††thanks: Corresponding author

# Energy storage and coherence in closed and open quantum batteries

Francesco Caravelli  Theoretical Division (T4),
Los Alamos National Laboratory, Los Alamos, New Mexico 87545, USA
<EMAIL_ADDRESS>Bin Yan  Center for Nonlinear Studies,
Los Alamos National Laboratory, Los Alamos, New Mexico 87545, USA  Theoretical
Division (T4),
Los Alamos National Laboratory, Los Alamos, New Mexico 87545, USA  Luis Pedro
García-Pintos  Joint Center for Quantum Information and Computer Science and
Joint Quantum Institute, NIST/University of Maryland, College Park, Maryland
20742, USA  Alioscia Hamma  Department of Physics, University of
Massachusetts, Boston, Massachusetts 02125, USA

###### Abstract

We study the role of coherence in closed and open quantum batteries. We obtain
upper bounds to the work performed or energy exchanged by both closed and open
quantum batteries in terms of coherence. Specifically, we show that the energy
storage can be bounded by the Hilbert-Schmidt coherence of the density matrix
in the spectral basis of the unitary operator that encodes the evolution of
the battery. We also show that an analogous bound can be obtained in terms of
the battery’s Hamiltonian coherence in the basis of the unitary operator by
evaluating their commutator. We apply these bounds to a 4-state quantum system
and the anisotropic XY Ising model in the closed system case, and the Spin-
Boson model in the open case.

##  I Introduction

With the improvement of technology able to manufacture and manipulate solid-
state devices, we are now able to harness the properties of physical systems
at the nanometer scale alicki0 ; Kosloff ; levy ; alicki ; LindenPRL2010 ;
campaioli . In this regime, their behavior can be affected by quantum
phenomena, and thermodynamic laws have to be investigated in detail
AlhambraPRX2016 ; 2019arXiv190202357A ; masanes2017general ;
workextractionBrandaoPRL13 ; gallego2016thermodynamic ; Huberreviewthermo2016
; Kosloff2014 ; horod ; sss ; AdC2014 ; Funo17 ; lostaglionature ;
Nieuwenhuizen ; janzig ; horodecki ; brandao ; brandao2 ; lostaglio2 ;
cwiklinski ; lostaglio3 ; goold ; bera ; wilming ; gour ; resourcethermo ;
kwonwork ; Markusmuller2018 . This includes physical systems designed to store
energy aaberg2013truly ; polini ; PoliniPRB2018 ; PoliniPRL2019 ;
PoliniPRB2019 ; andolinasyk ; Caravelli1 ; LewensteinBatteries18 ;
Modiarxiv2017 ; GelbwaserNJP2015 ; FrenzelPRE2014 ; HuberGaussianBatteries2017
; adc ; adc2 ; battp . Genuine quantum phenomena are connected to either some
interference pattern, or to the incompatibility of different observables. Such
notions are unified within the framework of quantum coherence
workextractionAbergPRL14 ; brandner2015coherence ; FrancicaPRE2019 ;
korzekwa2016extraction ; PetruccioneSciRep2019 ; lostaglio2015quantum ;
CoherenceRevMod2017 ; zanardicoh ; brac ; MarvianPRA2016 ; plenio , that, in
simple words, is a way of evaluating the anticommutability of a given
observable with the state of a system plenio . Quantum coherence can be also
described in term of resource theory anders2017focus ;
workextractionPopescuNatComm14 ; oppenheim ; winter . Since resource theories
serve a the bedrock of thermodynamics resourcethermo , it is natural that
quantum coherence and the role of entanglement sss ; erg ; lis ; hammaavg ;
hammaavg2 ; horod  have also been thoroughly studied in the context of quantum
thermodynamics lostaglio2015quantum .

Among the quantum devices capable of performing work, quantum batteries have a
special place. Quantum batteries are of fundamental importance, and are an
area of intense study alicki0 ; aaberg2013truly ; GelbwaserNJP2015 ;
AlhambraPRX2016 ; 2019arXiv190202357A ; levy ; AdCPRL2017cycleengine ;
masanes2017general ; CorreaPRE2013 ; anders2017focus ; LindenPRL2010 ;
workextractionPopescuNatComm14 , in thermodynamics campaioli ; alicki ; polini
; PoliniPRB2019 ; PoliniPRB2018 ; PoliniPRL2019 ; Modiarxiv2017 ;
LewensteinBatteries18 ; Caravelli1 . We model a quantum battery by a
Hamiltonian $H_{0}$ that gives a notion of energy, and a quantum state
$\rho_{t}$ evolving in time as $\mathcal{E}_{t}(\rho)=\rho_{t}$. Here the map
$\mathcal{E}_{t}$ is a generic quantum channel that needs not to be unitary,
as we consider also the possibility of open quantum systems
workextractionPopescuNatComm14 ; luis . The extracted or stored work results
from populating the levels of $H_{0}$ in a different way from the initial
state.

Previous work has shown the importance of quantum coherence in extracting work
from a quantum system. In kwonwork , the authors studied the amount of
coherence in the eigenbasis of the Hamiltonian that can be extracted by a
thermal process. Meanwhile, lostaglionature  shows how the behavior of quantum
coherence poses further constraints to thermodynamic processes, in particular,
coherence transformations are always irreversible. There are already other
various bounds on the performance of quantum batteries LewensteinBatteries18 ;
CorreaPRE2013 ; AdCPRL2017cycleengine ; levy ; luis  and recently some
appeared in which, specifically, coherence has been investigated Plastina ;
cakmak .

In this paper, we obtain quantitative bounds on work, extracted energy, and
power in both closed and open quantum systems in terms of the coherence of
both the state or the Hamiltonian $H_{0}$ in the eigenbasis of the time
evolution operator. By coherence of Hamiltonian in a given basis we mean the
amount of non commutativity of the Hamiltonian in that basis and discuss in
what regimes the bounds are tight. We show applications to several systems of
physical interests: arrays of closed quantum batteries, a spin chain after a
quantum quench, and an open system described in the spin-boson setting.

The paper is organized as follows. In Section II we introduce the coherence
operators which we use in the paper, and obtain upper bounds to the work,
power in terms of the coherence of the battery Hamiltonian and the density
matrix in the basis of the evolution operator. In Section III we show how
these results can be generalized to generic quantum channels, by finding
bounds in terms of both Kraus and Lindblad operators connecting the energy
exchange to the coherence in the basis of these operators. Finally, in Section
IV we show applications of these bounds. For the closed quantum system case,
we consider a 4-level system composed of two interacting spins, and a many
body system given by the anisotropic XY model. We show that our bounds are
fairly tight for small systems in which coherence is important. For the open
system case, we consider a spin-boson model, showing that the coherence bounds
give the right dependence at short times. Conclusions follow.

##  II Coherence and work in closed quantum batteries

###  II.1 Energy storage and coherence bounds

We describe a closed quantum battery by a quantum system on a Hilbert space
$\mathcal{H}\simeq\mathbb{C}^{n}$ with time dependent Hamiltonian
$H(t)=H_{0}+V(t)$. Time evolution generated by the Hamiltonian is unitary and
reads $\dot{\rho}_{t}=i[H,\rho_{t}]$, where we use units in which $\hbar=1$.
The energy of the battery is measured as the expectation value of $H_{0}$ so
that the work performed by or extracted from the battery is

$W(t)=\text{Tr}\Big{(}(\rho_{t}-\rho_{0})H_{0}\Big{)},$

where we used that no entropy change occurs for unitary dynamics. Since we
derive upper bounds to the energy, we also find maximum bounds on the
ergotropy, which is the maximum energy change when maximizing over unitary
operations campaioli , and in particular of the isospectral twirling of the
work, e.g. the spectral-preserving unitary average over the time evolution
Caravelli1 ; isospec1 ; isospec2 , or entangling power adpower ;
CoherenceRevMod2017 .

In fact, we are interested in obtaining bounds in terms of (generalizations
of) quantum coherence. There are different ways of defining quantum coherence
in a quantum system. First of all, there is the coherence of the state in a
given basis $\mathcal{B}=\\{|i\rangle\\}$. The norm-2 of coherence for the
state $\rho$ can be defined as the weight in norm 2 of the off diagonal
elements in the basis $\mathcal{B}$, namely $\mathbb{C}(\rho):=\sum_{i\neq
j}|\rho_{ij}|^{2}$ plenio ; zanardi2000entangling ; zanardi2001entanglement ;
zanardicoh ; styliaris2019coherence . If we define the dephasing super-
operator $\mathcal{D}(\cdot)=\sum_{i}\Pi_{i}\cdot\Pi_{i}$, with
$\Pi_{i}=|i\rangle\langle i|$ are rank one projectors, then in terms of the
dephasing superoperator, we have that

$\mathbb{C}(\rho)=\|\rho-\mathcal{D}(\rho)\|_{F}^{2},$ (1)

where $\|A\|_{F}=\sqrt{{\mbox{Tr\,}}{AA^{\dagger}}}$ is the norm-2 (Frobenius)
norm. A simple calculation shows that
$\mathbb{C}(\rho)={\mbox{Tr\,}}\rho^{2}-{\mbox{Tr\,}}(\mathcal{D}\rho)^{2}$.
As quantum coherence in a state is a basis dependent notion, it is important
to determine what is the relevant basis. In a typical quantum experiment,
quantum coherence is relevant in the basis of observables that will display
interference.

Here, we consider the instantaneous eigenbasis of the unitary evolution
operator $U_{t}$:

$U_{t}=\sum_{j=1}^{n}e^{i\theta_{j}(t)}\Pi_{j}(t),$ (2)

where $\rho_{t}=U_{t}\rho_{0}U_{t}^{\dagger}$. For instance, if one deals with
a time-independent Hamiltonian $H=\sum_{i}\lambda_{i}\Pi_{i}$, then
$U_{t}=\sum_{i}e^{-i\lambda_{i}t}\Pi_{i}$.

One can easily see that the work, expressed as the energy storage, can be
written as Caravelli1

$\displaystyle W\ =W_{A}$ $\displaystyle=$
$\displaystyle\underbrace{\text{Tr}(U_{t}^{\dagger}H_{0}[\rho_{0},U_{t}])}_{A},$
(3) $\displaystyle W\ =W_{B}$ $\displaystyle=$
$\displaystyle\underbrace{\text{Tr}(U_{t}[U_{t}^{\dagger}H_{0},\rho_{0}])}_{B},$
(4) $\displaystyle W\ =W_{C}$ $\displaystyle=$
$\displaystyle\underbrace{\text{Tr}(\rho_{0}[U_{t},U_{t}^{\dagger}H_{0}])}_{C}.$
(5)

Lower and upper bounds to work can be found by applying the Von Neumann’s
trace inequality to the energy storage, resulting in bounds in terms of the
singular values of the Hamiltonian. While these bounds are given for
completeness in App. A, the present paper focuses on bounds in terms of the
norm of the dephased operator.

The identities above show that the amount of work one can extract from a
quantum battery is related to how much the initial state $\rho_{0}$
anticommutes with the evolution operator $U_{t}$ (A), with the operator
$U_{t}^{\dagger}H_{0}$ (B), or how much these last two operators anticommute
with each other (C). The amount of non-commutativity between two operators is
related to how much off diagonal weight one operator has in the eigenbasis of
the other. For a state, this is the definition of quantum coherence. More
generally, we can define an operator-coherence in the basis $\mathcal{B}$ as

$\mathbb{C}(\hat{X}):=\frac{1}{2}\sum_{j}\|[\hat{X},\Pi_{j}]\|^{2}_{F}.$ (6)

We refer to the above super operator $\mathbb{C}(\cdot)$ as generalized
coherence operator and the quantity $\mathbb{C}(\hat{X})$ as the generalized
coherence of $\hat{X}$ in $\mathcal{B}$. In the following, all the definitions
of norms are provided in App. B. The proof of the identity relationship
between Eq. (6) and Eq. (1) is provided for completeness in App. C. For
quantum states, this reduces to the usual definition of norm-2 coherence. We
denote by $\mathbb{C}_{A}$ the operator coherence in the eigenbasis of the
unitary operator $A$. A similar notation applies also if $A$ is a Hermitian
operator. Given the notation above, let us now introduce some useful bounds
that we will use in the following, and proved in App. D.

In general, one has the following Frobenius inequality:

$\displaystyle|\text{tr}(A^{\dagger}B)|\leq\|A\|_{F}\|B\|_{F}.$ (7)

Also, following from Hölder inequality, one has (see App. B) for details):

$\displaystyle|\text{tr}(A^{\dagger}B)|\leq\|A\|\sqrt{\text{r}(B)}\|B\|_{F},$
(8)

where $\|A\|$ is the operator norm, that is, the maximum eigenvalue of the
operator $A$; and $\text{r}(B)$ is the rank of the operator $B$, i.e., the
number of its non-zero eigenvalues. Moreover, we will use the following
inequalities, which are proven in App. D.

Lemma 1 - Single Normal Coherence Inequality. Let $A$ be a normal operator and
let $B$ be an Hermitian operator. Then,

$\displaystyle\|[A,B]\|_{F}^{2}$ $\displaystyle\leq$ $\displaystyle
4\|A\|^{2}\mathbb{C}_{A}(B).$ (9)

A similar lemma applies also to the following commutator.

Lemma 2 - Double Normal Coherence Inequality. Assume A be a normal operator
and let B be Hermitian. Then,

$\|[A^{\dagger},BA]\|_{F}^{2}\leq 4\|A\|^{4}\mathbb{C}_{A}(B).$ (10)

The bound above applies in the case of open systems as we will see soon. Also,
according to Proposition 2 in App. D, one has
$\|[A^{\dagger},BA]\|_{F}^{2}=\|[A^{\dagger},AB]\|_{F}^{2}$. In the following
we will use also the subadditivity property of the rank, e.g. given two
operators $A:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$,
$B:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$, we have
$\text{r}(A+B)\leq\text{r}(A)+\text{r}(B)$. Clearly, however,
$\text{r}(A+B)\leq\text{min}\Big{(}\text{r}(A)+\text{r}(B),n\Big{)}$, and if
$U$ is a unitary operator, $\text{r}(U^{\dagger}AU)=\text{r}(A)$. Also, as a
side remark, we note that by using a different procedure, also bounds in terms
of $l_{1}$ coherence can be obtained (see for instance Lemma 2(b) in App. D).

###  II.2 Bounds in terms of the generalized coherence

Given the definitions above, we prove the following three upper bounds:

$\displaystyle|W(t)|$ $\displaystyle\underbrace{\leq}_{A}$ $\displaystyle
2\|H_{0}\|\sqrt{\text{min}\Big{(}2\text{r}(\rho_{0}),n\Big{)}\mathbb{C}_{U_{t}}(\rho_{0})}$
(11) $\displaystyle\underbrace{\leq}_{B}$ $\displaystyle
2\|\rho_{0}\|\sqrt{\mathbb{C}_{\rho_{0}}(U_{t}^{\dagger}H_{0})}$
$\displaystyle\underbrace{\leq}_{C}$ $\displaystyle
2\|\rho_{0}\|_{F}\sqrt{\mathbb{C}_{U_{t}}(H_{0})}$

with $n$ the dimension of the Hilbert space. We also assumed that
$\rho_{0}:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$. The expressions (A) and
(C) bound the energy storage in terms of the quantum coherence of $\rho_{0}$
and $H_{0}$ in the basis of $U$. One problem with the above bounds is that
both the rank of $H_{0}$ or the $2-$norm operator coherence $\mathbb{C}(X)$
may scale with the dimension $n$ of the Hilbert space. In this case, these
bounds can become very loose for a high-dimensional systems. On the other
hand, for low-dimensional systems, or for large $n$ systems with low
coherence, they are tighter and turn out to be useful.

We study specific models in Sec. IV. The bound (B) relates work to the purity
of the initial state, as $\|\rho\|_{F}^{2}={\mbox{Tr\,}}(\rho^{2})$. For a
pure initial state work is upper bounded by the operator norm. The bound is
saturated when the system is prepared in the highest excited state. One can
obtain a further bound by substituting the operator norm of $\rho$ with its
purity. As a side comment, it is interesting to note that since the bounds
(11) can be also obtained in the interaction representation, this is also true
if we use $\rho_{I}$ in the interacting representation
$\rho_{I}=U_{0}^{\dagger}\rho U_{0}$, $U_{0}=e^{-iH_{0}t}$ and $\Pi_{j}$’s are
in the interaction representation as well.

Inequality A.— We start with the inequality
$|\text{Tr}(AB)|\leq\sqrt{\text{r}(B)}\|A\|\|B\|_{F},$ where $\text{r}(X)$ is
the matrix rank (see App. B for a proof). We now pick $A=H_{0}$ and
$B=U_{t}^{\dagger}[\rho_{0},U_{t}]$. We have
$\|U_{t}^{\dagger}[\rho_{0},U_{t}]\|_{F}\leq\|[\rho_{0},U_{t}]\|_{F}$. Now
note that
$\text{r}(U_{t}^{\dagger}[\rho_{0},U_{t}])\leq\text{min}(2\text{r}(\rho_{0}),n)$
because of the subadditivity of the rank.

If $\rho_{0}$ has a smaller rank, e.g., $\text{r}(\rho_{0})<n/2$, then using
Eq. (9), we obtain

$\displaystyle|W_{A}|$ $\displaystyle\leq$
$\displaystyle\|H_{0}\|\sqrt{2\text{r}(\rho_{0})\|[\rho_{0},U_{t}]\|_{F}^{2}}$
(12) $\displaystyle=$
$\displaystyle\|H_{0}\|\sqrt{2\text{r}(\rho_{0})4\mathbb{C}_{U_{t}}(\rho_{0})}$
$\displaystyle=$ $\displaystyle
2\sqrt{2}\|H_{0}\|\sqrt{\text{r}(\rho_{0})\mathbb{C}_{U_{t}}(\rho_{0})},$

which is the upper bound we reported above.

