Comment on "Optimal convex approximations of quantum states"

TL;DR

This paper provides complete analytical solutions for the optimal convex approximation of qubit states, correcting and supplementing previous incomplete results in the literature.

Contribution

It offers new, comprehensive analytical solutions for approximating qubit states, addressing gaps in prior work.

Findings

Corrected previous analytical solutions for qubit state approximation

Provided complete formulas for optimal convex mixing

Enhanced understanding of quantum state approximation methods

Abstract

In a recent paper, M. F. Sacchi [Phys. Rev. A 96, 042325 (2017)] addressed the general problem of approximating an unavailable quantum state by the convex mixing of different available states. For the case of qubit mixed states, we show that the analytical solutions in some cases are invalid. In this Comment, we present complete analytical solutions for the optimal convex approximation. Our solutions can be viewed as correcting and supplementing the results in the aforementioned paper.

Equations

$$B_3=\{|0⟩,|1⟩,|2⟩=\frac{\sqrt{2}}{2}(|0⟩+|1⟩),|3⟩=\frac{\sqrt{2}}{2}(|0⟩-|1⟩),$$

$$|4⟩=\frac{\sqrt{2}}{2}(|0⟩+\sqrt{-1}|1⟩),|5⟩=\frac{\sqrt{2}}{2}(|0⟩-\sqrt{-1}|1⟩)\}.$$

$$\rho=\left(\begin{array}[]{cc}1-a&k\sqrt{a(1-a)}e^{-i\phi}\\ k\sqrt{a(1-a)}e^{i\phi}&a\\ \end{array}\right)\;$$

$$p_0=1-a-u-v-t_1-t_2,$$

$$p_1=a-u-v-t_1-t_2,$$

$$p_2=2u+t_1,$$

$$p_3=t_1,$$

$$p_4=2v+t_2,$$

$$p_5=t_2,$$

$$D_{B_3}(\rho )=\frac{\sqrt{3}}{3}(⟨{\sigma }_x⟩+⟨{\sigma }_y⟩+⟨{\sigma }_z⟩-1),$$

$$p_0=1-4a/3-2u/3-2v/3,$$

$$p_2=2a/3-2v/3+4u/3,$$

$$p_4=2a/3-2u/3+4v/3,$$

$$p_1=p_3=p_5=0.$$

$$D_{B_3}(\rho )=\sqrt{⟨{\sigma }_y{⟩}^2+\frac{1}{2}(⟨{\sigma }_x⟩+⟨{\sigma }_z⟩-1{)}^2},$$

$$p_0=1-a-u,$$

$$p_2=a+u,$$

$$p_1=p_3=p_4=p_5=0.$$

$$D_{B_3}(\rho )=\sqrt{⟨{\sigma }_x{⟩}^2+\frac{1}{2}(⟨{\sigma }_y⟩+⟨{\sigma }_z⟩-1{)}^2}.$$

$$p_0=1-a-v,$$

$$p_4=a+v,$$

$$p_1=p_2=p_3=p_5=0.$$

$$D_{B_3}(\rho )=\sqrt{⟨{\sigma }_z{⟩}^2+\frac{1}{2}(⟨{\sigma }_y⟩+⟨{\sigma }_x⟩-1{)}^2}$$

$$p_2=1/2+u-v,$$

$$p_4=1/2-u+v,$$

$$p_0=p_1=p_3=p_5=0.$$

$$Min\{2\sqrt{\mid Det(\rho -\sum _i{p}_i|i⟩⟨i|)\mid }\},$$

$$f(p_0,p_1,p_2,p_3,p_4,p_5)$$

$$-\sum _{i=0}^5{\lambda }_i{p}_i-\lambda \sum _{i=0}^5{p}_i.$$

$$\nabla f=0,{\lambda }_i{p}_i=0,{\lambda }_i\ge 0,p_i\ge 0,\sum _{j=0}^5{p}_j=1$$

