# On Entanglement Measures: Discrete Phase Space and Inverter-Chain Link
Viewpoint

Felix A. Buot
C&LB Reseaarch Institute, Carmen, Cebu 6005, Philippines;
LCFMNN, TCSE Group, Department of Physics,
University of San Carlos, Talamban, Cebu City 6000, Philippines

###### Abstract

In contrast to abstract statistical analyses in the literature, we present a
concrete physical diagrammatic model of entanglement characterization and
measure with its underlying discrete phase-space physics. This paper serves as
a pedagogical treatment of this complex subject of entanglement measures. We
review the important inherent concurrence property of entangled qubits, as
well as underscore its emergent qubit behavior. From the discrete phase space
point of view, concurrence translates to translation symmetry of entangled
binary systems in some quantitative measure of entanglement. Although the
focus is on bipartite system, the notion is readily extendable to multi-
partite system of qubits, as can easily be deduced from the physical inverter-
chain link model. A diagrammatic analysis of the entanglement of formation for
any multi-partite qubit system is given.

##  1 Introduction

Quantum entanglement has developed from a mere intellectual curiosity [1] of
the fundamental structure of quantum mechanics111Note that although Bell’s
theorem asserts the nonlocality of quantum mechanics, the EPR inquiry is still
not resolved, i.e., what is is still left unanswered is the mysterious ’link’
between qubits corresponding to our ”see-saw” or mechanical inverter-chain
link. (Note: Bell’s inequality theorem is widely discussed in the literature
and websites, we prefer not to cite specific reference). to become an
important and practical resource for quantum information processing in the
evolving theory of quantum information and ultra-fast computing. Thus, the
quantitative measure of entanglement has developed into one of the most active
fields of theoretical and experimental research. Here we will try to shed more
light on some of the important concepts in the quantification of quantum
entanglement by using a concrete simple mechanical model of a bipartite system
of qubits or chain of qubits. This treatment will be in contrast with mostly
abstract and statistical treatment of entanglement measure in the literature.

For the two-qubit entanglement, we will focus on the so-called entanglement of
formation and concurrence, two of the most important concepts to characterize
entanglement resource. Here we consider a qubit as a two-state system.
Moreover, we also consider an entangled qubit as effectively a two-state
system, an emergent qubit as depicted in our physical inverter-chain link
model. A two-state system has a unity entropy. Thus, a maximally entangled
state has entropy equal to $1$. From the discrete phase space point of view,
any two-state system can be considered to possess two lattice-site states.
Discrete Fourier transform or Hadamard transform implements unitary
superposition of the two lattice-site states (also referred to as ’Wannier
functions’) to yield a sort of crystal-momentum states (also referred to as
’Bloch functions’). For example, take the $\left|00\right\rangle$ and
$\left|11\right\rangle$ ’lattice-site states’, then the Hadamard bijective
discrete transformation gives the ’crystal-momentum states’, $\Phi^{+}$ and
$\Phi^{-}$, which are two of the Bell basis states (or ’Bloch function’
states).

##  2 Bell Basis Deduced from Inverter-Chain Link Model

We sketch here the derivation of the Bell basis states from our physical
model, as depicted in Fig. 1. The derivation is based in considering each
entangled diagram as a two-state (binary system) of emergent qubit. Clearly,
the entanglement of two bare qubits is divided into two orthogonal spaces of
triplet 222The use of the term ”triplet” is actually a misnomer here since the
entangled system is not free to assume a singlet or zero spin state. Thus,
this term is used here only as a label. and singlet entanglement states.

Figure 1: Physical diagrammatic model of ”triplet” (left) and singlet (right)
entanglement. By construction, each diagram is viewed as a two-state system,
respectively. The Bell basis are readily derived below each diagram, using the
Hadamard transformation. The actual physical implementation of the chain of
inverters may need frictionless male/female sliding tube coupling for large-
angle swing, but this is beside the point.

We also have the inverse relationship

$\left(\begin{array}[]{c}\left|00\right\rangle\\\
\left|11\right\rangle\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|\Phi^{+}\right\rangle\\\
\left|\Phi^{-}\right\rangle\end{array}\right)$ (1)
$\left(\begin{array}[]{c}\left|01\right\rangle\\\
\left|10\right\rangle\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\left|\Psi^{+}\right\rangle\\\
\left|\Psi^{-}\right\rangle\end{array}\right)$ (2)

We refer to $\left\\{\left|00\right\rangle,\left|11\right\rangle\right\\}$ as
the ”Wannier functions” space and the
$\left\\{\left|\Phi^{+}\right\rangle,\left|\Phi^{-}\right\rangle\right\\}$ as
the corresponding ”Bloch functions” space of a two-state triplet entanglement
system. Likewise,
$\left\\{\left|01\right\rangle,\left|10\right\rangle\right\\}$ as the ”Wannier
functions” space and the
$\left\\{\left|\Psi^{+}\right\rangle,\left|\Psi^{-}\right\rangle\right\\}$ as
the corresponding ”Bloch functions” space of a two-state singlet entanglement
system.

By virtue of this bijective relationship, any function of Wannier functions
will have a corresponding function of Bloch functions. For example, a
maximally mixed Wannier function states will generate a maximally mixed Bloch
function states or the so-called maximally mixed entanglement states. Mixed
and pure states will be further discussed below.

##  3  Entangled Qubits as an Emergent Qubit

The virtue of our inverter-chain link model is that the emergent two-state
property of entangled qubits is very transparent, since changing the state of
one of the qubits also immediately change the states of the rest of the
entangled partner(s) 333The idea that entanglement is due to conservation of
momentum does not hold for triplet entanglement since its two states give
opposing spin-angular momentum. However, one may interpret that quantum
superposition of the two-opposing angular momentum states conserves the
overall zero net spin-angular momentum, as supported by their eigenvalues
similar to a single qubit. On the other hand, for singlet entanglement, the
zero angular momentum is conserve in both two states. This seemingly apparent
physical difference of the triplet and singlet entanglements underscores the
importance of resolving the ”mysterious link” in the EPR inquiry, [1] in order
to further advance theoretical physics.. Here, we will rigorously justify our
claim that entangled qubits behave, as a whole, as an emergent qubit.

Here, we employ the matrix representation of states and operators. We now
represent $\left|\Phi^{+}\right\rangle$ and as

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

Then we have

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

proving the two eigenvalues, like a qubit property of the triplet system.

For the singlet, it is more convenient to change the phase of the second term
by $i$ so that we have

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

Then, we have

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

again proving the qubit property of the singlet system, where the two-state
aspect which is quite apparent in Fig. (1).

The above manipulation for the singlet is a bit round-about trick for the
singlet entanglement. The most straightforward manipulation is to recognize
that the singlet system as an independent system with its own two states which
is orthogonal to the triplet states. In this sense we can also represent the
singlet states just like that of the triplet states, namely,

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

Note that in the ’Bloch-state” space, $\sigma_{z}$ in the ”Wannier-state”
space is transfrom to $\sigma_{x}$ in entangled states,

$\displaystyle\sigma_{x}$ $\displaystyle=$ $\displaystyle H\sigma_{z}H$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right)\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}0&1\\\ 1&0\end{array}\right)$

Thus, we also have for the singlet states

$\displaystyle\left\langle\Psi^{+}\right|\sigma_{x}\left|\Psi^{+}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\begin{array}[]{cc}1&1\end{array}\right)\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)\left(\begin{array}[]{c}1\\\ 1\end{array}\right)=1$ (20)
$\displaystyle\left\langle\Psi^{-}\right|\sigma_{x}\left|\Psi^{-}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\begin{array}[]{cc}1&-1\end{array}\right)\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)\left(\begin{array}[]{c}1\\\ -1\end{array}\right)=-1$
(26)

Following this notion, Eqs. (20) and (26) can easily be extended to all
entangled qubits, bipartite or multi-partite qubit systems to yield an
emergent qubit. It is far more simpler to analyze the inverter-chain link
diagrams or employ diagrammatic analyses.

##  4 Translational/Shift Invariance of Bell Basis

The translational invariance of maximally entangled Bell basis states
demonstrates a unique characteristic of entangled qubits. This unique property
has been used to detect or measure how much entanglement is present in
arbitrary pure and mixed states. This is implemented in terms of the notion of
concurrence, to be elaborated below.

