# Notes on quantum coherence with $l_{1}$-norm and convex-roof $l_{1}$-norm

Jiayao Zhu1, Jian Ma1,2, Tinggui Zhang1,2,† 1School of Mathematics and
Statistics, Hainan Normal University, Haikou, 571158, China
2 Key Laboratory of Data Science and Smart Education, Ministry of Education,
Hainan Normal University, Haikou 571158, China
† Correspondence to<EMAIL_ADDRESS>
(17 July, 2021)

###### Abstract

Abstract In this work, we evaluate quantum coherence using the $l_{1}$-norm
and convex-roof $l_{1}$-norm and obtain several new results. First, we provide
some new general triangle-like inequalities of quantum coherence, with results
better than existing ones. Second, for some special three-dimensional quantum
states, a method for calculating the convex-roof $l_{1}$-norm is presented.
Lastly, we offer distinct upper bounds in the $l_{1}$-norm measure of
coherence based on the quantum state itself.

Keywords Quantum coherence; $l_{1}$-norm; Triangle-like inequalities; Convex-
roof $l_{1}$-norm

###### pacs:

03.67.-a, 02.20.Hj, 03.65.-w

##  I Introduction

Quantum coherence stems from the superposition of quantum states; it reflects
the ability of quantum states to display quantum coherence effects. It is the
most basic property distinguishing quantum physics from classical physics, and
it is also an important quantum mechanical property that is widely used in
quantum information processing and quantum computing. Consequently, quantum
information science and the derived quantum technology offer advantages that
the corresponding classical informatics and technology cannot provide, such as
the theoretical security of quantum communication1 , ultra-fast quantum
computing2 , and quantum precision measurements with accuracies exceeding the
classical limit3 . Additionally, quantum coherence is widely used in quantum
biology4 ; 5 , quantum thermodynamics6 ; 7 , and quantum computing8 ; 9 .
Therefore, quantifying the quantum coherence in a resource framework is
important and meaningful, and the study of quantum coherence plays an
important role in furthering the frontiers of physics research.

Quantum coherence and entanglement are two characteristics of the quantum
world. Quantum entanglement describes bipartite or multiple systems, whereas
quantum coherence is defined for single systems. With the rapid developments
in quantum information science, researchers have found that, similar to
entanglement, quantum coherence can be treated as a physical resource.
However, a core issue in this field is the quantification of coherence.
Adopting some of the methods used with entanglement, Baumgratz et al.
established a quantitative theory of coherence as a resource along with four
necessary conditions that should be satisfied by any appropriate measure of
coherence 10 .

Most of the measures of quantum coherence are based on distances, defined in
terms of the minimum distance between the selected quantum state and the given
set of incoherent states $\tilde{\mathcal{I}}$. These incoherent states are
quantum states that are diagonalizable under a certain reference basis
$\left\\{{|i\rangle}\right\\}_{i=1}^{d}$ in the Hilbert space, that is,
$\delta=\sum\nolimits_{i=1}^{d}{{\delta_{i}}|i\rangle\langle i|}$. A measure
of coherence is defined as follows:

${C_{D}}\left(\rho\right)=\mathop{\min}\limits_{\delta\in\tilde{\mathcal{I}}}D\left({\rho,\delta}\right),$

where $D\left({\rho,\delta}\right)$ denotes a certain measure of the distance
between $\rho$ and $\delta$.

Baumgratz and his coworkers found that the relative entropy of coherence

$C_{re}(\rho)=S(\rho_{diag})-S(\rho),$

where $S$ is the von Neumann entropy and ${\rho_{diag}}$ denotes the state
obtained from $\rho$ by deleting all off-diagonal elements and the
$l_{1}$-norm of coherence

${C_{{l_{1}}}}(\rho)=\sum\limits_{i\neq j}{\left|{{\rho_{ij}}}\right|},$

are both appropriate measures of coherence.

Thus far, many suitable measures have been reported, such as the robustness of
coherence11 , maximum relative entropy of coherence12 , entanglement-based
coherence measurement13 , average quantum coherencemjmt , and set coherence
sdru . However, the $l_{1}$-norm of coherence has continued to be significant
for research. Recently, it was discovered that the $l_{1}$-norm of coherence
can be used to describe wave-particle duality fcle .

Ref.14  showed that, if a rank-2 state $\rho$ can be expressed as a convex
combination of two pure states, i.e.

$\rho={p_{1}}|{\psi_{1}}\rangle\langle{\psi_{1}}|+{p_{2}}|{\psi_{2}}\rangle\langle{\psi_{2}}|,$

a triangle-like inequality can be established as follows:

$\displaystyle\left|{E\left({\sqrt{{p_{1}}}|{\psi_{1}}\rangle}\right)-E\left({\sqrt{{p_{2}}}|{\psi_{2}}\rangle}\right)}\right|\leq
E(\rho)$ $\displaystyle\qquad\qquad\qquad\qquad\leq
E\left({\sqrt{{p_{1}}}|{\psi_{1}}\rangle}\right)+E\left({\sqrt{{p_{2}}}|{\psi_{2}}\rangle}\right),$

where $E$ can be considered as either a coherence measure or an entanglement
concurrence.

Ref.15  provided a general triangle-like inequality satisfied by the
$l_{1}$-norm measure of coherence for a convex combination of $n$ arbitrary
pure states of a quantum state $\rho$, thus verifying the conclusions
presented in Ref.14 .

Both Ref.14  and Ref.15  estimated the coherence of a quantum state by using
state decomposition. We can also evaluate the coherence of a quantum state
based on its properties. As mentioned in Ref.16 , for a $d$-dimensional
quantum state, its $l_{1}$-norm is not more than $d-1$.

To date, many scholars have estimated the coherence of a quantum state,
providing lower and upper bounds in different forms. However, we find that
some of these estimates can be optimized, and we can measure the coherence of
a quantum state from different perspectives, thus affording distinct
conclusions. In this study, we select the $l_{1}$-norm and convex-roof
$l_{1}$-norm to quantify coherence. Based on the state decomposition of a
quantum state and its properties, we provide some new lower and upper bounds
for the $l_{1}$-norm and convex-roof $l_{1}$-norm.

##  II  Generalized triangular inequality for quantum coherence based on
$l_{1}$ norm

On basis of the state decomposition of a quantum state, Ref.14  provides the
following conclusion:

######  Lemma 1.

If a quantum state $\rho$ can be expressed as a convex combination of two
states, that is, $\rho=p_{1}\rho_{1}+p_{2}\rho_{2}$, then we have

$\left|{{p_{1}}{C_{{l_{1}}}}({\rho_{1}})-{p_{2}}{C_{{l_{1}}}}({\rho_{2}})}\right|\leq{C_{{l_{1}}}}(\rho).$

Moreover, Ref.15  provides a triangle-like inequality in the $l_{1}$-norm
measure of coherence.

