Testing the Monogamy Relations via Rank-2 Mixtures

Research into entanglement of quantum states has long history from the very beginning of quantum mechanicsepr-35 ; schrodinger-35 . At that time the main motivation for the study of entanglement was pure theoretical. It was to explore the non-local properties of quantum mechanics. Recent considerable attention to the quantum entanglementtext ; horodecki09 has both theoretical and practical aspects. While the former is for understanding of quantum information theories more deeply, the latter is for developing the quantum technology. As shown for last two decades quantum entanglement plays a central role in quantum teleportationteleportation , superdense codingsuperdense , quantum cloningclon , and quantum cryptographycryptography ; cryptography2 . It is also quantum entanglement, which makes the quantum computer¹¹1The current status of quantum computer technology was reviewed in Ref.qcreview . outperform the classical onecomputer . Thus, it is very important to understand how to quantify and how to characterize the entanglement. Still, however, this issue is not completely understood.

For bipartite quantum system many entanglement measures were constructed before such as distillable entanglementbenn96 , entanglement of formation (EOF)benn96 , and relative entropy of entanglement (REE)vedral-97-1 ; vedral-97-2 . Among them²²2Although there are a lot of attempts to derive the closed formula for REE, still we do not know how to compute the REE for the arbitrary two qubit mixtures except rare cases ree . the closed formula for the analytic computation of EOF for states of two qubits were found in Ref. woot-98 via the concurrence ${\cal C}$ as

$$E_{F}({\cal C})=h\left(\frac{1+\sqrt{1-{\cal C}^{2}}}{2}\right),$$

(1)

where $h(x)$ is a binary entropy function $h(x)=-x\ln x-(1-x)\ln(1-x)$. For two-qubit pure state $\lvert\psi\rangle_{AB}=\psi_{ij}\lvert ij\rangle_{AB}$ with $(i,j=0,1)$, the concurrence ${\cal C}_{A|B}$ between party $A$ and party $B$ is given by

$${\cal C}_{A|B}=|\epsilon_{i_{1}i_{2}}\epsilon_{j_{1}j_{2}}\psi_{i_{1}j_{1}}\psi_{i_{2}j_{2}}|=2|\psi_{00}\psi_{11}-\psi_{01}\psi_{10}|,$$

(2)

where the Einstein convention is understood and $\epsilon_{\mu\nu}$ is an antisymmetric tensor. For two-qubit mixed state $\rho_{AB}$ the concurrence ${\cal C}_{A|B}(\rho)$ can be computed by ${\cal C}_{A|B}=\max(\lambda_{1}-\lambda_{2}-\lambda_{3}-\lambda_{4},0)$, where $\{\lambda_{1}^{2},\lambda_{2}^{2},\lambda_{3}^{2},\lambda_{4}^{2}\}$ are eigenvalues of positive operator $\rho(\sigma_{y}\otimes\sigma_{y})\rho^{*}(\sigma_{y}\otimes\sigma_{y})$ with decreasing order. Thus, one can compute the EOF for all two-qubit states in principle.

Generalization to the multipartite entanglement is highly important and challenging issue in the context of quantum information theories. A seminal step toward this goal was initiated in Ref. ckw by examining the three-qubit pure states. Authors in Ref. ckw have shown analytically the monogamy relation

$${\cal C}^{2}_{q_{1}|(q_{2}q_{3})}\geq{\cal C}^{2}_{q_{1}|q_{2}}+{\cal C}^{2}_{q_{1}|q_{3}}.$$

(3)

This relation implies that the entanglement (measured by the squared concurrence) between $q_{1}$ and the remaining parties always exceeds entanglement between $q_{1}$ and $q_{2}$ plus entanglement between $q_{1}$ and $q_{3}$. This means that if $q_{1}$ and $q_{2}$ is maximally entangled, the whole system cannot have the tripartite entanglement. The inequality (3) is strong in a sense that the three-qubit W-statedur00

$$\lvert\mbox{W}_{3}\rangle=\frac{1}{\sqrt{3}}\left(\lvert 001\rangle+\lvert 010\rangle+\lvert 100\rangle\right)$$

(4)

saturates the inequality. Moreover, for three-qubit pure state $\lvert\psi\rangle_{ABC}=\psi_{ijk}\lvert ijk\rangle_{ABC}$ the leftover in the inequality

$$\tau_{A|B|C}={\cal C}^{2}_{A|(BC)}-\left({\cal C}^{2}_{A|B}+{\cal C}^{2}_{A|C}\right),$$

(5)

which we will call the residual entanglement³³3In this paper $\sqrt{\tau_{A|B|C}}$ is called the three-tangle., has following two expressions:

$\displaystyle\tau_{A|B|C}=\bigg{|}2\epsilon_{i_{1}i_{2}}\epsilon_{i_{3}i_{4}}\epsilon_{j_{1}j_{2}}\epsilon_{j_{3}j_{4}}\epsilon_{k_{1}k_{3}}\epsilon_{k_{2}k_{4}}\psi_{i_{1}j_{1}k_{1}}\psi_{i_{2}j_{2}k_{2}}\psi_{i_{3}j_{3}k_{3}}\psi_{i_{4}j_{4}k_{4}}\bigg{|}=4|d_{1}-2d_{2}+4d_{3}|$

