#  A Mechanical Implementation and Diagrammatic Calculation of Entangled Basis
States

F.A. Buot,1,2,3A.R. Elnar,2 G. Maglasang,2 and C.M. Galon2
1C&LB Research Institute, Carmen, Cebu 6005, Philippines
2CTCMP, Cebu Normal University, Cebu City 6000, Philippines
3LCFMNN, Department of Physics, University of San Carlos,
Talamban, Cebu 6000, Philippines

###### Abstract

We give for the first time a diagrammatic calculational tool of quantum
entanglement. We present a pedagogical and simple mechanical implementation of
quantum entanglement or ”spooky action at a distance” to give a tangible
realization of this weird quantum mechanical concept alien to classical
physics. When two or more particles are correlated in a certain way, no matter
how far apart they are in space, their states remain correlated. Their
correlation, which is instantaneous, does not seem to involve any
communication which is limited by the speed of light. The same mechanical
implementation demonstrates the fundamental physical limits of any
computational processes. The analytical derivations of calculational entangled
basis states are given and their corresponding diagrammatic representations
give an efficient aid in determining the calculational entangled basis states.
A quantum Fourier transform for the two-state diagrams representing entangled
basis states (’renormalized qubits’) can also be formulated. Our results seem
to advocate the idea that quantum entanglement generates the extra dimensions
of the gravitational theory, indeed quantum entanglement is related to deep
issues in the unification of general relativity and quantum mechanics. This
extra dimensions of spacetime entanglement are currently being speculated in
the literature.

##  1 INTRODUCTION

The principal characteristics of quantum mechanics, which are alien to
classical physics, are as follows. First is the notion of uncertainty, i.e.,
not all the classical physical observable properties of a system can be
simultaneously determined with exact precision. For example, position and
momentum cannot both be known at the same time. There may other sets of
observable properties that share the uncertainty properties. Second, the
notion of superposition, for example, a cat is a superposition of alive and
dead cat. In other words an arbitrary quantum state is a superposition of
several quantum sates that describe measurable events. Third, the notion of
entanglement, for example, if two photons or two particle spins become
entangled, i.e., they are allowed to interact initially so that they will
subsequently be defined by a single quantum state, then once they are
separated, they will still share the same quantum state. So measuring one will
determine the state of the other: for example, with a spin-zero entangled
state, if one particle is measured to be in a spin-up state, the other is
instantly forced to be in a spin-down state. In this paper, we would like to
address the weirdest aspect of quantum mechanics, i.e., the notion of quantum
entanglement.

Today, quantum entanglement forms the basis of several quantum technologies.
In quantum teleportation and quantum cryptography, entangled particles are
used to transmit signals that cannot be intercepted by an eavesdropper. The
preliminary viable quantum cryptography systems are now being used by banks.
And the exploding field of quantum computation uses superposition and
entangled quantum states to perform computational calculations in parallel and
at ultra-speed, so that some types of classically impractical calculations can
be done by quantum computers in reasonable length of time.

Historically, in a 1935 paper, Einstein, Boris Podolsky and Nathan Rosen argue
that quantum mechanics was not a complete physical theory. Known today as the
“EPR paradox,” the thought experiment was meant to demonstrate the inherent
conceptual, in terms of classical analogies, of quantum theory. Entanglement
claims that the result of a measurement on one particle of an entangled
quantum system can have an instantaneous effect on another particle,
regardless of their distance. But the EPR paradox did help deepen our
understanding of quantum mechanics by exposing the fundamentally non-classical
characteristics of the measurement process. Before that paper, most physicists
viewed a measurement as a physical disturbance inflicted directly on the
measured system: one shines light onto an electron to determine its position,
but this disturbs the electron and produces uncertainties. The EPR paradox
shows that a “measurement” can be performed on a particle without disturbing
it directly, by performing a measurement on a distant entangled particle.

Einstein postulated the existence of hidden variables, yet unknown local
properties of the system which should account for the discrepancy, so that no
instantaneous spooky action would be necessary. Bohr disagreed vehemently with
this view and defended the far stricter Copenhagen interpretation of quantum
mechanics. The two men were passionate about the subject, especially at the
Solvay Conferences of 1927 and 1930, neither Einstein nor Bohr conceded
defeat.

In this paper, we try to propose a simple mechanical model of a classical
analogy to the ’spooky at a distance’ property of quantum entanglement. This
is a simple mechanical model, an ideal classical analogy for Einstein
arguments. For what it is worth, it gives a simple classical/mechanical
implementation and pedagogical value to the concept of quantum entanglement,
which may revive interest in some intermediary medium (not hidden variables,
but part of the whole system, i.e., nonlocality and hence does not violate
Bell’s theorem) in quantum entanglement.

##  2 ANALYTICAL DERIVATION OF ENTANGLED BASIS STATES

First, we give formal derivation of calculational entangled basis states as
our foundation to the diagrammatic techniques. This was treated before in the
author’s book [1]. Since the treatment there is not widely annunciated, we
give it here for completeness. Consider the identity

$\left|p\right\rangle=\sum\limits_{q}\left\langle
q\right|\left.p\right\rangle\ \left|q\right\rangle,$

where the $\left\langle q\right|\left.p\right\rangle$ is the transformation
function. For discrete quantum mechanics, this is given by the discrete
Fourier transform function,

$\left\langle
q\right|\left.p\right\rangle=\frac{1}{\sqrt{N}}\exp\left(-\frac{i}{\hbar}p\cdot
q\right).$

Therefore,

$\left|p\right\rangle=\frac{1}{\sqrt{N}}\sum\limits_{q}\exp\left(-\frac{i}{\hbar}p\cdot
q\right)\ \left|q\right\rangle.$

The product state in the ’momentum’ basis is expanded in terms of the
’position’ basis

$\displaystyle\left|p^{\prime}\right\rangle\left|p^{\prime\prime}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{N}}\sum\limits_{q^{\prime}}\exp\left(-\frac{i}{\hbar}p^{\prime}\cdot
q^{\prime}\right)\
\left|q^{\prime}\right\rangle\frac{1}{\sqrt{N}}\sum\limits_{q^{\prime\prime}}\exp\left(-\frac{i}{\hbar}p^{\prime\prime}\cdot
q^{\prime\prime}\right)\ \left|q^{\prime\prime}\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{N}\sum\limits_{q^{\prime},q^{\prime\prime}}\exp\left(-\frac{i}{\hbar}p^{\prime}\cdot
q^{\prime}\right)\exp\left(-\frac{i}{\hbar}p^{\prime\prime}\cdot
q^{\prime\prime}\right)\ \left|q^{\prime}\right\rangle\
\left|q^{\prime\prime}\right\rangle.$

We can expressed the right hand side in terms of ’correlated’ (entangled)
product states by writing, $q^{\prime\prime}=q^{\prime}+m$, where $m$ is the
quantum-correlation ’distance’,

$\displaystyle\left|p^{\prime}\right\rangle\left|p^{\prime\prime}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{N}\sum\limits_{q^{\prime},m}\exp\left(-\frac{i}{\hbar}p^{\prime}\cdot
q^{\prime}\right)\exp\left(-\frac{i}{\hbar}p^{\prime\prime}\cdot\left(q^{\prime}+m\right)\right)\
\left|q^{\prime}\right\rangle\ \left|q^{\prime}+m\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{N}\sum\limits_{q^{\prime},m}\exp\left(-\frac{i}{\hbar}\left(p^{\prime}+p^{\prime\prime}\right)\cdot
q^{\prime}\right)\exp\left(-\frac{i}{\hbar}p^{\prime\prime}\cdot m\right)\
\left|q^{\prime}\right\rangle\ \left|q^{\prime}+m\right\rangle.$

Now writing $p^{\prime}+p^{\prime\prime}=p$, we have

$\displaystyle\left|p^{\prime}\right\rangle\left|p-p^{\prime}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{N}\sum\limits_{q^{\prime},m}\exp\left(-\frac{i}{\hbar}\left(p\right)\cdot
q^{\prime}\right)\exp\left(-\frac{i}{\hbar}\left(p-p^{\prime}\right)\cdot
m\right)\ \left|q^{\prime}\right\rangle\ \left|q^{\prime}+m\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{N}}\sum\limits_{m}\exp\left(-\frac{i}{\hbar}\left(p-p^{\prime}\right)\cdot
m\right)\left\\{\frac{1}{\sqrt{N}}\sum\limits_{q^{\prime}}\exp\left(-\frac{i}{\hbar}p\cdot
q^{\prime}\right)\ \left|q^{\prime}\right\rangle\
\left|q^{\prime}+m\right\rangle\right\\}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{N}}\sum\limits_{m}\exp\left(-\frac{i}{\hbar}\left(p-p^{\prime}\right)\cdot
m\right)\left|\psi_{p,m}\right\rangle.$

The inverse transformation gives $\left|\psi_{p,m}\right\rangle$ in terms of
the the ’momentum’ basis products

$\left|\psi_{p,m}\right\rangle=\frac{1}{\sqrt{N}}\sum\limits_{p^{\prime}}\exp\left(\frac{i}{\hbar}\left(p-p^{\prime}\right)\cdot
m\right)\left|p^{\prime}\right\rangle\left|p-p^{\prime}\right\rangle.$ (1)

Clearly, the correlated basis defined by $\left|\psi_{p,m}\right\rangle$ forms
orthonormal and complete set.