Inequality B.— If we instead start with the general bound frobnorm ; frobnorm2

$\|[A,B]\|_{F}^{2}\leq
2\|A\|_{F}\|B\|_{F}-2({\mbox{Tr\,}}{A^{\dagger}B})^{2},$ (13)

the bound would be loose, as $\|H_{0}\|_{F}$ is generally very large. We can
then apply the trace inequality of Eq. (7), and obtain the following upper
bound:

$|W_{B}||\text{Tr}(U_{t}[U_{t}^{\dagger}H_{0},\rho_{0}])|\leq\|U_{t}\|_{F}\sqrt{\|[U_{t}^{\dagger}H_{0},\rho_{0}])\|_{F}^{2}}.$
(14)

We can now use $\rho_{0}$ as the operator in which we perform the spectral
basis expansion. Then, using Lemma 1, Eq. (9), we have

$|\text{Tr}(U_{t}[U_{t}^{\dagger}H_{0},\rho_{0}])|\leq
2\|\rho_{0}\|\sqrt{\mathbb{C}_{\rho_{0}}(U_{t}^{\dagger}H_{0})}$ (15)

as claimed.

Inequality C.— We start with the trace inequality of Eq. (7), with
$|\text{Tr}(AB)|\leq\|A\|_{F}\|B\|_{F}$. Using the fact that
$\sqrt{\text{tr}(\rho^{2})}=\|\rho\|_{F}=\sqrt{\text{purity}(\rho)}\leq 1$, we
obtain

$|W_{C}|=\text{Tr}(\rho_{0}[U_{t},U_{t}^{\dagger}H_{0}])\leq\|\rho_{0}\|_{F}\|[U_{t},U_{t}^{\dagger}H_{0}]\|_{F}.$
(16)

Let us now focus on $\|[U_{t},U_{t}^{\dagger}H_{0}]\|_{F}^{2}$. Using the
spectral decomposition for the unitary operator $U_{t}$, we have

(17) $\displaystyle|[U^{\dagger},H_{0}A]\|_{F}^{2}\leq
4\|A\|^{4}\mathbb{C}_{A}(B).$ (18)

We see that the expression above is not in the form which can be directly
expressed in terms of coherence. Given the inequality above, then we have the
following inequality

$\displaystyle|W_{C}|$ $\displaystyle\leq$ $\displaystyle
2\|\rho_{0}\|_{F}\sqrt{\mathbb{C}_{U}(H_{0})},$ (19)

which is the inequality we reported above.

###  II.3 Power and OTOC for closed systems

Similar bounds to those we have obtained for the energy storage in the case of
the closed system can be obtained for the power, that is, for the rate of
change of the energy as a function of time. The energy can be written as

$\displaystyle W(t)$ $\displaystyle=$ $\displaystyle
W(0)+\int_{0}^{t}\frac{d}{dt^{\prime}}W(t^{\prime})dt^{\prime}=W(0)+\int_{0}^{t}P(t^{\prime})dt^{\prime}$
(20) $\displaystyle\leq$ $\displaystyle W(t)\ t\
\text{sup}_{t^{\prime}\in[0,t]}|P(t^{\prime})|.$

This inequality will be useful in later sections in particular in the case of
open systems.

Bounds for the power can then be useful for obtaining bounds on the energy. In
particular, as we will see below, the bounds on the power can be still
expressed in terms of the coherence operator $\mathbb{C}_{Q}(\cdot)$, but
where as we will see in a moment $Q$ is different from the unitary operator.

Let us then consider the power in a closed system. The power is given by

$\displaystyle P_{t}$ $\displaystyle=$
$\displaystyle|\frac{d}{dt}W|=|\text{Tr}\Big{(}H_{0}\dot{\rho}_{t}\Big{)}|$
(21) $\displaystyle=$
$\displaystyle|\text{Tr}\Big{(}H_{0}\Big{[}H_{0}+V,\rho_{t}\Big{]}\Big{)}|.$

Following the three formulations of the work formula as in the previous
sections, we have

$\displaystyle P_{t}=\frac{d}{dt}W$ $\displaystyle=$
$\displaystyle\text{Tr}\Big{(}H_{0}\dot{\rho}_{t}\Big{)}=\text{Tr}\Big{(}H_{0}\Big{[}H_{0}+V,\rho_{t}\Big{]}\Big{)}$
$\displaystyle\underbrace{=}_{A}$
$\displaystyle\text{Tr}\Big{(}H_{0}\Big{[}V,\rho_{t}\Big{]}\Big{)}$
$\displaystyle\underbrace{=}_{B}$
$\displaystyle\text{Tr}\Big{(}V\Big{[}\rho_{t},H_{0}\Big{]}\Big{)}$
$\displaystyle\underbrace{=}_{C}$
$\displaystyle\text{Tr}\Big{(}\rho_{t}\Big{[}H_{0},V\Big{]}\Big{)},$

where in the last equality we introduced the interaction picture operators
$V_{I}=U_{0}^{\dagger}VU_{0}$ and $\rho_{t}^{I}=U_{0}^{\dagger}\rho_{t}U_{0}$.

The bounds we obtain for the power are as follow:

$\displaystyle P$ $\displaystyle\underbrace{\leq}_{A}$ $\displaystyle
2\|H_{0}\|\cdot\|V\|\sqrt{\text{r}([V,\rho_{t}])\cdot\mathbb{C}_{V}(\rho_{t})},$
(22) $\displaystyle P$ $\displaystyle\underbrace{\leq}_{B}$ $\displaystyle
2\|H_{0}\|\cdot\|V\|\sqrt{\text{r}([\rho_{t},H_{0}])\cdot\mathbb{C}_{H_{0}}(\rho_{t})},$
(23) $\displaystyle P$ $\displaystyle\underbrace{\leq}_{C}$ $\displaystyle
2\|\rho_{0}\|_{F}\|V\|\sqrt{\mathbb{C}_{V}(H_{0})},\ $ (24)

where $\epsilon$ and $\bar{v}$ are the maximum eigenvalues of the operators
$H_{0}$ and $V$.

Proof. As for the case of the work, we see that the three different
formulations of the work lead to three different inequalities. We assume that
$H_{0}$ and $V$ are both Hermitian, and that we can write both the Hamiltonian
and the interaction operator in a spectral decomposition

$H_{0}=\sum_{j}\epsilon_{j}\Pi_{j}^{h},\ \ V=\sum_{j}v_{j}\Pi_{j}^{v},$ (25)

where $\Pi_{j}$’s are projector operators on the spectral basis of $H_{0}$ and
$V$ respectively. Let us focus on the first inequality. Using the
decomposition above, inequality A of Eq. (22) can be written as

$\displaystyle P$ $\displaystyle\leq$
$\displaystyle\|H_{0}\|\sqrt{\text{r}([V,\rho_{t}])\cdot\|[V,\rho_{t}]\|_{F}^{2}}.$

We now use Lemma 1, Eq. (9), using as the normal decomposition operator $V$.
We thus have

$P\leq
2\|H_{0}\|\cdot\|V\|\sqrt{\text{r}([V,\rho_{t}])\cdot\mathbb{C}_{V}(\rho_{t})},\\\
$ (26)

where $\mathbb{C}_{V}(\rho_{t})$ is the coherence of $\rho_{t}$ in the
eigenbasis of the interaction operator $V$. Since $\rho_{t}$ has the same rank
of $\rho_{0}$, and $\text{r}(AB)\leq\text{r}(A)$, we have $\text{
r}([V,\rho_{t}])\leq\text{r}(V)+\text{r}(\rho_{0})$. Thus, if $V$ the
interaction of a many body system it can be very large and thus reduce the
applicability or tightness of this bound.

Let us now look at Inequality B. We have

$\displaystyle P$ $\displaystyle\leq$
$\displaystyle\|V\|\sqrt{\text{r}([\rho_{t},H_{0}])\cdot\|[\rho_{t},H_{0}]\|_{F}^{2}}.$
(27)

If we expand $H_{0}=\sum_{k}\epsilon_{k}\Pi_{k}^{h}$, then following the steps
of the previous inequality we obtain

$\displaystyle P$ $\displaystyle\leq$ $\displaystyle 2\|H_{0}\|\
\|V\|\sqrt{\text{r}([\rho_{t},H_{0}])\cdot\mathbb{C}_{H_{0}}(\rho_{t})},$ (28)

where we see that also here we have
$\text{r}([\rho_{t},H_{0}])\leq\text{r}(\rho_{t})+\text{r}(H_{0})$, and for a
many body system such quantity can be very large.

At last, let us now focus on Inequality C of Eq. (23). We can use in this case
Eq. (7), and obtain

$\displaystyle P$ $\displaystyle\leq$ $\displaystyle
2\|\rho_{t}\|_{F}\sqrt{\|[H_{0},V]\|_{F}^{2}},$ (29)

and use Lemma 1 again, in both the eigenbasis of $H_{0}$ and $V$. Note that
since the evolution is unitary, we have $\|\rho_{t}\|_{F}=\|\rho_{0}\|_{F}$.
We thus obtain two equivalent inequalities

$\displaystyle P$ $\displaystyle\leq$ $\displaystyle
2\|\rho_{0}\|_{F}\|V\|\sqrt{\mathbb{C}_{V}(H_{0})}$ (30) $\displaystyle\leq$
$\displaystyle 2\|H_{0}\|\|\rho_{0}\|_{F}\sqrt{\mathbb{C}_{H_{0}}(V)},$

which shows that the the power depends also on the coherence of $H_{0}$ in the
basis of $V$ and viceversa, and on the purity of the state. Clearly, this is a
result of the fact that if the interaction $V$ commutes with $H_{0}$ no work
can be done. Bound $C$ is the most interesting, as $\text{purity}(\rho_{t})<1$
and the rank of $H_{0}$ or $V$ does not appear explicitly. This said, both
${\mathbb{C}}_{H_{0}}(V)$ or ${\mathbb{C}}_{V}(H_{0})$ can be large. $\square$

Interestingly, we note that the bounds for $P$ take the form of Out-of-Time-
Order-Correlators (OTOC) isospec1 ; webqc ; infoscrambling  at infinite
temperature, which grow fast in time for typical $V$ and $\rho$. Moreover,
note that inequalities A and B can be combined to obtain,

$\displaystyle\left(\frac{P}{\|H_{0}\|\cdot\|V\|}\right)^{2}\leq\text{min}\big{(}\mathcal{Q}(\rho_{t}),\mathcal{T}(\rho_{t})\big{)},$
(31)

where we defined
$\mathcal{Q}=\text{r}([\rho_{t},H_{0}])\|[\rho_{t},H_{0}]\|_{F}^{2}$ and
$\mathcal{T}=\text{r}([\rho_{t},V])\|[\rho_{t},V]\|_{F}^{2}$. Here, we see
observe that if $\rho_{t}$ commutes with all projectors, the bound collapses
to zero. Finally, inequality C shows that the power is zero for $[H_{0},V]=0$,
as we discussed for the case of the total work.

##  III Coherence and energy in open quantum batteries

In the previous sections we discussed the case of closed quantum batteries
undergoing unitary evolution. More generally, we can consider the case of non-
unitary quantum evolutions that describe, e.g., open quantum systems.

###  III.1 Bounds in terms of Kraus operators

A generic quantum operation can be described by a quantum channel, that is, a
completely positive, trace preserving map

$\mathcal{E}:\rho\mapsto\rho^{\prime}=\mathcal{E}(\rho)$ (32)

In the Kraus operator sum representation such a map can be described as

$\mathcal{E}(\rho)=\sum_{k}A_{k}\rho A_{k}^{\dagger}$ (33)

where the Kraus operators $A_{k}$ are linear operators satisfying
$\sum_{k}A_{k}^{\dagger}A_{k}=I$. This description can be used to model the
evolution of a subsystem interacting with other quantum systems. Indeed, for a
composite system $\mathcal{H}=\mathcal{H}_{\mathcal{S}}\otimes\mathcal{H}_{B}$
evolving unitarily with a Hamiltonian $H\in\mathcal{B}(\mathcal{H})$ by
$U=\exp(-iHt)$, the evolution of the reduced system is given by a completely
positive map $\rho_{\mathcal{S}}={\mbox{Tr\,}}_{B}(U\rho
U^{\dagger})\equiv\sum_{k}A_{k}\rho_{A}A_{k}^{\dagger}$. For open quantum
systems, or more generally for non unitary evolution, work cannot be
identified with the difference of energy but one has to consider also
variations of entropy. Here we focus thus just on the energy exchanged $\Delta
E(t)$.

Now we move on to generalize the bounds to energy exchange in terms of
coherence for the evolution map $\mathcal{E}_{t}$

Let us define $\rho_{a}(t)=A_{a}\rho(0)A_{a}^{\dagger}$. Also, we define
$\mathcal{E}$ as

$\displaystyle\mathcal{E}(X)=\sum_{a}A_{a}XA_{a}^{\dagger},\ \ \ \
\mathcal{E}_{*}(X)=\sum_{a}A_{a}^{\dagger}XA_{a}.$ (34)

First, we note that because of the Hilbert-Schmidt duality, we have

$\displaystyle\text{Tr}(\mathcal{E}(\rho)H_{0})=\text{Tr}(\rho\mathcal{E}_{*}(H_{0})).$
(35)

Then, we can rewrite

$\displaystyle\Delta E$ $\displaystyle=$
$\displaystyle\sum_{a}\text{Tr}\Big{(}\big{(}\rho_{\mathcal{S}}(0)-\rho_{a}(t)\big{)}H_{0}\Big{)}$
(36) $\displaystyle\underbrace{=}_{A}$
$\displaystyle\sum_{a}\text{Tr}\Big{(}A_{a}H_{0}[A_{a}^{\dagger},\rho_{\mathcal{S}}(0)]\Big{)}$
$\displaystyle\underbrace{=}_{B}$
$\displaystyle\sum_{a}\text{Tr}\Big{(}A_{a}^{\dagger}[\rho_{\mathcal{S}}(0),A_{a}H_{0}]\Big{)}$
$\displaystyle\underbrace{=}_{C}$
$\displaystyle\sum_{a}\text{Tr}\Big{(}\rho_{\mathcal{S}}(0)[A_{a}H_{0},A_{a}^{\dagger}]\Big{)},$

We can upper bound $|\Delta E|$ as in the case of the closed quantum system,
with the difference that now instead of a unique evolution operator, we have a
Kraus representation, which albeit satisfying
$\|\sum_{a}A_{a}^{\dagger}A_{a}\|=\|I\|=1$, are not unitary operators. We can
also write $\|A_{j}^{\dagger}+\sum_{a\neq
j}A_{a}^{\dagger}A_{a}A_{j}^{\dagger}\|=\|A_{j}^{\dagger}\|$.

From the equalities above, we obtain the three following inequalities,
assuming that $A_{a}$ is Hermitian:

$\displaystyle|\Delta E|$ $\displaystyle\underbrace{\leq}_{A}$ $\displaystyle
2\|H_{0}\|\sum_{a}\|A_{a}\|^{2}\sqrt{R_{a}^{1}\mathbb{C}_{a}(\rho_{\mathcal{S}}(0))},$
$\displaystyle|\Delta E|$ $\displaystyle\underbrace{\leq}_{B}$ $\displaystyle
2\|\rho_{S}(0)\|\|H_{0}\|\sum_{a}\|A_{a}\|\sqrt{R_{a}^{2}\
\mathbb{C}_{\rho_{\mathcal{S}}(0)}(A_{a}H_{0})},$ $\displaystyle|\Delta E|$
$\displaystyle\underbrace{\leq}_{C}$ $\displaystyle
2\|\rho_{\mathcal{S}}(0)\|_{F}\sum_{a}\|A_{a}\|^{2}\sqrt{{\mathbb{C}}_{a}(H_{0})},$
(37)

where $R_{a}^{1}=\text{r}([A_{a}^{\dagger},\rho_{\mathcal{S}}])$ and
$R_{a}^{2}=\text{r}([\rho_{\mathcal{S}}(0),A_{a}H_{0}])$;
$\mathbb{C}_{a}(\cdot)$ is the coherence operator in the spectral resolution
of $A_{a}$, while $\mathbb{C}_{\rho_{\mathcal{S}}(0)}(\cdot)$ is the
generalized coherence operator in the spectral basis of the reduced density
matrix.

Proof. As an example, let us consider first the latter equality in Eqs. (36).
We can write, for Equality C,

$\displaystyle\Delta E$ $\displaystyle=$
$\displaystyle\sum_{a}\text{Tr}\Big{(}\rho_{\mathcal{S}}(0)[A_{a}H_{0},A_{a}^{\dagger}]\Big{)}.$

We can thus write, using the trace inequality of Eq. (7), the following
expression:

$\displaystyle|\Delta
E|\leq\|\rho_{\mathcal{S}}(0)\|_{F}\sum_{a}\cdot\sqrt{\|[A_{a}H_{0},A_{a}^{\dagger}]\|_{F}^{2}}.$
(38)

We now make some assumptions about the Kraus operators, i.e., that these are
Hermitian operators, in order to obtain an upper bound for
$\|[A_{a}H_{0},A_{a}]\|_{F}$ in terms of the coherence of the Hamiltonian in
the $A_{a}$ basis. While this assumption is unphysical, it is a loss of
generality and extension of these bounds will be considered in the future.

Using Lemma 2 in Eq. (10), we have that if $A_{a}$ is Hermitian, then,

$\displaystyle|\Delta E|\leq
2\|\rho_{\mathcal{S}}(0)\|_{F}\sum_{a}\|A_{a}\|^{2}\sqrt{{\mathbb{C}}_{a}(H_{0})},$
(39)

where $\mathbb{C}_{a}(\cdot)$ is the coherence operator in the spectral basis
of $A_{a}$,

$\displaystyle\mathbb{C}_{a}(H_{0})=\frac{1}{2}\sum_{j}\|[H_{0},\hat{\Pi}_{a}^{j}]\|_{F}^{2}.$
(40)

The final upper bound above is given in terms of the sum of the coherences of
$\rho_{0}$ with respect to the Kraus operators $A_{a}$, with the assumption
that $\text{r}(H_{0})\gg 1$. While these bounds might be useful in some cases,
typically it is easier to work in terms of Lindblad operators, as we we do in
the next section.