(6)

where

	$\displaystyle d_{1}=\psi^{2}_{000}\psi^{2}_{111}+\psi^{2}_{001}\psi^{2}_{110}+\psi^{2}_{010}\psi^{2}_{101}+\psi^{2}_{100}\psi^{2}_{011},$		(7)
	$\displaystyle d_{2}=\psi_{000}\psi_{111}\psi_{011}\psi_{100}+\psi_{000}\psi_{111}\psi_{101}\psi_{010}+\psi_{000}\psi_{111}\psi_{110}\psi_{001}$
	$\displaystyle\hskip 28.45274pt+\psi_{011}\psi_{100}\psi_{101}\psi_{010}+\psi_{011}\psi_{100}\psi_{110}\psi_{001}+\psi_{101}\psi_{010}\psi_{110}\psi_{001},$
	$\displaystyle d_{3}=\psi_{000}\psi_{110}\psi_{101}\psi_{011}+\psi_{111}\psi_{001}\psi_{010}\psi_{100}.$

From first expression one can show that $\tau_{A|B|C}$ is invariant under a stochastic local operation and classical communication (SLOCC)bennet00 . From second expression one can show that $\tau_{A|B|C}$ is invariant under the qubit permutation. It was also shown in Ref. ckw that $\tau_{A|B|C}$ is an entanglement monotone. Thus, the residual entanglement (or three-tangle) can play a role as an important measure for the genuine three-way entanglement.

By making use of Eq. (6) one can compute the residual entanglement of all three-qubit pure states. For mixed state the residual entanglement is usually defined as a convex roof methodbenn96 ; uhlmann99-1

$$\tau_{A|B|C}(\rho)=\min\sum_{i}p_{i}\tau_{A|B|C}(\lvert\psi_{i}\rangle\langle\psi_{i}\lvert)$$

(8)

where the minimum is taken over all possible ensembles of pure states. The ensemble corresponding to the minimum of $\tau_{A|B|C}$ is called optimal decomposition. For given three-qubit mixed state it is highly difficult, in general, to find its optimal decomposition except very rare casestangle ⁴⁴4Recently, the three-tangle of the GHZ-symmetric stateselts12-1 has been computed analyticallysiewert12-1 ..

In order to find the entanglement measures in the multipartite system, there are two different approaches. First approach is to find the invariant monotones under the SLOCC transformation. As Ref.verst03 has shown, any linearly homogeneous positive function of a pure state that is invariant under determinant $1$ SLOCC operations is an entanglement monotone. Thus, the concurrence ${\cal C}_{A|B}$ and the three-tangle $\sqrt{\tau_{A|B|C}}$ are monotones. It is also possible to construct the SLOCC-invariant monotones in the higher-qubit systems. In the higher-qubit systems, however, there are many independent monotones, because the number of independent SLOCC-invariant monotones is equal to the degrees of freedom of pure quantum state minus the degrees of freedom induced by the determinant $1$ SLOCC operations. For example, there are $2(2^{n}-1)-6n$ independent monotones in $n$-qubit system. Thus, in four-qubit system there are six invariant monotones. Among them, it was shown in Ref. four-way by making use of the antilinearityuhlmann99-1 that there are following three independent monotones which measure the true four-way entanglement:

	$\displaystyle{\cal F}^{(4)}_{1}=(\sigma_{\mu}\sigma_{\nu}\sigma_{2}\sigma_{2})\bullet(\sigma^{\mu}\sigma_{2}\sigma_{\lambda}\sigma_{2})\bullet(\sigma_{2}\sigma^{\nu}\sigma^{\lambda}\sigma_{2})$
	$\displaystyle{\cal F}^{(4)}_{2}=(\sigma_{\mu}\sigma_{\nu}\sigma_{2}\sigma_{2})\bullet(\sigma^{\mu}\sigma_{2}\sigma_{\lambda}\sigma_{2})\bullet(\sigma_{2}\sigma^{\nu}\sigma_{2}\sigma_{\tau})\bullet(\sigma_{2}\sigma_{2}\sigma^{\lambda}\sigma^{\tau})$		(9)
	$\displaystyle{\cal F}^{(4)}_{3}=\frac{1}{2}(\sigma_{\mu}\sigma_{\nu}\sigma_{2}\sigma_{2})\bullet(\sigma^{\mu}\sigma^{\nu}\sigma_{2}\sigma_{2})\bullet(\sigma_{\rho}\sigma_{2}\sigma_{\tau}\sigma_{2})\bullet(\sigma^{\rho}\sigma_{2}\sigma^{\tau}\sigma_{2})\bullet(\sigma_{\kappa}\sigma_{2}\sigma_{2}\sigma_{\lambda})\bullet(\sigma^{\kappa}\sigma_{2}\sigma_{2}\sigma^{\lambda}),$

