Quantum Discord and Logarithmic Negativity in the Generalized n-qubit Werner State

Abstract

Quantum Discord (QD) is a measure of the total quantum non-local correlations of a quantum system. The formalism of quantum discord has been applied to various two-qubit mixed states and it has been reported that there is a non-zero quantum discord even when the states are unentangled. To this end, we have calculated the Quantum Discord for a higher than two qubit mixed state, that is, the generalized n-qubit Werner state with a bipartite split. We found that the QD saturates to a straight line with a unit slope in the thermodynamic limit. Qualitative studies of entanglement between the two subsystems using logarithmic negativity revealed that the entanglement content between them increases non-uniformly with the number of qubits leading to its saturation. We have proved the above claims both analytically and numerically.

Appendices

Appendix A: Convexity of Quantum Discord

It can be seen from Fig. 1 ($\mathcal {D}(A:B)$ versus p) that the quantum discord is a convex function over the range of mixing probability p. We can prove analytically as well as numerically that the expression for quantum discord written in (A.1) below is a convex function for 0 < p < 1. To this end, we will use results from differential calculus which says that the function $\mathcal {D}(A:B)$ is a convex function in the aforesaid range of p if $\frac {\partial ^{2} \mathcal {D}(A:B)}{\partial p^{2}} > 0$ for all values of n. We start by writing the analytical expression of $\mathcal {D}(A:B)$ as,

$$ \begin{array}{@{}rcl@{}} \mathcal{D}(A:B) &=& 1 + (2^{n} - 1) \left( \frac{1-p}{2^{n}}\right)\log_{2} \left( \frac{1-p}{2^{n}}\right)+\left( \frac{1+(2^{n} -1)p}{2^{n}}\right) \\ && \log_{2} \left( \frac{1+(2^{n} -1)p}{2^{n}}\right) - (2^{n-1} - 1) \left( \frac{1-p}{2^{n-1}}\right)\log_{2} \left( \frac{1-p}{2^{n-1}}\right) \\ &&-\left( \frac{1+(2^{n-1} -1)p}{2^{n-1}}\right) \log_{2} \left( \frac{1+(2^{n-1} -1)p}{2^{n-1}}\right). \end{array} $$

(A.1)

Now we will rearrange the above equation in terms of new variables x, y and z as shown below,

$$ \begin{array}{@{}rcl@{}} x&=& 1-p, \end{array} $$

(A.2)

$$ \begin{array}{@{}rcl@{}} y&=& 1+ (2^{n} -1)p, \end{array} $$

(A.3)

$$ \begin{array}{@{}rcl@{}} z&=& 1 + (2^{n-1}-1)p. \end{array} $$

(A.4)

Based on (A.2–A.4), we have the following relations between x, y and z.

$$ \begin{array}{@{}rcl@{}} 2z &=& x+y, \end{array} $$

(A.5)

$$ \begin{array}{@{}rcl@{}} y &=& x+ 2^{n}p. \end{array} $$

(A.6)

Now, $\mathcal {D}(A:B)$ can be written in terms of x, y and z as,

$$ \begin{array}{@{}rcl@{}} \mathcal{D}(A:B) &=& 1 + (2^{n} -1)\frac{x}{2^{n}}\log_{2}\left( \frac{x}{2^{n}}\right) + \frac{y}{2^{n}}\log_{2}\left( \frac{y}{2^{n}}\right) - (2^{n-1}-1) \\ &&\frac{x}{2^{n-1}}\log_{2} \left( \frac{x}{2^{n-1}}\right) - \frac{z}{2^{n-1}}\log_{2}\left( \frac{z}{2^{n-1}}\right). \end{array} $$

(A.7)

We can use the property, $\log (1/u) = -\log (u)$ (for u > 0) and $\log _{2}2^{n} = n$ for further simplification of (A.7) to get,

$$ \begin{array}{@{}rcl@{}} \mathcal{D}(A:B) &=&1 + \left( \frac{2^{n} -1}{2^{n}}\right) x \left( \log_{2} x - n\right) + \left( \frac{y}{2^{n}}\right)(\log_{2} y -n) - \left( \frac{2^{n-1}-1}{2^{n-1}}\right) \\ &&x(\log_{2} x - (n-1)) - \left( \frac{z}{2^{n-1}}\right)(\log_{2} z - (n-1)), \end{array} $$