If instead we start from Equality A and B in Eqs. (36), and using the same
procedure we used in the closed system, we obtain the bounds thanks to Lemmas
1 and 2, we obtain the following bounds:

$\displaystyle\|[A_{a}^{\dagger},\rho_{\mathcal{S}}]\|_{F}$
$\displaystyle\leq$ $\displaystyle
2\|A_{a}\|\sqrt{\mathbb{C}_{a}(\rho_{\mathcal{S}}(0))},$
$\displaystyle\|[\rho_{\mathcal{S}}(0),A_{a}H_{0}]\|_{F}$ $\displaystyle\leq$
$\displaystyle
2\|\rho_{S}(0)\|\sqrt{\mathbb{C}_{\rho_{\mathcal{S}}(0)}(A_{a}H_{0})}.$ (41)

If we combine these with the inequality of Eq. (8), we obtain

$\displaystyle|\Delta E|$ $\displaystyle\underbrace{\leq}_{A}$ $\displaystyle
2\|H_{0}\|\sum_{a}\|A_{a}\|^{2}\sqrt{R_{a}^{1}\mathbb{C}_{a}(\rho_{\mathcal{S}}(0))},$
$\displaystyle|\Delta E|$ $\displaystyle\underbrace{\leq}_{B}$ $\displaystyle
2\|\rho_{S}(0)\|\|H_{0}\|\sum_{a}\|A_{a}\|\sqrt{R_{a}^{2}\
\mathbb{C}_{\rho_{\mathcal{S}}(0)}(A_{a}H_{0})},$

where $R_{a}^{1}=\text{r}([A_{a}^{\dagger},\rho_{\mathcal{S}}])$ and
$R_{a}^{2}=\text{r}([\rho_{\mathcal{S}}(0),A_{a}H_{0}])$. $\square$

###  III.2 Bounds in terms of Lindblad operators

Consider the case of a quantum system $\mathcal{H}_{S}$ interacting with a
bath $\mathcal{H}_{B}$. Under standard assumptions such as Markovian evolution
and small couplings to the environment, the evolution of the density matrix
$\rho_{S}$ of the subsystem can be described by the Liouvillian operator
$\mathcal{L}$ through the master equation:

$\displaystyle\frac{d}{dt}\rho_{\mathcal{S}}(t)=\mathcal{L}(\rho_{\mathcal{S}}(t)).$
(42)

We can then write

$\displaystyle\rho_{\mathcal{S}}(t)=\lim_{n\rightarrow\infty}(1+\frac{\mathcal{L}t}{n})^{n}=e^{\mathcal{L}t}\rho_{\mathcal{S}}(0).$
(43)

Then if we assume that $\rho_{\mathcal{S}}(t)$ has a Kraus representation, at
the order $dt$ we must have

$\displaystyle\hat{A}_{0}$ $\displaystyle=$ $\displaystyle
I+dt(-i\hat{H}+\sum_{i=1}\hat{L}_{i}\hat{L}_{i}^{\dagger}),$ (44)
$\displaystyle\hat{A}_{i}$ $\displaystyle=$ $\displaystyle\sqrt{dt}\
\hat{L}_{i},i\geq 1.$ (45)

see, e.g., krauslindblad .

The Kraus operators in fact contain both the battery Hamiltonian and the
interaction with the bath. These enter in a non-trivial way into the density
matrix evolution for the battery in terms of Kraus operators. One way to
overcome this problem is via the fundamental theorem of calculus, which gives

$\displaystyle
E(t)-E(0)=\int^{t}_{0}\frac{dE}{dt}dt=\int^{t}_{0}P_{\tau}d\tau.$ (46)

In order to bound the energy difference for $t>0$, we can use the integral
inequality, e.g. that for arbitrary scalar functions $A(\tau)$ we have
$\int_{0}^{t}A(\tau)d\tau\leq\int_{0}^{t}|A(\tau)|d\tau\leq t\
\text{sup}_{\tau\in[0,t]}|A(\tau)|$. This bound works in particular well for
short times $t$. In an open system we have that

$\displaystyle\frac{d}{dt}E(t)$ $\displaystyle=$
$\displaystyle\text{tr}\left(\frac{d}{dt}\rho_{\mathcal{S}}(t)H_{0}\right)$
(47) $\displaystyle=$
$\displaystyle\text{tr}\bigg{(}\Big{(}\frac{i}{\hbar}[\rho_{\mathcal{S}},H_{0}+V]+\mathbb{L}(\rho_{t})\Big{)}H_{0}\bigg{)},$

where

$\displaystyle\mathbb{L}(\rho_{\mathcal{S}})=\sum_{n}\gamma_{n}\Big{(}L_{n}\rho_{\mathcal{S}}(t)L_{n}^{\dagger}-\frac{1}{2}\\{L_{n}^{\dagger}L_{n},\rho_{\mathcal{S}}(t)\\}\Big{)}$
(48)

with $\\{\cdot,\cdot\\}$ the anticommutators, and where the Lindblad operators
$L_{n}$ satisfy $\text{tr}[L_{p}L_{t}^{\dagger}]=\delta_{pt}$. We assume below
that $L_{n}$ is Hermitian that, while not being the most general case, does
include many physically relevant models.

Following the definitions above, then, we prove the following upper bound:

$\displaystyle\frac{\hbar|E(t)-E_{0}|}{\|H_{0}\|}\leq t(W_{A}+W_{B}),$ (49)

where the quantity $W_{A}$ can be expressed in the following forms:

$\displaystyle W_{A}=2\
\text{min}_{\tau\in[0,t]}\begin{cases}\|V\|\sqrt{\text{r}([\rho_{\mathcal{S}}(\tau),V])\mathbb{C}_{V}(\rho_{\mathcal{S}}(\tau))}\\\
\|\rho_{\mathcal{S}}(\tau)\|_{F}\sqrt{\mathbb{C}_{H_{0}}(V)}\\\
\|V\|\sqrt{\text{r}([\rho_{\mathcal{S}},H_{0}])\mathbb{C}_{H_{0}}(\rho_{\mathcal{S}}(\tau))},\end{cases}$
(50)

while

$\displaystyle W_{B}$ $\displaystyle=$ $\displaystyle
3\sum_{n}\gamma_{n}\text{sup}_{\tau\in[0,t]}\sqrt{r_{n}\mathbb{C}_{L_{n}}\big{(}\rho_{\mathcal{S}}(\tau)\big{)}},$
(51)

where $C_{L_{n}}(\rho_{\mathcal{S}})$ is the coherence in the spectral basis
of $L_{n}$ of $\rho_{\mathcal{S}}(\tau)$. Let us call $\Pi_{i}^{n}$ the
spectral basis of $L_{n}$. Then,

$\displaystyle
C_{L_{n}}(\rho_{\mathcal{S}})=\sum_{i}\|[\Pi_{i}^{n},\rho_{\mathcal{S}}]\|_{F}^{2}.$
(52)

The constant $r_{n}$ is given by
$r_{n}=\text{r}(\rho_{\mathcal{S}})+\text{r}(L_{n})$.

Proof. Following from the integral inequality, we begin with,

$\displaystyle\hbar|\Delta E(t)|$ $\displaystyle\leq$
$\displaystyle\int_{0}^{t}|\text{tr}([\rho_{\mathcal{S}}(\tau),V]H_{0})|d\tau$
(53) $\displaystyle+$
$\displaystyle\int_{0}^{t}|\text{tr}(\mathbb{L}\big{(}\rho_{\mathcal{S}}(\tau)\big{)}H_{0})|d\tau.$

The first term can be bounded as we did in the case of the closed system, thus
we have the identities

$\displaystyle\underbrace{|\text{tr}([\rho_{\mathcal{S}}(\tau),V]H_{0})|}_{W_{A}^{1}}=\underbrace{|\text{tr}([V,H_{0}]\rho_{\mathcal{S}}(\tau))|}_{W_{A}^{2}}=\underbrace{|\text{tr}([\rho_{\mathcal{S}},H_{0}]V)|}_{W_{A}^{3}}.$

It follows that can use, following the bounds of Sec. II.2, we immediately
write the three possible inequalities

$\displaystyle W_{A}^{1}$ $\displaystyle\leq$ $\displaystyle
2\|H_{0}\|\|V\|\sqrt{\text{r}([\rho_{\mathcal{S}}(\tau),V])\mathbb{C}_{V}(\rho_{0})},$
$\displaystyle W_{A}^{2}$ $\displaystyle\leq$ $\displaystyle
2\|\rho_{\mathcal{S}}\|_{F}\|H_{0}\|\sqrt{\mathbb{C}_{H_{0}}(V)},$
$\displaystyle W_{A}^{3}$ $\displaystyle\leq$ $\displaystyle
2\|V\|\|H_{0}\|\sqrt{\text{r}([\rho_{\mathcal{S}},H_{0}])\mathbb{C}_{H_{0}}(\rho_{0})}.$
(54)

In the expressions above we defined
$\mathbb{C}_{V}(\cdot)=\sum_{k}\|[\cdot,\Pi_{k}^{V}]\|_{F}^{2}$,, and we wrote
$\Pi_{k}^{V}$ is the spectral basis of the interaction operator $V$, and of
course similarly
$\mathbb{C}_{H_{0}}(\cdot)=\sum_{k}\|[\cdot,\Pi_{k}^{H_{0}}]\|_{F}^{2}$, with
$\Pi_{k}^{H_{0}}$ the spectral basis of the battery Hamiltonian. Let us now
focus on the Lindbladian. First, note that we have, because
$\\{A,B\\}=[A,B]+2BA$,

$\displaystyle\mathbb{L}\big{(}\rho_{\mathcal{S}}(\tau)\big{)}$
$\displaystyle=$
$\displaystyle\sum_{n}\gamma_{n}\big{(}L_{n}\rho_{\mathcal{S}}L_{n}^{\dagger}-\frac{1}{2}\\{L_{n}^{\dagger}L_{n},\rho_{\mathcal{S}}\\}\big{)}$
$\displaystyle=$
$\displaystyle\sum_{n}\gamma_{n}\Big{(}L_{n}\rho_{\mathcal{S}}L_{n}^{\dagger}-\frac{1}{2}[L_{n}^{\dagger}L_{n},\rho_{\mathcal{S}}]-\rho_{\mathcal{S}}L_{n}^{\dagger}L_{n}\Big{)}$
$\displaystyle=$
$\displaystyle-\sum_{n}\gamma_{n}\Big{(}[\rho_{\mathcal{S}}L_{n}^{\dagger},L_{n}]+\frac{1}{2}[L_{n}^{\dagger}L_{n},\rho_{\mathcal{S}}(\tau)]\Big{)}.$

We thus have

$\displaystyle\hbar|\Delta E(t)|$ $\displaystyle\leq$ $\displaystyle W_{A}$
(56) $\displaystyle+$
$\displaystyle\sum_{n}\frac{\gamma_{n}}{2}\int_{0}^{t}|\text{tr}\Big{(}[L_{n}^{\dagger}L_{n},\rho_{\mathcal{S}}(\tau)]H_{0}\Big{)}|d\tau$
$\displaystyle+$
$\displaystyle\sum_{n}\gamma_{n}\int_{0}^{t}|\text{tr}\Big{(}[\rho_{\mathcal{S}}(\tau)L_{n}^{\dagger},L_{n}]H_{0}\Big{)}|d\tau.$
(57)

Let us now make the assumption the Lindblads are Hermitian operators, e.g.
$L_{n}^{\dagger}=L_{n}$. This bound is the one we use in the following for the
case of the Spin-Boson model, for which such condition holds true.

For the term of Eq. (56), we use the trace inequality of Eq. (8), and the fact
that $L_{n}^{\dagger}L_{n}$ is an Hermitian operator, it has a spectral
decomposition of the form

$\displaystyle\tilde{L}_{n}^{2}=L_{n}^{\dagger}L_{n}=\sum_{k}|\eta^{n}_{k}|^{2}\Pi_{k}^{n}.$
(58)

We can now use the bounds we used in terms of coherence. We have, using Lemma
1 and Eq. (9), we have

$\displaystyle\|[L_{n}^{\dagger}L_{n},\rho_{\mathcal{S}}(\tau)]\|_{F}^{2}\leq
4\|L_{n}\|^{4}\mathbb{C}_{L_{n}}\big{(}\rho_{\mathcal{S}}(\tau)\big{)},$ (59)

where
$\mathbb{C}_{L_{n}}\big{(}\rho_{\mathcal{S}}(\tau)\big{)}=\sum_{k}\|[\Pi_{k}^{n},\rho_{\mathcal{S}}(\tau)]\|_{F}^{2}$.
Since $\text{tr}(L_{n}^{\dagger}L_{n})=1$, we have $\|L_{n}\|_{F}=1$ and
$\|L_{n}\|\leq 1$.

If we call
$R_{n}^{\prime}=\text{r}([L_{n}^{\dagger}L_{n},\rho_{\mathcal{S}}(\tau)])$,
then

$\displaystyle\int_{0}^{t}|\text{tr}\Big{(}[L_{n}^{\dagger}L_{n},\rho_{\mathcal{S}}(\tau)]H_{0}\Big{)}|d\tau$
$\displaystyle\leq$ $\displaystyle 2\ t\ \|H_{0}\|\cdot$ $\displaystyle\
\cdot\text{sup}_{\tau\in[0,t]}\sqrt{R_{n}^{\prime}\mathbb{C}_{L_{n}}\big{(}\rho_{\mathcal{S}}(\tau)\big{)}}$

For the second term, in Eq. (57), we need to use the fact that $L_{n}$ is
Hermitian.

If $L_{n}$ is an Hermitian operator, then it does have a spectral
decomposition of the form $L_{n}=\sum_{i}\eta_{i}^{n}\Pi_{k}^{n}$, and it
follows trivially that
$\tilde{L}_{n}^{2}=\sum_{i}|\eta_{i}^{n}|^{2}\Pi_{k}^{n}$. In this case, we
obtain directly from Eq. (57) that, calling
$R_{n}^{\prime\prime}=\text{r}([\rho_{\mathcal{S}}(\tau)L_{n}^{\dagger},L_{n}])$,
that

$|\text{tr}\Big{(}[\rho_{\mathcal{S}}(\tau)L_{n}^{\dagger},L_{n}]H_{0}\Big{)}|\leq\|H_{0}\|\
\sqrt{R_{n}^{\prime\prime}}\|[\rho_{\mathcal{S}}(\tau)L_{n}^{\dagger},L_{n}]\|_{F}.$
(60)

At this point, we can use the fact that $L_{n}$ is Hermitian. We expand
$L_{n}$ in the $\Pi$’s basis, obtaining from Lemma 2 of Eq. (10):

$\|[A,BA]\|_{F}^{2}\leq 4\|A\|^{4}\mathbb{C}_{A}(B)$

. Using the bound above, we have

$\displaystyle\|[\rho_{\mathcal{S}}(\tau)L_{n}^{\dagger},L_{n}]\|_{F}$
$\displaystyle\leq$ $\displaystyle
2\|L_{n}\|^{2}\sqrt{\mathbb{C}_{L_{n}}(\rho_{\mathcal{S}}(\tau))}.$ (61)

Thus,

$\displaystyle\|[\rho_{\mathcal{S}}(\tau)L_{n}^{\dagger},L_{n}]\|_{F}$
$\displaystyle\leq$ $\displaystyle
2\sqrt{\mathbb{C}_{L_{n}}(\rho_{\mathcal{S}}(\tau))}.$ (62)

We now note that both
$R_{n}^{\prime}\leq\text{r}(L_{n})+\text{r}(\rho_{\mathcal{S}})$, and
$R_{n}^{\prime\prime}\leq\text{r}(L_{n})+\text{r}(\rho_{\mathcal{S}})$. Let us
thus call $r_{n}=\text{r}(L_{n})+\text{r}(\rho_{\mathcal{S}})$.

Thus, we have,

$\displaystyle\frac{\hbar|\Delta E(t)|}{\|H_{0}\|t}$ $\displaystyle\leq$
$\displaystyle
W_{A}+3\sum_{n}\gamma_{n}\text{sup}_{\tau}\sqrt{r_{n}\mathbb{C}_{L_{n}}\big{(}\rho_{\mathcal{S}}(\tau)\big{)}},$

which is the final result which we use in the statement above. $\square$

The bound of Eq. (LABEL:eq:gron) is a Gronwall type of inequality. In the
following, we will apply the bounds for the case of the slave boson model.

