Determination of locally perfect discrimination for two-qubit unitary operations

Abstract

In the study of local discrimination for multipartite unitary operations, Duan et al. (Phys Rev Lett 100(2):020503, 2008) exhibited an ingenious expression: Any two different unitary operations $U_1$ and $U_2$ are perfectly distinguishable by local operations and classical communication in the single-run scenario if and only if 0 is in the local numerical range of $U_1^\dag U_2$. However, how to determine when 0 is in the local numerical range remains unclear. So it is generally hard to decide the local discrimination of nonlocal unitary operations with a single run. In this paper, for two-qubit diagonal unitary matrices V and their local unitary equivalent matrices, we present a necessary and sufficient condition for determining whether the local numerical range is a convex set or not. The result can be used to easily judge the locally perfect distinguishability of any two unitary operations $U_1$ and $U_2$ satisfying $U_1^\dag U_2=V$. Moreover, we design the corresponding protocol of local discrimination. Meanwhile, an interesting phenomenon is discovered: Under certain conditions with a single run, $U_1$ and $U_2$ such that $U_1^\dag U_2=V$ are locally distinguishable with certainty if and only if they are perfectly distinguishable by global operations.

Appendix

Here Lemma 6 of the main text will be proved.

Firstly, we show the sufficiency. It suffices to show that if $\phi <\theta <\psi$, then for any $y=a_1+a_2\hbox {e}^{i\phi }+a_3\hbox {e}^{i\psi }+a_4\hbox {e}^{i\theta }\in W(V)$ with $a_i\ge 0$ and $\sum \nolimits _{i=1}^4a_i=1$, there always exist $s,t\in [0,1]$ such that

$$\begin{aligned} st+s(1-t)\hbox {e}^{i\phi }+(1-s)t\hbox {e}^{i\psi }+(1-s)(1-t)\hbox {e}^{i\theta } =a_1+a_2\hbox {e}^{i\phi }+a_3\hbox {e}^{i\psi }+a_4\hbox {e}^{i\theta }. \end{aligned}$$

As to $\psi <\theta <\phi$, the proof is similar.

By the equal of the complex numbers, we can obtain

$$\begin{aligned}&s(1-t)\sin \phi +(1-s)t\sin \psi +(1-s)(1-t)\sin \theta \\&\quad = a_2\sin \phi +a_3\sin \psi +a_4\sin \theta , \\&\quad \qquad st+s(1-t)\cos \phi +(1-s)t\cos \psi +(1-s)(1-t)\cos \theta \\&\quad = a_1+a_2\cos \phi +a_3\cos \psi +a_4\cos \theta . \end{aligned}$$

They can be regarded as an equation about s and t.

$$\begin{aligned}&s(\sin \phi -\sin \theta )+t(\sin \psi -\sin \theta ) -st(\sin \phi +\sin \psi -\sin \theta )\\&\quad = a_2\sin \phi +a_3\sin \psi +(a_4-1)\sin \theta , \\&\quad \qquad s(\cos \phi -\cos \theta )+t(\cos \psi -\cos \theta ) +st(1-\cos \phi -\cos \psi +\cos \theta ) \\&\quad = a_1+a_2\cos \phi +a_3\cos \psi +(a_4-1)\cos \theta . \end{aligned}$$

Further, we have

$$\begin{aligned} As+Bt=C, \end{aligned}$$

where

$$\begin{aligned} A= & {} \sin \phi -\sin \theta +\sin (\psi -\phi )+\sin (\theta -\psi )\\= & {} -4\sin \frac{\psi }{2}\sin \frac{\phi -\theta }{2}\sin \frac{\phi -\psi +\theta }{2},\\ B= & {} \sin \psi -\sin \theta -\sin (\psi -\phi )+\sin (\theta -\phi )\\= & {} -4\sin \frac{\phi }{2}\sin \frac{\psi -\theta }{2}\sin \frac{\psi +\theta -\phi }{2},\\ C= & {} (a_1+a_2)\left[ \sin \phi +\sin (\theta -\psi )\right] +(a_1+a_3)\\&\left[ \sin \psi -\sin (\phi -\theta )\right] -(2a_1+a_2+a_3)\sin \theta \\&+\,(a_2-a_3)\sin (\psi -\phi ). \\ \end{aligned}$$

Case 1 If $A=0$, that is $\psi =\phi +\theta$, then $Bt=C$, which induces $\theta =\pi$ or $t=a_1+a_3$.

