Uncertainty Relation of Quantum Coherence Measure Based on Hellinger Distance

Abstract

In this paper, we focus on the uncertainty relations based on the quantum coherence measure of Hellinger distance. First, we discuss two kinds of quantum uncertainty relations for single qubit state based on two orthogonal bases. The lower bound of the coherence sum under two sets of orthogonal bases is related to the overlap of the two sets of bases and the mixedness of single-qubit state. We find that the lower bound $B_1$ decrease with the growing of the mixedness, and increase with the growing of the overlap of the two sets of bases. Second, we consider uncertainty relation in a subsystem for a two-qubit quantum state. The result shows that the lower bound increases with the growing of entanglement. In addition, we find that the sum of coherence measure will be affected by the existence of entanglement by numerical analysis. Finally, the uncertainty relations of d-dimensional quantum systems are studied.

Appendices

Appendix A: Proof of Theorem 1

Proof

By the Lemma 1. in Ref. [48], we can know that there is a value $\omega$ such that $|\langle x_{0}|\varphi \rangle |^{2}=\sin ^{2}(\theta +\omega )$. For convenience, replacing $\theta +\omega$ by $\theta '$, we obtain $|\langle x_{0}|\varphi \rangle |^{2}=\sin ^{2}(\theta ')$, $|\langle x_{0}|\varphi _{\perp }\rangle |^{2}=|\langle x_{1}|\varphi \rangle |^{2}=\cos ^{2}(\theta ')$ and $|\langle x_{1}|\varphi _{\perp }\rangle |^{2}=\sin ^{2}(\theta ')$. We have $\langle {x_0}\sqrt{\rho }|{x_0}\rangle = \sqrt{p}\sin ^2(\theta ')+\sqrt{1-p}\cos ^2(\theta ')$ and $\langle {x_1}|\sqrt{\rho }|{x_1}\rangle = \sqrt{p}\cos ^2(\theta ')+\sqrt{1-p}\sin ^2(\theta ')$, then

$$\begin{aligned}{} & {} \sum _{i}\langle {x_i}|\sqrt{\rho }|{x_i}\rangle ^2\nonumber \\{} & {} =\langle {x_0}|\sqrt{\rho }|{x_0}\rangle ^{2}+\langle {x_1}|\sqrt{\rho }|{x_1}\rangle ^{2}\nonumber \\{} & {} =[\sqrt{p}\sin ^2(\theta ')+\sqrt{1-p}\cos ^2(\theta ')]^2+[\sqrt{p}\cos ^2(\theta ')+\sqrt{1-p}\sin ^2(\theta ')]^2\nonumber \\{} & {} =1-\frac{1-2\sqrt{p(1-p)}}{2}\sin ^{2}(2\theta '). \end{aligned}$$

(22)

Replacing $\theta '$ by $\theta '+\varepsilon$, we can obtain $\sum _{j}\langle {y_j}|\sqrt{\rho }|{y_j}\rangle ^2=1-\frac{1-2\sqrt{p(1-p)}}{2}\sin ^{2}(2\theta '+2\varepsilon )$. Since $\sqrt{x}+\sqrt{y}\le \sqrt{2(x+y)}$,then

$$\begin{aligned} \begin{aligned}&\sqrt{\sum _{i}\langle {x_i}|\sqrt{\rho }|{x_i}\rangle ^2}+\sqrt{\sum _{j}^{}\langle {y_j}|\sqrt{\rho }|{y_j}\rangle ^2}\\&\le \sqrt{2(\sum _{i}\langle {x_i}|\sqrt{\rho }|{x_i}\rangle ^2+\sum _{j}\langle {y_j}|\sqrt{\rho }|{y_j}\rangle ^2)}\\&=\sqrt{4-[1-2\sqrt{p(1-p)}][\sin ^2(2\theta ')+\cos ^2(2\theta '+2\varepsilon )]}. \end{aligned} \end{aligned}$$

(23)