###  III.3 Ensemble of batteries

A setup of interest for practical applications is the case of an array of
batteries interacting through a potential $V$, and we want to show how the
coherence enters in this case. Here we assume that the system is closed, but
the formalism is analogous to the case of the open system as we discuss below.
We consider $M$ copies of the same system, a battery, coupled only via an
interaction potential. Formally, such situation can be represented as follows.
The $M$ copies of the same system are represented by a Hamiltonian of the form
$H=\bigoplus_{i=1}^{M}H_{i}=\bigoplus_{i=1}^{M}(H_{0}^{i}+V^{i})$, e.g. an
ensemble of batteries If we assume that the $M$ copies are not interacting,
since $\rho(t)=\otimes_{i=1}^{M}\rho_{i}(t)$, we have

$\displaystyle
W(t)=\sum_{i=1}^{M}\text{Tr}_{i}\Big{(}\big{(}\rho^{i}(t)-\rho^{i}(0)\big{)}H_{0}^{i}\Big{)}-E_{0}\leq
MW_{max},$

where $W_{max}$ is maximum work obtained for the single system
$H_{0}^{i}+V^{i}$. In this case we can apply the formalism of the closed
system for each subsystem and thus it is not so interesting . If however we
only have that
$H_{0}=\bigoplus_{i=1}^{M}H_{0}^{i}=\sum_{i=1}^{M}H_{0}^{i}\otimes
I_{\mathcal{S}\setminus i}$, while $V$ does have support on multiple batteries
single battery, then we have that the density matrix is in general not
factorizable, and the energy storage can be written in the form

$\displaystyle W(t)$ $\displaystyle=$
$\displaystyle\text{Tr}\big{(}\rho(t)(\sum_{i=1}^{M}H_{0}^{i}\otimes
I_{\mathcal{S}\setminus_{i}})\big{)}-E_{0}$ (65) $\displaystyle=$
$\displaystyle\sum_{i=1}^{M}\text{Tr}_{i}\Big{(}\text{Tr}_{\mathcal{S}\setminus
i}\big{(}\rho(t)\big{)}H_{0}^{i}\Big{)}-E_{0},$ (66)

where $\rho_{i}(t)=\text{Tr}_{\mathcal{S}\setminus i}\big{(}\rho(t)\big{)}$ is
the local density matrix relative to subsystem $i$. In this case, the time
evolution of the reduced density matrix can be written in terms of a positive
map petruccione ; schlosshauer . We can write
$\rho_{i}(t)=\sum_{a}A_{a}^{i}(t)\rho_{i}(0)A_{a}^{i}(t)$, with
$\sum_{a}A_{a}^{i}(t)(A_{a}^{i}(t))^{\dagger}=I$ where $A_{a}^{i}$ are Kraus
operators. It follows that we can write

$\displaystyle W(t)$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{M}\sum_{a}\text{Tr}_{i}\Big{(}\big{(}A_{a}^{i}(t)\rho_{i}(0)A_{a}^{i}(t)^{\dagger}-\rho_{i}(0)\big{)}H_{0}^{i}\Big{)}$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{M}\sum_{a}\text{Tr}_{i}\Big{(}\rho_{i}(0)[A_{a}^{i},H_{0}^{i}(A_{a}^{i})^{\dagger}]\Big{)},$

where in the second line we have used the Hilbert-Schmidt duality. Here, we
called $A_{a}^{i}$ the Kraus operators to avoid confusion with the open system
case, but the meaning is similar.

If the Kraus operators $(A^{i})^{\dagger}_{a}=A^{i}_{a}$ are Hermitian,
similarly to what we have done before, they can be expanded via a spectral
decomposition of the form
$A_{a}^{i}=\sum_{k}(\eta_{i}^{a})_{k}(\Pi_{i}^{a})_{k}$, and we can apply
Lemma 2 of Eq. (10), proven in App. D. Using the spectral basis decomposition,
then we have

$\displaystyle\|[A_{a}^{i},H_{0}^{i}A_{a}^{i}]\|_{F}^{2}$ $\displaystyle\leq$
$\displaystyle 4\|A_{a}^{i}\|^{4}\mathbb{C}_{ai}(H_{0}^{i}),$ (67)

where
$\mathbb{C}_{ai}(H_{0}^{i})=1/2\sum_{k}\|[(\Pi_{i}^{a})_{k},H_{0}^{i}]\|_{F}^{2}$.
We now use the trace inequality of Eq. (7), obtaining

$\displaystyle W(t)$ $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{M}\sum_{a}\|\rho_{i}(0)\|_{F}\|[A_{a}^{i},H_{0}^{i}(A_{a}^{i})^{\dagger}]\|_{F}.$
(68) $\displaystyle\leq$ $\displaystyle
2\sum_{i=1}^{M}\sum_{a}\|\rho_{i}(0)\|_{F}\|A_{a}^{i}\|^{2}\sqrt{\mathbb{C}_{ai}(H_{0}^{i})},$

from which it follows that the energy storage of the system can be reduced to
coherence bounds on the single subsystem.

##  IV Models study

In this section we study the work and charging power for some models. We
consider two closed quantum systems. The first one is a two-body spin model,
the second one a quantum many-body system described by a the anisotropic XY
spin chain. The third model studied is the open quantum system of a single
spin in a bosonic bath.

###  IV.1 A 2-body system example

As described previously, the bounds we have obtained can be tight for small
systems. Here we consider first a 4-level system given by two spins $s_{1}$
and $s_{2}$ interacting with two external field $\vec{G}(t)$ and $\vec{F}(t)$
and among themselves Bragov1 ; Bragov2 . The Hamiltonian of the system is
assumed to be $H=H_{0}+V$, with $H_{0}=2J\hat{s}_{1}\otimes\hat{s}_{2}$, where
$\hat{s}_{1}=\vec{\sigma}\otimes I$ and
$\hat{s}_{2}=I\otimes\frac{1}{2}\vec{\sigma}$. The external fields enter in
$V$, with $V=2\big{(}\hat{h}_{1}\otimes I+I\otimes\hat{h}_{2}\big{)}$, with
$\hat{h}_{1}=\vec{\sigma}\cdot\vec{G}$,
$\hat{h}_{2}=\vec{\sigma}\cdot\vec{F}$.

For simplicity, here we consider $\vec{F}$ and $\vec{G}$ to be aligned along
the $z$ direction, thus $\vec{G}=(0,0,B_{1})$ and $\vec{F}=(0,0,B_{2})$, which
is an exactly solvable model. In this system, the energy can be stored due to
the coupling between the two spins. If one introduces $B_{\pm}=B_{1}(t)\pm
B_{2}(t)$,the spinor $\Psi=(v_{1},v_{2},v_{3},v_{4})$ satisfies the equation

$i\partial_{t}\Psi=\hat{H}\Psi,$ (69)

with

$\displaystyle H$ $\displaystyle=$
$\displaystyle\begin{pmatrix}B_{+}+J/2&0&0&0\\\ 0&B_{-}-J/2&J&0\\\
0&J&B_{-}-J/2&0\\\ 0&0&0&J/2-B_{+}\end{pmatrix}$ (70) $\displaystyle=$
$\displaystyle H_{0}+V,$

where we assume $H_{0}$ to be dependent only on the $J$ coupling. The
components $1$ and $4$ of the spinor satisfy the solution, assuming that $J$
is constant,

$\displaystyle v_{1}(t)$ $\displaystyle=$ $\displaystyle
e^{-i\int_{0}^{t}B_{+}(t^{\prime})dt^{\prime}-i\frac{J}{2}t}v_{1}(0),$ (71)
$\displaystyle v_{4}(t)$ $\displaystyle=$ $\displaystyle
e^{+i\int_{0}^{t}B_{+}(t^{\prime})dt^{\prime}-i\frac{J}{2}t}v_{4}(0).$ (72)

If we define $\psi^{\prime}=(v_{2},v_{3})$, then these components satisfy the
equation

$i\partial_{t}\psi^{\prime}(t)=(\vec{\sigma}\cdot\vec{K}-\frac{J}{2})\tilde{\psi}^{\prime}(t),$
(73)

where $\vec{K}(t)=\Big{(}J,0,B_{-}(t)\Big{)}$.

Figure 1: The second protocol proposed for the closed 4-level system via local
fields with opposite orientation.

####  IV.1.1 Work and Power

Let us thus consider the following two quench protocols.

First protocol. In the first, we have $B_{1}(t)=B_{2}(t)=\theta(t-t_{0})B$.
Then, $B_{+}=2B\theta(t-t_{0})$ and $B_{-}(t)=0$. Thus, the spin components
$v_{2}$ and $v_{3}$ decouple from the external field. In this case, $\psi_{a}$
is a equivalent to a single spin interacting via a $\sigma_{z}$ Zeeman
coupling with external field $B_{+}$. The work is given by
$W(t)=\text{Tr}(\rho(t)H_{0})-\text{Tr}(\rho_{0}H_{0})$, and the components
$2$ are $3$ decoupled and do not change their populations. Thus, this protocol
while natural is not particularly interesting. Using Eq. (72) it is not hard
to see that $W(t)=0$. Another way of seeing this is through the inequality $B$
of Eq. (11). We have in fact that $\Pi_{V}$’s are identical to the projectors
on subspaces spanned by $H_{0}$, and thus $\mathbb{C}_{U_{t}}(H_{0})=0$.

Second protocol. We consider $B_{1}(t)=-B_{2}(t)=\theta(t-t_{0})B$, as shown
in Fig. 1. Then, $B_{-}=2B\theta(t-t_{0})$ and $B_{+}(t)=0$. Also in this case
the components $v_{1}$ and $v_{4}$ decoupled and do not contribute to the
change in energy. We follow all the steps in order to calculate the bounds
later. The Stone operator for the reduced system is given by

$\displaystyle U(t>t_{0})$ $\displaystyle=$ $\displaystyle
e^{-i\int_{t_{0}}^{t}(\vec{\sigma}\cdot\vec{K}(t)-\frac{J}{2})dt}$ (74)
$\displaystyle=$ $\displaystyle
e^{-i\Big{(}\sigma_{x}J(t-t_{0})+2B(t-t_{0})\sigma_{z}-\frac{J}{2}(t-t_{0})\Big{)}.}$

If we use the formula
$e^{ia\hat{n}\cdot\vec{\sigma}}=I\cos(a)+i\sin(a)\hat{n}\cdot\vec{\sigma}$,
then we can write, using $a=\sqrt{J^{2}+4B^{2}}$ and
$\hat{n}=(\frac{J}{a},0,\frac{2B}{a})$

$\displaystyle U(t>t_{0})$ $\displaystyle=$ $\displaystyle
e^{i\frac{J}{2}(t-t_{0})}\big{(}I\cos(a(t-t_{0}))$ $\displaystyle\ \ \ \ \ \ \
\ \ \ \ \
+i\sin(a(t-t_{0}))(\frac{J}{a}\hat{\sigma}_{x}+\frac{2B}{a}\hat{\sigma}_{z})\big{)}.$

If we define $\rho_{0}$ on the basis of $v_{2},v_{3}$, we have a base
Hamiltonian for the battery defined as $H_{0}=J\sigma_{x}-\frac{J}{2}I$ (with
eigenvalues $J/2,-3/2J$) and $V=2B\hat{\sigma}_{z}$ (with eigenvalues $\pm
2B$). We thus have $\|H_{0}\|=\frac{3}{2}J$ and $\|V\|=2B$. The eigenvalues
and eigenvectors of the unitary operator $U$ are given by

$\displaystyle\Lambda(U)=\\{e^{i(\frac{J}{2}\pm a)}\\},\ \ \ \ \ \ \
|v_{\pm}\rangle=(\frac{2B\mp a}{J},1),$ (75)

and thus we can easily obtain with projector operators
$\Pi_{\pm}=\frac{|v_{\pm}\rangle\langle v_{\pm}^{t}|}{\|v_{\pm}\|^{2}}$. It
follows that, using the identities

$\displaystyle[\sigma_{x},\sigma_{z}\sigma_{x}]$ $\displaystyle=-2\sigma_{z},$
(76) $\displaystyle[\sigma_{z},\sigma_{x}]$ $\displaystyle=2i\sigma_{y},$ (77)
$\displaystyle[\sigma_{z},\sigma_{z}\sigma_{x}]$ $\displaystyle=2\sigma_{x},$
(78)

and after some algebra, we obtain

$\displaystyle W(t\geq t_{0})$ $\displaystyle=$
$\displaystyle\text{Tr}(\rho_{0}[U_{t},U_{t}^{\dagger}H_{0}])$
$\displaystyle=$ $\displaystyle\text{Tr}(\rho_{0}\hat{\Delta H(t)}),$

where we defined the operator $\hat{\Delta H(t)}$ as

$\displaystyle\hat{\Delta H(t)}$ $\displaystyle=$
$\displaystyle-\frac{4BJ^{2}\sin^{2}\left(a(t-t_{0})\right)}{a^{2}}\hat{\sigma}_{z}$
$\displaystyle-$
$\displaystyle\frac{2BJ\sin\left(2a(t-t_{0})\right)}{a}\hat{\sigma}_{y}$
$\displaystyle+$
$\displaystyle\frac{4B^{2}J\left(1-\cos\left(2a(t-t_{0})\right)\right)}{a^{2}}\hat{\sigma}_{x}$

Since the system in this protocol has only two involved states, we can write

$\displaystyle\rho_{0}=\epsilon_{0}I+\epsilon_{1}\sigma_{x}+\epsilon_{2}\sigma_{y}+\epsilon_{3}\sigma_{z},$
(81)

with the constraints $\text{Tr}(\rho_{0})=1$. If we impose the constraint, we
must impose $\epsilon_{0}=\frac{1}{2}$, and write

$\displaystyle\rho_{0}=\frac{I}{2}+\epsilon_{1}\sigma_{x}+\epsilon_{2}\sigma_{y}+\epsilon_{3}\sigma_{z}.$
(82)

The purity of the state is then given by

$\displaystyle\text{Tr}(\rho^{2})=\frac{1}{2}+2(\epsilon_{1}^{2}+\epsilon_{2}^{2}+\epsilon_{3}^{2})=\frac{1}{2}+2\|\vec{\epsilon}\|^{2},$
(83)

from which it follows that $\|\epsilon\|^{2}\leq\frac{1}{4}.$ The work can
then be written as

$\displaystyle W(t)$ $\displaystyle=$ $\displaystyle
4BJ\Big{(}\frac{\left(4B\epsilon_{1}-2J\epsilon_{3}\right)\sin^{2}\left(at\right)}{a^{2}}$
(84) $\displaystyle-$
$\displaystyle\frac{2\epsilon_{2}\sin\left(at\right)\cos\left(at\right)}{a}\Big{)}.$

It is easy to see that work is zero only if $4B\epsilon_{1}-2J\epsilon_{3}=0$
and $\epsilon_{2}=0$.

Let us now focus on the upper bound. In order to calculate the upper bound, we
first evaluate

$\displaystyle\mathbb{C}_{U_{t}}(\rho_{0})$ $\displaystyle=$
$\displaystyle\|[\Pi_{+},\rho_{0}]\|^{2}+\|[\Pi_{-},\rho_{0}]\|^{2}$
$\displaystyle=$
$\displaystyle\frac{4\left(\epsilon_{2}^{2}\left(4B^{2}+J^{2}\right)+4B^{2}\epsilon_{1}^{2}-4BJ\epsilon_{3}\epsilon_{1}+J^{2}\epsilon_{3}^{2}\right)}{4B^{2}+J^{2}}.$

It follows that bound A for the work is given by

$\displaystyle|W(t)|\leq\sqrt{8J^{2}\frac{\left(\epsilon_{2}^{2}\left(4B^{2}+J^{2}\right)+4B^{2}\epsilon_{1}^{2}-4BJ\epsilon_{3}\epsilon_{1}+J^{2}\epsilon_{3}^{2}\right)}{4B^{2}+J^{2}}},$

from which we see that the upper bound is of the same order of magnitude as
the work itself. Also, it is easy to see that if
$\epsilon_{2}=4B\epsilon_{1}-2J\epsilon_{3}=0$, then
$\epsilon_{2}^{2}\left(4B^{2}+J^{2}\right)+4B^{2}\epsilon_{1}^{2}-4BJ\epsilon_{3}\epsilon_{1}+J^{2}\epsilon_{3}^{2}=0$,
which implies that the maximum work is also zero, showing that the bounds are
tight for this set of parameters.

We can also calculate the power. We observe that

$\displaystyle P(t)=\frac{dW(t)}{dt}$ $\displaystyle=$ $\displaystyle
8BJ\Big{(}\frac{\left(J\epsilon_{3}-2B\epsilon_{1}\right)\sin\left(2t\sqrt{4B^{2}+J^{2}}\right)}{\sqrt{4B^{2}+J^{2}}}$
(85) $\displaystyle\ \ \ \ \ \ \
-\epsilon_{2}\cos\left(2t\sqrt{4B^{2}+J^{2}}\right),\Big{)}$

while our bound A on the power is given by
$\|H_{0}\|\cdot\|V\|\sqrt{\text{r}([V,\rho_{t}])\cdot\mathbb{C}_{V}(\rho_{t})}$.
We have $\Pi_{V}^{1}=(1,0)^{t}\otimes(1,0)$ and
$\Pi_{V}^{2}=(0,1)^{t}\otimes(0,1)$, or

$\displaystyle\Pi^{1}_{V}=\begin{pmatrix}1&0\\\ 0&0\\\
\end{pmatrix},\Pi^{2}_{V}=\begin{pmatrix}0&0\\\ 0&1\\\ \end{pmatrix},$ (86)

It follows that

$\displaystyle\mathbb{C}_{V}(\rho_{0})=2(\epsilon_{1}^{2}+\epsilon_{2}^{2}).$
(87)

$P_{C}\leq
2*3/2J*2B*\sqrt{2(2\epsilon_{1}^{2}+2\epsilon_{2}^{2}}=12BJ\sqrt{\epsilon_{1}^{2}+\epsilon_{2}^{2})}$,
which is also of the same order as the power.

###  IV.2 The anisotropic XY model

With increasing experimental control over larger unitary quantum systems, it
becomes more pressing to study the thermodynamics of many-body quantum
systems. Theoretical studies have shown advantages by exploiting collective
effects in quantum batteries battp ; ModiPRL2017 ; polini ; adc ; marti2017 ;
ghosh . Here, we apply the bounds on work extraction to a simple many-body
spin system described by the XY model xymodel2 ; xymodel3 ; xymodel , which
can be investigated experimentally xymodelexp1 ; xymodelexp2 .

The (anisotropic) transverse-field XY model in one dimension is a well-known
spin model in Statistical Mechanics. One of its advantages is that properties
of the ground and excited states are known exactly. The transverse XY Ising
model has an interesting phase diagram in which also quantum phase transitions
are present. In recent years the transverse field XY model has been studied in
relation to quantities of interest in quantum information theory, such as
entanglement and quantum discord. Also, it has been shown that fidelity
measures present signatures of QPTs.

