# Entanglement Measure Based on Optimal Entanglement Witness

Nan Yang1 1College of Science, China University of Petroleum, 266580 Qingdao,
P.R. China  Jiaji Wu1 1College of Science, China University of Petroleum,
266580 Qingdao, P.R. China  Xianyun Dong1 1College of Science, China
University of Petroleum, 266580 Qingdao, P.R. China  Longyu Xiao1 1College of
Science, China University of Petroleum, 266580 Qingdao, P.R. China  Jing Wang1
<EMAIL_ADDRESS>1College of Science, China University of Petroleum,
266580 Qingdao, P.R. China  Ming Li1 1College of Science, China University of
Petroleum, 266580 Qingdao, P.R. China

###### Abstract

We introduce a new entanglement measure based on optimal entanglement witness.
First of all, we show that the entanglement measure satisfies some necessary
properties, including zero entanglements for all separable states, convexity,
continuity, invariance under local unitary operations and non-increase under
local operations and classical communication(LOCC). More than that, we give a
specific mathematical expression for the lower bound of this entanglement
measure for any bipartite mixed states. We further improve the lower bound for
2$\otimes$2 systems. Finally, we numerically simulate the lower bound of
several types of specific quantum states.

##  1 Introduction

Entanglement is one of the most remarkable features of quantum mechanics,
which has been recognized as an essential resource in quantum information
theory [1, 2]. In recent years, quantum entanglement is increasingly the focus
of people’s attention and is widely applied to quantum information processing
tasks [3].

Since entanglement is a newly discovered state of resources, it’s of vital
importance to discover the mathematical structure behind its theory.
Entanglement measure is the quantization of entanglement and various
entanglement measures have been well defined. Arindam Lala [4] has studied
various entanglement measures associated with certain non-conformal field
theories. And Jacob L. Beckey et.al. [5] have proposed entanglement measures
that are computable and operationally meaningful multipartite.

Entanglement witness(EW) gives a sufficient condition for detecting
entanglement, which is equivalent to making a hyperplane in the quantum state
space to separate some separable states from entangled states. In Ref. [6],
the authors have studied the optimal EW(OEW) based on local orthogonal
observables. And the optimal entanglement witness constructed by Ref. [7]
could effectively detect the entangled states produced by cooper pair
splitters.

In this paper, we introduce a new entanglement measure of bipartite quantum
states, which is closely related to optimal entanglement witness. In Sec.III,
we prove explicitly that this new measure satisfies many properties of
bipartite entanglement measures. In Sec.IV, we obtain the lower bound of this
measure for any bipartite mixed states. Furthermore, considering the property
that there is a double cover relationship between SU(2) group and SO(3) group,
we give a better lower bound on $2\otimes 2$ space. Finally, we present
numerical simulations of the lower bound for some specific forms of quantum
states.

##  2 Construction of Entanglement Measure

In Ref. [8], we know that EW could be the operator in the form of
$\alpha$$I+L$. Let $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ be arbitrary
finite-dimensional Hilbert spaces. If an EW $W_{E}$ acting on
$\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ can be written in the following form:

$\displaystyle W_{E}=\alpha(L)I-L,$ (1)

where $I$ is the identity operator, $L$ is a self-adjoint operator, and
$\alpha(L)$ is the maximum of expectation values taken over all separable pure
states $|\varphi\rangle$:

$\displaystyle\alpha=\max\limits_{|\varphi\rangle\in
sep}\langle\varphi|L|\varphi\rangle,$ (2)

then this EW is called an OEW. We use the results of this OEW to quantify
entanglement.

######  Definition 1.

Let $M_{a}$ be a positive semi-definite set

$M_{a}=\\{H\geq 0\ |\ H\in\mathcal{H}_{A}\otimes\mathcal{H}_{B},tr(H)=a,a\in
R^{+}\\}.$ (3)

The entanglement measure $W_{E}$ of a bipartite state
$\rho\in(\mathcal{H}_{A}\otimes\mathcal{H}_{B})$ is given by

$\displaystyle C_{w}(\rho)=\max\limits_{L\in M_{a}}\\{-tr(W_{E}\rho)\\}.$ (4)

We define $||M||_{KF}=Tr[\sqrt{MM^{\dagger}}]$,
$||M||_{HS}=\sqrt{Tr[MM^{\dagger}]}$ and $||M||_{2}=\sigma_{max}(M)$. They
respectively represent the trace norm, Frobenius norm and 2-norm of the matrix
$M$. Meanwhile, we could also get the following result according to the
relevant knowledge of EW.

$\displaystyle tr(W_{E}\rho)<0,$ (5) $\displaystyle tr(W_{E}\sigma)\geq 0,$
(6)

where $\rho$ is an arbitrary entangled state acting on
$\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ and $\sigma$ is an arbitrary separable
state acting on $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$. It’s not difficult to
see that $W_{E}$ is one type of OEW, and we could get the above inequalities
from the Ref. [9].

##  3 Properties of Entanglement Measure

For $\forall\rho\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$, an entanglement
measure is a functional $C$ defined on the set of density operators on the
Hilbert space. A well-defined measure of entanglement $C$ should satisfy the
following requirements:

(E1) $C(\rho)=0$, for any $\rho\in Sep$(zero for all separable states).

(E2) $C(\lambda\rho_{1}+(1-\lambda)\rho_{2})$$\leq\lambda
C(\rho_{1})+(1-\lambda)C(\rho_{2})$, for all $\rho_{1}$, $\rho_{2}$ and
$0\leq\lambda\leq 1$ (convexity).

(E3) Keeping invariant under local unitary transformations, i.e.,

$C((U_{1}\otimes U_{2})^{\dagger}\rho(U_{1}\otimes U_{2}))=C(\rho)$

holds for all unitary operators $U_{1}$ and $U_{2}$.

(E4) $C(\Lambda_{LOCC}(\rho))$$\leq C(\rho)$(nonincreasing under local
operations and classical communication (LOCC)).

(E5) When $\|\rho-\sigma\|\rightarrow 0$, $C(\rho)-C(\sigma)\rightarrow
0$(continuity).

Subsequently, we will prove that our entanglement measure defined in Eq.(4)
fulfills these requirements.

######  Propositon 1.

$C_{w}(\rho)$ satisfies property (E2).