(A.8)

$$ \begin{array}{@{}rcl@{}} &=& 1+\left( \frac{2^{n} -1 }{2^{n}} - \frac{2^{n-1}-1}{2^{n-1}}\right)x\log_{2} x - \frac{xn(2^{n} - 1)}{2^{n}} + \left( \frac{y}{2^{n}}\right)\log_{2} y \\ &&- \frac{ny}{2^{n}} +(n-1)\left( \frac{2^{n-1}-1}{2^{n-1}}\right)x - \left( \frac{z}{2^{n-1}}\right)\log_{2} z + \left( \frac{n-1}{2^{n-1}}\right)z. \end{array} $$

(A.9)

The above equation can be further rearranged as follows,

$$ \begin{array}{@{}rcl@{}} \mathcal{D}(A:B) &=& 1+ x\left[\frac{-n(2^{n}-1)}{2^{n}} + \frac{(n-1)(2^{n-1}-1)}{2^{n-1}} \right] -\frac{ny}{2^{n}} + \left( \frac{n-1}{2^{n}}\right) \\ &&(2z) + \left( \frac{x}{2^{n}}\right)\log_{2}x + \left( \frac{y}{2^{n}}\right)\log_{2} y - \left( \frac{z}{2^{n-1}}\right)\log_{2}z. \end{array} $$

(A.10)

Using (A.5), we can substitute for 2z in terms of x and y in (A.10) to get,

$$ \begin{array}{@{}rcl@{}} \mathcal{D}(A:B) &=& 1+ x\left[\frac{-n(2^{n}-1)}{2^{n}} + \frac{(n-1)(2^{n-1}-1)}{2^{n-1}} \right] -\frac{ny}{2^{n}} + \left( \frac{n-1}{2^{n}}\right) \\ &&(x+y) + \left( \frac{x}{2^{n}}\right)\log_{2}x + \left( \frac{y}{2^{n}}\right)\log_{2} y - \left( \frac{z}{2^{n-1}}\right)\log_{2}z, \end{array} $$

(A.11)

$$ \begin{array}{@{}rcl@{}} &=&1+ x\left[\frac{-n(2^{n}-1)}{2^{n}} + \frac{(n-1)(2^{n}-2)}{2^{n}} + \frac{(n-1)}{2^{n}}\right] -\frac{ny}{2^{n}} \\ &&+ \frac{ny}{2^{n}} - \frac{y}{2^{n}} + \left( \frac{x}{2^{n}}\right)\log_{2}x + \left( \frac{y}{2^{n}}\right)\log_{2} y - \left( \frac{z}{2^{n-1}}\right)\log_{2}z. \end{array} $$

(A.12)

Again using (A.6), we can write y in terms of x in (A.12) and therefore, the above equation becomes,

$$ \begin{array}{@{}rcl@{}} \mathcal{D}(A:B) &=& x\left[\frac{-n(2^{n}-1)}{2^{n}} + \frac{(n-1)(2^{n}-2)}{2^{n}} + \frac{(n-1)}{2^{n}}\right]- \frac{(x + 2^{n}p)}{2^{n}} + 1 \\ && + \left( \frac{x}{2^{n}}\right)\log_{2} x + \left( \frac{y}{2^{n}}\right)\log_{2} y - \left( \frac{z}{2^{n-1}}\right)\log_{2} z, \end{array} $$

(A.13)

$$ \begin{array}{@{}rcl@{}} &=&x\left[\frac{-n(2^{n}-1)}{2^{n}} + \frac{(n-1)(2^{n}-2)}{2^{n}} + \frac{(n-1)}{2^{n}}\right]- \frac{x}{2^{n}} - p + 1 \\ && + \left( \frac{x}{2^{n}}\right)\log_{2} x + \left( \frac{y}{2^{n}}\right)\log_{2} y - \left( \frac{z}{2^{n-1}}\right)\log_{2} z, \end{array} $$

(A.14)

$$ \begin{array}{@{}rcl@{}} &=&x\left[\frac{-n(2^{n}-1)}{2^{n}} + \frac{(n-1)(2^{n}-2)}{2^{n}} + \frac{(n-1)}{2^{n}} - \frac{1}{2^{n}} + 1\right] \\ && + \left( \frac{x}{2^{n}}\right)\log_{2} x + \left( \frac{y}{2^{n}}\right)\log_{2} y - \left( \frac{z}{2^{n-1}}\right)\log_{2}z. \end{array} $$

(A.15)