The Hamiltonian of the anisotropic XY spin chain reads

$H=-\frac{1}{2}\sum_{i=1}^{N}\left(\frac{1+\eta}{2}\sigma_{i}^{x}\sigma_{i+1}^{x}+\frac{1-\eta}{2}\sigma_{i}^{y}\sigma_{i+1}^{y}+h\sigma_{i}^{z}\right),$
(88)

where $h$ is the external magnetic field, and $\eta$ is the anisotropy
parameter. We also assume perodic boundary condition.

For any value of $\eta,h$, the XY model can be diagonalized using the Jordan-
Wigner transformation:

$\sigma_{i}^{z}=1-2c^{\dagger}_{i}c_{i},\
\sigma_{i}^{-}=c_{i}^{\dagger}e^{i\pi\sum_{i=1}^{i-1}c^{\dagger}_{j}c_{j}}.$
(89)

The Hamiltonian is then diagonalzed in the form

$H=\sum_{k>0}\Lambda_{k}\left(\gamma^{\dagger}_{k}\gamma_{k}+\gamma^{\dagger}_{-k}\gamma_{-k}-1\right).$
(90)

Here, the fermionic operator $\gamma_{k}$ is defined by the Bogoliubov
transformation of the Fourier transformed operators, $c_{k}$ i.e.

$c_{k}=\cos{\theta_{k}}\gamma_{k}+i\sin{\theta_{k}}\gamma^{\dagger}_{-k},$
(91)

with the dispersion ralation given by
$\Lambda_{k}=\sqrt{\epsilon_{k}^{2}+\eta^{2}\sin^{2}{k}}$,
$\epsilon_{k}=h-\cos{k}$, and the angle
$\theta_{k}=\tan^{-1}[(\eta\sin{k})/(\epsilon_{k}+\Lambda_{k})]$.

####  IV.2.1 Work and power

The battery internal Hamiltonian is assumed as the XY model Hamiltonain
without external field $h$. The initial state of the battery is prepared as
the ground state of the internal Hamiltonian. The charging process is achieved
by turning on a constant field. This is a standard quantum quench problem. In
the following, we use superscript $(1)$ to label the operators with initial
parameters, and $(2)$ (or no superscript for simplicity) to label the
operators with quenched parameters.

Figure 2: Work $W(t)$ (left) and power $P(t)$ (right) after a quantum quench
of the anisotropic XY model (88) for $N=1000$ spins, with the anisotropic
parameter $\eta=0.5$. The battery internal (unquenched) Hamiltonian has zero
field, $h^{(1)}=0$. Colors label various quenching field $h^{(2)}$ at both
non-critical (black dashed and blue dotted) and critical values (red solid).

The initial and quenched fermionic operators $\gamma_{k}$ are linked through

$\gamma_{k}^{(1)}=\cos{\chi_{k}}\gamma_{k}^{(2)}+i\sin{\chi_{k}}\gamma_{k}^{(2){\dagger}},$
(92)

where $\chi_{k}\equiv\theta_{k}^{(2)}-\theta_{k}^{(1)}$. This allows us to
express the initial Hamiltonian in terms of the operators with the quenched
parameters, i.e.,

$\displaystyle H^{(1)}$
$\displaystyle=\sum_{k>0}\Lambda_{k}^{(1)}[\cos^{2}{\chi_{k}}(\gamma_{k}^{\dagger}\gamma_{k}+\gamma_{-k}^{\dagger}\gamma_{-k})+$
(93) $\displaystyle\ \ \
+\sin^{2}{\chi_{k}}(\gamma_{k}\gamma^{\dagger}_{k}+\gamma_{-k}\gamma^{\dagger}_{-k})+$
$\displaystyle\ \ \
+2i\sin{\chi_{k}}\cos{\chi_{k}}(\gamma_{k}\gamma_{-k}+\gamma^{\dagger}_{k}\gamma^{\dagger}_{-k})-1].$

The initial ground state of the battery internal Hamiltonian can also be
written in terms of the quenched operators,

$|\psi(0)\rangle=\prod_{k>0}(\cos{\chi_{k}}-i\sin{\chi_{k}}\gamma_{k}^{\dagger}\gamma_{-k}^{\dagger})|0_{k}\rangle,$
(94)

where $|0_{k}\rangle$ is the vaccum state of $\gamma_{\pm k}$. The time
evolved state is then,

$|\psi(t)\rangle=\prod_{k>0}(\cos{\chi_{k}}e^{i\Lambda_{k}t}-i\sin{\chi_{k}}e^{-i\Lambda_{k}t}\gamma_{k}^{\dagger}\gamma_{-k}^{\dagger})|0_{k}\rangle.$
(95)

Now, the time-dependent work and power can be computed explicitly, i.e.,

$W(t)=2\sum_{k>0}\Lambda_{k}^{(1)}\sin^{2}{2\chi_{k}}\sin^{2}{\Lambda_{k}^{(2)}t},$
(96)

and

$P(t)=2\sum_{k>0}\Lambda_{k}^{(1)}\Lambda_{k}^{(2)}\sin^{2}{2\chi_{k}}\sin{2\Lambda_{k}^{(2)}t}.$
(97)

For non-zero anisotropic paramaters $\eta$, the XY model exhibits two regions
of criticality at $h=\pm 1$. Figure 2 depicts the charging work (96) and power
(97) with the battery Hamiltonian quenched to critical and non-critical
values. The battery reaches a more stable charging process and a higher work
with a critical quench. However, even though non-critical quenches finally
charges the battery to a lower work, the initial power is higher for larger
quench fields. Therefore, an optimal protocol could use an initial high field
to fast charge the battery, and then switch to a critical field to maintain
the maximal work.

####  IV.2.2 Upper bounds

We now apply the some of the upper bounds derived in the previous sections to
the work and power of the anisotropic XY model.

The first upper bound we are interested in is the bounds of work in terms of
the coherence of the initial state in the basis of the time evolution
operators. The the quench Hamiltonian is time-independent, hence the eigen-
basis of the unitary evolution operator $U$ is the basis of the Hamiltonian
itself, i.e.,

$\Pi_{\alpha}=\prod_{k>0}(\gamma_{k}^{\dagger})^{\alpha_{k}}(\gamma_{-k}^{\dagger})^{\alpha_{-k}}|0\rangle\langle
0|\gamma_{k}^{\alpha_{k}}\gamma_{-k}^{\alpha_{-k}}.$ (98)

Here, $\alpha_{k}$ is either $0$ or $1$, and the string
$\alpha=\\{\alpha_{k},\alpha_{-k}\\}_{k>0}$ labels posibble configurations of
all $\alpha_{k}$’s. There are $2^{N}$ possible configurations corresponding to
the $2^{N}$ projectors.

For each given projector, its average over the initial state
$\rho_{0}=|\psi(0)\rangle\langle\psi(0)|$ is

$\displaystyle\langle\psi(0)|\Pi_{\alpha}|\psi(0)\rangle$ (99)
$\displaystyle=$
$\displaystyle\prod_{k>0}\left(\cos^{2}{\chi_{k}}\delta_{\alpha_{k},0}\delta_{\alpha_{-k},0}+\sin^{2}{\chi_{k}}\delta_{\alpha_{k},1}\delta_{\alpha_{-k},1}\right).$

The coherence is then

$\displaystyle C_{U_{t}}(\rho_{0})=$
$\displaystyle\text{Tr}\rho_{0}^{2}-\sum_{\alpha}|\langle\psi(0)|\Pi_{\alpha}|\psi(0)\rangle|^{2}$
(100) $\displaystyle=$ $\displaystyle
1-\prod_{k>0}\left(\cos^{4}{\chi_{k}}+\sin^{4}{\chi_{k}}\right).$

The maximum eigenvalue of initial Hamiltonian is

$E^{(1)}_{max}=\sum_{k>0}\Lambda_{k}^{(1)}.$ (101)

Together with the coherence of the initial state, we get the upper bound (A)
of the work,

$\displaystyle W(t)$ $\displaystyle\leq$ $\displaystyle
2\sqrt{2}|\sum_{k>0}\Lambda_{k}^{(1)}|\sqrt{1-\prod_{l>0}\left(\cos^{4}{\chi_{l}}+\sin^{4}{\chi_{l}}\right)}$

Since $\chi_{k}\equiv\theta_{k}^{(2)}-\theta_{k}^{(1)}$ is cannot be zero for
all $k^{\prime}s$, unless the quenched Hamiltonian is the same as the initial
Hamiltonian, the product term in the above bound vanishes exponentially fast
with the system size. Hence, this upper bound of the work is only determined
by the largest eigenvalue of the internal Hamiltonia, i.e., its operator norm,
which scales linearly in the number of spins. For the example addressed in
Fig. 2, this upper bound estimates as $544$, independent of the quenching
parameters.

The second example we studied is the bound (C) for work, which involves the
coherence of the internal Hamiltonian $H_{0}=H^{(1)}$ in the basis of the
evolution operator. The coherence of $H_{0}$ can be computed as well, i.e.,

$\displaystyle C_{U_{t}}(H_{0})$
$\displaystyle=\sum_{\alpha}\left[\sum_{k>0}\delta_{\alpha_{k},\alpha_{-k}}(-1)^{\beta_{k}(\alpha)}\Lambda_{k}^{(1)}(1+\cos{(2\chi_{k})})\right.$
(103)
$\displaystyle\left.\times\sum_{k>0}\delta_{\alpha_{k},\alpha_{-k}}(-1)^{\beta_{k}(\alpha)}\Lambda_{k}^{(1)}(1-\cos{(2\chi_{k})})\right],$

where the outer summation ranges over all possible configurations of the
strings $\alpha=\\{\alpha_{k},\alpha_{-k}\\}_{k>0}$ and $\alpha_{\pm k}$ is
either $0$ or $1$. $\delta$ is the Kronecker delta. Also,

$\beta_{k}(\alpha)=\left\\{\begin{matrix}0,\ \text{if}\
\alpha_{k}=\alpha_{-k}=1,\\\ 1,\ \text{if}\
\alpha_{k}=\alpha_{-k}=0.\end{matrix}\right.$ (104)

This result gives a quantitative evaluation of the energy extracted in a
quantum many-body quench since the work extracted is basically upper bounded
by twice the square root of the coherence of $H_{0}$ (See Eq. (11)). As
confirmed by our numerical evaluation, this upper bound increases
exponentially fast in the number of spins. Consequently, for the quench
parameters in Fig. 2, this bound is larger than the exact work for a few
orders of magnitude. The exponential scaling is more apparent for the upper
bounds of the power. Hence, these upper bounds can be very loose for many-body
systems. However, they could be very useful for small systems which are more
sensitive to the structure of the system Hamiltonian as detected by the
coherence in corresponding the upper bounds.

###  IV.3 Spin-Boson model

As an application of the bounds in the open system, we consider the simplest
exactly solved model, the spin-boson model Vernon ; Caldeira ; Palma . The
spin-model model describes the precession of the two-level system of a spin in
an open environment. The open environment is described by a reservoir of
Harmonic oscillators. The model is well known because it serves the purpose of
describing decoherence in a simple exactly solvable setting schlosshauer .

The total Hamiltonian of the system is given by

$\displaystyle
H=\underbrace{\frac{1}{2}(\omega_{0}\sigma_{z}-\Delta_{0}\sigma_{x})}_{H_{0}}+\sum_{k}\omega_{k}b_{k}^{\dagger}b_{k}+\sum_{k}\sigma_{z}(g_{k}b_{k}^{\dagger}+g_{k}^{*}b_{k}).$

The constants $\omega_{0}$ and $\omega_{k}$ are the level spacing of the two-
level system and the frequencies of the bosonic degrees of freedom
respectively, with $[b_{k},b_{k^{\prime}}^{\dagger}]=\delta_{k,k^{\prime}}$.
The parameter $\Delta_{0}$ is associated to the tunneling between the two
states.

Here, we consider $\mathcal{S}$ to be the two-level system. Under the
condition of markovianity, the master equation for the spin-boson system can
be written in the form schlosshauer

$\displaystyle\frac{d}{dt}\rho_{\mathcal{S}}$ $\displaystyle=$
$\displaystyle\frac{i}{\hbar}[\rho_{\mathcal{S}},H_{0}]+\gamma\sigma_{z}\rho_{\mathcal{S}}\sigma_{z}^{\dagger}-\frac{\gamma}{2}\sigma_{z}^{\dagger}\sigma_{z}\rho_{\mathcal{S}}-\frac{\gamma}{2}\sigma_{z}^{\dagger}\sigma_{z}$
$\displaystyle=$
$\displaystyle\frac{i\omega_{0}}{2\hbar}[\rho_{\mathcal{S}},\sigma_{z}]-\frac{i\Delta_{0}}{2\hbar}[\rho_{\mathcal{S}},\sigma_{z}]+\gamma\sigma_{z}\rho_{\mathcal{S}}\sigma_{z}-\gamma\rho_{\mathcal{S}},$

where we used $\sigma_{z}^{2}=I$. where the single Lindblad operator are given
by $L_{1}=L_{1}^{\dagger}=\sigma_{z}$, and thus is Hermitian. In the
derivation of the reduced equation above the assumption of a thermal reservoir
is made, with the constant $\gamma>0$ containing the spectrum of the bath.

For the reduced system, it is easy to see that the energy storage is given by

$\displaystyle
E(t)=\frac{\omega_{0}}{2}\text{tr}(\rho_{\mathcal{S}}\sigma_{z})-\frac{\Delta_{0}}{2}\text{tr}(\rho_{\mathcal{S}}\sigma_{x}).$
(107)

If we define

$\displaystyle\rho_{\mathcal{S}}=\begin{pmatrix}\rho_{11}&\rho_{12}\\\
\rho_{21}&\rho_{22}\end{pmatrix}\rightarrow E(t)$ $\displaystyle=$
$\displaystyle\frac{\omega_{0}}{2}(\rho_{11}-\rho_{22})$ (108)
$\displaystyle-$ $\displaystyle\frac{\Delta_{0}}{2}(\rho_{12}+\rho_{21}).$

The evolution of the system is given by the following general dynamical
equations:

$\displaystyle\dot{\rho}_{11}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}i\Delta_{0}\left(\rho_{12}-\rho_{21}\right),$
$\displaystyle\dot{\rho}_{22}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}i\Delta_{0}\left(\rho_{21}-\rho_{12}\right),$
$\displaystyle\dot{\rho}_{12}$ $\displaystyle=$
$\displaystyle-\gamma\rho_{12}-\frac{1}{2}i\Delta_{0}\left(\rho_{11}-\rho_{22}\right)-\frac{1}{2}i\rho_{12}\omega_{0},$
$\displaystyle\dot{\rho}_{21}$ $\displaystyle=$
$\displaystyle-\gamma\rho_{21}-\frac{1}{2}i\Delta_{0}\left(\rho_{22}-\rho_{11}\right)+\frac{1}{2}i\rho_{21}\omega_{0},$

from which it is immediate to see that the trace of the density matrix is
preserved, but the off-diagonal terms change due to decoherence arising from
the interaction with the bath. Substituting the equations above into
$\frac{dE}{dt}$ we obtain

$\displaystyle\frac{dE}{dt}$ $\displaystyle=$
$\displaystyle-\frac{\Delta_{0}}{2}\Big{(}\rho_{12}\left(-\gamma-\frac{i\omega_{0}}{2}\right)+\rho_{21}\left(-\gamma+\frac{i\omega_{0}}{2}\right)\Big{)}$
(110) $\displaystyle=$
$\displaystyle\Delta_{0}\text{Re}\Big{(}\rho_{12}\left(\gamma+\frac{i\omega_{0}}{2}\right)\Big{)}$
$\displaystyle=$
$\displaystyle\Delta_{0}\gamma\text{Re}(\rho_{12})-\frac{\Delta_{0}\omega_{0}}{2}\text{Im}(\rho_{12}),$

from which we see instead that a change in energy is associated to the
tunneling, and is zero otherwise. It is also easy to see that the energy
storage changes when coherence is present, due to the proportionality of its
derivative with $\rho_{12}$.

We now wish to apply the bounds of Sec. III.2, in particular Eq. (49), in
which only term $W_{B}$ contributes:

$\displaystyle\frac{|\Delta E(t)|}{\|H_{0}\|}$ $\displaystyle\leq$
$\displaystyle 3\gamma
t\text{sup}_{\tau\in[0,t]}\sqrt{r_{n}\mathbb{C}_{L_{n}}\big{(}\rho_{\mathcal{S}}(\tau)\big{)}}.$

To this end, we first calculate various quantities that are related to the
bound. The eigenvalues of
$\Lambda(H_{0})=\\{\pm\frac{1}{2}\sqrt{\Delta_{0}^{2}+\omega_{0}^{2}}\\}$.
Thus we have
$\|H_{0}\|=\frac{1}{2}\sqrt{\Delta_{0}^{2}+\omega_{0}^{2}}\approx\Delta_{0}/2$,
while $\|L_{1}^{2}\|=1$ and $l_{1}=1$. Also, both $R_{n}\leq 4$ and $r_{n}\leq
2$. The spectral decomposition of $\sigma_{z}$ is given by

$\displaystyle\Pi_{1}=\begin{pmatrix}1&0\\\ 0&0\\\
\end{pmatrix},\Pi_{2}=\begin{pmatrix}0&0\\\ 0&1\\\ \end{pmatrix},$ (111)

from which, after an immediate calculation, we get

$\displaystyle\sqrt{\mathbb{C}_{L_{1}}(\rho_{\mathcal{S}})}=\sqrt{2(|\rho_{12}|^{2}+|\rho_{21}|^{2})}.$
(112)

Using the fact that $\rho_{21}=\rho_{12}^{*}$, assuming $\hbar=1$, and
applying the upper bound of Eq. (49), we obtain

$\displaystyle\frac{|E(t)-E_{0}|}{t}$ $\displaystyle\leq$
$\displaystyle\Delta_{0}3\sqrt{2}\gamma sup_{\tau\in[0,t]}|\rho_{12}(\tau)|$
(113) $\displaystyle\leq$ $\displaystyle
4.24(2)\Delta_{0}\gamma|\rho_{12}^{0}|.$

Let us now make some plots in the regime of strong decoherence and tunneling,
in which we have $\gamma\gg\Delta_{0}\gg\omega_{0}$. In this case we have
$\rho_{12}(t)=\rho_{21}^{*}(t)\approx\rho_{12}^{0}e^{-(i\frac{\omega_{0}}{2}+\gamma)t}$,
and thus,

$\displaystyle E(t)-E_{0}\approx|\rho_{12}^{0}|\Delta_{0}\frac{2e^{-\gamma
t}\left(-\cos\left(\frac{t\omega_{0}}{2}\right)+e^{\gamma t}\right)}{2}.$
(114)

At the first order in $t$, we have instead the exact result

$\displaystyle\frac{|E(t)-E_{0}|}{t}\approx
0.25\Delta_{0}\gamma|\rho_{12}^{0}|+O(t).$

We thus see that for short times the bound has the same functional form as the
actual first order approximation of the energy, and the two constants are both
of order one.