######  Proof.

Let $\rho=\lambda\rho_{1}+(1-\lambda)\rho_{2}$, where $0\leq\lambda\leq 1$. By
substituting it into $C_{w}(\rho)$ we have

$\displaystyle C_{w}(\rho)$ $\displaystyle=$ $\displaystyle
C_{w}(\lambda\rho_{1}+(1-\lambda)\rho_{2})$ (7) $\displaystyle=$
$\displaystyle\max\limits_{L}\\{-\alpha(L)+\lambda
tr(L\rho_{1})+(1-\lambda)tr(L\rho_{2})\\}$ $\displaystyle=$
$\displaystyle\max\limits_{L}\\{\lambda(-\alpha(L)+tr(L\rho_{1}))+(1-\lambda)(-\alpha(L)+tr(L\rho_{2}))\\}$
$\displaystyle\leq$ $\displaystyle\lambda
C_{w}(\rho_{1})+(1-\lambda)C_{w}(\rho_{2}).$

∎

######  Propositon 2.

$C_{w}(\rho)$ satisfies property (E1).

###### Proof.

Assume that $\rho$ is a separable and pure state acting on the space of
$\mathcal{H}_{A}\otimes\mathcal{H}_{B}$, i.e. $\rho$ could be written as
$\rho=\rho_{1}\otimes\rho_{2}=|\varphi_{1}\rangle\langle\varphi_{1}|\otimes|\varphi_{2}\rangle\langle\varphi_{2}|$,
where $|\varphi_{1}\rangle\in\mathcal{H}_{A}$,
$|\varphi_{2}\rangle\in\mathcal{H}_{B}$. Then we have

$\displaystyle tr(W_{E}\rho)$ $\displaystyle=$ $\displaystyle
tr(\alpha\rho-L\rho)$ (8) $\displaystyle=$ $\displaystyle
tr(\alpha)-tr(L\rho_{1}\otimes\rho_{2})$ $\displaystyle=$ $\displaystyle
tr(\max\langle\varphi^{A}|\otimes\langle\varphi^{B}|L|\varphi^{A}\rangle\otimes|\varphi^{B}\rangle)-tr(L\rho_{1}\otimes\rho_{2})$
$\displaystyle=$
$\displaystyle\max\limits_{\rho^{A*},\rho^{B*}}tr(L\rho^{A*}\otimes\rho^{B*})-tr(L\rho_{1}\otimes\rho_{2}),$

where $\rho^{A*}=|\varphi^{A*}\rangle\langle\varphi^{A*}|$ and
$\rho^{B*}=|\varphi^{B*}\rangle\langle\varphi^{B*}|$. According to the
definition and Eq.(6) we have $-tr(W_{E}\rho)\leq 0$. Let $L=\frac{a}{n}I$,
where $n$ is the dimension of $L$, we have $W_{E}=0$. Thus the entanglement
degree for the separable pure states is 0, which illustrate that

$\displaystyle C_{w}(\rho)=\max\limits_{L}\\{-tr(W_{E}\rho)\\}=0.$ (9)

The above-mentioned fact is discussed when $\rho$ is a separable pure state
acting on the space of $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$. If $\rho$ is a
separable mixed state, we have a decomposition of $\rho$:

$\displaystyle\rho=\sum_{i}p_{i}|\varphi_{1i}\rangle\langle\varphi_{1i}|\otimes|\varphi_{2i}\rangle\langle\varphi_{2i}|.$
(10)

According to property (E2), we know $C_{w}(\rho)$ satisfies

$\displaystyle
C_{w}(\rho)\leq\sum_{i}p_{i}C(|\varphi_{1i}\rangle\langle\varphi_{i}|\otimes|\varphi_{2i}\rangle\langle\varphi_{2i}|)=0.$
(11)

Note that when $L=\frac{a}{n}I$, we have $C_{w}(\rho)=0$. Thus for
$\forall\rho$ that is separable, $C_{w}(\rho)=0$. ∎

######  Propositon 3.

$C_{w}(\rho)$ has local unitary transformation invariance, satisfying
property(E3).

###### Proof.

For $U^{\dagger}\rho U$, we write the measure as:

$\displaystyle C_{w}(U^{\dagger}\rho U)$ $\displaystyle=$
$\displaystyle\max\limits_{L}\\{-tr(W_{E}U^{\dagger}\rho U)\\}$ (12)
$\displaystyle=$
$\displaystyle\max\limits_{L}\\{-\alpha(L)+tr(LU^{\dagger}\rho U)\\}$
$\displaystyle=$
$\displaystyle\max\limits_{L}\\{-\alpha(L)+tr(ULU^{\dagger}\rho)\\}.$

Since $L$ is positive semi-definite, we have
$\langle\varphi|L|\varphi\rangle\geq 0$ and
$\langle\varphi|{L}^{\prime}|\varphi\rangle=(\langle\varphi|U)L(U^{\dagger}|\varphi\rangle)\geq
0$ for any $|\varphi\rangle$, so ${L}^{\prime}$ is also positive semi-
definite. Furthermore, the unitary transformation does not change the
eigenvalues, so $tr({L}^{\prime})=tr(L)=a$. As $U$ is a local unitary matrix,
$|\varphi^{*}\rangle=U^{\dagger}|\varphi\rangle$ is still a separable pure
state. In addition, when $|\varphi\rangle$ takes all separable states,
$|\varphi^{*}\rangle$ could also take all the separable states owing to the
fact that $U^{\dagger}$ is reversible, then

$\displaystyle\alpha(ULU^{\dagger})$ $\displaystyle=$
$\displaystyle\max\limits_{|\varphi\rangle\in
sep}\langle\varphi|ULU^{\dagger}|\varphi\rangle$ (13) $\displaystyle=$
$\displaystyle\max\limits_{|\varphi\rangle\in
sep}(\langle\varphi|U)L(U^{\dagger}|\varphi\rangle)$ $\displaystyle=$
$\displaystyle\max\limits_{|\varphi\rangle^{*}\in
sep}\langle\varphi^{*}|L|\varphi^{*}\rangle$ $\displaystyle=$
$\displaystyle\max\limits_{|\varphi\rangle\in
sep}\langle\varphi|L|\varphi\rangle$ $\displaystyle=$
$\displaystyle\alpha(L).$

When $L$ takes all Hermitian matrices, $ULU^{\dagger}$ can also take all
Hermitian matrices. For $\forall
L^{\prime}\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$, we have
$L=U^{\dagger}L^{\prime}U\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ which
satisfies $L^{\prime}=ULU^{\dagger}$. Therefore

