_m\nonumber\\
&&P_n\longrightarrow{\cal R}_m\nonumber\\
&&L_+\longrightarrow D_+.
\end{eqnarray}

\begin{figure}[t]
\begin{center}\vspace{25mm}
\psfrag{CNO}[Bc][30][0.6][0]{CNOT}
\includegraphics[width=10cm,height=5cm]{fig54}
\caption{Schematic diagram of a reversal operator (${\cal R}$) in
the momentum space.} \label{fig54}
\end{center}
\end{figure}
Anyway, again it is relevant to see question how one can conceive
these operators in practice.  For ${\cal{P}}_m$ operator, it is
clear that it acts like a phase factor as follows:
\begin{eqnarray}
{\cal P}_m=e^{i\theta(\mp p_m)\pi}
\end{eqnarray}
which works the same as $N_n$ operator in the position space.
However, the design of the other operators is relatively
different. For instance, ${\cal{R}}$ operator can be considered to
operate as Fig.~\ref{fig54} shows.  In this scheme, in addition to
a spin-$\frac{1}{2}$ target which controls the two reflector
gates, we have introduced a spin-1 target for controlling the two
switch gates.  It is assumed that, the spin-1 target is initially
adjusted in a fixed state, e.g. $|0\rangle\equiv|s_z=-1\rangle$.
Each quantum interaction between the particle passing through the
channel and the spin-1 particle results in a quantum rotation for
the target's particle in the spin space, which can be represented
as
\begin{eqnarray}
J_{+}^{n}|j\rangle=|j+n\rangle_{{\rm mod}(3)};\hspace{1cm}j=0,1,2
\end{eqnarray}
where $J_+$ is a typical ladder operator in the spin space,
$|j\rangle$ is the target's spin state and $n$ shows number of the
quantum interactions.  Furthermore, $CJ_+$ operators work in the
sense that
\begin{figure}[t]
\begin{center}\vspace{35mm}
\psfrag{CNO}[10][15][0.6][0]{CNOT}
\includegraphics[width=11cm,height=5cm]{fig55}\caption{Schematic drawing of a filtering quantum circuit.  If in the first quantum interaction
box $t\geq t_m$, where $t$ is the time of interaction and
$t_m=d/v_m$, then the initial state of the lower
spin-$\frac{1}{2}$ target will change and the two related $S$-gate
will be turned ON. Moreover, if in the second quantum interaction
box $t\geq t_{m+1}$, then the initial state of the upper
spin-$\frac{1}{2}$ target will change and its related $S$-gates
will be turned ON. The G-gate can be considered any desired local
gate such as a phase shifter or the ${\cal R}$ operator.}
\label{fig55}
\end{center}
\end{figure}
\begin{eqnarray}
CJ_+|x,y\rangle=|x,x+y\rangle_{{\rm mod}(3)}\nonumber\\
\overline{CJ_+}|x,y\rangle=|x,\overline{x}+y\rangle_{{\rm mod}(3)}
\end{eqnarray}
where $x=0,1$ and $y=0,1,2$.  Here, we have assumed that, if $j=0
(1,2)$, then the two switch gates are turned OFF (ON), i.e., the
main channels are open (closed) and the side channels are closed
(open). For the spin-$\frac{1}{2}$ target, it is also assumed that
any change in the target's initial spin state corresponds to
turning ON the reflector gate. If this gate is turned OFF, it
permits particles to pass, and if it is turned ON, then the
particles are elastically reflected from it.


So far, we have introduced just ${\cal P}$ and ${\cal R}$
operators. But in our scheme, particularly in the reconstruction
process, we need ${\cal P}_m$ and ${\cal R}_m$ operators which
should act just on the particles having the momentum $p_m$.  Thus,
we should design a quantum filtering circuit which leads only the
particles with the momentum $p_m$ into the considered gate.
Figure~\ref{fig55} shows such helpful circuit.  In this scheme, we
have assumed that each quantum interaction between the particle
and the spin-$\frac{1}{2}$ targets changes the target's state only
if the time of interactions are taken long enough.

\begin{figure}[t]
\begin{center}\vspace{40mm}
\includegraphics[width=7cm,height=3cm]{fig56}\caption{A typical scheme fo $D_{+}^{n}$ operator.  In the first box from the right hand side, the entering charged particle
obtains some energy corresponding to an electric field $nE_q$. We
have assumed that applying $E_q$ changes $p_m$ to $p_{m+1}$. Now,
if in the quantum interaction box $t\geq t_M$, where $t_M=d/v_M$
and $t$ is the time of interaction between the particle and the
spin-$\frac{1}{2}$ target, then the initial state of target will
change, and consequently, the $S$-gate will be turned OFF.
Otherwise, the $S$-gate is turned ON and the opposite field $E_M$
acts as the mod($M$) function.} \label{fig56}
\end{center}
\end{figure}
For the $D_+$ operator, it is reasonable to consider that, for
example, an electric field can easily control the momentum of
charged particles.  Meantime, it should be noted that $D_+$
operator has a cyclic behavior as shown in Eq.~(\ref{PRD}).
Therefore, a conceivable scheme for this operator can be
considered as represented in Fig.~\ref{fig56}.



\subsection{Momentum Bell state measurement}


In the momentum space, we should also entangled the momentum of
the two particles and perform a BSM on them in the $z$-direction.
The same as the grand unitary operator ${\cal H}_x$ represented in
Eq.~(\ref{grandx}), one can consider another grand unitary
operator in the momentum space to perform a BSM as follows:
\begin{eqnarray}
{\cal{H}}_p=\sum_{r=1}^{2M}\sum_{q=1}^{M}|p_r,\pm
p_{f'_{q}(r)}\rangle\langle\varphi'_{(q^{\pm},r)}|
\end{eqnarray}
where
\begin{eqnarray}
|\varphi'_{(q^{\pm},r)}\rangle_z=\frac{1}{\sqrt{2M}}\sum_{m=1}^{2M}h_{r,m}|p_r,\pm
p_{f'_{q}(r)}\rangle
\end{eqnarray}
and
\begin{eqnarray}
f'_{q}(r)=(r+q-1)_{{\rm mod}(2M)}.
\end{eqnarray}
\begin{figure}[t]
\begin{center}\vspace{40mm}
\includegraphics[width=8cm,height=4cm]{fig57}\caption{Schematic diagram of a momentum controlled gate (MCR).  If the scanned particle has a momentum
in the opposite of the $z$-direction, then it interacts with the
spin-$\frac{1}{2}$ target and changes its initial state. This
change results turning OFF of the S-gates.} \label{fig57}
\end{center}
\end{figure}
We have seen that these grand unitary operators can be
theoretically constructed using our proposed conditional gates.
Thus, here, we need to introduce a general momentum controlled
gate (MCR) which acts like
\begin{eqnarray}\label{MCG}
{\rm MCR}|p_l,p_m\rangle=\theta(-p_l)(I\otimes {\cal
R})|p_l,p_m\rangle+\theta(p_l)|p_l,p_m\rangle.
\end{eqnarray}
Fortunately, for this operator we do not need any filtering
process, because the only important parameter is the direction of
particle's momentum.  So, the scheme of this gate, based on our
introduced basic gates and side channels, can be easily conceived
as shown in Fig.~\ref{fig57}.




 From the previous chapter remember that, in the BSM we have used
a position Hadamard operator represented in Eq.~(\ref{Hxn}). Here,
we introduce the same operator but in the momentum space, that is,
\begin{eqnarray}
H_{p_{m}}=\frac{1}{\sqrt{2}}({\cal R}_m+{\cal P}_m).
\end{eqnarray}
Since, similar to Eq.~(\ref{Hexplicit}), these kinds of operators
should be applied on all $p_m$'s, thus it is not necessary to
consider any filtering process at this stage.  So, the circuit
diagram of $H_{p_{1}}H_{p_{2}}\ldots H_{p_{M}}$ operator can be
shown as in Fig.~\ref{fig58}.
\begin{figure}[h]
\begin{center}\vspace{25mm}
\includegraphics[width=5cm,height=1.5cm]{fig58}\caption{Schematic diagram of a momentum Hadamard operator using two ${\cal R}$ and ${\cal P}$ basic gates.  Here, it is assumed that
the probability of passing the particle through the upper or lower
channel is the same.} \label{fig58}
\end{center}
\end{figure}


At the next stage, we need to design a quantum circuit for
$V_{(M)}$ operator which acts like $U_{(N)}$ operator introduced
in Eq.~(\ref{ghoul}), but in the momentum space.  Thus, we can
represent it as
\begin{eqnarray}\label{ghoulp}
V_{(M)}={\frac{1}{\sqrt{M}}} \sum_{m=0}^{M/2-1}\{D_{+}^{2m}\otimes
[\prod_{i=1}^{M}{\cal R}_{i}^{(h_{b,i}-h_{2n+i,i})/2}{\cal P}_{i}{\cal R}_{i}^{(h_{b,i}-h_{M-2m+i,i})/2}]D_{+}^{2m}\nonumber\\
+D_{+}^{2m+1}\otimes[\prod_{i=1}^{M}({\cal R}_{i}{\cal P}_{i}{\cal
R}_{i}{\cal P}_{i})^{(h_{b,i}-h_{M-2m-1+i,i})/2}]D_{+}^{2m+1}
\}.\nonumber\\
\end{eqnarray}
Now, if we rewrite this operator as
\begin{eqnarray}
V_{(M)}=\frac{1}{\sqrt{M}}\sum_{m=0}^{M-1}(D_{+}^{m}\otimes
B_{(m+1)}D_{+}^{m})
\end{eqnarray}
then it would be possible to design theoretically a quantum
circuit for this operator, very similar to the one performed for
$U_{(N)}$ operator in the previous chapter.  Figure ~\ref{fig59}
shows a conceivable quantum circuit of the $V_{(M)}$ operator for
just a typical T-channel, using the introduced momentum basic
gates.
\begin{figure}[t]
\begin{center}\vspace{40mm}
\includegraphics[width=11cm,height=5.5cm]{fig59}\caption{Quantum circuit diagram of $V_{(M)}$ operator fot the $n$-th T-channel.  Here, each branch of the
T-channel acts as a control channel for the $D_{+}^{n}$ operators
applied on the C-channels.} \label{fig59}
\end{center}
\end{figure}


\section{Conclusions}
In conclusion, we have proposed a theoretical scheme which
introduces $N$-level position state teleportation of a particle
having other degrees of freedom in three-dimensional space. We
have given representations for Bell states and necessary unitary
operators, using just symmetric normalized Hadamard matrices. This
scheme uses just one special EPR pair of particles and a planar
quantum scanner.  Therefore, if realized, this scheme can be used
to provide a complete wave function teleportation of a particle
including spin, $\psi_S(\mathbf{x})$, at large $N$'s, for the
first time.  We have shown that an entangled position-spin state
can be teleported using an EPR pair entangled in position and spin
variables separately, without the need to assume any kind of
entanglement between the two variables in the source. By the way,
as an advantage, the scheme needs only one kind of mechanism for
BSM in each spatial dimension.  So our scheme can be considered
economically better than the original one, which needed two kinds
of measurement on position and momentum spaces per spatial
dimension.  Furthermore, one can now think of moving towards
understanding teleportation of multi-particle objects, using EPR
sources which only emit the constituent particles of the object.
Today, the technology for teleporting states of individual atoms
is at hand.  For instance, the group led by Haroche \cite{Haroche}
in Paris has demonstrated entanglement of atoms.  The entanglement
of molecules and then their teleportation may reasonably be
expected within the next decades.  However, what happens beyond
that, is anybody's guess.