##  V Conclusions

In this paper, we studied and have made an explicit connection between work
and coherence in a quantum battery. In particular, we have provided a
quantitative framework for evaluating upper bounds to energy storage in
quantum batteries, based on the notion of Hilbert-Schmidt density matrix
coherence, and its generalization to other operators like the Hamiltonian used
to measure the energy of the system, and the internal Hamiltonian $H_{0}$. We
derived two key results. The first one is that the charging (or discharging)
of a quantum battery is governed by the amount of non commutativity of both
the density matrix of the system and the internal Hamiltonian in the basis
given by the spectral decomposition of the interaction. We provided bounds
which clarify the intricate interplay between coherence of the density matrix,
of the Hamiltonian and the interaction in order for work to be performed by a
quantum battery. The amount of non commutativity is quantified by using the
Frobenius norm, namely by summing the modulus squares of all off-diagonal
elements of an operator in the desired basis. For a state, this is a measure
of coherence, and is bounded by one. For a generic operator, e.g. $H_{0}$,
this form of coherence can be very large and scale with the size of the
Hilbert space in certain cases, rather than the number of particles in the
system. As such, some of our bounds, if taken as they are, can be very loose.
This implies that some of our bounds capture the behaviour of small systems
well. We have tested these bounds on two exactly solvable closed quantum
models, i.e. a 4-level system and the anisotropic XY model. In the former, our
bounds are fairly tight, while in the latter we show under which conditions
the operator coherence is small enough in order to obtain a tight bound.

The second result is the extension to the case of generic quantum channels,
e.g., open quantum systems, both in terms of Kraus and Lindblad operators. In
order to see how these bounds apply quantitively, we have studied ensembles of
quantum batteries and the Spin Boson model, showing the role of coherence in
charging such battery model. In the case of open systems we have focused on
Hermitian Kraus and Linblad operators, but some of these bounds can be
extended to non-Hermitian ones, with some technical challenges. This will be
the subject of future investigations. In perspective, we find that the results
of this paper open a certain number of interesting questions. Since in an open
quantum system coherence is typically exponentially suppressed, we are
interested in showing how decoherence free subspaces zanardidfs ; violadfs ;
lidardfs  can be used to obtain more efficient quantum batteries. In the
spirit of the typicality arguments used both in Caravelli1 ; zanardicoh ;
isospec2  we can ask how typical quantum maps can be used to exchange energy.
Finally, an important generalization would be to take in consideration the
entropy change in open quantum systems and extend these results to the free
energy available to a quantum battery.

## Acknowledgements

We acknowledge the support of NNSA for the U.S. DoE at LANL under Contract No.
DE-AC52-06NA25396. A.H. acknowledges partial support from NSF award number
2014000. FC was also financed via DOE-LDRD grants PRD20170660 and PRD20190195.
L.P.G.P. acknowledges partial support by AFOSR MURI project “Scalable
Certification of Quantum Computing Devices and Networks”, DoE ASCR FAR-QC
(award No. DE-SC0020312), DoE BES Materials and Chemical Sciences Research for
Quantum Information Science program (award No. DE-SC0019449), DoE ASCR Quantum
Testbed Pathfinder program (award No. DE-SC0019040), NSF PFCQC program, AFOSR,
ARO MURI, AFOSR MURI, and NSF PFC at JQI. B.Y. also acknowledges support from
the U.S. DoE, Office of Science, Basic Energy Sciences, Materials Sciences and
Engineering Division, Condensed Matter Theory Program, and partial support
from the Center for Nonlinear Studies.

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##  Appendix A Work and von Neumann’s trace inequality

Let us now consider another set of bound for the work, based on von Neumann’s
trace inequalities. Consider two operators $\mathcal{A}$ and $\mathcal{B}$.
Let $\alpha_{i}$ and $\beta_{i}$ be the ordered set of singular values of
$\mathcal{A}$ and $\mathcal{B}$, e.g. $\alpha_{i-1}\geq\alpha_{i}$, and
$\beta_{i-1}\geq\beta_{i}$. Then we have

$\displaystyle|{\mbox{Tr\,}}{\mathcal{A}\mathcal{B}}|\leq\sum_{i=1}^{n}\alpha_{i}\beta_{i},$
(115)

and if $\mathcal{A}$ and $\mathcal{B}$ are Hermitian, then

$\displaystyle\sum_{i=1}^{n}\alpha_{i}\beta_{n-i+1}\leq{\mbox{Tr\,}}{\mathcal{A}\mathcal{B}}\leq\sum_{i=1}^{n}\alpha_{i}\beta_{i}.$
(116)

It follows that, if we use the equalities from the previous section:

$W=\underbrace{\text{Tr}(U_{t}^{\dagger}H_{0}[\rho_{0},U_{t}])}_{A}=\underbrace{\text{Tr}(\rho[U_{t},U_{t}^{\dagger}H_{0})]}_{B}=\underbrace{\text{Tr}(U_{t}[U_{t}^{\dagger}H_{0},\rho_{0}])}_{C},$
(117)

we have three pairs of operators: $\mathcal{A}_{A}=U_{t}^{\dagger}H_{0}$,
$\mathcal{B}_{A}=[\rho,U_{t}]$; $\mathcal{A}_{B}=\rho$,
$\mathcal{B}_{B}=[U_{t},U_{t}^{\dagger}H_{0}]$ and $\mathcal{A}_{C}=U_{t}$,
$\mathcal{B}_{C}=[U_{t}^{\dagger}H_{0},\rho]$.

Let us thus consider the singular values of the three pairs. Given an operator
$Q$, the singular values square are the (square roots of the) eigenvalues of
the operator $Q_{2}=Q^{\dagger}Q$. Von Neumann’s trace inequality can be
applied to equality $C$ as both $\rho$ and
$[U_{t},U_{t}^{\dagger}H_{0}]=H_{0}-U_{t}^{\dagger}H_{0}U_{t}$ are Hermitian
operators. The singular values of $\rho$ are the eigenvalues of $\rho^{2}$.
Let $d_{i}$ be the singular values of $\rho$, and $\sigma_{i}(t)$ be the
singular values of $H_{0}-U_{t}^{\dagger}H_{0}U_{t}$. Then, it follows that

$\displaystyle\sum_{i=1}^{n}d_{n-i+1}\sigma_{i}(t)\leq|W(t)|\leq\sum_{i=1}^{n}d_{i}\sigma_{i}(t)$
(118)

We can perform another upper bound: $W(t)<\sum_{i}d_{i}\sigma_{i}(A+B)$ where
A is $H_{0}$ and $B=-U^{\dagger}H_{0}U$ now, we can use Weyl’s inequality for
the singular values of the sum of Hermitian matrices. This is

$W(t)<max\ d\sum_{i}\sigma_{i}(A+B)<\text{max}\
d\sum_{i}(\sigma_{i}(A)+\sigma_{i}(U^{\dagger}AU))<2\text{max}\
d\sum_{i}\sigma_{i}(A)<2nmax\ dmax\ \epsilon(H_{0})$

One interesting comment is that both $\rho$ and $[U_{t},U_{t}^{\dagger}H_{0}]$
are Hermitian operators. Thus, following von Neumann’s trace inequality, we
can both upper and lower bound the work with

$\displaystyle\underline{W}=\sum_{i=1}^{n}d_{n-i+1}\sigma_{i}(t)\leq|W(t)|\leq\sum_{i=1}^{n}d_{i}\sigma_{i}(t)=\overline{W}$
(119)

where $\sigma_{i}(t)$ are the singular values of
$\hat{\epsilon}=H_{0}-U_{t}H_{0}U_{t}^{\dagger}$, while $d_{i}$ are the
singular values of $\rho$. It can be immediately seen that the inequality of
Eq. (119) is tighter.

These bounds can be further simplified as follows. Consider the upper bound
$\overline{W}=\sum_{i=1}^{n}d_{i}\sigma_{i}(t)$. We can further upper bound it
via

$\displaystyle\bar{W}\leq\bar{d}\sum_{j}\sigma_{j}(t),$ (120)

where $\bar{d}=\text{max}_{i}d_{i}$. We note that numerically the upper bound
above is hard to calculate because of necessity to diagonalize 3 matrices.
Now, for the singular values of a sum of Hermitian matrices we have
$\sum_{j}\sigma_{j}(A+B)\leq\sum_{j}(\sigma_{j}(A)+\sigma_{j}(B))$ HJ91 . Now
note that $B=-U^{\dagger}AU$ where $U$ is a unitary transformation, and thus
$\sigma_{i}(U^{\dagger}AU)=\sigma_{i}(A)$. It follows that

$\displaystyle\bar{W}\leq 2\ n\ \bar{d}\ \bar{\epsilon}(H_{0})$ (121)

where $n$ is the dimension of the Hilbert space and
$\bar{\epsilon}(H_{0})=\text{max}_{i}\sigma_{i}(H_{0})$. We thus obtain the
upper bound

$\displaystyle|W(t)|\leq 2\ n\ \bar{d}\ \bar{\sigma}(H_{0}).$ (122)

which only requires to solve two maximum eigenvalues problems.

##  Appendix B Norms definitions and upper bounds

In the paper we use various norms, so for the sake of clarity we define the
following quantities. Given a matrix $A$, we define the 2-norm as
$\|A\|_{2}=\sup_{i}\sigma_{i}(A)$, where $\sigma_{i}(A)$ is the i-th singular
value. For a square matrix $A$ of size $n$, We then call the Frobenius norm
(or Hilbert-Schmidt) norm the following:

$\displaystyle\|A\|_{F}=\sqrt{\sum_{ij}|a_{ij}|^{2}}=\sqrt{{\mbox{Tr\,}}{A^{\dagger}A}}=\sqrt{\sum_{i=1}^{n}\sigma_{i}^{2}(A)}.$
(123)

A bound for singular value of two matrices will be the following. Consider two
matrices $A$ and $B$. First, let us prove a property of the Frobenius norm
that will turn useful in the following. If
$\text{tr}(A^{\dagger}B)=\text{tr}(B^{\dagger}A)=00$, then
$\|A+B\|_{F}^{2}=\|A\|_{F}^{2}+\|B\|_{F}^{2}$. In order to see this, notice
that
$\|A+B\|_{F}^{2}=\text{tr}(A+B)^{\dagger}(A+B)=\|A\|_{F}^{2}+\|B\|_{F}^{2}+\text{tr}(A^{\dagger}B+B^{\dagger}A)$
from which the statement follows.

In general, we have the inequalities

$\|A\|_{2}\leq\|A\|_{F}\leq\sqrt{n}\|A\|_{2}.$ (124)

Let us now prove the following statement. In general, one has that
$|\text{tr}(AB)|\leq\sqrt{\|A\|_{F}^{2}\|B\|_{F}^{2}}$. In this paper we do
use the notation $\|A\|_{F}$ for the Frobenius norm to avoid confusion with
the 2-norm $\|A\|_{2}$. A tighter series of inequalities can be however:

$\displaystyle|\text{tr}(A^{\dagger}B)|\leq\sigma_{1}(A)\sum_{j}\sigma_{j}(B)\leq\sigma_{1}(A)\sum_{j}\sigma_{j}(B)\leq\sqrt{n}\sigma_{1}(A)\sqrt{\sum_{j}\sigma_{j}^{2}(B)}.$
(125)

Since $\sum_{j}\sigma_{j}^{2}(B)=\|B\|_{F}^{2}$, the inequality above can be
tighter by a factor $\sqrt{n}$ if used properly.

The inequality follows from the following two statements, which can be proved
using the singular value decomposition. For any matrix $M=A^{\dagger}B$, we
have $|\text{tr}(M)|\leq\sum_{i}\sigma_{i}(M)$. Also,
$\sigma_{i}(A^{\dagger}B)\leq\sigma_{1}(A)\sigma_{i}(B)$, which can be proved
using the Fischer min-max theorem.

Another inequality in terms of the Frobenius norm for the trace can be derived
as follows. Consider again $|\text{tr}(A^{\dagger}B)|$. We can write the
following Hölder inequality

$\displaystyle|\text{tr}(A^{\dagger}B)|\leq|\text{tr}(|A^{\dagger}|^{p})|^{\frac{1}{p}}|\text{tr}(|B|^{q})|^{\frac{1}{q}}$
(126)

with $1/p+1/q=1$. Then, if we write $p\rightarrow\infty$ (or q), we can write

$\displaystyle|\text{tr}(A^{\dagger}B)|\leq\|A\|\|B\|_{1}$ (127)

where $\|A\|$ is the operator norm, and $\|B\|_{1}$ the 1-norm. We also make
use of the inequality

$\displaystyle\|B\|_{F}\leq\|B\|_{1}\leq\sqrt{\text{r}(B)}\|B\|_{F}$ (128)

where $\text{r}(B)$ is the rank of $B$. The right inequality holds when the
singular values of $B$ are all the same, while the left inequality when only
one singular value is nonzero. We can say something more if at least one of
the two matrices can be diagonalized, which is our case in the paper.

If $B=U_{t}^{\dagger}[\rho_{0},U_{t}]=\rho_{t}-\rho_{0}$, then because of the
subadditivity of the rank and the fact that a unitary transformation does not
change the rank, we have $\text{r}(\rho_{t}-\rho_{0})\leq
2\text{r}(\rho_{0})$. If the state $\rho_{0}$ is thus pure, we have that
$\text{r}(B)=2$, which is of order $1$.

A bound we will also use is the one for the square root. In fact we have that

$\displaystyle\sum_{i}\sqrt{\lambda_{i}a_{i}}\leq\sqrt{\sum_{i}\lambda_{i}}\sqrt{\sum_{i}a_{i}},$
(129)

which is due to the concavity of the square root.

##  Appendix C Relation between $\mathbb{C}(\cdot)$ and $\mathcal{D}(\cdot)$

Let us define the overlap in a certain basis defined by projector operators
$\Pi_{k}$ as

$\mathbb{C}(X)=\frac{1}{2}\sum_{j=1}^{n}\|[X,\Pi_{j}]\|_{F}^{2}.$ (130)

If the operator $X$ is the density matrix, $\mathbb{C}(\rho)$ is exactly the
coherence of $\rho$ in the basis given by $\Pi$’s.

First, we show that the function $\mathbb{C}(X)$ is twice the coherence. We
define the super operator $\mathcal{D}(\cdot)=\sum_{i}\Pi_{i}\cdot\Pi_{i}$,
and the coherence as

$\displaystyle\|X-\mathcal{D}(X)\|_{F}^{2}$ $\displaystyle=$
$\displaystyle\|X-\sum_{i}\Pi_{i}X\Pi_{i}\|_{F}^{2}$ (131) $\displaystyle=$
$\displaystyle\text{Tr}\Big{(}(X^{\dagger}-\sum_{j}\Pi_{j}X^{\dagger}\Pi_{j})(X-\sum_{i}\Pi_{i}X\Pi_{i})\Big{)}$
$\displaystyle=$
$\displaystyle\text{Tr}\Big{(}X^{\dagger}X-\sum_{j}\Pi_{j}X^{\dagger}\Pi_{j}X-X\sum_{i}\Pi_{i}X\Pi_{i}+\sum_{ij}\Pi_{j}X^{\dagger}\Pi_{j}\Pi_{i}X\Pi_{i}\Big{)}$
$\displaystyle=$
$\displaystyle\text{Tr}\Big{(}X^{\dagger}X-\sum_{i}X^{\dagger}\Pi_{i}X\Pi_{i}\Big{)}.$

Now we have

$\displaystyle\mathbb{C}(X)\equiv\mathbb{C}_{\Pi}(X)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{i}\|[X,\Pi_{i}]\|_{F}^{2}=\sum_{i}\text{Tr}\Big{(}(X\Pi_{i}-\Pi_{i}X)^{\dagger}(X\Pi_{i}-\Pi_{i}X)\Big{)}$
(132) $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{i}\text{Tr}\Big{(}\Pi_{i}X^{\dagger}X\Pi_{i}-\Pi_{i}X^{\dagger}\Pi_{i}X-X^{\dagger}\Pi_{i}X\Pi_{i}+X^{\dagger}\Pi_{i}\Pi_{i}X\Big{)}$
$\displaystyle=$ $\displaystyle
2\text{Tr}\Big{(}X^{\dagger}X-\sum_{i}X^{\dagger}\Pi_{i}X\Pi_{i}\Big{)}=\|X-\mathcal{D}(X)\|_{F}^{2}.$

It follows that $\mathbb{C}(\cdot)$ is the coherence operator in the basis of
$\Pi$’s. Also, we note that, notationally if we write $X_{ij}$ in the basis of
$\Pi_{i}$’s, in general $\sum_{i\neq j}|x_{ij}|^{2}=\mathbb{C}_{\Pi}(X)$. If a
certain operator $A$ has a spectral decomposition in terms of the projectors
$\Pi$’s, we will write with an abuse of notation
$\mathbb{C}_{\Pi}(\cdot)=\mathbb{C}_{A}(\cdot)$.