Let $\theta =\pi$, then $A=B=C=0$, which implies $s-t=a_2-a_3$, $s+t=1+a_1-a_4$. Thus $s=a_1+a_2,t=a_1+a_3$.

Let $t=a_1+a_3$, then $B\ne 0$ which induces

$$\begin{aligned} s=\frac{a_2(\sin \phi -\sin \theta )-a_1\sin (\phi +\theta )}{(a_2+a_4)(\sin \phi -\sin \theta )-(a_1+a_3)\sin (\phi +\theta )}. \end{aligned}$$

Since $0<\phi <\theta <\psi <2\pi$ and $0<\psi =\phi +\theta <2\pi$, then $\sin \frac{\phi -\theta }{2}<0,\sin \frac{\phi +\theta }{2}>0$. Further

$$\begin{aligned} (\sin \phi -\sin \theta )\sin (\phi +\theta ) =4\cos ^2\frac{\phi +\theta }{2}\sin \frac{\phi +\theta }{2}\sin \frac{\phi -\theta }{2}\le 0. \end{aligned}$$

Therefore $0\le s\le 1$. Now, we have shown that if $A=0$, then $s,t\in [0,1]$.

Case 2 Suppose that $A\ne 0$, then $s={(C-Bt)}/{A}$. Further we can get that

$$\begin{aligned} at^2+bt+c=0, \end{aligned}$$

where

$$\begin{aligned} a= & {} \sin (\phi -\psi )+\sin \psi -\sin (\phi -\theta )-\sin \theta \\= & {} -4\sin \frac{\phi }{2}\sin \frac{\theta -\psi }{2}\sin \frac{\phi -\psi -\theta }{2},\\ b= & {} -(a_1+a_2)\sin \phi +(-1+a_2-a_3)\sin (\phi -\psi )\\&-\,(a_1+a_3)\sin \psi +(1+a_1+a_3)\sin (\phi -\theta )\\&+\,(a_1+a_2-1)\sin (\psi -\theta )+(2a_1+a_2+a_3)\sin \theta , \\ c= & {} a_1\left[ \sin \phi -\sin \theta -\sin (\phi -\theta )\right] \\&+\,a_3\left[ \sin (\phi -\psi )-\sin (\phi -\theta )+\sin (\psi -\theta )\right] \\= & {} -4\sin \frac{\phi -\theta }{2}\left( a_1\sin \frac{\phi }{2}\sin \frac{\theta }{2}+a_3\sin \frac{\phi -\psi }{2}\sin \frac{\theta -\psi }{2}\right) , \\ d= & {} a+b+c\\= & {} -4\sin \frac{\psi }{2}\left( a_2\sin \frac{\phi }{2}\sin \frac{\psi -\phi }{2}+a_4\sin \frac{\theta }{2}\sin \frac{\psi -\theta }{2}\right) . \end{aligned}$$

If $0<\phi <\theta <\psi <2\pi$, then $c\ge 0$ and $a+b+c\le 0$. But both cannot be 0 at the same time, otherwise $\sum \nolimits _{i=1}^4a_i=0$, which is a contradiction.

Case 2.1 Let $a=0$. Since $c\ge 0$, $a+b+c\le 0$, and they cannot be 0 simultaneously; thus, $t=-c/b\in [0,1]$.

Case 2.2 Let $a\ne 0$, it need to be divided into three cases:

(I) $a>0,b\le 0,c\ge 0$; (II) $a<0,b\ge 0,c\ge 0$; (III) $a<0,b\le 0,c\ge 0$.

As to the cases (II) (III), $b^2-4ac\ge 0$. And for (I), $b^2-4ac=(d-a-c)^2-4ac\ge {(a-c)^2}\ge 0$.

In the following we denote $t_1=(-b+\sqrt{b^2-4ac})/2a$ and $t_2=(-b-\sqrt{b^2-4ac})/2a$. Further we will show that $t=t_2$ is the only solution in [0, 1].

According to (I)–(III), $t_2\ge 0$. Since $a+b+c\le 0$, we know that

if $a>0$, $a^2+ab+ac\le 0$, then $(2a+b)^2\le b^2-4ac$. If $2a+b\le \sqrt{b^2-4ac}$, then $t_1\ge 1$. If $-2a-b\le \sqrt{b^2-4ac}$, then $t_2\le 1$.

if $a<0$, then $a^2+ab+ac\ge 0$, and thus $(2a+b)^2\ge b^2-4ac$. Again, $2a+b=d+a-c<0$, then $t_2\le 1$.

