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Super quantum discord for general two qubit X states

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Abstract

The exact solutions of the super quantum discord are derived for general two qubit X states in terms of a one-variable function. Several exact solutions of the super quantum discord are given for the general X state over nontrivial regions of a seven-dimensional manifold.

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Acknowledgements

The research is partially supported by NSFC Grant (Nos. 11271138, 11531004) and Simons Foundation (No. 198129).

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Correspondence to Bing Yu.

Appendix

Appendix

Proof of Lemma 3.1

First we notice that \(G(\theta , z)\) is a strictly decreasing function of \(\theta \):

$$\begin{aligned} \frac{\partial G}{\partial \theta }=&-\frac{\tanh ^2x}{8R_+}\log _2\left( \frac{1+sz\tanh x+\sqrt{r^2+2rc_3z\tanh x+\theta \tanh ^2 x}}{1+sz\tanh x-\sqrt{r^2+2rc_3z\tanh x+\theta \tanh ^2x}}\right) \\&-\frac{\tanh ^2x}{8R_-}\log _2\left( \frac{1-sz\tanh x+\sqrt{r^2-2rc_3z\tanh x+\theta \tanh ^2x}}{1-sz\tanh x-\sqrt{r^2-2rc_3z\tanh x+\theta \tanh ^2x}}\right) \\&<0. \end{aligned}$$

Therefore, there are no interior critical points and extremal points must lie on the boundary of the domain. Since \(\frac{\partial G}{\partial \theta }<0\), we further conclude that \(\min G\) takes place at the largest value of \(\theta \) for some \(z\in [0, 1]\). As \(z_1^2+z_2^2+z^2=1\), we have that

$$\begin{aligned} \theta&=a^2+c_3^2z^2\\&=z_1^2|c_1|^2+z_2^2|c_2|^2+2z_1z_2|c_1\times c_2|+c_3^2z^2\\&\leqslant \left( \frac{|c_1|^2+|c_2|^2+\sqrt{(|c_1|^2-|c_2|^2)^2+4|c_1\times c_2|^2}}{2}\right) \left( z_1^2+z_2^2\right) +c_3^2z^2\\&=b^2+\left( c_3^2-b^2\right) z^2, \end{aligned}$$

where \(b^2=\frac{|c_1|^2+|c_2|^2+\sqrt{(|c_1|^2-|c_2|^2)^2+4|c_1\times c_2|^2}}{2}\).

For each fixed z, the maximum value \(b^2+(c_3^2-b^2)z^2\) can be achieved by appropriate \(z_1, z_2\). Therefore, \(\min G(z, \theta )=\min _{z\in [0,1]}G(z, b^2+(c_3^2-b^2)z^2)\), which is explicitly given in (3.14). \(\square \)

Proof of Theorem 3.2

We compute the derivative of F(z).

$$\begin{aligned} \begin{aligned}&F^{\prime }(z)\\&\quad =-\frac{1}{4}\left( s\tanh x\log _2\frac{1-A_+^2}{1-A_-^2} \quad \quad +\frac{rc_3\tanh x+(c_3^2-b^2)z\tanh ^2x}{1+sz\tanh x}\frac{1}{A_+}\log _2\frac{1+A_+}{1-A_+}\right. \\&\qquad +\left. \frac{-rc_3\tanh x+(c_3^2-b^2)z\tanh ^2x}{1-sz\tanh x}\frac{1}{A_-}\log _2\frac{1+A_-}{1-A_-}\right) \end{aligned} \end{aligned}$$
(5.1)

where \(H_{\pm }^{\prime }(z)=H_\pm ^{-1}(\pm rc_3\tanh x+(c_3^2-b^2)z\tanh ^2x)\) and \(\displaystyle A_{\pm }=\frac{H_\pm }{1\pm sz\tanh x}\in [0, 1]\). \(\square \)