$\displaystyle C_{w}(U^{\dagger}\rho U)$ $\displaystyle=$
$\displaystyle\max\limits_{L}\\{-\alpha(L)+tr(ULU^{\dagger}\rho)\\}$ (14)
$\displaystyle=$
$\displaystyle\max\limits_{U^{\dagger}LU}\\{-\alpha(ULU^{\dagger})+tr(ULU^{\dagger}\rho)\\}$
$\displaystyle=$
$\displaystyle\max\limits_{L^{\prime}}\\{-\alpha(L^{\prime})+tr(L^{\prime}\rho)\\}$
$\displaystyle=$ $\displaystyle C_{w}(\rho).$

∎

######  Propositon 4.

$C_{w}(\rho)$ is entangled monotone with respect to separable operation
$\varepsilon(\rho)$, where

$\displaystyle\varepsilon(\rho)=\sum_{i}(E_{i}\otimes
F_{i})\rho(E_{i}\dagger\otimes F_{i}\dagger),$ (15)

and operator completeness is satisfied, which means
$\sum_{i}E_{i}E_{i}\dagger\otimes
F_{i}F_{i}\dagger=I_{\mathcal{H}_{A}\otimes\mathcal{H}_{B}}$.

###### Proof.

We have

$\displaystyle C_{w}(\varepsilon(\rho))$ $\displaystyle=$
$\displaystyle\max\limits_{L}\\{-\alpha(L)+tr(L\varepsilon(\rho))\\}$ (16)
$\displaystyle=$
$\displaystyle\max\limits_{L}\\{-\alpha(L)+tr(L\sum_{i}(E_{i}\otimes
F_{i})\rho(E_{i}\dagger\otimes F_{i}\dagger))\\}$ $\displaystyle=$
$\displaystyle\max\limits_{L}\\{-\alpha(L)+tr(\sum_{i}(E_{i}\dagger\otimes
F_{i}\dagger)L(E_{i}\otimes F_{i})\rho)\\}.$

Firstly, we prove $\sum_{i}(E_{i}\dagger\otimes F_{i}\dagger)L(E_{i}\otimes
F_{i}))$ conforms to the definition of $L$, i.e. it satisfies the properties
of $L$ in the definition

$\displaystyle tr(\sum_{i}(E_{i}\dagger\otimes F_{i}\dagger)L(E_{i}\otimes
F_{i}))=tr(\sum_{i}(E_{i}E_{i}\dagger\otimes F_{i}F_{i}\dagger)L)=tr(L)=a>0.$
(17)

$\sum_{i}(E_{i}\dagger\otimes F_{i}\dagger)L(E_{i}\otimes F_{i}))$ is
obviously a self-conjugate positive semi-definite matrix. So
$\sum_{i}(E_{i}\dagger\otimes F_{i}\dagger)L(E_{i}\otimes F_{i}))$ meets the
definition of $L$. Next we consider the relationship between $\alpha(L)$ and
$\alpha(\sum_{i}(E_{i}\dagger\otimes F_{i}\dagger)L(E_{i}\otimes F_{i}))$:

$\displaystyle\alpha(\sum_{i}(E_{i}\dagger\otimes F_{i}\dagger)L(E_{i}\otimes
F_{i}))$ $\displaystyle=$ $\displaystyle\max\limits_{|\varphi\rangle\in
sep}\langle\varphi|\sum_{i}(E_{i}\dagger\otimes F_{i}\dagger)L(E_{i}\otimes
F_{i})|\varphi\rangle$ (18) $\displaystyle=$
$\displaystyle\max\limits_{|\varphi\rangle\in
sep}\sum_{i}\langle\varphi|(E_{i}\dagger\otimes F_{i}\dagger)L(E_{i}\otimes
F_{i})|\varphi\rangle.$

Let $|\varphi^{*}_{i}\rangle=(E_{i}\otimes
F_{i})|\varphi\rangle/\sqrt{p_{i}}$, where
$p_{i}=tr(\langle\varphi|(E_{i}\dagger\otimes F_{i}\dagger)(E_{i}\otimes
F_{i})|\varphi\rangle)$, and $\sum_{i}p_{i}=1$. We have

$\displaystyle\alpha(\sum_{i}(E_{i}\dagger\otimes F_{i}\dagger)L(E_{i}\otimes
F_{i}))$ $\displaystyle=$ $\displaystyle\max\limits_{|\varphi\rangle\in
sep}\sum_{i}p_{i}\langle\varphi^{*}_{i}|L|\varphi^{*}_{i}\rangle$ (19)
$\displaystyle=$ $\displaystyle\sum_{i}p_{i}\max\limits_{|\varphi\rangle\in
sep}\langle\varphi^{*}_{i}|L|\varphi^{*}_{i}\rangle$ $\displaystyle\leq$
$\displaystyle\sum_{i}p_{i}\max\limits_{|\varphi*_{i}\rangle\in
sep}\langle\varphi^{*}_{i}|L|\varphi^{*}_{i}\rangle$ $\displaystyle\leq$
$\displaystyle\alpha(L).$

We may clearly arrive at the conclusion that

$\displaystyle C_{w}(\varepsilon(\rho))$ $\displaystyle=$
$\displaystyle-\alpha(L)+tr(L\varepsilon(\rho))$ (20) $\displaystyle=$
$\displaystyle-\alpha(L)+tr(\sum_{i}(E_{i}\dagger\otimes
F_{i}\dagger)L(E_{i}\otimes F_{i})\rho)$ $\displaystyle\leq$
$\displaystyle-\alpha(\sum_{i}(E_{i}\dagger\otimes F_{i}\dagger)L(E_{i}\otimes
F_{i}))+tr(\sum_{i}(E_{i}\dagger\otimes F_{i}\dagger)L(E_{i}\otimes
F_{i})\rho)$ $\displaystyle=$
$\displaystyle-\alpha(L^{\prime})+tr(L^{\prime}\rho)$ $\displaystyle=$
$\displaystyle C_{w}(\rho).$

∎

######  Corollary 1.

$C_{w}(\rho)$ is entangled monotone with respect to LOCC (E4).

###### Proof.

LOCC belongs to separable operation, and we have proved $C_{w}(\rho)$ is
entangled monotone with respect to separable operation. ∎

######  Propositon 5.

$C_{w}(\rho)$ is continuous for $\rho$, satisfying the property (E5).