The translational or shift operation (also rreferred to as the flip operation
in the literature) is demonstrated here for the Bell Basis states and their
superpositions. First, let us consider all the Bell basis states. We have

$\Phi^{+}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle+\left|1\right\rangle\left|1\right\rangle\right)$

Upon applying the translation (shift by
$+1\left(\mathop{\mathrm{m}od}2\right)$) operator, $T\left(+1\right)$, we
obtain

$\displaystyle T\left(+1\right)\Phi^{+}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|0+1\right\rangle\left|0+1\right\rangle+\left|1+1\right\rangle\left|1+1\right\rangle\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|1\right\rangle\left|1\right\rangle+\left|0\right\rangle\left|0\right\rangle\right)$
$\displaystyle=$ $\displaystyle\Phi^{+}$

where addition obeys modular arithmetic ($mod$ $2$). Similarly, we have

$\displaystyle T\left(+1\right)\Phi^{-}$ $\displaystyle=$
$\displaystyle-\Phi^{-}$ $\displaystyle T\left(+1\right)\Psi^{-}$
$\displaystyle=$ $\displaystyle-\Psi^{-}$ $\displaystyle
T\left(+1\right)\Psi^{+}$ $\displaystyle=$ $\displaystyle\Psi^{+}$

We still consider the $-\Phi^{-}$ and $-\Psi^{-}$as invariant since it differs
from the unshifted Bell states by a global phase factor. In the literature
this translation operation is the so-called flipping operation first used by
Wooters, et al, [2, 3]

###  4.1 Translational property of maximal superposition of Bell basis

Now let us consider the superposition of maximally entangled Bell basis
states. We can readily see that applying the Hadamard transformation to two
Bloch states will yield the Wannier states. We have

$\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&1\\\
1&-1\end{array}\right)\left(\begin{array}[]{c}\Phi^{+}\\\
\Phi^{-}\end{array}\right)=\left(\begin{array}[]{c}\left|0\right\rangle\left|0\right\rangle\\\
\left|1\right\rangle\left|1\right\rangle\end{array}\right)$ (27)

Thus,

$\displaystyle\frac{1}{\sqrt{2}}\left(\Phi^{+}+\Phi^{-}\right)$
$\displaystyle=$ $\displaystyle\left|0\right\rangle\left|0\right\rangle$
$\displaystyle\frac{1}{\sqrt{2}}\left(\Phi^{+}-\Phi^{-}\right)$
$\displaystyle=$ $\displaystyle\left|1\right\rangle\left|1\right\rangle$ (28)

The thing to notice is that although the superposition is made up of two
maximally entangled Bell basis, the results are not shift invariant (i.e.,
generation of another Wannier state yields Wannier state located at another
site, which is orthogonal to the unshifted one). This means that the
superposition given above yields unentangled states (product states) in Eq.
(28). In what follows, we will see that only the following combination of
entangled basis states yields another entangled basis states as long as they
both belong to either ’even’ or ’odd’ spaces, i.e., by diagrammatic
construction, we have,

$\displaystyle triplet^{\pm}\otimes triplet^{\pm}$ $\displaystyle=$
$\displaystyle triplet^{\pm}\text{ or }singlet^{\pm}$ (29) $\displaystyle
triplet^{\pm}\otimes singlet^{\pm}$ $\displaystyle=$ $\displaystyle
singlet^{\pm}\text{ or }triplet^{\pm}$ (30) $\displaystyle
singlet^{\pm}\otimes singlet^{\pm}$ $\displaystyle=$ $\displaystyle
triplet^{\pm}\text{ or }singlet^{\pm}$ (31)

Equation (30) is quite interesting because it holds on a complete expansion of
a direct product of two qubit states. These relations can easily be deduced
from the physical diagrammatic model, see Fig. 1. For example, if we define
the operation as a superposition such as

$\displaystyle\frac{1}{\sqrt{2}}\left(\Phi^{+}+\Psi^{+}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(\left|1\right\rangle\left|1\right\rangle+\left|0\right\rangle\left|0\right\rangle\right)+\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|0\right\rangle\right)\right)$
(32) $\displaystyle=$
$\displaystyle\frac{1}{2}\left[\left(\left|1\right\rangle\left|1\right\rangle+\left|0\right\rangle\left|0\right\rangle\right)+\left(\left|0\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|0\right\rangle\right)\right]$
$\displaystyle=$
$\displaystyle\left[\frac{1}{\sqrt{2}}\left(\left|1\right\rangle+\left|0\right\rangle\right)\otimes\frac{1}{\sqrt{2}}\left(\left|1\right\rangle+\left|0\right\rangle\right)\right]$

We see that Eq. (32) is a direct product state of two qubits. We will see in
what follows that this is an entangled state and corresponds to Eq. (30) of
the $singlet^{+}$ or $triplet^{+}$ of the model diagrams depending on the
actual linkage. Similarly, we have,

$\displaystyle\frac{1}{\sqrt{2}}\left(\Phi^{-}+\Psi^{-}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(\left|1\right\rangle\left|1\right\rangle-\left|0\right\rangle\left|0\right\rangle\right)+\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|0\right\rangle\right)\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left[\left(\left|1\right\rangle\left|1\right\rangle-\left|0\right\rangle\left|0\right\rangle\right)+\left(\left|0\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|0\right\rangle\right)\right]$
$\displaystyle=$
$\displaystyle\left[\frac{1}{\sqrt{2}}\left(\left|1\right\rangle+\left|0\right\rangle\right)\otimes\frac{1}{\sqrt{2}}\left(\left|1\right\rangle-\left|0\right\rangle\right)\right]$

corresponds to Eq. (30) of the $singlet$ or $triplet$ of the model diagrams,
depending on the actual linking.

However, the following combinations of ’even’ and ’odd’ entangled states
result in unentangled states, namely,

$\displaystyle\frac{1}{\sqrt{2}}\left(\Phi^{+}+\Psi^{-}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(\left|1\right\rangle\left|1\right\rangle+\left|0\right\rangle\left|0\right\rangle\right)+\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|0\right\rangle\right)\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left[\left(\left|1\right\rangle\left|1\right\rangle+\left|0\right\rangle\left|0\right\rangle\right)+\left(\left|0\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|0\right\rangle\right)\right]$

and

$\displaystyle\frac{1}{\sqrt{2}}\left(\Phi^{-}+\Psi^{+}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left(\left|1\right\rangle\left|1\right\rangle-\left|0\right\rangle\left|0\right\rangle\right)+\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|0\right\rangle\right)\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left[\left(\left|1\right\rangle\left|1\right\rangle-\left|0\right\rangle\left|0\right\rangle\right)+\left(\left|0\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|0\right\rangle\right)\right]$

by virtue of the failure to have global sign factors. All these claims are
justified through the concept of concurrence, an inherent property of
entangled qubits, to be discussed in what follows. Moreover, this feature of
failing to have global sign factor is also reflected in the failure to
represent by our inverter-chain link diagrams.

## Bipartite System

Let the Hilbert spaces of a bipartite system consisting of $A$ and $B$ be
denoted by $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$, respectively. A bipartite
system is a system with Hilbert space equal to the direct product of
$\mathcal{H}_{A}$ and $\mathcal{H}_{B}$, i.e.,

$\mathcal{H}_{AB}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}$

Let the density matrix for the whole system be denoted by $\rho$. Then the
reduced density matrix of a subsystem $A$ is given by the partial trace,

$\rho_{A}=Tr_{{}_{B}}\ \rho$

The entanglement entropy, $S_{A}$, is the von Newmann entropy of the reduced
density matrix, $\rho_{A}$,

$S_{A}=-Tr\ \rho_{A}\log\rho_{A}$

###### Example 1

Let $\Omega$ be the number of distinct states in subsystem $A.$ Assume a
uniform distribution among states, hence $\rho_{A}$ has eigenvalues
$\frac{1}{\Omega}$, i.e., $\rho_{A}$ can be represented by a diagonal
$\Omega\times\Omega$ matrix with identical matrix elements given by
$\frac{1}{\Omega}$. Thus, in taking the trace we can use the eigenvalues of
the reduced density matrix operator, $\rho_{A}$. Therefore, we have

$\displaystyle S_{A}$ $\displaystyle=$ $\displaystyle-
Tr\frac{1}{\Omega}\log\frac{1}{\Omega}$ $\displaystyle=$
$\displaystyle-\Omega\left(\frac{1}{\Omega}\log\frac{1}{\Omega}\right)$
$\displaystyle=$ $\displaystyle-\log\frac{1}{\Omega}$ $\displaystyle=$
$\displaystyle\log\ \Omega$

Upon multiplying by the Boltzmann constant, $k_{B}$, we obtain

$k_{B}S_{A}=k_{B}\log\ \Omega$

which is the Boltzmann thermodynamic entropy, based on ergodic theorem.

###### Example 2

Two qubit system.