$$p_1+p_2/2+p_3/2+p_4/2+p_5/2+{\lambda }_0+\lambda -a=0,$$

$$p_0+p_2/2+p_3/2+p_4/2+p_5/2+{\lambda }_1+\lambda -1+a=0,$$

$$p_0/2+p_1/2+p_3+p_4/2+p_5/2+{\lambda }_2+\lambda -1/2+u,$$

$$p_0/2+p_1/2+p_2+p_4/2+p_5/2+{\lambda }_3+\lambda -1/2-u=0,$$

$$p_0/2+p_1/2+p_2/2+p_3/2+p_5+{\lambda }_4+\lambda -1/2+v=0,$$

$$p_0/2+p_1/2+p_2/2+p_3/2+p_4+{\lambda }_5+\lambda -1/2-v=0,$$

$${\lambda }_i{p}_i=0,{\lambda }_i\ge 0,p_i\ge 0,i=0,1,2,3,4,5,$$

$${\Sigma }_i{p}_i=1,$$

$$p_0=1-a-u-v-t_1-t_2,$$

$$p_1=a-u-v-t_1-t_2,$$

$$p_2=2u+t_1,$$

$$p_3=t_1,$$

$$p_4=2v+t_2,$$

$$p_5=t_2,$$

Full text

In Sec. III of Ref. Sacchi15 , the problem of optimally generating a desired quantum state $\rho$ by the given set of the eigenstates of all Pauli matrices was provided. Namely, consider the optimal convex approximation of a quantum state with respect to the set

[TABLE]

The optimal convex approximation of $\rho$ with respect to $B_{3}$ is defined as $D_{B_{3}}(\rho)=min\{\parallel\rho-\sum_{i}p_{i}\rho_{i}\parallel_{1}\}$, where $\rho_{i}=|i\rangle\langle i|$, $0\leq p_{i}\leq 1$, $\sum_{i}p_{i}=1$, the minimum is taken over all possible probability weights $\{p_{i}\}$, and $\parallel A\parallel_{1}$ denotes the trace norm of $A$, that is, $\parallel A\parallel_{1}=Tr\sqrt{A^{\dagger}A}=\sum_{i}s_{i}(A)$ with $\{s_{i}(A)\}$ representing the singular values of $A$. The optimal convex approximate set is given by $S(\rho^{opt})=\{\rho^{opt}|D_{B_{3}}(\rho)=\parallel\rho-\rho^{opt}\parallel_{1}\}$.

Here we point out that the analytical solution given in [Phy. Rev. A 96, 042325(2017)] is invalid in some cases. We first provide a simple example. Consider the target qubit $\rho$ given by

[TABLE]

with $a\in[0,1]$, $\phi\in[0,2\pi]$, and $k\in[0,1]$. If we set $a=1/2$, $k=1$, $\phi=\pi/4$, it is easily verified that the point belongs to the region of case (i) in Ref. Sacchi15 , that is, $k_{th}\equiv a/({\sqrt{a(1-a)}(\cos\phi+\sin\phi)})<k\leq a/({\sqrt{a(1-a)}})$. Then the optimal convex approximation and the corresponding optimal weights are given by Eq. (18) and (19) in Sacchi15 , respectively. However, if one substitutes $a=1/2$, $k=1$ and $\phi=\pi/4$ into Eq. (19) in Sacchi15 , one has $p_{0}=1-4a/3-2k\sqrt{a(1-a)}(\cos\phi+\sin\phi)/3=(1-\sqrt{2})/3<0$, which implies that the optimal probability is negative and this solution is invalid.

In the following, in terms of the method used in liang16 (see also the Karush-Kuhn-Tucker theorem and its conclusion in Forst , p46-60), we provide the complete analytical solution for the optimal convex approximation of a quantum state under $B_{3}$ distance and the corresponding optimal weights.

For simplicity, we denote $u=k\sqrt{a(1-a)}\cos\phi$, $v=k\sqrt{a(1-a)}\sin\phi$, where $k\in[0,1]$, $a\in[0,\frac{1}{2}]$ and $\phi\in[0,\pi/2]$. When $a-u-v\geq 0$, one has $D_{B_{3}}(\rho)=0$. The pertaining weights corresponding to $\rho_{i}$ are given by

[TABLE]

where $t_{1}$ and $t_{2}$ are arbitrary non-negative arguments such that $p_{1}\geq 0$. If $t_{1}=t_{2}=0$, then Eq. (Comment on “Optimal convex approximations of quantum states”) reduce to Eq. (14) in Ref. Sacchi15 . However, if one sets $t_{1}=a-u-v,t_{2}=0$ in (Comment on “Optimal convex approximations of quantum states”), one gets $p_{0}=1-2a$, $p_{1}=0$, $p_{2}=a+u-v$, $p_{3}=a-u-v$, $p_{4}=2v$ and $p_{5}=0$. This is another kind of decomposition which is different from the one in Ref. [1]. Thus, our decompositions can be viewed as a complete supplement to the results in Ref. Sacchi15 .