where $\sigma_{0}=\openone_{2}$, $\sigma_{1}=\sigma_{x}$, $\sigma_{2}=\sigma_{y}$, $\sigma_{3}=\sigma_{z}$, and the Einstein convention is introduced with a metric $g^{\mu\nu}=\mbox{diag}\{-1,1,0,1\}$. The solid dot in Eq. (I) is defined as follows. Let $\lvert\psi\rangle$ be a four-qubit state. Then, for example, ${\cal F}^{(4)}_{1}$ of $\lvert\psi\rangle$ is defined as

$${\cal F}^{(4)}_{1}(\psi)=\bigg{|}\langle\psi^{*}\lvert\sigma_{\mu}\otimes\sigma_{\nu}\otimes\sigma_{2}\otimes\sigma_{2}\lvert\psi\rangle\langle\psi^{*}\lvert\sigma^{\mu}\otimes\sigma_{2}\otimes\sigma_{\lambda}\otimes\sigma_{2}\lvert\psi\rangle\langle\psi^{*}\lvert\sigma_{2}\otimes\sigma^{\nu}\otimes\sigma^{\lambda}\otimes\sigma_{2}\lvert\psi\rangle\bigg{|}.$$

(10)

Other measures can be computed similarly. Furthermore, it was shown in Ref. oster06-1 that there are following three maximally entangled states in four-qubit system:

	$\displaystyle\lvert\mbox{GHZ}_{4}\rangle=\frac{1}{\sqrt{2}}\big{(}\lvert 0000\rangle+\lvert 1111\rangle\big{)}$
	$\displaystyle\lvert\Phi_{2}\rangle=\frac{1}{\sqrt{6}}\left(\sqrt{2}\lvert 1111\rangle+\lvert 1000\rangle+\lvert 0100\rangle+\lvert 0010\rangle+\lvert 0001\rangle\right)$		(11)
	$\displaystyle\lvert\Phi_{3}\rangle=\frac{1}{2}\big{(}\lvert 1111\rangle+\lvert 1100\rangle+\lvert 0010\rangle+\lvert 0001\rangle\big{)}.$

The measures ${\cal F}^{(4)}_{1}$, ${\cal F}^{(4)}_{2}$, and ${\cal F}^{(4)}_{3}$ of $\lvert\mbox{GHZ}_{4}\rangle$, $\lvert\Phi_{2}\rangle$, $\lvert\Phi_{3}\rangle$, and

$$\lvert\tilde{\mbox{W}}_{4}\rangle=\frac{1}{2}\big{(}\lvert 0111\rangle+\lvert 1011\rangle+\lvert 1101\rangle+\lvert 1110\rangle\big{)}$$

(12)

are summarized in Table I. Recently, ${\cal F}^{(4)}_{j}\hskip 5.69046pt(j=1,2,3)$ and the corresponding linear monotones⁵⁵5The linear monotone means a monotone of homogeneous degree $2$. Thus, if ${\cal F}^{(4)}_{j}$ is a measure of homogeneous degree $D$, the corresponding one is ${\cal G}^{(4)}_{j}=\left({\cal F}^{(4)}_{j}\right)^{2/D}$. ${\cal G}^{(4)}_{j}\hskip 5.69046pt(j=1,2,3)$ for the rank-$2$ mixtures consist of one of the maximally entangled state and $\lvert\tilde{\mbox{W}}_{4}\rangle$ are explicitly computedeylee15-1 .

Second approach is to find the monogamy relations in the multipartite system. As Ref. osborne06-1 has shown analytically the following monogamy relation

$${\cal C}^{2}_{q_{1}|(q_{2}\cdots q_{n})}\geq{\cal C}^{2}_{q_{1}|q_{2}}+{\cal C}^{2}_{q_{1}|q_{3}}+\cdots+{\cal C}^{2}_{q_{1}|q_{n}}$$

(13)

holds in the $n$-qubit pure-state system. However, the leftover of Eq. (13) is not entanglement monotone. The authors in Ref. bai07-1 ; bai08-1 conjectured that in four-qubit system the following quantity

$$t_{1}=\frac{\pi_{A}+\pi_{B}+\pi_{C}+\pi_{D}}{4}$$

(14)

is a monotone, where $\pi_{A}={\cal C}^{2}_{A|(BCD)}-({\cal C}^{2}_{A|B}+{\cal C}^{2}_{A|C}+{\cal C}^{2}_{A|D})$ and other ones are obtained by changing the focusing qubit. Even though $t_{1}$ might be an entanglement monotone, it is obvious that it is not a true four-way measure because it detects the three-way entanglement. For example, $t_{1}(g_{3})=3/4$, where $\lvert g_{3}\rangle=(\lvert 0000\rangle+\lvert 1110\rangle)/\sqrt{2}$.