Case (a) Since

$$\begin{aligned} \begin{aligned}&A_+^2-A_-^2=\frac{H_+^2}{(1+sz\tanh x)^2}-\frac{H_-^2}{(1-sz\tanh x)^2}\\&\quad =\frac{(1+s^2z^2\tanh ^2x)4rc_3z\tanh x-4sz\tanh x[(r^2+c_3^2z^2\tanh ^2x)+b^2\tanh ^2x(1-z^2)]}{(1-sz\tanh x)^2(1+sz\tanh x)^2} \end{aligned} \end{aligned}$$
(5.2)

Then the first term of \(F'(z)\le 0\) iff \(-s\tanh x(A_+^2-A_-^2)\geqslant 0\), which holds if \(s\tanh x\geqslant 0\) and \(rc_3\tanh x\leqslant 0\) or \(s\tanh x\leqslant 0\) and \(rc_3\tanh x\geqslant 0\).

Note that \(\displaystyle g(x)=\frac{1}{x}\ln \frac{1+x}{1-x}\) is a strictly increasing function on (0, 1), as

$$\begin{aligned} g'(x)&=-\frac{1}{x^2}\log _2\frac{1+x}{1-x}+\frac{2}{x\ln 2}\frac{1}{1-x^2}\\&=\frac{2}{x\ln 2}\left( \sum _{n=0}^{\infty }\frac{-x^{2n}}{2n+1}+\sum _{n=0}^{\infty }x^{2n}\right) >0. \end{aligned}$$

Therefore, \(A_+\geqslant A_-\) iff

$$\begin{aligned} \frac{1}{A_+}\ln \frac{1+A_+}{1-A_+}\geqslant \frac{1}{A_-}\ln \frac{1+A_-}{1-A_-}. \end{aligned}$$
(5.3)

(i) If \(s\tanh x\geqslant 0,rc_3\tanh x\leqslant 0\) and \(c_3^2-b^2\geqslant 0\), it follows from (5.2) that \(A_+\leqslant A_-\), then (5.3) implies that

$$\begin{aligned} F'(z)&\leqslant -\frac{1}{4}\left( \frac{rc_3\tanh x+(c_3^2-b^2)z\tanh ^2x}{1+sz\tanh x}\frac{1}{A_+}\log _2\frac{1+A_+}{1-A_+}\right. \\&\left. \qquad +\frac{-rc_3\tanh x+(c_3^2-b^2)z\tanh ^2x}{1-sz\tanh x}\frac{1}{A_-}\log _2\frac{1+A_-}{1-A_-}\right) \\&\leqslant -\frac{1}{4}\left( \frac{rc_3\tanh x+(c_3^2-b^2)z\tanh ^2x}{1+sz\tanh x}+\frac{-rc_3\tanh x+(c_3^2-b^2)z\tanh ^2x}{1+sz\tanh x}\right) \\&\quad \quad \frac{1}{A_+}\log _2\frac{1+A_+}{1-A_+} =-\frac{1}{2}\frac{(c_3^2-b^2)z\tanh ^2x}{1+sz\tanh x}\frac{1}{A_+}\log _2\frac{1+A_+}{1-A_+}\leqslant 0. \end{aligned}$$

(ii) If \(s\tanh x\geqslant 0,rc_3\tanh x\leqslant 0\) and \(src_3\le c_3^2-b^2\leqslant 0\) , we have that

$$\begin{aligned}&F'(z)\leqslant -\frac{1}{4}\left( \frac{rc_3\tanh x+(c_3^2-b^2)z\tanh ^2x}{1+sz\tanh x}\frac{1}{A_+}\log _2\frac{1+A_+}{1-A_+}\right. \\&\left. \quad \quad \qquad +\frac{-rc_3\tanh x+(c_3^2-b^2)z\tanh ^2x}{1-sz\tanh x}\frac{1}{A_-}\log _2\frac{1+A_-}{1-A_-}\right) \\&\quad \leqslant -\frac{1}{4}\left( \frac{rc_3\tanh x+(c_3^2-b^2)z\tanh ^2x}{1+sz\tanh x}+\frac{-rc_3\tanh x+(c_3^2-b^2)z\tanh ^2 x}{1-sz\tanh x}\right) \\&\quad \quad \quad \frac{1}{A_-}\log _2\frac{1+A_-}{1-A_-}=-\frac{1}{2}\frac{(-rc_3s+c_3^2-b^2)z\tanh ^2x}{(1-sz\tanh x)(1+sz\tanh x)}\frac{1}{A_-}\log _2\frac{1+A_-}{1-A_-}\leqslant 0. \end{aligned}$$