%\section*{Conclusions and Outlook}
%\addcontentsline{toc}{section}{Conclusions and Outlook}



%-----------------------------AppendixA.tex--------------------------------
{\appendix

\chapter{A clarification on the definition of center of mass coordinate of the EPR pair}

Here we want to show that the two different definitions for the
center of mass position of the two entangled particles, that is,
$y_1-y_2$ in Eq.~(\ref{incident}) and $y_1+y_2$ in
Eq.~(\ref{eqmotion}), are two consistent representations for the
center of mass coordinate.

At first, Concerning the center of mass coordinate $y_1-y_2$
represented in Eq.~(\ref{incident}), we can consider that
\begin{eqnarray}
&&{\bf{k}_1}.{\bf{y}_1}=|{\bf{k}_1}||{\bf{y}_1}|cos(\widehat{{\bf{k}_1},{\bf{y}_1}})=\pm|{\bf{k}_1}||{\bf{y}_1}|\nonumber\\
&&{\bf{k}_1}.{\bf{y}_2}=|{\bf{k}_1}||{\bf{y}_2}|cos(\widehat{{\bf{k}_1},{\bf{y}_2}})=\pm|{\bf{k}_1}||{\bf{y}_2}|.
\end{eqnarray}
In addition, we should assume that
$cos(\widehat{{\bf{k}_1},{\bf{y}_1}})=cos(\widehat{{\bf{k}_1},{\bf{y}_2}})$.
Because this guarantees that one of the entangled particles moves
into the positive and the other one into the negative direction,
which is a necessary constraint in the EPR
experiment.{\footnote{It is well known that a plane wave with
representation $e^{ikx}$ goes out from the origin of coordinate
while $e^{-ikx}$ goes into the origin.  When the two particles of
an EPR pair are considered at the one side of the origin on the
$x$-axis, one of them goes away from the origin and another
approaches to it.  On the other hand, if the origin is considered
between the two entangled particles, then the two particles either
go away from the origin or go into it.}} Then, the $y$ component
of Eq.~(\ref{incident}) can be rewritten in a clearer form, that
is
\begin{eqnarray}
\psi(y_1,y_2)&=&\int e^{i\bf{k}_1.\bf{y}_1}
e^{i\bf{k}_2.(\bf{y}_2+\bf{y}_0)} \delta ({\bf{k}}_1+{\bf{k}}_2)
d{\bf k}_1d{\bf k}_2\nonumber\\
&=&\int
e^{i\bf{k}_1.\bf{y}_1}e^{-i{\bf{k}}_1.(\bf{y}_2+\bf{y}_0)}d{\bf k}_1\nonumber\\
&=&\int^{+\infty}_{-\infty}e^{ik_1(|\bf{y}_1|-|\bf{y}_2|-|\bf{y}_0|)}dk_1=2\pi\delta
(|\bf{y}_1|-|\bf{y}_2|-|\bf{y}_0|)\nonumber\\
\end{eqnarray}
where $|\bf{y}_0|=|\bf{y}_1|-|\bf{y}_2|$ obviously represents the
center of mass position of the two particles (with a 1/2 factor)
if either $y_1>0$ and $y_2<0$, or vice versa. This means that it
is possible to have simultaneous eigenfunctions for the total
momentum and the center of mass coordinate for the two entangled
particles, as was previously shown by Einstein {\it et al.}
\cite{EPR}. Now, It is proper to select y$_0=0$. Therefore, in
Eq.~(\ref{incident}), $y_1-y_2$ really represents
$|\bf{y}_1|-|\bf{y}_2|$ which is clearly the center of mass
coordinate.  These arguments are also completely consistent with
the common realization of the EPR experiment.

On the other hand, in Eq.~(\ref{psiabt}) and consequently in
Eqs.~(\ref{eqmotion}) and (\ref{ppsi}), the $y$ parameters are
variables or vectors in one dimension, not absolute values. Thus,
in these equations, $\bf{y}_1+\bf{y}_2$$\equiv$$y_1(t)+y_2(t)$
naturally represents the center of mass coordinate of the two
particles. Again, one can show that having simultaneous
eigenfunction for the total momentum $\bf{p}_{y_1}+\bf{p}_{y_2}$
and the center of mass position $\bf{y}_1+\bf{y}_2$ for the two
entangled particles is feasible. To see this, consider the
following properties of our considered particles
\begin{eqnarray}
&&{\bf{p}_{y_1}}=p_{y_1}\widehat{j},\hspace{6mm}
{\bf{p}_{y_2}}=p_{y_2}\widehat{j}=-p_{y_1}\widehat{j}\nonumber\\
&&{\bf{y}}_1=y_1\widehat{j},\hspace{1cm}
{\bf{y}}_2=-y_2\widehat{j}
\end{eqnarray}
where similar to the EPR's wave function the center of mass
position is represented by
${\bf{y}}_1+{\bf{y}}_2=(y_1-y_2)\widehat{j}$, in which the values
of $y_1$ and $y_2$ have the same signs.  Now, it is easy to show
that
\begin{eqnarray}
[{\bf{p}_{y_1}}+{\bf{p}_{y_2}},{\bf{y}_1}+{\bf{y}_2}]=[p_{y_1}+p_{y_2},y_1-y_2]=0
\end{eqnarray}
which means it is possible to determine the total momentum and the
center of mass coordinate, simultaneously, for the two entangled
particles.


\chapter{Details on preparing and measuring processes
for some initial cases}

In this appendix, we have presented some details on preparing
(encoding) and measuring (decoding) processes in the dense coding
scheme introduced in Chapter 3 for $N=1$, 2, 4, 8 and 12 special
cases in order to clarify the procedure of obtaining some
relatively complex relations.  The explicit forms of the encoding
and decoding operators represented in the text using the basic
gates, are principally generalizations of the following
calculations.

For the $N=1$ case, there are four mutually entangled orthogonal
states as
\begin{eqnarray}
&&|\psi^{(1)}_{1,2}\rangle=\frac{1}{\sqrt{2}}[|1,-1\rangle\pm|-1,1\rangle] \label{psi1x}\\
&&|\psi^{(1)}_{3,4}\rangle=\frac{1}{\sqrt{2}}[|1,1\rangle\pm|-1,-1\rangle]
\label{phi1x}
\end{eqnarray}
where all required processes are clearly very similar to what can
be done for the well known system of two spin-$\frac{1}{2}$
particles.  The preparation and measuring processes for this case
are summarized in Tables \ref{tab1} and \ref{tab2}, respectively,
\begingroup
\begin{table}[h]\small
%\begin{ruledtabular}
\begin{center}
\caption{Alice's preparation process for $N=1$.}
\begin{tabular}{l@{}c}
\hline
 Transformation & New state\\
\hline
$(I\otimes I){|\psi^{(1)}_{1}\rangle}$&$|{\psi^{(1)}_{1}} \rangle$\\
$(N_{1}I_{s}\otimes I){|\psi^{(1)}_{1}\rangle}$&$|{\psi^{(1)}_{2}} \rangle$ \\
$(P_{1}I_{s}\otimes I){|\psi^{(1)}_{1}\rangle}$&$|{\psi^{(1)}_{3}} \rangle$\\
$(N_{1}P_{1}I_{s}\otimes I){|\psi^{(1)}_{1}\rangle}$&$|{\psi^{(1)}_{4}} \rangle$\\
\hline
\end{tabular}{\label{tab1}}
\end{center}
%\end{ruledtabular}
\end{table}
\begin{table}[h]\small
\begin{center}
\caption{Bob's measurement process for $N=1$.}
%\begin{ruledtabular}
\begin{tabular}{cl@{\hspace{5mm}}lc}
\hline
Initial state & PCS & $ H_{x_1}\otimes I $ & Renamed \\
\hline
${|\psi^{(1)}_{1}\rangle}$&${\frac{1}{2}}(|1\rangle+|-1\rangle)|-1\rangle $&$|1,-1\rangle$&01\\
${|\psi^{(1)}_{2}\rangle}$&${\frac{1}{2}}(|1\rangle-|-1\rangle)|-1\rangle $&$|-1,-1\rangle$&11\\
${|\psi^{(1)}_{3}\rangle}$&${\frac{1}{2}}(|1\rangle+|-1\rangle)|1\rangle $&$|1,1\rangle$&00\\
${|\psi^{(1)}_{4}\rangle}$&${\frac{1}{2}}(|1\rangle-|-1\rangle)|1\rangle $&$|-1,1\rangle$&10\\
\hline
\end{tabular}
\label{tab2}
%\end{ruledtabular}
\end{center}
\end{table}
\endgroup
where we have applied the renaming convention
\begin{eqnarray}{\label{conven}}
|n\rangle\equiv 0,~|-n\rangle\equiv 1
\end{eqnarray}
in the last column of Tab. \ref{tab2}.