In Ref. regula14-1 another following multipartite monogamy relation is derived:

$${\cal C}^{2}_{q_{1}|(q_{2}\cdots q_{n})}\geq\underbrace{\sum_{j=2}^{n}{\cal C}^{2}_{q_{1}|q_{j}}}_{2-\mbox{partite}}+\underbrace{\sum_{k>j=2}^{n}\left[\tau_{q_{1}|q_{j}|q_{k}}\right]^{\mu_{3}}}_{3-\mbox{partite}}+\cdots+\underbrace{\sum_{\ell=2}^{n}\left[\tau_{q_{1}|q_{2}|\cdots|q_{\ell-1}|q_{\ell+1}|\cdots|q_{n}}\right]^{\mu_{n-1}}}_{(n-1)-\mbox{partite}}.$$

(15)

In Eq. (15) the power factors $\left\{\mu_{m}\right\}_{m=3}^{n-1}$ are included to regulate the weight assigned to the different $m$-partite contributions. If all power factors $\mu_{m}$ go to infinity, Eq. (15) reduces to Eq. (13). Especially, the authors in Ref. regula14-1 have conjectured $\mu_{3}=3/2$. Thus, in four-qubit system one can construct another possible candidate of the tangle-based entanglement measure

$$t_{2}=\frac{\sigma_{A}+\sigma_{B}+\sigma_{C}+\sigma_{D}}{4},$$

(16)

where $\sigma_{A}={\cal C}^{2}_{A|(BCD)}-\left({\cal C}^{2}_{A|B}+{\cal C}^{2}_{A|C}+{\cal C}^{2}_{A|D}\right)-\left(\left[\tau_{A|B|C}\right]^{\mu}+\left[\tau_{A|B|D}\right]^{\mu}+\left[\tau_{A|C|D}\right]^{\mu}\right)$, and others are obtained by changing the focusing qubit. One can show easily $t_{2}(g_{3})=0$. Thus, the measure $t_{2}$ cannot be excluded as a true four-way entanglement measure.

In Ref. jin15-1 ; karmakar16-1 two different negativity-based monogamy relations have been examined. From these relations one can construct the following candidates of the four-party entanglement measures:

$$n_{1}=\frac{u_{A}+u_{B}+u_{C}+u_{D}}{4}$$

(17)

where $u_{A}={\cal N}_{A|(BCD)}-\left({\cal N}_{A|B}+{\cal N}_{A|C}+{\cal N}_{A|D}\right)-\left(\left[{\cal N}_{A||B|C}\right]^{\nu_{1}}+\left[{\cal N}_{A||B|D}\right]^{\nu_{1}}+\left[{\cal N}_{A||C|D}\right]^{\nu_{1}}\right)$ with ${\cal N}_{I||J|K}\equiv{\cal N}_{I|(JK)}-{\cal N}_{I|J}-{\cal N}_{I|K}$ and

$$n_{2}=\frac{v_{A}+v_{B}+v_{C}+v_{D}}{4}$$

(18)

where $v_{A}={\cal N}^{2}_{A|(BCD)}-\left({\cal N}^{2}_{A|B}+{\cal N}^{2}_{A|C}+{\cal N}^{2}_{A|D}\right)-\left(\left[{\cal N}^{2}_{A||B|C}\right]^{\nu_{2}}+\left[{\cal N}^{2}_{A||B|D}\right]^{\nu_{2}}+\left[{\cal N}^{2}_{A||C|D}\right]^{\nu_{2}}\right)$ with ${\cal N}_{I||J|K}^{2}\equiv{\cal N}_{I|(JK)}^{2}-{\cal N}_{I|J}^{2}-{\cal N}_{I|K}^{2}$. The negativity ${\cal N}$ is defined as vidal02 ; soojoon03

$${\cal N}(\rho_{AB})=||\rho_{AB}^{T_{A}}||-1$$

(19)

where $||X||\equiv\mbox{tr}(\sqrt{XX^{\dagger}})$ and the superscript $T_{A}$ means the partial transposition of $A$-qubit. Of course other quantities can be obtained by changing the focusing qubit.

The purpose of this paper is to test $t_{1}$, $t_{2}$, $n_{1}$, and $n_{2}$ by computing them for the rank-$2$ mixture

$$\rho_{4}=p\lvert\mbox{GHZ}_{4}\rangle\langle\mbox{GHZ}_{4}\lvert+(1-p)\lvert\mbox{W}_{4}\rangle\langle\mbox{W}_{4}\lvert$$

