# Witnessing quantum coherence with prior knowledge of observables

Mao-Sheng Li  School of Mathematics, South China University of Technology,
Guangzhou 510641, China  Wen Xu  School of Mathematics, South China University
of Technology, Guangzhou 510641, China  Shao-Ming Fei<EMAIL_ADDRESS>School of Mathematical Sciences, Capital Normal University, Beijing 100048,
China  Max-Planck-Institute for Mathematics in the Sciences, 04103 Leipzig,
Germany  Zhu-Jun Zheng<EMAIL_ADDRESS>School of Mathematics, South China
University of Technology, Guangzhou 510641, China  Yan-Ling Wang
<EMAIL_ADDRESS>School of Computer Science and Technology, Dongguan
University of Technology, Dongguan, 523808, China

(August 28, 2024)

###### Abstract

Quantum coherence is the key resource in quantum technologies including faster
computing, secure communication and advanced sensing. Its quantification and
detection are, therefore, paramount within the context of quantum information
processing. Having certain priori knowledge on the observables may enhance the
efficiency of coherence detection. In this work, we posit that the trace of
the observables is a known quantity. Our investigation confirms that this
assumption indeed extends the scope of coherence detection capabilities.
Utilizing this prior knowledge of the trace of the observables, we establish a
series of coherence detection criteria. We investigate the detection
capabilities of these coherence criteria from diverse perspectives and
ultimately ascertain the existence of four distinct and inequivalent criteria.
These findings contribute to the deepening of our understanding of coherence
detection methodologies, thereby potentially opening new avenues for
advancements in quantum technologies.

##  I Introduction

Quantum coherence Streltsov17 ; Hu2018  is a fundamental phenomenon in quantum
mechanics, enabling quantum systems to exist in superpositions of multiple
states simultaneously. It plays a pivotal role in various quantum
technologies, from quantum computing Hillery16 , quantum metrology
Giovannetti04 ; Giovannetti11 , to nanoscale thermodynamics Gour15 ;
Lostaglio15 ; Narasimhachar15 ; Francica19  and energy transport in biological
systems Lloyd11 ; Huelga13 ; Romero14 . As such, coherence represents a
critical resource for various quantum information processing tasks.

Significant strides have been made toward the measurement of quantum coherence
based on numerous innovative approaches Baumgratz14 ; Aberg14 ; Yuan2015 ;
Streltsov2015 ; Du2015 ; Winter2016 ; Napoli16 ; Piani16 ; Qi2017 ; Bu2017 ;
Luo2017 ; Jin2018 ; Fang2018 ; Ringbauer18 ; Xi2018 ; Ma19 ; Xi2019 ;
Bischof2019 ; Yu2020 ; Xu2020 ; Cui2020 ; Li2021 ; Bischof2021 ; Sun2021 ;
Kim2021 ; Ray2022 ; Sun2022 ; Basiri2022 ; Sun-Yu2022 ; Yu-Yu2022 ; Liu2023 .
Quantum resource theory Aberg14 ; Winter2016 ; Napoli16 ; Bu2017 ; Bischof2019
has garnered significant attention, proving instrumental in advancing our
understanding of coherence. Under the quantum resource theory framework,
various coherence monotones and measures have been introduced, including the
relative entropy of coherence, the $\ell_{1}$ norm of coherence, the coherence
of formation, the geometric measure of coherence and the robustness of
coherence.

Coherence witnesses, akin to their entanglement witness Guhne09 ; Horodecki09
; Wu07 ; Hou10a ; Wang18  counterparts, have emerged as potent tools for
coherence detection in experimental settings Ren2017 ; Wang17 ; Zheng18 ;
Nie19  and coherence quantification in theoretical contexts. In contrast to
conventional methods that rely on state tomography, coherence witnesses can
directly identify the coherent states Ma2021 ; Wang21 ; Wang21b . It is worth
noting that Napoli et al. Napoli16  have made significant contributions to
estimating the lower bound on the robustness of coherence by utilizing the
coherence witnesses, emphasizing the importance of this concept in deepening
our understanding of quantum coherence. More recently, Wang et al. Wang21 ;
Wang21b  conducted a relatively comprehensive exploration of coherence
witnesses. Nevertheless, research on coherence witnesses remains relatively
limited. In practical applications, we often possess valuable prior knowledge
about the observables involved. This naturally raises an important question:
can such prior knowledge enhance the accuracy and effectiveness of coherence
detection?

In this paper, we systematically investigate coherence witnesses within the
framework of prior knowledge about observables. In Section II, we provide an
overview of coherence witnesses and unveiling a series of coherence criteria
under varying degrees of prior knowledge concerning the traces of observables.
In Section III, we analyze these coherence criteria from various perspectives,
including completeness, finite completeness, finite intersection and
inclusion. Finally, in Section IV we summarize our findings and outline
potential avenues for future researches.

##  II Coherence witnesses with prior knowledge of observables

We denote $\mathcal{H}$ a $d$-dimensional Hilbert space with computational
basis $\mathcal{B}:=\\{|i\rangle,\,i=1,2,\cdots,d\\}$, $\mathbb{H}$ the set of
all $d\times d$ Hermitian matrices, and $\mathbb{D}$ the set of all density
matrices (self-adjoint positive semidefinite matrices with trace 1). Let
$\Delta$ be the set of incoherent states with respect to the basis
$\mathcal{B}$, namely, $\Delta$ comprises all density matrices of the form,

$\delta=\sum_{i=1}^{d}\delta_{i}|i\rangle\langle i|.$ (1)

Hence, all the states within $\mathbb{D}\setminus\Delta$ are referred to as
coherent states.

By definition the set $\Delta$ of incoherent states is convex and compact.
From the Hahn-Banach theorem Edwards65 , there must exist a hyperplane which
separates an arbitrary given coherent state from the set of incoherent states.
A coherence witness is an Hermitian operator $W\in\mathbb{H}$ such that (i)
$\mathrm{Tr}(W\delta)\geq 0$ for all incoherent states $\delta\in\Delta$, and
(ii) there exists a coherent state $\rho$ such that $\mathrm{Tr}(W\rho)<0$.
The first condition implies that all the diagonals of $W$ are nonnegative,
while the second implies that $W$ has some negative eigenvalue. We denote
$\mathbb{H}_{\geq}$ the set of $d\times d$ Hermitian matrices with nonnegative
diagonal elements and $\Lambda_{-}$ the set with some negative eigenvalue. The
general coherence witnesses are then generally given by
$\mathbb{H}_{\geq}\cap\Lambda_{-}.$

In addition to the above coherent witnesses, there are other kinds of coherent
witnesses defined by strengthening the condition (i) to
$\mathrm{Tr}(W\delta)=0$ for all $\delta\in\Delta$, but relaxing the condition
(ii) to $\mathrm{Tr}(W\rho)\neq 0$ for some state $\rho$ Napoli16 ; Ren2017 ;
Wang21 . Here, the strengthen condition implies that all the diagonals of $W$
must be zero, while the relaxed condition allows for detecting more coherent
states by this witness. In other words, prior knowledge about the observables
may enable one to detect better the coherence.

For a usual coherence witness $W\in\mathbb{H}_{\geq}\cap\Lambda_{-}$, the
condition $\mathrm{Tr}(W\delta)=0$ for all incoherent states $\delta\in\Delta$
is equivalent to $\mathrm{Tr}[W]=0.$ This motivates us to define the set of
all coherence witnesses with the same trace. Therefore, for each real number
$x\geq 0,$ we introduce the set:

$\mathbb{W}_{x}:=\\{W\in\mathbb{H}_{\geq}\ \big{|}\ \mathrm{Tr}[W]=x\\}.$

The prior knowledge on the trace of witness can enhance the ability of
detecting coherence. For fixed $x>0$ we have for any
$W=(w_{ij})\in\mathbb{W}_{x}$ and
$\delta=\sum_{i=1}^{d}\delta_{i}|i\rangle\langle i|\in\Delta$,

$0\leq\mathrm{Tr}[W\delta]=\sum_{i=1}^{d}w_{ii}\delta_{i}\leq\sum_{i=1}^{d}w_{ii}=x.$

Therefore, if $W=(w_{ij})$ is priorly known in the set $\mathbb{W}_{x}$, then
either $\mathrm{Tr}[W\rho]<0$ or $\mathrm{Tr}[W\rho]>x$ implies the coherence
of $\rho$. Moreover, there do exist some $W=(w_{ij})\in\mathbb{W}_{x}$ and
density matrix $\rho$ such that $\mathrm{Tr}[W\rho]>x.$ In fact, for any
$0<r<x$, let

$W=(x-r)|1\rangle\langle 1|+\sqrt{rx}\big{(}|1\rangle\langle
2|+|2\rangle\langle 1|\big{)}+r|2\rangle\langle 2|$

and $W=\lambda_{+}|v_{+}\rangle\langle v_{+}|+\lambda_{-}|v_{-}\rangle\langle
v_{-}|$ be its spectral decomposition. As
$\lambda_{+}\lambda_{-}=(x-r)r-(\sqrt{rx})^{2}=-r^{2}<0$, we can always assume
$\lambda_{+}>0$ and $\lambda_{-}<0$. For $\rho=|v_{+}\rangle\langle v_{+}|$ we
have

$\mathrm{Tr}[W\rho]=\lambda_{+}=(\lambda_{+}+\lambda_{-})-\lambda_{-}=\mathrm{Tr}[W]-\lambda_{-}=x-\lambda_{-}>x.$

As a consequence, the prior knowledge on the trace of the observables indeed
extends the scope of coherent state detections (see Fig. 1).