###  2.1 Bell Basis

Thus, for $N=2$, we have the standard momentum product states in terms of the
so-called Bell basis, $\left|\psi_{p,m}\right\rangle$, for example,

$\left|k^{\prime}\right\rangle\left|k-k^{\prime}\right\rangle=\frac{1}{\sqrt{2}}\sum\limits_{m=0}^{1}\exp\left(-\pi
i\left(k-k^{\prime}\right)\cdot m\right)\left|\psi_{k,m}\right\rangle,$

which yields

$\left|0\right\rangle\left|k\right\rangle=\frac{1}{\sqrt{2}}\sum\limits_{m=0}^{1}\exp\left(-\pi
ik\cdot m\right)\left|\psi_{k,m}\right\rangle,$

and using the identities

$\displaystyle\left|\psi_{00}\right\rangle$ $\displaystyle=$
$\displaystyle\left|\Phi^{+}\right\rangle,$
$\displaystyle\left|\psi_{01}\right\rangle$ $\displaystyle=$
$\displaystyle\left|\Psi^{+}\right\rangle,$
$\displaystyle\left|\psi_{10}\right\rangle$ $\displaystyle=$
$\displaystyle\left|\Phi^{-}\right\rangle,$
$\displaystyle\left|\psi_{11}\right\rangle$ $\displaystyle=$
$\displaystyle\left|\Psi^{-}\right\rangle,$

we have,

$\displaystyle\left|0\right\rangle\left|0\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\sum\limits_{m=0}^{1}\exp\left(-\pi ik\cdot
m\right)\left|\psi_{0,m}\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|\psi_{00}\right\rangle+\left|\psi_{01}\right\rangle\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|\Phi^{+}\right\rangle+\left|\Psi^{+}\right\rangle\right),$
$\displaystyle\left|0\right\rangle\left|1\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\sum\limits_{m=0}^{1}\exp\left(-\pi i1\cdot
m\right)\left|\psi_{1,m}\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|\psi_{1,0}\right\rangle-\left|\psi_{1,1}\right\rangle\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|\Phi^{-}\right\rangle-\left|\Psi^{-}\right\rangle\right),$

We also have

$\left|1\right\rangle\left|k+1\right\rangle=\frac{1}{\sqrt{2}}\sum\limits_{m=0}^{1}\exp\left(-\pi
i\left(k+1\right)\cdot
m\right)\left|\psi_{k,m}\right\rangle\left(\mathop{\mathrm{m}od}2\right),$

which yields

$\displaystyle\left|1\right\rangle\left|1\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\sum\limits_{m=0}^{1}\exp\left(-\pi
i\left(1\right)\cdot m\right)\left|\psi_{0,m}\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|\psi_{0,0}\right\rangle-\left|\psi_{0,1}\right\rangle\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|\Phi^{+}\right\rangle-\left|\Psi^{+}\right\rangle\right),$
$\displaystyle\left|1\right\rangle\left|1+1\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\sum\limits_{m=0}^{1}\exp\left(-\pi
i\left(1+1\right)\cdot
m\right)\left|\psi_{1,m}\right\rangle\left(\mathop{\mathrm{m}od}2\right),$
$\displaystyle\left|1\right\rangle\left|0\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|\psi_{1,0}\right\rangle+\left|\psi_{1,1}\right\rangle\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|\Phi^{-}\right\rangle+\left|\Psi^{-}\right\rangle\right).$

Therefore, for $N=2$, we have the following transformation from the maximally
entangled Bell basis to the standard ’momentum’-state product basis given by

$\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{p}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}1&1&0&0\\\
0&0&1&-1\\\ 0&0&1&1\\\
1&-1&0&0\end{array}\right)\left(\begin{array}[]{c}\left|\Phi^{+}\right\rangle\\\
\left|\Psi^{+}\right\rangle\\\ \left|\Phi^{-}\right\rangle\\\
\left|\Psi^{-}\right\rangle\end{array}\right)_{Bell}.$

Hence,

$\displaystyle\left(\begin{array}[]{c}\left|\Phi^{+}\right\rangle\\\
\left|\Psi^{+}\right\rangle\\\ \left|\Phi^{-}\right\rangle\\\
\left|\Psi^{-}\right\rangle\end{array}\right)_{Bell}$ $\displaystyle=$
$\displaystyle\left[\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}1&1&0&0\\\
0&0&1&-1\\\ 0&0&1&1\\\
1&-1&0&0\end{array}\right)\right]^{-1}\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{p}$ (14)
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}1&0&0&1\\\
1&0&0&-1\\\ 0&1&1&0\\\
0&-1&1&0\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{p}.$ (23)

The above result also follows from the general inverse expression, Eq. (1), in
terms of ’momentum’ product basis,

$\left|\psi_{p,m}\right\rangle=\frac{1}{\sqrt{N}}\sum\limits_{p^{\prime}}\exp\left(\frac{i}{\hbar}\left(p-p^{\prime}\right)\cdot
m\right)\left|p^{\prime}\right\rangle\left|p-p^{\prime}\right\rangle,$

so that for $N=2$,

$\left|\psi_{k,m}\right\rangle=\frac{1}{\sqrt{2}}\sum\limits_{k^{\prime}}\exp\left(i\pi\left(k-k^{\prime}\right)\cdot
m\right)\left|k^{\prime}\right\rangle\left|k-k^{\prime}\right\rangle,$

which yields

$\displaystyle\left|\psi_{00}\right\rangle$ $\displaystyle=$
$\displaystyle\left|\Phi^{+}\right\rangle=\frac{1}{\sqrt{2}}\sum\limits_{k^{\prime}}\exp\left(i\pi\left(0-k^{\prime}\right)\cdot
0\right)\left|k^{\prime}\right\rangle\left|0-k^{\prime}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle+\left|1\right\rangle\left|1\right\rangle\right),$
$\displaystyle\left|\psi_{01}\right\rangle$ $\displaystyle=$
$\displaystyle\left|\Psi^{+}\right\rangle=\frac{1}{\sqrt{2}}\sum\limits_{k^{\prime}}\exp\left(i\pi\left(k^{\prime}\right)\cdot
1\right)\left|k^{\prime}\right\rangle\left|k^{\prime}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle-\left|1\right\rangle\left|1\right\rangle\right),$
$\displaystyle\left|\psi_{10}\right\rangle$ $\displaystyle=$
$\displaystyle\left|\Phi^{-}\right\rangle=\frac{1}{\sqrt{2}}\sum\limits_{k^{\prime}}\exp\left(i\pi\left(1-k^{\prime}\right)\cdot
0\right)\left|k^{\prime}\right\rangle\left|1-k^{\prime}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|0\right\rangle\right),$
$\displaystyle\left|\psi_{1,1}\right\rangle$ $\displaystyle=$
$\displaystyle\left|\psi_{k,m}\right\rangle=\frac{1}{\sqrt{2}}\sum\limits_{k^{\prime}}\exp\left(i\pi\left(1-k^{\prime}\right)\cdot
1\right)\left|k^{\prime}\right\rangle\left|1-k^{\prime}\right\rangle$
$\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|0\right\rangle\right).$

