Comment on 'Monogamy of multi-qubit entanglement using Rényi entropy'

In 2010, Kim et al [1] introduced the concept of Rényi-α entanglement as a generalization of the entanglement of formation, and it reduces to the standard entanglement of formation when α tends to 1 [1, 2]. Recently, it has been proved [3] that the squared Rényi-α entanglement obeys a general monogamy inequality in an arbitrary N-qubit mixed state. Correspondingly, we can construct the multipartite entanglement indicators in terms of Rényi-α entanglement which still work well even when the indicators based on the concurrence and entanglement of formation lose their efficacy. In [1] Kim et al first derived an analytic formula for the bipartite entanglement measure based on Rényi-α entanglement in two-qubit systems for $\alpha \geqslant 1$. Based on this analytic formula, they further established the monogamy inequalities of multi-qubit entanglement in terms of Rényi-α entanglement for all $\alpha\geqslant 2$. In [1] theorems 2 and 3 are fundamental results in establishing the monogamy inequalities in terms of Rényi-α entanglement. However, we find the proofs of theorems 2 and 3 contain some errors and we also present an alternative derivation to overcome this flaw. It is shown that the main conclusions remain unchanged despite the invalidity of the proofs of theorems 2 and 3.

During the proof of theorem 2 in [1], the author claim that equation (35) holds for any $\alpha>1$ due to equation (33) in [1]. However, this derivation is incorrect because equation 33 does not hold for any $\alpha>1$. Notice that the binomial series for $A_1^\alpha$ and $A_2^\alpha$ are given by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{q1} A_1^\alpha =& ~\left( {1 + \sqrt {1 - x^2 } } \right)^\alpha \nonumber \\ =&~ 1 + {\alpha}\sqrt {1 - x^2 } + \frac{{{\alpha}\left( {\alpha - 1} \right)}}{{2!}}\left( {\sqrt {1 - x^2 } } \right)^2 + \frac{{{\alpha}\left( {\alpha - 1} \right)\left( {\alpha - 2} \right)}}{{3!}}\left( {\sqrt {1 - x^2 } } \right)^3 \nonumber \\ &+ \frac{{{\alpha}\left( {\alpha - 1} \right)\left( {\alpha - 2} \right)\left( {\alpha - 3} \right)}}{{4!}}\left( {\sqrt {1 - x^2 } } \right)^4 + \cdots \nonumber \end{align} \tag{ 1 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{q2} A_2^\alpha =&~ \left( {1 - \sqrt {1 - x^2 } } \right)^\alpha \nonumber \\ =&~ 1 - {\alpha }\sqrt {1 - x^2 } + \frac{{{\alpha }\left( {\alpha - 1} \right)}}{{2!}}\left( {\sqrt {1 - x^2 } } \right)^2 - \frac{{{\alpha }\left( {\alpha - 1} \right)\left( {\alpha - 2} \right)}}{{3!}}\left( {\sqrt {1 - x^2 } } \right)^3 \nonumber \\ &+ \frac{{{\alpha }\left( {\alpha - 1} \right)\left( {\alpha - 2} \right)\left( {\alpha - 3} \right)}}{{4!}}\left( {\sqrt {1 - x^2 } } \right)^4 + \cdots \nonumber \\ \nonumber \end{align} \tag{ 2 }$$

and thus

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{q3} A_1^\alpha - A_2^\alpha \geqslant 2\alpha \sqrt {1 - x^2 } + C_1 \nonumber \end{align} \tag{ 3 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{q4} A_1^\alpha + A_2^\alpha \geqslant 2 + C_2. \nonumber \end{align} \tag{ 4 }$$