Figure 1:  Classical coherence witness vs coherence witness with prior
knowledge of trace.

However, if we just know that the observables could not be traceless, i.e.,
such observables are choosing from the set
$\mathbb{W}_{>}:=\\{W\in\mathbb{H}_{\geq}\ \big{|}\ \mathrm{Tr}[W]>0\\}$, one
finds that such little prior knowledge of the trace of observables could not
help to enhance the coherence detection. That is, to ensure that the witness
$W$ detects the coherence of $\rho$, we still need to observe the violation of
$\mathrm{Tr}[W\rho]\geq 0$ as usual. For proving this, we only need to show
that

$\\{\mathrm{Tr}[W\delta]\ \big{|}\
W\in\mathbb{W}_{>},\delta\in\Delta\\}=\\{x\in\mathbb{R}\mid x\geq 0\\},$

that is, for any real number $x\geq 0$, there always exist
$W\in\mathbb{W}_{>}$ and $\delta\in\Delta$ such that $\mathrm{Tr}[W\delta]=x$.
If $x>0$, choosing any $W\in\mathbb{W}_{>}$ with $\mathrm{Tr}[W]=dx$ and
$\delta=\frac{\mathbb{I}_{d}}{d}$, we have

$\mathrm{Tr}[W\delta]=\frac{1}{d}\mathrm{Tr}[W]=x.$

If $x=0$, setting $W=|1\rangle\langle 1|+|1\rangle\langle 2|+|2\rangle\langle
1|$ and $\rho=|2\rangle\langle 2|$, we have $\mathrm{Tr}[W\rho]=0.$

Throughout this paper, we set $\mathcal{X}:=\\{r\in\mathbb{R}|r\geq
0\\}\cup\\{>,\geq\\}.$ For each $x\in\mathcal{X},$ we have set
$I_{x}:=\\{\mathrm{Tr}[W\delta]\ \big{|}\
W\in\mathbb{W}_{x},\delta\in\Delta\\}$ and $D_{x}:=\\{\mathrm{Tr}[W\rho]\
\big{|}\ W\in\mathbb{W}_{x},\rho\in\mathbb{D}\\}.$ For each $x\in\mathcal{X}$,
a corresponding coherence witness criterion can be formulated as follows.

Coherence Criterion: $(\mathbb{W}_{x},\ D_{x},\ I_{x})$. For any
$\rho\in\mathbb{D},$ if there exists some $W\in\mathbb{W}_{x},$ such that
$\mathrm{Tr}[W\rho]\in D_{x}\setminus I_{x},$ then $\rho$ is a coherent state.

For any $x\in\mathcal{X},$ it is directly to verify that $D_{x}=\mathbb{R}$.
From the above discussion, the set $I_{x}$ is classified into following three
classes, see Fig. 2:

  1. (a)

$I_{x}=\\{r\in\mathbb{R}\ \big{|}0\leq r\leq x\\}$ when $x$ is a positive real
number;

  2. (b)

$I_{x}=\\{r\in\mathbb{R}\ \big{|}\ r\geq 0\\}$ when $x$ is either $\geq$ or
$>$;

  3. (c)

$I_{0}=\\{0\\}$ when $x=0$.

Figure 2:  An intuitive view of the range of $I_{x}$ for
$x\in\\{r\in\mathbb{R}|r\geq 0\\}\cup\\{>,\geq\\}$.

To classify coherence criteria we define $\mathbb{C}_{x}[W]$ to be the set of
all coherent states that can be witnessed by $W$ in the setting
$(\mathbb{W}_{x},\ D_{x},\ I_{x})$, that is,
$\mathbb{C}_{x}[W]=\\{\rho\in\mathbb{D}\ |\ \mathrm{Tr}[W\rho]\in
D_{x}\setminus I_{x}\\}.$ Moreover, we denote
$\mathbb{W}_{x}^{c}:=\\{W\in\mathbb{W}_{x}\ \big{|}\
\mathbb{C}_{x}[W]\neq\emptyset\\}$, i.e., the set of all nontrivial coherence
witnesses of coherence criterion $(\mathbb{W}_{x},D_{x},I_{x})$. One finds
that $\mathbb{W}_{x}^{c}=\mathbb{W}_{x}\cap\Lambda_{-}$ (see the proof in
appendix A).

For $W\in\mathbb{H}_{\geq},$ if we set $x=\mathrm{Tr}[W]$, then $W$ belongs to
$\mathbb{W}_{x}$ and $\mathbb{W}_{\geq}.$ One finds that

$\mathbb{C}_{x}[W]=\mathbb{C}_{\geq}[W]\cup\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W].$

It is easy to check that $\mathbb{C}_{\geq}[W]$ is a convex set, i.e., if
$\mathrm{Tr}[W\rho]<0$ and $\mathrm{Tr}[W\sigma]<0$, then
$\mathrm{Tr}[W\left(t\rho+(1-t)\sigma\right)]<0$ for $t\in[0,1].$ However, if
$\mathbb{C}_{\geq}[W]\neq\emptyset$ and
$\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W]\neq\emptyset$, the set
$\mathbb{C}_{x}[W]$ is not a convex set. In this case, $\mathbb{C}_{x}[W]$ is
a disjoint union of two convex sets, see cases (A) and (C) of Fig. 3. In fact,
for any $\rho\in\mathbb{C}_{\geq}[W]$ and
$\sigma\in\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W]$, we must have
$\mathrm{Tr}[W\rho]<0$ and $\mathrm{Tr}[W\sigma]>x\geq 0$. Therefore, there
are must some $t_{*}\in(0,1)$ such that
$\mathrm{Tr}[W\left(t_{*}\rho+(1-t_{*})\sigma\right)]=0$, which indicates that
$t_{*}\rho+(1-t_{*})\sigma\notin\mathbb{C}_{x}[W].$

Figure 3:  An intuitive view of the detection range of each type of coherence
witnesses. Here the red line represents the set of incoherent states. The deep
orange area represents the coherent states that could be detected by classical
witness, i.e., $\mathbb{C}_{\geq}[W]$. The deep purple area represents the
coherent states that could be detected by the condition
$\mathrm{Tr}[W\rho]>x,$ namely, $\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W]$.

We say that two coherence criteria $(\mathbb{W}_{x},D_{x},I_{x})$ and
$(\mathbb{W}_{y},D_{y},I_{y})$ are equivalent if there is a bijection
$F:\mathbb{W}_{x}^{c}\rightarrow\mathbb{W}_{y}^{c}$ such that
$\mathbb{C}_{x}[W]=\mathbb{C}_{y}[F(W)]$ for any $W\in\mathbb{W}_{x}^{c}.$ It
is an intuition that the coherence criteria arising from knowing different
values of traces may be almost the same in detecting coherence. For any
positive real number $x,y$, the two criterions $(\mathbb{W}_{x},D_{x},I_{x})$
and $(\mathbb{W}_{y},D_{y},I_{y})$ are equivalent in the above sense. In fact,
we can define a map $F$ from $\mathbb{W}_{x}^{c}$ to $\mathbb{W}_{y}^{c}$ by
sending $W\in\mathbb{W}_{x}^{c}$ to $\frac{y}{x}W.$ Clearly,
$\frac{y}{x}W\in\mathbb{W}_{y}^{c}$ as $W$ has nonnegative diagonals and
$\mathrm{Tr}[\frac{y}{x}W]=\frac{y}{x}\mathrm{Tr}[W]=\frac{y}{x}x=y.$ It is
easy to verify that it is a bijection. For any $\rho\in\mathbb{D},$
$0\leq\mathrm{Tr}[W\rho]\leq x$ is equivalent to
$0\leq\mathrm{Tr}[\frac{y}{x}W\rho]\leq y.$ Therefore, we always have
$\mathbb{C}_{x}[W]=\mathbb{C}_{y}[F(W)].$

However, to fully classify the previous mentioned coherence criteria, we need
to study more properties on these criteria. After that, we will go back to
tackle with this problem.

##  III Properties and classification of coherence criteria

For each $x\in\mathcal{X}$, we have established the coherence criterion
$(\mathbb{W}_{x},\ D_{x},\ I_{x})$. It is naturally to ask whether these
criteria are complete in the sense that they detect all the coherent states,
namely, the following relation holds,

$\mathbb{D}\setminus\Delta=\bigcup_{W\in\mathbb{W}_{x}}\mathbb{C}_{x}[W].$

Concerning the completeness of our coherence criteria, we have the following
conclusion whose proof can be seen in appendix B.

######  Theorem 1 (Completeness).

For each $x\in\mathcal{X},$ the coherence criterion
($\mathbb{W}_{x},D_{x},I_{x}$) is complete.