######  Lemma 2.

If the state $\rho$ can be expressed as a convex combination of $n(n\geq 2)$
states, that is, $\rho=\sum\nolimits_{i=1}^{n}{{p_{i}}{\rho_{i}}}$, then
${C_{{l_{1}}}}\left(\rho\right)$ satisfies the following triangle-like
inequality

$\frac{1}{n}\sum\limits_{k=1}^{n}{\left|{G_{k}^{(n-1)}-{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}\right|}\leq{C_{{l_{1}}}}\left(\rho\right)\leq\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}({\rho_{k}})},$

where $G_{k}^{(n-1)},k=1,2,\cdots,n$ are the lower bounds of
$\sum\nolimits_{i\neq
k}^{n}{{p_{i}}{C_{{l_{1}}}}}\left({\frac{{\sum\nolimits_{j\neq
k}^{n}{{p_{j}}{\rho_{j}}}}}{{\sum\nolimits_{t\neq k}^{n}{{p_{t}}}}}}\right)$.

We can further develop the conclusion of Lemma 2. However, we first introduce
a useful lemma followed by a new triangle-like inequality regarding the
$l_{1}$-norm considering the state decomposition of a quantum state.

######  Lemma 3.

Let ${a_{i}}\geq 0(i=1,2,\cdots,n)$ and not be complete zeros. If
$b\geq{a_{i}},i=1,2,\cdots,n$, then we obtain

$\displaystyle
b\geq\frac{{\sum\nolimits_{i=1}^{n}{a_{i}^{2}}}}{{\sum\nolimits_{i=1}^{n}{a_{i}}}}\geq\frac{1}{n}\sum\nolimits_{i=1}^{n}{a_{i}}.$

Proof: Based on these assumptions, we have

$\displaystyle\sum\nolimits_{i=1}^{n}{a_{i}^{2}}\leq\sum\nolimits_{i=1}^{n}{a_{i}}b=b\sum\nolimits_{i=1}^{n}{a_{i}}.$

Note that ${a_{1}},\cdots,{a_{n}}$ are not complete zeros; thus,
$\sum\nolimits_{i=1}^{n}{a_{i}}>0$. Therefore, it holds that

$\displaystyle
b\geq\frac{{\sum\nolimits_{i=1}^{n}{a_{i}^{2}}}}{{\sum\nolimits_{i=1}^{n}{a_{i}}}}.$

Moreover, using the inequality

$\displaystyle{({x_{1}}+{x_{2}}+\cdots+{x_{n}})^{2}}\leq
n(x_{\rm{1}}^{\rm{2}}+x_{\rm{2}}^{\rm{2}}+\cdots+x_{n}^{2}),$

we can directly obtain

$\displaystyle\frac{{\sum\nolimits_{i=1}^{n}{a_{i}^{2}}}}{{\sum\nolimits_{i=1}^{n}{a_{i}}}}\geq\frac{1}{n}\sum\nolimits_{i=1}^{n}{a_{i}}.$

Next, we present the general triangle-like inequality based on the
$l_{1}$-norm measure of coherence.

######  Theorem 1.

If the state $\rho$ can be expressed as a convex combination of $n(n\geq 2)$
states, that is, $\rho=\sum\nolimits_{i=1}^{n}{{p_{i}}{\rho_{i}}}$, then
${C_{{l_{1}}}}(\rho)$ satisfies

$\frac{{\sum\nolimits_{i=1}^{n}{A_{i}^{2}}}}{{\sum\nolimits_{i=1}^{n}{{A_{i}}}}}\leq{C_{{l_{1}}}}(\rho)\leq\frac{{\rm{1}}}{n}\sum\limits_{k=1}^{n}{\left[{G_{k}^{(n-1)}+{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}\right]},$
(1)

where

$G_{k}^{(n-1)}=(1-{p_{k}}){C_{{l_{1}}}}({\textstyle{\frac{\rho-{p_{k}}{\rho_{k}}}{1-{p_{k}}}}}),k=1,2,\cdots,,n;$
${A_{i}}=\left|{G_{i}^{(n-1)}-{p_{i}}{C_{{l_{1}}}}({\rho_{i}})}\right|.$

It should be noted that we only consider the case where

$G_{k}^{(n-1)}-{p_{k}}{C_{{l_{1}}}}({\rho_{k}}),k=1,2,\cdots,n$

are not all zeros. If

$G_{k}^{(n-1)}-{p_{k}}{C_{{l_{1}}}}({\rho_{k}})=0,k=1,2,\cdots,n,$

then 0 can act as a lower bound of ${C_{{l_{1}}}}(\rho)$.

Proof: First, we analyze the right-hand side of Eq.(1). In the case of $n=2$,
it holds that

$G_{1}^{(1)}={p_{2}}{C_{{l_{1}}}}({\rho_{2}}),G_{2}^{(1)}={p_{1}}{C_{{l_{1}}}}({\rho_{1}}).$
(2)

Therefore, from Lemma 2, we have

$\displaystyle\quad\
\frac{1}{2}\left[{(G_{1}^{(1)}+{p_{1}}{C_{{l_{1}}}}({\rho_{1}}))+(G_{2}^{(1)}+{p_{2}}{C_{{l_{1}}}}({\rho_{2}}))}\right]$
$\displaystyle=\frac{1}{2}\left[{2({p_{1}}{C_{{l_{1}}}}({\rho_{1}})+{p_{2}}{C_{{l_{1}}}}({\rho_{2}}))}\right]$
$\displaystyle={p_{1}}{C_{{l_{1}}}}({\rho_{1}})+{p_{2}}{C_{{l_{1}}}}({\rho_{2}})\geq{C_{{l_{1}}}}(\rho).$

Hence, the conclusion is true when $n=2$.