We also have

$\left(\begin{array}[]{c}\left|\Phi^{+}\right\rangle\\\
\left|\Psi^{+}\right\rangle\\\ \left|\Phi^{-}\right\rangle\\\
\left|\Psi^{-}\right\rangle\end{array}\right)_{Bell}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}1&0&0&1\\\
0&1&1&0\\\ 1&0&0&-1\\\
0&1&-1&0\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{q}.$ (24)

Therefore, we have the identity

$\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}1&0&0&1\\\ 1&0&0&-1\\\
0&1&1&0\\\
0&-1&1&0\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{p}=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}1&0&0&1\\\
0&1&1&0\\\ 1&0&0&-1\\\
0&1&-1&0\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{q},$

which gives

$\displaystyle\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{p}$
$\displaystyle=$
$\displaystyle\left[\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}1&0&0&1\\\
1&0&0&-1\\\ 0&1&1&0\\\
0&-1&1&0\end{array}\right)\right]^{-1}\left(\begin{array}[]{cccc}1&0&0&1\\\
0&1&1&0\\\ 1&0&0&-1\\\
0&1&-1&0\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{q}$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}1&1&0&0\\\
0&0&1&-1\\\ 0&0&1&1\\\
1&-1&0&0\end{array}\right)\frac{1}{\sqrt{2}}\left(\begin{array}[]{cccc}1&0&0&1\\\
0&1&1&0\\\ 1&0&0&-1\\\
0&1&-1&0\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{q}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\begin{array}[]{cccc}1&1&0&0\\\ 0&0&1&-1\\\
0&0&1&1\\\ 1&-1&0&0\end{array}\right)\left(\begin{array}[]{cccc}1&0&0&1\\\
0&1&1&0\\\ 1&0&0&-1\\\
0&1&-1&0\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{q}.$

Therefore, ’momentum’ basis is given by the transformation of the ’position’
basis as

$\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{p}=\frac{1}{2}\left(\begin{array}[]{cccc}1&1&1&1\\\
1&-1&1&-1\\\ 1&1&-1&-1\\\
1&-1&-1&1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)_{q}.$

###  2.2 Three-Qubit Entangled Basis

As in the two-qubit, the product state in the ’momentum’ basis is expanded in
terms of the ’position’ basis

$\displaystyle\left|p^{\prime}\right\rangle\left|p^{\prime\prime}\right\rangle\left|p^{\prime\prime\prime}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{N^{\frac{3}{2}}}\sum\limits_{q^{\prime}}\exp\left(-\frac{i}{\hbar}p^{\prime}\cdot
q^{\prime}\right)\
\left|q^{\prime}\right\rangle\sum\limits_{q^{\prime\prime}}\exp\left(-\frac{i}{\hbar}p^{\prime\prime}\cdot
q^{\prime\prime}\right)\
\left|q^{\prime\prime}\right\rangle\sum\limits_{q^{\prime\prime\prime}}\exp\left(-\frac{i}{\hbar}p^{\prime\prime\prime}\cdot
q^{\prime\prime\prime}\right)\ \left|q^{\prime\prime\prime}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{N^{\frac{3}{2}}}\sum\limits_{q^{\prime},q^{\prime\prime},q^{\prime\prime\prime}}\exp\left(-\frac{i}{\hbar}p^{\prime}\cdot
q^{\prime}\right)\exp\left(-\frac{i}{\hbar}p^{\prime\prime}\cdot
q^{\prime\prime}\right)\exp\left(-\frac{i}{\hbar}p^{\prime\prime\prime}\cdot
q^{\prime\prime\prime}\right)\ \ \left|q^{\prime}\right\rangle\
\left|q^{\prime\prime}\right\rangle\left|q^{\prime\prime\prime}\right\rangle.$

We can expressed the right hand side in terms of ’correlated’ (entangled)
product states by writing, $q^{\prime\prime}=q^{\prime}+m$,
$q^{\prime\prime\prime}=q^{\prime}+l$ where $m$ and $l$ are the quantum-
correlation ’distances’,

$\displaystyle\left|p^{\prime}\right\rangle\left|p^{\prime\prime}\right\rangle\left|p^{\prime\prime\prime}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{N^{\frac{3}{2}}}\sum\limits_{q^{\prime},m,l}\exp\left(-\frac{i}{\hbar}p^{\prime}\cdot
q^{\prime}\right)\exp\left(-\frac{i}{\hbar}p^{\prime\prime}\cdot\left(q^{\prime}+m\right)\right)$
$\displaystyle\times\exp\left(-\frac{i}{\hbar}p^{\prime\prime\prime}\cdot\left(q^{\prime}+l\right)\right)\
\left|q^{\prime}\right\rangle\
\left|q^{\prime}+m\right\rangle\left|q^{\prime}+l\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{N^{\frac{3}{2}}}\sum\limits_{q^{\prime},m,l}\exp\left(-\frac{i}{\hbar}\left(p^{\prime}+p^{\prime\prime}+p^{\prime\prime\prime}\right)\cdot
q^{\prime}\right)$
$\displaystyle\times\exp\left(-\frac{i}{\hbar}p^{\prime\prime}\cdot
m\right)\exp\left(-\frac{i}{\hbar}p^{\prime\prime\prime}\cdot l\right)\
\left|q^{\prime}\right\rangle\
\left|q^{\prime}+m\right\rangle\left|q^{\prime}+l\right\rangle.$

Now writing $p^{\prime}+p^{\prime\prime}+p^{\prime\prime\prime}=p$, we have

$\displaystyle\left|p^{\prime}\right\rangle\left|p-p^{\prime}-p^{\prime\prime\prime}\right\rangle\left|p^{\prime\prime\prime}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\left(N\right)^{\frac{3}{2}}}\sum\limits_{m,l}\exp\left(-\frac{i}{\hbar}\left(p-p^{\prime}\right)\cdot
m\right)\exp\left(-\frac{i}{\hbar}p^{\prime\prime\prime}\cdot\left(l-m\right)\right)$
$\displaystyle\times\sum\limits_{q^{\prime}}\exp\left(-\frac{i}{\hbar}p\cdot
q^{\prime}\right)\left|q^{\prime}\right\rangle\
\left|q^{\prime}+m\right\rangle\left|q^{\prime}+l\right\rangle\ $
$\displaystyle=$
$\displaystyle\frac{1}{N}\sum\limits_{m,l}\exp\left(-\frac{i}{\hbar}\left(p-p^{\prime}\right)\cdot
m\right)\exp\left(-\frac{i}{\hbar}p^{\prime\prime\prime}\cdot\left(l-m\right)\right)$
$\displaystyle\times\left|\psi_{p,m,l}\right\rangle,$

where

$\left|\psi_{p,m,l}\right\rangle=\frac{1}{\sqrt{N}}\sum\limits_{q^{\prime}}\exp\left(-\frac{i}{\hbar}p\cdot
q^{\prime}\right)\left|q^{\prime}\right\rangle\
\left|q^{\prime}+m\right\rangle\left|q^{\prime}+l\right\rangle,$

is the three-qubit correlated (entangled) state. Substituting $N=2$, and
$p=\pi\hbar k$, we have

$\left|\psi_{k,m,l}\right\rangle=\frac{1}{\sqrt{2}}\sum\limits_{q^{\prime}=0}^{1}\exp\left(-i\pi
k\cdot q^{\prime}\right)\left|q^{\prime}\right\rangle\
\left|q^{\prime}+m\right\rangle\left|q^{\prime}+l\right\rangle,$

and obtain

$\left|\psi_{0,0,0}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle\left|0\right\rangle+\left|1\right\rangle\left|1\right\rangle\left|1\right\rangle\right)=\Theta_{3}^{+},$

which is the well-known Greenberger-Horne-Zeilinger state,
$\left|GHZ\right\rangle$, which is the maximally entangled state of three
qubits since $m=0$, and $l=0$ (a complete correlation). We give the
expressions for the rest of $\left|\psi_{k,m,l}\right\rangle$,

$\left|\psi_{0,0,1}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|1\right\rangle\left|0\right\rangle\right)=\Gamma^{+},$
$\left|\psi_{0,1,0}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle\left|0\right\rangle+\left|1\right\rangle\left|0\right\rangle\left|1\right\rangle\right)=\Omega^{+},$
$\left|\psi_{0,1,1}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|0\right\rangle\left|0\right\rangle\right)=\Xi^{+},$
$\left|\psi_{1,0,0}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle\left|0\right\rangle-\left|1\right\rangle\left|1\right\rangle\left|1\right\rangle\right)=\Theta_{3}^{-},$
$\left|\psi_{1,0,0}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle\left|0\right\rangle+\left|1\right\rangle\left|1\right\rangle\left|1\right\rangle\right)=\Theta_{3}^{+},$
$\left|\psi_{1,0,1}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|1\right\rangle\left|0\right\rangle\right)=\Gamma^{-},$
$\left|\psi_{1,1,0}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle\left|0\right\rangle-\left|1\right\rangle\left|0\right\rangle\left|1\right\rangle\right)=\Omega^{-},$
$\left|\psi_{1,1,1}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|0\right\rangle\left|0\right\rangle\right)=\Xi^{-}.$

Just like the Bell basis, the set of three-qubit correlated states,
$\left\\{\left|\psi_{k,m,l}\right\rangle\right\\}$, is a complete and
orthonormal basis set, since each element is connected to the complete
orthonormal product states by unitary transformations. In what follows, we
will to $\left\\{\left|\psi_{k,m,l}\right\rangle\right\\}$ as the three-
particle correlated basis (TPCB).