A qubit is simply a quantum bit whose number of distinct eigenstates is $2$.
We denote these eigenstates as $\left|0\right\rangle$ and
$\left|1\right\rangle$, i.e., a two-state system. If each subsystem $A$ or $B$
is a single qubit, then the Hilbert space of the whole system is span by the
following $4$ direct product states, namely,

$\left|00\right\rangle\text{, }\left|01\right\rangle\text{,
}\left|10\right\rangle\text{, }\left|11\right\rangle$

Here, the first bit refers to subsystem $A$ and the second bit refers to
subsystem $B$, where we write

$\left|ij\right\rangle=\left|i\right\rangle_{A}\ \left|j\right\rangle_{B}\
\epsilon\ \mathcal{H}_{A}\otimes\mathcal{H}_{B}$

Now, the density matrix of a pure state is,

$\rho=\left|\psi\right\rangle\left\langle\psi\right|$

then

$\displaystyle\rho^{2}$ $\displaystyle=$
$\displaystyle\left|\psi\right\rangle\left|\psi\right\rangle\left\langle\psi\right|\left|\psi\right\rangle\left\langle\psi\right|$
$\displaystyle=$ $\displaystyle\left|\psi\right\rangle\left\langle\psi\right|$
$\displaystyle=$ $\displaystyle\rho$

An operator whose square is equal to itself must have an eigenvalue equal to
unity. Let us write for pure state of the two qubits as,

$\displaystyle\left|\psi\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left|00\right\rangle+\left|11\right\rangle\right)$
$\displaystyle\left\langle\psi\right|$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\left\langle 00\right|+\left\langle
11\right|\right)$
$\displaystyle\left|\psi\right\rangle\left\langle\psi\right|$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\left|00\right\rangle+\left|11\right\rangle\right)\left(\left\langle
00\right|+\left\langle 11\right|\right)$

We have

$\displaystyle\left\langle\psi\right|\left|\psi\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\left\langle 00\right|+\left\langle
11\right|\right)\left(\left|00\right\rangle+\left|11\right\rangle\right)$
$\displaystyle=$ $\displaystyle\frac{1}{2}\left(\left\langle
00\right|\left|00\right\rangle+\left\langle
11\right|\left|11\right\rangle\right)$ $\displaystyle=$ $\displaystyle 1$

Thus, indeed,

$\displaystyle\rho^{2}$ $\displaystyle=$
$\displaystyle\left|\psi\right\rangle\left\langle\psi\right|\left|\psi\right\rangle\left\langle\psi\right|$
$\displaystyle=$ $\displaystyle\left|\psi\right\rangle\left\langle\psi\right|$
$\displaystyle=$ $\displaystyle\rho$

We refer to $\left|\psi\right\rangle$ as a maximally entangled state in the
sense that the first qubit is exactly ”mapped” to the second qubit. The
reduced density matrix for subsystem $A$ is

$\displaystyle\rho_{A}$ $\displaystyle=$ $\displaystyle Tr_{B}\ \rho$
$\displaystyle=$ $\displaystyle\frac{1}{2}\left\langle
0_{B}\right|\left(\left|00\right\rangle+\left|11\right\rangle\right)\left(\left\langle
00\right|+\left\langle 11\right|\right)\left|0_{B}\right\rangle$
$\displaystyle+\frac{1}{2}\left\langle
1_{B}\right|\left(\left|00\right\rangle+\left|11\right\rangle\right)\left(\left\langle
00\right|+\left\langle 11\right|\right)\left|1_{B}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\left|0_{A}\right\rangle\left\langle
0_{A}\right|+\left|1_{A}\right\rangle\left\langle 1_{A}\right|\right)$

Now clearly

$\displaystyle\rho_{A}^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\left(\left|0_{A}\right\rangle\left\langle
0_{A}\right|+\left|1_{A}\right\rangle\left\langle
1_{A}\right|\right)\left(\left|0_{A}\right\rangle\left\langle
0_{A}\right|+\left|1_{A}\right\rangle\left\langle 1_{A}\right|\right)$
$\displaystyle=$
$\displaystyle\frac{1}{4}\left(\left|0_{A}\right\rangle\left\langle
0_{A}\right|+\left|1_{A}\right\rangle\left\langle 1_{A}\right|\right)$
$\displaystyle=$ $\displaystyle\frac{1}{2}\rho_{A}$

so that $\rho_{A}$ is not a pure state but mixed, i.e., a mixture of two pure
states, $\left|0_{A}\right\rangle$ and $\left|1_{A}\right\rangle$. Note that
the mixed states density matrix do not possess off-diagonal elements. From the
last identity, the eigenvalues of $\rho_{A}$ is $\frac{1}{2}$ and since the
subsystem $A$ is a single qubit with two distinct states, then eigenvalues of
$\rho_{A}$ corresponds to $\frac{1}{\Omega}$ of our first example above. Thus
we refer to $\rho_{A}$ as ”uniformly’” mixed often referred to as ”maximally
mixed” with the initial state $\left|\psi\right\rangle$ as ”maximally
entangled”.

##  5 Entanglement of Formation of Multi-Partite Qubit Systems

The entanglement entropy of subsystem $A$ can be calculated using the
eigenvalues of the $2\times 2$ matrix of $\rho_{A}$, which is
$\lambda_{A}=\frac{1}{2}$. Therefore we have for the entanglement entropy or
entanglement of formation given by,

$\displaystyle S_{A}$ $\displaystyle=$ $\displaystyle-Tr\rho_{A}\log\rho_{A}$
(33) $\displaystyle=$ $\displaystyle-Tr\frac{1}{2}\log\frac{1}{2}$
$\displaystyle=$ $\displaystyle-2\left(\frac{1}{2}\log\frac{1}{2}\right)$
$\displaystyle=$ $\displaystyle\log 2\equiv\log 2^{1}$

where exponent base $2$, here $1$, is the number of qubits that is entangled
with system $B$. This will be made clear in the next example.

###### Example 3

A four qubit system:

If each subsystem $A$ or $B$ has two qubits, then the Hilbert space of the
whole system is span by the following $16$ direct product states,i.e.,
$2^{4}=16$ states. The maximally entangled pure state is determined by the
following eight diagrams [4], see Fig.2, and their flipped or translated
states,

Figure 2: Two-state eight diagrams for entangled four qubits. The so-called
flip operation yields the second state for each of the above diagrams. The
entangled basis is constructed by the superposition, via the Hadamard
transformation, of each diagram and its corresponding flipped diagram.

Any combination or superposition of both ’even’ triplet and singlet states
comprised a maximally entangled state, i.e.,

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

Thus, from the states given above, we have for example the maximally entangled
state,

$\displaystyle\left|\psi\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{4}}\left(\left|00,00\right\rangle+\left|01,01\right\rangle+\left|10,10\right\rangle+\left|11,11\right\rangle\right)$
(34) $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\Phi_{1}^{+}+\Phi_{8}^{+}\right)$

and similarly, we can also form another entangled state,

$\displaystyle\left|\psi\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{4}}\left(\left|00,00\right\rangle+\left|01,01\right\rangle-\left|10,10\right\rangle-\left|11,11\right\rangle\right)$
(35) $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\Phi_{1}^{-}+\Phi_{8}^{-}\right)$

where the first two bits belongs to subsystem $A$ and the second two bits
belong to subsystem $B$. We note that

$\displaystyle\left\langle\psi\right|\left|\psi\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{4}\left(\left\langle 00,00\right|+\left\langle
01,01\right|+\left\langle 10,10\right|+\left\langle
11,11\right|\right)\left(\left|00,00\right\rangle+\left|01,01\right\rangle+\left|10,10\right\rangle+\left|11,11\right\rangle\right)$
$\displaystyle=$ $\displaystyle 1$

Therefore

$\displaystyle\rho^{2}$ $\displaystyle=$
$\displaystyle\left(\left|\psi\right\rangle\left\langle\psi\right|\right)^{2}$
$\displaystyle=$
$\displaystyle\left|\psi\right\rangle\left\langle\psi\right|=\rho$

So $\left|\psi\right\rangle$ is a pure state.

In general, the following maximally entangled state corresponds to a chain of
entangled basis states, namely,

$\left|\psi^{+}\right\rangle=\frac{1}{\sqrt{8}}\left(\Phi_{1}^{+}+\Phi_{2}^{+}+\Phi_{3}^{+}+\Phi_{4}^{+}+\Phi_{5}^{+}+\Phi_{6}^{+}+\Phi_{7}^{+}+\Phi_{8}^{+}\right)$

as well as

$\left|\psi^{-}\right\rangle=\frac{1}{\sqrt{8}}\left(\Phi_{1}^{-}+\Phi_{2}^{-}+\Phi_{3}^{-}+\Phi_{4}^{-}+\Phi_{5}^{-}+\Phi_{6}^{-}+\Phi_{7}^{-}+\Phi_{8}^{-}\right)$

of a four qubit system.