Assuming that the conclusion is true when $n=m$, let us analyze the case of
$n=m+1$. For $\forall k\in\\{1,2,\cdots,m+1\\}$,

$\displaystyle\sum\limits_{i\neq
k}{{\textstyle{\frac{{p_{i}}}{1-{p_{k}}}}}{\rho_{i}}}$

remains a density matrix. According to Lemma 2, we have

$\displaystyle{C_{{l_{1}}}}(\rho)$
$\displaystyle={C_{{l_{1}}}}((1-{p_{k}})(\sum\limits_{i\neq
k}{{\textstyle{\frac{{p_{i}}}{1-{p_{k}}}}}{\rho_{i}}})+{p_{k}}{\rho_{k}})$
$\displaystyle\leq(1-{p_{k}}){C_{{l_{1}}}}(\sum\limits_{i\neq
k}{{\textstyle{{\frac{p_{i}}{1-{p_{k}}}}}{\rho_{i}}}})+{p_{k}}{C_{{l_{1}}}}({\rho_{k}})$
$\displaystyle=G_{k}^{(m)}+{p_{k}}{C_{{l_{1}}}}({\rho_{k}}).$

For any arbitrary $k$, it holds that

$\displaystyle(m+1){C_{{l_{1}}}}(\rho)=\sum\limits_{k=1}^{m+1}{{C_{{l_{1}}}}(\rho)}\leq\sum\limits_{k=1}^{m+1}(G_{k}^{(m)}+{p_{k}}{C_{{l_{1}}}}({\rho_{k}})).$

In other words,

$\displaystyle{C_{{l_{1}}}}(\rho)\leq\frac{1}{{m+1}}\sum\limits_{k=1}^{m+1}(G_{k}^{(m)}+{p_{k}}{C_{{l_{1}}}}({\rho_{k}})).$

Hence, the inequality also holds when $n=m+1$. Thus the right side of Eq.(1)
is true.

Next, we consider the left side of Eq.(1).

First, we analyze the case of $n=2$. From Eq.$(\ref{2})$, we have

$\displaystyle\frac{{{{\left|{{p_{2}}{C_{{l_{1}}}}({\rho_{2}})-{p_{1}}{C_{{l_{1}}}}({\rho_{1}})}\right|}^{2}}+{{\left|{{p_{1}}{C_{{l_{1}}}}({\rho_{1}})-{p_{2}}{C_{{l_{1}}}}({\rho_{2}})}\right|}^{2}}}}{{2\left|{{p_{1}}{C_{{l_{1}}}}({\rho_{1}})-{p_{2}}{C_{{l_{1}}}}({\rho_{2}})}\right|}}$
$\displaystyle=\left|{{p_{1}}{C_{{l_{1}}}}({\rho_{1}})-{p_{2}}{C_{{l_{1}}}}({\rho_{2}})}\right|\leq{C_{{l_{1}}}}(\rho).$

Thus, the conclusion holds when $n=2$.

Assuming that the conclusion holds when $n=m$, we now consider the case of
$n=m+1$.

For $\forall k\in\\{1,2,\cdots,m+1\\}$, according to Lemma 1, we have

$\displaystyle{C_{{l_{1}}}}(\rho)$
$\displaystyle={C_{{l_{1}}}}((1-{p_{k}})(\sum\limits_{i\neq
k}{{\textstyle{\frac{{p_{i}}}{1-{p_{k}}}}}{\rho_{i}}})+{p_{k}}{\rho_{k}})$
$\displaystyle\geq\left|{(1-{p_{k}}){C_{{l_{1}}}}(\sum\limits_{i\neq
k}{{\textstyle{\frac{{p_{i}}}{1-{p_{k}}}}}{\rho_{i}}})-{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}\right|$
$\displaystyle=\left|{G_{k}^{(m)}-{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}\right|.$

Based on Lemma 3, the conclusion is true for the case of $n=m+1$.

Combined with the above analysis, Eq.(1) is true for any arbitrary natural
number $n$.

Furthermore, we can prove that our estimate of coherence is more accurate than
that provided by Lemma 2.

######  Theorem 2.

If $\rho$ can be expressed as a convex combination of $n$ states, that is,
$\rho=\sum\nolimits_{i=1}^{n}{{p_{i}}{\rho_{i}}}$, then

$\frac{1}{n}\sum\limits_{k=1}^{n}{\left|{G_{k}^{(n-1)}-{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}\right|}\leq\frac{{\sum\nolimits_{i=1}^{n}{A_{i}^{2}}}}{{\sum\nolimits_{i=1}^{n}{{A_{i}}}}},$
(3)
$\frac{1}{n}\sum\limits_{k=1}^{n}(G_{k}^{(n-1)}+{p_{k}}{C_{{l_{1}}}}({\rho_{k}}))\leq\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}({\rho_{k}})},$
(4)

where
${A_{i}}=|G_{i}^{(n-1)}-{p_{i}}{C_{{l_{1}}}}({\rho_{i}})|,i=1,2,\cdots,n$.

Proof: Assume that $\rho$ can be expressed as a convex combination of $n$
states, i.e., $\rho=\sum\nolimits_{i=1}^{n}{{p_{i}}{\rho_{i}}}$. Based on
Lemma 3, Eq.(3) is true.

We also have

$\displaystyle\quad\
\frac{1}{n}\sum\limits_{k=1}^{n}(G_{k}^{(n-1)}+{p_{k}}{C_{{l_{1}}}}({\rho_{k}}))$
$\displaystyle=\frac{1}{n}\sum\limits_{k=1}^{n}{G_{k}^{(n-1)}}+\frac{1}{n}\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}$
$\displaystyle=\frac{1}{n}\sum\limits_{k=1}^{n}{(1-{p_{k}}){C_{{l_{1}}}}(\sum\limits_{i\neq
k}{{\textstyle{\frac{{p_{i}}}{1-{p_{k}}}}}{\rho_{i}}})}+\frac{1}{n}\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}$
$\displaystyle\leq\frac{1}{n}\sum\limits_{k=1}^{n}{(1-{p_{k}})\sum\limits_{i\neq
k}{{\textstyle{\frac{{p_{i}}}{1-{p_{k}}}}}}{C_{{l_{1}}}}({\rho_{i}})}+\frac{1}{n}\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}$
$\displaystyle=\frac{1}{n}\sum\limits_{k=1}^{n}{\sum\limits_{i\neq
k}p_{i}{{C_{{l_{1}}}}({\rho_{i}})}}+\frac{1}{n}\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}$
$\displaystyle=\frac{(n-1)}{n}\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}+\frac{1}{n}\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}$
$\displaystyle=\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}({\rho_{k}})}.$

Thus, Eq.(4) holds.

##  III  Generalized triangular inequality for quantum coherence based on
convex-roof $l_{1}$-norm

The convex-roof $l_{1}$-norm is another method for measuring coherence17 . The
convex-roof $l_{1}$-norm of a mixed state is defined as

$\displaystyle\mathaccent
869{{C_{{l_{1}}}}}(\rho)=\mathop{\min}\limits_{\\{{p_{i}},|{\psi_{i}}\rangle\\}}\sum\limits_{i}{{p_{i}}{C_{{l_{1}}}}(|{\psi_{i}}\rangle)}.$

However, we are unfamiliar with the convex-roof $l_{1}$-norm owing to the
difficulty in its calculation. To date, we can only calculate the convex-roof
$l_{1}$-norm of a two-dimensional state17 ; 18  and some special high-
dimensional states19 . In this study, we prove that, for certain three-
dimensional quantum states, ${C_{{l_{1}}}}\left(\rho\right)=\mathaccent
869{{C_{{l_{1}}}}}\left(\rho\right)$ holds. We also provide a distinct
triangle-like inequality in the convex-roof $l_{1}$-norm of coherence. To this
end, we begin with the two lemmas mentioned in Ref.17  and Ref.14 .