###### Remark 1

There is another three-qubit entangled state, the so-called $W$ state. A $W$
state is a collective spin state with one ‘excitation’ as compared to a
coherent spin state which is fully aligned. With all but one spin aligned in
the x-direction, the state can be written:

$\left|W\right\rangle=\frac{1}{\sqrt{N}}\left(\left|\downarrow\uparrow\uparrow...\uparrow\right\rangle_{1}+\left|\uparrow\downarrow\uparrow...\uparrow\right\rangle_{2}+....\left|\uparrow\uparrow\uparrow...\downarrow\right\rangle_{N}\right)$

The $\left|W\right\rangle$ maybe viewed as a ’zero momentum’ superposition of
’position coordinates’ indicated by the position of an excitation in an array
of fully-aligned spin

$\displaystyle\left|p\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{N}}\sum\limits_{i=1}^{N}e^{\frac{i}{\hbar}p\cdot
R_{i}}\left|R_{i}\right\rangle$ $\displaystyle\left|W\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{N}}\sum\limits_{i=1}^{N}e^{\frac{i}{\hbar}0\cdot
R_{i}}\left|R_{i}\right\rangle=\frac{1}{\sqrt{N}}\sum\limits_{i=1}^{N}\left|R_{i}\right\rangle$

where the ’position’ coordinate, $\left|R_{i}\right\rangle$, correspond to the
different position of one-excitation, $\downarrow$, in a spin array
$\left|spin\ \downarrow array\right\rangle_{i}$ as indicated above (more
appropriately in a spin ring). However, the ’position’ coordinates used is
only a small subspace of the complete and orthonormal product Hilbert space of
an array of spin, moreover $\left|W\right\rangle$ corresponds only to
$\left|p\right\rangle=\left|p=0\right\rangle$, i.e., zero-phase superposition.
In general there are also $N$ values of $p$ producing other
$\left|W\right\rangle$-type entangled states of array of spins with one
excitation. Furthermore, corresponding to the superposition of position
eigenstates to produce a Schroedinger wavefunction, one can also form an
arbitrary superposition of spin array $\left|R_{i}\right\rangle$ to form other
correlated (entangled) state, for exmple,

$\left|\Psi\right\rangle=\sum\limits_{i=1}^{N}\psi\left(R_{i}\right)\left|R_{i}\right\rangle$

where $\psi\left(R_{i}\right)$ may no longer be obtained from Schroedinger
equation, but from some optimization criteria, say for optimal condition of a
universal cloning machine, which for three-particle entangled state, in the
form (Phys. Rev. A 57, 2368 (1998)

$\left|\Psi\right\rangle=\sqrt{\frac{2}{3}}\left|100\right\rangle-\sqrt{\frac{1}{6}}\left|010\right\rangle-\sqrt{\frac{1}{6}}\left|001\right\rangle$

It is worth pointing out that the method of generating entangled states using
the product of ’momentum’ basis states, $\left|p\right\rangle$, in the manner
given above automatically yield complete orthonormal entangled basis states.

###  2.3 A qubit Teleportation Using Three-Particle Entanglement

Here we will used as the quantum channel shared by Alice and Bob to be the GHZ
state

$\left|GHZ\right\rangle=\Theta_{3}^{+}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_{1}\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left|1\right\rangle_{1}\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right).$

As before Alice wants to send to Bob the unknown state,
$\left|\psi\right\rangle=\alpha\left|0\right\rangle_{U}+\beta\left|1\right\rangle_{U}$.
To set up the teleportation, let us assume that particles $1$ and $2$ are kept
by Alice and particle $3$ is sent to Bob. So, Alice has three particles ($U$,
the one she wants to teleport, and particles $1$ and $2$, two of the entangled
three qubits), and Bob has one particle $3$. In the total system, the state of
these four particles is given by

$\displaystyle\left|\psi\right\rangle\otimes\left|\Phi_{3}^{+}\right\rangle$
$\displaystyle=$
$\displaystyle\left(\alpha\left|0\right\rangle_{U}+\beta\left|1\right\rangle_{U}\right)\otimes\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_{1}\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left|1\right\rangle_{1}\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)$
$\displaystyle=$
$\displaystyle\frac{\alpha}{\sqrt{2}}\left(\left[\left|0\right\rangle_{U}\left|0\right\rangle_{1}\left|0\right\rangle_{2}\right]\left|0\right\rangle_{3}+\left[\left|0\right\rangle_{U}\left|1\right\rangle_{1}\left|1\right\rangle_{2}\right]\left|1\right\rangle_{3}\right)$
$\displaystyle+\frac{\beta}{\sqrt{2}}\left(\left[\left|1\right\rangle_{U}\left|0\right\rangle_{1}\left|0\right\rangle_{2}\right]\left|0\right\rangle_{3}+\left[\left|1\right\rangle_{U}\left|1\right\rangle_{1}\left|1\right\rangle_{2}\right]\left|1\right\rangle_{3}\right),$

where we have enclosed in square bracket the particles belonging to Alice. We
will now express Alice particle states in terms of the TPCB basis. We have

$\left|0\right\rangle_{U}\left|0\right\rangle_{1}\left|0\right\rangle_{2}=\frac{1}{\sqrt{2}}\left(\Theta_{3}^{+}+\Theta_{3}^{-}\right)_{U12},$
$\left|1\right\rangle_{U}\left|1\right\rangle_{1}\left|1\right\rangle_{2}=\frac{1}{\sqrt{2}}\left(\Theta_{3}^{+}-\Theta_{3}^{-}\right)_{U12},$
$\left|0\right\rangle_{U}\left|1\right\rangle_{1}\left|1\right\rangle_{2}=\frac{1}{\sqrt{2}}\left(\Xi^{+}+\Xi^{-}\right)_{U12},$
$\left|1\right\rangle_{U}\left|0\right\rangle_{1}\left|0\right\rangle_{2}=\frac{1}{\sqrt{2}}\left(\Xi^{+}-\Xi^{-}\right)_{U12}.$

Therefore we can write

$\displaystyle\left|\psi\right\rangle\otimes\left|\Phi_{3}^{+}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{\alpha}{\sqrt{2}}\left(\left[\frac{1}{\sqrt{2}}\left(\Theta_{3}^{+}+\Theta_{3}^{-}\right)_{U12}\right]\left|0\right\rangle_{3}+\left[\frac{1}{\sqrt{2}}\left(\Xi^{+}+\Xi^{-}\right)_{U12}\right]\left|1\right\rangle_{3}\right)$
$\displaystyle+\frac{\beta}{\sqrt{2}}\left(\left[\frac{1}{\sqrt{2}}\left(\Xi^{+}-\Xi^{-}\right)_{U12}\right]\left|0\right\rangle_{3}+\left[\frac{1}{\sqrt{2}}\left(\Theta_{3}^{+}-\Theta_{3}^{-}\right)_{U12}\right]\left|1\right\rangle_{3}\right),$
$\displaystyle\left|\psi\right\rangle\otimes\left|\Phi_{3}^{+}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\Theta_{3}^{+}\right)_{U12}\left(\alpha\left|0\right\rangle_{3}+\beta\left|1\right\rangle_{3}\right)+\frac{1}{2}\left(\Xi^{+}\right)_{U12}\left(\alpha\left|1\right\rangle_{3}+\beta\left|0\right\rangle_{3}\right)$
$\displaystyle+\frac{1}{2}\left(\Theta_{3}^{-}\right)_{U12}\left(\alpha\left|0\right\rangle_{3}-\beta\left|1\right\rangle_{3}\right)+\frac{1}{2}\left(\Xi^{-}\right)_{U12}\left(\alpha\left|1\right\rangle_{3}-\beta\left|0\right\rangle_{3}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\Theta_{3}^{+}\right)_{U12}I\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right)+\frac{1}{2}\left(\Xi^{+}\right)_{U12}\sigma_{x}\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right)$
$\displaystyle+\frac{1}{2}\left(\Theta_{3}^{-}\right)_{U12}\sigma_{z}\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right)+\frac{1}{2}\left(\Xi^{-}\right)_{U12}\left(-i\sigma_{y}\right)\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right).$