It can be observed from (A.15) that the coefficient of x is independent of p and it is equal to zero for any value of n. Therefore, we can write the simplified expression for $\mathcal {D}(A:B)$ as follows,

$$ \begin{array}{@{}rcl@{}} \mathcal{D}(A:B) = \frac{1}{2^{n}}\left\{x\log_{2} x + y\log_{2} y - 2z\log_{2} z \right\}. \end{array} $$

(A.16)

The first order derivative of $\mathcal {D}(A:B)$ in (A.16) with respect to p is calculated to be,

$$ \begin{array}{@{}rcl@{}} \frac{\partial \mathcal{D}(A:B)}{\partial p} &=& \frac{1}{2^{n}}\left\{x\times\frac{1}{x\ln2}\left( \frac{\partial x}{\partial p}\right) + \left( \frac{\partial x}{\partial p}\right)\log_{2}x + y\times\frac{1}{y\ln2}\left( \frac{\partial y}{\partial p} \right)+\right. \\ &&\left. \left( \frac{\partial y}{\partial p}\right)\log_{2}y-2\left( z\times\frac{1}{z\ln2}\left( \frac{\partial z}{\partial p} \right)+\left( \frac{\partial z}{\partial p}\right)\log_{2}z\right)\right\}, \end{array} $$

(A.17)

$$ \begin{array}{@{}rcl@{}} &=&\frac{1}{2^{n}}\left\{\left( \frac{\partial x}{\partial p}\right)\left( \frac{1}{\ln2} + \log_{2}x\right) + \left( \frac{\partial y}{\partial p}\right)\left( \frac{1}{\ln2} + \log_{2}y\right) \right. \\ &&\left.- 2 \left( \frac{\partial z}{\partial p}\right)\left( \frac{1}{\ln2} + \log_{2}z\right) \right\}. \end{array} $$

(A.18)

The second order derivative of $\mathcal {D}(A:B)$ can be calculated using (A.18) as follows,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} \mathcal{D}(A:B)}{\partial p^{2} } &=& \frac{1}{2^{n}}\left\{ \left( \frac{1}{\ln2}+\log_{2}x\right)\left( \frac{\partial^{2}x}{\partial p^{2}}\right) + \left( \frac{\partial x}{\partial p}\right)\times \frac{1}{x\ln2}\left( \frac{\partial x}{\partial p}\right) +\right. \\ && \left. \left( \frac{1}{\ln2}+\log_{2}y\right)\left( \frac{\partial^{2}y}{\partial p^{2}}\right)+\left( \frac{\partial y}{\partial p}\right)\times \frac{1}{y\ln2}\left( \frac{\partial y}{\partial p}\right) -2\right. \\ && \left. \left( \frac{1}{\ln2}+\log_{2}z\right)\left( \frac{\partial^{2} z}{\partial p^{2}}\right) -2 \left( \frac{\partial z}{\partial p}\right)\times \frac{1}{z\ln2}\left( \frac{\partial z}{\partial p}\right) \right\}. \end{array} $$

(A.19)

Recall (A.2–A.4), we can compute the first and second order derivatives of x, y and z with respect to p occurring in (A.19) as,

$$ \begin{array}{@{}rcl@{}} \frac{\partial x}{\partial p} = -1 \qquad\qquad \text{and} \qquad\qquad \frac{\partial^{2} x}{\partial p^{2}} = 0, \end{array} $$

(A.20)

$$ \begin{array}{@{}rcl@{}} \frac{\partial y}{\partial p} = (2^{n} -1) \qquad\qquad \text{and} \qquad\qquad \frac{\partial^{2} y}{\partial p^{2}} = 0, \end{array} $$

(A.21)

$$ \begin{array}{@{}rcl@{}} \frac{\partial z}{\partial p} = (2^{n-1}-1) \qquad\qquad \text{and} \qquad\qquad \frac{\partial^{2} z}{\partial p^{2}}=0. \end{array} $$

(A.22)

Since all x, y and z are linear functions of p, it can be observed that all the second order partial derivatives of x, y and z in (A.19) are zero as shown in (A.20–A.22). Considering the above facts, (A.19) can be rewritten as,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} \mathcal{D}(A:B)}{\partial p^{2} } &=& \frac{1}{2^{n}\ln2}\left\{\frac{1}{x}\left( \frac{\partial x}{\partial p}\right)^{2}+\frac{1}{y}\left( \frac{\partial y}{\partial p}\right)^{2} -\frac{2}{z}\left( \frac{\partial z}{\partial p}\right)^{2}\right\}. \end{array} $$

(A.23)