######  Lemma 4.

For any quantum state $\rho$, we have

${C_{{l_{1}}}}(\rho)\leq\mathaccent 869{{C_{{l_{1}}}}}(\rho).$

######  Lemma 5.

If $\rho$ is a pure state, then

${C_{{l_{1}}}}(\rho)=\mathaccent 869{{C_{{l_{1}}}}}(\rho).$

According to Lemma 4, we obtain the following theorem.

######  Theorem 3.

Let $\rho$ be a three-dimensional quantum state with $rank(\rho)=3$. If the
matrix of $\rho$ under reference basis
$\left\\{{|i\rangle}\right\\}_{i=1}^{d}$ has the following form

$\displaystyle\rho=\left({\begin{array}[]{*{20}{c}}{{\rho_{11}}}&{{\rho_{12}}}&0\\\
{\overline{{\rho_{12}}}}&{{\rho_{22}}}&0\\\
0&0&{{\rho_{33}}}\end{array}}\right),$

then

${C_{{l_{1}}}}(\rho)=\mathaccent 869{{C_{{l_{1}}}}}(\rho).$

Proof: Select $|{\psi_{1}}\rangle,|{\psi_{2}}\rangle,|{\psi_{3}}\rangle$, and
${p_{1}},{p_{2}},{p_{3}}$ as follows:

$|{\psi_{1}}\rangle=\frac{1}{{\sqrt{{p_{1}}}}}\left({\sqrt{{\rho_{11}}}|1\rangle+\frac{{\left|{{\rho_{12}}}\right|}}{{\sqrt{{\rho_{11}}}}}{e^{-iarg{\rho_{12}}}}|2\rangle}\right),$
$|{\psi_{2}}\rangle=|2\rangle,|{\psi_{3}}\rangle=|3\rangle,\\\ $
${p_{1}}={\rho_{11}}+\frac{{{{\left|{{\rho_{12}}}\right|}^{2}}}}{{{\rho_{11}}}},{p_{2}}={\rho_{22}}-\frac{{{{\left|{{\rho_{12}}}\right|}^{2}}}}{{{\rho_{11}}}},{p_{3}}={\rho_{33}}.$

Thus,

$\rho={p_{1}}|{\psi_{1}}\rangle\langle{\psi_{1}}|+{p_{2}}|{\psi_{2}}\rangle\langle{\psi_{2}}|+{p_{3}}|{\psi_{3}}\rangle\langle{\psi_{3}}|.$

and

${p_{1}}{C_{{l_{1}}}}\left({|{\psi_{1}}\rangle}\right)+{p_{2}}{C_{{l_{1}}}}\left({|{\psi_{2}}\rangle}\right)+{p_{3}}{C_{{l_{1}}}}\left({|{\psi_{3}}\rangle}\right)={C_{{l_{1}}}}\left(\rho\right).$

According to the definition of the convex-roof $l_{1}$-norm, we have

${C_{{l_{1}}}}(\rho)\geq\mathaccent 869{{C_{{l_{1}}}}}(\rho).$

Thus, it holds that

${C_{{l_{1}}}}(\rho)=\mathaccent 869{{C_{{l_{1}}}}}(\rho).$

Ref.15  provides the following triangle-like inequality regarding the convex-
roof $l_{1}$-norm by using pure state decomposition.

######  Lemma 6.

If $\rho$ can be expressed as a convex combination of $n$ linearly independent
pure states, that is,
$\rho=\sum\nolimits_{i=1}^{n}{{p_{i}}|{\psi_{i}}\rangle\langle{\psi_{i}}|}$,
then the convex-roof $l_{1}$-norm satisfies the following triangle-like
inequality

$\frac{1}{n}\sum\limits_{k=1}^{n}{\left|{\mathaccent
869{G}_{k}^{(n-1)}-{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}\left({|{\psi_{k}}\rangle}\right)}\right|\leq}\mathaccent
869{{C_{{l_{1}}}}}\left(\rho\right)\leq\sum\limits_{i}{{p_{i}}\mathaccent
869{{C_{{l_{1}}}}}\left({|{\psi_{i}}\rangle}\right)},$ (5)

where $\mathaccent
869{{C_{{l_{1}}}}}\left({|{\psi_{i}}\rangle}\right)={C_{{l_{1}}}}\left({|{\psi_{i}}\rangle}\right)$
is the convex-roof $l_{1}$-norm of the pure state
$|{\psi_{i}}\rangle\left({1\leq i\leq n}\right)$ and $\mathaccent
869{G}_{k}^{(n-1)}\left({1\leq k\leq n}\right)$ is the lower bound of
$\sum\nolimits_{i\neq k}^{n}{{p_{i}}\mathaccent
869{{C_{{l_{1}}}}}}\left({\frac{{\sum\nolimits_{j\neq
k}^{n}{{p_{j}}{\rho_{j}}}}}{{\sum\nolimits_{t\neq k}^{n}{{p_{t}}}}}}\right)$.

More importantly, while the correctness of the left side of Eq.(5) can be
discussed, the right side of Eq.(5) can be improved.

Next, we present a new general triangle-like inequality for the convex-roof
$l_{1}$-norm measure of coherence.

######  Theorem 4.

Let $\rho$ be a mixed state that can be expressed as a convex combination of
$n\left({n\geq 2}\right)$ linearly independent pure states:
$|{\psi_{1}}\rangle,|{\psi_{2}}\rangle,\cdots,|{\psi_{n}}\rangle$, i.e.
$\rho=\sum\nolimits_{i=1}^{n}{{p_{i}}|{\psi_{i}}\rangle\langle{\psi_{i}}|}$.
If we know some off-diagonal element $\rho{}_{st}$ of $\rho$, then

$\frac{{{{\left|{{\rho_{st}}}\right|}^{2}}}}{B}\leq\mathaccent
869{{C_{{l_{1}}}}}(\rho)\leq\frac{1}{n}\sum\limits_{k=1}^{n}{\left[{\mathaccent
869{G}_{k}^{(n-1)}+{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}(|{\psi_{k}}\rangle)}\right]},$ (6)

where

$B={{n(\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}(|{\psi_{k}}\rangle)}+1)}},$
$\mathaccent 869{G}_{k}^{(n-1)}=(1-{p_{k}})\mathaccent
869{{C_{{l_{1}}}}}(\sum\limits_{i\neq
k}{{\textstyle{\frac{{p_{i}}}{1-{p_{k}}}}}|{\psi_{k}}\rangle)},k=1,\cdots,n.$