Hence, it follows that regardless of the unknown state
$\left|\psi\right\rangle$, by using the maximally-entangled quantum channel,
$\left|GHZ\right\rangle$, Alice can still perform only four measurements with
outcomes equally likely with probability equal to $\frac{1}{4}$. After Alice
measurement, Bob’s particle $3$ will have been projected to one of the four
pure states, after which Bob has to perform the necessary transformation to
regain the original state that Alice has, as indicated in the following table,

Alice measurement | Bob’s particle state | Bob’s transformation
---|---|---
$\left(\Theta_{3}^{+}\right)_{U12}$ | $\left(\alpha\left|0\right\rangle_{3}+\beta\left|1\right\rangle_{3}\right)$ | $I$
$\left(\Theta_{3}^{-}\right)_{U12}$ | $\left(\alpha\left|0\right\rangle_{3}-\beta\left|1\right\rangle_{3}\right)$ | $\sigma_{z}$
$\left(\Xi^{+}\right)_{U12}$ | $\left(\alpha\left|1\right\rangle_{3}+\beta\left|0\right\rangle_{3}\right)$ | $\sigma_{x}$
$\left(\Xi^{-}\right)_{U12}$ | $\left(\alpha\left|1\right\rangle_{3}-\beta\left|0\right\rangle_{3}\right)$ | $i\sigma_{y}$

To transmit Alice’s classical (measurement) result to Bob, Alice needs a two-
bit classical channel to transmit which one of the four equally likely results
. This is in contrast with other proposed more complex teleportation scheme
using the same maximally entangled $\left|GHZ\right\rangle$ quantum channel
which employs Bob as an ancilla to transmit the unknown state to Cliff, and
use two classical bit between Alice and Bob and one classical bit between Bob
and Cliff. Here, whether we use the Bell basis or TPCB basis to transmit an
unknown quantum state, only two-bit classical channel is simply needed
directly between the sending and receiving parties, and without the use of
intermediary ancilla and only one local measurement is involved.

###  2.4 Teleportation Using Three-Particle Entanglement and an Ancilla

Instead of keeping the particles $1$ and $2$ as done above, Alice only keep
particle $1$ and send particle $2$ to Bob and particle $3$ to Cliff. The idea
here is that Alice will locally only measure one of the orthonormal two-
particle entangled Bell states, $\left|\Phi^{\pm}\right\rangle$,
$\left|\Psi^{\pm}\right\rangle$, instead of one of the complete orthonormal
three-particle correlated (entangled) states given above. The price to pay in
this scheme is that there is a need to have two bits of classical
communication channel between Alice and Bob and one more bit of classical
communication channel between Bob and Cliff. Moreover, two local measurements
has to be done, one local measurement by Alice on her Bell states and one
local measurement by Bob on his qubit, to implement the teleportation of one
qubit from Alice to Cliff. Thus, Bob becomes an ancilla in this teleportation
scheme.

Again, using the Alice wants to send to Bob the unknown state,
$\left|\psi\right\rangle=\alpha\left|0\right\rangle_{U}+\beta\left|1\right\rangle_{U}$,
using the $\left|GHZ\right\rangle$ shared quantum channel. To set up the
teleportation, Alice keep particle $1$ and send particle $2$ to Bob, and
particle $3$ to Cliff. So, Alice has two particles ($U$, the unknown qubit she
wants to teleport and particle $1$, one of the entangled three
$\left|GHZ\right\rangle$ qubits). In the total system, the state of these four
particles is given by

$\displaystyle\left|\psi\right\rangle\otimes\left|GHZ\right\rangle$
$\displaystyle=$
$\displaystyle\left(\alpha\left|0\right\rangle_{U}+\beta\left|1\right\rangle_{U}\right)\otimes\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_{1}\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left|1\right\rangle_{1}\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)$
$\displaystyle=$
$\displaystyle\frac{\alpha}{\sqrt{2}}\left(\left[\left|0\right\rangle_{U}\left|0\right\rangle_{1}\right]\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left[\left|0\right\rangle_{U}\left|1\right\rangle_{1}\right]\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)$
$\displaystyle+\frac{\beta}{\sqrt{2}}\left(\left[\left|1\right\rangle_{U}\left|0\right\rangle_{1}\right]\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left[\left|1\right\rangle_{U}\left|1\right\rangle_{1}\right]\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right),$

where the particles enclosed in square brackets belongs to Alice. Upon
changing the Alice product basis states to Bell basis, we have

$\displaystyle\left|\psi\right\rangle\otimes\left|GHZ\right\rangle$
$\displaystyle=$
$\displaystyle\left(\alpha\left|0\right\rangle_{U}+\beta\left|1\right\rangle_{U}\right)\otimes\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_{1}\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left|1\right\rangle_{1}\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)$
$\displaystyle=$
$\displaystyle\frac{\alpha}{\sqrt{2}}\left(\left[\frac{1}{\sqrt{2}}\left(\left|\Phi^{+}\right\rangle_{U1}+\left|\Phi^{-}\right\rangle_{U1}\right)\right]\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\frac{1}{\sqrt{2}}\left(\left|\Psi^{+}\right\rangle_{U1}+\left|\Psi^{-}\right\rangle_{U1}\right)\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)$
$\displaystyle+\frac{\beta}{\sqrt{2}}\left(\left[\frac{1}{\sqrt{2}}\left(\left|\Psi^{+}\right\rangle_{U1}-\left|\Psi^{-}\right\rangle_{U1}\right)\right]\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left[\frac{1}{\sqrt{2}}\left(\left|\Phi^{+}\right\rangle_{U1}-\left|\Phi^{-}\right\rangle_{U1}\right)\right]\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right),$
$\displaystyle\left|\psi\right\rangle\otimes\left|GHZ\right\rangle$
$\displaystyle=$
$\displaystyle\left(\alpha\left|0\right\rangle_{U}+\beta\left|1\right\rangle_{U}\right)\otimes\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_{1}\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left|1\right\rangle_{1}\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\begin{array}[]{c}\left|\Phi^{+}\right\rangle_{U1}\left(\alpha\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\beta\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)\\\
+\left|\Psi^{+}\right\rangle_{U1}\left(\alpha\left|1\right\rangle_{2}\left|1\right\rangle_{3}+\beta\left|0\right\rangle_{2}\left|0\right\rangle_{3}\right)\end{array}\right)$
$\displaystyle+\frac{1}{2}\left(\begin{array}[]{c}\left|\Phi^{-}\right\rangle_{U1}\left(\alpha\left|0\right\rangle_{2}\left|0\right\rangle_{3}-\beta\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)\\\
+\left|\Psi^{-}\right\rangle_{U1}\left(\alpha\left|1\right\rangle_{2}\left|1\right\rangle_{3}-\beta\left|0\right\rangle_{2}\left|0\right\rangle_{3}\right)\end{array}\right).$

A local measurement of one of the four Bell states, which have equal
probability equal to $\frac{1}{4}$, will project the joint state of the
particles $2$ and $3$ possessed by Bob and Cliff, respectively, into one of
the entangled states shown above. Assume that Alice local measurement yields
$\left|\Phi^{+}\right\rangle_{U1}$, which is communicated to Bob through a
classical two-bit channel. Then the state of particles $2$ and $3$ is

$\left|\psi_{23}\right\rangle=\alpha\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\beta\left|1\right\rangle_{2}\left|1\right\rangle_{3}.$

In order to effect a transfer of the unknown qubit to Cliff, Bob must now
unentangle his qubit from that of Cliff. To do this Bob has to make a local
measurement on his qubit $2$.