The density matrix operator for the whole $4$-qubit system can be written as,

$\rho=\frac{1}{\sqrt{2}}\left(\left|\Phi_{1}^{-}+\Phi_{8}^{-}\right\rangle\right)\frac{1}{\sqrt{2}}\left(\left\langle\Phi_{1}^{-}+\Phi_{8}^{-}\right|\right)$
$\rho=\frac{1}{4}\left[\left|00,00\right\rangle+\left|01,01\right\rangle+\left|10,10\right\rangle+\left|11,11\right\rangle\right]\left[\left\langle
00,00\right|+\left\langle 01,01\right|+\left\langle 10,10\right|+\left\langle
11,11\right|\right]$

The reduced density matrix operator for subsystem $A$ is again obtained by
taking partial trace with respect to subsystem $B$.

$\displaystyle\rho_{A}$ $\displaystyle=$ $\displaystyle Tr_{B}\ \rho$
$\displaystyle=$ $\displaystyle\left\langle
00_{B}\right|\rho\left|00_{B}\right\rangle+\left\langle
01_{B}\right|\rho\left|01_{B}\right\rangle+\left\langle
10_{B}\right|\rho\left|10_{B}\right\rangle+\left\langle
11_{B}\right|\rho\left|11_{B}\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{4}\left|00_{A}\right\rangle\left\langle
00_{A}\right|+\left|01_{A}\right\rangle\left\langle
01_{A}\right|+\left|10_{A}\right\rangle\left\langle
10_{A}\right|+\left|11_{A}\right\rangle\left\langle 11_{A}\right|$

To determine the eigenvalues for $\rho_{A}$, we take its square,

$\displaystyle\rho_{A}^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{16}\left(\left|00_{A}\right\rangle\left\langle
00_{A}\right|+\left|01_{A}\right\rangle\left\langle
01_{A}\right|+\left|10_{A}\right\rangle\left\langle
10_{A}\right|+\left|11_{A}\right\rangle\left\langle 11_{A}\right|\right)$
$\displaystyle\times\left(\left|00_{A}\right\rangle\left\langle
00_{A}\right|+\left|01_{A}\right\rangle\left\langle
01_{A}\right|+\left|10_{A}\right\rangle\left\langle
10_{A}\right|+\left|11_{A}\right\rangle\left\langle 11_{A}\right|\right)$
$\displaystyle=$
$\displaystyle\frac{1}{16}\left(\left|00_{A}\right\rangle\left\langle
00_{A}\right|+\left|01_{A}\right\rangle\left\langle
01_{A}\right|+\left|10_{A}\right\rangle\left\langle
10_{A}\right|+\left|11_{A}\right\rangle\left\langle 11_{A}\right|\right)$
$\displaystyle=$ $\displaystyle\frac{1}{16}\rho_{A}$

So we have

$\rho_{A}^{2}=\frac{1}{16}\rho_{A}$

and the eigenvalues of $\rho_{A}$ is $\frac{1}{4}$. We can now calculate the
entanglement entropy of subsystem $A.$ We have

$\displaystyle S_{A}$ $\displaystyle=$ $\displaystyle-Tr\rho_{A}\log\rho_{A}$
$\displaystyle=$ $\displaystyle-4\left[\frac{1}{4}\log\frac{1}{4}\right]$
$\displaystyle=$ $\displaystyle-\log\frac{1}{4}$ $\displaystyle=$
$\displaystyle\log 2^{2}$

The exponent $2$ correspond to the number of qubits that is entangled with
subsystem $B$. In general, for maximally entangled bipartite system $A$ and
$B$, each having $k$ number of qubits $S_{A}$ is given by

$S_{A}=\log 2^{k}$

Now of course, for this bipartite system,

$S_{A}=S_{B}$

which simply means a complete matching of configurations of each system $A$
and $B$, respectively, otherwise some degrees of freedom will be hanging, not
matched or cannot be entangled.

###### Example 4

Tripartite sytem of three qubits:

The following diagrams represent the entangled tripartite system of qubits.

Figure 3: Two-state four diagrams for entangled three qubits of a tripartite
system. Flip operations yield the respective second states.

The entangled basis states are as follows,

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

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

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

A chain or superposition of any choice of entangled even basis states also
form a maximally entangled state, e.g.,

$\displaystyle\left|\Psi^{4}\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{4}}\left[\Xi^{+}+\Theta_{3}^{+}+\Omega^{+}+\Gamma^{+}\right]$
$\displaystyle\left|\Psi^{3}\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{3}}\left[\Xi^{+}+\Theta_{3}^{+}+\Omega^{+}\right]$
$\displaystyle\left|\Psi^{2}\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left[\Xi^{+}+\Theta_{3}^{+}\right]$
$\displaystyle\left|\Psi^{1}\right\rangle$ $\displaystyle=$
$\displaystyle\Xi^{+}$

are all, respectively, maximally entangled states with emergent two-state or
qubit properties.

###### Example 5

Here, we need to define what we mean by entanglement of formation as the
entanglement entropy of unentangling the three-qubit tripartite system into
three separate one-qubit systems. It is important to point out that any
entangled qubits, irrespective of their number, identically behave like a
single qubit, i.e., behaving exactly like a two-state system. Thus, the
natural order of disentangling is as follows: First, one entangles one qubit
from the remaining two entangled qubits. The entanglement of formation is
equivalent to one qubit. Next, one disentangled the remaining entangled two
qubits. This further give an entanglement of formation of one qubit. Thus, the
total entanglement of formation is $1+1=2$ qubits.

###  5.1 On the monogamy inequality in multi-partite entanglement

The reasoning we have given above yields exact equality of the monogamy,
usually given as inequality in the literature as deduced from statistical
analysis, e.g., one given by Kim [5], The exact equality relation for the
entanglement of formation is diagrammatically shown in Fig. 4

Figure 4: Figure reproduced from Ref.[5]. In our diagrammatic analysis the
right-hand side of the figure yields $1+1=2$ as the entanglement of formation,
being the sum of two two-qubit entanglements. Thus, in our diagrammatic
analysis, we obtained exact monogamy equality not inequality.

The exact equality comes about by construction, since on both sides of Fig. 4,
two $2$-qubit entangled states are being untangled to obtain the entanglement
of formation of this multi-partite system. This comes about through the
knowledge that any multi-partite entangled qubits behave as an emergent qubit.

Similarly, for an entangled $4$-partite system, the entanglement of formation
is determined schematically by the diagrams of Fig.5, where the unentangling
operations to be done on the left side are itemized on the unentangling
operations of three entangled $2$-qubits on the right side of the equality
sign,

Figure 5: In our diagrammatic analysis the right-hand side of the figure
yields $1+1+1=3$ as the entanglement of formation, being the sum of three two-
qubit entanglements. Thus, in our diagrammatic analysis, we obtained exact
equality not inequality.

This sort of diagrammatic analysis of the entropy of entanglement of formation
can straightforwardly be employed to all multi-partite qubit systems by a
simple counting argument as is done here. Although our diagrammatic analysis
is based on maximally entangled qubits, we believe the diagrammatic
construction holds also for any arbitrarily entangled multi-partite qubit
systems. For example, for $18$ multi-partite entangled qubits in Fig. 6, we
have the entanglement of formation schematically depicted by the equality
where there is $17$ number of monogamy in the right-hand side yielding $17$
qubits of entropy of entanglement of formation.

Figure 6: In our diagrammatic analysis the right hand side of the figure
yields $1+1+1,......1=17$ qubits as the entanglement of formation, being the
sum of $17$ two-qubit entanglements of formation. There are $18$ multi-party
entangled qubits in the left-hand side.