So each coherence criterion presented here is complete. As consequences, we
can restate the coherence criteria by the following four classes (see also
Fig. 3):

  1. (A):

$(\mathbb{W}_{r},D_{r},I_{r})$ with $r$ a positive real number. A state
$\rho\in\mathbb{D}$ is coherent if and only if there exists
$W\in\mathbb{W}_{r}$ such that $\mathrm{Tr}[W\rho]<0$ or
$\mathrm{Tr}[W\rho]>r.$

  2. ($\mathrm{B}_{1}$):

$(\mathbb{W}_{>},D_{>},I_{>})$. A state $\rho\in\mathbb{D}$ is coherent if and
only if there exists $W\in\mathbb{W}_{>}$ such that $\mathrm{Tr}[W\rho]<0.$

  3. ($\mathrm{B}_{2}$):

$(\mathbb{W}_{\geq},D_{\geq},I_{\geq})$. A state $\rho\in\mathbb{D}$ is
coherent if and only if there exists $W\in\mathbb{W}_{\geq}$ such that
$\mathrm{Tr}[W\rho]<0.$

  4. (C):

$(\mathbb{W}_{0},D_{0},I_{0})$. A state $\rho\in\mathbb{D}$ is coherent if and
only if there exists $W\in\mathbb{W}_{0}$ such that $\mathrm{Tr}[W\rho]\neq
0.$

A coherence criterion ($\mathbb{W}_{x},D_{x},I_{x}$) is called finitely
completable if all the coherent states could be detected by a finite set of
coherent witnesses in $\mathbb{W}_{x}^{c}.$ That is, there exists a finite set
$\\{W_{i}\\}_{i=1}^{n}\subseteq\mathbb{W}_{x}^{c}$ such that

$\mathbb{D}\setminus\Delta=\bigcup_{i=1}^{n}\mathbb{C}_{x}[W_{i}].$

Otherwise, we call it finitely incompletable.

######  Theorem 2 (Characterization of finite completeness).

Let $x$ be an element in $\mathcal{X}.$ The coherence criterion
($\mathbb{W}_{x},D_{x},I_{x}$) is finitely completable if and only if $x$ is
$0$ or $\geq$.

See appendix C for the proof. From Theorem 2, only those criteria that include
all observables with traces zero are finitely completable. From this sense,
the types $(A)$ and $(B_{1})$ are quite different from the types $(B_{2})$ and
$(C).$ Although the criteria or the figures (presented in Fig. 3) of types
$(B_{1})$ and $(B_{2})$ seem the same, they are in fact inequivalent.

In the following we discuss the conditions when the finite intersections of
$\mathbb{C}_{x}[{W}]$ maybe empty, for which the case with $x$ being $\geq$
has been discussed in Ref. Wang21 .

######  Theorem 3 (Property of finite intersection).

Let $x\in\mathcal{X}$ and
$\mathcal{W}:=\\{W_{i}\\}_{i=1}^{n}\subseteq\mathbb{W}_{x}^{c}.$ Then the
following statements hold.

  1. (a)

For the case $x\in(0,\infty)$ and any subset $\mathcal{W}^{\prime}$ with
cardinality $|\mathcal{W}^{\prime}|\geq\lceil\frac{n}{2}\rceil$, if there
always exist $t_{W}>0$ for each $W\in\mathcal{W}^{\prime}$ and
$\sum_{W\in\mathcal{W}^{\prime}}t_{W}=1$ such that
$\sum_{W\in\mathcal{W}^{\prime}}t_{W}W\geq\boldsymbol{0}$, then
$\bigcap_{i=1}^{n}\mathbb{C}_{x}[{W_{i}}]=\emptyset.$

  2. (b)

If $x$ is $>$ or $\geq,$ then
$\bigcap_{i=1}^{n}\mathbb{C}_{x}[{W_{i}}]=\emptyset$ if and only if there is
some nonnegative $t_{i}\geq 0$ and $\sum_{i=1}^{n}t_{i}=1$ such that
$\sum_{i=1}^{n}t_{i}W_{i}$ is positive semidefinite.

  3. (c)

If $x=0,$ there exists a common coherent state that can be detected by all of
$W_{i}$ via criterion of type (C). That is, we always have

$\mathcal{C}_{0}[{\mathcal{W}}]:=\bigcap_{i=1}^{n}\mathbb{C}_{0}[{W_{i}}]\neq\emptyset.$

Moreover, the cardinality of $\mathcal{C}_{0}[{\mathcal{W}}]$ is infinity.
That is, given any set of finitely many coherent witnesses, they share with
infinitely many common coherent states.

The proof is in appendix D. From Theorem 3, we conclude that the coherence
criterion of type (C) is different from the other types. Moreover, we can also
conclude that the coherence criterion of type (A) is also different from type
(B) (that is, type $(\mathrm{B}_{1})$ and type $(\mathrm{B}_{2})$) by the
following examples.

Firstly, there do exist $W_{i}$’s $\in\mathbb{W}_{x}^{c}$ which satisfy all
the conditions of (a) in Theorem 3. Take $d=n=3$ for example,

$W_{1}=\left[\begin{array}[]{ccc}\frac{x}{2}&0&\frac{x}{6}\\\\[5.69054pt]
0&\frac{x}{2}&\frac{x}{6}\\\\[5.69054pt]
\frac{x}{6}&\frac{x}{6}&0\\\\[5.69054pt]
\end{array}\right],\leavevmode\nobreak\
W_{2}=\left[\begin{array}[]{ccc}0&\frac{x}{6}&\frac{x}{6}\\\\[5.69054pt]
\frac{x}{6}&\frac{x}{2}&0\\\\[5.69054pt]
\frac{x}{6}&0&\frac{x}{2}\\\\[5.69054pt]
\end{array}\right],\leavevmode\nobreak\
W_{3}=\left[\begin{array}[]{ccc}\frac{x}{2}&\frac{x}{6}&0\\\\[5.69054pt]
\frac{x}{6}&0&\frac{x}{6}\\\\[5.69054pt]
0&\frac{x}{6}&\frac{x}{2}\\\\[5.69054pt] \end{array}\right].$

One easily checks that $\frac{1}{2}W_{1}+\frac{1}{2}W_{2},$
$\frac{1}{2}W_{2}+\frac{1}{2}W_{3},$ $\frac{1}{2}W_{1}+\frac{1}{2}W_{3}$ and
$\frac{1}{3}W_{1}+\frac{1}{3}W_{2}+\frac{1}{3}W_{3}$ satisfy the assumed
conditions. Hence
$\mathbb{C}_{x}[W_{1}]\cap\mathbb{C}_{x}[W_{2}]\cap\mathbb{C}_{x}[W_{3}]=\emptyset.$
Moreover, for the case $d=n=2$, if we set

$W_{1}=\left[\begin{array}[]{cc}x&-\frac{x}{2}\\\\[5.69054pt]
-\frac{x}{2}&0\end{array}\right],\
W_{2}=\left[\begin{array}[]{cc}0&\frac{x}{2}\\\\[5.69054pt]
\frac{x}{2}&x\end{array}\right],\
\rho=\left[\begin{array}[]{cc}\frac{1}{4}&\frac{1}{3}\\\\[5.69054pt]
\frac{1}{3}&\frac{3}{4}\end{array}\right],$

then $\frac{1}{2}W_{1}+\frac{1}{2}W_{2}=x\mathbb{I}_{2}.$ Hence
$\mathbb{C}_{>}[W_{1}]\cap\mathbb{C}_{>}[W_{2}]=\mathbb{C}_{\geq}[W_{1}]\cap\mathbb{C}_{\geq}[W_{2}]=\emptyset.$
However, $\mathrm{Tr}[W_{1}\rho]=-\frac{x}{12}<0$ and
$\mathrm{Tr}[W_{2}\rho]=\frac{13}{12}x>x.$ Therefore,
$\rho\in\mathbb{C}_{x}[W_{1}]\cap\mathbb{C}_{x}[W_{2}].$

Now we are studying more about the inclusion and identity relations among the
family of sets $\mathbb{C}_{x}[{W}]$ where $W\in\mathbb{W}_{x}$.

######  Theorem 4 (Property of inclusion or identity).

Let $x\in\mathcal{X}$ and $W_{1},W_{2}\in\mathbb{W}_{x}^{c}.$ The following
statements hold.

  1. (a)

If $x\in(0,\infty),$ then $\mathbb{C}_{x}[{W_{1}}]=\mathbb{C}_{x}[{W_{2}}]$ if
and only if $W_{1}=W_{2}$ or $W_{1}+W_{2}=x\mathbb{I}_{d}$ (the statement is
true only if $d=2$).

  2. (b)

If $x$ is $>$ or $\geq$, $\mathbb{C}_{x}[{W_{1}}]=\mathbb{C}_{x}[{W_{2}}]$ if
and only if there exists $r\in\mathbb{R}_{+}:=\\{r\in\mathbb{R}\ \big{|}\
r>0\\}$ such that $W_{2}=rW_{1}.$ Moreover,
$\mathbb{C}_{x}[{W_{2}}]\subseteq\mathbb{C}_{x}[{W_{1}}]$ if and only if there
exists $a\in\mathbb{R}_{+}$ and a positive semidefintie operator $P$ such that
$W_{2}=aW_{1}+P$.