(20)

where $\lvert\mbox{GHZ}_{4}\rangle$ is defined in Eq. (I) and $\lvert\mbox{W}_{4}\rangle=(\sigma_{x}\otimes\sigma_{x}\otimes\sigma_{x}\otimes\sigma_{x})\lvert\tilde{\mbox{W}}_{4}\rangle$. In section II we compute $t_{1}$, $t_{2}$, $n_{1}$, and $n_{2}$ for the maximal entangled pure states (I) and $\lvert\mbox{W}_{4}\rangle$. The results are summarized in Table II. It is shown that the negativity-based measures $n_{1}$ and $n_{2}$ become negative for $\lvert\Phi_{3}\rangle$. In section III we try to compute $t_{1}$ and $t_{2}$ for $\rho_{4}$ by finding the optimal decompositions. For $t_{1}$ it turns out that Eq. (20) itself is an optimal decomposition. However, we fail to compute $t_{2}$ because analytic computation of the residual entanglement is extremely difficult. In section IV we compute $n_{1}$ and $n_{2}$ for $\rho_{4}$ in the range $\nu_{1*}\leq\nu_{1}$ and $\nu_{2*}\leq\nu_{2}$ by finding the optimal decompositions, where

$$\nu_{1*}=\frac{\ln\left(\frac{3-3\sqrt{2}+\sqrt{3}}{6}\right)}{\ln\left(\frac{3}{2}-\sqrt{2}\right)}=1.02053\hskip 14.22636pt\nu_{2*}=\frac{\ln\left(\frac{\sqrt{2}-1}{2}\right)}{\ln\left(\sqrt{2}-\frac{5}{4}\right)}=0.871544.$$

(21)

In this region $n_{1}(\mbox{W}_{4})$ and $n_{2}(\mbox{W}_{4})$ become non-negative. In section V a brief conclusion is given. In appendix we try to explain why the computation of the residual entanglement is highly difficult.

In this section we compute $t_{1}$, $t_{2}$, $n_{1}$, and $n_{2}$ for four-qubit maximally entangled states (I) and W-state $\lvert\mbox{W}_{4}\rangle$. The most special case is $\lvert\mbox{GHZ}_{4}\rangle$, which gives $t_{1}=t_{2}=n_{1}=n_{2}=1$. Since $\lvert\mbox{W}_{4}\rangle$ saturates the monogamy relations (13) and (15), $t_{1}$ and $t_{2}$ of $\lvert\mbox{W}_{4}\rangle$ are exactly zero. However, $\lvert\mbox{W}_{4}\rangle$ does not saturate the negativity-based monogamy relations. It is straightforwardkarmakar16-1 to show that $n_{1}$ and $n_{2}$ of $\lvert\mbox{W}_{4}\rangle$ are

	$\displaystyle n_{1}(\mbox{W}_{4})=\frac{3+\sqrt{3}-3\sqrt{2}}{2}-3\left(\frac{3-2\sqrt{2}}{2}\right)^{\nu_{1}}$		(22)
	$\displaystyle n_{2}(\mbox{W}_{4})=\frac{3}{2}(\sqrt{2}-1)-3\left(\frac{4\sqrt{2}-5}{4}\right)^{\nu_{2}}.$

Thus, as we commented, $n_{1}(\mbox{W}_{4})$ and $n_{2}(\mbox{W}_{4})$ become non-negative when $\nu_{1*}\leq\nu_{1}$ and $\nu_{2*}\leq\nu_{2}$.

For $\lvert\Phi_{2}\rangle$ it is easy to show that ${\cal C}_{I|(JKL)}=1$ and ${\cal C}_{I|J}=0$ for all $\{I,J,K,L\}=\{A,B,C,D\}$. The tripartite states derived from $\lvert\Phi_{2}\rangle\langle\Phi_{2}\lvert$ by tracing over any one-party is given by

$$\rho_{3}^{(2)}=\frac{1}{2}\lvert\psi_{1}\rangle\langle\psi_{1}\lvert+\frac{1}{2}\lvert\mbox{W}_{3}\rangle\langle\mbox{W}_{3}\lvert$$

(23)

where $\lvert\psi_{1}\rangle=(\lvert 000\rangle+\sqrt{2}\lvert 111\rangle)/\sqrt{3}$ and $\lvert\mbox{W}_{3}\rangle=(\lvert 001\rangle+\lvert 010\rangle+\lvert 100\rangle)/\sqrt{3}$. In order to compute the residual entanglement of $\rho_{3}^{(2)}$ let us consider the quantum state $p\lvert gGHZ\rangle\langle gGHZ\lvert+(1-p)\lvert gW\rangle\langle gW\lvert$, where $\lvert gGHZ\rangle=a\lvert 000\rangle+b\lvert 111\rangle$ and $\lvert gW\rangle=c\lvert 001\rangle+d\lvert 010\rangle+f\lvert 100\rangle$. As the second reference of Ref. tangle has shown, the residual entanglement of this state is exactly zero when $p\leq p_{0}\equiv s^{2/3}/(1+s^{2/3})$ where $s=4cdf/(a^{2}b)$. Since, for our case, $p_{0}=2/3$ is larger than $1/2$, the residual entanglement of $\rho_{3}^{(2)}$ is zero. Thus, $t_{1}$ and $t_{2}$ of $\lvert\Phi_{2}\rangle$ are

$$t_{1}(\Phi_{2})=t_{2}(\Phi_{2})=1.$$

(24)

Various negativities of $\lvert\Phi_{2}\rangle$ can be directly computed and the final expressions are

$${\cal N}_{I|(JKL)}=1\hskip 28.45274pt{\cal N}_{I|(JK)}=\frac{2}{3}\hskip 28.45274pt{\cal N}_{I|J}=0$$

