[1,2]Surajit Sen

[1]Physics Department, Guru Charan College,
Silchar, 788004, India

2] Centre of Advanced Studies and Innovation Lab,
18/27 Kali Mohan Road, Tarapur, Silchar, 788003, India

# An Inequality for Entangled Qutrits in SU(3) basis

<EMAIL_ADDRESS>Tushar Kanti Dey<EMAIL_ADDRESS>*  [

###### Abstract

It is well-known from the representation theory of particle physics that the
tensor product of two fundamental representation of SU(2) and SU(3) group can
be decomposed to obtain the desired spectrum of the physical states. In this
paper, we apply this tenet in case of two non-local qubits and qutrits, which
leads the complete spectrum of their entangled states in their respective
basis. For qutrit system, the study of their properties reveals the existence
of a new $\sqrt{2}$ inequality, in addition to usual Bell-CHSH type
$2\sqrt{2}$ inequality, which is significant from the experimental point of
view.

###### keywords:

Entanglement, SU(3) Group, Qutrit states, Inequality for Qutrits

##  1 Introduction

Quantum entanglement is considered as one of the most exquisite traits of the
quantum world [1]. During the early years of its development, Einstein-
Podolsky-Rosen (EPR) raised their doubt about the ability of quantum theory to
explain the nature in a way that is consistent with both the special theory of
relativity and the objective realism [2]. Their stance necessitates the
postulation of the concept of hidden variables in support of their perspective
and also to address some compatibility issues related to the foundational
aspects of quantum mechanics. However, soon after that, Bohr refuted their
viewpoint by emphasizing on the crucial role of the observer in the
measurement process in the quantum realm, which invalidates the proposition of
such variables [3]. Later, the issue became more popular when Bohm
reformulated it in a more pragmatic footing [4], enabling Bell [5, 6], and
thereafter Clauser-Horne-Shimony-Holt (CHSH) [7] to point out the existence of
an experimentally testable identity. Finally, it was verified experimentally
by Aspect and his coworkers [9], which was further refined by fixing several
loopholes [10, 11, 12]. Although these studies have laid the foundation of the
modern paradigm of quantum information science, the study of entanglement
beyond qubit particularly in high dimension, always remains at the center
stage for harnessing faster information processing.

The simplest representative of an entangled system is the two-qubit system,
formally known as the Bell states. This set of entangled states is constructed
by the successive operations of the Hadamard and CNOT gates, some variant of
the Pauli gates, on the two-qubit state which is intricately linked with the
Bell-CHSH inequality [7, 8]. However, in a three-dimensional Hilbert space,
the straightforward extension to an entangled qutrit system is not available
for various reasons. Recently significant efforts have been made to understand
various aspect of qutrits from different perspective. These endeavors include
the exploration of the entanglement scenario of the qutrit system [13, 14, 15,
16, 17, 18, 19], the Bloch space structure of qutrits [20, 21, 22, 23, 24, 25,
26] etc. In addition, since the qutrit system is inherently a three-level
system, several quantum-optical phenomena have been investigated within the
framework of the SU(3) group [28, 27, 31, 29, 30]. Although these studies
revealed the crucial role of the SU(3) group to understand various properties
of the qutrit system, the information about all conceivable SU(3) based
entangled qutrit states is still unknown. The prime hindrance of it is due to
the absence of experimentally verifiable inequality for the qutrits,
nonavailability of suitable representation of the qutrit gates, and added to
them, the complexity arises due to our incomplete understanding of the
entanglement scenario of high dimensional Hilbert space.

There is another reason to study the qutrit system. Soon after the
experimental breakthrough of the Bell-CHSH inequality [7, 9], it was well
understood that an inequality could serve as a powerful tool to verify the
non-local character of the quantum correlation. In that context, the qutrit
based inequality emerges as a natural choice to address the entanglement
scenario of the high dimensional system. In the recent past, the Collins-
Gisin-Linden-Massar-Popescu (CGLMP) inequality has garnered significant
attention as a multi-partite entangled system, including the qutrit system
[32, 33, 34, 35]. The evaluation of the magnitude of violation from this
inequality necessitates the calculation of the expectation value of the CGLMP
joint probability operator using an appropriate qutrit state. In this context,
the Maximally Entangled State (MES), namely,
$\psi_{MES}=\frac{1}{\sqrt{3}}\sum_{j=0}^{2}\ket{j}_{A}\otimes\ket{j}_{B}$,
which is a singlet state of SU(3) group, has been studied in the recent past
[14, 13, 32, 36, 37]. Therefore, it is intriguing to explore other conceivable
representations of the two-qutrit states using SU(3) group and then to look
for the associated inequality. The measurement setting of the CGLMP operator
is determined by various factors such as, the number of parties involved, the
number of observable per party, and the number of outcomes per observable
[32]. Such requirement imposes a stringent constraint on the geometry of the
experimental setup, making it quite difficult for functional implementation.
Therefore, it is worthwhile to search for an inequality for the qutrit system
in a simpler framework which is closely parallel to the conventional Bell-CHSH
setup.