###### Proof.

Set $\rho^{\prime}=\rho+v_{n}$ and $\rho^{\prime}=\rho-v_{n}$, we could have

$\displaystyle|C_{w}(\rho)-C_{w}(\rho^{\prime})|\leq\max\limits_{L}|tr(Lv_{n})|,$
(21)

where $v_{n}$ is a Hermitian matrix. The spectral decompose of $v_{n}$ is
$\sum_{i}\lambda_{i}|\varphi_{i}\rangle\langle\varphi_{i}|$, and
$\sum_{i}\lambda_{i}=0$. Thus

$\displaystyle
tr(Lv_{n})=\sum_{i}\lambda_{i}\langle\varphi_{i}|L|\varphi_{i}\rangle,$ (22)

where
$\langle\varphi_{i}|L|\varphi_{i}\rangle=\langle\varphi_{i}^{*}|\Lambda|\varphi_{i}^{*}\rangle\leq\lambda_{max}(L)$,
$L=U^{\dagger}\Lambda U$, $U$ is a unitary matrix, $\Lambda$ is a real
diagonal matrix with the diagonal elements being the eigenvalues of $L$,
$|\varphi^{*}\rangle=U|\varphi\rangle$. According to the properties of $L$,
$|tr(Lv_{n})|$ satisfies

$\displaystyle|tr(Lv_{n})|\leq
a\sum_{i}|\lambda_{i}|=atr(\sqrt{v_{n}^{\dagger}v_{n}})=a||v_{n}||_{KF}.$ (23)

When $\rho^{\prime}\rightarrow\rho$, $v_{n}\rightarrow 0$, i.e.
$||v_{n}||_{KF}\rightarrow 0$, we have

$\displaystyle
0\leq|C_{w}(\rho)-C_{w}(\rho^{\prime})|\leq\max\limits_{L}|tr(Lv_{n})|\leq
a||v_{n}||_{KF}\rightarrow 0,$ (24)

which proves that $C_{w}(\rho)$ is continuous. ∎

##  4 Lower Bound

###  4.1 Theoretical Derivation

We present a lower bound using the idea of finding a specific $L$ to simplify
the calculation. Without loss of generality, we discuss the situation of $L\in
M_{1}$. Then we can set $L^{{}^{\prime}}=aL$ to satisfy $L^{{}^{\prime}}\in
M_{a}$. In the following, we discuss the lower bound in two cases: $\rho$ is a
pure state and $\rho$ is a mixed state.

####  4.1.1 Pure State

We will divide the lower bound of $C_{w}(\rho)$ into two steps: calculating
$\alpha(L)$ and $tr(L\rho)$. If we are capable to obtain the estimate of
$\alpha(L)$ in advance, the original problem will be simplified. As a matter
of fact, J.Sperling and W.Vogel[9] have gotten the estimate of $\alpha(L)$. We
also demonstrate this result by using Schmidt decomposition rather than the
Lagrange multiplier.

######  Theorem 1.

When $rank(L)=1$, we could regard $L$ as a pure state and we have

$\displaystyle\alpha(L)=\max\limits_{|\varphi\rangle\in
sep}\langle\varphi|L|\varphi\rangle=\mu_{1},$ (25)

where $\sqrt{\mu_{1}}$ is the maximum Schmidt coefficient of $L$.

###### Proof.

For
$\forall|\varphi_{pure}\rangle\in\mathcal{H}^{(m)}_{A}\otimes\mathcal{H}^{(n)}_{B}(m\leq
n)$, we have

$\displaystyle|\varphi_{pure}\rangle=\sum_{i}\sqrt{\mu_{i}}|i_{A}i_{B}\rangle,$
(26)

where $\sqrt{\mu_{i}}$ ($i=1,...m$) are the Schmidt coefficients, and
$\mu_{1}\geq\mu_{2}\geq...\geq\mu_{m}\geq 0$, $\sum_{i}\mu_{i}=1$.
$|i_{A}\rangle$ and $|i_{B}\rangle$ are the orthonormal basis in
$\mathcal{H}^{(m)}_{A}$ and $\mathcal{H}^{(n)}_{B}$ respectively. Then
$L=|\varphi_{pure}\rangle\langle\varphi_{pure}|$ could be written as
$\sum_{i,j}\sqrt{\mu_{i}\mu_{j}}|i_{A}i_{B}\rangle\langle j_{A}j_{B}|$.

Similarly, we could write separable pure state $|\varphi_{sep}\rangle$ in form
of $\sum_{i,j}a_{i}b_{j}|i_{A}j_{B}\rangle$. By using these representations,
we could conclude that

$\displaystyle\langle\varphi_{sep}|L|\varphi_{sep}\rangle$ $\displaystyle=$
$\displaystyle\langle\varphi_{A}|\otimes\langle\varphi_{B}|L|\varphi_{A}\rangle\otimes|\varphi_{B}\rangle$
(27) $\displaystyle=$
$\displaystyle\sum_{i,j}\sqrt{\mu_{i}\mu_{j}}a_{i}b_{i}a_{j}^{*}b_{j}^{*}$
$\displaystyle\leq$ $\displaystyle(\sum_{i}\sqrt{\mu_{i}}|a_{i}||b_{i}|)^{2}$
$\displaystyle=$
$\displaystyle(\sum_{i}\sqrt[4]{\mu_{i}}|a_{i}|\sqrt[4]{\mu_{i}}|b_{i}|)^{2}$
$\displaystyle\leq$
$\displaystyle(\sum_{i}\sqrt{\mu_{i}}|a_{i}|^{2})(\sum_{j}\sqrt{\mu_{j}}|b_{j}|^{2}).$

The first inequality holds by applying the absolute value inequality and the
second one holds due to the Cauchy-Schwarz inequality. Note that
$\sum_{i}|a_{i}|^{2}=1$ and $\sum_{j}|b_{j}|^{2}=1$. We can obtain
$\langle\varphi_{sep}|L|\varphi_{sep}\rangle\leq\mu_{1}$ and the equal sign
works if and only if it meets the condition that $|a_{1}|=1$, $|b_{1}|=1$,
$a_{i}=0$, $b_{i}=0$ $(i>1)$. Ultimately, we have $\alpha(L)=\mu_{1}$. ∎

After getting the estimate of $\alpha(L)$, we may substitute it into
$C_{w}(\rho)$. Thus the original problem will be transformed into estimating
of $tr(L\rho)$. In the pure states, we obtain the following result.