(25)

for all $\{I,J,K,L\}=\{A,B,C,D\}$. Thus, it is easy to show

$$n_{1}(\Phi_{2})=1-3\left(\frac{2}{3}\right)^{\nu_{1}}\hskip 28.45274ptn_{2}(\Phi_{2})=1-3\left(\frac{4}{9}\right)^{\nu_{2}}.$$

(26)

For $\lvert\Phi_{3}\rangle$ one can show ${\cal C}_{I|(JKL)}=1$ and ${\cal C}_{I|J}=0$ for all $\{I,J,K,L\}$. The tripartite states derived from $\lvert\Phi_{3}\rangle\langle\Phi_{3}\lvert$ are

	$\displaystyle\rho_{ACD}^{(3)}=\rho_{BCD}^{(3)}=\frac{1}{2}\lvert\phi_{1}\rangle\langle\phi_{1}\lvert+\frac{1}{2}\lvert\phi_{2}\rangle\langle\phi_{2}\lvert$		(27)
	$\displaystyle\rho_{ABC}^{(3)}=\rho_{ABD}^{(3)}=\frac{1}{2}\lvert g_{1}\rangle\langle g_{1}\lvert+\frac{1}{2}\lvert g_{2}\rangle\langle g_{2}\lvert$

where

	$\displaystyle\lvert\phi_{1}\rangle=\frac{1}{\sqrt{2}}(\lvert 100\rangle+\lvert 111\rangle)\hskip 28.45274pt\lvert\phi_{2}\rangle=\frac{1}{\sqrt{2}}(\lvert 001\rangle+\lvert 010\rangle)$		(28)
	$\displaystyle\lvert g_{1}\rangle=\frac{1}{\sqrt{2}}(\lvert 000\rangle+\lvert 111\rangle)\hskip 28.45274pt\lvert g_{2}\rangle=\frac{1}{\sqrt{2}}(\lvert 001\rangle+\lvert 110\rangle).$

It is easy to show that the residual entanglements of $\rho_{ACD}^{(3)}$ and $\rho_{BCD}^{(3)}$ are zero because $\lvert\phi_{1}\rangle$ and $\lvert\phi_{2}\rangle$ are bi-separable. In order to compute the residual entanglement of $\rho_{ABC}^{(3)}$ and $\rho_{ABD}^{(3)}$ let us consider the quantum state $p\lvert g_{1}\rangle\langle g_{1}\lvert+(1-p)\lvert g_{2}\rangle\langle g_{2}\lvert$. As the last reference of Ref. tangle has shown, the residual entanglement of this state is $(2p-1)^{2}$. Thus, the residual entanglements of $\rho_{ABC}^{(3)}$ and $\rho_{ABD}^{(3)}$ are also zero, all of which yields

$$t_{1}(\Phi_{3})=t_{2}(\Phi_{3})=1.$$

(29)

Various negativities of $\lvert\Phi_{3}\rangle$ can be computed directly and the final expressions are

$${\cal N}_{I|(JKL)}=1\hskip 28.45274pt{\cal N}_{I|J}=0$$

(30)

for all $\{I,J,K,L\}=\{A,B,C,D\}$ and

	$\displaystyle{\cal N}_{A|(CD)}={\cal N}_{B|(CD)}={\cal N}_{C|(AB)}={\cal N}_{D|(AB)}=0$		(31)
	$\displaystyle{\cal N}_{A|(BC)}={\cal N}_{A|(BD)}={\cal N}_{B|(AC)}={\cal N}_{B|(AD)}={\cal N}_{C|(AD)}={\cal N}_{C|(BD)}={\cal N}_{D|(AC)}={\cal N}_{D|(BC)}=1,$

all of which yields

$$n_{1}(\Phi_{3})=n_{2}(\Phi_{3})=-1.$$

(32)

All results are summarized in Table II.

In this section we try to compute $t_{1}$ and $t_{2}$ for the rank-$2$ mixture $\rho_{4}$. For computation of $t_{1}$ and $t_{2}$ we have to find an optimal decomposition. In order to find the optimal decompositions for $t_{1}$ and $t_{2}$ we define

$$\lvert Z_{4}(p,\varphi)\rangle=\sqrt{p}\lvert\mbox{GHZ}_{4}\rangle-e^{i\varphi}\sqrt{1-p}\lvert\mbox{W}_{4}\rangle.$$

(33)

Then, one can show that all reduced bipartite states from $\lvert Z_{4}(p,\varphi)\rangle\langle Z_{4}(p,\varphi)\lvert$ are equal to

$\displaystyle\rho_{IJ}=\frac{1}{2}\left(\begin{array}[]{cccc}1&-\sqrt{\frac{p(1-p)}{2}}e^{-i\varphi}&-\sqrt{\frac{p(1-p)}{2}}e^{-i\varphi}&0\\ -\sqrt{\frac{p(1-p)}{2}}e^{i\varphi}&\frac{1-p}{2}&\frac{1-p}{2}&0\\ -\sqrt{\frac{p(1-p)}{2}}e^{i\varphi}&\frac{1-p}{2}&\frac{1-p}{2}&0\\ 0&0&0&p\end{array}\right)\hskip 14.22636pt(I,J\in\{A,B,C,D\})$