The aim of this paper is to generate the complete spectrum of entangled qutrit
states within framework of SU(3) group and to find the associated inequality.
To accomplish this objective, the paper is structured as follows: In Section
II, we develop a heuristic scheme to derive the Bell states from the SU(2)
representation theory and review its connection with the Bell-CHSH-Tsirelson
bound [7, 8]. In Section III, we extend this methodology to construct all
qutrit states using the SU(3) as the basis group and then, discuss their
properties in IV. In Section V, we demonstrate how these states can be
utilized to develop a pair of inequalities, which we refer as Qutrit
Inequality. Finally, we conclude by summarizing the key findings of the paper
and discuss the future prospects.

##  2  SU(2) Group and Bell States

To construct the Bell states, let ${q}_{i}^{\alpha}$
($q_{i}^{1}=u_{i},q_{i}^{2}=d_{i}$) be the fundamental representation of
$SU(2)$ group in two-dimensional Hilbert space $\mathcal{H}_{i}^{2}$ with site
index $i$ ($i=A,B$). Following the decomposition of the SU(2) representation
of particle physics [38, 39], the tensor product of two non-local doublets at
site $A$ and $B$ can be expressed as their direct sum,

$\displaystyle\\{{\bf{\bar{2}}}\\}_{A}\otimes\\{{\bf 2}\\}_{B}={\bf
3}_{AB}\oplus{\bf 1}_{AB},$ (1)

where ${\bf 3}_{AB}$ and ${\bf 1}_{AB}$ represent three triplet and one
singlet states forming a composite entangled system $AB$ in Hilbert space
$\mathcal{H}^{2\otimes 2}_{AB}$.

To construct Bell states, we adopt a heuristic approach and define the
following $\it star$ $\it product$ ($\star$) between the non-local doublets,

$\displaystyle{q_{A}^{\alpha}}^{T}\star\sigma_{0}\star q_{B}^{\alpha}$
$\displaystyle:={u_{A}}\star u_{B}+{d_{A}}\star d_{B},$ (2a)
$\displaystyle{q_{A}^{\alpha}}^{T}\star\sigma_{1}\star q_{B}^{\alpha}$
$\displaystyle:={u_{A}}\star d_{B}+{d_{A}}\star u_{B},$ (2b)
$\displaystyle{q_{A}^{\alpha}}^{T}\star i\sigma_{2}\star q_{B}^{\alpha}$
$\displaystyle:={u_{A}}\star d_{B}-{d_{A}}\star u_{B},$ (2c)
$\displaystyle{q_{A}^{\alpha}}^{T}\star\sigma_{3}\star q_{B}^{\alpha}$
$\displaystyle:={u_{A}}\star u_{B}-{d_{A}}\star d_{B},$ (2d)

where $\sigma_{i}(i=0,1,2,3)$ is the Pauli matrix. Now our method involves two
steps: a) promote the star product to tensor product of the basis states
($\star\rightarrow\otimes$), i.e.,

$\displaystyle{u_{A}}\star u_{B}+{d_{A}}\star d_{B}\rightarrow$
$\displaystyle\mid\sigma_{0}\rangle_{AB}:=u_{A}\otimes u_{B}+{d_{A}}\otimes
d_{B},$ (3a) $\displaystyle{u_{A}}\star d_{B}+{d_{A}}\star u_{B}\rightarrow$
$\displaystyle\mid\sigma_{1}\rangle_{AB}:=u_{A}\otimes d_{B}+{d_{A}}\otimes
u_{B},$ (3b) $\displaystyle{u_{A}}\star d_{B}-{d_{A}}\star u_{B}\rightarrow$
$\displaystyle\mid\sigma_{2}\rangle_{AB}:=u_{A}\otimes d_{B}-{d_{A}}\otimes
u_{B},$ (3c) $\displaystyle{u_{A}}\star u_{B}-{d_{A}}\star d_{B}\rightarrow$
$\displaystyle\mid\sigma_{3}\rangle_{AB}:=u_{A}\otimes u_{B}-{d_{A}}\otimes
d_{B},$ (3d)

and then, b) identify the components of the doublets with the standard qubit
basis, namely, $u_{i}:=\ket{0}_{i}=(1,0)_{i}^{T}$,
$d_{i}:=\ket{1}_{i}=(0,1)_{i}^{T}$. Thus the normalized Bell states obtained
from Eq.(3) are given by [1, 6] (For notational convenience, we write
$\ket{0}_{A}\otimes\ket{0}_{B}=\ket{00}$,
$\ket{1}_{A}\otimes\ket{0}_{B}=\ket{10}$ etc and drop local indices $i=A,B$),

$\displaystyle\ket{\Phi^{+}}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\ket{00}+\ket{11}\big{)},$ (4a)
$\displaystyle\ket{\Psi^{+}}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\ket{01}+\ket{10}\big{)},$ (4b)
$\displaystyle\ket{\Psi^{-}}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\ket{01}-\ket{10}\big{)},$ (4c)
$\displaystyle\ket{\Phi^{-}}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\ket{00}-\ket{11}\big{)},$ (4d)

where $\ket{\sigma_{0}}_{AB}:=\ket{\Phi^{+}}$,
$\ket{\sigma_{1}}_{AB}:=\ket{\Psi^{+}}$,
$\ket{\sigma_{2}}_{AB}:=\ket{\Psi^{-}}$ and
$\ket{\sigma_{3}}_{AB}:=\ket{\Phi^{-}}$.

