Tomographic Witnessing and Holographic Quantifying of Coherence

Abstract

The detection and quantification of quantum coherence play significant roles in quantum information processing. We present an efficient way of tomographic witnessing for both theoretical and experimental detection of coherence. We prove that a coherence witness is optimal if and only if all of its diagonal elements are zero. Naturally, we obtain a bona fide homographic measure of coherence given by the sum of the absolute values of the real and the imaginary parts of the non-diagonal entries of a density matrix, together with its interesting relations with other coherence measures like $l_1$ norm coherence and robust of coherence.

Appendix

1.1 Proof of Corollary 1

(If) For all $\rho \in D_{W_1}$ we have that $0>tr(W_1\rho )= (1-\epsilon )tr(W_2\rho )+\epsilon tr(P\rho )$, which implies $tr(W_2\rho )<0$ and hence $\rho \in D_{W_2}$.

(Only if) We have $D_{W_1}\subseteq D_{W_2}$ if $W_2$ is finer than $W_1$. If $D_{W_1}=D_{W_2}$, then Lemma 1 (e) gives rise to that $W_1=W_2$ (i.e., $\epsilon =0$). If $D_{W_1}\subset D_{W_2}$, we have $W_1 \ne W_2$ due to $D_{W_1}\ne D_{W_2}$ by using Lemma 1 (e). Hence $tr(W_2\rho )\ne tr(W_1\rho )$. For all $\rho \in D_{W_1}$, we have $tr(W_1\rho )<0$. Combining with Lemma 1 (b), we get that $tr(W_2\rho )<tr(W_1\rho )$ and $\xi > 1$. Denote $P=(\xi -1)^{-1}(\xi W_1-W_2)$ and $\epsilon =1-1/\xi >0$. We have that $W_1=(1-\epsilon )W_2+\epsilon P$. It only remains to be shown that $P\ge 0$. But this follows from Lemma 1 (a-c) and the definition of $\xi$. One easily verifies that P is either not finer than $W_1$ or it is incoherent. $\Box$

1.2 Proof of Corollary 2

(If) According to Corollary 1, there is no coherence witness which is finer than W. Therefore, W is optimal.

(Only if) If $W'$ is a coherence witness, then according to Corollary 1 W is not optimal. $\Box$

1.3 Proof of Theorem 2

Clearly, $W_{\rho }$ is Hermitian and not positive. The diagonal elements of $W_{\rho }$ are 0, and $tr(W_{\rho }\delta )\ge 0$ (factually $tr(W_{\rho }\delta )=0$) for any incoherent state $\delta$.

Since $\sum _{i=0}^{d-1}\langle i|\rho |i\rangle |i\rangle \langle i|$ is a matrix with the same diagonal elements as $\rho$ and other elements 0, we have $tr(\rho ^2)>tr(\rho \sum _{i=0}^{d-1}\langle i|\rho |i\rangle |i\rangle \langle i|)$ and

$$ tr(\rho W_{\rho })=-tr(\rho ^2)+tr(\rho \sum _{i=0}^{d-1}\langle i|\rho |i\rangle |i\rangle \langle i|)<0. $$

$\square$

1.4 Proof of Theorem 3

A proper coherence measure $\mathscr {C}(\rho )$ for a quantum state $\rho$ should satisfy the following conditions:

• (C1) Nonnegativity. $\mathscr {C}(\rho )\ge 0$ for any quantum state $\rho$, and $\mathscr {C}(\rho )=0$ if and only if $\rho$ is an incoherent state.

• (C2a) Monotonicity. $\mathscr {C}(\rho )\ge \mathscr {C}(\Phi (\rho ))$ for all incoherent completely positive and trace-preserving (ICPTP) maps $\Phi (\rho )=\sum _n \hat{K}_n\rho \hat{K}_n^\dag$, where $\{\hat{K}_n\}$ is a set of Kraus operators, $\sum _n\hat{K}_n^\dag \hat{K}_n=\mathbb {I}$ and $\hat{K}_n\mathcal {I}\hat{K}_n^\dag \subset \mathcal {I}$.

• (C2b) Strong monotonicity (under selective measurements on average). $\mathscr {C}(\rho )\ge \sum _np_n\mathscr {C}(\rho _n)$, where $\rho _n=\frac{\hat{K}_n\rho \hat{K}_n^\dag }{p_n}$, $p_n=tr(\hat{K}_n\rho \hat{K}_n^\dag )$, $\sum _n\hat{K}_n^\dag \hat{K}_n=\mathbb {I}$ and $K_n\mathcal {I}K_n^\dag \subset \mathcal {I}$.