In the limit of $x \to 1^ -$, C₁ and C₂ can be written as $C_1 =$$\frac{{{\alpha}\left( {\alpha - 1} \right)\left( {\alpha - 2} \right)}}{3}\left( {\sqrt {1 - x^2 } } \right){\hspace{0pt}}^3 + o\left( {\left( {\sqrt {1 - x^2 } } \right){\hspace{0pt}}^{n} } \right), C_2 = {\alpha}\left( {\alpha - 1} \right)\left( {\sqrt {1 - x^2 } } \right){\hspace{0pt}}^2 + o\left( {\left( {\sqrt {1 - x^2 } } \right){\hspace{0pt}}^{m} } \right)$ with $n=5, 7, 9, \ldots, m=4, 6, 8, \ldots$. It is obviously that C₁ is non-negative under the condition ${{\alpha }\left( {\alpha - 1} \right)\left( {\alpha - 2} \right)}\geqslant 0$, and C₂ is non-negative under the condition ${\alpha }\left( {\alpha - 1} \right) \geqslant 0$. Thus we conclude that equation (35) in [1] does not hold for $2<\alpha<3$ since C₁ is a negative value in this case. Correspondingly, theorem 2 cannot be proved using the method in [1]. Furthermore, it is shown in figure 1 that $(A_1^{\alpha - 1} - A_2^{\alpha - 1})-2{\alpha} \sqrt {1 - x^2 } < 0$ for $2<\alpha<3, 0\leqslant x\leqslant 1$. For $\alpha\geqslant1, x \to 1^ -$, we have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{q5} & \Gamma _1 + \Gamma _2 = \frac{{(A_1^{\alpha - 1} - A_2^{\alpha - 1})(A_1^\alpha + A_2^\alpha)}}{{\sqrt {1 - x^2 } }} + x^2 (A_1^{\alpha - 1} - A_2^{\alpha - 1})^2 \nonumber \\ & = \frac{1}{{\sqrt {1 - x^2 } }}(2(\alpha - 1)\sqrt {1 - x^2 } + \frac{{(\alpha - 1)(\alpha - 2)(\alpha - 3)}}{3} (\sqrt {1 - x^2 })^3 + o(\sqrt {1 - x^2 })^{n})\nonumber \\ &\quad\times(2 + \alpha (\alpha - 1)(\sqrt {1 - x^2 })^2 + o(\sqrt {1 - x^2 })^{m}) \nonumber \\ &\quad + x^2 (2(\alpha - 1)\sqrt {1 - x^2 } + \frac{{(\alpha - 1)(\alpha - 2)(\alpha - 3)}}{3}(\sqrt {1 - x^2 })^3 + o(\sqrt {1 - x^2 })^{n})^2 \nonumber \\ &= 4(\alpha - 1) + 4x^2 (\alpha - 1)^2 (1 - x^2) + \frac{{\rm{4}}}{{\rm{3}}}{\rm{(}}{\alpha} - 1{\rm{) (2}}{\alpha}^2 - 4{\rm{}}{\alpha}{\rm{}} + 3{\rm{)}}(\sqrt {1 - x^2 })^2 + o(\sqrt {1 - x^2 })^n \nonumber \\ &\quad\geqslant 4(\alpha - 1) + 4x^2 (\alpha - 1)^2 (1 - x^2) \nonumber \end{align} \tag{ 5 }$$

where $n\geqslant4$. Thus we have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{q6} \frac{{{\rm d}^2 g_\alpha (x)}}{{{\rm d}x^2 }} =&~ B(\Gamma _1 + \Gamma _2 - 4(\alpha - 1)x^{2\alpha - 2}) \nonumber \\ &\geqslant B(4(\alpha - 1) + 4x^2 (\alpha - 1)^2 (1 - x^2) - 4(\alpha - 1)x^{2\alpha - 2})\nonumber \\ &\geqslant 0. \nonumber \end{align} \tag{ 6 }$$

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Below we present a strict proof to show that ${{\rm d}^2 g_\alpha (x)}/{rm dx^2 }$ is non-negative and we summarize our result as the following lemma.

Lemma 1. The second-order derivative of $g_\alpha (x)$ is a non-negative function for $0 \leqslant x \leqslant 1, \alpha\geqslant 1$.