Case (b) Is treated in two subcases, (i) Suppose that \(s\tanh x\leqslant 0, rc_3\tanh x\geqslant 0\), and \(c_3^2-b^2\geqslant 0\), (ii) Suppose that \(s\tanh x\le 0,rc_3\tanh x\geqslant 0\) and \(src_3\le c_3^2-b^2\le 0\). We can solve the problem by the same method as Case (a).

Case (c) If \(s=r=0\), then

$$\begin{aligned} F'(z)= -2H_+^{-1}(c_3^2-b^2)z\tanh ^2x\log _2\frac{1+H_+}{1-H_+}. \end{aligned}$$

Therefore, \(\min F(z)\) is F(0) according to \(c_3^2\le b^2\) or not. Thus, the minimum of F(z) on \(z\in [0,1]\) is F(0).

Case (d) If \(s=rc_3, b^2=c_3^2\), and \(r^2+c_3^2\tanh ^x\pm rc_3\tanh x\ge 1\), it follows from (5.2) that the first term \(F'(z)\ge 0\).

Let \(k(z)=\frac{rc_3\tanh x}{H(z)}\log _2\frac{1+sz\tanh x z+H(z)}{1+sz\tanh x z -H(z)}\), where \(H(z)=\sqrt{r^2+b^2\tanh ^2x+2sz\tanh x }\). Then

$$\begin{aligned} F'(z)&\ge -\frac{rc_3\tanh x}{4H_+(z)}\log _2\frac{1+sz\tanh x +H_+}{1+sz\tanh x -H_+}\nonumber \\&\quad \;-\frac{-rc_3\tanh x}{4H_-(z)}\log _2\frac{1-sz\tanh x +H_-}{1-sz\tanh x -H_-}\nonumber \\&=-\frac{1}{4}(k(z)-k(-z)). \end{aligned}$$
(5.4)

As a function of z, we have that \(H'(z)=\frac{s\tanh x}{H(z)}, H''(z)=-\frac{s^2\tanh ^2x}{H^3(z)}\) and

$$\begin{aligned} k'(z)&=H''(z)\log _2\frac{1+sz\tanh x +H(z)}{1+sz\tanh x -H(z)}\\&\quad +\frac{s\tanh x}{H(z)\ln 2}\left( \frac{s\tanh x+H'(z)}{1+sz\tanh x +H(s)}-\frac{s\tanh x-H'(z)}{1+sz\tanh x -H(z)}\right) \\&=-\frac{s^2\tanh ^2x}{H^3(z)}\log _2\frac{1+sz\tanh x +H(z)}{1+sz\tanh x -H(z)}\\&\quad +\frac{2s^2\tanh ^2x}{H^2(z)\ln 2}\frac{1-sz\tanh x -r^2-b^2\tanh ^2x}{(1+sz\tanh x )^2-H^2(z)}\le 0, \end{aligned}$$

the inequality holds because

$$\begin{aligned} r^2+c_3^2\tanh ^2x\pm rc_3\tanh x\ge 1. \end{aligned}$$

Similarly \(k'(-z)\le 0\), thus \(-\frac{1}{4}(k'(z)+k'(-z))\ge 0\), which implies that \(F'(z)\ge 0\).

Therefore, the minimum of F(z) on \(z\in [0,1]\) is F(0).

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Jing, N., Yu, B. Super quantum discord for general two qubit X states. Quantum Inf Process 16, 99 (2017). https://doi.org/10.1007/s11128-017-1547-5

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