• (C3) Convexity. $\sum _np_n \mathscr {C}(\rho _n)\ge \mathscr {C}(\sum _n p_n\rho _n)$ for any ensemble of $\{p_n,\rho _n\}$ of a state $\rho$.

For the convinience of proof, we define a so-called holographic matrix norm. Denote $M_n$ the $n\times n$ complex matrices $M_{n,n}(C)$. We define the holographic norm of $A\in M_n$ by

$$\begin{aligned} \parallel A\parallel _h=\sum _{l,m=1}^n|a_{lm}^R|+\sum _{l,m=1}^n|a_{lm}^I|, \end{aligned}$$

(15)

where $a_{lm}^R$ and $a_{lm}^I$ are the real and the imaginary parts of $a_{lm}$, respectively. We clarify that (15) is not a true matrix norm but a pseudo matrix norm. Recall that a matrix norm $\Vert \cdot \Vert$ is a function satisfying the following properties: for $A,B\in M_n$,

$$\begin{aligned}&\text{(1) }&\Vert A\Vert \ge 0,~ \Vert A\Vert =0 \text{ iff } A=0 \text{(Nonnegative) } \\&\text{(2) }&\Vert c A\Vert =|c|\Vert A\Vert \text{ for } \text{ all } \text{ complex } \text{ c } \text{(Homoheneous) }\\&\text{(3) }&\Vert A+B\Vert \le \Vert A\Vert +\Vert B\Vert \text{(Triangle } \text{ inequality) } \\&\text{(4) }&\Vert A B\Vert \le \Vert A\Vert \Vert B\Vert \text{(Submultiplicative). } \end{aligned}$$

One can prove that $\Vert A\Vert _{h}$ defined in (15) satisfies the properties (1), (3) and (4), as well as the modified property (2a): $\Vert c A\Vert _{h}=|c|\Vert A\Vert _{h}$ for all real and pure imaginary numbers c, but not for arbitrary complex scalars c. We remark that we do not use this property for arbitrary complex scalars c in our proof. We call $\Vert A\Vert _{h}$ a pseudo matrix norm.

Properties (1), (2a) and (3) are easy to prove. We prove the property (4) as follows:

$$\begin{aligned} \Vert AB\Vert _{h}\le & {} \sum _{l,m=1}^{n}\left| \left( \sum _{j=1}^{n} a_{lj}b_{jm}\right) ^{R}\left| +\sum _{l,m=1}^{n}\right| \left( \sum _{j=1}^{n}a_{lj}b_{jm}\right) ^{I}\right| \\= & {} \sum _{l,m=1}^{n}\left| \sum _{j=1}^{n}\left( a_{lj}^{R} b_{jm}^{R}-a_{lj}^{I}b_{jm}^{I}\right) \left| +\sum _{l,m=1}^{n}\right| \sum _{j=1}^{n}\left( a_{lj}^{R}b_{jm}^{I}+a_{lj}^{I}b_{jm}^{R}\right) \right| \\\le & {} \sum _{l,m=1}^{n}\sum _{j=1}^{n}(|a_{lj}^{R}| |b_{jm}^{R}|+|a_{lj}^{I}| |b_{jm}^{I}|+|a_{lj}^{R}||b_{jm}^{I}|+|a_{lj}^{I}||b_{jm}^{R}|)\\= & {} \sum _{l,m=1}^{n}\sum _{j=1}^{n} (|a_{lj}^{R}|+ |a_{lj}^{I}|)(|b_{jm}^{R}|+|b_{jm}^{I}|)\\\le & {} \left( \sum _{l,j=1}^{n}(|a_{lj}^{R}|+|a_{lj}^{I}|)\right) \left( \sum _{j,m=1}^{n}(|b_{jm}^{R}|+|b_{jm}^{I}|)\right) \nonumber \\= & {} \Vert A\Vert _{h}\Vert B\Vert _{h}. \end{aligned}$$

Now we prove that the holographic measure of coherence $\mathscr {C}_h(\rho )$ satisfies the required conditions.

• (C1) Obviously, $\mathscr {C}_{h}(\rho )\ge 0$ for any quantum state $\rho$ and $\mathscr {C}_{h}(\rho )=0$ if and only if $\rho$ is incoherent.