For the $N=2$ case, we have 16 entangled orthonormal spatial
states as follows:
\begingroup
\small
\begin{table}[h]
\begin{center}
\caption{entangled orthonormal position states for $N=2$.}
\begin{tabular}{clllc}
\hline
$|\psi^{(2)}_{1,2}\rangle=\frac{1}{2}[(|1,-1\rangle\pm|-1,1\rangle)+(|2,-2\rangle\pm|-2,2\rangle)]$\\
$|\psi^{(2)}_{3,4}\rangle=\frac{1}{2}[(|1,-1\rangle\pm|-1,1\rangle)-(|2,-2\rangle\pm|-2,2\rangle)]$\\
\hline
$|\psi^{(2)}_{5,6}\rangle=\frac{1}{2}[(|1,1\rangle\pm|-1,-1\rangle)+(|2,2\rangle\pm|-2,-2\rangle)]$\\
$|\psi^{(2)}_{7,8}\rangle=\frac{1}{2}[(|1,1\rangle\pm|-1,-1\rangle)-(|2,2\rangle\pm|-2,-2\rangle)]$\\
\hline
$|\psi^{(2)}_{9,10}\rangle=\frac{1}{2}[(|2,-1\rangle\pm|-2,1\rangle)+(|1,-2\rangle\pm|-1,2\rangle)]$\\
$|\psi^{(2)}_{11,12}\rangle=\frac{1}{2}[(|2,-1\rangle\pm|-2,1\rangle)-(|1,-2\rangle\pm|-1,2\rangle)]$\\
\hline
$|\psi^{(2)}_{13,14}\rangle=\frac{1}{2}[(|2,1\rangle\pm|-2,-1\rangle)+(|1,2\rangle\pm|-1,-2\rangle)]$\\
$|\psi^{(2)}_{15,16}\rangle=\frac{1}{2}[(|2,1\rangle\pm|-2,-1\rangle)-(|1,2\rangle\pm|-1,-2\rangle)]$\\
\hline
\end{tabular}
\label{N=2x}
\end{center}
\end{table}
\endgroup
which form our Bell basis in the $N=2$ case.  Now, it is proper to
introduce the notion of family of states. By a family we mean sets
of states which share the same kets in their structure and at most
differ in their signs.  So in Tab.~\ref{N=2x}, four families can
be distinguished.  Tables ~\ref{AN2} and~\ref{BN2} illustrate all
preparation processes that can be selected and done by Alice, and
also Bob's measurement process, for each family, respectively.
\begingroup
\begin{table}[h]\small
\begin{center}
\caption{Alice's preparation process for $N=2$.}
%\begin{ruledtabular}
\begin{tabular}{lc}
\hline
Transformation&New state\\
\hline
$(I\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{1}\rangle}$\\
$(N_{1}N_{2}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{2}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{3}\rangle}$\\
$(P_{2}N_{1}N_{2}P_{2}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{4}\rangle}$\\
\hline
$(P_{1}P_{2}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{5}\rangle}$\\
$(N_{1}N_{2}P_{1}P_{2}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{6}\rangle}$\\
$(N_{2}P_{1}P_{2}N_{2}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{7}\rangle}$\\
$(N_{1}P_{1}P_{2}N_{2}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{8}\rangle}$\\
\hline
$(L_{+}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{9}\rangle}$\\
$(N_{1}N_{2}L_{+}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{10}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}L_{+}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{11}\rangle}$\\
$(P_{2}N_{1}N_{2}P_{2}L_{+}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{12}\rangle}$\\
\hline
$(P_{1}P_{2}L_{+}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{13}\rangle}$\\
$(N_{1}N_{2}P_{1}P_{2}L_{+}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{14}\rangle}$\\
$(N_{2}P_{1}P_{2}N_{2}L_{+}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{15}\rangle}$\\
$(N_{1}P_{1}P_{2}N_{2}L_{+}\otimes I){|\psi^{(2)}_{1}\rangle}$&${|\psi^{(2)}_{16}\rangle}$\\
\hline
\end{tabular}
\label{AN2}
%\end{ruledtabular}
\end{center}
\end{table}
\endgroup
%%%%%%%%%%%%%%%%%%%%%%%%%
\begingroup
\begin{table}[h]
\tiny
\begin{center}
\caption{Bob's measurement process for $N=2$}
%\begin{ruledtabular}
\begin{tabular}{clllc}
\hline Initial state&PCS&$H_{x_{1}}H_{x_{2}}\otimes
I$&$U_{(2)}$&Renamed\\
\hline
${|\psi^{(2)}_{1}\rangle}$&${1\over
2}[(|1\rangle+|-1\rangle)|-1\rangle+(|2\rangle+|-2\rangle)|-2\rangle]$&${1
\over
 \sqrt{2}}[|1,-1\rangle+|2,-2\rangle]$&$|1,-1\rangle$&0$\sqcup$1$\sqcup$\\
${|\psi^{(2)}_{2}\rangle}$&${1\over
 2}[(|1\rangle-|-1\rangle)|-1\rangle+(|2\rangle-|-2\rangle)|-2\rangle]$&${1 \over
 \sqrt{2}}[|-1,-1\rangle+|-2,-2\rangle]$&$|-1,-1\rangle$&1$\sqcup$1$\sqcup$\\
${|\psi^{(2)}_{3}\rangle}$&${1\over
 2}[(|1\rangle+|-1\rangle)|-1\rangle-(|2\rangle+|-2\rangle)|-2\rangle]$&${1 \over
 \sqrt{2}}[|1,-1\rangle-|2,-2\rangle]$&$|2,-2\rangle$&$\sqcup$0$\sqcup$1\\
${|\psi^{(2)}_{4}\rangle}$&${1\over
 2}[(|1\rangle-|-1\rangle)|-1\rangle-(|2\rangle-|-2\rangle)|-2\rangle]$&${1 \over
 \sqrt{2}}[|-1,-1\rangle-|-2,-2\rangle]$&$|-2,-2\rangle$&$\sqcup$1$\sqcup$1\\
\hline
${|\psi^{(2)}_{5}\rangle}$&${1\over
 2}[(|1\rangle+|-1\rangle)|1\rangle+(|2\rangle+|-2\rangle)|2\rangle]$&${1 \over
 \sqrt{2}}[|1,1\rangle+|2,2\rangle]$&$|2,2\rangle$&$\sqcup$0$\sqcup$0\\
${|\psi^{(2)}_{6}\rangle}$&${1\over
 2}[(|1\rangle-|-1\rangle)|1\rangle+(|2\rangle-|-2\rangle)|2\rangle]$&${1 \over
 \sqrt{2}}[|-1,1\rangle+|-2,2\rangle]$&$|-2,2\rangle$&$\sqcup$1$\sqcup$0\\
${|\psi^{(2)}_{7}\rangle}$&${1\over
 2}[(|1\rangle+|-1\rangle)|1\rangle-(|2\rangle+|-2\rangle)|2\rangle]$&${1 \over
 \sqrt{2}}[|1,1\rangle-|2,2\rangle]$&$|1,1\rangle$&0$\sqcup$0$\sqcup$\\
${|\psi^{(2)}_{8}\rangle}$&${1\over
 2}[(|1\rangle-|-1\rangle)|1\rangle-(|2\rangle-|-2\rangle)|2\rangle]$&${1 \over
 \sqrt{2}}[|-1,1\rangle-|-2,2\rangle]$&$|-1,1\rangle$&1$\sqcup$0$\sqcup$\\
\hline
${|\psi^{(2)}_{9}\rangle}$&${1\over
 2}[(|1\rangle+|-1\rangle)|-2\rangle+(|2\rangle+|-2\rangle)|-1\rangle]$&${1 \over
 \sqrt{2}}[|2,-1\rangle+|1,-2\rangle]$&$|2,-1\rangle$&$\sqcup$01$\sqcup$\\
${|\psi^{(2)}_{10}\rangle}$&${1\over
 2}[(|1\rangle-|-1\rangle)|-2\rangle+(|2\rangle-|-2\rangle)|-1\rangle]$&${1 \over
 \sqrt{2}}[|-2,-1\rangle+|-1,-2\rangle]$&$|-2,-1\rangle$&$\sqcup$11$\sqcup$\\
${|\psi^{(2)}_{11}\rangle}$&${1\over
 2}[(|1\rangle+|-1\rangle)|-2\rangle-(|2\rangle+|-2\rangle)|-1\rangle]$&${1 \over
 \sqrt{2}}[|2,-1\rangle-|1,-2\rangle]$&$|1,-2\rangle$&0$\sqcup\sqcup$1\\
${|\psi^{(2)}_{12}\rangle}$&${1\over
 2}[(|1\rangle-|-1\rangle)|-2\rangle-(|2\rangle-|-2\rangle)|-1\rangle]$&${1 \over
 \sqrt{2}}[|-2,-1\rangle-|-1,-2\rangle]$&$|-1,-2\rangle$&1$\sqcup\sqcup$1\\
\hline
${|\psi^{(2)}_{13}\rangle}$&${1\over
 2}[(|1\rangle+|-1\rangle)|2\rangle+(|2\rangle+|-2\rangle)|1\rangle]$&${1 \over
 \sqrt{2}}[|2,1\rangle+|1,2\rangle]$&$|1,2\rangle$&0$\sqcup\sqcup$0\\
${|\psi^{(2)}_{14}\rangle}$&${1\over
 2}[(|1\rangle-|-1\rangle)|2\rangle+(|2\rangle-|-2\rangle)|1\rangle]$&${1 \over
 \sqrt{2}}[|-2,1\rangle+|-1,2\rangle]$&$|-1,2\rangle$&1$\sqcup\sqcup$0\\
${|\psi^{(2)}_{15}\rangle}$&${1\over
 2}[(|1\rangle+|-1\rangle)|2\rangle-(|2\rangle+|-2\rangle)|1\rangle]$&${1 \over
 \sqrt{2}}[|2,1\rangle-|1,2\rangle]$&$|2,1\rangle$&$\sqcup$00$\sqcup$\\
${|\psi^{(2)}_{16}\rangle}$&${1\over
 2}[(|1\rangle-|-1\rangle)|2\rangle-(|2\rangle-|-2\rangle)|1\rangle]$&${1 \over
 \sqrt{2}}[|-2,1\rangle-|-1,2\rangle]$&$|-2,1\rangle$&$\sqcup$10$\sqcup$\\
\hline
\end{tabular}
\label{BN2}
%\end{ruledtabular}
\end{center}
\end{table}
\endgroup
In Tab.~\ref{BN2} it is seen what differs from the $N=1$ case is
that applying just $H_{x_{1}}H_{x_{2}}$ does not produce Bob's
desired orthonormal spatial states.  Therefore, we have applied a
new unitary operator in the form
\begin{eqnarray}{\label{grandH}}
U_{(2)}=\frac{1}{\sqrt{2}}[(I\otimes
P_{1}N_{1}N_{2}P_{1})+(L_{+}\otimes L_{+})]
\end{eqnarray}
which transforms the processed orthonormal states to the spatial
bases which are manageable by Bob.  Furthermore, in the last
column of Tab.~\ref{BN2}, the following renaming convention is
added to the previous one in Eq.~(\ref{conven}).  At first,
according to Fig.~{\ref{dfig7}}, Bob prepares a sequence of blank
spaces the number of which is
 equal to his receivers, that is, $2N$.  Now, the first half of these spaces belong to the
particle
 received from Alice and the second half belong to his own particle. Then, he observes the outcomes of all C-channels.
 If one of them, e.g. $i$-th C-channel, is ON ($i\in (-N,\ldots, N) $), he fills the $|i|-$th space with 1
 or 0, according to Eq.~(\ref{conven}), otherwise all the remaining spaces are left blank;
 $\sqcup$. Then, he scans all T-channels and in the same manner he
 fills one of the blank spaces.  Finally, Bob obtains a $2N$
 character strip which obtains 2 digits including 0 and/or 1 as
 well as $(2N-2)$ $\sqcup$'s.

\begin{figure}[h]
\begin{center}\vspace{15mm}
\includegraphics[height=1.5cm,width=7cm,angle=0]{dFIG7.EPS}
\caption{ Initial (OFF) state of Bob's detector.} \label{dfig7}
\end{center}
\end{figure}

At the next stage, it is natural to consider the $N=3$ case. Here,
one may expect that the foregoing mechanism works for this case,
too. As we have seen, the first step is to find all entangled
orthonormal states, like Eq.~(\ref{N=2x}). At the first glance,
one may consider
\begin{eqnarray}{\label{Psi3}}
|\psi^{(3)}_{1,2}\rangle=
\frac{1}{\sqrt{6}}[(|1,-1\rangle\pm|-1,1\rangle)+(|2,-2\rangle\pm|-2,2\rangle)+(|3,-3\rangle\pm|-3,3\rangle)]\nonumber\\
\end{eqnarray}
 as examples of two entangled orthonormal states of a family. But, the number of sets of
  parenthesis, i.e. ( ), in them is odd, so that constructing more entangled orthonormal states for this family,
  by permuting $\pm$ signs, is impossible.  So, it is seen that there is not any
other member
 for it.  This means that the size of this family is smaller than the case of $N=2$,
 in Tab.~\ref{N=2x}.  Thus, we do not have sufficient entangled orthonormal states to
 perform, say, dense coding process in this case. In other words, we only have
12 entangled orthonormal position states. A similar argumentation
also works for the case of odd $N$.