It has been proved in (11) of Ref. [48] that $\sin ^{2}(2\theta ')+\sin ^{2}(2\theta '+2\varepsilon )$ have the minimum value $2(1-c)$. Therefore, we have

$$\begin{aligned} C_H^{X}(\rho )+C_H^{Y}(\rho )= & {} 2\left( 1-\sqrt{\sum _{i}\langle {x_i}|\sqrt{\rho }|{x_i}\rangle ^2}\right) +2\left( 1-\sqrt{\sum _{j}\langle {y_j}|\sqrt{\rho }|{y_j}\rangle ^2}\right) \nonumber \\= & {} 4-2\left( \sqrt{\sum _{i}\langle {x_i}|\sqrt{\rho }|{x_i}\rangle ^2}+\sqrt{\sum _{j}\langle {y_j}|\sqrt{\rho }|{y_j}\rangle ^2}\right) \nonumber \\\ge & {} 4-2\sqrt{4-[1-2\sqrt{p(1-p)}][\sin ^2(2\theta ')+\cos ^2(2\theta '+2\varepsilon )]}\nonumber \\\ge & {} 4-2\sqrt{4-2[1-2\sqrt{p(1-p)}](1-c)}=B_1. \end{aligned}$$

(24)

It is easy to see that the first inequality becomes equality if and only if $\sin ^2(2\theta ')=\sin ^2(2\theta '+2\varepsilon )$, namely, $\theta '=\frac{\pi }{4}-\frac{\varepsilon }{2}$ or $\theta '=\frac{\pi }{2}-\frac{\varepsilon }{2}$. If $\varepsilon \le \frac{\pi }{4}$, the second inequality becomes equality when $\theta '=\frac{\pi }{2}-\frac{\varepsilon }{2}$; if $\varepsilon \ge \frac{\pi }{4}$, $\sin ^2(2\theta ')+\cos ^2(2\theta '+2\varepsilon )$ achieves a unique minimum value $2(1-c)$, when $\theta '=\frac{\pi }{4}-\frac{\varepsilon }{2}$. In summary, the inequality (11) becomes equality $\theta '=\frac{\pi }{2}-\frac{\varepsilon }{2}$ when $\varepsilon \le \frac{\pi }{4}$, or $\theta '=\frac{\pi }{4}-\frac{\varepsilon }{2}$ when $\varepsilon \ge \frac{\pi }{4}$.$\square$

Appendix B: Proof of Theorem 2

Proof

Let $p_{ij}=\langle {X_{ij}}|\sqrt{\rho _{AB}}|{X_{ij}}\rangle =\langle {X_{ij}}|{\psi _{AB}}\rangle \langle {\psi _{AB}}|{X_{ij}}\rangle =|\langle {X_{ij}}|{\psi _{AB}}\rangle |^2$. After calculation, we can get

$$\begin{aligned} {\begin{matrix} &{}p_{00}=|\langle {X_{00}}|{\psi _{AB}}\rangle |^2=|\sqrt{\lambda }\cos (\theta _1)\cos (\theta _2)+\sqrt{1-\lambda }\sin (\theta _1)\sin (\theta _2)|^2,\\ &{}p_{01}=|\langle {X_{01}}|{\psi _{AB}}\rangle |^2=|-\sqrt{\lambda }\cos (\theta _1)\sin (\theta _2)+\sqrt{1-\lambda }\sin (\theta _1)\cos (\theta _2)|^2,\\ &{}p_{10}=|\langle {X_{10}}|{\psi _{AB}}\rangle |^2=|-\sqrt{\lambda }\sin (\theta _1)\cos (\theta _2)+\sqrt{1-\lambda }\cos (\theta _1)\sin (\theta _2)|^2,\\ &{}p_{11}=|\langle {X_{11}}|{\psi _{AB}}\rangle |^2=|\sqrt{\lambda }\sin (\theta _1)\sin (\theta _2)+\sqrt{1-\lambda }\cos (\theta _1)\cos (\theta _2)|^2.\\ \end{matrix}} \end{aligned}$$

(25)

According to the properties of quantum states, we can get $\sum _{ij}p_{ij}=1$. The quantum coherence measure of $\psi _{AB}$ based on X can be written as

$$\begin{aligned} C_H^{X}(|{\psi _{AB}}\rangle )=2(1-\sqrt{\sum _{ij}p_{ij}^2}). \end{aligned}$$

(26)

Let $f(\theta _1,\theta _2)=\sum _{ij}p_{ij}^2$. We have

$$\begin{aligned} (\sum _{ij}p_{ij})^2=\sum _{ij}p_{ij}^2+2p_{00}p_{01}+2p_{10}p_{11}+2(p_{00}+p_{01})(p_{10}+p_{11})=1. \end{aligned}$$