Substituting the values of the first order derivatives of x, y and z from (A.20–A.22) in (A.23), we get,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} \mathcal{D}(A:B)}{\partial p^{2}} &=& \frac{1}{2^{n}\ln2}\left\{\frac{(-1)^{2}}{(1-p)} + \frac{(2^{n} -1)^{2}}{(1+(2^{n}-1)p)} - \frac{2(2^{n-1}-1)^{2}}{(1+(2^{n-1}-1)p)}\right\}. \end{array} $$

(A.24)

The task now reduces to show that the RHS of (A.24) is greater than zero. To this effect, the quantity inside the flower brackets in RHS of (A.24) can be expanded out as shown below,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} \mathcal{D}(A:B)}{\partial p^{2}}&=& \frac{\alpha}{\gamma}, \end{array} $$

(A.25)

where, α the numerator of (A.25) is given by,

$$ \begin{array}{@{}rcl@{}} \alpha &=& \left[(1+(2^{n}-1)p)(1+(2^{n-1}-1)p) + (2^{n} -1)^{2}(1-p)(1+(2^{n-1}-1)p)\right. \\ &&\left.- 2(2^{n-1}-1)^{2}(1-p)(1+(2^{n}-1)p)\right], \end{array} $$

(A.26)

and γ, the denominator of (A.25) is given by,

$$ \begin{array}{@{}rcl@{}} \gamma &=& 2^{n}\ln2(1-p)(1+(2^{n}-1)p)(1+(2^{n-1}-1)p). \end{array} $$

(A.27)

The denominator γ of (A.25) contains terms, (1 − p), (1 + (2ⁿ − 1)p) and (1 + (2^n− 1 − 1)p) which are positive in the range of p considered for any n, therefore, the denominator is a positive quantity. Since we need to prove that $\frac {\partial ^{2} \mathcal {D}(A:B)}{\partial p^{2}}>0$, which implies, we have to show that the numerator α should also be greater than zero. To this end, we can rewrite the numerator of (A.25) as a quadratic expression in p by grouping the appropriate terms as,

$$ \begin{array}{@{}rcl@{}} \left[(2^{n}-1)(2^{n-1}-1) - (2^{n} -1)^{2}(2^{n-1}-1) + 2(2^{n-1}-1)^{2}(2^{n}-1)\right]p^{2} + \\ \left[(2^{n} -1) + (2^{n-1}-1) - (2^{n} -1)^{2} + (2^{n}-1)^{2}(2^{n-1}-1) + 2(2^{n-1}-1)^{2} \right. \\ \left.- 2(2^{n-1}-1)^{2}(2^{n}-1)\right] p + 1 + (2^{n} -1)^{2} - 2(2^{n-1}-1)^{2}. \end{array} $$

(A.28)

Here we will first simplify the coefficient of p², then p and finally the constant term individually. Therefore, considering the coefficient of p² in (A.28) as shown below,

$$ \begin{array}{@{}rcl@{}} \lefteqn{\left[(2^{n}-1)(2^{n-1}-1) - (2^{n} -1)^{2}(2^{n-1}-1) + 2(2^{n-1}-1)^{2}(2^{n}-1)\right]} \\ &=& (2^{n}-1)(2^{n-1}-1)\left( 1-(2^{n}-1)+2(2^{n-1}-1)\right) \\ &=& (2^{n}-1)(2^{n-1}-1)\left( 1-2^{n} +1 +2^{n} -2\right) \\ &=& 0. \end{array} $$

(A.29)

Therefore, the coefficient of p² term is zero, now we consider the coefficient of p in (A.28) as follows,

$$ \begin{array}{@{}rcl@{}} \lefteqn{\left[(2^{n} -1) + (2^{n-1}-1) - (2^{n} -1)^{2} + (2^{n}-1)^{2}(2^{n-1}-1) + 2(2^{n-1}-1)^{2} \right.} \\ &&\left.- 2(2^{n-1}-1)^{2}(2^{n}-1)\right] \\ &=&\left[2^{n} -1 + 2^{3n-1} + 2^{n-1} - 2^{2n} - 2^{2n+1} - 2 + 2^{n+2} + 2^{n-1} -1 - 2^{3n-1} \right. \\ &&\left.- 2^{n+1} + 2^{2n+1}+2^{2n} + 2^{2} - 2^{n+2}\right] \\ &=& [2^{n} + 2.2^{n-1} - 2^{n+1}] = [2.2^{n} = 2^{n+1}] = [2^{n+1} - 2^{n+1}] \\ &=& 0. \end{array} $$