For the next case, consider $N=4$.  All of the 64 entangled
orthonormal position states, classified in 8 families, are listed
in Tab.~{\ref{tab6}}.
\begingroup
\begin{table}\tiny
\begin{center}
\caption{Entangled orthogonal position states for $N=4$.}
%\begin{ruledtabular}
\begin{tabular}{lc}
\hline\hline
%\begin{ruledtabular}
%\begin{tabular}{c}
%\endfirsthead
%\caption[]{(continued)}\\
%\hline\hline
%\endhead
$|\psi^{(4)}_{1,2}\rangle=\frac{1}{\sqrt{8}}[(|1,-1\rangle\pm|-1,1\rangle)+(|2,-2\rangle\pm|-2,2\rangle)+(|3,-3\rangle\pm|-3,3\rangle)+(|4,-4\rangle\pm|-4,4\rangle)] $\\
$|\psi^{(4)}_{3,4}\rangle=\frac{1}{\sqrt{8}}[(|1,-1\rangle\pm|-1,1\rangle)-(|2,-2\rangle\pm|-2,2\rangle)-(|3,-3\rangle\pm|-3,3\rangle)+(|4,-4\rangle\pm|-4,4\rangle)]$\\
$|\psi^{(4)}_{5,6}\rangle=\frac{1}{\sqrt{8}}[(|1,-1\rangle\pm|-1,1\rangle)-(|2,-2\rangle\pm|-2,2\rangle)+(|3,-3\rangle\pm|-3,3\rangle)-(|4,-4\rangle\pm|-4,4\rangle)] $\\
$|\psi^{(4)}_{7,8}\rangle=\frac{1}{\sqrt{8}}[(|1,-1\rangle\pm|-1,1\rangle)+(|2,-2\rangle\pm|-2,2\rangle)-(|3,-3\rangle\pm|-3,3\rangle)-(|4,-4\rangle\pm|-4,4\rangle)]$ \\
\hline
$|\psi^{(4)}_{9,10}\rangle=\frac{1}{\sqrt{8}}[(|1,1\rangle\pm|-1,-1\rangle)+(|2,2\rangle\pm|-2,-2\rangle)+(|3,3\rangle\pm|-3,-3\rangle)+(|4,4\rangle\pm|-4,-4\rangle)]$ \\
$|\psi^{(4)}_{11,12}\rangle=\frac{1}{\sqrt{8}}[(|1,1\rangle\pm|-1,-1\rangle)-(|2,2\rangle\pm|-2,-2\rangle)-(|3,3\rangle\pm|-3,-3\rangle)+(|4,4\rangle\pm|-4,-4\rangle)]$ \\
$|\psi^{(4)}_{13,14}\rangle=\frac{1}{\sqrt{8}}[(|1,1\rangle\pm|-1,-1\rangle)-(|2,2\rangle\pm|-2,-2\rangle)+(|3,3\rangle\pm|-3,-3\rangle)-(|4,4\rangle\pm|-4,-4\rangle)]$ \\
$|\psi^{(4)}_{15,16}\rangle=\frac{1}{\sqrt{8}}[(|1,1\rangle\pm|-1,-1\rangle)+(|2,2\rangle\pm|-2,-2\rangle)-(|3,3\rangle\pm|-3,-3\rangle)-(|4,4\rangle\pm|-4,-4\rangle)] $\\
\hline
$|\psi^{(4)}_{17,18}\rangle=\frac{1}{\sqrt{8}}[(|1,-2\rangle\pm|-1,2\rangle)+(|2,-3\rangle\pm|-2,3\rangle)+(|3,-4\rangle\pm|-3,4\rangle)+(|4,-1\rangle\pm|-4,1\rangle)] $\\
$|\psi^{(4)}_{19,20}\rangle=\frac{1}{\sqrt{8}}[(|1,-2\rangle\pm|-1,2\rangle)-(|2,-3\rangle\pm|-2,3\rangle)-(|3,-4\rangle\pm|-3,4\rangle)+(|4,-1\rangle\pm|-4,1\rangle)] $\\
$|\psi^{(4)}_{21,22}\rangle=\frac{1}{\sqrt{8}}[(|1,-2\rangle\pm|-1,2\rangle)-(|2,-3\rangle\pm|-2,3\rangle)+(|3,-4\rangle\pm|-3,4\rangle)-(|4,-1\rangle\pm|-4,1\rangle)] $\\
$|\psi^{(4)}_{23,24}\rangle=\frac{1}{\sqrt{8}}[(|1,-2\rangle\pm|-1,2\rangle)+(|2,-3\rangle\pm|-2,3\rangle)-(|3,-4\rangle\pm|-3,4\rangle)-(|4,-1\rangle\pm|-4,1\rangle)] $\\
\hline
$|\psi^{(4)}_{25,26}\rangle=\frac{1}{\sqrt{8}}[(|1,2\rangle\pm|-1,-2\rangle)+(|2,3\rangle\pm|-2,-3\rangle)+(|3,4\rangle\pm|-3,-4\rangle)+(|4,1\rangle\pm|-4,-1\rangle)]$\\
$|\psi^{(4)}_{27,28}\rangle=\frac{1}{\sqrt{8}}[(|1,2\rangle\pm|-1,-2\rangle)-(|2,3\rangle\pm|-2,-3\rangle)-(|3,4\rangle\pm|-3,-4\rangle)+(|4,1\rangle\pm|-4,-1\rangle)]$\\
$|\psi^{(4)}_{29,30}\rangle=\frac{1}{\sqrt{8}}[(|1,2\rangle\pm|-1,-2\rangle)-(|2,3\rangle\pm|-2,-3\rangle)+(|3,4\rangle\pm|-3,-4\rangle)-(|4,1\rangle\pm|-4,-1\rangle)]$\\
$|\psi^{(4)}_{31,32}\rangle=\frac{1}{\sqrt{8}}[(|1,2\rangle\pm|-1,-2\rangle)+(|2,3\rangle\pm|-2,-3\rangle)-(|3,4\rangle\pm|-3,-4\rangle)-(|4,1\rangle\pm|-4,-1\rangle)]$\\
\hline
$|\psi^{(4)}_{33,34}\rangle=\frac{1}{\sqrt{8}}[(|1,-3\rangle\pm|-1,3\rangle)+(|2,-4\rangle\pm|-2,4\rangle)+(|3,-1\rangle\pm|-3,1\rangle)+(|4,-2\rangle\pm|-4,2\rangle)] $\\
$|\psi^{(4)}_{35,36}\rangle=\frac{1}{\sqrt{8}}[(|1,-3\rangle\pm|-1,3\rangle)-(|2,-4\rangle\pm|-2,4\rangle)-(|3,-1\rangle\pm|-3,1\rangle)+(|4,-2\rangle\pm|-4,2\rangle)] $\\
$|\psi^{(4)}_{37,38}\rangle=\frac{1}{\sqrt{8}}[(|1,-3\rangle\pm|-1,3\rangle)-(|2,-4\rangle\pm|-2,4\rangle)+(|3,-1\rangle\pm|-3,1\rangle)-(|4,-2\rangle\pm|-4,2\rangle)]$ \\
$|\psi^{(4)}_{39,40}\rangle=\frac{1}{\sqrt{8}}[(|1,-3\rangle\pm|-1,3\rangle)+(|2,-4\rangle\pm|-2,4\rangle)-(|3,-1\rangle\pm|-3,1\rangle)-(|4,-2\rangle\pm|-4,2\rangle)] $\\
\hline
$|\psi^{(4)}_{41,42}\rangle=\frac{1}{\sqrt{8}}[(|1,3\rangle\pm|-1,-3\rangle)+(|2,4\rangle\pm|-2,-4\rangle)+(|3,1\rangle\pm|-3,-1\rangle)+(|4,2\rangle\pm|-4,-2\rangle)] $\\
$|\psi^{(4)}_{43,44}\rangle=\frac{1}{\sqrt{8}}[(|1,3\rangle\pm|-1,-3\rangle)-(|2,4\rangle\pm|-2,-4\rangle)-(|3,1\rangle\pm|-3,-1\rangle)+(|4,2\rangle\pm|-4,-2\rangle)]$ \\
$|\psi^{(4)}_{45,46}\rangle=\frac{1}{\sqrt{8}}[(|1,3\rangle\pm|-1,-3\rangle)-(|2,4\rangle\pm|-2,-4\rangle)+(|3,1\rangle\pm|-3,-1\rangle)-(|4,2\rangle\pm|-4,-2\rangle)]$\\
$|\psi^{(4)}_{47,48}\rangle=\frac{1}{\sqrt{8}}[(|1,3\rangle\pm|-1,-3\rangle)+(|2,4\rangle\pm|-2,-4\rangle)-(|3,1\rangle\pm|-3,-1\rangle)-(|4,2\rangle\pm|-4,-2\rangle)]$\\
\hline
$|\psi^{(4)}_{49,50}\rangle=\frac{1}{\sqrt{8}}[(|1,-4\rangle\pm|-1,4\rangle)+(|2,-1\rangle\pm|-2,1\rangle)+(|3,-2\rangle\pm|-3,2\rangle)+(|4,-3\rangle\pm|-4,3\rangle)] $\\
$|\psi^{(4)}_{51,52}\rangle=\frac{1}{\sqrt{8}}[(|1,-4\rangle\pm|-1,4\rangle)-(|2,-1\rangle\pm|-2,1\rangle)-(|3,-2\rangle\pm|-3,2\rangle)+(|4,-3\rangle\pm|-4,3\rangle)]$\\
$|\psi^{(4)}_{53,54}\rangle=\frac{1}{\sqrt{8}}[(|1,-4\rangle\pm|-1,4\rangle)-(|2,-1\rangle\pm|-2,1\rangle)+(|3,-2\rangle\pm|-3,2\rangle)-(|4,-3\rangle\pm|-4,3\rangle)] $\\
$|\psi^{(4)}_{55,56}\rangle=\frac{1}{\sqrt{8}}[(|1,-4\rangle\pm|-1,4\rangle)+(|2,-1\rangle\pm|-2,1\rangle)-(|3,-2\rangle\pm|-3,2\rangle)-(|4,-3\rangle\pm|-4,3\rangle)]$ \\
\hline
$|\psi^{(4)}_{57,58}\rangle=\frac{1}{\sqrt{8}}[(|1,4\rangle\pm|-1,-4\rangle)+(|2,1\rangle\pm|-2,-1\rangle)+(|3,2\rangle\pm|-3,-2\rangle)+(|4,3\rangle\pm|-4,-3\rangle)] $\\
$|\psi^{(4)}_{59,60}\rangle=\frac{1}{\sqrt{8}}[(|1,4\rangle\pm|-1,-4\rangle)-(|2,1\rangle\pm|-2,-1\rangle)-(|3,2\rangle\pm|-3,-2\rangle)+(|4,3\rangle\pm|-4,-3\rangle)] $\\
$|\psi^{(4)}_{61,62}\rangle=\frac{1}{\sqrt{8}}[(|1,4\rangle\pm|-1,-4\rangle)-(|2,1\rangle\pm|-2,-1\rangle)+(|3,2\rangle\pm|-3,-2\rangle)-(|4,3\rangle\pm|-4,-3\rangle)] $\\
$|\psi^{(4)}_{63,64}\rangle=\frac{1}{\sqrt{8}}[(|1,4\rangle\pm|-1,-4\rangle)+(|2,1\rangle\pm|-2,-1\rangle)-(|3,2\rangle\pm|-3,-2\rangle)-(|4,3\rangle\pm|-4,-3\rangle)] $\\
\hline\hline
%\end{tabular}
{\label{tab6}}
\end{tabular}
%\end{ruledtabular}
\end{center}
\end{table}
\endgroup
Now, Alice can apply proper combinations of the basic gates on her
particle to produce other members of the Bell bases.  Table
\ref{tab7} shows all preparation that can be selected and
performed by Alice for each family.
\begingroup
\begin{table}
\vskip-9mm\tiny
\begin{center}
\caption{Alice's preparation process for $N=4$.}
%\begin{ruledtabular}
\begin{tabular}{lc}
\hline
Transformation&New state\\
\hline
$(I\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{1}\rangle}$\\
$(N_{1}N_{2}N_{3}N_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{2}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{3}N_{3}P_{3}N_{3}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{3}\rangle}$\\
$(N_{1}N_{4}P_{2}N_{2}P_{2}P_{3}N_{3}P_{3}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{4}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{4}N_{4}P_{4}N_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{5}\rangle}$\\
$(N_{1}N_{3}P_{2}N_{2}P_{2}P_{4}N_{4}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{6}\rangle}$\\
$(P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{7}\rangle}$\\
$(N_{1}N_{2}P_{3}N_{3}P_{3}P_{4}N_{4}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{8}\rangle}$\\
\hline
$(P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{9}\rangle}$\\
$(N_{1}N_{2}N_{3}N_{4}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{10}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{3}N_{3}P_{3}N_{3}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{11}\rangle}$\\
$(N_{1}N_{4}P_{2}N_{2}P_{2}P_{3}N_{3}P_{3}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{12}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{4}N_{4}P_{4}N_{4}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{13}\rangle}$\\
$(N_{1}N_{3}P_{2}N_{2}P_{2}P_{4}N_{4}P_{4}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{14}\rangle}$\\
$(P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{15}\rangle}$\\
$(N_{1}N_{2}P_{3}N_{3}P_{3}P_{4}N_{4}P_{4}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{16}\rangle}$\\
\hline
$(L^{3}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{17}\rangle}$\\
$(N_{1}N_{2}N_{3}N_{4}L^{3}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{18}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{3}N_{3}P_{3}N_{3}L^{3}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{19}\rangle}$\\
$(N_{1}N_{4}P_{2}N_{2}P_{2}P_{3}N_{3}P_{3}L^{3}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{20}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{4}N_{4}P_{4}N_{4}L^{3}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{21}\rangle}$\\
$(N_{1}N_{3}P_{2}N_{2}P_{2}P_{4}N_{4}P_{4}L^{3}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{22}\rangle}$\\
$(P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}L^{3}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{23}\rangle}$\\
$(N_{1}N_{2}P_{3}N_{3}P_{3}P_{4}N_{4}P_{4}L^{3}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{24}\rangle}$\\
\hline
$(L^{3}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{25}\rangle}$\\
$(N_{1}N_{2}N_{3}N_{4}L^{3}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{26}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{3}N_{3}P_{3}N_{3}L^{3}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{27}\rangle}$\\
$(N_{1}N_{4}P_{2}N_{2}P_{2}P_{3}N_{3}P_{3}L^{3}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{28}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{4}N_{4}P_{4}N_{4}L^{3}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{29}\rangle}$\\
$(N_{1}N_{3}P_{2}N_{2}P_{2}P_{4}N_{4}P_{4}L^{3}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{30}\rangle}$\\
$(P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}L^{3}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{31}\rangle}$\\
$(N_{1}N_{2}P_{3}N_{3}P_{3}P_{4}N_{4}P_{4}L^{3}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{32}\rangle}$\\
\hline
$(L^{2}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{33}\rangle}$\\
$(N_{1}N_{2}N_{3}N_{4}L^{2}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{34}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{3}N_{3}P_{3}N_{3}L^{2}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{35}\rangle}$\\
$(N_{1}N_{4}P_{2}N_{2}P_{2}P_{3}N_{3}P_{3}L^{2}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{36}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{4}N_{4}P_{4}N_{4}L^{2}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{37}\rangle}$\\
$(N_{1}N_{3}P_{2}N_{2}P_{2}P_{4}N_{4}P_{4}L^{2}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{38}\rangle}$\\
$(P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}L^{2}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{39}\rangle}$\\
$(N_{1}N_{2}P_{3}N_{3}P_{3}P_{4}N_{4}P_{4}L^{2}_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{40}\rangle}$\\
\hline
$(L^{2}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{41}\rangle}$\\
$(N_{1}N_{2}N_{3}N_{4}L^{2}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{42}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{3}N_{3}P_{3}N_{3}L^{2}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{43}\rangle}$\\
$(N_{1}N_{4}P_{2}N_{2}P_{2}P_{3}N_{3}P_{3}L^{2}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{44}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{4}N_{4}P_{4}N_{4}L^{2}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{45}\rangle}$\\
$(N_{1}N_{3}P_{2}N_{2}P_{2}P_{4}N_{4}P_{4}L^{2}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{46}\rangle}$\\
$(P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}L^{2}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{47}\rangle}$\\
$(N_{1}N_{2}P_{3}N_{3}P_{3}P_{4}N_{4}P_{4}L^{2}_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{48}\rangle}$\\
\hline
$(L_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{49}\rangle}$\\
$(N_{1}N_{2}N_{3}N_{4}L_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{50}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{3}N_{3}P_{3}N_{3}L_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{51}\rangle}$\\
$(N_{1}N_{4}P_{2}N_{2}P_{2}P_{3}N_{3}P_{3}L_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{52}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{4}N_{4}P_{4}N_{4}L_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{53}\rangle}$\\
$(N_{1}N_{3}P_{2}N_{2}P_{2}P_{4}N_{4}P_{4}L_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{54}\rangle}$\\
$(P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}L_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{55}\rangle}$\\