Proof: We use mathematical induction to prove this theorem. First, we consider
the case of $n=2$. Assume that $\rho={p_{1}}{\rho_{1}}+{p_{2}}{\rho_{2}}$,
where ${\rho_{1}}and{\rho_{2}}$ are two linearly independent pure states.
According to the definition of the convex-roof $l_{1}$-norm, we have

$\displaystyle{\mathaccent 869{C}_{{l_{1}}}}(\rho)\leq{p_{1}}\mathaccent
869{{C_{{l_{1}}}}}({\rho_{1}})+{p_{2}}\mathaccent
869{{C_{{l_{1}}}}}({\rho_{2}}).$

According to the definition, we can directly obtain

$\mathaccent 869{G}_{1}^{(1)}={p_{2}}\mathaccent
869{{C_{{l_{1}}}}}({\rho_{2}}),\mathaccent 869{G}_{2}^{(1)}={p_{1}}\mathaccent
869{{C_{{l_{1}}}}}({\rho_{1}}).$

Furthermore, we have

$\displaystyle{\mathaccent 869{C}_{{l_{1}}}}(\rho)$
$\displaystyle\leq{p_{1}}\mathaccent
869{{C_{{l_{1}}}}}({\rho_{1}})+{p_{2}}\mathaccent
869{{C_{{l_{1}}}}}({\rho_{2}})$ $\displaystyle=\frac{1}{2}\left[{(\mathaccent
869{G}_{1}^{(1)}+{p_{1}}\mathaccent
869{{C_{{l_{1}}}}}({\rho_{1}}))+(\mathaccent
869{G}_{2}^{(1)}+{p_{2}}\mathaccent 869{{C_{{l_{1}}}}}({\rho_{2}}))}\right].$

Therefore, the conclusion is true when $n=2$.

Assuming that the conclusion holds in the case of $n=m$, we now analyze the
case of $n=m+1$.

For $\forall k\in\\{1,2,\cdots,m+1\\}$, we have

$\displaystyle\mathaccent 869{{C_{{l_{1}}}}}(\rho)$
$\displaystyle\leq(1-{p_{k}})\mathaccent 869{{C_{{l_{1}}}}}(\sum\limits_{i\neq
k}{{\textstyle{\frac{{p_{i}}}{1-{p_{k}}}}}{\rho_{i}}})+{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}({\rho_{k}})$ (7) $\displaystyle=\mathaccent
869{G}_{k}^{(n-1)}+{p_{k}}\mathaccent 869{{C_{{l_{1}}}}}({\rho_{k}}).$

Summing over all the $k$ in Eq.(7), we have

$\displaystyle\mathaccent
869{{C_{{l_{1}}}}}(\rho)\leq\frac{1}{{m+1}}\sum\limits_{k=1}^{m+1}(\mathaccent
869{G}_{k}^{(m)}+{p_{k}}\mathaccent 869{{C_{{l_{1}}}}}({\rho_{k}})).$

In other words, the conclusion is true in the case of $n=m+1$. Thus, the right
side of Eq.(6) holds.

Next, we analyze the left side of Eq. (6). Assume that
$\left\\{{{q_{k}},|{\phi_{k}}\rangle}\right\\}$ is the optimal decomposition
of $\rho$, i.e.,

$\displaystyle\mathaccent
869{{C_{{l_{1}}}}}(\rho)=\sum\limits_{k=1}^{n}{{q_{k}}{C_{{l_{1}}}}(|{\phi_{k}}\rangle)}.$

Let
$|{\phi_{k}}\rangle=\sum\nolimits_{j=1}^{d}{{\phi_{kj}}{e^{i{\theta_{kj}}}}|j\rangle},k=1,2,\cdots,n$,
where ${\phi_{kj}}$ are non-negative real numbers and ${\theta_{kj}}$ are real
numbers. Thus,

$\displaystyle|{\phi_{k}}\rangle\langle{\phi_{k}}|$
$\displaystyle=\left({\sum\nolimits_{j=1}^{d}{{\phi_{kj}}{e^{i{\theta_{kj}}}}|j\rangle}}\right)\left({\sum\nolimits_{t=1}^{d}{{\phi_{kt}}{e^{-i{\theta_{kt}}}}\langle
t|}}\right)$
$\displaystyle=\left({\sum\nolimits_{j=1}^{d}{\sum\nolimits_{t=1}^{d}{{\phi_{kj}}{\phi_{kt}}{e^{i({\theta_{kj}}-{\theta_{kt}})}}|j\rangle\langle
t|}}}\right).$

Therefore,

${C_{{l_{1}}}}(|{\phi_{k}}\rangle)=\sum\limits_{i\neq
j}{{\phi_{ki}}{\phi_{kj}}}.$ (8)

Then, we have

$\displaystyle\mathaccent 869{{C_{{l_{1}}}}}(\rho)$
$\displaystyle=\sum\limits_{k=1}^{n}{{q_{k}}{C_{{l_{1}}}}(|{\phi_{k}}\rangle)}=\sum\limits_{k=1}^{n}{{q_{k}}\sum\limits_{i\neq
j}{{\phi_{ki}}{\phi_{kj}}}}$
$\displaystyle\geq\sum\limits_{k=1}^{n}{{q_{k}}\frac{{\sum\limits_{i\neq
j}{{\phi_{ki}}{\phi_{kj}}}}}{{1+\sum\limits_{i\neq
j}{{\phi_{ki}}{\phi_{kj}}}}}}\geq\frac{{\sum\limits_{k=1}^{n}{q_{k}^{2}\sum\limits_{i\neq
j}{{\phi_{ki}}{\phi_{kj}}}}}}{{\sum\limits_{k=1}^{n}{{q_{k}}(1+\sum\limits_{i\neq
j}{{\phi_{ki}}{\phi_{kj}}})}}}$
$\displaystyle\geq\frac{{\sum\limits_{k=1}^{n}{q_{k}^{2}\sum\limits_{i\neq
j}{{\phi_{ki}}{\phi_{kj}}}}}}{{\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}(|{\psi_{k}}\rangle)}+1}}=\frac{{\sum\limits_{k=1}^{n}{\sum\limits_{i\neq
j}{{{\left({\sqrt{q_{k}{\phi_{ki}}{\phi_{kj}}}}\right)}^{2}}}}}}{{\sum\limits_{k=1}^{n}{{p_{k}}{C_{{l_{1}}}}(|{\psi_{k}}\rangle)}+1}}$
$\displaystyle\geq\frac{{\sum\limits_{i\neq
j}{{{\left({\sum\limits_{k=1}^{n}{\sqrt{q_{k}{\phi_{ki}}{\phi_{kj}}}}}\right)}^{2}}}}}{B}\geq\frac{{\sum\limits_{i\neq
j}{{{\left|{{\rho_{ij}}}\right|}^{2}}}}}{B}\geq\frac{{{{\left|{{\rho_{st}}}\right|}^{2}}}}{B}.$

We can also prove that the upper bound provided by theorem 4 is better than
that provided by Lemma 6.