After developing the Bell states, we shall very briefly recall its connection
with the Bell-CHSH inequality to understand the subsequent development of the
paper. The correlation tensor in $\mathcal{H}^{2\otimes 2}_{AB}$ read-off from
this inequality is given by [1, 7, 8],

$\displaystyle\hat{\mathcal{C}}^{2\otimes
2}=\hat{Q}\otimes\hat{S}+\hat{R}\otimes\hat{S}+\hat{R}\otimes\hat{T}-\hat{Q}\otimes\hat{T},$
(5)

where $\hat{Q}=\hat{\sigma}_{3}$,
$\hat{S}=-\frac{1}{\sqrt{2}}(\hat{\sigma}_{3}+\hat{\sigma}_{1})$,
$\hat{R}=\hat{\sigma}_{1}$, and
$\hat{T}=\frac{1}{\sqrt{2}}(\hat{\sigma}_{3}-\hat{\sigma}_{1})$. This setting
can be equivalently expressed as,

$\displaystyle\hat{\mathcal{C}}^{2\otimes
2}=\left(\begin{array}[]{cccc}-\sqrt{2}&0&0&-\sqrt{2}\\\
0&\sqrt{2}&-\sqrt{2}&0\\\ 0&-\sqrt{2}&\sqrt{2}&0\\\ -\sqrt{2}&0&0&-\sqrt{2}\\\
\end{array}\right).$ (10)

It is easy to show that the $X$-structured correlation operator is equal to
the outer product of two of the Bell states or, equivalently, the tensor
product of the Pauli matrices (For detail see, Note added in proof),

$\displaystyle\hat{\mathcal{C}}^{2\otimes 2}$
$\displaystyle=2\sqrt{2}\big{(}\ket{\Psi^{-}}\bra{\Psi^{-}}-\ket{\Phi^{+}}\bra{\Phi^{+}}\big{)}$
$\displaystyle=-\sqrt{2}(\hat{\sigma}_{3}\otimes\hat{\sigma}_{3}+\hat{\sigma_{1}}\otimes\hat{\sigma_{1}}).$
(11)

This ensures the non-zero expectation value giving Bell-CHSH-Tsirelson bound
[8, 40],

$\displaystyle\langle\Phi^{+}\mid\hat{\mathcal{C}}^{2\otimes
2}\mid\Phi^{+}\rangle$ $\displaystyle=-2\sqrt{2},$ (12a)
$\displaystyle\langle\Psi^{-}\mid\hat{\mathcal{C}}^{2\otimes
2}\mid\Psi^{-}\rangle$ $\displaystyle=2\sqrt{2},$ (12b)

while for remaining two Bell states it vanishes. Eq.(12) is the celebrated
bound for the states $\ket{\psi^{i}}=\\{\ket{\Phi^{+}},\ket{\Psi^{-}}\\}$
known as the Bell-CHSH inequality [1]

${\mid\langle}\psi^{i}\mid\hat{\mathcal{C}}^{2\otimes
2}\mid\psi^{i}\rangle\mid\leq 2\sqrt{2}.$ (13)

Thus, it is worth noting that even in the absence of specific information
about quantum gates, e.g., Hadamard and CNOT gates, it is still possible to
derive the complete spectrum of the entangled states and the inequality
associated with it. We now proceed to extend above procedure to derive an
inequality for the entangled qutrit system.

##  3  SU(3) Group and Entangled Qutrit States

In three-dimensional Hilbert space $\mathcal{H}^{3}_{i}$ ($i=A$ or $B$), a
qutrit system consists of three orthogonal basis states,

$\displaystyle\ket{\psi}=c_{0}\ket{{0}}+c_{1}\ket{{1}}+c_{2}\ket{{2}},$ (14)

where the standard basis are given by,

$\begin{split}\ket{{0}}=\left(\begin{array}[]{c}1\\\ 0\\\ 0\\\
\end{array}\right),\quad\ket{{1}}=\left(\begin{array}[]{c}0\\\ 1\\\ 0\\\
\end{array}\right),\quad\ket{{2}}=\left(\begin{array}[]{c}0\\\ 0\\\ 1\\\
\end{array}\right),\end{split}$ (15)

with the amplitude normalized as $\sum_{i=0}^{2}|c_{i}|^{2}=1$. To obtain all
possible entangled states in composite Hilbert space
$\mathcal{H}^{{3}\otimes{3}}_{AB}$, we first consider the Gell-Mann matrices
which in terms of the qutrit basis are given by,

$\displaystyle\lambda_{0}$
$\displaystyle=\ket{0}\bra{0}+\ket{1}\bra{1}+\ket{2}\bra{2},\qquad\lambda_{1}=\ket{0}\bra{1}+\ket{1}\bra{0}$
(16) $\displaystyle\lambda_{2}$
$\displaystyle=i(\ket{1}\bra{0}-\ket{0}\bra{1}),\qquad\lambda_{3}=\ket{0}\bra{0}-\ket{1}\bra{1},$
$\displaystyle\lambda_{4}$
$\displaystyle=\ket{0}\bra{2}+\ket{2}\bra{0},\quad\qquad\lambda_{5}=i(\ket{2}\bra{0}-\ket{0}\bra{2}),$
$\displaystyle\lambda_{6}$
$\displaystyle=\ket{1}\bra{2}+\ket{2}\bra{1},\qquad\lambda_{7}=i(\ket{2}\bra{1}-\ket{1}\bra{2}),$
$\displaystyle\lambda_{8}$
$\displaystyle=\frac{1}{\sqrt{3}}(\ket{0}\bra{0}+\ket{1}\bra{1}-2\ket{2}\bra{2}).$