  3. (c)

If $x=0,$ then $\mathbb{C}_{0}[W_{1}]=\mathbb{C}_{0}[W_{2}]$ if and only if
there exist some $r\in\mathbb{R}\setminus\\{0\\}$ such that $W_{2}=rW_{1}$ (we
call such two witnesses $\mathbb{R}$ equivalent). If $W_{1}$ and $W_{2}$ are
not $\mathbb{R}$ equivalent, then
$\mathbb{C}_{0}[W_{1}]\not\subseteq\mathbb{C}_{0}[W_{2}]$ and
$\mathbb{C}_{0}[W_{2}]\not\subseteq\mathbb{C}_{0}[W_{1}].$

The proof is presented in appendix E. With the above results, we are ready to
give a full classification of the variant coherence criteria.

######  Theorem 5.

There are exactly four inequivalent classes among the variant coherence
criteria $\\{(\mathbb{W}_{x},D_{x},I_{x})\ \big{|}\ x\in\mathcal{X}\\}.$

###### Proof.

Firstly, we have shown that the criteria of type (A) are all equivalent.

Secondly, the criterion of type (C) is not equivalent to other types. In fact,
in the settings of (A), ($\mathrm{B}_{1}$) and ($\mathrm{B}_{2}$), there are
always two witnesses $W_{1}$ and $W_{2}$ such that
$\mathbb{C}_{x}[W_{1}]\cap\mathbb{C}_{x}[W_{2}]=\emptyset.$ However, this is
not the case when $x=0$.

Thirdly, criterion of type ($\mathrm{B}_{2}$), i.e.,
$(\mathbb{W}_{\geq},D_{\geq},I_{\geq})$, is not equivalent to the types (A)
and ($\mathrm{B}_{1}$). In fact, criterion of type ($\mathrm{B}_{2}$) is
finitely completable but the other twos not.

Moreover, $\mathbb{C}_{>}[W]$ are always convex for each
$W\in\mathbb{W}_{>}^{c}$. But this fact may not be true for
$\mathbb{C}_{x}[W]$ when $x>0$. Therefore, criteria of type (A) and ($B_{1}$)
are inequivalent. To conclude we have exactly four inequivalent classes of
coherence criteria. ∎

##  IV Conclusion and Discussions

Based on prior knowledge of observables, we have presented a series coherence
criteria which detect coherence better than the usual ones without prior
knowledge of observables. Moreover, through a systematic and rigorous study on
the properties of criteria such as completeness, finite completeness, finite
intersection and inclusion, we have singled out four classes of inequivalent
coherence criteria. These results help to the deepening of our understanding
of coherence detection methodologies, and thereby highlight on advancements in
quantum technologies.

There are also some interesting problems left such as the condition of
inclusion of $\mathbb{C}_{x}[W]$ when $x>0$ for the part (a) of Theorem 4.
Further exploration of coherence detection enriched by prior knowledge may
promise to unlock even greater potential. It would be also appealing to extend
our scheme to deal with the case of entanglement witnesses. Our research may
serve as a catalyst for future investigations on quantum coherence detection,
as well as detections of other resources like quantum correlations.

###### Acknowledgements.

This work is supported by National Natural Science Foundation of China (Grant
Nos. 12371458, 62072119, 12075159, 12171044), the Guangdong Basic and Applied
Basic Research Foundation under Grants No. 2023A1515012074, Key Research,
Development Project of Guangdong Province under Grant No.2020B0303300001, Key
Lab of Guangzhou for Quantum Precision Measurement under Grant
No.202201000010, the Academician Innovation Platform of Hainan Province, and
the Science and Technology Planning Project of Guangzhou under Grants No.
2023A04J1296.

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##  Appendix A Two Lemmas and a proof of the claim:
$\mathbb{W}_{x}^{c}=\mathbb{W}_{x}\cap\Lambda_{-}$

In order to prove the claim
$\mathbb{W}_{x}^{c}=\mathbb{W}_{x}\cap\Lambda_{-}$, we need the following
lemma.

######  Lemma 1.

If $W\in\mathbb{H}_{\geq}\setminus\\{\mathbf{0}\\}$ is not positive
semidefinite, then there exists some density matrices $\rho\in\mathbb{D}$ such
that $\mathrm{Tr}[W\rho]<0$. Moreover, there also exists positive density
matrix $\sigma\in\mathbb{D}$ such that $\mathrm{Tr}[W\sigma]=0.$

###### Proof.

By the given assumption, the matrix $W$ has both positive and negative
eigenvalues. Suppose the spectral decomposition of $W$ is

$W=\sum_{\lambda_{i}>0}\lambda_{i}|e_{i}\rangle\langle
e_{i}|+\sum_{\lambda_{j}<0}\lambda_{j}|e_{j}\rangle\langle
e_{j}|+\sum_{\lambda_{k}=0}\lambda_{k}|e_{k}\rangle\langle e_{k}|,$

where
$\\{|e_{i}\rangle\\}_{\lambda_{i}>0}\cup\\{|e_{j}\rangle\\}_{\lambda_{j}<0}\cup\\{|e_{k}\rangle\\}_{\lambda_{k}=0}$
is an orthonormal basis of the system $\mathcal{H}.$ For any fixed $j$ such
that $\lambda_{j}<0,$ the state $\rho=|e_{j}\rangle\langle e_{j}|$ satisfies
$\mathrm{Tr}[W\rho]=\lambda_{j}<0.$ Set
$\alpha:=\sum_{\lambda_{i}>0}\lambda_{i}$ and
$\beta:=\sum_{\lambda_{j}<0}|\lambda_{j}|.$ Clearly, $\alpha,\beta>0.$ We
define

$Q=\sum_{\lambda_{i}>0}\beta|e_{i}\rangle\langle
e_{i}|+\sum_{\lambda_{j}<0}\alpha|e_{j}\rangle\langle
e_{j}|+\sum_{\lambda_{k}=0}|e_{k}\rangle\langle e_{k}|.$

Then $Q$ is a positive matrix. Moreover, we have

$\mathrm{Tr}[WQ]=\beta\sum_{\lambda_{i}>0}\lambda_{i}+\alpha\sum_{\lambda_{j}<0}\lambda_{j}=\beta\alpha+\alpha\times(-\beta)=0.$

Then $\sigma=Q/\mathrm{Tr}[Q]$ is a density matrix we wanted. That is,
$\sigma$ is a positive matrix and satisfies $\mathrm{Tr}[W\sigma]=0.$ ∎

The proof of the claim: $\mathbb{W}_{x}^{c}=\mathbb{W}_{x}\cap\Lambda_{-}$.
Clearly, both sets are contains in $\mathbb{W}_{x}\setminus\\{\mathbf{0}\\}.$
For any $W\in\mathbb{W}_{x}\cap\Lambda_{-}$, $W$ has some negative eigenvalue.
By Lemma 1, there exists $\rho$ such that $\mathrm{Tr}[W\rho]<0$. Hence,
$\mathbb{C}_{x}[W]\neq\emptyset$ for such $W.$ Therefore,
$\mathbb{W}_{x}\cap\Lambda_{-}\subseteq\mathbb{W}_{x}^{c}$.

Now suppose that $W\in\mathbb{W}_{x}\setminus\\{\mathbf{0}\\}$ but
$W\notin\mathbb{W}_{x}\cap\Lambda_{-}$. So $W$ must be positive semidefinite
(which is impossible when $x=0$) which implies that $\mathrm{Tr}[W\rho]\geq 0$
for all $\rho\in\mathbb{D}.$ Therefore, when $x$ is $\geq$ or $>$,
$\mathbb{C}_{x}[W]=\emptyset.$ If $x$ is a positive real number, all the
eigenvalues of $W$ must be less or equal to $x.$ Hence we have
$\boldsymbol{0}\leq W\leq x\mathbb{I}_{d}.$ Therefore,
$0\leq\mathrm{Tr}[W\rho]\leq x$ for all $\rho\in\mathbb{D}$ and
$\mathbb{C}_{x}[W]=\emptyset$ for such $W.$ No matter what $x$ is, the set
$\mathbb{C}_{x}[W]=\emptyset$. This implies that such $W$ is not an effective
coherence witness, i.e., $W\notin\mathbb{W}_{x}^{c}.$ Hence we must have
$\mathbb{W}_{x}\cap\Lambda_{-}=\mathbb{W}_{x}^{c}$. ∎

In the following, we present another lemma that will be reused multiple times
throughout this paper.

######  Lemma 2.

Let $x\in\mathcal{X}$ and $W\in\mathbb{W}_{x}^{c}$ be a coherent witness.
Define $S_{W}:=\\{\rho\in\mathbb{D}|\ \mathrm{Tr}[W\rho]=0\\}$,
$H_{W}:=\\{H\in\mathbb{H}|\ \mathrm{Tr}[WH]=0\\}$ and
$M_{W}:=\\{X\in\mathrm{Mat}_{d}(\mathbb{C})|\ \mathrm{Tr}[WX]=0\\}.$ Then the
following sets are equal,
$\mathrm{span}_{\mathbb{C}}(S_{W})=\mathrm{span}_{\mathbb{C}}(H_{W})=M_{W}$.