(38)

in the computational basis. In Eq. (38) the parties $I$ and $J$ can be chosen any two different parties from $\{A,B,C,D\}$. Although $\rho_{IJ}$ is a rank-$3$ mixture, one can compute its concurrence analytically by following Wootters procedure:

$${\cal C}_{I|J}=\sqrt{\Lambda}-\sqrt{\Lambda_{+}}-\sqrt{\Lambda_{-}}$$

(39)

where

	$\displaystyle\Lambda=\frac{1}{12}\left[(1+p^{2})+2(\alpha^{2}+\beta^{2})^{1/6}\cos\theta\right]$		(40)
	$\displaystyle\Lambda_{\pm}=\frac{1}{12}\left[(1+p^{2})-2(\alpha^{2}+\beta^{2})^{1/6}\cos\left(\frac{\pi}{3}\pm\theta\right)\right]$

and

	$\displaystyle\alpha=1-9p+39p^{2}-90p^{3}+\frac{231}{2}p^{4}-81p^{5}+\frac{47}{2}p^{6}$
	$\displaystyle\beta=\frac{3p^{2}(1-p)}{2}\sqrt{3p(4-28p+96p^{2}-147p^{3}+110p^{4}-31p^{5})}$		(41)
	$\displaystyle\theta=\frac{1}{3}\tan^{-1}\left(\frac{\beta}{\alpha}\right).$

It is interesting to note that ${\cal C}_{I|J}=0$ at $p=1/3$. Furthermore, it is worthwhile noting that ${\cal C}_{I|J}$ is independent of the phase factor $\varphi$. On the contrary, the corresponding concurrence ${\cal C}_{I|J}^{(3)}$ derived from the three-qubit state $\lvert Z_{3}(p,\varphi)\rangle=\sqrt{p}\lvert\mbox{GHZ}_{3}\rangle-e^{i\varphi}\sqrt{1-p}\lvert\mbox{W}_{3}\rangle$ is explicitly dependent on $\varphi$. The $p$-dependence of ${\cal C}_{I|J}$ and ${\cal C}_{I|J}^{(3)}$ are plotted in Fig. 1. The $\varphi$-dependence of ${\cal C}_{I|J}^{(3)}$ makes the three-qubit rank-$2$ mixture $\rho_{3}=p\lvert\mbox{GHZ}_{3}\rangle\langle\mbox{GHZ}_{3}\lvert+(1-p)\lvert\mbox{W}_{3}\rangle\langle\mbox{W}_{3}\lvert$ have the nontrivial residual entanglementtangle . As we will show shortly, the $\varphi$-independence of ${\cal C}_{I|J}$ makes $t_{1}$ of $\rho_{4}$ to be trivial.

The single qubit states derived from $\lvert Z_{4}(p,\varphi)\rangle\langle Z_{4}(p,\varphi)\lvert$ are all equal to

$\displaystyle\rho_{J}=\left(\begin{array}[]{cc}\frac{3-p}{4}&-\frac{1}{2}\sqrt{\frac{p(1-p)}{2}}e^{-i\varphi}\\ -\frac{1}{2}\sqrt{\frac{p(1-p)}{2}}e^{i\varphi}&\frac{1+p}{4}\end{array}\right)\hskip 14.22636pt(J\in\{A,B,C,D\})$

(44)

in the computational basis. Thus, ${\cal C}^{2}_{I|(JKL)}$ for $\lvert Z_{4}(p,\varphi)\rangle$ is

$${\cal C}^{2}_{I|(JKL)}=4\mbox{det}\rho_{I}=\frac{3+p^{2}}{4}$$

(45)

for all $I,J,K,L$. The corresponding results derived from $\lvert Z_{3}(p,\varphi)\rangle$ is $(8-4p+5p^{2})/9$.

Thus, $t_{1}\left[Z_{4}(p,\varphi)\right]$ is independent of $\varphi$ as

$$t_{1}\left[Z_{4}(p,\varphi)\right]=\frac{3+p^{2}}{4}-3{\cal C}_{I|J}^{2}.$$

(46)

The corresponding residual entanglement $\tau\left[Z_{3}(p,\varphi)\right]$ for $\lvert Z_{3}(p,\varphi)\rangle$ is dependent on $\varphi$ due to ${\cal C}_{I|J}^{(3)}$. The $p$-dependence of $t_{1}\left[Z_{4}(p,\varphi)\right]$ and $\tau\left[Z_{3}(p,\varphi)\right]$ is plotted in Fig. 2(a) and Fig. 2(b) respectively.