Here the $\lambda_{i}$ matrices are normalized as
$\lambda_{l}\lambda_{m}=\delta_{lm}+d_{lmn}\lambda_{n}+f_{lmp}\lambda_{p}$
with $d_{lmn}$ and $f_{lmp}$ ($l,m,n,p=1,2,\dots,8$) as the completely
symmetric and completely antisymmetric structure constants [39]. Similar to
the previous section, we now consider two non-local triplet
$q_{i}^{\alpha}=(u_{i},d_{i},s_{i})^{T}\in\mathcal{H}_{i}^{3}$ and promote the
star product into tensor product,

$\displaystyle{q_{A}^{\alpha}}^{T}\star\lambda_{0}\star{q_{B}^{\alpha}}\rightarrow$
$\displaystyle\mid\lambda_{0}\rangle_{AB}:=u_{A}\otimes u_{B}+d_{A}\otimes
d_{B}+s_{A}\otimes s_{B},$ (17a)
$\displaystyle{q_{A}^{\alpha}}^{T}\star\lambda_{1}\star{q_{B}^{\alpha}}\rightarrow$
$\displaystyle\mid\lambda_{1}\rangle_{AB}:=s_{A}\otimes v_{B}+v_{B}\otimes
s_{A},$ (17b) $\displaystyle{q_{A}^{\alpha}}^{T}\star
i\lambda_{2}\star{q_{B}^{\alpha}}\rightarrow$
$\displaystyle\mid\lambda_{2}\rangle_{AB}:=-s_{A}\otimes v_{B}+v_{B}\otimes
s_{A},$ (17c)
$\displaystyle{q_{A}^{\alpha}}^{T}\star\lambda_{3}\star{q_{B}^{\alpha}}\rightarrow$
$\displaystyle\mid\lambda_{3}\rangle_{AB}:=-v_{A}\otimes v_{B}+s_{A}\otimes
s_{B},$ (17d)
$\displaystyle{q_{A}^{\alpha}}^{T}\star\lambda_{4}\star{q_{B}^{\alpha}}\rightarrow$
$\displaystyle\mid\lambda_{4}\rangle_{AB}:=u_{A}\otimes s_{B}+s_{B}\otimes
u_{A},$ (17e) $\displaystyle{q_{A}^{\alpha}}^{T}\star
i\lambda_{5}\star{q_{B}^{\alpha}}\rightarrow$
$\displaystyle\mid\lambda_{5}\rangle_{AB}:=-u_{A}\otimes s_{B}+s_{B}\otimes
u_{A},$ (17f)
$\displaystyle{q_{A}^{\alpha}}^{T}\star\lambda_{6}\star{q_{B}^{\alpha}}\rightarrow$
$\displaystyle\mid\lambda_{6}\rangle_{AB}:=u_{A}\otimes v_{B}+v_{B}\otimes
u_{A},$ (17g) $\displaystyle{q_{A}^{\alpha}}^{T}\star
i\lambda_{7}\star{q_{B}^{\alpha}}\rightarrow$
$\displaystyle\mid\lambda_{7}\rangle_{AB}:=-u_{A}\otimes v_{B}+v_{B}\otimes
u_{A},$ (17h)
$\displaystyle{q_{A}^{\alpha}}^{T}\star\lambda_{8}\star{q_{B}^{\alpha}}\rightarrow$
$\displaystyle\mid\lambda_{8}\rangle_{AB}:=-2u_{A}\otimes u_{B}+d_{A}\otimes
d_{B}+s_{A}\otimes s_{B}.$ (17i)

If we identify the states $\\{u_{i},d_{i},s_{i}\\}$ with the standard basis in
Eq.(16), i.e., $u_{i}=\ket{0}_{i}$, $d_{i}=\ket{1}_{i}$, $s_{i}=\ket{2}_{i}$,
Eq.(17) leads to the normalized entangled qutrit states,

$\displaystyle\ket{\psi_{00}}$
$\displaystyle=\frac{1}{\sqrt{3}}\big{(}\ket{00}+\ket{11}+\ket{22}\big{)},$
(18a) $\displaystyle\ket{\psi_{21}^{+}}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\ket{21}+\ket{12}\big{)},$ (18b)
$\displaystyle\ket{\psi_{21}^{-}}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\ket{21}-\ket{12}\big{)},$ (18c)
$\displaystyle\ket{\psi_{11}}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}-\ket{11}+\ket{22}\big{)},$ (18d)
$\displaystyle\ket{\psi_{20}^{+}}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\ket{20}+\ket{02}\big{)}$ (18e)
$\displaystyle\ket{\psi_{20}^{-}}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\ket{20}-\ket{02}\big{)},$ (18f)
$\displaystyle\ket{\psi_{10}^{+}}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\ket{10}+\ket{01}\big{)},$ (18g)
$\displaystyle\ket{\psi_{10}^{-}}$
$\displaystyle=\frac{1}{\sqrt{2}}\big{(}\ket{10}-\ket{01}\big{)},$ (18h)
$\displaystyle\ket{\psi_{22}}$
$\displaystyle=\frac{1}{\sqrt{6}}\big{(}-2\ket{00}+\ket{11}+\ket{22}\big{)}.$
(18i)