###### Proof.

Notice that each $W\in\mathbb{W}_{x}^{c}$ is not positive semidefinite. By
Lemma 1, there is a positive state $\rho\in\mathbb{D}$ such that
$\mathrm{Tr}[W\rho]=0.$ So $\rho\in S_{W}$ by definition.

Clearly, $S_{W}\subseteq H_{W}\subseteq M_{W}$ by definitions. Therefore,
$\mathrm{span}_{\mathbb{C}}(S_{W})\subseteq\mathrm{span}_{\mathbb{C}}(H_{W})\subseteq
M_{W}.$ First, we show that $M_{W}\subseteq\mathrm{span}_{\mathbb{C}}(H_{W}).$
In fact, for any $X\in M_{W},$ we have $\mathrm{Tr}[WX]=0.$ Therefore,
$\mathrm{Tr}[WX^{\dagger}]=\mathrm{Tr}[X^{\dagger}W]=\mathrm{Tr}[X^{\dagger}W^{\dagger}]=\mathrm{Tr}[(WX)^{\dagger}]=\mathrm{Tr}[WX]^{\dagger}=0.$
Then we have $\mathrm{Tr}[W(X+X^{\dagger})]=0$ and $\mathrm{Tr}[W(iX-
iX^{\dagger})]=0.$ However, since $H_{1}=(X+X^{\dagger})$ and
$H_{2}=i(X-X^{\dagger})\in\mathbb{H}$, they both belong to $H_{W}$. Notice
that $X=(H_{1}-iH_{2})/2$. Hence it belongs to
$\mathrm{span}_{\mathbb{C}}(H_{W}).$

Now we prove that
$\mathrm{span}_{\mathbb{C}}(H_{W})\subseteq\mathrm{span}_{\mathbb{C}}(S_{W}).$
It suffices to show that $H_{W}\subseteq\mathrm{span}_{\mathbb{C}}(S_{W}).$
For any $H\in H_{W},$ we have $H\in\mathbb{H}$ and $\mathrm{Tr}[WH]=0.$ We can
find small enough $\epsilon\in(0,1)$ such that $P_{\epsilon}:=\epsilon
H+(1-\epsilon)\rho$ is positive. Set $N=\mathrm{Tr}(P_{\epsilon}).$ Then
$\rho_{\epsilon}:=P_{\epsilon}/N=\frac{\epsilon}{N}H+\frac{1-\epsilon}{N}\rho\in\mathbb{D}.$
Moreover, we have the equality,

$\mathrm{Tr}[W\rho_{\epsilon}]=\frac{\epsilon}{N}\mathrm{Tr}[WH]+\frac{1-\epsilon}{N}\mathrm{Tr}[W\rho]=0.$

Therefore, $\rho_{\epsilon}\in S_{W}.$ Note that $H$ can be written as a
linear combination of $\rho_{\epsilon}$ and $\rho.$ Hence,
$H_{W}\subseteq\mathrm{span}_{\mathbb{C}}(S_{W})$, which completes the proof.
∎

##  Appendix B Proof of Theorem 1

######  Theorem 1 (Completeness).

For each $x\in\mathcal{X},$ the coherence criterion
($\mathbb{W}_{x},D_{x},I_{x}$) is complete.

###### Proof.

Suppose $\rho$ is a coherent state. Without loss of generality, we assume
$\rho_{mn}=\langle m|\rho|n\rangle\neq 0$, where $m\neq n$. We need to show
that there always exist some $W\in\mathbb{W}_{x}$ such that
$\mathrm{Tr}[W\rho]\in D_{x}\setminus I_{x}.$ As $x\in\mathcal{X}$, the
interval $(-\infty,0)\subseteq D_{x}\setminus I_{x}$. It is sufficient to
prove that there always exist some $W\in\mathbb{W}_{x}$ such that
$\mathrm{Tr}[W\rho]<0$. We prove the theorem according to following four
cases:

Case (i): $x=0$. We define $W_{j,k}^{R}:=(|j\rangle\langle k|+|k\rangle\langle
j|)/2$ and $W_{j,k}^{I}:=\sqrt{-1}(|j\rangle\langle k|-|k\rangle\langle j|)/2$
for each $1\leq j<k\leq d.$ Note that we always have
$\mathrm{Tr}[W_{j,k}^{R}\rho]=\Re(\rho_{jk})$ and
$\mathrm{Tr}[W_{j,k}^{I}\rho]=\Im(\rho_{jk})$, where $\rho_{jk}=\langle
j|\rho|k\rangle$, $\Re(z)=\alpha$ and $\Im(z)=\beta$ for any complex number
$z=\alpha+\sqrt{-1}\beta\in\mathbb{C}.$ Clearly,
$\mathcal{W}_{0}:=\\{W_{j,k}^{R},W_{j,k}^{I}\ \big{|}\ 1\leq j<k\leq
d\\}\subseteq\mathbb{W}_{0}^{c}.$ As $\rho_{mn}\neq 0$, either
$\mathrm{Tr}[W_{m,n}^{R}\rho]\neq 0$ or $\mathrm{Tr}[W_{m,n}^{I}\rho]\neq 0.$
That is, $\rho$ is in $\mathbb{C}_{0}[W_{m,n}^{R}]$ or
$\mathbb{C}_{0}[W_{m,n}^{I}].$ Therefore,

$\mathbb{D}\setminus\Delta=\bigcup_{W\in\mathcal{W}_{0}}\mathbb{C}_{0}[W].$
(1)

Without loss of generality, we assume that $\mathrm{Tr}[W_{m,n}^{R}\rho]\neq
0.$ Since $\mathrm{Tr}[W_{m,n}^{R}\rho]$ is a real number,
$\mathrm{Tr}[W_{m,n}^{R}\rho]$ or $\mathrm{Tr}[-W_{m,n}^{R}\rho]$ is negative
and both $\pm W_{m,n}^{R}$ are in $\mathbb{W}_{0}.$ Hence, there always exist
some $W\in\mathbb{W}_{0}$ such that $\mathrm{Tr}[W\rho]<0.$

Case (ii): $x$ is $\geq$. Denote $\mathcal{W}_{\geq}:=\\{\pm W_{j,k}^{R},\pm
W_{j,k}^{I}\ \big{|}\ 1\leq j<k\leq d\\}$. By definition, we have
$\mathcal{W}_{\geq}\subseteq\mathbb{W}_{\geq}^{c}.$ Similar to the argument in
above case, there always exist some $W\in\mathcal{W}_{\geq}$ such that
$\mathrm{Tr}[W\rho]<0.$ Therefore,

$\mathbb{D}\setminus\Delta=\bigcup_{W\in\mathcal{W}_{\geq}}\mathbb{C}_{\geq}[W].$
(2)

Case (iii): $x$ is $>.$ From the case (ii), there is some
$W\in\mathcal{W}_{\geq}$ such that $\mathrm{Tr}[W\rho]<0.$ Note that the
diagonals of $W$ are all zeros. For each $\epsilon>0$, set
$W_{\epsilon}=W+\epsilon\mathbb{I}_{d}$. Clearly, $W_{\epsilon}$ has positive
diagonals and $\mathrm{Tr}[W_{\epsilon}]=d\epsilon>0$. Hence,
$W_{\epsilon}\in\mathbb{W}_{>}$. Specially, choosing
$\epsilon=-\frac{\mathrm{Tr}[W\rho]}{2}$ we have

$\mathrm{Tr}[W_{\epsilon}\rho]=\mathrm{Tr}[W\rho]+\epsilon\mathrm{Tr}[\rho]=\mathrm{Tr}[W\rho]+\epsilon=\frac{\mathrm{Tr}[W\rho]}{2}<0.$

Case (iv): $x$ is a positive real number. By the case (iii), there is some
$W_{\epsilon}\in\mathcal{W}_{>}$ such that $\mathrm{Tr}[W_{\epsilon}\rho]<0.$
Set $W_{x}=\frac{x}{\mathrm{Tr}[W_{\epsilon}]}W_{\epsilon}.$ Clearly,
$W_{x}\in\mathbb{W}_{x}$ and

$\mathrm{Tr}[W_{x}\rho]=\frac{x}{\mathrm{Tr}[W_{\epsilon}]}\mathrm{Tr}[W_{\epsilon}\rho]<0.$

The above four cases together complete the proof. ∎

##  Appendix C Proof of Theorem 2

######  Theorem 2 (Characterization of finite completeness).

Let $x$ be an element in $\mathcal{X}.$ The coherence criterion
($\mathbb{W}_{x},D_{x},I_{x}$) is finitely completable if and only if $x$ is
$0$ or $\geq$.

###### Proof.

By Eqs. (1) and (2) and the definition of finitely completable, both coherence
criteria ($\mathbb{W}_{0},D_{0},I_{0}$) and
($\mathbb{W}_{\geq},D_{\geq},I_{\geq}$) are finitely completable.