• (C2b) A general state can be written as $\rho =\rho ^R+i\rho ^I$, where $\rho ^R$ and $\rho ^I$ are the real and imaginary parts of $\rho$, respectively. The holographic norm of coherence of $\rho$ can be written as

  $$\begin{aligned} \mathscr {C}_{h}(\rho )={\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|[\rho ^R]_{i,j}|+{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|[\rho ^I]_{i,j}|. \end{aligned}$$

  Therefore, we have

  $$\begin{aligned} \sum _np_n\mathscr {C}_{h}({\rho }_n)= & {} \sum _{n}p_{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}\bigg (|[\rho _{n}^{R}]_{i,j}|+|[\rho _{n}^{I}]_{i,j}|\bigg )\nonumber \\= & {} \sum _{n}p_{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|[\rho _{n}^{R}]_{i,j}|+\sum _{n}p_{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|[\rho _{n}^{I}]_{i,j}|\nonumber \\= & {} \sum _{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|p_{n}[\rho _{n}^{R}]_{i,j}|+\sum _{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|p_{n}[\rho _{n}^{I}]_{i,j}|\nonumber \\= & {} \sum _{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|[\hat{K}_{n}\rho _{n}^{R}\hat{K}_{n}^{\dag }]_{i,j}|+\sum _{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|[\hat{K}_{n}\rho _{n}^{I}\hat{K}_{n}^{\dag }]_{i,j}|\nonumber \\= & {} \sum _{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|\sum _{k,l}[\hat{K}_{n}]_{i,k}[\rho _{n}^{R}]_{k,l}[\hat{K}_{n}^{\dag }]_{l,j}|\nonumber \\&\quad +\sum _{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|\sum _{k,l}[\hat{K}_{n}]_{i,k}[\rho _{n}^{I}]_{k,l}[\hat{K}_{n}^{\dag }]_{l,j}|. \end{aligned}$$

  (16)

Note that for $k=l$ in (16), we have $\sum _{k}[\hat{K}_{n}]_{i,k}\rho _{k,k}^{R}[\hat{K}_{n}^{\dag }]_{k,j}=\sum _{k}\langle i|\hat{K}_{n}|k\rangle \rho _{k,k}^{R}\langle k|\hat{K}_{n}^{\dag }|j\rangle =\langle i|\hat{K}_{n}\Delta (\rho ^R)\hat{K}_{n}^{\dag }|j\rangle$, where $\Delta (\rho ^R)=\sum _{k}\rho _{k,k}^{R}|k\rangle \langle k|$. Due to the fact that $\Delta (\rho ^{R})\in \mathcal {I}$ and $\hat{K}_{n} \Delta (\rho ^{R}) \hat{K}_{n}^{\dagger }\in \mathcal {I}$, we obtain $\sum _{k}[\hat{K}_{n}]_{i,k}\rho _{k,k}^{R}[\hat{K}_{n}^{\dag }]_{k,j}=\delta _{i,j}(\hat{K}_{n}\Delta (\rho ^R)\hat{K}_{n}^{\dag })$. Hence, for the case $i\ne j$, we only need to consider the case of $k\ne l$. Then the term of the real part in (16) can be written as

$$\begin{aligned}&\sum _{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|\sum _{k, l}[\hat{K}_{n}]_{i, k}[\rho _{n}^{R}]_{k, l}[\hat{K}_{n}^{\dag }]_{l, j}|\nonumber \\&\quad =\sum _{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|{\sum }{\begin{array}{c} {k, l} \\ {k\ne l} \end{array}}[\hat{K}_{n}]_{i,k}[\rho _{n}^{R}]_{k,l}[\hat{K}_{n}^{\dag }]_{l,j}|\nonumber \\&\quad \le {\sum }{\begin{array}{c} {k, l} \\ {k\ne l} \end{array}}|[\rho _{n}^{R}]_{k,l}|\sum _{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|[\hat{K}_{n}]_{i,k}[\hat{K}_{n}^{\dag }]_{l,j}|\nonumber \\&\quad \le {\sum }{\begin{array}{c} {k, l} \\ {k\ne l} \end{array}}|[\rho _{n}^{R}]_{k,l}|\sum _{n}\sum _{i}|[\hat{K}_{n}]_{i,k}|\sum _{j}|[\hat{K}_{n}^{\dag }]_{l,j}|. \end{aligned}$$