Before we calculate $t_{1}(\rho_{4})$ it seems to be helpful to review briefly how to compute $\tau(\rho_{3})$ for $\rho_{3}=p\lvert\mbox{GHZ}_{3}\rangle\langle\mbox{GHZ}_{3}\lvert+(1-p)\lvert\mbox{W}_{3}\rangle\langle\mbox{W}_{3}\lvert$. As Fig. 2(b) shows, when $\varphi=0$ $\tau\left[Z_{3}(p,\varphi)\right]$ has nontrivial zero at $p=p_{0}$ with $p_{0}\sim 0.627$. Furthermore, $\tau\left[Z_{3}(p,\varphi=0)\right]$ is not convex in the regions $0\leq p\leq p_{0}$ and $p_{1}\leq p\leq 1$ with $p_{1}\sim 0.826$. Since $\tau\left[Z_{3}(p,\varphi)\right]$ depends on $\varphi$ through only $\cos 3\varphi$, $\tau\left[Z_{3}(p_{0},\varphi)\right]=0$ for $\varphi=0,2\pi/3,4\pi/3$. Thus, at the small concave region it is possible to convexify the residual entanglement by making use of $\{\lvert\mbox{W}_{3}\rangle,\lvert Z_{3}(p,\frac{2\pi}{3}j)\rangle\hskip 5.69046pt(j=0,1,2)\}$. At the large concave region it is also possible to convexify it by making use of $\{\lvert\mbox{GHZ}_{3}\rangle,\lvert Z_{3}(p,\frac{2\pi}{3}j)\rangle\hskip 5.69046pt(j=0,1,2)\}$.

Now, let us return to the four-qubit case. As Fig. 2(a) shows $t_{1}\left[Z_{4}(p,\varphi)\right]$ is not convex at $0\leq p\leq p_{0}$ and $p_{1}\leq p\leq 1$ with $p_{0}\sim 0.279$ and $p_{1}\sim 0.936$. As the three-qubit case it is possible to convexify the entanglement by making use of $\{\lvert\mbox{GHZ}_{4}\rangle,\lvert Z_{4}(p,0)\rangle,\lvert Z_{4}(p,\pi)\rangle\}$ in the large $p$-region. However, it is impossible to convexify it in the small $p$-region because $t_{1}\left[Z_{4}(p_{0},0)\right]\neq 0$. The only way to obtain the convex result in the entire range of $p$ is

$$t_{1}(\rho_{4})=p.$$

(47)

As Fig. 2(a) shows obviously as a dashed line this is a convex hull of $t_{1}\left[Z_{4}(p,\varphi)\right]$. Thus the optimal decomposition for $t_{1}$ is nothing but the spectral decomposition (20) itself.

In order to compute $t_{2}(\rho_{4})$ we should compute the residual entanglement for the three-qubit states reduced from $\lvert Z_{4}(p,\varphi)\rangle$. One can show that all tripartite states derived by tracing over single qubit are equal to

$$\rho_{IJK}=\lambda\lvert\psi_{+}\rangle\langle\psi_{+}\lvert+(1-\lambda)\lvert\psi_{-}\rangle\langle\psi_{-}\lvert\hskip 14.22636pt(I,J,K\in\{A,B,C,D\})$$

(48)

where

	$\displaystyle\lambda=\frac{2+\sqrt{1-p^{2}}}{4}$		(49)
	$\displaystyle\lvert\psi_{\pm}\rangle=\frac{1}{N_{\pm}}\left[\mu_{\pm}\lvert 000\rangle-e^{-i\varphi}\lvert 111\rangle-\nu_{\pm}e^{i\varphi}\left(\lvert 001\rangle+\lvert 010\rangle+\lvert 100\rangle\right)\right]$

with

	$\displaystyle N_{\pm}^{2}=\frac{(1+p)(3-p)\pm(3+p)\sqrt{1-p^{2}}}{2p^{2}}\hskip 14.22636pt\mu_{\pm}=\frac{2(1-p)\pm\sqrt{1-p^{2}}}{\sqrt{2p(1-p)}}$		(50)
	$\displaystyle\hskip 85.35826pt\nu_{\pm}=\frac{(3+p)(1-p)\pm(3-p)\sqrt{1-p^{2}}}{2p(2\pm\sqrt{1-p^{2}})}.$

The residual entanglement for $\lvert\psi_{\pm}\rangle$ is

$$\tau(\psi_{\pm})=\frac{4}{N_{\pm}^{4}}\sqrt{\mu_{\pm}^{4}+16\nu_{\pm}^{6}+8\mu_{\pm}^{2}\nu_{\pm}^{3}\cos 4\varphi}.$$

(51)

Thus, the spectral decomposition (48) indicates that the residual entanglement for $\rho_{IJK}$ satisfies

$$\tau(\rho_{IJK})\leq\lambda\tau(\psi_{+})+(1-\lambda)\tau(\psi_{-}).$$

(52)

However, the analytic computation of the residual entanglement for $\rho_{IJK}$ is highly difficult even though it is rank-$2$ tensor. In appendix we try to describe why it is highly difficult. Therefore, we fail to compute $t_{2}(\rho_{4})$ analytically.