(27)

Due to $x^2+y^2\ge \frac{(x+y)^2}{2}$,we can obtain $2p_{00}p_{01}+2p_{10}p_{11}\ge (\sqrt{p_{00}p_{01}}+\sqrt{p_{10}p_{11}})^2\ge (1-2\lambda )^2\sin ^2(\theta _2)\cos ^2(\theta _2)$. Then

$$\begin{aligned} \sum _{ij}p_{ij}^2+(\sqrt{p_{00}p_{01}}+\sqrt{p_{10}p_{11}})^2+2(p_{00}+p_{01})(p_{10}+p_{11})\le 1. \end{aligned}$$

(28)

Through calculation, we can directly obtain

$$\begin{aligned} \begin{aligned}&f(\theta _1,\theta _2)=\sum _{ij}p_{ij}^2\\&\le 1-(\sqrt{p_{00}p_{01}}+\sqrt{p_{10}p_{11}})^2-2(p_{00}+p_{01})(p_{10}+p_{11})\\&\le 1-(1-2\lambda )^2\sin ^2(\theta _2)\cos ^2(\theta _2)-2\lambda (1-\lambda )-2(1-2\lambda )^2\sin ^2(\theta _1)\cos ^2(\theta _1). \end{aligned} \end{aligned}$$

(29)

We replace $\theta _1$ by $\theta _1+\varepsilon$ and obtain

$$\begin{aligned} \begin{aligned} f(\theta _1+\varepsilon ,\theta _2)\le&1-(1-2\lambda )^2\sin ^2(\theta _2)\cos ^2(\theta _2)-\\&2\lambda (1-\lambda )-2(1-2\lambda )^2\sin ^2(\theta _1+\varepsilon )\cos ^2(\theta _1+\varepsilon ). \end{aligned} \end{aligned}$$

(30)

According to (29) and (30), we can obtain

$$\begin{aligned} \begin{aligned}&f(\theta _1,\theta _2)+f(\theta _1+\varepsilon ,\theta _2)\\&\le 2-4\lambda (1-\lambda )-2(1-2\lambda )^2\sin ^2(\theta _2)\cos ^2(\theta _2)-(1-2\lambda )^2(1-c)\\&=1+(1-2\lambda )^2[c-\frac{1}{2}\sin ^2(2\theta _2)]. \end{aligned} \end{aligned}$$

(31)

The second inequality comes from the proof of the Lemma 1. in Ref [48],i.e., $\sin ^2(2\theta )+\sin ^2(2\theta +2\varepsilon )\ge 2(1-c)$. Due to $\sqrt{x}+\sqrt{y}\le \sqrt{2(x+y)}$, we have

$$\begin{aligned} \begin{aligned} \sqrt{f(\theta _1,\theta _2)}+\sqrt{f(\theta _1+\varepsilon ,\theta _2)}&\le \sqrt{2(f(\theta _1,\theta _2)+f(\theta _1+\varepsilon ,\theta _2))}\\&\le \sqrt{2+[2c-\sin ^2(2\theta _2)](1-2\lambda )^2}. \end{aligned} \end{aligned}$$

(32)

According to the above discussion, one has

$$\begin{aligned} \begin{aligned} C_H^{X}(|{\psi _{AB}}\rangle )+C_H^{Y}(|{\psi _{AB}}\rangle )&=2[2-(\sqrt{f(\theta _1,\theta _2)}+\sqrt{f(\theta _1+\varepsilon ,\theta _2)})]\\&\ge 4-2\sqrt{2+[2c-\sin ^2(2\theta _2)](1-2\lambda )^2}=B_3. \end{aligned} \end{aligned}$$

(33)

$\square$

Appendix C: Proof of Theorem 3

A few lemmas are to be introduced before completing the proof of Theorem 3. Łukasz Rudnicki et al. [7] have has proved $\textbf{p}\oplus \textbf{q}\prec 1\oplus \textbf{w}$.

Lemma 1

For probability vectors p and q defined in (18) the following majorization relation holds

$$\begin{aligned} (p_1,p_2,\dots ,p_d,q_1,q_2,\dots ,q_d)\prec (1,s_1,s_2-s_1,\cdots s_d-s_{d-1}). \end{aligned}$$

(34)

That the right hand side of the above relation concerns the vector $\textbf{w}$ present in (17).