##  6 Expansion of Product States in Bell Basis

Let us consider $\left|00\right\rangle$, $\left|01\right\rangle$,
$\left|10\right\rangle$, $\left|11\right\rangle$ as our direct product basis
in expanding $\left|\psi\right\rangle$,

$\displaystyle\left|\psi\right\rangle$ $\displaystyle=$
$\displaystyle\alpha_{0}\left|00\right\rangle+\alpha_{1}\left|01\right\rangle+\alpha_{2}\left|10\right\rangle+\alpha_{3}\left|11\right\rangle$
$\displaystyle=$ $\displaystyle\sum\limits_{i}\beta_{i}e_{i}$

where $e_{i}$ denotes the Bell basis. We have the Bell basis given by

$\displaystyle e_{1}$ $\displaystyle=$
$\displaystyle\left|\Phi^{+}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle+\left|1\right\rangle\left|1\right\rangle\right)$
$\displaystyle e_{2}$ $\displaystyle=$
$\displaystyle\left|\Phi^{-}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle-\left|1\right\rangle\left|1\right\rangle\right)$
$\displaystyle e_{3}$ $\displaystyle=$
$\displaystyle\left|\Psi^{+}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|0\right\rangle\right)$
$\displaystyle e_{4}$ $\displaystyle=$
$\displaystyle\left|\Psi^{-}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|0\right\rangle\right)$
(36)

Standard basis can be express in terms of Bell basis, a re-statement of Eqs.
(1) and (2), namely,

$\displaystyle\left|0\right\rangle\left|0\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(e_{1}+e_{2}\right)$
$\displaystyle\left|1\right\rangle\left|1\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(e_{1}-e_{2}\right)$
$\displaystyle\left|0\right\rangle\left|1\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(e_{3}+e_{4}\right)$
$\displaystyle\left|1\right\rangle\left|0\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(e_{3}-e_{4}\right)$

so we can also expand $\left|\psi\right\rangle$ as

$\displaystyle\left|\psi\right\rangle$ $\displaystyle=$
$\displaystyle\beta_{1}e_{1}+\beta_{2}e_{2}+\beta_{3}e_{3}+\beta_{4}e_{4}$
$\displaystyle=$
$\displaystyle\beta_{1}\left(\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle+\left|1\right\rangle\left|1\right\rangle\right)\right)+\beta_{2}\left(\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|0\right\rangle-\left|1\right\rangle\left|1\right\rangle\right)\right)$
$\displaystyle+\beta_{3}\left(\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|0\right\rangle\right)\right)+\beta_{4}\left(\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|0\right\rangle\right)\right)$
$\displaystyle\left|\psi\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\beta_{1}+\beta_{2}\right)\left|0\right\rangle\left|0\right\rangle+\frac{1}{\sqrt{2}}\left(\beta_{1}-\beta_{2}\right)\left|1\right\rangle\left|1\right\rangle$
$\displaystyle+\frac{1}{\sqrt{2}}\left(\beta_{3}+\beta_{4}\right)\left|0\right\rangle\left|1\right\rangle+\frac{1}{\sqrt{2}}\left(\beta_{3}-\beta_{4}\right)\left|1\right\rangle\left|0\right\rangle$

Therefore, equating coefficients, we have the equality,

$\displaystyle\left|\psi\right\rangle$ $\displaystyle=$
$\displaystyle\alpha_{0}\left|00\right\rangle+\alpha_{1}\left|01\right\rangle+\alpha_{2}\left|10\right\rangle+\alpha_{3}\left|11\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\beta_{1}+\beta_{2}\right)\left|0\right\rangle\left|0\right\rangle+\frac{1}{\sqrt{2}}\left(\beta_{1}-\beta_{2}\right)\left|1\right\rangle\left|1\right\rangle$
$\displaystyle+\frac{1}{\sqrt{2}}\left(\beta_{3}+\beta_{4}\right)\left|0\right\rangle\left|1\right\rangle+\frac{1}{\sqrt{2}}\left(\beta_{3}-\beta_{4}\right)\left|1\right\rangle\left|0\right\rangle$

yielding

$\displaystyle\alpha_{0}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\beta_{1}+\beta_{2}\right)$
$\displaystyle\alpha_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\beta_{3}+\beta_{4}\right)$
$\displaystyle\alpha_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\beta_{3}-\beta_{4}\right)$
$\displaystyle\alpha_{3}$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}\left(\beta_{1}-\beta_{2}\right)$

Then

$\displaystyle\sum\limits_{0}^{3}\left|\alpha_{i}\right|^{2}$ $\displaystyle=$
$\displaystyle\sum\limits_{1}^{4}\left|\beta_{i}\right|^{2}=1$
$\displaystyle=$
$\displaystyle\left|\frac{1}{\sqrt{2}}\left(\beta_{1}+\beta_{2}\right)\right|^{2}+\left|\frac{1}{\sqrt{2}}\left(\beta_{1}-\beta_{2}\right)\right|^{2}+\left|\frac{1}{\sqrt{2}}\left(\beta_{3}+\beta_{4}\right)\right|^{2}+\left|\frac{1}{\sqrt{2}}\left(\beta_{3}-\beta_{4}\right)\right|^{2}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(2\left(\beta_{1}^{2}+\beta_{2}^{2}\right)\right)+\frac{1}{2}\left(2\left(\beta_{3}^{2}+\beta_{4}^{2}\right)\right)=\sum\limits_{1}^{4}\left|\beta_{i}\right|^{2}=1$

Summarizing, the bijective relation between the Bell basis states and the
product states given in Eq. (36) reflects the discrete phase-space physics of
these two-state systems.

##  7 Chiral Degrees of Freedom and Entangled Qubits

Observe that the pseudo-spin variables in semiconductor Bloch equation is
defined by the following expressions,

$\displaystyle S_{0}$ $\displaystyle=$
$\displaystyle\left(\rho_{cc}+\rho_{vv}\right)$ $\displaystyle S_{x}$
$\displaystyle=$ $\displaystyle\rho_{vc}+\rho_{cv}$ $\displaystyle S_{y}$
$\displaystyle=$ $\displaystyle-i\left(\rho_{vc}-\rho_{cv}\right)$
$\displaystyle S_{z}$ $\displaystyle=$
$\displaystyle\left(\rho_{cc}-\rho_{vv}\right)$

This maps to

$\displaystyle e_{0}$ $\displaystyle=$
$\displaystyle\left|\Phi^{+}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|1\right\rangle\left|1\right\rangle+\left|0\right\rangle\left|0\right\rangle\right)$
$\displaystyle e_{x}$ $\displaystyle=$
$\displaystyle\left|\Psi^{+}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle+\left|1\right\rangle\left|0\right\rangle\right)$
$\displaystyle ie_{y}$ $\displaystyle=$ $\displaystyle
i\left|\Psi^{-}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle\left|1\right\rangle-\left|1\right\rangle\left|0\right\rangle\right)$
$\displaystyle e_{z}$ $\displaystyle=$
$\displaystyle\left|\Phi^{-}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|1\right\rangle\left|1\right\rangle-\left|0\right\rangle\left|0\right\rangle\right)$

We refer to the basis $\left(e_{0},e_{x},e_{y},e_{z}\right)$ as the pseudo-
spin component basis, which differ from the Bell basis in
$\left|\Psi^{-}\right\rangle$ by the factor $i$ in
$\left|\Psi^{-}\right\rangle$. An important observation that follows from this
is that the entangling of chiral degrees of freedom creates another chiral
degrees of freedom, e.g., Eq. (35).

This is a very important observation. This will aid in understanding the
entanglement-induced delocalization, for example, in topological insulators
where entanglement occurs at the edge of the sample. This is treated by the
quantum transport approach in Ref.[6].

##  8 The Concurrence Concept and Emergent Qubit

By virtue of our diagrammatic construction, coupled with discrete phase space
Hadamard transformation, in deriving the entangled basis states, the
concurrence concept defined by Wooters [2, 3], as well as the two-state
(qubit) properties of entangled multi-partite qubits, naturally coincides with
the mathematical description of our physical model of qubit entanglement. In
other words, the essence of the quantum description of an entangled qubits in
Fig. 1 consists of the superposition of product ’site states’ and their
corresponding translated ’site states’ of all qubits. This means a
superposition of the two states of the entangled two qubits of Fig.1. One
observe that irrespective of how many qubits are entangled, the resulting
entangled state is an emergent two-state system and therefore behave just like
one qubit with two-states, namely, the first state and its translated state
[4]. Therefore its associated entropy is just one, Eq. (33).444This is also
deduced when concurrence $C=1$ [3].

The concept of concurrence is basically contained by construction of our
physical model of qubit entanglement. It is defined by

$\left|\left\langle\Psi\right|\left|\tilde{\Psi}\right\rangle\right|=C$ (37)

where $C$ is the quantitative value of concurrence. This number lies between
$0$ and $1,$ $0$ $\leq C\leq$ $1$. Here $\tilde{\Psi}$ is the corresponding
translated qubits of $\Psi$.

##  9 A Natural Measure of Entanglement

The two properties of any entangled qubits, namely, concurrence and its
emergent qubit behavior, lead Wooters [2, 3] to introduce a measure of
entanglement, as incorporated in the two formulas,

$E\left(C\right)=H\left(\frac{1}{2}+\frac{1}{2}\sqrt{1-C^{2}}\right)$ (38)
$H\left(x\right)=-x\ln x-\left(1-x\right)\ln\left(1-x\right)$ (39)

Equation (38) basically say that if there is complete concurrence, i.e.,
$C=1$, then Eq. (39) says that the system behave as an emergent qubit or two-
state system, as depicted clearly in our diagrams. Thus, for maximally
entangled multi-partite qubits, we have $C=1$,

$\displaystyle E\left(C\right)$ $\displaystyle=$ $\displaystyle
H\left(\frac{1}{2}\right)$ $\displaystyle H\left(\frac{1}{2}\right)$
$\displaystyle=$ $\displaystyle\frac{1}{2}\ln 2+\frac{1}{2}\ln 2$
$\displaystyle=$ $\displaystyle 1$

affirming that entangled qubits behave as a two-state system or as an emergent
qubit, yielding entropy equals one.