Now we show that for any positive real number $x$, the criterion
($\mathbb{W}_{x},D_{x},I_{x}$) is finitely incompletable. If not, there exists
some $x>0$ and a set
$\mathcal{W}:=\\{W_{i}\\}_{i=1}^{n}\subseteq\mathbb{W}_{x}^{c}$ such that
$\mathbb{D}\setminus\Delta=\cup_{i=1}^{n}\mathbb{C}_{x}[W_{i}]$. That is, for
any coherent states $\rho$, there always exists some $W\in\mathcal{W}$ such
that $\mathrm{Tr}[W\rho]<0$ or $\mathrm{Tr}[W\rho]>x.$ On the other hand, set
$\rho_{\epsilon}=\frac{\mathbb{I}_{d}}{d}+\epsilon H$, where
$H=|1\rangle\langle 2|+|2\rangle\langle 1|$ and $\epsilon\in(0,\frac{1}{d})$,
and $M=\max\\{\big{|}\mathrm{Tr}[(W-\Delta(W)H]\big{|}:W\in\mathcal{W}\\}+1.$
Note that

$\mathrm{Tr}[W\rho_{\epsilon}]=\frac{x}{d}+\epsilon\mathrm{Tr}[(W-\Delta(W))H].$

If we set $\epsilon=\min\\{\frac{x}{2dM},\frac{1}{2d}\\},$ then we have
$0<\frac{1}{2d}x\leq\mathrm{Tr}[W\rho_{\epsilon}]\leq\frac{3}{2d}x<x$ for
every $W\in\mathcal{W}.$ However, $\rho_{\epsilon}$ is a coherent state which
can not be detected by all the witnesses in $\mathcal{W}.$

Now we show that the criterion ($\mathbb{W}_{>},D_{>},I_{>}$) is also finitely
incompletable. For any finite set
$\mathcal{W}_{>}:=\\{W_{i}\\}_{i=1}^{n}\subseteq\mathbb{W}_{>}^{c}$, we define
$M_{1}:=\max\\{\big{|}\mathrm{Tr}[(W-\Delta(W)H]\big{|}:W\in\mathcal{W}\\}+1,$
and $m_{1}:=\min\\{\mathrm{Tr}[W]:W\in\mathcal{W}_{>}\\}>0.$ If we set
$\epsilon_{1}=\min\\{\frac{m_{1}}{2dM_{1}},\frac{1}{2d}\\},$ then
$\mathrm{Tr}[W\rho_{\epsilon_{1}}]\geq\frac{m_{1}}{2d}>0$ for every
$W\in\mathcal{W}_{>}.$ That is, $\rho_{\epsilon_{1}}$ is a coherent state
whose coherence can not be detected by the witnesses in $\mathcal{W}_{>}.$ ∎

##  Appendix D Proof of Theorem 3

######  Theorem 3 (Property of finite intersection).

Let $x\in\mathcal{X}$ and
$\mathcal{W}:=\\{W_{i}\\}_{i=1}^{n}\subseteq\mathbb{W}_{x}^{c}.$ Then the
following statements hold.

  1. (a)

For the case $x\in(0,\infty)$ and any subset $\mathcal{W}^{\prime}$ with
cardinality $|\mathcal{W}^{\prime}|\geq\lceil\frac{n}{2}\rceil$, if there
always exist $t_{W}>0$ for each $W\in\mathcal{W}^{\prime}$ and
$\sum_{W\in\mathcal{W}^{\prime}}t_{W}=1$ such that
$\sum_{W\in\mathcal{W}^{\prime}}t_{W}W\geq\boldsymbol{0}$, then
$\bigcap_{i=1}^{n}\mathbb{C}_{x}[{W_{i}}]=\emptyset.$

  2. (b)

If $x$ is $>$ or $\geq,$ then
$\bigcap_{i=1}^{n}\mathbb{C}_{x}[{W_{i}}]=\emptyset$ if and only if there is
some nonnegative $t_{i}\geq 0$ and $\sum_{i=1}^{n}t_{i}=1$ such that
$\sum_{i=1}^{n}t_{i}W_{i}$ is positive semidefinite.

  3. (c)

If $x=0,$ there exists a common coherent state that can be detected by all of
$W_{i}$ via criterion of type (C). That is, we always have

$\mathcal{C}_{0}[{\mathcal{W}}]:=\bigcap_{i=1}^{n}\mathbb{C}_{0}[{W_{i}}]\neq\emptyset.$

Moreover, the cardinality of $\mathcal{C}_{0}[{\mathcal{W}}]$ is infinity.
That is, given any set of finitely many coherent witnesses, they share with
infinitely many common coherent states.

###### Proof.

(a). If $\rho\in\bigcap_{i=1}^{n}\mathbb{C}_{x}[{W_{i}}]$, then
$\mathrm{Tr}[W_{i}\rho]<0$ or $\mathrm{Tr}[W_{i}\rho]>x.$ Let
$\mathcal{I}_{0}:=\\{i\ \big{|}\ \mathrm{Tr}[W_{i}\rho]<0,\ 1\leq i\leq n\\}$
and $\mathcal{I}_{x}:=\\{i\ \big{|}\ \mathrm{Tr}[W_{i}\rho]>x,\ 1\leq i\leq
n\\}.$ Without loss of generality, we assume the set $\mathcal{I}_{x}$ is
larger than $\mathcal{I}_{0}$. Then correspondingly the cardinality must be
greater or equal to $\lceil\frac{n}{2}\rceil.$ Define
$\mathcal{W}^{\prime}:=\\{W_{i}|i\in\mathcal{I}_{x}\\}.$ By assumption, there
are $t_{i}>0$ for $i\in\mathcal{I}_{x}$ such that
$\sum_{i\in\mathcal{I}_{x}}t_{i}=1$ and
$W:=\sum_{i\in\mathcal{I}_{x}}t_{i}W_{i}\geq\boldsymbol{0}.$ Therefore, all
eigenvalues of $W$ are nonnegative. Moreover, as $\mathrm{Tr}[W]=x$, all the
eigenvalues of $W$ are less or equal to $x$. Hence, $W\leq x\mathbb{I}_{d}$,
which implies $\mathrm{Tr}[W\rho]\leq x.$ On the other hand, for each
$i\in\mathcal{I}_{x},$ we have $\mathrm{Tr}[W_{i}\rho]>x$, which leads to

$\mathrm{Tr}[W\rho]=\sum_{i\in\mathcal{I}_{x}}t_{i}\mathrm{Tr}[W_{i}\rho]>x\sum_{i\in\mathcal{I}_{x}}t_{i}=x,$

and thus a contradiction. Hence
$\bigcap_{i=1}^{n}\mathbb{C}_{x}[{W_{i}}]=\emptyset.$

(b). The argument for the case that $x$ is $\geq$ can be directly referred to
reference Wang21 . When $x$ is $>$, as the $D_{>}$ and $I_{>}$ are the same as
$D_{\geq}$ and $I_{\geq}$, the proof goes similarly to the one given in Wang21
.

(c). We prove the statement by induction on $n.$ First, we consider the case
$n=2.$ As both $\mathbb{C}_{0}[{W_{1}}]$ and $\mathbb{C}_{0}[{W_{2}}]$ are
nonempty, we assume $\rho\in\mathbb{C}_{0}[{W_{1}}]$ and
$\sigma\in\mathbb{C}_{0}[{W_{2}}].$ We may assume that
$\rho,\sigma\notin\mathbb{C}_{0}[{W_{1}}]\cap\mathbb{C}_{0}[{W_{2}}]$, i.e.,
$\rho\notin\mathbb{C}_{0}[{W_{2}}]$ and $\sigma\notin\mathbb{C}_{0}[{W_{1}}].$
Equivalently, we have $\mathrm{Tr}[W_{2}\rho]=\mathrm{Tr}[W_{1}\sigma]=0.$
Then for any $\lambda\in(0,1),$ the matrix
$\pi_{\lambda}:=\lambda\rho+(1-\lambda)\sigma\in\mathbb{D}.$ Moreover,
$\mathrm{Tr}[W_{1}\pi_{\lambda}]=\lambda\mathrm{Tr}[W_{1}\rho]\neq 0$ and
$\mathrm{Tr}[W_{2}\pi_{\lambda}]=(1-\lambda)\mathrm{Tr}[W_{2}\sigma]\neq 0$.
That is, for each $\lambda\in(0,1),$ the state $\pi_{\lambda}$ belongs to
$\mathbb{C}_{0}[{W_{1}}]\cap\mathbb{C}_{0}[{W_{2}}]$. Now suppose that the
statement holds when cardinality of the set of coherence witnesses is equal to
$n-1.$ Let $\mathcal{W}=\\{W_{1},W_{2},\cdots,W_{n}\\}.$ We define
$\mathcal{W}_{1}=\\{W_{1},\cdots,W_{n-1}\\}$ and
$\mathcal{W}_{2}=\\{W_{2},\cdots,W_{n}\\}.$ By induction, the following two
sets