######  Theorem 5.

If a mixed state $\rho$ can be expressed as a convex combination of $n$ pure
states, $|{\psi_{1}}\rangle,|{\psi_{2}}\rangle,\cdots,|{\psi_{n}}\rangle$,
that is,

$\rho=\sum\nolimits_{i=1}^{n}{{p_{i}}|{\psi_{i}}\rangle\langle{\psi_{i}}|},$

then it holds that

$\displaystyle\frac{1}{n}\sum\limits_{k=1}^{n}{\left[{\mathaccent
869{G}_{k}^{(n-1)}+{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}(|{\psi_{k}}\rangle)}\right]}\leq\sum\limits_{k=1}^{n}{{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}(|{\psi_{k}}\rangle)}.$

Proof: Assume that a mixed state $\rho$ can be expressed as a convex
combination of $n$ pure states
$|{\psi_{1}}\rangle,|{\psi_{2}}\rangle,\cdots,|{\psi_{n}}\rangle$, that is,

$\rho=\sum\nolimits_{i=1}^{n}{{p_{i}}|{\psi_{i}}\rangle\langle{\psi_{i}}|}.$

Thus, we have

$\displaystyle\quad\ \frac{1}{n}\sum\limits_{k=1}^{n}{\left[{\mathaccent
869{G}_{k}^{(n-1)}+{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}(|{\psi_{k}}\rangle)}\right]}$
$\displaystyle=\frac{1}{n}\sum\limits_{k=1}^{n}{\mathaccent
869{G}_{k}^{(n-1)}}+\frac{1}{n}\sum\limits_{k=1}^{n}{{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}(|{\psi_{k}}\rangle)}$
$\displaystyle=\frac{1}{n}\sum\limits_{k=1}^{n}{(1-{p_{k}})\mathaccent
869{{C_{{l_{1}}}}}(\sum\limits_{i\neq
k}{{\textstyle{\frac{{p_{i}}}{1-{p_{k}}}}}|{\psi_{i}}\rangle})}+\frac{1}{n}\sum\limits_{k=1}^{n}{{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}(|{\psi_{k}}\rangle)}$
$\displaystyle\leq\frac{1}{n}\sum\limits_{k=1}^{n}{(1-{p_{k}})\sum\limits_{i\neq
k}{{\textstyle{\frac{{p_{i}}}{1-{p_{k}}}}}\mathaccent
869{{C_{{l_{1}}}}}(|{\psi_{i}}\rangle)}}+\frac{1}{n}\sum\limits_{k=1}^{n}{{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}(|{\psi_{k}}\rangle)}$
$\displaystyle=\frac{{(n-1)}}{n}\sum\limits_{k=1}^{n}{{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}(|{\psi_{k}}\rangle)}+\frac{1}{n}\sum\limits_{k=1}^{n}{{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}(|{\psi_{k}}\rangle)}$
$\displaystyle=\sum\limits_{k=1}^{n}{{p_{k}}\mathaccent
869{{C_{{l_{1}}}}}(|{\psi_{k}}\rangle)}.$

##  IV Different upper bounds of coherence

In this section, we detail the evaluation of the coherence of a quantum state
using its properties. Ref.16 , using the dimension of a quantum state,
recorded the following conclusion:

######  Lemma 7.

For any $d$-dimensional quantum state $\rho$, it holds that
${C_{{l_{1}}}}(\rho)\leq d-1$. Moreover, ${C_{{l_{1}}}}(\rho)$ reaches the
upper bound when $\rho$ is in the maximally mixed state:

$|\psi\rangle=\frac{1}{{\sqrt{d}}}\sum\nolimits_{i=1}^{d}{|i\rangle}.$

However, in some cases, the evaluation provided by Lemma 7 is highly
inaccurate. For example,

$\rho=diag\\{{\lambda_{1}},{\lambda_{2}},\cdots,{\lambda_{d}}\\}.$

With additional conditions, Lemma 7 can be improved. To this end, we introduce
the following lemma 20 .

######  Lemma 8.

Consider that $A=[{a_{ij}}]$ is positive semi-definite and that for some
$k\in\\{1,\cdots,n\\}$, it holds that ${a_{kk}}=0$. Then, for each
$i\in\\{1,\cdots,n\\}$, we have ${a_{ik}}={a_{ki}}=0$.

######  Theorem 6.

Let $\rho$ be a $d$-dimensional quantum state. If there exists $r$ null
elements in the main diagonal of $\rho$, then

$\displaystyle{C_{{l_{1}}}}(\rho)\leq d-r-1.$ (9)

Proof: Assume that $\rho$ is a $d$-dimensional quantum state satisfying

${\rho_{{k_{i}},{k_{i}}}}\neq
0,i=1,2,\cdots,d-r({k_{1}}<{k_{2}}<\cdots<k_{d-r}).$

Let the matrix of $\rho$ under reference basis
$\left\\{{|i\rangle}\right\\}_{i=1}^{d}$ be

$\rho=\left({\begin{array}[]{*{20}{c}}{{\rho_{11}}}&{{\rho_{12}}}&\cdots&{{\rho_{1d}}}\\\
{{\rho_{21}}}&{{\rho_{22}}}&\cdots&{{\rho_{2d}}}\\\
\vdots&\vdots&{}\hfil&\vdots\\\
{{\rho_{d1}}}&{{\rho_{d2}}}&\cdots&{{\rho_{dd}}},\end{array}}\right).$

Define

$\rho^{\prime}=\left({\begin{array}[]{*{20}{c}}{{\rho_{{k_{1}},{k_{1}}}}}&*&\cdots&*\\\
&{{\rho_{{k_{2}},{k_{2}}}}}&{}\hfil&*\\\ \vdots&{}\hfil&\ddots&\vdots\\\
&*&\cdots&{{\rho_{{k_{d-r}},{k_{d-r}}}}}\end{array}}\right).$

where $\rho^{\prime}$ is obtained via the permutation of each element of
$\rho$ whose row index or column index is $k_{i},i=1,2,\cdots,d-r$ in the same
order. Furthermore, all the principal minors of $\rho^{\prime}$ are the
principal minors of $\rho$. Therefore, all the principal minors of
$\rho^{\prime}$ are non-negative, given that $\rho$ is positive semi-definite.

In addition,

$tr(\rho^{\prime})=tr(\rho)=1.$

Hence, $\rho^{\prime}$ is a density matrix, the dimension of which is $d-r$.

According to Lemma 7 and Lemma 8, we can directly obtain

${C_{{l_{1}}}}(\rho)={C_{{l_{1}}}}({\rho^{\prime}})\leq d-r-1.$

Furthermore, we provide another two distinct upper bounds of the $l_{1}$-norm.