$(N_{1}N_{2}P_{3}N_{3}P_{3}P_{4}N_{4}P_{4}L_{+}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{56}\rangle}$\\
\hline
$(L_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{57}\rangle}$\\
$(N_{1}N_{2}N_{3}N_{4}L_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{58}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{3}N_{3}P_{3}N_{3}L_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{59}\rangle}$\\
$(N_{1}N_{4}P_{2}N_{2}P_{2}P_{3}N_{3}P_{3}L_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{60}\rangle}$\\
$(P_{2}N_{2}P_{2}N_{2}P_{4}N_{4}P_{4}N_{4}L_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{61}\rangle}$\\
$(N_{1}N_{3}P_{2}N_{2}P_{2}P_{4}N_{4}P_{4}L_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{62}\rangle}$\\
$(P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}L_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{63}\rangle}$\\
$(N_{1}N_{2}P_{3}N_{3}P_{3}P_{4}N_{4}P_{4}L_{+}P_{1}P_{2}P_{3}P_{4}\otimes I){|\psi^{(4)}_{1}\rangle}$&${|\psi^{(4)}_{64}\rangle}$\\
\hline \hline \label{tab7}
\end{tabular}
%\end{ruledtabular}
\end{center}
\end{table}
\endgroup
After receiving Alice's processed particle by Bob, he perform a
BSM on the two particle, as is concisely shown in Tab.~\ref{tab8}.
\begingroup
\begin{table}
\vskip-9mm \tiny
\begin{center}
\caption{Bob's measurement process for $N=4$.}
%\begin{ruledtabular}
\begin{tabular}{clllc}
\hline Initial state&$(H_{x_{1}}H_{x_{2}}H_{x_{3}}H_{x_{4}}\otimes
I)$PCS&$U_{(4)}$&Renamed\\
\hline
${|\psi^{(4)}_{1}\rangle}_{x}$&${1 \over 2}[|1,-1\rangle+|2,-2\rangle+|3,-3\rangle+|4,-4\rangle]$&$|1,-1\rangle$&$0 \sqcup\sqcup\sqcup 1 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{2}\rangle}_{x}$&${1 \over 2}[|-1,-1\rangle+|-2,-2\rangle+|-3,-3\rangle+|-4,-4\rangle]$&$|-1,-1\rangle$&$1\sqcup\sqcup\sqcup 1 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{3}\rangle}_{x}$&${1 \over 2}[|1,-1\rangle-|2,-2\rangle-|3,-3\rangle+|4,-4\rangle]$&$|4,-4\rangle$&$\sqcup\sqcup\sqcup 0 \sqcup\sqcup\sqcup 1$\\
${|\psi^{(4)}_{4}\rangle}_{x}$&${1 \over 2}[|-1,-1\rangle-|-2,-2\rangle-|-3,-3\rangle+|-4,-4\rangle]$&$|-4,-4\rangle$&$\sqcup\sqcup\sqcup 1 \sqcup\sqcup\sqcup 1$\\
${|\psi^{(4)}_{5}\rangle}_{x}$&${1 \over 2}[|1,-1\rangle-|2,-2\rangle+|3,-3\rangle-|4,-4\rangle]$&$|2,-2\rangle$&$\sqcup 0 \sqcup\sqcup\sqcup 1 \sqcup\sqcup$\\
${|\psi^{(4)}_{6}\rangle}_{x}$&${1 \over 2}[|-1-,1\rangle-|-2,-2\rangle+|-3,-3\rangle-|-4,-4\rangle]$&$|-2,-2\rangle$&$\sqcup 1 \sqcup\sqcup\sqcup 1 \sqcup\sqcup$\\
${|\psi^{(4)}_{7}\rangle}_{x}$&${1 \over 2}[|1,-1\rangle+|2,-2\rangle-|3,-3\rangle-|4,-4\rangle]$&$|3,-3\rangle$&$\sqcup\sqcup 0 \sqcup\sqcup\sqcup 1 \sqcup$\\
${|\psi^{(4)}_{8}\rangle}_{x}$&${1 \over 2}[|-1,-1\rangle+|-2,-2\rangle-|-3,-3\rangle-|-4,-4\rangle]$&$|-3,-3\rangle$&$\sqcup\sqcup 1 \sqcup\sqcup\sqcup 1 \sqcup$\\
\hline
${|\psi^{(4)}_{9}\rangle}_{x}$&${1 \over 2}[|1,1\rangle+|2,2\rangle+|3,3\rangle+|4,4\rangle]$&$|2,2\rangle$&$\sqcup 0 \sqcup\sqcup\sqcup 0 \sqcup\sqcup$\\
${|\psi^{(4)}_{10}\rangle}_{x}$&${1 \over 2}[|-1,1\rangle+|-2,2\rangle+|-3,3\rangle+|-4,4\rangle]$&$|-2,2\rangle$&$\sqcup 1 \sqcup\sqcup\sqcup 0 \sqcup\sqcup$\\
${|\psi^{(4)}_{11}\rangle}_{x}$&${1 \over 2}[|1,1\rangle-|2,2\rangle-|3,3\rangle+|4,4\rangle]$&$|3,3\rangle$&$\sqcup\sqcup 0 \sqcup\sqcup\sqcup 0 \sqcup$\\
${|\psi^{(4)}_{12}\rangle}_{x}$&${1 \over 2}[|-1,1\rangle-|-2,2\rangle-|-3,3\rangle+|-4,4\rangle]$&$|-3,3\rangle$&$\sqcup\sqcup 1 \sqcup\sqcup\sqcup 0 \sqcup$\\
${|\psi^{(4)}_{13}\rangle}_{x}$&${1 \over 2}[|1,1\rangle-|2,2\rangle+|3,3\rangle-|4,4\rangle]$&$|1,1\rangle$&$0 \sqcup\sqcup\sqcup 0 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{14}\rangle}_{x}$&${1 \over 2}[|-1,1\rangle-|-2,2\rangle+|-3,3\rangle-|-4,4\rangle]$&$|-1,1\rangle$&$1 \sqcup\sqcup\sqcup 0 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{15}\rangle}_{x}$&${1 \over 2}[|1,1\rangle+|2,2\rangle-|3,3\rangle-|4,4\rangle]$&$|4,4\rangle$&$\sqcup\sqcup\sqcup 0 \sqcup\sqcup\sqcup 0$\\
${|\psi^{(4)}_{16}\rangle}_{x}$&${1 \over 2}[|-1,1\rangle+|-2,2\rangle-|-3,3\rangle-|-4,4\rangle]$&$|-4,4\rangle$&$\sqcup\sqcup\sqcup 1 \sqcup\sqcup\sqcup 0$\\
\hline
${|\psi^{(4)}_{17}\rangle}_{x}$&${1 \over 2}[|1,-2\rangle+|2,-3\rangle+|3,-4\rangle+|4,-1\rangle]$&$|4,-1\rangle$&$\sqcup\sqcup\sqcup 0 1 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{18}\rangle}_{x}$&${1 \over 2}[|-1,-2\rangle+|-2,-3\rangle+|-3,-4\rangle+|-4,-1\rangle]$&$|-4,-1\rangle$&$\sqcup\sqcup\sqcup 1 1 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{19}\rangle}_{x}$&${1 \over 2}[|1,-2\rangle-|2,-3\rangle-|3,-4\rangle+|4,-1\rangle]$&$|2,-3\rangle$&$\sqcup0 \sqcup\sqcup\sqcup\sqcup 1 \sqcup$\\
${|\psi^{(4)}_{20}\rangle}_{x}$&${1 \over 2}[|-1,-2\rangle-|-2,-3\rangle-|-3,-4\rangle+|-4,-1\rangle]$&$|-2,-3\rangle$&$\sqcup 1 \sqcup\sqcup\sqcup\sqcup 1 \sqcup$\\
${|\psi^{(4)}_{21}\rangle}_{x}$&${1 \over 2}[|1,-2\rangle-|2,-3\rangle+|3,-4\rangle-|4,-1\rangle]$&$|1,-2\rangle$&$0 \sqcup\sqcup\sqcup\sqcup 1 \sqcup\sqcup$\\
${|\psi^{(4)}_{22}\rangle}_{x}$&${1 \over 2}[|-1,-2\rangle-|-2,-3\rangle+|-3,-4\rangle-|-4,-1\rangle]$&$|-1,-2\rangle$&$1 \sqcup\sqcup\sqcup\sqcup 1 \sqcup\sqcup$\\
${|\psi^{(4)}_{23}\rangle}_{x}$&${1 \over 2}[|1,-2\rangle+|2,-3\rangle-|3,-4\rangle-|4,-1\rangle]$&$|3,-4\rangle$&$\sqcup\sqcup 0 \sqcup\sqcup\sqcup\sqcup 1$\\
${|\psi^{(4)}_{24}\rangle}_{x}$&${1 \over 2}[|-1,-2\rangle+|-2,-3\rangle-|-3,-4\rangle-|-4,-1\rangle]$&$|-3,-4\rangle$&$\sqcup\sqcup 1 \sqcup\sqcup\sqcup\sqcup 1$\\
\hline
${|\psi^{(4)}_{25}\rangle}_{x}$&${1 \over 2}[|1,2\rangle+|2,3\rangle+|3,4\rangle+|4,1\rangle]$&$|1,2\rangle$&$0 \sqcup\sqcup\sqcup\sqcup 0 \sqcup\sqcup$\\
${|\psi^{(4)}_{26}\rangle}_{x}$&${1 \over 2}[|-1,2\rangle+|-2,3\rangle+|-3,4\rangle+|-4,1\rangle]$&$|-1,2\rangle$&$1 \sqcup\sqcup\sqcup\sqcup 0 \sqcup\sqcup$\\
${|\psi^{(4)}_{27}\rangle}_{x}$&${1 \over 2}[|1,2\rangle-|2,3\rangle-|3,4\rangle+|4,1\rangle]$&$|3,4\rangle$&$\sqcup\sqcup 0 \sqcup\sqcup\sqcup\sqcup0$\\
${|\psi^{(4)}_{28}\rangle}_{x}$&${1 \over 2}[|-1,2\rangle-|-2,3\rangle-|-3,4\rangle+|-4,1\rangle]$&$|-3,4\rangle$&$\sqcup\sqcup 1 \sqcup\sqcup\sqcup\sqcup0$\\
${|\psi^{(4)}_{29}\rangle}_{x}$&${1 \over 2}[|1,2\rangle-|2,3\rangle+|3,4\rangle-|4,1\rangle]$&$|4,1\rangle$&$\sqcup\sqcup\sqcup 0 0 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{30}\rangle}_{x}$&${1 \over 2}[|-1,2\rangle-|-2,3\rangle+|-3,4\rangle-|-4,1\rangle]$&$|-4,1\rangle$&$\sqcup\sqcup\sqcup 1 0 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{31}\rangle}_{x}$&${1 \over 2}[|1,2\rangle+|2,3\rangle-|3,4\rangle-|4,1\rangle]$&$|2,3\rangle$&$\sqcup 0 \sqcup\sqcup\sqcup\sqcup 0 \sqcup$\\
${|\psi^{(4)}_{32}\rangle}_{x}$&${1 \over 2}[|-1,2\rangle+|-2,3\rangle-|-3,4\rangle-|-4,1\rangle]$&$|-2,3\rangle$&$\sqcup 1 \sqcup\sqcup\sqcup\sqcup 0 \sqcup$\\
\hline
${|\psi^{(4)}_{33}\rangle}_{x}$&${1 \over 2}[|1,-3\rangle+|2,-4\rangle+|3,-1\rangle+|4,-2\rangle]$&$|3,-1\rangle$&$\sqcup\sqcup 0 \sqcup 1 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{34}\rangle}_{x}$&${1 \over 2}[|-1,-3\rangle+|-2,-4\rangle+|-3,-1\rangle+|-4,-2\rangle]$&$|-3,1\rangle$&$\sqcup\sqcup 1 \sqcup 1 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{35}\rangle}_{x}$&${1 \over 2}[|1,-3\rangle-|2,-4\rangle-|3,-1\rangle+|4,-2\rangle]$&$|2,-4\rangle$&$\sqcup 0 \sqcup\sqcup\sqcup\sqcup\sqcup 1$\\
${|\psi^{(4)}_{36}\rangle}_{x}$&${1 \over 2}[|-1,-3\rangle-|-2,-4\rangle-|-3,-1\rangle+|-4,-2\rangle]$&$|-2,-4\rangle$&$\sqcup 1 \sqcup\sqcup\sqcup\sqcup\sqcup 1$\\
${|\psi^{(4)}_{37}\rangle}_{x}$&${1 \over 2}[|1,-3\rangle-|2,-4\rangle+|3,-1\rangle-|4,-2\rangle]$&$|4,-2\rangle$&$\sqcup\sqcup\sqcup 0 \sqcup 1 \sqcup\sqcup$\\
${|\psi^{(4)}_{38}\rangle}_{x}$&${1 \over 2}[|-1,-3\rangle-|-2,-4\rangle+|-3,-1\rangle-|-4,-2\rangle]$&$|-4,-2\rangle$&$\sqcup\sqcup\sqcup 1 \sqcup 1 \sqcup\sqcup$\\
${|\psi^{(4)}_{39}\rangle}_{x}$&${1 \over 2}[|1,-3\rangle+|2,-4\rangle-|3,-1\rangle-|4,-2\rangle]$&$|1,-3\rangle$&$0 \sqcup\sqcup\sqcup\sqcup\sqcup1 \sqcup$\\
${|\psi^{(4)}_{40}\rangle}_{x}$&${1 \over 2}[|-1,-3\rangle+|-2,-4\rangle-|-3,-1\rangle-|-4,-2\rangle]$&$|-1,-3\rangle$&$1 \sqcup\sqcup\sqcup\sqcup\sqcup 1 \sqcup$\\
\hline
${|\psi^{(4)}_{41}\rangle}_{x}$&${1 \over 2}[|1,3\rangle+|2,4\rangle+|3,1\rangle+|4,2\rangle]$&$|4,2\rangle$&$\sqcup\sqcup\sqcup 0 \sqcup 0 \sqcup\sqcup$\\
${|\psi^{(4)}_{42}\rangle}_{x}$&${1 \over 2}[|-1,3\rangle+|-2,4\rangle+|-3,1\rangle+|-4,2\rangle]$&$|-4,2\rangle$&$\sqcup\sqcup\sqcup 1 \sqcup0 \sqcup\sqcup$\\
${|\psi^{(4)}_{43}\rangle}_{x}$&${1 \over 2}[|1,3\rangle-|2,4\rangle-|3,1\rangle+|4,2\rangle]$&$|1,3\rangle$&$0 \sqcup\sqcup\sqcup\sqcup\sqcup 0 \sqcup$\\
${|\psi^{(4)}_{44}\rangle}_{x}$&${1 \over 2}[|-1,3\rangle-|-2,4\rangle-|-3,1\rangle+|-4,2\rangle]$&$|-1,3\rangle$&$1 \sqcup\sqcup\sqcup\sqcup\sqcup0 \sqcup$\\
${|\psi^{(4)}_{45}\rangle}_{x}$&${1 \over 2}[|1,3\rangle-|2,4\rangle+|3,1\rangle-|4,2\rangle]$&$|3,1\rangle$&$\sqcup\sqcup 0 \sqcup 0\sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{46}\rangle}_{x}$&${1 \over 2}[|-1,3\rangle-|-2,4\rangle+|-3,1\rangle-|-4,2\rangle]$&$|-3,1\rangle$&$\sqcup\sqcup 1 \sqcup 0 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{47}\rangle}_{x}$&${1 \over 2}[|1,3\rangle+|2,4\rangle-|3,1\rangle-|4,2\rangle]$&$|2,4\rangle$&$\sqcup0 \sqcup\sqcup\sqcup\sqcup\sqcup 0$\\
${|\psi^{(4)}_{48}\rangle}_{x}$&${1 \over 2}[|-1,3\rangle+|-2,4\rangle-|-3,1\rangle-|-4,2\rangle]$&$|-2,4\rangle$&$\sqcup 1 \sqcup\sqcup\sqcup\sqcup\sqcup 0$\\
\hline
${|\psi^{(4)}_{49}\rangle}_{x}$&${1 \over 2}[|1,-4\rangle+|2,-1\rangle+|3,-2\rangle+|4,-3\rangle]$&$|2,-1\rangle$&$\sqcup 0 \sqcup\sqcup 1 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{50}\rangle}_{x}$&${1 \over 2}[|-1,-4\rangle+|-2,-1\rangle+|-3,-2\rangle+|-4,-3\rangle]$&$|-2,-1\rangle$&$\sqcup 1 \sqcup\sqcup 1 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{51}\rangle}_{x}$&${1 \over 2}[|1,-4\rangle-|2,-1\rangle-|3,-2\rangle+|4,-3\rangle]$&$|4,-3\rangle$&$\sqcup\sqcup\sqcup 0 \sqcup\sqcup 1 \sqcup$\\
${|\psi^{(4)}_{52}\rangle}_{x}$&${1 \over 2}[|-1,-4\rangle-|-2,-1\rangle-|-3,-2\rangle+|-4,-3\rangle]$&$|-4,-3\rangle$&$\sqcup\sqcup\sqcup 1 \sqcup\sqcup 1 \sqcup$\\
${|\psi^{(4)}_{53}\rangle}_{x}$&${1 \over 2}[|1,-4\rangle-|2,-1\rangle+|3,-2\rangle-|4,-3\rangle]$&$|3,-2\rangle$&$\sqcup\sqcup 0 \sqcup\sqcup 1 \sqcup\sqcup$\\
${|\psi^{(4)}_{54}\rangle}_{x}$&${1 \over 2}[|-1,-4\rangle-|-2,-1\rangle+|-3,-2\rangle-|-4,-3\rangle]$&$|-3,-2\rangle$&$\sqcup\sqcup 1 \sqcup\sqcup 1\sqcup\sqcup$\\
${|\psi^{(4)}_{55}\rangle}_{x}$&${1 \over 2}[|1,-4\rangle+|2,-1\rangle-|3,-2\rangle-|4,-3\rangle]$&$|1,-4\rangle$&$0 \sqcup\sqcup\sqcup\sqcup\sqcup\sqcup 1$\\
${|\psi^{(4)}_{56}\rangle}_{x}$&${1 \over 2}[|-1,-4\rangle+|-2,-1\rangle-|-3,-2\rangle-|-4,-3\rangle]$&$|-1,-4\rangle$&$1 \sqcup\sqcup\sqcup\sqcup\sqcup\sqcup 1$\\
\hline
${|\psi^{(4)}_{57}\rangle}_{x}$&${1 \over 2}[|1,4\rangle+|2,1\rangle+|3,2\rangle+|4,3\rangle]$&$|3,2\rangle$&$\sqcup\sqcup 0 \sqcup\sqcup 0 \sqcup\sqcup$\\
${|\psi^{(4)}_{58}\rangle}_{x}$&${1 \over 2}[|-1,4\rangle+|-2,1\rangle+|-3,2\rangle+|-4,3\rangle]$&$|-3,2\rangle$&$\sqcup\sqcup 1\sqcup\sqcup 0 \sqcup\sqcup$\\
${|\psi^{(4)}_{59}\rangle}_{x}$&${1 \over 2}[|1,4\rangle-|2,1\rangle-|3,2\rangle+|4,3\rangle]$&$|1,4\rangle$&$0 \sqcup\sqcup\sqcup\sqcup\sqcup\sqcup 0$\\
${|\psi^{(4)}_{60}\rangle}_{x}$&${1 \over 2}[|-1,4\rangle-|-2,1\rangle-|-3,2\rangle+|-4,3\rangle]$&$|-1,4\rangle$&$1 \sqcup\sqcup\sqcup\sqcup\sqcup\sqcup 0$\\
${|\psi^{(4)}_{61}\rangle}_{x}$&${1 \over 2}[|1,4\rangle-|2,1\rangle+|3,2\rangle-|4,3\rangle]$&$|2,1\rangle$&$\sqcup 0 \sqcup\sqcup 0 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{62}\rangle}_{x}$&${1 \over 2}[|-1,4\rangle-|-2,1\rangle+|-3,2\rangle-|-4,3\rangle]$&$|-2,1\rangle$&$\sqcup 1 \sqcup\sqcup 0 \sqcup\sqcup\sqcup$\\
${|\psi^{(4)}_{63}\rangle}_{x}$&${1 \over 2}[|1,4\rangle+|2,1\rangle-|3,2\rangle-|4,3\rangle]$&$|4,3\rangle$&$\sqcup\sqcup\sqcup0\sqcup\sqcup 0 \sqcup$\\
${|\psi^{(4)}_{64}\rangle}_{x}$&${1 \over 2}[|-1,4\rangle+|-2,1\rangle-|-3,2\rangle-|-4,3\rangle]$&$|-4,3\rangle$&$\sqcup\sqcup\sqcup 1 \sqcup\sqcup 0 \sqcup$\\
\hline\hline
\end{tabular}
 \label{tab8}
%\end{ruledtabular}
\end{center}
\end{table}
\endgroup
In this case, the last unitary operator for BSM can be considered
like
\begin{eqnarray}{\label{H4}}
U_{(4)}=&{1\over{\sqrt{4}}}&[I\otimes
P_{1}P_{4}N_{1}N_{2}N_{3}N_{4}P_{1}P_{4}+L_{+}\otimes
P_{4}N_{4}P_{4}N_{4}L_{+}\nonumber\\
&+&L_{+}^{2}\otimes P_{1}P_{3}N_{1}
N_{2}N_{3}N_{4}P_{1}P_{3}L_{+}^{2}+L_{+}^{3}\otimes
P_{3}N_{3}P_{3}N_{3}L_{+}^{3}].\nonumber\\
\end{eqnarray}