The previous complete analytical solution can be classified into the following four cases, see proof in Supplemental Material:

$i)$ If $a<u+v\leq(3-4a)/2$, $a-v+2u\geq 0$ and $a-u+2v\geq 0$, the optimal convex approximation of $\rho$ is given by

[TABLE]

with corresponding optimal weights

[TABLE]

$ii)$ If $a<u+v\leq(3-4a)/2$, $a-v+2u\geq 0$ and $a-u+2v<0$, the optimal convex approximation of $\rho$ is given by

[TABLE]

with the corresponding optimal weights

[TABLE]

$iii)$ If $a<u+v\leq(3-4a)/2$, $a-v+2u<0$ and $a-u+2v\geq 0$, the optimal convex approximation of $\rho$ is given by

[TABLE]

The related optimal weights are given by

[TABLE]

$iv)$ If $u+v>(3-4a)/2$, we have

[TABLE]

with the pertaining optimal weights

[TABLE]

Up to now, we have refined the conclusions in Sec. III of Ref. Sacchi15 . Particularly, we have added the case $iv)$ as a valid supplement. Moreover, we point out that the Fig. 2 in Ref. Sacchi15 is inaccurate in some areas. In the following, we plot the accurate $D_{B_{3}}(\rho)$ for fixed value of the phase parameter $\phi=\frac{\pi}{3}$, see Fig. 1.

As another related example, consider $k=1$ and $a=1/2$. According to Eq. (18) in Ref. Sacchi15 , we get that $D_{B_{3}}(\rho)$ is about 0.2113. From Eq. (19) of Sacchi15 , $p_{0}=(\sqrt{3}-1)/3<0$. In fact, according to (Comment on “Optimal convex approximations of quantum states”), the accurate $D_{B_{3}}(\rho)\approx 0.2588$, and the corresponding probability is $p_{0}=p_{1}=p_{3}=p_{5}=0$, $p_{2}=(1+\sqrt{3})/4$ and $p_{4}=(3-\sqrt{3})/4$. We plot the difference of Fig. 1 in our paper and Fig. 2 in Sacchi15 , see Fig. 2.

In summary, we have derived the complete solution for the optimal convex approximation of a qubit mixed state under $B_{3}$ distance. We have revised the problem related to the result for $a<u+v$ in Ref. Sacchi15 . In addition, if $a\geq u+v$, our decompositions are the complete supplement to the representative decompositions in Ref. Sacchi15 .

We would like to say that the idea of looking for the least distinguishable states is nice, and the condition Eq. (13) in Ref. Sacchi15 for exact convex decomposition is also correct. We would also point out that the discussion in the last section of Ref. Sacchi15 , on the case of many copies of quantum states, the non-additivity of the distance, and the role of correlations, maintain general validity.

I Appendix

We now provide a detail proof of the state classification in the main text. To find $D_{B_{3}}(\rho)=min\parallel\rho-\sum_{i}p_{i}\rho_{i}\parallel_{1}$ is equivalent to search for the solution of the following minimum,

[TABLE]

such that $p_{i}\geq 0$ and $\sum_{j}p_{j}=1$. Denote

[TABLE]

According to the Karush-Kuhn-Tucker Theorem Forst and the related conclusion (see page 46-60 in Forst ), the above question is equal to

[TABLE]

for $i=0,1,2,3,4,5$. One then obtains the following equations and inequalities

[TABLE]

where $u=k\sqrt{a(1-a)}\cos\phi$ and $v=k\sqrt{a(1-a)}\sin\phi$. From (I) we have