######  Theorem 2.

Let $\rho$ be a pure quantum state acting on the space of
$\mathcal{H}_{A}^{(d_{1})}\otimes\mathcal{H}_{B}^{(d_{2})}$. We have

$\displaystyle C_{w}(\rho)\geq\frac{\|\rho^{T_{A}}\|_{KF}-1}{d},$ (28)

where $d=min\\{d_{1},d_{2}\\}$.

###### Proof.

It is easy to deduce that
$\rho=\sum_{i,j}\sqrt{\mu_{i}\mu_{j}}|i_{A}i_{B}\rangle\langle j_{A}j_{B}|$.
Then we take a special $L^{\prime}$, which has the standard Schmidt form
$L^{\prime}=\sum_{i,j}\sqrt{m_{i}m_{j}}|i_{A}i_{B}\rangle\langle j_{A}j_{B}|$,
where $m_{1}\geq m_{2}\geq...\geq m_{d}\geq 0$ and $\sum_{i}m_{i}=1$. We can
notice that $L$ is a pure state. Then

$\displaystyle C_{w}(\rho)$ $\displaystyle\geq
tr(L^{\prime}\rho)-\alpha(L^{\prime})=(\sum_{i}\sqrt{m_{i}\mu_{i}})^{2}-m_{1}.$
(29)

By designating $m_{i}=\frac{1}{d}$( it meets the above restraint conditions
obviously), we can obtain

$\displaystyle
C_{w}(\rho)\geq\frac{(\sum_{i}\sqrt{\mu_{i}})^{2}-1}{d}=\frac{\|\rho^{T_{A}}\|_{KF}-1}{d}.$
(30)

From the Ref. [10], we know that $(\sum_{i}\sqrt{\mu_{i}})^{2}$ is equal to
$\|R(\rho)\|_{KF}$ and $\|\rho^{T_{A}}\|_{KF}$ for a pure state $\rho$, where
$R(\rho)$ is the realigned matrix of $\rho$. Here we write
$(\sum_{i}\sqrt{\mu_{i}})^{2}$ as $\|\rho^{T_{A}}\|_{KF}$. Hence, the
inequality in the previous equation holds. ∎

In this way and by means of Schmidt decomposition we succeeded in obtaining
the lower bound of $C_{w}(\rho)$ for pure states. For mixed states, we still
attain the expression of $\alpha(L)$ firstly.

####  4.1.2 Mixed State

To solve the problem for the mixed state, we introduce a new form of density
operator. An arbitrary state $\rho$ acting on
$\mathcal{H}_{A}^{(d_{1})}\otimes\mathcal{H}_{B}^{(d_{2})}$ could be Bloch
represented as follows

$\displaystyle\rho=\frac{1}{d_{1}d_{2}}(I\otimes
I+\sum_{i}{s_{i}\lambda_{i}^{A}\otimes I}+\sum_{j}{I\otimes
t_{j}\lambda_{j}^{B}+\sum_{i,j}{r_{ij}\lambda_{i}^{A}\otimes\lambda_{j}^{B}}}),$
(31)

where $I$ stands for identity operator, and
$\\{\lambda_{i}^{A}\\}_{i=1}^{d_{1}^{2}-1}$,
$\\{\lambda_{i}^{B}\\}_{i=1}^{d_{2}^{2}-1}$ are generalized Gell-Mann
matrices, which satisfy $tr(\lambda_{i}\lambda_{j})=2\delta_{ij}$,
$\lambda_{i}^{\dagger}=\lambda_{i}$. It’s clear that $s_{i}$, $t_{i}$ and
$r_{ij}$ are all real. Hence, the value of $\alpha(L)$ could be estimated
again when $L$ is a mixed state.

######  Theorem 3.

Let $s_{i}=0$, $t_{i}=0$, $L$ is written as Bloch representation as follows

$\displaystyle L=\frac{1}{d_{1}d_{2}}(I\otimes
I+\sum_{i,j}{r_{ij}\lambda_{i}^{A}\otimes\lambda_{j}^{B}}).$ (32)

Then the estimate of $\alpha(L)$ is

$\displaystyle\alpha(L)=\frac{1}{d_{1}d_{2}}+\frac{4}{d_{1}^{2}d_{2}^{2}}\|R_{L}\|_{2},$
(33)

where $R_{L}=(r_{ij})$.

###### Proof.

For Hermitian matrix $L$, $L$ can be written as

$\displaystyle L=\frac{1}{d_{1}d_{2}}(I\otimes
I+\sum_{i}{s_{i}\lambda_{i}^{A}\otimes I}+\sum_{j}{I\otimes
t_{j}\lambda_{j}^{B}+\sum_{i.j}{r_{ij}\lambda_{i}^{A}\otimes\lambda_{j}^{B}}}).$
(34)

For separable state $\rho$, it can be written as same as $L$:

$\displaystyle\rho=\frac{1}{d_{1}d_{2}}(I\otimes
I+\sum_{i}{s_{i}^{{}^{\prime}}\lambda_{i}^{A}\otimes I}+\sum_{j}{I\otimes
t_{j}^{{}^{\prime}}\lambda_{j}^{B}+\sum_{i,j}{r_{ij}^{{}^{\prime}}\lambda_{i}^{A}\otimes\lambda_{j}^{B}}}).$
(35)

Assume $S$, $S_{1}$, $T$, $T_{1}$ are vectors constituted by $s_{i}$,
$s_{i}^{\prime}$, $t_{j}$, $t_{j}^{\prime}$ respectively, $R_{L}$ and $R_{1}$
are the correlation matrices of $L$ and $\rho$. We could prove
$R_{1}=S_{1}T_{1}^{T}$ as $\rho$ is a separable state, then

$\displaystyle\alpha(L)=\frac{1}{d_{1}^{2}d_{2}^{2}}(d_{1}d_{2}+\max\limits_{\|S_{1}\|_{2}=1,\|T_{1}\|_{2}=1}(2d_{2}S_{1}^{T}S+2d_{1}T_{1}^{T}T+4S_{1}^{T}R_{L}T_{1})).$
(36)

Let $L=\frac{1}{d_{1}d_{2}}(I\otimes
I+\sum{r_{ij}\lambda_{i}^{A}\otimes\lambda_{j}^{B}})$. We may simplify the
above formula as

$\displaystyle\alpha(L)=\frac{1}{d_{1}d_{2}}+\max\limits_{\|S_{1}\|_{2}=1,\|T_{1}\|_{2}=1}\frac{4}{d_{1}^{2}d_{2}^{2}}S_{1}^{T}R_{L}T_{1}.$
(37)