$\mathcal{C}_{0}[{\mathcal{W}_{1}}]:=\bigcap_{W\in\mathcal{W}_{1}}\mathbb{C}_{0}[{W}],\
\
\mathcal{C}_{0}[{\mathcal{W}_{2}}]:=\bigcap_{W\in\mathcal{W}_{2}}\mathbb{C}_{0}[{W}],$

are nonempty. Suppose $\rho\in\mathcal{C}_{0}[{\mathcal{W}_{1}}]$ and
$\sigma\in\mathcal{C}_{0}[{\mathcal{W}_{2}}],$ i.e., $\mathrm{Tr}[W_{i}\
\rho]\neq 0$ and $\mathrm{Tr}[W_{i+1}\sigma]\neq 0$ for all
$i=1,2,\cdots,(n-1).$ If $\mathrm{Tr}[W_{n}\rho]\neq 0$ (or
$\mathrm{Tr}[W_{1}\sigma]\neq 0$), we have
$\rho\in\mathcal{C}_{0}[{\mathcal{W}}]$ (or
$\sigma\in\mathcal{C}_{0}[{\mathcal{W}}]$) which proves the nonempty of
$\mathcal{C}_{0}[{\mathcal{W}}]$. Therefore, we might assume
$\mathrm{Tr}[W_{n}\rho]=0$ and $\mathrm{Tr}[W_{1}\sigma]=0$. Similarly, for
each $\lambda\in(0,1),$ we define
$\pi_{\lambda}:=\lambda\rho+(1-\lambda)\sigma\in\mathbb{D}.$ Then we have
$\mathrm{Tr}[W_{1}\pi_{\lambda}]=\lambda\mathrm{Tr}[W_{1}\rho]\neq 0$ and
$\mathrm{Tr}[W_{n}\pi_{\lambda}]=(1-\lambda)\mathrm{Tr}[W_{n}\sigma]\neq 0.$
Moreover, for each integer $i\in[2,n-1],$ we have
$a_{i}:=\mathrm{Tr}[W_{i}\rho]\neq 0$ and $b_{i}:=\mathrm{Tr}[W_{i}\sigma]\neq
0.$ Define $f_{i}(\lambda):=\mathrm{Tr}[W_{i}\pi_{\lambda}]=\lambda
a_{i}+(1-\lambda)b_{i}=(a_{i}-b_{i})\lambda+b_{i}$ and
$f(\lambda):=\prod_{i=2}^{n-1}f_{i}(\lambda).$ If $a_{i}=b_{i}$ for all
$i\in[2,n-1],$ we have $\mathrm{Tr}[W_{i}\pi_{\lambda}]=b_{i}\neq 0.$ Then
$\pi_{\lambda}\in\mathcal{C}_{0}[{\mathcal{W}}].$ If not, $f(\lambda)$ is a
nontrivial (not a constant) polynomial of $\lambda.$ Therefore, there are only
finite $\lambda$’s, say $\lambda_{j}$ $(j=1,2,\cdots,N)$ such that
$f(\lambda_{j})=0.$ Therefore, for each
$\lambda\in(0,1)\setminus\\{\lambda_{j}\\}_{j=1}^{N},$ we always have
$f(\lambda)\neq 0.$ Hence $\mathrm{Tr}[W_{i}\pi_{\lambda}]=f_{i}(\lambda)\neq
0$ for all integer $i\in[2,n-1]$, which also holds for $i=1,n.$ Therefore,
they all belong to the set $\mathcal{C}_{0}[{\mathcal{W}}].$

For each coherent state $\rho$, there exists witness $W_{\rho}$ which can not
detect the coherence of $\rho$, i.e., $\rho\notin\mathbb{C}_{0}[{W_{\rho}}].$
If $\mathcal{C}_{0}[{\mathcal{W}}]$ is finite, suppose
$\mathcal{C}_{0}[{\mathcal{W}}]=\\{\rho_{j}\\}_{j=1}^{m}$. Then one has

$(\cap_{i=1}^{n}\mathbb{C}_{0}[{W_{i}}])\bigcap(\cap_{j=1}^{m}\mathbb{C}_{0}[{W_{\rho_{j}}}])=\emptyset,$

which leads to a contradiction. ∎

##  Appendix E  Proof of Theorem 4

######  Theorem 4 (Property of inclusion or identity).

Let $x\in\mathcal{X}$ and $W_{1},W_{2}\in\mathbb{W}_{x}^{c}.$ The following
statements hold.

  1. (a)

If $x\in(0,\infty),$ then $\mathbb{C}_{x}[{W_{1}}]=\mathbb{C}_{x}[{W_{2}}]$ if
and only if $W_{1}=W_{2}$ or $W_{1}+W_{2}=x\mathbb{I}_{d}$ (the statement is
true only if $d=2$).

  2. (b)

If $x$ is $>$ or $\geq$, $\mathbb{C}_{x}[{W_{1}}]=\mathbb{C}_{x}[{W_{2}}]$ if
and only if there exists $r\in\mathbb{R}_{+}:=\\{r\in\mathbb{R}\ \big{|}\
r>0\\}$ such that $W_{2}=rW_{1}.$ Moreover,
$\mathbb{C}_{x}[{W_{2}}]\subseteq\mathbb{C}_{x}[{W_{1}}]$ if and only if there
exists $a\in\mathbb{R}_{+}$ and a positive semidefintie operator $P$ such that
$W_{2}=aW_{1}+P$.

  3. (c)

If $x=0,$ then $\mathbb{C}_{0}[W_{1}]=\mathbb{C}_{0}[W_{2}]$ if and only if
there exist some $r\in\mathbb{R}\setminus\\{0\\}$ such that $W_{2}=rW_{1}$ (we
call such two witnesses $\mathbb{R}$ equivalent). If $W_{1}$ and $W_{2}$ are
not $\mathbb{R}$ equivalent, then
$\mathbb{C}_{0}[W_{1}]\not\subseteq\mathbb{C}_{0}[W_{2}]$ and
$\mathbb{C}_{0}[W_{2}]\not\subseteq\mathbb{C}_{0}[W_{1}].$

###### Proof.

We first prove the statements (b) and (c).

(b). For any Hermitian matrix $W\in\mathbb{H}$, define
$S_{W}:=\\{\rho\in\mathbb{D}|\ \mathrm{Tr}[W\rho]=0\\}$ and
$M_{W}:=\\{X\in\mathrm{Mat}_{d}(\mathbb{C})|\ \mathrm{Tr}[WX]=0\\}.$ Suppose
that $\mathbb{C}_{x}[W_{1}]=\mathbb{C}_{x}[W_{2}].$ We claim that
$S_{W_{1}}=S_{W_{2}}.$ Otherwise, without loss of generality, we assume that
$\rho\in S_{W_{1}}$ and $\rho\notin S_{W_{2}}$, i.e.,
$\mathrm{Tr}[W_{1}\rho]=0$ and $\mathrm{Tr}[W_{2}\rho]\neq 0.$ If
$\mathrm{Tr}[W_{2}\rho]<0,$ then $\rho\in\mathbb{C}_{x}[W_{2}]$ and
$\rho\notin\mathbb{C}_{x}[W_{1}]$ which contradicts to the assumption.
Therefore, $\mathrm{Tr}[W_{2}\rho]>0.$ Now choosing any
$\sigma\in\mathbb{C}_{x}[W_{1}],$ we have $\mathrm{Tr}[W_{1}\sigma]<0$ and
$\mathrm{Tr}[W_{2}\sigma]<0.$ For each $\epsilon\in(0,1),$ define
$\rho_{\epsilon}=\epsilon\sigma+(1-\epsilon)\rho\in\mathbb{D}.$ Then for small
enough $\epsilon$, we have $\mathrm{Tr}[W_{1}\rho_{\epsilon}]<0$ but
$\mathrm{Tr}[W_{2}\rho_{\epsilon}]>0$, which implies that
$\rho_{\epsilon}\in\mathbb{C}_{x}[W_{1}]$ but
$\rho_{\epsilon}\notin\mathbb{C}_{x}[W_{2}].$ Hence we obtain again a
contradiction. Therefore, $S_{W_{1}}=S_{W_{2}}.$ By Lemma 2,
$M_{W_{1}}=M_{W_{2}}$ implies $W_{2}=cW_{1}$ for some nonzero complex number.
With this at hand, it is easy to show that $c$ is in fact a positive real
number. The other direction of the first statement in (b) is obvious.

Now we prove the second statement. Suppose that there are $a\in\mathbb{R}_{+}$
and a positive semidefintie operator $P$ such that $W_{2}=aW_{1}+P.$ For any
$\rho\in\mathbb{C}_{x}[W_{2}],$ we have
$0>\mathrm{Tr}[W_{2}\rho]=a\mathrm{Tr}[W_{1}\rho]+\mathrm{Tr}[P\rho].$ This
implies that $\mathrm{Tr}[W_{1}\rho]<0$ as we always have
$\mathrm{Tr}[P\rho]\geq 0$ for positive semidefinite $P.$ Therefore,
$\rho\in\mathbb{C}_{x}[W_{1}]$ and
$\mathbb{C}_{x}[W_{2}]\subseteq\mathbb{C}_{x}[W_{1}].$

Suppose that $\mathbb{C}_{x}[W_{2}]\subseteq\mathbb{C}_{x}[W_{1}].$ We prove
that there exist $a\in\mathbb{R}_{+}$ and a positive semidefinite operator $P$
such that $W_{2}=aW_{1}+P.$ Set $t_{1}=\mathrm{Tr}[W_{1}]$ and
$t_{2}=\mathrm{Tr}[W_{2}].$ Clearly, $t_{1},t_{2}\geq 0$ as
$W_{1},W_{2}\in\mathbb{H}_{\geq}$. We prove the conclusion according to the
following four cases.