Proof. We first define a function $h_{\alpha}(x)={{\rm d}^2 g_\alpha (x)}/{{\rm d}x^2 }$ on the domain $\mathcal{D} = \left\{{\left( {\alpha, x} \right)\vert 0 \leqslant x \leqslant 1, \alpha\geqslant 1 } \right\}$. For the proof of the non-negativity of $h_{\alpha}(x)$, it is sufficient to analyze its maximal or minimal value on the domain $\mathcal{D}$. The critical points of $h_{\alpha}(x)$ satisfy the condition

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{q7} \nabla h_\alpha (x) = \left( {\frac{{\partial h_\alpha }}{{\partial \alpha }},\frac{{\partial h_\alpha }}{{\partial x}}} \right) = 0. \nonumber \end{align} \tag{ 7 }$$

In figures 2(a) and (b) we plot the solution to equations $\partial h_\alpha /\partial \alpha =0$ and $\partial h_\alpha /\partial {x} =0$, respectively. We can see clearly from figure 2 that they have no common solutions on the domain $\mathcal{D}$. Therefore, there are no critical points on the domain $\mathcal{D}$ and the maximal or minimal value of $h_{\alpha}(x)$ can arise only on the boundary of $\mathcal{D}$. Next, we consider the function value of $h_{\alpha}(x)$ on the boundaries of $\mathcal{D}$ and we divide the proof into the following cases.

• Case 1.  
  When x  =  0, we have $h_{\alpha}(x) =\alpha/2 \geqslant 0$.
• Case 2.  
  When $x \to 1$, we have $h_{\alpha}(x) =\alpha(\alpha-1)(2\alpha^2+ 2\alpha-3)/3$ which is always a non-negative monotonically increasing function of α for $\alpha \geqslant 1$.
• Case 3.  
  When $\alpha=1$, we have $h_{\alpha}(x) = 0$.Combining the above results, we have complete the proof of the lemma. □

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Similarly, the proof of theorem 3 in [1] is also incorrect because equation 49 does not hold for $2<\alpha<3$ as shown in figure 1. Thus the proof of theorem 3 in [1] is incorrect for all $\alpha\geqslant2, 0\leqslant x\leqslant1$. For the proof of the non-negative of the first-order derivative of $l_\alpha (x)$ defined in [1], we first consider the following function in the limit of $x \to 1^ -$. According to equation (1) and (2), we have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{q7:2} &\frac{{(A_1^{\alpha - 1} - A_2^{\alpha - 1})(A_1^\alpha + A_2^\alpha)}}{{\sqrt {1 - x^2 } }} + (A_1^{\alpha - 1} - A_2^{\alpha - 1})^2 \nonumber \\ & = \frac{1}{{\sqrt {1 - x^2 } }}(2(\alpha - 1)\sqrt {1 - x^2 } + \frac{{(\alpha - 1)(\alpha - 2)(\alpha - 3)}}{3}(\sqrt {1 - x^2 })^3 + o(\sqrt {1 - x^2 })^{n}) \nonumber \\ &\quad\times (2 + \alpha (\alpha - 1)(\sqrt {1 - x^2 })^2 + o(\sqrt {1 - x^2 })^{m}) \nonumber \\ &\quad+ (2(\alpha - 1)\sqrt {1 - x^2 } + \frac{{(\alpha - 1)(\alpha - 2)(\alpha - 3)}}{3}(\sqrt {1 - x^2 })^3 + o(\sqrt {1 - x^2 })^{n})^2 \nonumber \\ &= 4(\alpha - 1) + 4(\alpha - 1)^2 (\sqrt {1 - x^2 })^2 \nonumber \\ & \quad + \frac{{\rm{4}}}{{\rm{3}}}{\rm{(}}\alpha - 1{\rm{) (2}}\alpha ^2 - 4{\rm{}}\alpha {\rm{}} + 3{\rm{)}}(\sqrt {1 - x^2 })^2 + o(\sqrt {1 - x^2 })^n \nonumber \\ &\quad\geqslant 4(\alpha - 1) + 4(\alpha - 1)^2 (1 - x^2) \nonumber \end{align} \tag{ 8 }$$