Assume that the singular value decomposition(SVD) form of $R_{L}$ is $U\Lambda
V^{\dagger}$, $\Lambda=diag(\sigma_{i})(\sigma_{1}\geq\sigma_{2}\geq...)$. Set
$S_{1}^{{}^{\prime}T}=S_{1}^{T}U$, $T_{1}^{{}^{\prime}}=V^{\dagger}T_{1}$,
$S_{1}^{T}R_{L}T_{1}$ can be written as $S_{1}^{{}^{\prime}T}\Lambda
T_{1}^{{}^{\prime}}$. We have the following inequality

$\displaystyle S_{1}^{{}^{\prime}T}\Lambda T_{1}^{{}^{\prime}}$
$\displaystyle=\sum
s_{i}t_{j}\sigma_{i}\leq\sqrt{(\sum|s_{i}|^{2}\sigma_{i})(\sum|t_{i}|^{2}\sigma_{i})}\leq\sigma_{1}.$
(38)

Obviously, the upper bound in Eq(38) is saturated. From Eq(37) and Eq(38), we
could conclude that

$\displaystyle\alpha(L)$ $\displaystyle=$
$\displaystyle\frac{1}{d_{1}d_{2}}+\frac{4}{d_{1}^{2}d_{2}^{2}}\sigma_{max}(R_{L})$
(39) $\displaystyle=$
$\displaystyle\frac{1}{d_{1}d_{2}}+\frac{4}{d_{1}^{2}d_{2}^{2}}\|R_{L}\|_{2}.$
(40)

∎

After getting the estimate of $\alpha(L)$ when $L$ is a special mixed state,
we may firstly discuss the lower bound of $C_{w}(\rho)$ when $\rho$ acts on
the $\mathcal{H}^{d_{1}}\otimes\mathcal{H}^{d_{2}}$ space. Then we improve it
on the $\mathcal{H}^{2}\otimes\mathcal{H}^{2}$ space.

######  Theorem 4.

Considering the special $L$ which can be written as the following Bloch
representation

$\displaystyle L=\frac{1}{d_{1}d_{2}}(I\otimes
I+\sum{r_{ij}\lambda_{i}\otimes\lambda_{j}}).$ (41)

Assume $R_{L}=(r_{ij})$, when $R_{\rho}\neq 0$, we have

$\displaystyle
C_{w}(\rho)\geq\frac{2(\|R_{\rho}\|_{KF}-1)}{d_{1}^{2}d_{2}^{2}\sqrt{rank(R_{\rho})}}.$
(42)

When $R_{\rho}=0$, we have $C_{w}(\rho)\geq 0$.

###### Proof.

From Theorem 3, the above $L$ could derive
$\alpha(L)=\frac{1}{d_{1}d_{2}}+\max\limits_{\|S_{1}\|_{2}=1,\|T_{1}\|_{2}=1}\frac{4}{d_{1}^{2}d_{2}^{2}}S_{1}^{T}R_{L}T_{1}$.
Then

$\displaystyle C_{w}(\rho)\geq
tr(L\rho)-\alpha(L)=\frac{4}{d_{1}^{2}d_{2}^{2}}(tr(R_{L}^{T}R_{\rho})-\|R_{L}\|_{2}).$
(43)

Let $R_{\rho}=U\Sigma V^{T}$ be the singular value decomposition of
$R_{\rho}$, where $U$ and $V$ are orthogonal matrices. Assume
$\Sigma^{\prime}$ is a diagonal matrix which satisfies $tr(\Sigma^{\prime
T}\Sigma)=tr(\Sigma)=\|R_{\rho}\|_{KF}$, with the diagonal element to be
$sgn(\sigma_{i})$, where $\sigma_{i}$ is the element of $\Sigma$ in
corresponding position. Assume $R_{L}=\frac{1}{c}U\Sigma^{\prime}V^{T}$, then
we have

$\displaystyle
C_{w}(\rho)\geq\frac{4(\|R_{\rho}\|_{KF}-1)}{d_{1}^{2}d_{2}^{2}c}.$ (44)

Next we will prove that when $c=2\sqrt{rank(R_{\rho})}$, $L\geq 0$. When
$c=2\sqrt{rank(R_{\rho})}$, and $rank(R_{\rho})\neq 0$, we have

$\displaystyle\|\sum{r_{ij}\lambda_{i}\otimes\lambda_{j}}\|_{HS}$
$\displaystyle=$
$\displaystyle\sqrt{tr(\sum{r_{ij}r_{sk}\lambda_{i}\lambda_{s}\otimes\lambda_{j}\lambda_{k}})}$
(45) $\displaystyle=$ $\displaystyle\sqrt{4\sum{r_{ij}^{2}}}$ $\displaystyle=$
$\displaystyle 2\|R_{L}\|_{HS}$

and

$\displaystyle\|R_{L}\|_{HS}$ $\displaystyle=$
$\displaystyle\sqrt{tr(\frac{1}{c^{2}}V\Sigma^{\prime
T}U^{T}U\Sigma^{\prime}V)}$ (46) $\displaystyle=$
$\displaystyle\frac{1}{c}\sqrt{tr(\Sigma^{\prime T}\Sigma^{\prime})}$
$\displaystyle=$ $\displaystyle\frac{1}{c}\sqrt{rank(R_{\rho})}.$

Then $\|\sum{r_{ij}\lambda_{i}\otimes\lambda_{j}}\|_{HS}=2\|R_{L}\|_{HS}=1$.
So we have
$\lambda_{max}(\sum{r_{ij}\lambda_{i}\otimes\lambda_{j}})\leq\|\sum{r_{ij}\lambda_{i}\otimes\lambda_{j}}\|_{HS}=1$.
Thus $L=\frac{1}{d_{1}d_{2}}(I\otimes
I+\sum{r_{ij}\lambda_{i}\otimes\lambda_{j}})\geq 0$. So we get

$\displaystyle
C_{w}(\rho)\geq\frac{2(\|R_{\rho}\|_{KF}-1)}{d_{1}^{2}d_{2}^{2}\sqrt{rank(R_{\rho})}}.$
(47)

When $rank(R_{\rho})=0$, i.e. $R_{\rho}=0$, we have $C_{w}(\rho)\geq 0$, which
ends the proof. ∎

######  Theorem 5.

For $\rho\in\mathcal{H}^{2}\otimes\mathcal{H}^{2}$, we have

$C_{w}(\rho)\geq\frac{1}{4}\max\\{0,m_{1}+m_{2}-sgn(|R_{\rho}|)m_{3}-1,\frac{m_{1}+m_{2}+sgn(|R_{\rho}|)m_{3}-1}{3}\\},$
(48)

where $R_{\rho}$ is the correlation matrix of $\rho$, and $m_{i}$ ($m_{1}\geq
m_{2}\geq m_{3}\geq 0$) is the singular value of $R_{\rho}$.