###  9.1 Entropic distance in entanglement measure

From the above developments, one can introduce an entanglement entropic
distance by the formula

$\mathcal{E}=\left|Tr\left(\rho\ln\rho\right)-1\right|$

where $Tr\left(\rho\ln\rho\right)$ is evaluated using Eqs. (38) and (39). This
distance has a range: $0\leq\mathcal{E\leq}1.$ When the concurrence, $C=1$,
$Tr\left(\rho\ln\rho\right)=1=H\left(\frac{1}{2}\right)$, then we have the
entropy-distance from maximally entangled state, $\mathcal{E}=0$. When $C=0$,
$Tr\left(\rho\ln\rho\right)=0=H\left(1\right)$, then we have the entropy-
distance from maximally entangled state, $\mathcal{E}=1$.

An example where the $C<1$ may occur in the following singlet state, with less
entanglement,

$\left|\psi\right\rangle=\alpha\left|01\right\rangle+\beta\left|10\right\rangle$

where

$\alpha^{2}+\beta^{2}=1$

Then

$\displaystyle C$ $\displaystyle=$
$\displaystyle\left|\left\langle\psi\right|\left|\tilde{\psi}\right\rangle\right|=\left(\alpha\left\langle
01\right|+\beta\left\langle
10\right|\right)\left(\alpha\left|10\right\rangle+\beta\left|01\right\rangle\right)$
$\displaystyle=$ $\displaystyle\alpha\beta+\beta\alpha\leq 1$

##  10 Mixed States and Pure States

In discussing mixed and pure states, one makes use of density-matrix
operators. This is the domain of abstract statistical treatment usually found
in the literature, perhaps following the statistical tradition of Bell’s
theorem. To elucidate the basic physics, we will here avoid abstract
statistical treatment and only discuss specific situations and examples.

###  10.1 Mixed states and mixed entanglements

Consider the maximally mixed state of a triplet system,

$\hat{\rho}_{W}=\frac{1}{2}\left(\left|00\right\rangle\left\langle
00\right|+\left|11\right\rangle\left\langle 11\right|\right)$

From Eqs. (1), we obtain the mixed entanglements given by

$\hat{\rho}_{B}=\frac{1}{2}\left(\left|\Phi^{+}\Phi^{+}\right\rangle\left\langle\Phi^{+}\Phi^{+}\right|+\left|\Phi^{-}\Phi^{-}\right\rangle\left\langle\Phi^{-}\Phi^{-}\right|\right)$

Similarly, consider the maximally mixed state of a singlet system,

$\hat{\rho}_{W}=\frac{1}{2}\left(\left|01\right\rangle\left\langle
01\right|+\left|10\right\rangle\left\langle 10\right|\right)$

From Eqs. (2), we obtain the mixed entanglements given by

$\hat{\rho}_{B}=\frac{1}{2}\left(\left|\Psi^{+}\Psi^{+}\right\rangle\left\langle\Psi^{+}\Psi^{+}\right|+\left|\Psi^{-}\Psi^{-}\right\rangle\left\langle\Psi^{-}\Psi^{-}\right|\right)$

###  10.2 Mixed state from pure state (entangled state)

Consider the example of a bipartite of two qubit system. Consider a pure
entangled state,

$\left|\psi\right\rangle=\frac{1}{\sqrt{2}}\left(\left|00\right\rangle+\left|11\right\rangle\right).$

Then the density matrix is,

$\rho=\frac{1}{2}\left(\left|00\right\rangle+\left|11\right\rangle\right)\left(\left\langle
00\right|+\left\langle 11\right|\right),$ (40)

where the first qubit belongs to party $A$ and the second qubit belongs to
party $B$.555What we mean by party $A$ and $B$ is in the general sense since
any entangled number of qubits behave as an emergent qubit. We have

$\rho^{2}=\rho$

so that Eq. (40) is a pure state. Again, we have, by tracing the party $B$ we
obtain,

$\displaystyle\rho_{A}$ $\displaystyle=$ $\displaystyle Tr_{B}\ \rho$
$\displaystyle=$ $\displaystyle\frac{1}{2}\left\langle
0_{B}\right|\left(\left|00\right\rangle+\left|11\right\rangle\right)\left(\left\langle
00\right|+\left\langle 11\right|\right)\left|0_{B}\right\rangle$
$\displaystyle+\frac{1}{2}\left\langle
1_{B}\right|\left(\left|00\right\rangle+\left|11\right\rangle\right)\left(\left\langle
00\right|+\left\langle 11\right|\right)\left|1_{B}\right\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\left|0_{A}\right\rangle\left\langle
0_{A}\right|+\left|1_{A}\right\rangle\left\langle 1_{A}\right|\right)$

Now clearly

$\displaystyle\rho_{A}^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\left(\left|0_{A}\right\rangle\left\langle
0_{A}\right|+\left|1_{A}\right\rangle\left\langle
1_{A}\right|\right)\left(\left|0_{A}\right\rangle\left\langle
0_{A}\right|+\left|1_{A}\right\rangle\left\langle 1_{A}\right|\right)$
$\displaystyle=$
$\displaystyle\frac{1}{4}\left(\left|0_{A}\right\rangle\left\langle
0_{A}\right|+\left|1_{A}\right\rangle\left\langle 1_{A}\right|\right)$
$\displaystyle=$ $\displaystyle\frac{1}{2}\rho_{A}$

so that $\rho_{A}$ is not a pure state but mixed, i.e., a mixture of two pure
states, $\left|0_{A}\right\rangle$ and $\left|1_{A}\right\rangle$.666If the
parties $A$ and $B$ are entangled states, then what we have obtain are also
mixed entanglement through the inverse relationships, Eq. (1) and (2). Note
that the mixed states density matrix do not possess off-diagonal elements. The
tracing operation basically eliminates the contribution of the ”off-diagonal”
terms. From the last identity, the eigenvalues of $\rho_{A}$ is $\frac{1}{2}$
and since the subsystem $A$ is an emergent qubit, in general with two distinct
states, then eigenvalues of $\rho_{A}$ corresponds to $\frac{1}{\Omega}$ of
our first example above. Thus we refer to $\rho_{A}$ as ”uniformly’” mixed
often referred to as ”maximally mixed” with the inital state
$\left|\psi\right\rangle$ as ”maximally entangled”.

##  11 Concluding Remarks

The discrete phase-space viewpoint for qubit entanglement has been very
fruitful in analytically deriving the entanglement basis states for any number
of entangled qubits as demonstrated by the author in his book [7], as well as
in Ref. [4]. Moreover, the inverter-chain mechanical model of entanglement
link has been demonstrated to faithfully implement the discrete phase space
viewpoint [4]. The crucial observation that arise from this inverter-chain
link model is that any multi-partite qubit entangled system has the property
of an emergent qubit, which has been rigorously justified. This aspect of the
model readily lead us to the equality relations of the entropy of entanglement
formation in the so-called monogamy inequality of entanglement formation,
using statistical arguments [5] discussed in the literature. In this paper,
mixed states and mixed entanglements are related by the Hadamard
transformations. In addition, we show that mixed state can be extracted from
pure entangled state, where the party $A$ or $B$ need not be a single qubit
themselves but can each be an entangled multi-partite qubit system,
respectively, by virtue of emergent qubit property of entangled qubit systems.

The natual measure of entanglement should be based on entropy of entanglement
formation, concurrence, and entropic distance from maximally entangled
reference.

## References

  * [1] A. Einstein, B. Podolsky and N. Rosen, Can Quantum-Mechanical Description of Physical Reality be Considered Complete? Phys. Rev. 47, 777 (1935).
  * [2] S. Hill and W.K. Wooters, Entanglement of a Pair of Quantum Bits, Phys. Rev. Letts. 78, 5022 (1977).
  * [3] W.K. Wooters, Entanglement of a Pair of Quantum Bits, Phys. Rev. Letts. 80, 2245 (1998).
  * [4] F.A. Buot, A.R. Elnar, G. Maglasang, C.M. Galon, A Mechanical Implementation and Diagrammatic Calculation of Entangled Basis States, arXiv:2112.10291 [cond-mat.mes-hall] (Dec. 2021)
  * [5] J.S. Kim, Entanglement of formation and monogamy of multi-party quantum entanglement, Scientific Reports 11:2364 (2021); https://doi.org/10.1038/s41598-021-82052-3.
  * [6] F.A. Buot, K.B. Rivero, R.E.S. Otadoy, Generalized nonequilibrium quantum transport of spin and pseudospins: Entanglements and topological phases, Physica B 559, 42-61 (2019).
  * [7] F.A. Buot, Nonequilibrium Quantum Transport Physics in Nanosystem (World Scientific, 2009), and references therein.