with $n\geqslant4$. Then we can prove that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{q8} \frac{{\rm d}l_\alpha (x)}{{\rm d}x} =& ~B[\frac{{(A_1^{\alpha - 1} - A_2^{\alpha - 1})(A_1^\alpha + A_2^\alpha)}}{{\sqrt {1 - x^2 } }} + (A_1^{\alpha - 1} - A_2^{\alpha - 1})^2 ] - 4B(\alpha - 1)x^{2\alpha - 4} \nonumber \\ & \geqslant 4B[(\alpha - 1) + (\alpha - 1)^2 (1 - x^2) - (\alpha - 1)x^{2\alpha - 4} ] \nonumber \\ &\geqslant 4B[(\alpha - 1) + (\alpha - 1)(1 - x^2) - (\alpha - 1)] \nonumber \\ =&~ 4B(\alpha - 1)(1 - x^2) \geqslant 0. \nonumber \end{align} \tag{ 9 }$$

In order to prove that ${{\rm d}l_\alpha (x)}/{{\rm d}x} \geqslant 0$ for whole $0\leqslant x\leqslant 1$, we define a two-variable function $m_\alpha (x)={dl_\alpha (x)}/{{\rm d}x}$ on the domain $\mathcal{D}_0 = \left\{{\left( {\alpha, x} \right)\vert 0 \leqslant x \leqslant 1, \alpha\geqslant 2 } \right\}$. The critical points of $m_\alpha (x)$ satisfy the condition $\nabla m_\alpha (x) = (\partial m_\alpha (x) /\partial \alpha, \partial m_\alpha (x)/\partial x) = 0$. It is shown in figures 3(a) and (b) that they have no common solutions on the domain $\mathcal{D}_0$ which indicate that $m_\alpha (x)$ has no critical points on the domain $\mathcal{D}_0$. Then we consider the value of $m_\alpha (x)$ on the boundary of $\mathcal{D}_0$.

• Case 1.  
  When x  =  0, we have $m_{\alpha}(x) = 0$.
• Case 2.  
  When $x \to 1$, we have $m_{\alpha}(x) =2(\alpha^3- 4\alpha + 3)/3$ which is always a positive monotonically increasing function of α for $\alpha\geqslant2$.
• Case 3.  
  When $\alpha=2$, we have $m_{\alpha}(x) = 2x/(x^2-2){\hspace{0pt}}^2\geqslant 0$ for $0\leqslant x\leqslant1$.

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Thus we have proved that ${{\rm d}l_\alpha (x)}/{{\rm d}x} \geqslant 0$ on the domain $\mathcal{D}_0$, and we can complete the proof of theorem 3 with a similar procedure as Kim et al in [1].

We should note that the present manuscript concern different topics with [5] although their errors both come from the incorrect binomial series. [1] is mainly focused on the monogamy relation of multi-qubit systems while [4] is focused on the polygamy property. Furthermore, lemma 2 in [4] is not a generalization of theorem 3 in [1] and the details of the proofs are different. Correspondingly, their results imply different physical meanings since monogamy and polygamy properties are different kinds of entanglement constraint in multi-qubit systems.

In conclusion, we have pointed out that the proof of theorems 2 and 3 in [1] contain some errors and we also present an improved method to overcome this flaw. It should be stressed that the main conclusions of the original paper [1] remains unchanged.

This work was supported by the Anhui Provincial Natural Science Foundation under Grant Nos 1708085MA12, 1708085MA10, the Key Program of the Education Department of Anhui Province under Grant Nos KJ2016A583, KJ2017A922, the discipline top-notch talents Foundation of Anhui Provincial Universities under Grants Nos gxbjZD2017024, gxbjZD2016078, the Key Program for Excellent Young Talents in the University of Anhui Province No gxyqZD2016206.