###### Proof.

From Theorem 3, when $L^{\prime}=\frac{1}{4}(I\otimes
I+\sum{r_{ij}^{\prime}\lambda_{i}\otimes\lambda_{j}})$, we have
$\alpha(L^{\prime})=\frac{1}{4}(1+\|R_{L^{\prime}}\|_{2})$,
$R_{L^{\prime}}=(r_{ij}^{\prime})$. For
$\rho\in\mathcal{H}^{2}\otimes\mathcal{H}^{2}$, $\rho=\frac{1}{4}(I\otimes
I+\vec{s}\cdot\vec{\lambda}\otimes I+I\otimes\vec{t}\cdot\vec{\lambda}+\sum
r_{ij}\lambda_{i}\otimes\lambda_{j})$ and $R_{\rho}=(r_{ij})$, we can select
an appropriate local unitary transformation $U_{1}\otimes U_{2}$ on $\rho$ to
diagonalize $R_{\rho}$, then the transformed $R_{\rho}$ is
$\Sigma_{\rho}=O_{1}^{T}R_{\rho}O_{2}=diag(m_{1},m_{2},sgn(|R_{\rho}|)m_{3})$
($m_{1}\geq m_{2}\geq m_{3}\geq 0$), where $O_{1}$ and $O_{2}$ are real
orthogonal matrices satisfying $|O_{1}|=|O_{2}|=1$. We can construct
$L^{\prime}=\frac{1}{4}(I\otimes
I+\sum{r_{ij}}^{\prime}\lambda_{i}\otimes\lambda_{j})$, where
${R}^{\prime}=({r_{ij}}^{\prime})=O_{1}^{T}\Sigma_{L^{\prime}}O_{2}$,
$\Sigma_{L^{\prime}}=diag(r_{1},r_{2},r_{3})$, then

$\displaystyle C_{w}(\rho)$ $\displaystyle\geq$
$\displaystyle\max\limits_{L^{\prime}\geq
0}\\{tr(L^{\prime}\rho)-\alpha(L^{\prime})\\}$ (49) $\displaystyle=$
$\displaystyle\max\limits_{L^{\prime}\geq
0}\\{\frac{1}{4}(tr(R_{L}^{T}R_{\rho})-\max\limits_{i}|r_{i}|)\\}$ (50)
$\displaystyle=$ $\displaystyle\max\limits_{L^{\prime}\geq
0}\\{\frac{1}{4}(\textbf{r}\cdot\textbf{m}-\max\limits_{i}|r_{i}|)\\},$ (51)

where $\textbf{r}=(r_{1},r_{2},r_{3})$ and
$\textbf{m}=(m_{1},m_{2},sgn(|R_{\rho}|)m_{3})$. From Ref. [11], we know that
the above matrix $L^{\prime}$ is positive semi-definite if and only if r
belongs to the tetrahedron $\Gamma$ with vertices $\textbf{t}_{1}=(-1,-1,-1)$,
$\textbf{t}_{2}=(1,1,-1)$, $\textbf{t}_{3}=(1,-1,1)$,
$\textbf{t}_{4}=(-1,1,1)$. Since tetrahedron $\Gamma$ is symmetric about three
axes and three coordinate planes, r still belongs to tetrahedron $\Gamma$ when
we change the order of elements or the sign of the two elements.

When r satisfies
$\frac{1}{4}(\textbf{r}\cdot\textbf{m}-\max\limits_{i}|r_{i}|)=\max\limits_{L^{\prime}\geq
0}\\{\frac{1}{4}(\textbf{r}\cdot\textbf{m}-\max\limits_{i}|r_{i}|)\\}$, we
have $r_{1}\geq r_{2}\geq|r_{3}|\geq 0$. Otherwise, For
$\forall\textbf{r}=(r_{1},r_{2},r_{3})\in\Gamma$, we can change r to
$\textbf{r}^{\prime}=(r^{\prime}_{1},r^{\prime}_{2},r^{\prime}_{3})\in\Gamma(r^{\prime}_{1}\geq
r^{\prime}_{2}\geq|r^{\prime}_{3}|)$ and
$\textbf{r}^{\prime}-\textbf{r}=(c_{1},-c_{1}+c_{2},-c_{2}+c_{3})(c_{i}\geq
0,i=1,2,3)$, then

$\displaystyle\frac{1}{4}(\textbf{r}^{\prime}\cdot\textbf{m}-\max\limits_{i}|r_{i}|)-\frac{1}{4}(\textbf{r}\cdot\textbf{m}-\max\limits_{i}|r^{\prime}_{i}|)$
$\displaystyle=$
$\displaystyle\frac{1}{4}(\textbf{r}^{\prime}-\textbf{r})\cdot\textbf{m}$ (52)
$\displaystyle=$
$\displaystyle\frac{1}{4}[c_{1}(m_{1}-m_{2})+c_{2}(m_{2}-m_{3})+c_{3}m_{3}]$
(53) $\displaystyle\geq$ $\displaystyle 0.$ (54)

We can simplify the problem to

$\displaystyle
C_{w}(\rho)\geq\max\limits_{\textbf{r}\in\Gamma}\frac{1}{4}(\textbf{r}\cdot\textbf{b}),$
(55)

where $\textbf{b}=(m_{1}-1,m_{2},sgn(|R_{\rho}|)m_{3})$ and $r_{1}\geq
r_{2}\geq|r_{3}|$. Hence, r is in a new tetrahedron $\Gamma^{\prime}$ with
vertices $\textbf{t}_{1}^{\prime}=(0,0,0)$,
$\textbf{t}_{2}^{\prime}=(1,1,-1)$, $\textbf{t}_{3}^{\prime}=(1,0,0)$,
$\textbf{t}_{4}^{\prime}=(1/3,1/3,1/3)$. Since the maximum value of
$\textbf{r}\cdot\textbf{b}$ can be obtained on the corners of the tetrahedron,
and
$\textbf{t}_{2}^{\prime}\cdot\textbf{b}\geq\textbf{t}_{3}^{\prime}\cdot\textbf{b}$
always holds. Then we have

$\displaystyle
C_{w}(\rho)\geq\frac{1}{4}\max\\{0,m_{1}+m_{2}-sgn(|R_{\rho}|)m_{3}-1,\frac{m_{1}+m_{2}+sgn(|R_{\rho}|)m_{3}-1}{3}\\}.$
(56)

∎

###  4.2 Numerical Simulation

The measurable lower bound can be used to check whether the quantum state is
entangled. Next we discuss the lower bound for quantum states in systems with
different dimensions.