where $\ket{\lambda_{0}}_{AB}:=\ket{\psi_{00}}$,
$\ket{\lambda_{1}}_{AB}:=\ket{\psi_{21}^{+}}$ etc. Thus we note that,
analogous to $SU(2)$ theory, two qutrits follow SU(3) decomposition, namely,
$\\{{\bf{\bar{3}}}\\}_{A}\otimes\\{{\bf{3}}\\}_{B}={\bf 8}_{AB}\oplus{\bf
1}_{AB}$ in Hilbert space $\mathcal{H}^{3\otimes 3}_{AB}$ which leads to eight
octet-like and one singlet-like entangled states. Among them, only the singlet
state $\ket{\psi_{00}}$ has been recently investigated both theoretically [14,
32, 34] and experimentally [36, 37], while other states remain unexplored.

##  4  Properties of Entangled Qutrit States

Before delivering the inequalities associated with these states, we shall
touch upon their properties. We note that, on swapping positions, out of the
nine states,
$\ket{\psi_{00}},\ket{\psi_{10}^{+}},\ket{\psi_{20}^{+}},\ket{\psi_{21}^{+}}$
are symmetric states,
$\ket{\psi_{11}},\ket{\psi_{10}^{-}},\ket{\psi_{20}^{-}},\ket{\psi_{21}^{-}}$
are antisymmetric states, while $\ket{\psi_{22}}$ does not conform to this
pattern.

To see whether their subsystems are pure or mixed state, it is customary to
consider the reduced density matrix of the system. Taking the partial trace
over one subsystem (say, B-subsystem) we obtain,

$\displaystyle\rho_{00}^{A}$
$\displaystyle=\frac{1}{3}\big{(}\ket{0}_{A}\bra{0}+\ket{1}_{A}\bra{1}+\ket{2}_{A}\bra{2}\big{)},$
$\displaystyle{\rho_{10}^{+}}^{A}$
$\displaystyle={\rho_{10}^{-}}^{A}=\frac{1}{2}(\ket{0}_{A}\bra{1}+\ket{1}_{A}\bra{0}\big{)},$
(19) $\displaystyle{\rho_{20}^{+}}^{A}$
$\displaystyle={\rho_{20}^{-}}^{A}=\frac{1}{2}\big{(}\ket{0}_{A}\bra{2}+\ket{2}_{A}\bra{0}\big{)},$
$\displaystyle{\rho_{21}^{+}}^{A}$
$\displaystyle={\rho_{21}^{-}}^{A}=\rho_{11}^{A}=\frac{1}{2}\big{(}\ket{1}_{A}\bra{2}+\ket{2}_{A}\bra{1}\big{)},$
$\displaystyle\rho_{22}^{A}$
$\displaystyle=\frac{1}{6}\big{(}4\ket{0}_{A}\bra{0}+\ket{1}_{A}\bra{1}+\ket{2}_{A}\bra{2}\big{)},$

We note that the reduced density matrices satisfy the properties
$Tr[\rho^{A}]=1$ and $Tr[{\rho^{A}}^{2}]<1$, indicating that they are
essentially a mixed state. The maximally mixed state can be corroborated by
noting the entropy, $S^{A}_{00}=\log_{2}3:=1.585$ for the singlet state in
comparison to the other states, $S^{A}_{ij}=1$ and $S^{A}_{22}=1.2516$.

For completeness, we finally discuss the change of basis of the bipartite
qutrit system. In the computational basis, most general entangled two-qutrit
system is given by,

$\displaystyle\ket{\Psi_{T}}$
$\displaystyle=c_{00}\ket{00}+c_{01}\ket{01}+c_{02}\ket{02}$ (20)
$\displaystyle+c_{10}\ket{10}+c_{11}\ket{11}+c_{12}\ket{12}$
$\displaystyle+c_{20}\ket{20}+c_{21}\ket{21}+c_{22}\ket{22}.$

where the amplitudes are normalized as $\sum_{i,j}|c_{ij}|^{2}=1$. Plucking
back $\ket{00}$, $\ket{01}$ etc from Eq.(18) into Eq.(20), the wave function
can be transformed into $SU(3)$ basis,

$\displaystyle\ket{\Psi_{SU(3)}}$
$\displaystyle=b_{00}\ket{\psi_{00}}+b_{21}^{+}\ket{\psi_{21}^{+}}+b_{21}^{-}\ket{\psi_{21}^{-}},$
(21)
$\displaystyle+b_{11}\ket{\psi_{11}}+b_{20}^{+}\ket{\psi_{20}^{+}}+b_{20}^{-}\ket{\psi_{02}^{-}},$
$\displaystyle+b_{10}^{+}\ket{\psi_{10}^{+}}+b_{10}^{-}\ket{\psi_{10}^{-}}+b_{22}\ket{\psi_{22}},$