# Learning from Unlabelled Data with Transformers: Domain Adaptation for
Semantic Segmentation of High Resolution Aerial Images

###### Abstract

Data from satellites or aerial vehicles are most of the times unlabelled.
Annotating such data accurately is difficult, requires expertise, and is
costly in terms of time. Even if Earth Observation (EO) data were correctly
labelled, labels might change over time. Learning from unlabelled data within
a semi-supervised learning framework for segmentation of aerial images is
challenging. In this paper, we develop a new model for semantic segmentation
of unlabelled images, the Non-annotated Earth Observation Semantic
Segmentation (NEOS) model. NEOS performs domain adaptation as the target
domain does not have ground truth masks. The distribution inconsistencies
between the target and source domains are due to differences in acquisition
scenes, environment conditions, sensors, and times. Our model aligns the
learned representations of the different domains to make them coincide. The
evaluation results show that it is successful and outperforms other models for
semantic segmentation of unlabelled data.

Index Terms—  Semantic segmentation, Unlabelled data

##  1 Introduction

Importance and overview. Remote Sensing (RS) images from satellites or aerial
vehicles can be used to map trees and land cover classes [1]. While both RS
technology and AI for data analysis continue to advance [2], the integration
and use of airplanes and drones for localized studies is nowadays also
increasing. Supervised learning has shown good performance for classification
and segmentation. However, it requires high-quality handcrafted large labelled
datasets. Learning from unlabelled data is challenging as the performance of
models depends highly on the size and quality of the data. However, for real-
world applications [3], labelling large datasets is laborious, expensive, and
time-consuming. This holds for Earth Observation (EO), where huge amounts of
data are produced daily. Also, data from satellites or aircrafts usually
require domain expertise. Furthermore, labels for specific geographical
regions may change over time (task of change detection) due to nature
(seasonality), man-induced changes, and natural hazards (volcano eruptions).
Also, for specific regions, some labels might be incorrect (task of learning
from noisy labels). Because many satellites and aerial images are unlabelled,
it is challenging to effectively use these data. Developing semi-supervised
learning methods is crucial to improve generalization performance. Semi-
supervised learning, which involves training on both a labelled dataset, where
both images and their annotations are provided, and on an unlabelled set, with
only image data, is a more realistic setting than supervised learning, as in
RS, unlabelled data are plentiful, while labelled data can be hard to find.
This holds for semantic segmentation (pixel-level labels) [1], which requires
assigning a class label to each pixel [4, 5] by understanding its semantics.
This task is crucial for several applications, including land cover mapping
and urban change detection. In this work, we propose a method to perform
semantic segmentation on unlabelled datasets, and we evaluate it on the
unlabelled Cross-View USA (CVUSA) dataset [6]. To the best of the authors’
knowledge, accurate semantic segmentation on the CVUSA aerial dataset, which
is used for cross-view aerial-ground matching [7, 8] and has no land cover
label annotations, has not yet been performed.

Domain adaptation. In this paper, we develop a model for semantic segmentation
of aerial images, the Non-annotated Earth Observation Semantic Segmentation
(NEOS) model. NEOS makes the learned representations of the different domains
to coincide. This is achieved by minimizing the distribution differences of
the different domains. Hence, we enforce the model to be able to work well
with the different datasets that have distribution inconsistencies due to
differences in acquisition scenes, environment conditions, sensors, and times.
Our model performs semantic segmentation on the unlabelled dataset CVUSA.
During training, a loss function is minimized that makes the network to align
the latent features of the different domains to minimize distribution
differences. Our main contribution is the development of a novel model for
semantic segmentation of aerial images that do not have ground truth
segmentation masks, also performing domain adaptation.

##  2 Related Work

Domain adaptation methods in deep learning, as well as in RS, have been
developed recently. Furthermore, domain adaptation methods for classification
and segmentation have also been developed. Models trained on data from one
domain may not generalize well on other domains. Even in one domain, accurate
semantic segmentation is challenging. There may be a loss in accuracy when
deploying a model on unseen data due to a shift between the distributions in
the source and target domains [9]. Domain adaptation tries to overcome this
[10, 11]. Domain gaps are common in aerial images [12, 13], e.g. region
change. In [8], no off-the-shelf semantic segmentation model transferred well/
accurately on the aerial CVUSA dataset.

Architectures. Several models have been developed recently for semantic
segmentation, including Transformer (e.g. SegFormer [14]), encoder-decoder
like SegNet or U-Net with skip connections, Fully Convolutional Network (FCN),
and dilated convolutions with larger receptive field to capture long-range
information. SegFormer [14] for semantic segmentation combines the Transformer
with an efficient Multi-Layer Perceptron (MLP) decoder and outputs multi-scale
features. The decoder combines these multi-scale feature maps, which use local
and global attention. UNetFormer is a UNet-like architecture for segmentation
based on a ResNet encoder and a Transformer decoder [15]. It uses efficient
attention in the decoder to model global and local information. The Segment
Anything Model (SAM) [16] has also been recently proposed for instance
segmentation, but not semantic segmentation.

Unsupervised adaptation. In the Unsupervised Domain Adaptation (UDA) setting,
the model is trained on both labelled and unlabelled data from the source and
target domains, respectively. Accurate real-world semantic segmentation
without labels is non-trivial. Domain Symmetric Networks (SymNet) [17, 12]
design the source and target domains classifier symmetrically to learn domain-
invariant features for effective domain adaptation. The model Source
Hypothesis Transfer (SHOT) [18] uses hypothesis transfer, training only the
backbone and making the classifier of the network non-trainable. We focus on
the scenario in which the number of classes and the classes themselves are the
same in the source and target domains, the closed-set setting [12, 19]. The
aim is to achieve good performance during inference on the target domain test
dataset, as well as on the source domain test dataset [10, 20].

Semi-supervised learning methods for segmentation in deep learning, as well as
in RS, have been developed. Semi-supervision [5, 21], which is halfway between
supervised and unsupervised learning, deals with settings where labelled sets
of data and their targets are provided and unlabelled sets with data only are
available. Unlabelled data help the learning process to improve performance.
To improve generalization [21, 22], models should be able to handle labelled
and unlabelled data, as well as operate within a multi-task optimization
framework [5] performing unsupervised minimization tasks.

Fig. 1: Flowchart of NEOS for semantic segmentation using domain adaptation on
datasets with no ground truth labels.

##  3 Proposed Methodology

Flowchart. NEOS is presented in Fig. 1. The input is the image from labelled
and unlabelled datasets. The output is the estimated semantic segmentation
mask. NEOS is based on the SegFormer B5 [14] architecture and uses a second
output for the feature misalignment loss term for domain adaptation.

Loss function. NEOS minimizes a loss comprising the terms: (a) cross-entropy
for pixel-level classification which is computed using the input labelled
images and their corresponding ground truth segmentation masks, (b) ($1$ \-
Dice score) for segmentation, and (c) the features misalignment loss. The
latter is for domain adaptation to enforce the model to be able to work well
with the different datasets. Here, an architecture with two heads is used. The
first two loss terms control the first output head for accurate segmentation
and classification on the different labelled datasets, while the features
misalignment loss, which controls the second output head, enforces the network
to reduce the distance between the latent representations of the labelled and
unlabelled data in the cross-entropy metric. This forces the features to
achieve manifold alignment of the embeddings of the different domains. We
denote the data by x and the ground truth masks by y. Now, the cost function
is:

$\displaystyle\text{argmin}_{f}\ L\text{, }\ \ \
L=L_{0}(\textbf{x},\textbf{y})+\lambda_{1}L_{1}(\textbf{x},\textbf{y})+\lambda_{2}L_{2}(\textbf{x})\text{,}$
(1)

where we denote our model by $f(\cdot)$, the first, second, and third loss
terms by $L_{0}$, $L_{1}$, and $L_{2}$, respectively, and the hyperparameters
of the second and third losses by $\lambda_{1}$ and $\lambda_{2}$.