####  4.2.1 $\rho\in\mathcal{H}^{2}\otimes\mathcal{H}^{2}$

When $\rho\in\mathcal{H}^{2}\otimes\mathcal{H}^{2}$, we have two measures to
discuss the lower bound of the quantum state. Firstly, let’s concentrate on
this pure quantum state[12]

$\displaystyle\rho=|\varphi\rangle\langle\varphi|,$ (57)

where $|\varphi\rangle=(a,0,0,1/\sqrt{2})^{T}/\sqrt{a^{2}+1/2}$. We can get a
group of the lower bound of this quantum state by changing the value of $a$,
as shown Fig.1.

Figure 1: Lower bounds of $C_{w}(\rho)$, where
$\rho\in\mathcal{H}^{2}\otimes\mathcal{H}^{2}$ and $\rho$ is pure.

It is clear that Theorem 2 and Theorem 5 are equivalent in the pure state when
$\rho\in\mathcal{H}^{2}\otimes\mathcal{H}^{2}$. The reason is that the
particular $L$ in Theorem 2 is the maximally entangled state corresponding to
the pure state. And Theorem 5 is equivalent to finding the lower bound by
checking the maximally entangled state. The difference between them is that
Theorem 2 uses the measure of Schmidt decomposition and Theorem 5 uses the
measure of Bloch representation. Similar to the pure state, we have two ways
to get the value of the lower bound of mixed state. Focusing on the following
mixed quantum state:

$\displaystyle\rho=\frac{x}{4}I+(1-x)|\varphi\rangle\langle\varphi|,$ (58)

where $|\varphi\rangle$ has the same definition as the upper one. For $x$=0.1,
we have the lower bounds for $C_{w}(\rho)$, as shown Fig.2.

Figure 2: Lower bounds of $C_{w}(\rho)$, where
$\rho\in\mathcal{H}^{2}\otimes\mathcal{H}^{2}$ and $\rho$ is mixed for
$x$=0.1.

Aiming to compare the different quantum state’s influence on the same measure,
we figure the similar picture with $x$=0.01, as shown Fig.3.

Figure 3: Lower bounds of $C_{w}(\rho)$, where
$\rho\in\mathcal{H}^{2}\otimes\mathcal{H}^{2}$ and $\rho$ is mixed for x=0.01.

We can see that the lower bounds of $C_{w}(\rho)$ increase as $x$ gets lower.
And Theorem 5 gives us a better result than Theorem 4 when we discuss the
quantum state at $2\otimes 2$ Hilbert spaces.

####  4.2.2 $\rho\in\mathcal{H}^{3}\otimes\mathcal{H}^{3}$

In the same way, we can analyze the quantum state which is higher-dimensional.
To avoid the hassle, we’ll just cover the quantum states that belong to
$\mathcal{H}^{3}\otimes\mathcal{H}^{3}$. Consider the following mixed quantum
state.

$\displaystyle\rho=\frac{x}{9}I+(1-x)|\varphi\rangle\langle\varphi|,$ (59)

where
$|\varphi\rangle=(a,0,0,0,1/\sqrt{3},0,0,0,1/\sqrt{3})^{T}/\sqrt{a^{2}+2/3}$.
For $x$=0.01 and $x$=0.1, we can also get these mixed quantum’s lower bound
utilizing the Theorem 4, as shown Fig.4.

Figure 4: Lower bounds of $C_{w}(\rho)$, where
$\rho\in\mathcal{H}^{3}\otimes\mathcal{H}^{3}$ and $\rho$ is mixed.

When we limit it to $tr(L)=1$, we can see that the lower bounds form Theorem 4
are relatively small when we discuss the higher dimensional quantum state. But
we can deal with it by limiting the $L\in M_{a}$ and $a$ is higher than 1. If
we cut off the unitary part of the quantum state, we can obtain the pure state
belonging to $\mathcal{H}^{3}\otimes\mathcal{H}^{3}$, we can get the
corresponding lower bound, as shown Fig.5.

Figure 5: Lower bounds of $C_{w}(\rho)$, where
$\rho\in\mathcal{H}^{3}\otimes\mathcal{H}^{3}$ and $\rho$ is pure.

##  5 Multipartite Entanglement

So far we focused our attention on Hilbert spaces of two finite-dimensional
Hilbert spaces. Now we consider the Hilbert space
$\mathcal{H}=\mathcal{H}_{1}\otimes\cdots\otimes\mathcal{H}_{n}$. The
generalization of the definition of $C_{w}(\rho)$ is

$\displaystyle C_{w}(\rho)=\max\limits_{L}\\{-tr(W_{E}\rho)\\},$

and it satisfies:

$\displaystyle L\in M_{a},$ $\displaystyle W_{E}=\alpha I-L,$
$\displaystyle\alpha=\max\limits_{|\varphi\rangle\in
sep}\langle\varphi|L|\varphi\rangle.$

It is obvious that there must be a Hermitian matrix $L$ which satisfies
$tr(W_{E}\rho)<0$, $tr(W_{E}\sigma)\geq 0$, where $\rho$ is an entangled
state, $\sigma$ is a separable state.

##  6 Conclusion

In this paper, we define an entanglement measure based on entanglement witness
and extend its definition to the multipartite entanglement of quantum states
in arbitrary dimensional systems. This measure satisfies some necessary
properties of an entanglement measure and we estimate its lower bound on
bipartite system. The rationality of the entanglement measure is verified by
the estimation of its lower bound by discussing pure and mixed states
separately and the corresponding numerical simulation.

## Acknowledgments

This work is supported by the Shandong Provincial Natural Science Foundation
for Quantum Science No.ZR2021LLZ002, and the Fundamental Research Funds for
the Central Universities No.22CX03005A.

## Data availability statement

All data generated or analysed during this study are included in this
published article.

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