where the amplitudes are given by,

$\displaystyle b_{00}=$
$\displaystyle\frac{1}{\sqrt{3}}(c_{00}-\sqrt{2}c_{22}),\quad
b_{21}^{+}=\frac{1}{\sqrt{2}}({c_{01}}+{c_{02}}),\quad
b_{21}^{-}=\frac{1}{\sqrt{2}}({c_{01}}-{c_{02}}),$ (22) $\displaystyle
b_{11}=$
$\displaystyle\frac{1}{\sqrt{6}}(\sqrt{2}c_{20}-\sqrt{3}c_{02}+c_{22}),\quad
b_{20}^{+}=\frac{1}{\sqrt{2}}({c_{20}}+{c_{21}}),\quad
b_{20}^{-}=\frac{1}{\sqrt{2}}({c_{20}}-{c_{21}}),$ $\displaystyle b_{10}^{+}=$
$\displaystyle\frac{1}{\sqrt{2}}({c_{11}}+{c_{12}}),\quad
b_{10}^{-}=\frac{1}{\sqrt{2}}({c_{11}}-{c_{12}}),\quad
b_{22}=\frac{1}{\sqrt{6}}(\sqrt{2}c_{00}+\sqrt{3}c_{10}+c_{22}).$

In this new basis, the amplitudes once again are normalized, i.e.,
$\sum_{i,j}|b_{ij}|^{2}=1$, indicating the consistency of our construction of
all entangled qutrit states. In both basis, the density matrix satisfies the
pure state condition, $\rho^{2}_{T}=\rho_{T}$ and
$\rho^{2}_{SU(3)}=\rho_{SU(3)}$.

##  5  Qutrit Inequality

To this end, we are finally in position to discuss the inequality associated
with the states obtained above. To construct the correlation tensor using
SU(3) as the basis group, we adopt the methodology developed above. Similar to
Eq.(10), the correlation tensor for the qutrit system should satisfy the
following properties: i) it must be a symmetric matrix, ii) it should be
diagonal in the qutrit basis states $\ket{\psi_{ij}}$ and, iii) it should be
expressed as the tensor product of the $SU(3)$ matrices [40]. After a lengthy
but straight forward calculation yields,

$\displaystyle\hat{\mathcal{C}}^{3\otimes 3}$
$\displaystyle=\left(\begin{array}[]{ccccccccc}0&0&0&0&\sqrt{2}&0&0&0&\sqrt{2}\\\
0&\sqrt{2}&0&-\sqrt{2}&0&0&0&0&0\\\ 0&0&-\sqrt{2}&0&0&0&\sqrt{2}&0&0\\\
0&-\sqrt{2}&0&\sqrt{2}&0&0&0&0&0\\\ \sqrt{2}&0&0&0&\sqrt{2}&0&0&0&0\\\
0&0&0&0&0&-\sqrt{2}&0&\sqrt{2}&0\\\ 0&0&\sqrt{2}&0&0&0&-\sqrt{2}&0&0\\\
0&0&0&0&0&\sqrt{2}&0&-\sqrt{2}&0\\\ \sqrt{2}&0&0&0&0&0&0&0&\sqrt{2}\\\
\end{array}\right).$ (32)

It is worth mentioning here that, similar to Bell system, the diagonalization
of above correlation matrix precisely gives the Bell-like qutrit states
Eq.(18) as its eigenvectors [40]. Finally it is easy to see that the matrix
can be decomposed in terms of their outer product (See, Note added in proof),

$\displaystyle\hat{\mathcal{C}}^{3\otimes 3}$
$\displaystyle=\sqrt{2}\big{(}\ket{\psi_{11}}\bra{\psi_{11}}-\ket{\psi_{22}}\bra{\psi_{22}}\big{)}$
$\displaystyle+2\sqrt{2}\big{(}\ket{\psi_{00}}\bra{\psi_{00}}+\ket{\psi_{10}^{-}}\bra{\psi_{10}^{-}}-\ket{\psi_{12}^{-}}\bra{\psi_{12}^{-}}-\ket{\psi_{20}^{-}}\bra{\psi_{20}^{-}}\big{)}$
(33)

which can be further expressed as the tensor product of GellMann matrices,

$\displaystyle\hat{\mathcal{C}}^{3\otimes 3}$
$\displaystyle=\sqrt{2}(\lambda_{4}\otimes\lambda_{4}-\lambda_{2}\otimes\lambda_{2})+\frac{1}{\sqrt{2}}(\lambda_{6}\otimes\lambda_{6}+\lambda_{7}\otimes\lambda_{7})$
$\displaystyle-\frac{1}{2\sqrt{2}}(\lambda_{3}\otimes\lambda_{3}-5\lambda_{8}\otimes\lambda_{8})+\frac{1}{\sqrt{6}}(\lambda_{8}\otimes\lambda_{0}+\lambda_{0}\otimes\lambda_{8})$
(34)
$\displaystyle-\frac{1}{2\sqrt{6}}(\lambda_{3}\otimes\lambda_{8}+\lambda_{8}\otimes\lambda_{3})-\frac{1}{3\sqrt{2}}(\lambda_{0}\otimes\lambda_{3}+\lambda_{3}\otimes\lambda_{0}).$

The basis in which the correlation matrix is diagonal leads to the following
bounds,