 Now, consider the $N=6$ case.  Again, at first glance, it seems that a similar
procedure works for this case. Thus, imagine that one has
succeeded to perform the protocol and produce a set of final
entangled orthonormal states from a complete set of initial
entangled orthonormal states (Bell bases). Among them, there
should be a state like {\begin{eqnarray}{\label{x}}
\frac{1}{\sqrt{6}}(|1,1\rangle+|2,2\rangle+|3,3\rangle+|4,4\rangle+|5,5\rangle+|6,6\rangle).
\end{eqnarray}}
The first obvious entangled state which is orthogonal to this
state can be considered to be
\begin{eqnarray}{\label{X}}
\frac{1}{\sqrt{6}}(|1,1\rangle-|2,2\rangle+|3,3\rangle-|4,4\rangle+|5,5\rangle-|6,6\rangle).
\end{eqnarray}
But, it can be tested that no other states of this family, formed
by changing kets' signs, may be found which are orthogonal to both
of them, simultaneously. So, the primitive assumption about the
existence of such initial entangled orthonormal states is not true
and we do not have enough initial orthonormal states for the
procedure. Thus, there does not exist any related efficient dense
coding scheme for the $N=6$ case. It can be seen that a similar
argument also works for the case of $N=3$, which was explained
before. The important point is that one can see by a similar
method as above that this scheme dose not work for odd $N$'s and
$N$'s which have odd halves. This inability to extend the scheme
for the mentioned $N$'s is in accordance with the known fact
concerning Hadamard matrices, which says that the order of a
Hadamard matrix is 1, 2 or $4k$, with $k$ being a positive integer
{\cite{Wallis,Agaian}}. At this stage, we understood that there
might be a kind of extension that takes any Hadamard matrix design
and turns it into a general framework  for representing Bell
states and the required
 unitary operators which are used in the preparation and measurement processes.  For example,
 using the coefficients of kets for each family represented in
 Tables.~\ref{N=2x} and \ref{tab6}, one can obtain 4 and
 8-dimensional normalized Hadamard matrices, respectively.  Hence,
 it is possible to represent the Bell bases of each family and also the unitary operators which transform them together
 using this kind of Hadamard matrices, as we have shown generally
 in Eqs.~(\ref{Bell}) and (\ref{O_j}).