Suppose Bob’s measurement apparatus collapses his qubit state to two possible
outcomes, namely $k_{1}$ and $k_{2}$. Then Bob can decompose his incoming
state to the new basis states, namely, $\left|k_{1}\right\rangle$ and
$\left|k_{2}\right\rangle$, and write the unitary/orthogonal transformation
from the old basis to the new basis as

$\displaystyle\left|0\right\rangle_{2}$ $\displaystyle=$
$\displaystyle\sin\theta\ \left|k_{1}\right\rangle+\cos\theta\
\left|k_{2}\right\rangle,$ $\displaystyle\left|1\right\rangle_{2}$
$\displaystyle=$ $\displaystyle\cos\theta\
\left|k_{1}\right\rangle-\sin\theta\ \left|k_{2}\right\rangle.$

Then $\left|\psi_{23}\right\rangle$ in terms of Bob’s new basis is

$\displaystyle\left|\psi_{23}\right\rangle$ $\displaystyle=$
$\displaystyle\left(\sin\theta\ \left|k_{1}\right\rangle+\cos\theta\
\left|k_{2}\right\rangle\right)\alpha\left|0\right\rangle_{3}+\left(\cos\theta\
\left|k_{1}\right\rangle-\sin\theta\
\left|k_{2}\right\rangle\right)\beta\left|1\right\rangle_{3}$ $\displaystyle=$
$\displaystyle\sin\theta\
\alpha\left|0\right\rangle_{3}\left|k_{1}\right\rangle+\cos\theta\
\alpha\left|0\right\rangle_{3}\left|k_{2}\right\rangle+\cos\theta\
\beta\left|1\right\rangle_{3}\left|k_{1}\right\rangle-\sin\theta\beta\left|1\right\rangle_{3}\
\left|k_{2}\right\rangle$ $\displaystyle=$ $\displaystyle\left(\sin\theta\
\alpha\left|0\right\rangle_{3}+\cos\theta\
\beta\left|1\right\rangle_{3}\right)\left|k_{1}\right\rangle+\left(\cos\theta\
\alpha\left|0\right\rangle_{3}-\sin\theta\beta\left|1\right\rangle_{3}\right)\
\left|k_{2}\right\rangle.$

In general, the new basis and the old basis are mutually-unbiased
basis111Example of mutually-unbiased bases are the position basis states and
momentum basis states., which means $\sin\theta=\cos\theta$ or
$\theta=\frac{\pi}{4}$. Then $\left|\psi_{23}\right\rangle$ becomes

$\left|\psi_{23}\right\rangle=\left(\ \alpha\left|0\right\rangle_{3}+\
\beta\left|1\right\rangle_{3}\right)\frac{1}{\sqrt{2}}\left|k_{1}\right\rangle+\left(\
\alpha\left|0\right\rangle_{3}-\beta\left|1\right\rangle_{3}\right)\
\frac{1}{\sqrt{2}}\left|k_{2}\right\rangle.$

Whatever Bob’s measurement outcome is, Bob has to communicate to Cliff his
result through a one-bit classical channel. If the outcome of Bob’s local
measurement is $k_{1}$ then Cliff has the unknown qubit from Alice and do
nothing. On the other hand if Bob’s measurement yields $k_{2}$, then Cliff has
to perform a $\sigma_{z}$ transformation on his qubit to obtain the unknown
qubit from Alice.

###  2.5 Two-Qubit Teleportation Using Three-Particle Entanglement

We will now show that a full use of the capability of the GHZ quantum channel
is achieved when teleporting two qubit of information. As before, the quantum
channel shared by Alice and Bob to be the GHZ state

$\left|GHZ\right\rangle=\Theta_{3}^{+}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_{1}\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left|1\right\rangle_{1}\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right).$

But now Alice wants to send an unknown two-particle bits (triplet) to Bob,
namely,

$\left|\psi\right\rangle=\alpha\left|0\right\rangle_{U1}\left|0\right\rangle_{U2}+\delta\left|1\right\rangle_{U1}\left|1\right\rangle_{U2},$

which we may write in vector form as

$\left|\psi\right\rangle=\left(\begin{array}[]{c}\alpha\left|0\right\rangle_{U1}\left|0\right\rangle_{U2}\\\
\delta\left|1\right\rangle_{U1}\left|1\right\rangle_{U2}\end{array}\right)\Rightarrow\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right),$

where $\left|\alpha\right|^{2}+\left|\delta\right|^{2}=1$. To set up the two-
bit teleportation, particle $1$ is kept by Alice and particle $2$ are$3$ are
sent to Bob. In the total system, the state of these three particles is given
by

$\displaystyle\left|\psi\right\rangle\otimes\left|GHZ\right\rangle$
$\displaystyle=$
$\displaystyle\left(\alpha\left|0\right\rangle_{U1}\left|0\right\rangle_{U2}+\delta\left|1\right\rangle_{U1}\left|1\right\rangle_{U2}\right)$
$\displaystyle\otimes\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_{1}\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left|1\right\rangle_{1}\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\begin{array}[]{c}\alpha\left(\left|0\right\rangle_{U1}\left|0\right\rangle_{U2}\left|0\right\rangle_{1}\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left|0\right\rangle_{U1}\left|0\right\rangle_{U2}\left|1\right\rangle_{1}\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)\\\
+\delta\left(\left|1\right\rangle_{U1}\left|1\right\rangle_{U2}\left|0\right\rangle_{1}\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\left|1\right\rangle_{U1}\left|1\right\rangle_{U2}\left|1\right\rangle_{1}\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)\end{array}\right).$

Now we change basis using the complete orthonormal set of three-particle
correlated (entangled) states

$\left|\psi_{0,0,0}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle\left|0\right\rangle+\left|1\right\rangle\left|1\right\rangle\left|1\right\rangle\right)=\Theta_{3}^{+},$
$\left|\psi_{0,0,1}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|1\right\rangle\left|0\right\rangle\right)=\Gamma^{+},$
$\left|\psi_{0,1,0}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle\left|0\right\rangle+\left|1\right\rangle\left|0\right\rangle\left|1\right\rangle\right)=\Omega^{+},$
$\left|\psi_{0,1,1}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|0\right\rangle\left|0\right\rangle\right)=\Xi^{+},$
$\left|\psi_{1,0,0}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle\left|0\right\rangle-\left|1\right\rangle\left|1\right\rangle\left|1\right\rangle\right)=\Theta_{3}^{-},$
$\left|\psi_{1,0,1}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|1\right\rangle\left|0\right\rangle\right)=\Gamma^{-},$
$\left|\psi_{1,1,0}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle\left|0\right\rangle-\left|1\right\rangle\left|0\right\rangle\left|1\right\rangle\right)=\Omega^{-},$
$\left|\psi_{1,1,1}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|0\right\rangle\left|0\right\rangle\right)=\Xi^{-},$

to obtain

$\left|0\right\rangle_{U1}\left|0\right\rangle_{U2}\left|0\right\rangle_{1}=\frac{1}{\sqrt{2}}\left(\Theta_{3}^{+}+\Theta_{3}^{-}\right),$
$\left|0\right\rangle_{U1}\left|0\right\rangle_{U2}\left|1\right\rangle_{1}=\frac{1}{\sqrt{2}}\left(\Gamma^{+}+\Gamma^{-}\right),$
$\left|0\right\rangle_{U1}\left|1\right\rangle_{U2}\left|0\right\rangle_{1}=\frac{1}{\sqrt{2}}\left(\Omega^{+}+\Omega^{-}\right),$
$\left|0\right\rangle_{U1}\left|1\right\rangle_{U2}\left|1\right\rangle_{1}=\frac{1}{\sqrt{2}}\left(\Xi^{+}+\Xi^{-}\right),$
$\left|1\right\rangle_{U1}\left|0\right\rangle_{U2}\left|0\right\rangle_{1}=\frac{1}{\sqrt{2}}\left(\Xi^{+}-\Xi^{-}\right),$
$\left|1\right\rangle_{U1}\left|0\right\rangle_{U2}\left|1\right\rangle_{1}=\frac{1}{\sqrt{2}}\left(\Omega^{+}-\Omega^{-}\right),$
$\left|1\right\rangle_{U1}\left|1\right\rangle_{U2}\left|0\right\rangle_{1}=\frac{1}{\sqrt{2}}\left(\Gamma^{+}-\Gamma^{-}\right),$
$\left|1\right\rangle_{U1}\left|1\right\rangle_{U2}\left|1\right\rangle_{1}=\frac{1}{\sqrt{2}}\left(\Theta_{3}^{+}-\Theta_{3}^{-}\right).$