$\displaystyle\langle\psi_{ij}\mid\hat{\mathcal{C}}^{3\otimes
3}\mid\psi_{ij}\rangle$
$\displaystyle=\sqrt{2}\quad\textrm{for}\quad\\{\ket{\psi_{11}}\\},$ (35)
$\displaystyle\langle\psi_{ij}\mid\hat{\mathcal{C}}^{3\otimes
3}\mid\psi_{ij}\rangle$
$\displaystyle=2\sqrt{2}\quad\textrm{for}\quad\\{\ket{\psi_{00}},\ket{\psi_{10}^{-}}\\},$
$\displaystyle\langle\psi_{ij}\mid\hat{\mathcal{C}}^{3\otimes
3}\mid\psi_{ij}\rangle$
$\displaystyle=-\sqrt{2}\quad\textrm{for}\quad\ket{\\{\psi_{22}}\\},$
$\displaystyle\langle\psi_{ij}\mid\hat{\mathcal{C}}^{3\otimes
3}\mid\psi_{ij}\rangle$
$\displaystyle=-2\sqrt{2}\quad\textrm{for}\quad\\{\ket{\psi_{12}^{-}},\ket{\psi_{20}^{-}}\\}.$

while for the remaining states, namely,
$\\{\ket{\psi_{12}^{+}},\ket{\psi_{20}^{+}},\ket{\psi_{10}^{+}}\\}$, it
vanishes. Thus we have two distinct set of inequalities for the entangled
qutrit system,

$\displaystyle\mid\langle\psi_{ij}\mid\hat{\mathcal{C}}^{3\otimes
3}\mid\psi_{ij}\rangle\mid$ $\displaystyle\leq
2\sqrt{2}\quad\textrm{for}\quad\\{\ket{\psi_{00}},\ket{\psi_{12}^{-}},\ket{\psi_{20}^{-}},\ket{\psi_{20}^{-}}\\},$
(36) $\displaystyle\mid\langle\psi_{ij}\mid\hat{\mathcal{C}}^{3\otimes
3}\mid\psi_{ij}\rangle\mid$
$\displaystyle\leq\sqrt{2},\quad\textrm{for}\quad\\{\ket{\psi_{11}},\ket{\psi_{22}}\\}.$

These are desired Qutrit Inequality for the qutrit system. The emergence of
the Bell-CHSH-like $2\sqrt{2}$ inequality in the sideline of our treatment
shows the consistency of the SU(3) based approach of which SU(2) is an
subgroup.

##  6  Conclusion

In this paper, using the representation theory of the SU(2) and SU(3) group,
we present a method to derive the Bell states for qubit and Bell-like
entangled states for qutrit systems. Our approach, which is inspired by
particle physics, uses a heuristic star product method to generate the entire
spectrum of entangled states without prior information about the quantum
gates. After discussing the key properties of these states and their special
attributes, we proceed to find the correlation function of such system which
can be expressed in terms of the tensor product of the Gell-Mann matrices. It
is noteworthy that the appearance of a new $\sqrt{2}$ inequality, apart from
the well-known Bell-CHSH $2\sqrt{2}$ inequality, without prior knowledge of
the detector setting, is a natural outcome of our analysis, which may have
experimental importance. In the ever-expanding landscape of quantum
technology, the qutrits may play a significant role in shaping the future
quantum computers with improved algorithms, augmented communication ability
and multifaceted security.

## Acknowledgement

We thank Professor Bijan Bagchi for fruitful discussion.

##  Note added in proof:

Here we give an outline of the derivation of Eq.(2) and Eq.(5) from the
entangled states of the Bell and Bell-like Qutrit systems, which is crucial to
understand the derivation of the corresponding inequalities.

To derive the correlation matrix for the Bell states, we start with the linear
combination of all possible density matrix operators obtained by taking the
outer product of the Bell states,

$\displaystyle
a_{1}\rho_{\Phi^{+}}+a_{2}\rho_{\Phi^{-}}+a_{3}\rho_{\Psi^{+}}+a_{4}\rho_{\Psi^{-}}$
$\displaystyle=$ $\displaystyle
a_{1}\ket{\Phi^{+}}\bra{\Phi^{+}}+a_{2}\ket{\Phi^{-}}\bra{\Phi^{-}}+a_{3}\ket{\Psi^{+}}\bra{\Psi^{+}}+a_{4}\ket{\Psi^{-}}\bra{\Psi^{-}}$
$\displaystyle=\begin{pmatrix}\frac{1}{2}(a_{1}+a_{2})&0&0&\frac{1}{2}(a_{1}-a_{2})\\\
0&\frac{1}{2}(a_{3}+a_{4})&\frac{1}{2}(a_{3}-a_{4})&0\\\
0&\frac{1}{2}(a_{3}-a_{4})&\frac{1}{2}(a_{3}+a_{4})&0\\\
\frac{1}{2}(a_{1}-a_{2})&0&0&\frac{1}{2}(a_{1}+a_{2})\end{pmatrix},$ (37)

where $a_{i}$ ($i=1,2,3,4$) be constants to be determined. Comparing Eq.(Note
added in proof:) with Eq.(10), the constants are found to be
$a_{1}=-2\sqrt{2}$, $a_{2}=a_{3}=0$, $a_{4}=2\sqrt{2}$ and we have,

$\displaystyle
a_{1}\rho_{\Phi^{+}}+a_{2}\rho_{\Phi^{-}}+a_{3}\rho_{\Psi^{+}}+a_{4}\rho_{\Psi^{-}}$
$\displaystyle=2\sqrt{2}\big{(}\ket{\Psi^{-}}\bra{\Psi^{-}}-\ket{\Phi^{+}}\bra{\Phi^{+}}\big{)}$
$\displaystyle=-\sqrt{2}(\hat{\sigma}_{3}\otimes\hat{\sigma}_{3}+\hat{\sigma_{1}}\otimes\hat{\sigma_{1}})$
$\displaystyle=\hat{\mathcal{C}}^{2\otimes 2}\quad Q.E.D.$ (38)

In the last step we used the fact that the outer product of the Bells states
is numerically equal to the tensor product of the Pauli matrices.