The first term of our model’s objective loss function is:

$\displaystyle
L_{0}=-\dfrac{1}{NWH}\sum_{j=1}^{N}\sum_{i=1}^{W}\sum_{l=1}^{H}\log\dfrac{\exp(f_{y_{j,i,l}}(\textbf{x}_{j}))}{\sum_{k=1}^{K}\exp(f_{k,i,l}(\textbf{x}_{j}))}\text{,}$
(2)

where $W$ and $H$ are the width and height of the images, $i$ and $l$ the
indices for the width and height, $N$ the number of samples, $j$ their index,
$K$ the number of classes, and $k$ their index. In (2), the $N$ training
samples originate from $Q$ labelled datasets. This affects both
$\textbf{x}_{j}$ and $y_{j,i,l}$. The features before the normalized
exponential, i.e. softmax, are denoted by $f_{y_{j,i,l}}(\textbf{x}_{j})$. The
model computes the estimated probability of the labels for the cross-entropy
loss to reward correct classification, penalizing deviation from the correct
class. The estimated semantic segmentation mask, $\hat{y}_{j,i,l}$ is the
pixel-wise class label. Here, for the labelled data, to perform accurate
classification, NEOS minimizes (2), which is the pixel-wise cross-entropy
loss.

Fig. 2: Evaluation of NEOS in accuracy (Acc), F1-score (F1) and IoU on the
dataset Potsdam with the class Clutter [12].

Next, the second term of the loss function, $L_{1}$ in (1), is:

$\displaystyle 1\ -\ \dfrac{2\
\sum_{j=1}^{N}\sum_{i=1,l=1}^{W,H}g_{j,i,l}s_{j,i,l}}{\sum_{j=1}^{N}\sum_{i=1,l=1}^{W,H}g_{j,i,l}+\sum_{j=1}^{N}\sum_{i=1,l=1}^{W,H}s_{j,i,l}}\text{,}$
(3)

where $g_{j,i,l}$ is the true binary indicator of the class label, and
$s_{j,i,l}$ is the estimated semantic segmentation probability outputted by
the model. The $N$ samples originate from $Q$ labelled datasets. This affects
both $g_{j,i,l}$ and $s_{j,i,l}$. To perform accurate segmentation on the
labelled data, NEOS minimizes (3), the Dice loss. Next, the third term of the
loss function, $L_{2}$, is:

$\displaystyle L_{2}\ \ =\ \ \dfrac{1}{J}\
\sum_{j=1}^{J}\log\dfrac{\exp(f_{z_{j}}(\textbf{x}_{j}))}{\sum_{m=1}^{M}\exp(f_{m}(\textbf{x}_{j}))}\text{,}$
(4)

where the true domain label is denoted by $z_{j}$, and the number of domain
labels by $M$. The samples originate from $Q$ labelled and $R$ unlabelled
datasets. For domain adaptation, NEOS minimizes (4). To improve generalization
(better performance), we perform data augmentation to incorporate invariances
into the model. We regularize and enforce the model to generalize and be
robust to data transformations, and we also perform downsampling, having
inputs at different multi-scale levels.

##  4 Evaluation and Results

Labelled and unlabelled datasets. We train NEOS on the aerial image datasets:
labelled Potsdam and Vaihingen, and unlabelled CVUSA by performing domain
adaptation. We test NEOS on Potsdam and Vaihingen, as well as on CVUSA. We
also compare our model to other baseline models, including SAM [16]. For the
labelled datasets, the classes are: Buildings (blue colour), Trees (green),
Cars (yellow), Low vegetation (cyan), Roads (white), and Clutter (red) [23,
24]. Also, for the different datasets and the domain adaptation loss term, as
well as the tags for the different datasets, Tag A is used for the dataset
Potsdam, Tag B for Vaihingen, and C for CVUSA.

Evaluation of NEOS on Potsdam. We evaluate NEOS on the dataset Potsdam in Fig.
2. In this experiment, the class Clutter is considered in the evaluation.
Here, the evaluation is based on the accuracy, F1-score, and Intersection over
Union (IoU) metrics. It can be observed in Fig. 2 that the proposed model
outperforms all the other baseline models [23]. In Fig. 3, we evaluate the
per-class F1-score performance of NEOS on Potsdam. In Fig. 4, we present NEOS
qualitative results.

Fig. 3: Per-class F1-score evaluation (in $\%$) of NEOS on the Potsdam dataset
including the class Clutter in the evaluation.

a) Input

b) NEOS (Ours)

c) Ground truth

Fig. 4: Semantic segmentation masks by NEOS on Potsdam.

Evaluation of NEOS on Vaihingen. We evaluate NEOS in Fig. 5 in accuracy,
F1-score and IoU. Here, the results of NEOS on Vaihingen are comparable to
other models. In Fig. 6, we also evaluate the F1-score of NEOS for each class
[15]. For Roads, Buildings and Cars, NEOS outperforms other models.

Evaluation on the unlabelled dataset CVUSA. We evaluate NEOS on the unlabelled
dataset aerial CVUSA. The testing is done on a dataset that does not have
ground truth masks. In Fig. 7, we present the qualitative results of NEOS. We
observe in (b) and (e) that NEOS performs semantic segmentation and is able to
recognize effectively classes such as Roads and Low vegetation. This holds for
NEOS for the vertical roads that have shadows (occlusion) in (a)-(b). In the
next paragraphs, we present a numerical evaluation of NEOS on the unlabelled
CVUSA dataset and a further comparison to other models.

Fig. 5: Evaluation of NEOS on Vaihingen in Acc, F1 and IoU.

Fig. 6: Per-class F1-score evaluation of NEOS on Vaihingen.

a) Input

b) NEOS (Ours)

c) samgeoHQ

d) Input

e) NEOS (Ours)

f) samgeoHQ

Fig. 7: Qualitative evaluation of NEOS on the unlabelled CVUSA aerial dataset,
and comparison to samgeoHQ [25].

In Fig. 7, we examine the qualitative results of NEOS, and we also do this at
large scale, automating the process. We assess the performance of NEOS on many
images, and to capture the big picture, we evaluate NEOS numerically by
computing the Segments of Predictions and Inputs Error (SPIE). We first
perform detection of segments on the estimated mask and input images, and for
this, we use a variant of SAM [26]. Then, the error is calculated and
normalized. For perfect segmentation without considering semantic information
(no labels), SPIE is zero. For completely inaccurate segmentation, SPIE is
equal to one. SPIE is the mean residual image where the residual is between
the estimated mask and the input after being modified by a detection of
segments algorithm, and its definition is:

$\displaystyle\text{SPIE}\ \ =\ \ \dfrac{1}{R}\
\sum_{j=1}^{R}g(f(\textbf{x}_{j}))-g(\textbf{x}_{j})$ (5)

where $f(\cdot)$ is NEOS from (1) (or another model), $R$ the number of
evaluation samples, and $g(\cdot)$ a detection of segments algorithm. We use
SPIE as an indicator of good performance and as an empirical metric that works
in practice. Using SPIE in (5), the numerical results match the qualitative
results we obtain. In addition, we have also included the code in [27].

Table 1: Evaluation of NEOS on the CVUSA dataset, on both the aerial (Aer) and street (Str), and the improvement (I) over the base model. SPIE for aerial & street |  Aer |  I Aer |  Str |  I Str
---|---|---|---|---
NEOS (Ours) |  $0.047$ |  $32\%$ |  $0.041$ |  $21\%$
Base model, SegFormer |  $0.069$ |  N/A |  $0.052$ |  N/A
CNN-based using Eq. (1) |  $0.064$ |  $7.2\%$ |  $0.049$ |  $5.8\%$

We evaluate NEOS numerically on CVUSA in Table 1 using SPIE, and we compare
with other models for semantic segmentation. The improvement of NEOS for
aerial images over the base model, SegFormer [14], is $32\%$. NEOS also
outperforms [14] for street images when we use the labelled dataset CityScapes
[28] for the domain adaptation loss in Sec. 3.

Further comparison of NEOS to other models. In Fig. 7, we examine the results
of NEOS and its comparison to SAM [16]. NEOS performs joint classification and
segmentation, while SAM does not consider semantics [29]. There are several
variants of SAM: Geospatial SAM [25] fine-tunes [16] on aerial data. SAM-HQ
[30] achieves improved accuracy (FastSAM [31], speed). Semantic SAM (SSAM)
[32, 33] modifies [16] to perform semantic (rather than instance)
segmentation, but not for aerial data. We examined the performance of SAM and
its variants on the unlabelled CVUSA dataset. In Fig. 7, samgeoHQ [25], which
does not perform semantic segmentation, is sensitive to even small changes in
the scene. For scenes with details, we need to adjust the several tunable
parameters of SAM to control how dense the estimated masks are. For NEOS, we
do not need to adjust its parameters, and this can potentially lead to
improved user convenience and ease of use.

##  5 Conclusion

We have developed a semantic segmentation method that is effective for
unlabelled datasets. The results show that NEOS outperforms other models. We
have used the unlabelled aerial CVUSA dataset, where accurate semantic
segmentation has not yet been performed, to the best of the authors’
knowledge, and we plan to also use the results for cross-view geo-location
matching to accurately match aerial and street images [34, 35].

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