######  Theorem 7.

Assume that $\rho$ is a $d$-dimensional quantum state. For convenience, the
matrix of $\rho$ under the reference basis
$\left\\{{|i\rangle}\right\\}_{i=1}^{d}$ is still denoted by $\rho$. All
integers $k(2\leq k\leq d)$ are assembled such that
${{\rho_{k,1}},{\rho_{k+1,2}},\cdots{\rho_{d,d-k+1}}}$ are not complete zeros
in a set ${S_{\rho}}$. Defining the number of elements in ${S_{\rho}}$ as $t$,
we have

${C_{{l_{1}}}}(\rho)\leq 2t.$

Proof: Assume that $\rho$ is a $d$-dimensional quantum state. Let

${S_{\rho}}=\left\\{{r_{1},r_{2},\cdots,r_{t}}\right\\}.$

All the principal minors of $\rho$ are non-negative because $\rho$ is positive
semi-definite, that is,

$\displaystyle\rho\left({\begin{array}[]{*{20}{c}}i&j\\\
i&j\end{array}}\right)=\left|{\begin{array}[]{*{20}{c}}{{\rho_{ii}}}&{{\rho_{ij}}}\\\
{{\rho_{ji}}}&{{\rho_{jj}}}\end{array}}\right|={\rho_{ii}}{\rho_{jj}}-{\left|{{\rho_{ij}}}\right|^{2}}\geq
0.$

Furthermore, we have

$\displaystyle\left|{{\rho_{ij}}}\right|\leq\sqrt{{\rho_{ii}}{\rho_{jj}}}\leq\frac{1}{2}({\rho_{ii}}+{\rho_{jj}}).$

It should be noted that $\rho$ is Hermite; hence, it follows that

$\displaystyle{C_{{l_{1}}}}(\rho)$ $\displaystyle=\sum\limits_{i\neq
j}{\left|{{\rho_{ij}}}\right|}=2\sum\limits_{i=1}^{t}{\sum\limits_{j=1}^{d-{r_{i}}+1}{\left|{{\rho_{{r_{i}}+j-1,j}}}\right|}}$
$\displaystyle\leq
2\sum\limits_{i=1}^{t}{\sum\limits_{j=1}^{d-{r_{i}}+1}{\frac{{\left({{\rho_{{r_{i}}+j-1,{r_{i}}+j-1}}+{\rho_{jj}}}\right)}}{2}}}$
$\displaystyle\leq
2{\sum\limits_{j=1}^{d}{{\textstyle{\frac{{\rho_{jj}}}{2}}}}}+2\sum\limits_{i=1}^{t}{\sum\limits_{j=1}^{d-{r_{i}}+1}{\frac{{{\rho_{{r_{i}}+j-1,{r_{i}}+j-1}}}}{2}}}$
$\displaystyle=t+\sum\limits_{i=1}^{t}{\sum\limits_{j=1}^{d-{r_{i}}+1}{{\rho_{{r_{i}}+j-1,{r_{i}}+j-1}}}}.$

We declare that, among the following $t(d+1)-({r_{1}}+{r_{2}}+\cdots+{r_{t}})$
numbers

$\displaystyle\rho_{{r_{1}},{r_{1}}},\rho_{{r_{1}}+1,{r_{1}}+1},\cdots,\rho_{d,d},\cdots,$
$\displaystyle\quad\rho_{{r_{t}},{r_{t}}},\rho_{{r_{t}}+1,{r_{t}}+1},\cdots,\rho_{d,d},$

the same one appears $t$ times at most; thus,

$\displaystyle{C_{{l_{1}}}}(\rho)$ $\displaystyle\leq
t+\sum\limits_{i=1}^{t}{\sum\limits_{j=1}^{d-{r_{i}}+1}{{\rho_{{r_{i}}+j-1,{r_{i}}+j-1}}}}$
$\displaystyle\leq t+t({\rho_{11}}+\cdots+{\rho_{dd}})=2t.$

######  Theorem 8.

Let $\rho$ be a $d$-dimensional $(d\geq 2)$ quantum state. If there exists
${\rho_{ij}}$ such that $|{\rho_{ij}}|=0$, then we have

$\displaystyle{C_{{l_{1}}}}(\rho)\leq\sqrt{\left[{d(d-1)-1}\right]\mu}\leq
d-1,$ (10)

where
$\mu=\sum\limits_{k=1}^{d}{\lambda_{k}^{2}}-\sum\limits_{k=1}^{d}{\rho_{{}_{kk}}^{2}}$
and ${\lambda_{k}}\left({k=1,2,\cdots,d}\right)$ are the eigenvalues of
$\rho$.

Proof: Let $\rho$ be a $d$-dimensional quantum state. According to the
spectral decomposition theorem, there exists a unitary matrix $U$ such that

$\displaystyle\rho=U\left({\begin{array}[]{*{20}{c}}{{\lambda_{1}}}&{}\hfil&{}\hfil&{}\hfil\\\
{}\hfil&{{\lambda_{2}}}&{}\hfil&{}\hfil\\\ {}\hfil&{}\hfil&\ddots&{}\hfil\\\
{}\hfil&{}\hfil&{}\hfil&{{\lambda_{d}}},\end{array}}\right){U^{+}}.$

Then, $\rho^{2}$ can be expressed as

$\displaystyle{\rho^{2}}$
$\displaystyle=U\left({\begin{array}[]{*{20}{c}}{{\lambda_{1}}}&{}\hfil&{}\hfil\\\
{}\hfil&\ddots&{}\hfil\\\
{}\hfil&{}\hfil&{{\lambda_{d}}}\end{array}}\right){U^{+}}U\left({\begin{array}[]{*{20}{c}}{{\lambda_{1}}}&{}\hfil&{}\hfil\\\
{}\hfil&\ddots&{}\hfil\\\
{}\hfil&{}\hfil&{{\lambda_{d}}}\end{array}}\right){U^{+}}$
$\displaystyle=U\left({\begin{array}[]{*{20}{c}}{\lambda_{\rm{1}}^{\rm{2}}}&{}\hfil&{}\hfil\\\
{}\hfil&\ddots&{}\hfil\\\
{}\hfil&{}\hfil&{\lambda_{d}^{2}}\end{array}}\right){U^{+}}.$

Following the expression of $\rho^{2}$ established above, we obtain

$\displaystyle
tr({\rho^{2}})=\sum\limits_{i=1}^{d}{\sum\limits_{j=1}^{d}{|{\rho_{ij}}{|^{2}}}}=\sum\limits_{k=1}^{d}{\lambda_{k}^{2}}.$

Furthermore, we have

$\displaystyle\sum\limits_{i\neq
j}{|{\rho_{ij}}{|^{2}}}=\sum\limits_{k=1}^{d}{\lambda_{k}^{2}}-\sum\limits_{k=1}^{d}{\rho_{{}_{kk}}^{2}}.$

Now, we introduce the modified Cauchy inequality 21 . Assume that $V$ is a
product space; thus, for any $x,y,z\in V$, where $\|z\|=1$, it holds that

$\displaystyle{\left|{(x,y)}\right|^{2}}\leq{\left\|x\right\|^{2}}{\left\|y\right\|^{2}}-G(x,y,z),$

where

$G(x,y,z)={\left({\left\|x\right\|(y,z)-\left\|y\right\|(x,z)}\right)^{2}}.$

The equality holds only if $x,y,z$ are linearly dependent.