In a word, if $A\ne 0$, there always exists a solution $t=-c/b$ or $t=t_2$ in [0, 1].

Similarly, when $0<\phi <\theta <\psi <2\pi$, there is also a solution s in [0, 1]. Note that if $B\ne 0$, we can get a equation about s,

$$\begin{aligned} a^{\prime }s^2+b^{\prime }s+c^{\prime }=0, \end{aligned}$$

where

$$\begin{aligned} a^{\prime }= & {} \sin \phi -\sin (\phi -\psi )-\sin (\psi -\theta )-\sin \theta \\= & {} -4\sin \frac{\psi }{2}\sin \frac{\phi -\theta }{2}\sin \frac{\phi -\psi +\theta }{2}, \\ b^{\prime }= & {} -(a_1+a_2)\sin \phi +(1+a_2-a_3)\sin (\phi -\psi )\\&-\,(a_1+a_3)\sin \psi -(1-a_1-a_3)\sin (\phi -\theta )\\&+\,(1+a_1+a_2)\sin (\psi -\theta )+(2a_1+a_2+a_3)\sin \theta , \\ c^{\prime }= & {} a_1\left[ \sin \psi -\sin \theta -\sin (\psi -\theta )\right] \\&+\,a_2\left[ \sin (\phi -\theta )-\sin (\phi -\psi )-\sin (\psi -\theta )\right] \\= & {} -4\sin \frac{\psi -\theta }{2}\left( a_1\sin \frac{\psi }{2}\sin \frac{\theta }{2}+a_2\sin \frac{\phi -\psi }{2}\sin \frac{\phi -\theta }{2}\right) , \\ d^{\prime }= & {} a^{\prime }+b^{\prime }+c^{\prime } \\= & {} -4\sin \frac{\phi }{2}\left( a_3\sin \frac{\psi }{2}\sin \frac{\phi -\psi }{2}+a_4\sin \frac{\theta }{2}\sin \frac{\phi -\theta }{2}\right) . \end{aligned}$$

Now we turn to consider the necessity. We only need to show that when $\phi <\psi <\theta , \psi <\phi <\theta , \theta <\phi <\psi$ and $\theta <\psi <\phi$, $W^{\mathrm{local}}(V)\subset W(V)$. Namely, there always exists a point, such as $\frac{1}{2}+\frac{1}{2}\hbox {e}^{i\theta }$, such that $\frac{1}{2}+\frac{1}{2}\hbox {e}^{i\theta }\in W(V)$, but $\frac{1}{2}+\frac{1}{2}\hbox {e}^{i\theta }\notin W^{\mathrm{local}}(V)$. First we will prove that when $0<\phi <\psi <\theta <2\pi$, $\frac{1}{2}+\frac{1}{2}\hbox {e}^{i\theta }\notin W^{\mathrm{local}}(V)$. And if $0<\theta <\psi <\phi <2\pi$, the proof is similar.

When $0<\phi <\psi <\theta <2\pi$, $0<\frac{\phi -\psi +\theta }{2}<\pi$. So $A>0$. Thus the equation about t must be obtained.

$$\begin{aligned} at^2+bt+c=0, \end{aligned}$$

where

$$\begin{aligned} a= & {} -4\sin \frac{\phi }{2}\sin \frac{\theta -\psi }{2}\sin \frac{\phi -\psi -\theta }{2}, \\ b= & {} -\frac{1}{2}\sin \phi -\sin (\phi -\psi )-\frac{1}{2}\sin \psi \\&+\,\frac{3}{2}\sin (\phi -\theta )-\frac{1}{2}\sin (\psi -\theta )+\sin \theta ,\\ c= & {} -2\sin \frac{\phi }{2}\sin \frac{\theta }{2}\sin \frac{\phi -\theta }{2}, \\ d= & {} a+b+c=-2\sin \frac{\psi }{2}\sin \frac{\theta }{2}\sin \frac{\psi -\theta }{2}. \end{aligned}$$

And $c>0,a+b+c>0$.

Case 1 If $a=0$, then $bt=-c$. When $0<\phi <\psi <\theta <2\pi$, $a=0$ is equivalent to $\psi +\theta =\phi +2\pi$. So $b=4\sin \frac{\psi }{2}\cos \frac{\psi }{2}\sin ^2\frac{\theta }{2}$. Thus b can be positive, negative and zero. Again $b+c>0,c>0$, it is easy to see that if $b>0,t=-c/b<0$; if $-c<b<0, t=-c/b>1$; if $b=0$, there is no t.