  1. (i)

Both $t_{1}$ and $t_{2}$ are positive. Note that
$\mathbb{C}_{x}[W_{1}]=\mathbb{C}_{x}[W_{1}/{t_{1}}]$ and
$\mathbb{C}_{x}[W_{2}]=\mathbb{C}_{x}[W_{2}/{t_{2}}].$ Following the proof of
Lemma 1 and the Corollary 1 in Ref. Wang21b  (which is correct under the
assumption that the traces of the observables are 1), we have
$W_{2}/{t_{2}}=a_{1}W_{1}/{t_{1}}+Q$ for some positive $a_{1}$ and positive
semidefinite $Q.$ Therefore, $W_{2}=aW_{1}+P$, where $a=a_{1}{t_{2}}/{t_{1}}$
and $P={t_{2}}Q$.

  2. (ii)

$t_{1}=0$ but $t_{2}>0.$ It is easily verified that
$\mathbb{C}_{x}[W_{2}]\subseteq\mathbb{C}_{x}[W_{1}+W_{2}]\subseteq\mathbb{C}_{x}[W_{1}].$
Similarly, we have

$W_{2}/{t_{2}}=a_{1}(W_{1}+W_{2})/{t_{2}}+Q$ (3)

for some positive $a_{1}$ and positive semidefinite $Q.$ From the trace of Eq.
(3), we obtain $1=a_{1}+\mathrm{Tr}[Q].$ Therefore, $0<a_{1}\leq 1.$ Clearly,
$a_{1}\neq 1,$ otherwise, $\mathrm{Tr}[Q]=0$ implies that $Q=\mathbf{0}$ and
$W_{1}=-t_{2}/a_{1}Q=\mathbf{0}$ is not a nontrivial coherent witness. Hence,
$0<a_{1}<1.$ Eq. (3) can be reexpressed as

$W_{2}=\frac{a_{1}}{1-a_{1}}W_{1}+\frac{t_{2}}{1-a_{1}}Q.$

  3. (iii)

$t_{1}>0$ but $t_{2}=0.$ One easily shows that
$\mathbb{C}_{x}[W_{2}]\subseteq\mathbb{C}_{x}[W_{1}+W_{2}]\subseteq\mathbb{C}_{x}[W_{1}].$
Similarly, one has

$(W_{1}+W_{2})/{t_{1}}=a_{1}W_{1}/{t_{1}}+Q$ (4)

for some positive $a_{1}$ and positive semidefinite $Q.$ The trace of Eq. (4)
gives rise to $1=a_{1}+\mathrm{Tr}[Q].$ Therefore, $0<a_{1}\leq 1.$ Clearly,
$a_{1}\neq 1,$ otherwise, $W_{2}=t_{1}Q=\mathbf{0}$ can not be a nontrivial
coherent witness. We can rewrite Eq. (4) as $(1-a_{1})W_{1}+W_{2}=t_{1}Q$ and
obtain that
$\mathbb{C}_{x}[W_{2}]\subseteq\mathbb{C}_{x}[(1-a_{1})W_{1}+W_{2}]\subseteq\mathbb{C}_{x}[W_{1}]$.
On the other hand, $\mathbb{C}_{x}[t_{1}Q]=\emptyset$. Therefore, this case
can not happen as $\mathbb{C}_{x}[W_{2}]\neq\emptyset.$

  4. (iv)

$t_{1}=t_{2}=0.$ For this case, we show that this could happen only when
$W_{2}=aW_{1}.$ In fact, for small enough $\epsilon>0,$
$\mathbb{C}_{x}[W_{2}+\epsilon\mathbb{I}_{d}]\subseteq\mathbb{C}_{x}[W_{2}]\subseteq\mathbb{C}_{x}[W_{1}].$
By the argument of case (ii), we have $a_{\epsilon}>0$ and positive
semidefinite operator $P_{\epsilon}$ such that
$W_{2}+\epsilon\mathbb{I}_{d}=a_{\epsilon}W_{1}+P_{\epsilon}$, which implies
that the diagonals of $P_{\epsilon}$ are all $\epsilon$. Moreover, as
$P_{\epsilon}$ is positive semidefinite, the module of each offdiagonal
elements must be smaller than $\epsilon$. The matrix $P_{\epsilon}$ converges
to zero matrix (up to entries wise) when $\epsilon\rightarrow 0$. Taking the
limit $\epsilon\rightarrow 0$, we have $W_{2}=aW_{1}.$

(c). If $W_{2}=rW_{1}$ for $r\neq 0,$ we always have
$\mathrm{Tr}[W_{2}\rho]=r\mathrm{Tr}[W_{1}\rho]$ for all $\rho\in\mathbb{D}.$
In this case, $\mathrm{Tr}[W_{2}\rho]=0$ if and only if
$\mathrm{Tr}[W_{1}\rho]=0$. Hence
$\mathbb{C}_{0}[W_{2}]=\mathbb{C}_{0}[W_{1}].$ Conversely, if
$\mathbb{C}_{0}[W_{1}]=\mathbb{C}_{0}[W_{2}]$, we have equivalently
$S_{W_{1}}=S_{W_{2}}.$ By Lemma 2, we get $M_{W_{1}}=M_{W_{2}}.$ By
definition, $M_{W_{1}}$ is just the orthogonal complement space to the vector
$W_{1}$. Hence the space has dimensional $d^{2}-1.$ So the vectors that are
orthogonal to each elements in $M_{W_{1}}$ are just $\mathbb{C}W_{1}.$
Therefore, there is nonzero $c\in\mathbb{C}$ such that $W_{2}=cW_{1}.$ Since
$W_{2}$ is Hermitian, we have $c\in\mathbb{R}.$

We prove the second statement in (c) by contradiction. Without loss of
generality, we may assume that
$\mathbb{C}_{0}[W_{1}]\subseteq\mathbb{C}_{0}[W_{2}].$ Then we have
$M_{W_{1}}\subseteq M_{W_{2}}.$ As both are linear space of the same dimension
$d^{2}-1,$ they must be equal, i.e., $M_{W_{1}}=M_{W_{2}}.$ Similar to the
above derivations, we deduce that $W_{2}$ and $W_{1}$ are $\mathbb{R}$
equivalent, which contradicts to the assumption.

(a). If $x$ is positive real and $W\in\mathbb{W}_{x}^{c}$, we have
$\mathbb{C}_{x}[W]=\mathbb{C}_{\geq}[W]\cup\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W].$
If $x\mathbb{I}_{d}-W$ is not positive semidefinite, then
$\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W]\neq\emptyset.$ In such setting,
$\mathbb{C}_{x}[W]$ is a disjoint unions of two convex sets.

Suppose that $\mathbb{C}_{x}[W_{1}]=\mathbb{C}_{x}[W_{2}]$. If
$\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W_{1}]=\emptyset,$ we must have
$\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W_{2}]=\emptyset$, which yields
$\mathbb{C}_{\geq}[W_{1}]=\mathbb{C}_{\geq}[W_{2}].$ By the case (b),
$W_{2}=rW_{1}$ for some positive $r$. Taking into account the trace, we must
have $r=1$. If $\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W_{1}]\neq\emptyset,$ we
have $\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W_{2}]\neq\emptyset.$ Moreover, either
$\mathbb{C}_{\geq}[W_{1}]=\mathbb{C}_{\geq}[W_{2}]$ or
$\mathbb{C}_{\geq}[W_{1}]=\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W_{2}].$ For the
former case, we have shown that $W_{1}=W_{2}$. For the latter case, we also
have $\mathbb{C}_{\geq}[W_{2}]=\mathbb{C}_{\geq}[x\mathbb{I}_{d}-W_{1}]$, form
which we can deduce $(d-1)W_{1}=x\mathbb{I}_{d}-W_{2}$ and
$(d-1)W_{2}=x\mathbb{I}_{d}-W_{1}.$ If $d>2$, we have
$W_{1},W_{2}\propto\mathbb{I}_{d}$, which contradicts with
$W_{1},W_{2}\in\mathbb{W}_{x}^{c}.$ While for the case $d=2$, the two
equations merge into $W_{1}+W_{2}=x\mathbb{I}_{d}.$

Moreover, for $d=2$ one has $\mathrm{Tr}[W_{1}\rho]+\mathrm{Tr}[W_{2}\rho]=x$
for any $\rho\in\mathbb{D}.$ Hence, $\mathrm{Tr}[W_{1}\rho]<0$ if and only if
$\mathrm{Tr}[W_{2}\rho]>x$, and $\mathrm{Tr}[W_{1}\rho]>x$ if and only if
$\mathrm{Tr}[W_{2}\rho]<0.$ Therefore,
$\mathbb{C}_{x}[{W_{1}}]=\mathbb{C}_{x}[{W_{2}}].$ ∎