Now, we attempt to determine an upper bound of ${C_{{l_{1}}}}(\rho)$ using
this useful tool. Let $V={\mathbb{R}^{n}}(n\geq 2)$ and the standard inner
product be taken as the fixed product. Subsequently, $x,y,z$ is selected in
the following manner:

$\displaystyle
x=({x_{1}},{x_{2}},\cdots,{x_{n}}),{\rm{}}y=(\frac{1}{{\sqrt{n}}},\frac{1}{{\sqrt{n}}},\cdots,\frac{1}{{\sqrt{n}}}),$
$\displaystyle\qquad\qquad{\rm{}}z=(0,0,\cdots,1,0,\cdots,0).$

where the $i$th element of the $n$-tuple $z$ is 1. Owing to the definition of
the standard inner product, we have

$\displaystyle(x,y)=\frac{1}{{\sqrt{n}}}\sum\limits_{i=1}^{n}{{x_{i}}},\left\|x\right\|=\sqrt{\sum\limits_{i=1}^{n}{x_{i}^{2}}},\left\|y\right\|=1,$
$\displaystyle\qquad\qquad(y,z)=\frac{1}{{\sqrt{n}}},(x,z)={x_{i}}.$

Using the modified Cauchy inequality, we have

$\displaystyle\frac{1}{n}{\left({\sum\limits_{i=1}^{n}{{x_{i}}}}\right)^{2}}\leq\sum\limits_{i=1}^{n}{x_{i}^{2}}.{\left({\sqrt{\sum\limits_{i=1}^{n}{x_{i}^{2}}}\frac{1}{{\sqrt{n}}}-{x_{i}}}\right)^{2}}.$

For the convenience of narration, we introduce the following notations:

$\displaystyle\sum\limits_{i=1}^{n}{{x_{i}}}:=u,\sum\limits_{i=1}^{n}{x_{i}^{2}}:=\lambda.$

Then,

$\displaystyle\frac{1}{n}{u^{2}}\leq\lambda-{\left({\sqrt{\lambda}\frac{1}{{\sqrt{n}}}-{x_{i}}}\right)^{2}}.$

In other words,

$\displaystyle\frac{1}{n}{u^{2}}-\lambda+\frac{1}{n}\lambda\leq\frac{2}{{\sqrt{n}}}\sqrt{\lambda}{x_{i}}-x_{i}^{2}.$
(11)

Note that Eq.(11) holds for each $i$; hence,

$\frac{1}{n}{u^{2}}-\lambda+\frac{1}{n}\lambda\leq\min\left\\{{\left.{\sqrt{\frac{{{\rm{4}}\lambda}}{n}}{x_{i}}-x_{i}^{2}}\right|i=1,\
ldots,n}\right\\}.$

If $j\in\\{1,2,\cdots,n\\}$, then ${x_{j}}=0$. Therefore,

$\displaystyle\frac{1}{n}{u^{2}}-\lambda+\frac{1}{n}\lambda\leq 0.$

In other words,

$\displaystyle u\leq\sqrt{(n-1)\lambda}.$

In summary, it holds that

$\displaystyle{C_{{l_{1}}}}(\rho)=\sum\limits_{i\neq
j}{|{\rho_{ij}}|}\leq\sqrt{\left[{d(d-1)-1}\right]\mu},$

where
$\mu=\sum\limits_{k=1}^{d}{\lambda_{k}^{2}}-\sum\limits_{k,=1}^{d}{\rho_{{}_{kk}}^{2}}.$

Next, we focus on

$\displaystyle\sqrt{\left[{d(d-1)-1}\right]\mu}\leq d-1.$

According to the properties of density matrix, we have

$\displaystyle\ \
\sum\limits_{k=1}^{d}{\lambda_{k}^{2}}\leq{\left({\sum\limits_{k=1}^{d}{\lambda_{k}}}\right)^{2}}=1,$
$\displaystyle\sum\limits_{k=1}^{d}{\rho_{{}_{kk}}^{2}}\geq\frac{1}{d}{\left({\sum\limits_{k=1}^{d}{\rho_{{}_{kk}}}}\right)^{\rm{2}}}=\frac{1}{d}.$

Thus, $\mu$ can be expressed as

$\displaystyle\mu\leq 1-\frac{1}{d}=\frac{{d-1}}{d}.$ (12)

From Eq.(12), we can directly obtain

$\displaystyle\sqrt{\left[{d(d-1)-1}\right]\mu}$
$\displaystyle\leq\sqrt{\left[{d(d-1)-1}\right]\frac{{d-1}}{d}}$
$\displaystyle=\sqrt{{{\left({d-1}\right)}^{2}}-\frac{{d-1}}{d}}\leq d-1.$

##  V Conclusion and discussion

We studied the properties of quantum coherence based on the $l_{1}$-norm and
convex-roof $l_{1}$-norm, thus realizing a better generalized triangular
inequality. Furthermore, we proved that, for a certain type of three-
dimensional quantum state $\rho$, it holds that
${C_{{l_{1}}}}\left(\rho\right)=\mathaccent
869{{C_{{l_{1}}}}}\left(\rho\right)$. In addition, we offer a few upper bounds
for the $l_{1}$-norm in different forms according to the properties of a
quantum state.

For future research, we propose the following conjecture: Let $\rho$ be a
three-dimensional quantum state with $rank(\rho)\neq 2$; then,

${C_{{l_{1}}}}(\rho)=\mathaccent 869{{C_{{l_{1}}}}}(\rho).$

The difficulty lies in calculating the convex-roof $l_{1}$-norm of more
general quantum states.

We presented certain previously unreported facts, which we believe will be
helpful in the quantitative estimation of quantum coherence and other quantum
resources.

Acknowledgments: This work is supported by Hainan Provincial Natural Science
Foundation of China under Grant Nos.121RC539 and 121MS030; and by the National
Natural Science Foundation of China under Grant No.11861031. This project is
also supported by the Innovation Platform for Academinicians of Hainan
Province

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