Note that $\|[X,\Pi_{i}]\|_{F}^{2}=2(\sum_{k}|a_{ik}|^{2}-|a_{ii}|^{2})$.
Also, note that

$\displaystyle\text{Tr}\Big{(}X^{\dagger}X-\sum_{i}X^{\dagger}\Pi_{i}X\Pi_{i}\Big{)}$
$\displaystyle=$
$\displaystyle\text{Tr}\Big{(}X^{\dagger}X\sum_{i}\Pi_{i}-\sum_{i}X^{\dagger}\Pi_{i}X\Pi_{i}\Big{)}$
(133) $\displaystyle=$
$\displaystyle\sum_{i}\text{Tr}\Big{(}X^{\dagger}X\Pi_{i}-X^{\dagger}\Pi_{i}X\Pi_{i}\Big{)}=\sum_{i}\text{Tr}(X^{\dagger}[X\Pi_{i},\Pi_{i}])$

which is an expression we will use later.

##  Appendix D Coherence bounds

In this section we prove various auxiliary propositions which enter into the
coherence bounds lemmas proved below and reported in the main text.

First, we prove the following:

Proposition 1. Assuming that $\Pi_{i}\Pi_{j}=\delta_{ij}\Pi_{i}$, we have

$\text{tr}([\Pi_{i},A\Pi_{i}]^{\dagger}[\Pi_{j},A\Pi_{j}])=\delta_{ij}\|[\Pi_{j},A\Pi_{j}]\|_{F}^{2}$

.

Proof. Let us now prove some auxiliary properties related to the spectral
decomposition. Note that $\Pi_{i}^{2}=I$, and $\Pi_{i}\Pi_{j}=0$ if $i\neq j$.
Then we have

$\displaystyle\text{tr}([\Pi_{i},A\Pi_{i}]^{\dagger}[\Pi_{j},A\Pi_{j}])=0$
(134)

for any operator $A$. In fact

$\displaystyle\text{tr}([\Pi_{i},A\Pi_{i}]^{\dagger}[\Pi_{j},A\Pi_{j}])$
$\displaystyle=$
$\displaystyle\text{tr}\Big{(}(\Pi_{i}A\Pi_{i}-A\Pi_{i})^{\dagger}(\Pi_{j}A\Pi_{j}-A\Pi_{j})\Big{)}$
(135) $\displaystyle=$
$\displaystyle-\text{tr}\Big{(}(\Pi_{i}A\Pi_{i}-\Pi_{i}A)(\Pi_{j}A\Pi_{j}-A\Pi_{j})\Big{)}$
$\displaystyle=$
$\displaystyle-\text{tr}\Big{(}\Pi_{i}A\Pi_{i}\Pi_{j}A\Pi_{j}-\Pi_{i}A\Pi_{i}A\Pi_{j}-\Pi_{i}A\Pi_{j}A\Pi_{j}+\Pi_{i}A\Pi_{i}\Pi_{j}A\Pi_{j}\Big{)}$
$\displaystyle=$
$\displaystyle-\text{tr}\Big{(}\Pi_{i}A\Pi_{i}\Pi_{j}A\Pi_{j}-\Pi_{j}\Pi_{i}A\Pi_{i}A-\Pi_{j}\Pi_{i}A\Pi_{j}A+\Pi_{j}\Pi_{i}A\Pi_{i}\Pi_{j}A\Big{)}.$

Since in every term there is a product of the form $\Pi_{i}\Pi_{j}$, the trace
is given by

$\displaystyle\text{tr}([\Pi_{i},A\Pi_{i}]^{\dagger}[\Pi_{j},A\Pi_{j}])=\delta_{ij}\|[\Pi_{j},A\Pi_{j}]\|_{F}^{2},$
(136)

which completes the proof. $\square$

Proposition 2 If $\Pi_{i}\Pi_{j}=\Pi_{i}\delta_{ij}$, then

$\displaystyle\|[\Pi_{i},B\Pi_{j}]\|_{F}^{2}=\|[\Pi_{i},\Pi_{j}B]\|_{F}^{2}$
(137)

Proof. Let us evaluate both sides of the equality. On the right hand side we
have

$\displaystyle\text{tr}([\Pi_{i},\Pi_{j}B]^{\dagger}[\Pi_{i},\Pi_{j}B])$
$\displaystyle=$
$\displaystyle\text{tr}\Big{(}(\Pi_{i}\Pi_{j}B-\Pi_{j}B\Pi_{i})^{\dagger}(\Pi_{i}\Pi_{j}B-\Pi_{j}B\Pi_{i})\Big{)}$
(138) $\displaystyle=$
$\displaystyle\text{tr}\Big{(}(B\Pi_{i}\Pi_{j}-\Pi_{i}B\Pi_{j})(\Pi_{i}\Pi_{j}B-\Pi_{j}B\Pi_{i})\Big{)}$
$\displaystyle=$
$\displaystyle\text{tr}(B^{2}\Pi_{i}\Pi_{j}\Pi_{i}\Pi_{j}-B\Pi_{i}\Pi_{j}\Pi_{j}B\Pi_{i}-\Pi_{i}B\Pi_{j}\Pi_{i}\Pi_{j}B+\Pi_{i}B\Pi_{j}\Pi_{j}B\Pi_{i})$
$\displaystyle=$
$\displaystyle\text{tr}\Big{(}\delta_{ij}B^{2}\Pi_{i}-2\delta_{ij}B\Pi_{i}B\Pi_{i}+\Pi_{i}B\Pi_{j}B\Big{)}.$

On the left hand side instead we have

$\displaystyle\text{tr}([\Pi_{i},B\Pi_{j}]^{\dagger}[\Pi_{i},B\Pi_{j}])$
$\displaystyle=$
$\displaystyle\text{tr}\Big{(}(\Pi_{i}B\Pi_{j}-B\Pi_{j}\Pi_{i})^{\dagger}(\Pi_{i}B\Pi_{j}-B\Pi_{j}\Pi_{i})\Big{)}$
(139) $\displaystyle=$
$\displaystyle\text{tr}\Big{(}(\Pi_{j}B\Pi_{i}-\Pi_{i}\Pi_{j}B)(\Pi_{i}B\Pi_{j}-B\Pi_{j}\Pi_{i})\Big{)}$
$\displaystyle=$
$\displaystyle\text{tr}\Big{(}\Pi_{j}B\Pi_{i}\Pi_{i}B\Pi_{j}-\Pi_{j}B\Pi_{i}B\Pi_{j}\Pi_{i}-\Pi_{i}\Pi_{j}B\Pi_{i}B\Pi_{j}+\Pi_{i}\Pi_{j}B^{2}\Pi_{j}\Pi_{i}\Big{)}$
$\displaystyle=$
$\displaystyle\text{tr}\Big{(}B\Pi_{i}B\Pi_{j}-2\delta_{ij}B\Pi_{i}B\Pi_{j}+\delta_{ij}B^{2}\Pi_{i}\Big{)}$

from which we see that the equality applies. $\square$

Proposition 3 Let $\Pi_{i}\Pi_{j}=\delta_{ij}\Pi_{i}$. Then, if $A$ is
Hermitian, we have

$\displaystyle\|[\Pi_{i},A\Pi_{i}]\|_{F}^{2}=\frac{1}{2}\|[\Pi_{i},A]\|_{F}^{2}.$
(140)

Proof. Note that, for $A$ Hermitian

$\displaystyle\text{tr}[\Pi_{i},A\Pi_{i}]^{\dagger}[\Pi_{i},A\Pi_{i}]$
$\displaystyle=$
$\displaystyle\text{tr}(\Pi_{i}A\Pi_{i}-\Pi_{i}A)(\Pi_{i}A\Pi_{i}-A\Pi_{i})$
(141) $\displaystyle=$
$\displaystyle\text{tr}(\Pi_{i}A\Pi_{i}A\Pi_{i}-\Pi_{i}A\Pi_{i}A\Pi_{i}-\Pi_{i}A\Pi_{i}A\Pi_{i}+\Pi_{i}A^{2}\Pi_{i})$
$\displaystyle=$ $\displaystyle\text{tr}(A^{2}\Pi_{i}-\Pi_{i}A\Pi_{i}A).$

A rapid calculation shows that instead

$\displaystyle\text{tr}[\Pi_{i},A]^{\dagger}[\Pi_{i},A]$ $\displaystyle=$
$\displaystyle-\text{tr}\Big{(}\Pi_{i}A\Pi_{i}A-A^{2}\Pi_{i}-\Pi_{i}A^{2}+A\Pi_{i}A\Pi_{i}\Big{)}=2\text{tr}(A^{2}\Pi_{i}-\Pi_{i}A\Pi_{i}A),$
(142)

which completes the proof. $\square$

From the Proposition above, it follows that
$\sum_{i}\|[\Pi_{i},A\Pi_{i}]\|_{F}^{2}=\mathbb{C}_{\Pi}(A)$.

Proposition 4 Let $\Pi_{i}\Pi_{j}=\delta_{ij}\Pi_{i}$ and $B$ Hermitian. Then

$\displaystyle\text{tr}([\Pi_{i},\Pi_{j}B]^{\dagger}[\Pi_{a},\Pi_{b}B])=\delta_{ja}\delta_{ab}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{a}]\Big{)}+(\delta_{ia}\delta_{jb}-\delta_{ji}\delta_{jb})\text{tr}(B\Pi_{j}B\Pi_{a})$
(143)

Proof. We have

$\displaystyle\text{tr}([\Pi_{i},\Pi_{j}B]^{\dagger}[\Pi_{a},\Pi_{b}B])$
$\displaystyle=$
$\displaystyle\text{tr}\Big{(}(\Pi_{i}\Pi_{j}B-\Pi_{j}B\Pi_{i})^{\dagger}(\Pi_{a}\Pi_{b}B-\Pi_{b}B\Pi_{a})\Big{)}$
(144) $\displaystyle=$
$\displaystyle\text{tr}\Big{(}(B\Pi_{j}\Pi_{i}-\Pi_{i}B\Pi_{j})(\Pi_{a}\Pi_{b}B-\Pi_{b}B\Pi_{a})\Big{)}$
$\displaystyle=$
$\displaystyle\text{tr}\Big{(}B^{2}\Pi_{j}\Pi_{i}\Pi_{a}\Pi_{b}-\Pi_{i}B\Pi_{j}\Pi_{a}\Pi_{b}B-B\Pi_{j}\Pi_{i}\Pi_{b}B\Pi_{a}+\Pi_{i}B\Pi_{j}\Pi_{b}B\Pi_{a}\Big{)}$
$\displaystyle=$
$\displaystyle\text{tr}\Big{(}\underbrace{\delta_{ja}\delta_{ia}\delta_{ab}B^{2}\Pi_{i}}_{A}-\underbrace{\delta_{ja}\delta_{ab}B\Pi_{a}B\Pi_{i}}_{B}-\underbrace{\delta_{ji}\delta_{jb}B\Pi_{j}B\Pi_{a}}_{C}+\underbrace{\delta_{ia}\delta_{jb}B\Pi_{j}B\Pi_{a}}_{D}\Big{)}$

Now we note that terms $A+B$ can be written as

$\displaystyle\text{tr}\Big{(}\delta_{ja}\delta_{ia}\delta_{ab}B^{2}\Pi_{i}-\delta_{ja}\delta_{ab}B\Pi_{a}B\Pi_{i}\Big{)}$
$\displaystyle=$
$\displaystyle\text{tr}\Big{(}\delta_{ja}\delta_{ia}\delta_{ab}B^{2}\Pi_{i}-\delta_{ja}\delta_{ab}B\Pi_{a}B\Pi_{i}\Big{)}$
$\displaystyle=$
$\displaystyle\delta_{ja}\delta_{ab}\text{tr}\Big{(}BB\Pi_{i}\Pi_{a}-B\Pi_{a}B\Pi_{i}\Big{)}$
$\displaystyle=$
$\displaystyle\delta_{ja}\delta_{ab}\text{tr}\Big{(}B(B\Pi_{i}\Pi_{a}-\Pi_{a}B\Pi_{i})\Big{)}$
$\displaystyle=$
$\displaystyle\delta_{ja}\delta_{ab}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{a}]\Big{)}$

Let us now consider the terms $C+D$. We have that these can be written as

$\displaystyle(\delta_{ia}\delta_{jb}-\delta_{ji}\delta_{jb})\text{tr}(B\Pi_{j}B\Pi_{a})=(\delta_{ia}\delta_{jb}-\delta_{ji}\delta_{jb})|b_{ja}|^{2}$
(147)

which is the final result. $\square$

Corollary 1 We have

$\displaystyle\|[\Pi_{i},\Pi_{j}B]\|_{F}^{2}=\delta_{ij}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{i}]\Big{)}+(1-\delta_{ij})\text{tr}(B\Pi_{j}B\Pi_{i})$
(148)

Proof. From Proposition 4 we set $a=i$ and $b=j$.

Corollary 2 We have

$\displaystyle\sum_{ij}\|[\Pi_{i},\Pi_{j}B]\|_{F}^{2}=2\mathbb{C}_{\Pi}(B)$
(149)

Proof. From Corollary 1, we have

$\displaystyle\sum_{ij}\|[\Pi_{i},\Pi_{j}B]\|_{F}^{2}$ $\displaystyle=$
$\displaystyle\sum_{ij}\Big{(}\delta_{ij}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{i}]\Big{)}+(1-\delta_{ji})\text{tr}(B\Pi_{j}B\Pi_{i})\Big{)}$
(150) $\displaystyle=$
$\displaystyle\sum_{ij}\delta_{ij}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{i}]\Big{)}+\sum_{ij}(1-\delta_{ji})\text{tr}(B\Pi_{j}B\Pi_{i})$
$\displaystyle=$
$\displaystyle\sum_{i}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{i}]\Big{)}+\sum_{ij}|b_{ij}|^{2}-\sum_{i}|b_{ii}|^{2}$
$\displaystyle=$
$\displaystyle\sum_{i}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{i}]\Big{)}+\mathbb{C}_{\Pi}(B)$
$\displaystyle=$ $\displaystyle 2\mathbb{C}_{\Pi}(B)$

where we used Eq. (133). The final equation is what we claim in the statement
above. $\square$

Lemma 1 - Single Normal Coherence Inequality

$\displaystyle\|[U,A]\|_{F}^{2}$ $\displaystyle\leq$ $\displaystyle
4\|A\|^{2}\sum_{i\neq j}|a_{ij}|^{2}=4\|A\|^{2}\mathbb{C}_{U}(A)$ (151)

Proof. Let $U=\sum_{i}\eta_{i}\Pi_{i}$, with
$\Pi_{i}\Pi_{j}=\delta_{ij}\Pi_{i}$. Then, we have

$\displaystyle\|[U,A]\|_{F}^{2}$ $\displaystyle=$
$\displaystyle\sum_{ij}\eta_{i}\eta_{j}^{*}\text{tr}([\Pi_{j},A]^{\dagger}[\Pi_{i},A])$
(152) $\displaystyle=$ $\displaystyle
2\sum_{ij}\eta_{i}\eta_{j}^{*}\text{tr}\Big{(}A^{2}\Pi_{i}\delta_{ij}-A\Pi_{i}A\Pi_{j}\Big{)}$
$\displaystyle=$ $\displaystyle
2\sum_{i}|\eta_{i}|^{2}\text{tr}\Big{(}A^{2}\Pi_{i}-A\Pi_{i}A\Pi_{i}\Big{)}-2\sum_{i\neq
j}\eta_{i}\eta_{j}^{*}\text{tr}\Big{(}A\Pi_{i}A\Pi_{j}\Big{)}$

We note that
$2\sum_{i}|\eta_{i}|^{2}\text{tr}\Big{(}A^{2}\Pi_{i}-A\Pi_{i}A\Pi_{i}\Big{)}\leq
2\text{sup}_{i}|\eta_{i}^{2}\mathbb{C}_{U}(A)$. Now note that

$-2\sum_{i\neq
j}\eta_{i}\eta_{j}^{*}\text{tr}\Big{(}A\Pi_{i}A\Pi_{j}\Big{)}\leq
2\sum_{ij}|\eta_{i}\eta_{j}^{*}|\ |a_{ij}|^{2}\leq
2|\eta_{i}^{2}\mathbb{C}_{U}(A).$ (153)

It follows that we have

$\|[U,A]\|_{F}^{2}\leq 4\|A\|^{2}\mathbb{C}_{U}(A)$

as claimed. $\square$

Corollary For a unitary operator $U$, we have

$\displaystyle\|[U,A]\|_{F}^{2}$ $\displaystyle\leq$ $\displaystyle
4\sum_{i\neq j}|a_{ij}|^{2}=4\mathbb{C}_{U}(A)$ (154)

Proof. It follows from Lemma 1 and the fact that $\|U\|=1$.