A real-valued function $\phi$ defined on a set $I\subset \mathbb {R}^n$ is said to be Schur-convex on I if $x\prec y$ on I $\Longrightarrow \phi (x)\le \phi (y)$. It has been proposed in the [49] that the determining theorem for Schur’s convex functions.

Lemma 2

Let real-valued function $F(x_1\cdots x_n):\mathbb {R}^n\rightarrow \mathbb {R}$ be continuously differentiable. Necessary and sufficient conditions for F to be Schur-convex on $\mathbb {R}^n$ are that F is symmetric on $\mathbb {R}^n$, and for all $i\ne j,$

$$\begin{aligned} (x_i-x_j)(\frac{\partial F}{\partial x_i}-\frac{\partial F}{\partial x_j})\ge 0. \end{aligned}$$

(35)

Let us now prove Theorem 3.

Proof

We define the function $f(\textbf{u})=\sum _{i}u_i^2$, where $\textbf{u}=(u_1,\dots , u_d)$ is any d-dimensional vector. $f(\textbf{u})$ can easily be determined to be symmetry and for any $i,j\in \{1,2,\dots ,d\}$, we have:

$$(u_i-u_j)(\frac{\partial f}{\partial u_i}-\frac{\partial f}{\partial u_j})=(u_i-u_j)(2u_i-2u_j)\ge 0.$$

Hence, $f(\textbf{u})$ is a Schur-convex function [49] and $\textbf{p}\oplus \textbf{q}\prec 1\oplus \textbf{w}$ [6, 7]. We can obtain

$$\begin{aligned} r_X+r_Y=f(\textbf{p}\oplus \textbf{q})\le f(1\oplus \textbf{w})=r_w, \end{aligned}$$

(36)

where $\textbf{p}\oplus \textbf{q}=(p_1,p_2,\dots ,p_d,q_1,q_2,\dots ,q_d)$ and $1\oplus \textbf{w}=(1,s_1,s_2-s_1,\cdots s_d-s_{d-1}).$ So,

$$\begin{aligned} C_H^X(|{\psi }\rangle )+C_H^Y(|{\psi }\rangle )= & {} 2(2-\sqrt{r_X}-\sqrt{r_Y})\nonumber \\\ge & {} 2(2-\sqrt{2(r_X+r_Y)})\nonumber \\\ge & {} 2(2-\sqrt{2r_w})=B_{d_1}. \end{aligned}$$

(37)

The first inequality holds because $\sqrt{x}+\sqrt{y}\le \sqrt{2(x+y)}$.$\square$

Appendix D: Proof of Theorem 4

Proof

Let $p_k^i=|\langle {x_i}|{\psi _k}\rangle |^2,q_k^j=|\langle {y_j}|{\psi _k}\rangle |^2$. According to the convexity of $x^2$ and Jensen’s inequality, we can obtain

$$\begin{aligned} {\begin{matrix} &{}r_X=\sum _{i}\left( \sum _{k}\lambda _k^{\frac{1}{2}}p_k^i\right) ^2\le \sum _{k}\lambda \lambda _k^{\frac{1}{2}}\sum _{i}(p_k^i)^2,\\ &{}r_Y=\sum _{j}\left( \sum _{k}\lambda _k^{\frac{1}{2}}q_k^j\right) ^2\le \sum _{k}\lambda \lambda _k^{\frac{1}{2}}\sum _{i}(q_k^j)^2,\\ &{}r_X+r_Y\le \sum _{k}\lambda \lambda _k^{\frac{1}{2}}\sum _{ij}[(p_k^i)^2+(q_k^j)^2]\le \lambda ^2r_w. \end{matrix}} \end{aligned}$$

(38)

Therefore

$$\begin{aligned} \begin{aligned} C_H^X(\rho )+C_H^Y(\rho )=&2(2-\sqrt{r_X}-\sqrt{r_Y})\\ \ge&2(2-\sqrt{2(r_X+r_Y)})\\ \ge&2(2-\sqrt{2\lambda ^2r_w})=B_{d_2}. \end{aligned} \end{aligned}$$

(39)

$\square$