To derive the correlation tensor for the two-qutrit system, similar to the
previous case, we consider the linear combination of the density matrices from
all nine two-qutrit states derived in Eq.(18),

$\displaystyle a_{00}\rho_{\psi_{00}}$
$\displaystyle+a_{12}^{+}\rho_{\psi_{12}^{+}}+a_{12}^{-}\rho_{\psi_{12}^{-}}+a_{11}\rho_{\psi_{11}}$
$\displaystyle+a_{20}^{+}\rho_{\psi_{20}^{+}}+a_{20}^{-}\rho_{\psi_{20}^{-}}+a_{10}^{+}\rho_{\psi_{10}^{+}}+a_{10}^{-}\rho_{\psi_{10}^{-}}+a_{22}\rho_{\psi_{22}}$
$\displaystyle=a_{00}\ket{\psi_{00}}\bra{\psi_{00}}+a_{12}^{+}\ket{\psi_{12}^{+}}\bra{\psi_{12}^{+}}+a_{12}^{-}\ket{\psi_{12}^{-}}\bra{\psi_{12}^{-}}$
$\displaystyle+a_{11}\ket{\psi_{11}}\bra{\psi_{11}}+a_{20}^{+}\ket{\psi_{20}^{+}}\bra{\psi_{20}^{+}}+a_{20}^{-}\ket{\psi_{20}^{-}}\bra{\psi_{20}^{-}}$
$\displaystyle+a_{10}^{+}\ket{\psi_{10}^{+}}\bra{\psi_{10}^{+}}+a_{10}^{-}\ket{\psi_{10}^{-}}\bra{\psi_{10}^{-}}+a_{22}\ket{\psi_{22}}\bra{\psi_{22}}$
(39)

In absence of the correlation matrix for the qutrit system, aforesaid
procedure of evaluating constants by simple comparison does not work. We
therefore adopt a procedure where the diagonalization of that correlation
matrix must give all two-qutrit states as its eigen states [40]. This
requirement, supplemented by rigorous algebraic manipulation of a set of
coupled linear equations gives, $a_{00}=2\sqrt{2}$, $a_{12}^{+}=0$,
$a_{12}^{-}=-2\sqrt{2}$, $a_{11}=\sqrt{2}$, $a_{20}^{+}=0$,
$a_{20}^{-}=-2\sqrt{2}$, $a_{10}^{+}=0$, $a_{10}^{-}=2\sqrt{2}$,
$a_{22}=-\sqrt{2}$. Plugging them back in Eq.(Note added in proof:), we obtain
the qutrit correlation matrix given by Eq.(5), which can be also written as
the tensor product of the GellMann matrices,

$\displaystyle
a_{00}\rho_{\psi_{00}}+a_{12}^{+}\rho_{\psi_{12}^{+}}+a_{12}^{-}\rho_{\psi_{12}^{-}}+a_{11}\rho_{\psi_{11}}$
$\displaystyle+a_{20}^{+}\rho_{\psi_{20}^{+}}+a_{20}^{-}\rho_{\psi_{20}^{-}}+a_{10}^{+}\rho_{\psi_{10}^{+}}+a_{10}^{-}\rho_{\psi_{10}^{-}}+a_{22}\rho_{\psi_{22}}$
(40)
$\displaystyle=\sqrt{2}\big{(}\ket{\psi_{11}}\bra{\psi_{11}}-\ket{\psi_{22}}\bra{\psi_{22}}\big{)}$
$\displaystyle+2\sqrt{2}\big{(}\ket{\psi_{00}}\bra{\psi_{00}}+\ket{\psi_{10}^{-}}\bra{\psi_{10}^{-}}-\ket{\psi_{12}^{-}}\bra{\psi_{12}^{-}}-\ket{\psi_{20}^{-}}\bra{\psi_{20}^{-}}\big{)}$
$\displaystyle=\sqrt{2}(\lambda_{4}\otimes\lambda_{4}-\lambda_{2}\otimes\lambda_{2})+\frac{1}{\sqrt{2}}(\lambda_{6}\otimes\lambda_{6}+\lambda_{7}\otimes\lambda_{7})$
$\displaystyle-\frac{1}{2\sqrt{2}}(\lambda_{3}\otimes\lambda_{3}-5\lambda_{8}\otimes\lambda_{8})+\frac{1}{\sqrt{6}}(\lambda_{8}\otimes\lambda_{0}+\lambda_{0}\otimes\lambda_{8})$
$\displaystyle-\frac{1}{2\sqrt{6}}(\lambda_{3}\otimes\lambda_{8}+\lambda_{8}\otimes\lambda_{3})-\frac{1}{2\sqrt{3}}(\lambda_{0}\otimes\lambda_{3}+\lambda_{3}\otimes\lambda_{0}),$
$\displaystyle=\hat{\mathcal{C}}^{3\otimes 3}\qquad\qquad Q.E.D.$ (28)

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