Therefore, if $a=0$, there is no solution t in [0, 1].

Case 2 Let $a\ne 0$, it can be divided into four cases:

(I) $a>0,b\ge 0,c>0$; (II) $a>0,b\le 0,c>0$;

(III) $a<0,b\ge 0,c>0$; (IV) $a<0,b\le 0,c>0$.

The notations $t_1=(-b+\sqrt{b^2-4ac})/2a$ and $t_2=(-b-\sqrt{b^2-4ac})/2a$ will also be used as follows.

Case 2.1 As to (III) (IV), $b^2-4ac>0$ and only $t_2>0$. Since $a+b+c>0$ and $a<0$, then $(2a+b)^2<b^2-4ac$. So $-2a-b<\sqrt{b^2-4ac}$, thus $t_2>1$. Therefore, for (III) (IV), there is no solution t in [0, 1].

As to (I) (II), if $b^2-4ac<0$, there is no t. So let $b^2-4ac\ge 0$. For (I), there is no solution t in [0, 1]. For (II), since $a+b+c>0$ and $a>0$, then $(2a+b)^2>b^2-4ac$. If $2a+b<0$, then $t_2>1$, which does not meet our requirement. If $2a+b>0$, then $t_1<1$.

For (I) (III) (IV), we have shown that there is no solution t in [0, 1]. For (II), we will suppose that both s and t have the solution in [0, 1] and find the contradiction as follows.

Case 2.2 If $a>0,b\le 0,c>0$. Assume that $b^2-4ac\ge 0$ and $2a+b>0$, then $t=t_1\in [0,1]$ and $a>c$. And let $0\le s\le 1$, where $s={(C-Bt)}/{A}, A>0, B=a>0$, and $C=c+d=a+b+2c>0$. So $C-A\le Bt\le C$. By $Bt\le C$, then $\sqrt{{{b}^{2}}-4ac}\le 2a+3b+4c$. Since $2a+3b+4c\ge 0$, then $0\ge b\ge (-2a-4c)/3$, thus ${{b}^{2}}-4ac\le 4(a-4c)(a-c)/9<0$, which is due to that $a-c>0$ and

$$\begin{aligned} a-4c=4\sin \frac{\phi }{2}\left( \sin \frac{\psi }{2}\sin \frac{\phi -\psi }{2}+\sin \frac{\theta }{2}\sin \frac{\phi -\theta }{2}\right) <0. \end{aligned}$$

This is a contradiction.

Therefore, when $a\ne 0$, not both s and t have the solution in [0, 1]. That is at least one of s and t has no solution in [0, 1].

Above all, if $0<\phi <\psi <\theta <2\pi$, then $\frac{1}{2}+\frac{1}{2}\hbox {e}^{i\theta }\in W(V)$, but $\frac{1}{2}+\frac{1}{2}\hbox {e}^{i\theta }\notin W^{\mathrm{local}}(V)$. That is $W^{\mathrm{local}}(V)\subset W(V)$.

Similarly, when $0<\psi <\phi <\theta <2\pi$ and $0<\theta <\phi <\psi <2\pi$, $\frac{1}{2}+\frac{1}{2}\hbox {e}^{i\theta }\in W(V)$ and $\frac{1}{2}+\frac{1}{2}\hbox {e}^{i\theta }\notin W^{\mathrm{local}}(V)$. But it should be noted that in these two situations we must consider the equation about s,

$$\begin{aligned} a^{\prime }s^2+b^{\prime }s+c^{\prime }=0, \end{aligned}$$

where

$$\begin{aligned} a^{\prime }= & {} -4\sin \frac{\psi }{2}\sin \frac{\phi -\theta }{2}\sin \frac{\phi -\psi +\theta }{2},\\ b^{\prime }= & {} -\frac{1}{2}\sin \phi +\sin (\phi -\psi )-\frac{1}{2}\sin \psi -\frac{1}{2}\sin (\phi -\theta )\\&+\,\frac{3}{2}\sin (\psi -\theta )+\sin \theta , \\ c^{\prime }= & {} -2\sin \frac{\psi }{2}\sin \frac{\theta }{2}\sin \frac{\psi -\theta }{2},\\ d^{\prime }= & {} -2\sin \frac{\phi }{2}\sin \frac{\theta }{2}\sin \frac{\phi -\theta }{2}. \end{aligned}$$

To sum up, we have shown that $W^{\mathrm{local}}(V)=W(V)$ if and only if $\phi <\theta <\psi$ or $\psi <\theta <\phi$. $\square$