 To confirm our results as well as to obtain an explicit form for
 $U_{(N)}$ operator, we have checked our procedure, step by step, for
 $N=8$ and 12 cases.  Here, just, we have shown the results
 obtained for $U_{(N)}$ operator at these two cases.  For the
 $N=8$ case, for example, we have obtained
\begin{eqnarray}{\label{H8}}
U_{(8)} = & {1\over{ \sqrt{8}}} & [( I\otimes
P_{1}P_{3}P_{6}P_{8}N_{1}\ldots
N_{8}P_{1}P_{3}P_{6}P_{8})\nonumber\\&+& (L_{+}\otimes
P_{5}N_{5}P_{5}N_{5}P_{6}N_{6}P_{6}N_{6}P_{8}N_{8}P_{8}N_{8}L_{+}) \nonumber\\
& + & (L_{+}^{2}\otimes P_{1}P_{3}P_{6}P_{7}N_{1}\ldots
N_{8}P_{1}P_{3}P_{6}P_{7}L_{+}^{2})\nonumber\\&+&(L_{+}^{3}\otimes
P_{3}N_{3}P_{3}N_{3}P_{6}N_{6}P_{6}N_{6}
P_{7}N_{7}P_{7}N_{7}L_{+}^{3})  \nonumber\\
& + & (L_{+}^{4}\otimes P_{1}P_{4}P_{5}P_{8}N_{1}\ldots
N_{8}P_{1}P_{4}P_{5}P_{8}L_{+}^{4})\nonumber\\&+&(L_{+}^{5}\otimes
P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}P_{8}N_{8}P_{8}N_{8}L_{+}^{5}) \nonumber\\
& + & (L_{+}^{6}\otimes P_{1}P_{4}P_{5}P_{7}N_{1}\ldots
N_{8}P_{1}P_{4}P_{5}P_{7}L_{+}^{6})\nonumber\\&+&(L_{+}^{7}\otimes
P_{4}N_{4}P_{4}N_{4}P_{5}N_{5}P_{5}N_{5}P_{7}N_{7}P_{7}N_{7}L_{+}^{7})]
\end{eqnarray}
where $N_{1}\ldots
N_{8}=N_{1}N_{2}N_{3}N_{4}N_{5}N_{6}N_{7}N_{8}$.  Meantime, for
$N=12$, the $U_{(N)}$ operator can be written in the following
form
\begin{eqnarray}{\label{H12}} &&U_{(12)}  =\nonumber\\
&&{1\over{\sqrt{12}}} [(I\otimes
P_{1}P_{6}P_{7}P_{8}P_{11}P_{12}N_{1}\ldots
N_{12}P_{1}P_{6}P_{7}P_{8}P_{11}P_{12}) \nonumber\\
&& + (L_{+}\otimes
P_{6}N_{6}P_{6}N_{6}P_{7}N_{7}P_{7}N_{7}P_{9}N_{9}P_{9}N_{9}P_{11}N_{11}P_{11}N_{11}P_{12}N_{12}P_{12}N_{12}L_{+})
 \nonumber\\
&& +(L_{+}^{2}\otimes P_{1}P_{3}P_{5}P_{9}P_{10}N_{1}\ldots
N_{12}P_{1}P_{3}P_{5}P_{9}P_{10}L_{+}^{2}) \nonumber\\
&& +(L_{+}^{3}\otimes
P_{6}N_{6}P_{6}N_{6}P_{7}N_{7}P_{7}N_{7}P_{8}N_{8}P_{8}N_{8}P_{10}N_{10}P_{10}N_{10}P_{12}N_{12}P_{12}N_{12}L_{+}^{3})
\nonumber\\
&& +(L_{+}^{4}\otimes
P_{1}P_{3}P_{4}P_{5}P_{8}P_{9}P_{10}N_{1}\ldots
N_{12}P_{1}P_{3}P_{4}P_{5}P_{8}P_{9}P_{10}L_{+}^{4}) \nonumber\\
&& +(L_{+}^{5}\otimes
P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}P_{5}N_{5}P_{5}N_{5}P_{8}N_{8}P_{8}N_{8}P_{9}N_{9}P_{9}N_{9}L_{+}^{5})
\nonumber\\
&& +(L_{+}^{6}\otimes P_{1}P_{4}P_{6}P_{7}P_{10}P_{12}N_{1}\ldots
N_{12}P_{1}P_{4}P_{6}P_{7}P_{10}P_{12}L_{+}^{6}) \nonumber\\
&& +(L_{+}^{7}\otimes
P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}P_{10}N_{10}P_{10}N_{10}P_{11}N_{11}P_{11}N_{11}P_{12}N_{12}P_{12}N_{12}L_{+}^{7})
\nonumber\\
&& +(L_{+}^{8}\otimes
P_{1}P_{4}P_{5}P_{6}P_{9}P_{11}P_{12}N_{1}\ldots
N_{12}P_{1}P_{4}P_{5}P_{6}P_{9}P_{11}P_{12}L_{+}^{8}) \nonumber\\
&& +(L_{+}^{9}\otimes
P_{3}N_{3}P_{3}N_{3}P_{4}N_{4}P_{4}N_{4}P_{5}N_{5}P_{5}N_{5}P_{7}N_{7}P_{7}N_{7}P_{9}N_{9}P_{9}N_{9}L_{+}^{9})
\nonumber\\
&& +(L_{+}^{10}\otimes P_{1}P_{3}P_{7}P_{8}P_{11}N_{1}\ldots
N_{12}P_{1}P_{3}P_{7}P_{8}P_{11}L_{+}^{10}) \nonumber\\
&& +(L_{+}^{11}\otimes
P_{5}N_{5}P_{5}N_{5}P_{6}N_{6}P_{6}N_{6}P_{8}N_{8}P_{8}N_{8}P_{10}N_{10}P_{10}N_{10}P_{11}N_{11}P_{11}N_{11}L_{+}^{11})].\nonumber\\
\end{eqnarray}
In fact, these representations, in addition to the previous ones,
as examples, help one to obtain and understand the general form
presented for the $U_{(N)}$ operator in Eq.~(\ref{ghoul}).