(1) If $p_{0}\neq 0,p_{1}\neq 0$, from $\lambda_{1}=\lambda_{2}=0$ we have $\lambda=0$ and $\lambda_{i}=0$ ($i=2,3,4,5$). Similarly, if $p_{2}\neq 0,~{}p_{3}\neq 0$ or $p_{4}\neq 0,~{}p_{5}\neq 0$ or at least four of $\{p_{i}\}$ are nonzero, (I) is equivalent to $\nabla f=0,$ $\lambda=0,$ $\lambda_{i}=0$ ($i=0,1,2,3,4,5$), $\Sigma_{i}p_{i}=1$. Thus we have

[TABLE]

where $t_{1}$ and $t_{2}$ are arbitrary non-negative numbers such that $p_{1}\geq 0$. In this case, $D_{B_{3}}(\rho)=0$, and the condition $p_{i}\geq 0$, $i=0,1,2,3,4,5$, is transformed to $a-u-v\geq 0$.

(2) Only three numbers of $\{p_{i}\}$ are nonzero. According to (1), $D_{B_{3}}(\rho)=0$ for $a-u-v\geq 0$. For $a-u-v<0$, that only three numbers of $\{p_{i}\}$ are nonzero results in the following cases, $(i)$: $p_{0}\neq 0$, $p_{2}\neq 0$, $p_{4}\neq 0$; $(ii)$: $p_{0}\neq 0$, $p_{2}\neq 0$, $p_{5}\neq 0$; $(iii)$: $p_{0}\neq 0$, $p_{3}\neq 0$, $p_{5}\neq 0$; $(iv)$: $p_{1}\neq 0$, $p_{2}\neq 0$, $p_{4}\neq 0$; $(v)$: $p_{1}\neq 0$, $p_{2}\neq 0$, $p_{5}\neq 0$; $(vi)$: $p_{1}\neq 0$, $p_{3}\neq 0$, $p_{5}\neq 0$. However, the solution of (I) exists only for the case $(i)$, that is,

[TABLE]

We obtain that $a<u+v\leq(3-4a)/2$, $a-v+2u\geq 0$ and $a-u+2v\geq 0$.

Now we illustrate that for the case $(iv)$, no solution exists. According to the assumption, $\lambda_{1}=0$, $\lambda_{2}=0$ and $\lambda_{4}=0$. Then $p_{1}=4a/3-1/3-2u/3-2v/3$, $p_{2}=2/3-2a/3+4u/3-2v/3$ and $p_{4}=2/3-2a/3-2u/3+4v/3$. Notice that $p_{1}=4a/3-1/3-2u/3-2v/3\geq 0$. One obtains $a<u+v\leq 2a-1/2$, namely, $a>1/2$, which results in a contradiction. One can similarly show that for the cases $(ii)$, $(iii)$, $(v)$ and $(vi)$, also no solutions exist.

(3) Now consider the region $a<u+v\leq(3-4a)/2$.

Case $i$: $a<u+v\leq(3-4a)/2$, $a-v+2u\geq 0$ and $a-u+2v<0$. For $p_{0}\neq 0$ and $p_{2}\neq 0$, the equation (7) has the following solution,

[TABLE]

Case $ii$: $a<u+v\leq(3-4a)/2$, $a-v+2u<0$ and $a-u+2v\geq 0$, that is, $p_{0}\neq 0$ and $p_{4}\neq 0$, the solution of (7) is given by

[TABLE]

(4) The last case, $u+v>(3-4a)/2$ and only two numbers of $\{p_{i}\}$ are nonzero. In this case, only for $p_{2}\neq 0$ and $p_{4}\neq 0$, one has the following solution,

[TABLE]

For the other 6 cases with only two nonzero $\{p_{i}\}$, there do not exist solutions. For example, let us consider the case $p_{1}\neq 0$ and $p_{2}\neq 0$. By $\lambda_{1}=0$ we have $\lambda\leq 0$. Thus, $p_{1}=a-u$, $p_{2}=1-a+u$, $\lambda=1/2-a/2-u/2$, which leads to a contradiction, $1-a\leq u\leq a$ and $p_{1}=0$.

Acknowledgments This work is supported by NSFC (11765016,11675113), Jiangxi Education Department Fund (KJLD14088) and Key Project of Beijing Municipal Commission of Education under No. KZ201810028042.