Upon changing the basis of Alice particles, we have

$\left|\psi\right\rangle\otimes\left|GHZ\right\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}[]{c}\alpha\left(\frac{1}{\sqrt{2}}\left(\Theta_{3}^{+}+\Theta_{3}^{-}\right)\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\frac{1}{\sqrt{2}}\left(\Gamma^{+}+\Gamma^{-}\right)\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)\\\
+\delta\left(\frac{1}{\sqrt{2}}\left(\Gamma^{+}-\Gamma^{-}\right)\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\frac{1}{\sqrt{2}}\left(\Theta_{3}^{+}-\Theta_{3}^{-}\right)\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)\end{array}\right),$
$\displaystyle\left|\psi\right\rangle\otimes\left|GHZ\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\begin{array}[]{c}\Theta_{3}^{+}\left(\alpha\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\delta\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)\\\
+\Theta_{3}^{-}\left(\alpha\left|0\right\rangle_{2}\left|0\right\rangle_{3}-\delta\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)\\\
+\Gamma^{+}\left(\delta\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\alpha\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)\\\
+\Gamma^{-}\left(-\delta\left|0\right\rangle_{2}\left|0\right\rangle_{3}+\alpha\left|1\right\rangle_{2}\left|1\right\rangle_{3}\right)\end{array}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\begin{array}[]{c}\Theta_{3}^{+}I\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right)\\\
+\Theta_{3}^{-}\sigma_{z}\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right)\\\
+\Gamma^{+}\sigma_{x}\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right)\\\
+\Gamma^{-}\left(-i\sigma_{y}\right)\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right)\end{array}\right).$

Thus, depending on the result of Alice measurement using the equi-probable,
with probability$=\frac{1}{4}$, three-particle entanglement basis,
$\Theta_{3}^{+}$, $\Theta_{3}^{-}$, $\Gamma^{+}$, and $\Gamma^{-}$, which is
communicated to Bob via classical two-bit channel, Bob will then use the
inverse transformation to recover the original triplet sent by Alice.

Alice can also send an unknown singlet to Bob, namely,

$\left|\psi\right\rangle=\alpha\left|0\right\rangle_{U1}\left|1\right\rangle_{U2}+\delta\left|1\right\rangle_{U1}\left|0\right\rangle_{U2}.$

However, the quantum channel shared by Alice and Bob must now be chosen to be
given by

$\left|Q_{ch}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_{1}\left|1\right\rangle_{2}\left|0\right\rangle_{3}+\left|1\right\rangle_{1}\left|0\right\rangle_{2}\left|1\right\rangle_{3}\right)$

instead of $\left|GHZ\right\rangle$ used above. To set up the two-bit
teleportation, particle $1$ is kept by Alice and particle $2$ are$3$ are sent
to Bob. In the total system, the state of these three particles is given by

$\displaystyle\left|\psi\right\rangle\otimes\left|Q_{ch}\right\rangle$
$\displaystyle=$
$\displaystyle\left\\{\alpha\left|0\right\rangle_{U1}\left|1\right\rangle_{U2}+\delta\left|1\right\rangle_{U1}\left|0\right\rangle_{U2}\right\\}\otimes\left\\{\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_{1}\left|1\right\rangle_{2}\left|0\right\rangle_{3}+\left|1\right\rangle_{1}\left|0\right\rangle_{2}\left|1\right\rangle_{3}\right)\right\\}$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\begin{array}[]{c}\alpha\left(\left|0\right\rangle_{U1}\left|1\right\rangle_{U2}\left|0\right\rangle_{1}\left|1\right\rangle_{2}\left|0\right\rangle_{3}+\left|0\right\rangle_{U1}\left|1\right\rangle_{U2}\left|1\right\rangle_{1}\left|0\right\rangle_{2}\left|1\right\rangle_{3}\right)\\\
+\delta\left(\left|1\right\rangle_{U1}\left|0\right\rangle_{U2}\left|0\right\rangle_{1}\left|1\right\rangle_{2}\left|0\right\rangle_{3}+\left|1\right\rangle_{U1}\left|0\right\rangle_{U2}\left|1\right\rangle_{1}\left|0\right\rangle_{2}\left|1\right\rangle_{3}\right)\end{array}\right).$

Upon changing the basis to three-particle entanglement basis, we have

$\displaystyle\left|\psi\right\rangle\otimes\left|Q_{ch}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\begin{array}[]{c}\alpha\left(\frac{1}{\sqrt{2}}\left(\Omega^{+}+\Omega^{-}\right)\left|1\right\rangle_{2}\left|0\right\rangle_{3}+\frac{1}{\sqrt{2}}\left(\Xi^{+}+\Xi^{-}\right)\left|0\right\rangle_{2}\left|1\right\rangle_{3}\right)\\\
+\delta\left(\frac{1}{\sqrt{2}}\left(\Xi^{+}-\Xi^{-}\right)\left|1\right\rangle_{2}\left|0\right\rangle_{3}+\frac{1}{\sqrt{2}}\left(\Omega^{+}-\Omega^{-}\right)\left|0\right\rangle_{2}\left|1\right\rangle_{3}\right)\end{array}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\begin{array}[]{c}\Omega^{+}\left(\delta\left|0\right\rangle_{2}\left|1\right\rangle_{3}+\alpha\left|1\right\rangle_{2}\left|0\right\rangle_{3}\right)\\\
+\Xi^{+}\left(\alpha\left|0\right\rangle_{2}\left|1\right\rangle_{3}+\delta\left|1\right\rangle_{2}\left|0\right\rangle_{3}\right)\\\
+\Omega^{-}\left(-\delta\left|0\right\rangle_{2}\left|1\right\rangle_{3}+\alpha\left|1\right\rangle_{2}\left|0\right\rangle_{3}\right)\\\
+\Xi^{-}\left(\alpha\left|0\right\rangle_{2}\left|1\right\rangle_{3}-\delta\left|1\right\rangle_{2}\left|0\right\rangle_{3}\right)\end{array}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\begin{array}[]{c}\Omega^{+}\sigma_{x}\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right)\\\ +\Xi^{+}I\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right)\\\
+\Omega^{-}\left(-i\sigma_{y}\right)\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right)\\\
+\Xi^{-}\sigma_{z}\left(\begin{array}[]{c}\alpha\\\
\delta\end{array}\right)\end{array}\right).$

Thus, depending on the result of Alice measurement using the equi-probable,
with probability$=\frac{1}{4}$, three-particle entanglement basis,
$\Omega^{+}$, $\Omega^{-}$, $\Xi^{+}$, and $\Xi^{-}$, which is communicated to
Bob via classical two-bit channel, Bob will then use the inverse
transformation to recover the original singlet sent by Alice.

##  3 MECHANICAL MODELING OF QUANTUM ENTANGLEMENTS

Here, we now give a simple and pedagogical implementation of quantum
entanglement using a chain of mechanical inverters, see-saws or swings. This
mechanical model has been previously used by the author in investigating the
fundamental physical limits of computational processes [2].

###  3.1 Mechanical model for entangled two qubits

A schematic picture of an entangled two qubits is shown in Figs. 1 and 2. The
frictionless guides and fulcrum would make the inverter chain reversible. What
is more important though, from the point of view of entanglement, is the
presence of rigid coupling. Rigid coupling allows for simultaneous events to
take place irrespective of the length of the intermediary chain, thus
capturing the essence of quantum entanglement.

No communication is involved since the whole process is a reconfiguration of
the whole chain system. No hidden variables and the deterministic
reconfiguration process does not violate Bell’s inequality. In other words,
this mechanical model is conceptually an ideal model of quantum entanglement.
One is tempted to speculate whether spacetime provides rigid coupling in
realistic situations, indeed due to spacetime entanglement [3].

Figure 1 shows a configuration for a triplet entanglement.

Figure 1: An entangled spin triplet.

Another independent configuration of the chain embodies the spin singlet
configuration shown in Fig. 2

Figure 2: An entangled spin singlet.

##  4 Diagrammatic Techniques

In the following diagrammatic analyses, the inverter chains of Figs. 1 and 2
are abstracted into line segment representations.