\chapter{A comment on dense coding in pairwise entangled case}


Recently, Lee {\em et al.} {\cite{Lee}} have considered two
different multiqubit schemes for dense coding and compared their
efficiency. In the pairwise scheme, Alice and Bob share $N$
separated maximally entangled pairs, so this scheme is equivalent
to $N$ separate dense coding schemes.  On the other hand, in the
maximally entangled scheme, Alice has $N$ qubits and Bob possesses
one, which are prepared in $N+1$ maximally entangled qubits. The
number of different unitary operators which can be constructed in
the first scheme is $4^N$ corresponding to the number of different
messages
 which can be sent. Since if we have $M$ different messages we can
codify them in ${\log_{2} M}$ bits of information, then
${{\log}_{2} (4^N)}=2N$ bits of information can be transferred by
using the pairwise scheme form Alice to Bob, by sending $N$
particles.  However, Lee {\it et al.} {\cite{Lee}} performed an
essential mistake just here by which they have claimed that number
of bits is as much as $2^N$, which is unfounded.\footnote{It is
curious to note, though, that they still claim that the $2^N$
result is correct \cite{PraEditor}.} This, of course, is
inconsistent with the Holevo bound \cite{Holevo79,Holevo98}, which
states that $N$ qubits can at most carry $N$ bits of information.
Furthermore, another criticism on their work can be found in
\cite{Wojcik}.

Now, the rates of information gain (bits per unit time) that they
have deduced in their Eqs. (6) and (7) for the pairwise and
maximally entangled schemes must change into
\begin{eqnarray}
&&r_p=\frac{2N}{N(t_h+t_c)}\nonumber\\
&&r_m=\frac{N+1}{t_h+Nt_c}
\end{eqnarray}
where $t_c$ and $t_h$ are the operation times for CNOT  and
Hadamard gates, respectively. If $t_c$ and $t_h$ are assumed
equal, it is seen that $r_p=r_m=\frac{1}{t_c}$, and consequently,
there is no exponential efficiency in the pairwise scheme.
Therefore, considering $N$ different Alices or combining them as a
sole Alice (which changes the number of parties involved but not
the number of sent particles) cannot lead into a more efficient
protocol as they have claimed. This is in accordance with the
result of Bose {\em et al.} {\cite{Bose}}.

Meantime, it should be noted that a more reasonable measure of
efficiency of the scheme can be the number of bits transferred in
the protocol per needed time per sent particles. By considering
this we have
\begin{eqnarray}
&&r_p= \frac{2}{N(t_h+t_c)}\nonumber\\
&&r_m=\frac{N+1}{N(t_h+Nt_c)}.
\end{eqnarray}
 Now these quantities give more better sense of efficiency of the
protocols, though, the sent particles are the same here. This
concept of efficiency for these schemes has been also applied by
Bose {\em et al.} {\cite{Bose}}.








\newpage

\chapter{Curriculum Vitae}
%\addcontentsline{toc}{section}{Curriculum Vitae}

{\Large Omid Akhavan}
\begin{center}
\begin{tabular}{c p{10cm}}
1974       & Born in (Vanak) Tehran-Iran on July 8.\\
1980-1985  & Elementary school (Ferdows-Barin, Zafar) in Tehran.\\
1985-1988  & Middle school (Razi, Mirdamad) in Tehran.\\
1988-1992  & High school (Razi, Mirdamad) in Tehran.\\
1992-1996  & Studied physics at Uroumieh University, First Department Honors, Uroumieh, Iran.\\
1996       & Summer student at Institute for Advanced Studies in Basic Sciences, Zanjan, Iran.\\
1996       & National Physics Olympiad Competition, Ranking 12th, Iran.\\
1996-1998  & Obtained M. Sc. degree in physics under supervising
of Prof. A.Z. Moshfegh at Department of Physics, Sharif University
of Technology, Tehran.
Experimental work on physics of thin films.\\
1998-2003  & PhD student in Physics under supervising of Prof. M.
Golshani at Department of Physics, Sharif University of
Technology. Theoretical work on the quantum theory of motion and
hidden variables, quantum communication schemes such as
teleportation, dense coding and key distribution. Participation in
Workshop on ``Realizability of Quantum
Computation: Entanglement at the Nanoscale", ICTP, Italy.\\
2003 & Invited to Physics Department of Sharif University of
Technology for a postdoc position.\\
\end{tabular}
\end{center}




\backmatter

\section*{Papers and Manuscripts by the Author}


\begin{enumerate}
\addcontentsline{toc}{section}{Papers and Manuscripts by the
Author}

\item {\em{PVD deposition and characterization of Cu/Si(111) and
Cu/Ta/Si(111) multilayers,}} A.Z. Moshfegh, M.M. Shafiee, and O.
Akhavan, Proceeding of the 11-th Congress of the International
Federation for Heat Treatment and Surface Engineering (Florence,
Italy) Vol. II, 1998, 69.

\item {\em{$YBa_{2}Cu_{3}O_{7-\delta}-Ag$ sputtered thin films on
MgO(100) and LaAlO$_{3}$(100) biased and unbiased substrates,}}
A.Z. Moshfegh, O. Akhavan, H. Salamati, P. Kameli, and M. Akhavan,
Proceeding of the 1st Regional Conference on Magnetic and
Superconducting Materials, Tehran (World Scientific, Singapore,
1999) p. 585.

\item {\em{Bias sputtered Ta modified diffusion barrier in
Cu/Ta(V$_{b}$)/Si(111) structures,}} A.Z. Moshfegh and O. Akhavan,
{\it{Thin Solid Films}} {\bf 370}, 10 (2000).

\item {\em{Retardation of Ta silicidation by bias sputtering in
Cu/Ta/Si(111) thin films,}} A.Z. Moshfegh and O. Akhavan, {\it{J.
Phys. D}} {\bf 34}, 2103 (2001).

\item {\em{A two-slit experiment which distinguishes between
standard and Bohmian quantum mechanics,}} M. Golshani and O.
Akhavan, quant-ph/0009040.

\item {\em{Experiment can decide between standard and Bohmian
quantum mechanics,}} M. Golshani and O. Akhavan, quant-ph/0103100.

\item {\em{Bohmian prediction about a two double-slit experiment
and its disagreement with standard quantum mechanics,}} M.
Golshani and O. Akhavan, {\it{J. Phys. A}} {\bf 34}, 5259 (2001),
quant-ph/0103101.

\item {\em{On the experimental incompatibility between standard
and Bohmian quantum mechanics}}, M. Golshani and O. Akhavan,
quant-ph/0110123.

\item {\em{Reply to: Comment on ``Bohmian prediction about a two
double-slit experiment and its disagreement with SQM"}} O. Akhavan
and M. Golshani, quant-ph/0305020.

\item {\em{Study of cobalt silicides formation in Co/Ta-W/Si(100)
multilayer systems}}, A.Z. Moshfegh, S.J. Hashemifar, O. Akhavan,
and R. Azimirad, {\it{Thin Solid Films}} {\bf 433}, 298 (2003).

\item {\em{Quantum dense coding by spatial state entanglement,}}
O. Akhavan, A.T. Rezakhani, and M. Golshani, {\it{Phys. Lett. A}}
{\bf 313}, 261 (2003); quant-ph/0305118.

\item {\em{Comment on ``Dense coding in entangled states",}} O.
Akhavan and A.T. Rezakhani, {\it{Phys. Rev. A}} {\bf 68}, 016302
(2003); quant-ph/0306148.

\item {\em{A calculation of diffusion parameters for Cu/Ta and
Ta/Si interfaces in Cu/Ta/Si(111) structure,}} A.Z. Moshfegh and
O. Akhavan, {\it{Mater. Sci. Semicond. Proc.}} {\bf 6}, 165
(2003).

\item {\em{The growth of $CoSi_2$ thin films by Co/W/Si(100)
multilayer structure}}, A.Z. Moshfegh, S.J. Hashemifar, and O.
Akhavan, {\it{Solid State Communication}} {\bf 128}, 239 (2003).

\item {\em{A scheme for spatial wave function teleportation in
three dimensions}}, O. Akhavan, A.T. Rezakhani, and M. Golshani,
{\it{J. Quant. Inf. Comp.}}, submitted.
\end{enumerate}





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\end{thebibliography}

%\frontmatter



\end{document}

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