(17)

It is easy to see that

$$\begin{aligned}&|\sum _{n}\sum _{i}|[\hat{K}_{n}]_{i,k}|\sum _{j}|[\hat{K}_{n}^{\dag }]_{l,j}|\nonumber \\&\quad \le \sqrt{\sum _{n}(\sum _{i}|[\hat{K}_{n}]_{i,k}|)^{2}\sum _{m}(\sum _{j}|[\hat{K}_{n}^{\dag }]_{l,j}|)^{2}}. \end{aligned}$$

(18)

Since

$$\begin{aligned}&\sum _{n}(\sum _{i}|[\hat{K}_{n}]_{i,k}|)^{2}\nonumber \\&\quad =\sum _{n}\sum _{i,j}|[\hat{K}_{n}]_{i,k}[\hat{K}_{n}^{\dag }]_{k,j}|\nonumber \\&\quad =\sum _{n}\sum _{i,j}|\langle i|\hat{K}_{n}|k\rangle \langle k|\hat{K}_{n}^{\dag }|j\rangle |\nonumber \\&\quad =\sum _{n}\sum _{i}|\langle i|\hat{K}_{n}|k\rangle \langle k|\hat{K}_{n}^{\dag }|j\rangle |\nonumber \\&\quad =\sum _{n}\sum _{i}|\langle k|\hat{K}_{n}^{\dag }|j\rangle \langle i|\hat{K}_{n}|k\rangle |\nonumber \\&\quad =1, \end{aligned}$$

(19)

and similarly, $\sum _{m}(\sum _{j}|[\hat{K}_{n}^{\dag }]_{l,j}|)^{2}=1$, we obtain

$$\begin{aligned} |\sum _{n}\sum _{i}|[\hat{K}_{n}]_{i,k}|\sum _{j}|[\hat{K}_{n}^{\dag }]_{l,j}|\le 1. \end{aligned}$$

(20)

From (17) and (20), we have

$$\begin{aligned} \sum _{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|\sum _{k,l}[\hat{K}_{n}]_{i,k}[\rho _{n}^{R}]_{k,l}[\hat{K}_{n}^{\dag }]_{l,j}|\le {\sum }{\begin{array}{c} {k, l} \\ {k\ne l} \end{array}}|[\rho _{n}^{R}]_{k,l}|. \end{aligned}$$

(21)

In a similar way, one can aslo prove that

$$\begin{aligned} \sum _{n}{\sum }{\begin{array}{c} {i, j} \\ {i\ne j} \end{array}}|\sum _{k,l}[\hat{K}_{n}]_{i,k}[\rho _{n}^{I}]_{k,l}[\hat{K}_{n}^{\dag }]_{l,j}|\le {\sum }{\begin{array}{c} {k, l} \\ {k\ne l} \end{array}}|[\rho _{n}^{I}]_{k,l}|. \end{aligned}$$

(22)

Substituting (21) and (22) into (16), we get

$$\begin{aligned} \sum _{n}p_{n}\mathscr {C}_{h}(\rho _{n})\le {\sum }{\begin{array}{c} {k, l} \\ {k\ne l} \end{array}}|[\rho ^R]_{i,j}|+{\sum }{\begin{array}{c} {k, l} \\ {k\ne l} \end{array}}|[\rho ^I]_{i,j}|=\mathscr {C}_{h}(\rho ), \end{aligned}$$

(23)

which completes the proof.

• (C3) That the holographic norm of coherence may only decrease under mixing can be proved directly by the triangle inequality $\parallel A+B\parallel _h\le \parallel A\parallel _h+\parallel B\parallel _h$ satisfied by the holographic norm.

• (C2a) By (C2b) and (C3), we have

  $$\begin{aligned} \mathscr {C}_h(\Phi ({\rho }))=\mathscr {C}_h(\sum _np_n{\rho }_n){\mathop {\le }\limits ^{\text {(C3)}}}\sum _np_n\mathscr {C}_h({\rho }_n) {\mathop {\le }\limits ^{\text {(C2b)}}}\mathscr {C}_{h}({\rho }), \end{aligned}$$

  where ${\rho }_n=\hat{K}_n{\rho } \hat{K}_n^\dag /p_n$ and $p_n=tr(\hat{K}_n{\rho } \hat{K}_n^\dag )$. $\square$