(A.30)

Therefore, the coefficient of p is also zero. Now we consider the constant term in (A.28), which is,

$$ \begin{array}{@{}rcl@{}} \lefteqn{1 + (2^{n} -1)^{2} - 2(2^{n-1}-1)^{2}} \\ &=& 1 + 2^{2n} +1 - 2^{n+1} - 2^{2n-1} - 2 + 2^{n+1} \\ &=& 2^{2n} - \frac{2^{2n}}{2} = 2^{2n-1}. \end{array} $$

(A.31)

Finally, the only term that remains in the numerator of (A.25) is 2^2n− 1, which is positive for the considered range of p as n ≥ 2 and where n is an integer. Hence, we have proved that,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} \mathcal{D}(A:B)}{\partial p^{2}} >0. \end{array} $$

(A.32)

Therefore, $\mathcal {D}(A:B)$ is a convex function in the considered range of p. The above proof is completely analytical, however, to reinforce our claim and get deeper insights, we have done numerical analysis of the convexity property of $\mathcal {D}(A:B)$ for any n and p in the considered range. To this end, the second order derivative $\frac {\partial ^{2} \mathcal {D}(A:B)}{\partial p^{2}}$ was calculated numerically for different values of n in the considered range of p and the plots in Fig. 3 show conclusively that $\mathcal {D}(A:B)$ is indeed a convex function.

Fig. 3

Numerical plot for the variation of the second order derivative of $\mathcal {D}(A:B)$ with respect to p for different values of n. The insets show more closely the convexity of $\mathcal {D}(A:B)$ in the split ranges of p as shown

Appendix B: Concavity of Logarithmic Negativity

Now we will prove that the logarithmic negativity, as given below in (B.1) is a concave function. Again by using differential calculus in the range of p such that $\frac {1}{1+2^{n-1}}<p<1$, if $\frac {\partial ^{2} \mathcal {N}_{L}}{\partial p^{2}} <0$ for all values of n, then $\mathcal {N}_{L}$ is a concave function. Here we have logarithmic negativity $\mathcal {N}_{L}$ dependent upon p in the range $\frac {1}{1+2^{n-1}}<p<1$. Considering the expression for logarithmic negativity in the above range of p we have,

$$ \begin{array}{@{}rcl@{}} \mathcal{N}_{L} = \log_{2}\left( \frac{(2^{n-1}+1)p + (2^{n-1}-1)}{2^{n-1}}\right). \end{array} $$

(B.1)

We have to calculate the second order derivative of $\mathcal {N}_{L}$ with respect to p. To this end, we already know the first order derivative of $\mathcal {N}_{L}$ with respect to p from (61) as,

$$ \begin{array}{@{}rcl@{}} \frac{\partial \mathcal{N}_{L}}{\partial p} &=&\frac{1}{\ln2\left( p+\left( \frac{2^{n-1}-1}{2^{n-1}+1}\right)\right)}. \end{array} $$

(B.2)

We can now differentiate (B.2) with respect to p, to obtain second order derivative of $\mathcal {N}_{L}$ with respect to p as,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} \mathcal{N}_{L}}{\partial p^{2}} &=& \frac{1}{\ln2}\left( \frac{-1}{\left( p + \left( \frac{2^{n-1}-1}{2^{n-1}+1}\right)\right)^{2}}\right). \end{array} $$

(B.3)

Since the denominator of the RHS in (B.3) is always positive for the considered range of p, it is transparent to see that the RHS of (B.3) is always negative. Therefore, it can be concluded that,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} \mathcal{N}_{L}}{\partial p^{2}} <0, \end{array} $$

(B.4)

for all values of p in the range $\frac {1}{1+2^{n-1}}<p<1$ and for any value of n. Thus the logarithmic negativity is proved to be a concave function. The above result is analytical, we also calculated numerically the second order derivative of (B.1) with respect to p, to verify our analytical results. We have plotted the numerical results in Fig. 4 above, from which it can be seen that the second order derivative of logarithmic negativity with respect to p is always negative for all values of p such that $\frac {1}{1+2^{n-1}}<p<1$ and for any value of n. Therefore, the analytical result is backed by the numerical results.

Fig. 4

Numerical plot for the variation of the second order derivative of logarithmic negativity ($\mathcal {N}_{L}$) with respect to p such that $\frac {1}{1+2^{n-1}}<p<1$, for different values of n, showing clearly the concavity of $\mathcal {N}_{L}$