Lemma $1^{\prime}$ \- Single Unitary coherence inequality

Let $U$ be a unitary operator and $A$ an Hermitian operator. Then

$\displaystyle\|[U,A]\|_{F}^{2}$ $\displaystyle\leq$ $\displaystyle
2\sqrt{2}\|A\|\Big{(}2^{-\frac{1}{4}}\sqrt{n\|A\|}\mathbb{C}^{2}_{U}(A)+\mathbb{C}^{1}_{U}(A)\Big{)}$
(155)

Proof. Let $U=\sum_{i}\eta_{i}\Pi_{i}$, with
$\Pi_{i}\Pi_{j}=\delta_{ij}\Pi_{i}$. Then, we have

$\displaystyle\|[U,A]\|_{F}^{2}$ $\displaystyle=$
$\displaystyle\sum_{ij}\eta_{i}\eta_{j}^{*}\text{tr}([\Pi_{j},A]^{\dagger}[\Pi_{i},A])$
$\displaystyle=$
$\displaystyle-\sum_{ij}\eta_{i}\eta_{j}^{*}\text{tr}\Big{(}(\Pi_{j}A-A\Pi_{j})(\Pi_{i}A-A\Pi_{i})\Big{)}$
$\displaystyle=$
$\displaystyle-2\sum_{ij}\eta_{i}\eta_{j}^{*}\text{tr}(\Pi_{j}A\Pi_{i}A-\delta_{ij}A^{2}\Pi_{i})$
$\displaystyle=$ $\displaystyle
2\sum_{ij}\eta_{i}\eta_{j}^{*}\text{tr}(A[A\Pi_{j},\Pi_{i}])$

We can now upper bound the quantity above with the absolute value, and using
the identity
$|\text{tr}(A[A\Pi_{j},\Pi_{i}])|=|\text{tr}(A[\Pi_{j}A,\Pi_{i}])|$, we have

$\displaystyle\|[U,A]\|_{F}^{2}$ $\displaystyle\leq$ $\displaystyle
2\sum_{ij}|\text{tr}(A[\Pi_{j}A,\Pi_{i}])|$ (157) $\displaystyle=$
$\displaystyle 2\sum_{ij}|\text{tr}(A[\Pi_{j}A,\Pi_{i}])|$ $\displaystyle\leq$
$\displaystyle
2\|A\|\sum_{ij}\sqrt{\text{r}([\Pi_{j}A,\Pi_{i}])\|[\Pi_{j}A,\Pi_{i}]\|_{F}^{2}}$

We note now that $\text{r}([\Pi_{j}A,\Pi_{i}])\leq 2$. Then

$\displaystyle\|[U,A]\|_{F}^{2}$ $\displaystyle\leq$ $\displaystyle
2\sqrt{2}\|A\|\sum_{ij}\sqrt{\|[\Pi_{j}A,\Pi_{i}]\|_{F}^{2}}$ (158)

We now use the fact that, from Corollary 1,

$\displaystyle\|[U,A]\|_{F}^{2}$ $\displaystyle\leq$ $\displaystyle
2\sqrt{2}\|A\|\sum_{ij}\sqrt{\delta_{ij}\text{tr}\Big{(}A[A\Pi_{i},\Pi_{i}]\Big{)}+(1-\delta_{ij})\text{tr}(A\Pi_{j}A\Pi_{i})}$
$\displaystyle=$ $\displaystyle
2\sqrt{2}\|A\|\Big{(}\sum_{i}\sqrt{\text{tr}\Big{(}A[A\Pi_{i},\Pi_{i}]\Big{)}}+\sum_{i\neq
j}\sqrt{\text{tr}(A\Pi_{j}A\Pi_{i})}\Big{)}$ $\displaystyle=$ $\displaystyle
2\sqrt{2}\|A\|\Big{(}\sum_{i}\sqrt{\text{tr}\Big{(}A[A\Pi_{i},\Pi_{i}]\Big{)}}+\sum_{i\neq
j}|a_{ij}|\Big{)}$ $\displaystyle\leq$ $\displaystyle
2\sqrt{2}\|A\|\Big{(}\sqrt{\sqrt{2}\|A\|}\sum_{i}\sqrt{\frac{1}{2}\|[\Pi_{i},A]\|_{F}^{2}}+\sum_{i\neq
j}|a_{ij}|\Big{)}$ $\displaystyle\leq$ $\displaystyle
2\sqrt{2}\|A\|\Big{(}2^{-\frac{1}{4}}\sqrt{\|A\|}\sum_{i}\sqrt{\
\|[\Pi_{i},A]\|_{F}^{2}}+\sum_{i\neq j}|a_{ij}|\Big{)}$ $\displaystyle=$
$\displaystyle
2\sqrt{2}\|A\|\Big{(}2^{-\frac{1}{4}}\sqrt{\|A\|}\sum_{i}(\sum_{k\neq
i}|a_{ik}|^{2})^{\frac{1}{2}}+\sum_{i\neq j}|a_{ij}|\Big{)}$

The bound above is the first step of the bound. We now use the fact that the
square root is concave, and thus
$\sum_{ij}\sqrt{\lambda_{ij}a_{ij}}\leq\sqrt{\sum_{ij}\lambda_{ij}}\sqrt{\sum_{ij}a_{ij}}$.
Then

$\displaystyle\|[U,A]\|_{F}^{2}$ $\displaystyle\leq$ $\displaystyle
2\sqrt{2}\|A\|\Big{(}2^{-\frac{1}{4}}\sqrt{\|A\|}\sqrt{n}\sqrt{\sum_{i\neq
j}|a_{ij}|^{2}}+\sum_{i\neq j}|a_{ij}|\Big{)}$ (160)

It follows that

$\displaystyle\|[U,A]\|_{F}^{2}$ $\displaystyle\leq$ $\displaystyle
2\sqrt{2}\|A\|\Big{(}2^{-\frac{1}{4}}\sqrt{n\|A\|}\mathbb{C}^{2}_{U}(A)+\mathbb{C}^{1}_{U}(A)\Big{)}$
(161)

which concludes the proof. $\square$

Lemma 2 - Double Normal Coherence Inequality Assume A be a normal operator and
let B be Hermitian. Then,

$\|[A^{\dagger},BA]\|_{F}^{2}\leq 4\|A\|^{4}\mathbb{C}_{A}(B)$

Proof. For simplicity, we assume that $A$ is a normal operator, e.g. we have
$A=\sum_{i}\Pi_{i}\eta_{i}$, where $\eta_{i}$ are the eigenvalues of $A$. If
the operator is normal, then $A$ and $A^{\dagger}$ have an identical spectral
decomposition and the proof goes along similar steps, which is Lemma 3.

Then

$\displaystyle\|[A^{\dagger},BA]\|_{F}^{2}$ $\displaystyle=$
$\displaystyle\|\sum_{ij}\eta_{i}^{*}\eta_{j}[\Pi_{i},B\Pi_{j}]\|_{F}^{2}=\sum_{ij}\sum_{ab}\eta_{i}\eta_{j}^{*}\eta_{a}^{*}\eta_{b}\text{tr}([\Pi_{i},B\Pi_{j}]^{\dagger}[\Pi_{a},B\Pi_{b}]).$

Now using Proposition 4, we have

$\displaystyle\|[A,BA]\|_{F}^{2}$ $\displaystyle=$
$\displaystyle\|\sum_{ij}\eta_{i}\eta_{j}[\Pi_{i},B\Pi_{j}]\|_{F}^{2}=\sum_{ij}\sum_{ab}\eta_{i}\eta_{j}^{*}\eta_{a}^{*}\eta_{b}\text{tr}([\Pi_{i},B\Pi_{j}]^{\dagger}[\Pi_{a},B\Pi_{b}])$
(163) $\displaystyle=$
$\displaystyle\sum_{ij}\sum_{ab}\eta_{i}\eta_{j}^{*}\eta_{a}^{*}\eta_{b}\Big{(}\delta_{ja}\delta_{ab}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{a}]\Big{)}+(\delta_{ia}\delta_{jb}-\delta_{ji}\delta_{ib})\text{tr}(B\Pi_{j}B\Pi_{a})\Big{)}$
$\displaystyle=$
$\displaystyle\sum_{ij}\eta_{i}\eta_{j}^{*}|\eta_{j}|^{2}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{j}]\Big{)}+\sum_{i\neq
j}(|\eta_{i}|^{2}|\eta_{j}|^{2}-|\eta_{i}|^{2}\eta_{i}\eta_{j}^{*})||b_{ij}|^{2}$
$\displaystyle=$
$\displaystyle\sum_{ij}\eta_{i}\eta_{j}^{*}|\eta_{j}|^{2}\delta_{ij}\text{tr}\Big{(}B^{2}\Pi_{i}-B\Pi_{i}B\Pi_{i}\Big{)}-\sum_{i\neq
j}\eta_{i}\eta_{j}^{*}|\eta_{j}|^{2}\text{tr}\Big{(}B\Pi_{i}B\Pi_{j}\Big{)}+\sum_{i\neq
j}\eta_{i}\eta_{j}^{*}(\eta_{i}^{*}\eta_{j}-|\eta_{i}|^{2})||b_{ij}|^{2}$
$\displaystyle\leq$
$\displaystyle\text{sup}_{k}|\eta_{k}|^{4}\sum_{i}\text{tr}\Big{(}B^{2}\Pi_{i}-B\Pi_{i}B\Pi_{i}\Big{)}+\sum_{i\neq
j}|\eta_{i}\eta_{j}^{*}(\eta_{i}^{*}\eta_{j}-|\eta_{i}|^{2}-|\eta_{j}|^{2})|||b_{ij}|^{2}$
$\displaystyle\leq$ $\displaystyle\text{sup}_{k}|\eta_{k}|^{4}\sum_{i\neq
j}||b_{ij}|^{2}+\sum_{i\neq
j}|\eta_{i}\eta_{j}(\eta_{i}^{*}\eta_{j}-|\eta_{i}|^{2}-|\eta_{j}|^{2})|||b_{ij}|^{2}$
$\displaystyle\leq$ $\displaystyle 4\ \text{sup}_{k}|\eta_{k}|^{4}\sum_{i\neq
j}||b_{ij}|^{2}=4\|A\|^{4}\mathbb{C}_{A}(B)$

which is what we stated we would prove. $\square$

There is another inequality that can be obtained for a normal operator, in
terms of mixed $l_{1}$ coherence. We report it below for a Hermitian operator,
but a similar bound applies to general normal operator.

Lemma 2(b) - Double Hermitian Coherence Inequality Assume $A$ and $B$ are
Hermitian operators. Then,

$\|[A,BA]\|_{F}^{2}\leq\|A^{2}\|^{2}\Big{(}\|B\|\big{(}2^{-\frac{1}{4}}\sqrt{\|B\|}\sqrt{n}\sqrt{\sum_{k\neq
i}|b_{ik}|^{2}}+\sum_{i\neq j}|b_{ij}|\big{)}+2\mathbb{C}_{A}(B)\Big{)}.$

where $b_{ij}$ are the elements of the matrix $B$ in the basis of $A$.
Proof. For simplicity, we assume that $A$ is Hermitian, we have
$A=\sum_{i}\Pi_{i}\eta_{i}$, where $\eta_{i}$ are the eigenvalues of $A$ and
are real. If the operator is normal, then $A$ and $A^{\dagger}$ have an
identical spectral decomposition and the proof goes along similar steps, which
is Lemma 3.

Then

$\displaystyle\|[A,BA]\|_{F}^{2}$ $\displaystyle=$
$\displaystyle\|\sum_{ij}\eta_{i}\eta_{j}[\Pi_{i},B\Pi_{j}]\|_{F}^{2}=\sum_{ij}\sum_{ab}\eta_{i}\eta_{j}\eta_{a}\eta_{b}\text{tr}([\Pi_{i},B\Pi_{j}]^{\dagger}[\Pi_{a},B\Pi_{b}]).$

Now using Proposition 4, we have

$\displaystyle\|[A,BA]\|_{F}^{2}$ $\displaystyle=$
$\displaystyle\|\sum_{ij}\eta_{i}\eta_{j}[\Pi_{i},B\Pi_{j}]\|_{F}^{2}=\sum_{ij}\sum_{ab}\eta_{i}\eta_{j}\eta_{a}\eta_{b}\text{tr}([\Pi_{i},B\Pi_{j}]^{\dagger}[\Pi_{a},B\Pi_{b}])$
(165) $\displaystyle=$
$\displaystyle\sum_{ij}\sum_{ab}\eta_{i}\eta_{j}\eta_{a}\eta_{b}\Big{(}\delta_{ja}\delta_{ab}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{a}]\Big{)}+(\delta_{ia}\delta_{jb}-\delta_{ji}\delta_{ib})\text{tr}(B\Pi_{j}B\Pi_{a})\Big{)}$
$\displaystyle=$
$\displaystyle\sum_{ij}\eta_{i}\eta_{j}|\eta_{j}|^{2}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{j}]\Big{)}+\sum_{ij}(\eta_{i}^{2}\eta_{j}^{2}-\eta_{i}^{3}\eta_{j})||b_{ij}|^{2}$
$\displaystyle=$
$\displaystyle\sum_{ij}\eta_{i}\eta_{j}|\eta_{j}|^{2}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{j}]\Big{)}+\sum_{i\neq
j}\eta_{i}\eta_{j}(\eta_{i}\eta_{j}-\eta_{i}^{2})||b_{ij}|^{2}$
$\displaystyle\leq$ $\displaystyle\sum_{ij}|\eta_{i}\eta_{j}|\ |\eta_{j}|^{2}\
|\text{tr}\Big{(}B[B\Pi_{i},\Pi_{j}]\Big{)}|+\sum_{i\neq
j}|\eta_{i}\eta_{j}(\eta_{i}\eta_{j}-\eta_{i}^{2})|\ ||b_{ij}|^{2}$

Here we wrote $b_{ij}$ as the elements $ij$ of the operator $B$ in the basis
of $\Pi$’s. Let us now look at the two expressions separately. The exact
expression for first term is given by Frobenius norm is thus bounded by
another Frobenius norm as follows: First, we note that
$|\text{tr}([B[B\Pi_{i},\Pi_{j}])|=|\text{tr}([B[\Pi_{i}B,\Pi_{j}])|$. Then,
using $|\text{tr}(AB)|\leq\|A\|\sqrt{\text{r}(B)}\|B\|_{F}$,

$\displaystyle\sum_{ij}|\eta_{i}\eta_{j}|\ |\eta_{j}|^{2}\
|\text{tr}\Big{(}B[B\Pi_{i},\Pi_{j}]\Big{)}|$ $\displaystyle=$
$\displaystyle\sum_{ij}|\eta_{i}\eta_{j}|\eta_{i}^{2}\text{tr}([B[\Pi_{i}B,\Pi_{j}])|\leq\|B\|\sum_{ij}|\eta_{i}\eta_{j}|\eta_{i}^{2}\sqrt{\text{r}([\Pi_{i}B,\Pi_{j}])\|[\Pi_{i}B,\Pi_{j}]]\|_{F}^{2}}$

Now note that
$\text{r}([\Pi_{i}B,\Pi_{j})=\text{r}(\Pi_{i}B\Pi_{j}-\Pi_{j}\Pi_{i}B)\leq 2$,
and using Propositions 1, 2 and 3, we have

$\displaystyle\|[\Pi_{i}B,\Pi_{j}]]\|_{F}^{2}=\delta_{ij}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{i}]\Big{)}+(1-\delta_{ij})\text{tr}(B\Pi_{j}B\Pi_{i}).$
(167)

The first term can then be bounded by

$\displaystyle\sum_{ij}|\eta_{i}\eta_{j}|\ |\eta_{j}|^{2}\
|\text{tr}\Big{(}B[B\Pi_{i},\Pi_{j}]\Big{)}|$ $\displaystyle\leq$
$\displaystyle\|B\|\sum_{ij}|\eta_{i}|^{4}\sqrt{\delta_{ij}\text{tr}\Big{(}B[B\Pi_{i},\Pi_{i}]\Big{)}+(1-\delta_{ij})\text{tr}(B\Pi_{j}B\Pi_{i})}$
$\displaystyle=$ $\displaystyle\|B\|\
\|A^{2}\|^{2}\Big{(}\sum_{i}\sqrt{\text{tr}\Big{(}B[B\Pi_{i},\Pi_{i}]\Big{)}}+\sum_{i\neq
j}\sqrt{\text{tr}(B\Pi_{j}B\Pi_{i})}\Big{)}$ $\displaystyle=$
$\displaystyle\|B\|\
\|A^{2}\|^{2}\Big{(}2^{-\frac{1}{4}}\sqrt{\|B\|}\sum_{i}(\sum_{k\neq
i}|b_{ik}|^{2})^{\frac{1}{4}}+\sum_{i\neq j}|b_{ij}|\Big{)}$
$\displaystyle\leq$ $\displaystyle\|B\|\
\|A^{2}\|^{2}\Big{(}2^{-\frac{1}{4}}\sqrt{\|B\|}\sqrt{n}\sqrt{\sum_{k\neq
i}|b_{ik}|^{2}}+\sum_{i\neq j}|b_{ij}|\Big{)}$

For the second term, we have

$\displaystyle\sum_{i\neq j}|\eta_{i}\eta_{j}(\eta_{i}\eta_{j}-\eta_{i}^{2})|\
|b_{ij}|^{2}\leq\text{sup}_{ij}|\eta_{i}\eta_{j}(\eta_{i}\eta_{j}-\eta_{i}^{2})|\sum_{i\neq
j}|b_{ij}|^{2}.$ (169)

Now note that

$\displaystyle\text{sup}_{ij}|\eta_{i}\eta_{j}(\eta_{i}\eta_{j}-\eta_{i}^{2})|\leq
2\|A^{2}\|^{2}$ (170)

while $\sum_{i\neq j}|b_{ij}|^{2}=\frac{1}{2}\mathbb{C}_{A}(B)$. The second
term can then be bounded by

$\displaystyle\sum_{i\neq
j}|\eta_{i}\eta_{j}(\eta_{i}\eta_{j}-\eta_{i}^{2})||b_{ij}|^{2}\leq
2\|A^{2}\|^{2}\mathbb{C}_{A}(B)$ (171)

We thus obtain the final bound

$\displaystyle\|[A,BA]\|_{F}^{2}\leq\|A^{2}\|^{2}\Big{(}\|B\|\big{(}2^{-\frac{1}{4}}\sqrt{\|B\|}\sqrt{n}\sqrt{\sum_{k\neq
i}|b_{ik}|^{2}}+\sum_{i\neq j}|b_{ij}|\big{)}+2\mathbb{C}_{A}(B)\Big{)}$ (172)

which is what we stated we would prove. $\square$