###  4.1 Two-level diagram representations of entangled two qubits

The line diagrams below is an abstractions of the inverter chain of Fig. 1 and
Fig. 2, respectively. The first line of Fig. 3 represents a two-state diagram,
with entangled basis $\left|0\right\rangle\left|0\right\rangle$ and
$\left|1\right\rangle\left|1\right\rangle$. The second line represents a two-
state diagram, with entangled basis $\left|0\right\rangle\left|1\right\rangle$
and $\left|1\right\rangle\left|0\right\rangle$.

Figure 3: Two two-state diagrams for an entangled two qubits. The
calculational entangled basis is obtained using Hadamard transformation or
discrete Fourier transform for two-state systems.

####  4.1.1 Hadamard transformation for calculational entangled basis states

The Hadamard transform of the entangled basis states of two qubits given above
is what we referred to as the calculational entangled basis states. These are
the Bell entangled basis states derived formally before. For the first diagram
of Fig. 3, we have,

$\left(\begin{array}[]{c}\left|\Phi^{+}\right\rangle\\\
\left|\Phi^{-}\right\rangle\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)$

and for the second diagram,

$\left(\begin{array}[]{c}\left|\Psi^{+}\right\rangle\\\
\left|\Psi^{-}\right\rangle\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\end{array}\right)$

###  4.2 Mechanical modeling of entangled three qubits

For entangled $n$ qubits, there are $\frac{2^{n}}{2}$ integer number of two-
state entangled basis diagrams. Thus, for three qubits the four diagrams shown
in Fig. 4.

Figure 4: Two-state four diagrams for entangled three qubits.

From the diagrams of Fig. 4, and using the Hadamard transformation, we can
immediately write down the calculational entangled basis states for three
qubits.

$\left(\begin{array}[]{c}\Xi^{+}\\\
\Xi^{-}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|1\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\left|0\right\rangle\end{array}\right)$

$\left(\begin{array}[]{c}\Theta_{3}^{+}\\\
\Theta_{3}^{--}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\left|1\right\rangle\end{array}\right)$

$\left(\begin{array}[]{c}\Omega^{+}\\\
\Omega^{-}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\left|1\right\rangle\end{array}\right)$

$\left(\begin{array}[]{c}\Gamma^{+}\\\
\Gamma^{-}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\left|0\right\rangle\end{array}\right)$

###  4.3 Mechanical modeling of entangled four qubits

We can consider eight two-entangled-state diagrams for entangled four qubits.
These are shown in Fig. 5

Figure 5: Two-state eight diagrams for entangled four qubits.

Following the same procedure using Hadamard Fourier transformation, we can
also immediately write down the calculational entangled basis states. Here, we
use our own notations, $\Phi_{j}^{\pm}$, for the transformed calculational
basis states. Following the order of diagrams in Fig. 5, we have

$\left(\begin{array}[]{c}\Phi_{1}^{+}\\\
\Phi_{1}^{-}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\left|0\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\left|1\right\rangle\left|1\right\rangle\end{array}\right)$
$\left(\begin{array}[]{c}\Phi_{2}^{+}\\\
\Phi_{2}^{-}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\left|1\right\rangle\left|0\right\rangle\end{array}\right)$
$\left(\begin{array}[]{c}\Phi_{3}^{+}\\\
\Phi_{3}^{-}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\left|0\right\rangle\left|1\right\rangle\end{array}\right)$
$\left(\begin{array}[]{c}\Phi_{4}^{+}\\\
\Phi_{4}^{-}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\left|1\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\left|0\right\rangle\left|0\right\rangle\end{array}\right)$
$\left(\begin{array}[]{c}\Phi_{5}^{+}\\\
\Phi_{5}^{-}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|1\right\rangle\left|0\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\left|1\right\rangle\left|1\right\rangle\end{array}\right)$
$\left(\begin{array}[]{c}\Phi_{6}^{+}\\\
\Phi_{6}^{-}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|1\right\rangle\left|1\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\left|0\right\rangle\left|1\right\rangle\end{array}\right)$
$\left(\begin{array}[]{c}\Phi_{7}^{+}\\\
\Phi_{7}^{-}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|1\right\rangle\left|1\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\left|0\right\rangle\left|0\right\rangle\end{array}\right)$
$\left(\begin{array}[]{c}\Phi_{8}^{+}\\\
\Phi_{8}^{-}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|0\right\rangle\left|1\right\rangle\left|0\right\rangle\left|1\right\rangle\\\
\left|1\right\rangle\left|0\right\rangle\left|1\right\rangle\left|0\right\rangle\end{array}\right)$

In general, one can systematically write down the calculational entangled
basis states for any number, $n$, of qubits by using the physically meaningful
diagrammatic techniques prescribed here. One does not even need to draw the
diagrams themselves, but simply list the $\frac{2^{n}}{2}$ entangled basis
basis to deduce the calculational entangled basis states by Hadamard
transformation of the two configurations of imagined entangled diagrams.

##  5 Concluding Remarks

The apparently perfect scheme of the mechanical model in describing quantum
entanglement seems to advocate the existence of entangled spacetime-medium to
take the place of the chain of inverters in some unknown forms. An interesting
analogy is the bound/entangled pair of vortices and antivortex on the surface
of fluids which are really the ends of a vortex chain (tube) forming a U-shape
underneath the surface. For example, in $3$-D the superfluid quantized
vortices form a metastable closed ring or open chain ending at the surface. A
vortex chain with both ends ending at the same surface appears as a bound
vortex/anti-vortex pair at the surface. Thus, it seems extra dimensions are
needed in spacetime to have a theory of entanglement. The work of Ooguri [5]
and collaborators shows that this quantum entanglement generates the extra
dimensions of the gravitational theory. ”It appears that it seems possible to
generate a geometric connection between entangled qubits, even though there is
no direct interaction between the two systems [4]. Furthermore, the solid and
reliable structure of spacetime is due to the ghostly features of
entanglement.

Could it be that besides the geometrical spacetime aspects of gravity, there
is a purely quantum mechanical aspect of spacetime geometry with extra
dimensons that give rise to entanglement? A hint along this idea is also given
by Malcedona [4] when he stated that ”One can consider, therefore, a pair of
black holes where all the microstates are “entangled.” Namely, if we observe
one of the black holes in one particular microstate, then the other has to be
in exactly the same microstate. A pair of black holes in this particular EPR
entangled state would develop a wormhole, or Einstein-Rosen bridge, connecting
them through the inside. The geometry of this wormhole is given by the fully
extended Schwarzschild geometry. It is interesting that both wormholes and
entanglement naively appear to lead to a propagation of signals faster than
light.” ”It was known that quantum entanglement is related to deep issues in
the unification of general relativity and quantum mechanics, such as the black
hole information paradox and the firewall paradox,” says Hirosi Ooguri” [5].

In any case, from the computational point of view, deriving the computational
entangled basis states have now been given for the first-time a systematic and
physically meaningful procedure using a diagrammatic foundation. One can also
define a closed group of entanglement diagrams. For example, a direct product
of entangled diagrams in Fig. 3 generates the ordered fusion algebra:

$\displaystyle\text{triplet}\otimes\text{triplet}$ $\displaystyle=$ triplet
$\displaystyle\text{triplet}\otimes\text{singlet}$ $\displaystyle=$ singlet
$\displaystyle\text{singlet}\otimes\text{singlet}$ $\displaystyle=$ triplet

## References

  * [1] Felix A. Buot, ”Nonequilbrium Quantum Transport Physics in Nanosystems” (World Scientific, 2009) and references therein.
  * [2] F.A. Buot, C. Scott, I. Mack, and K.J. Sieger, “Fundamental Physical Limits of Computation: Rigid Coupling Model, Broken Symmetry and Self-Induced Transmission in a Computer Medium,” in F.L. Carter, R.A. Sietkowski, and H. Wohltjen (eds.) Molecular Electronic Devices: Proceedings of the 3rd International Symposium on Molecular Electronic Devices (North Holland, Amsterdam 1989).
  * [3] https://vixra.org/pdf/1505.0206v1.pdf ”Spacetime is built by Quantum Entanglement”
  * [4] J. Maldacena, https://www.ias.edu/ideas/2013/maldacena-entanglement
  * [5] https://phys.org/news/2015-05-spacetime-built-quantum-entanglement.html