# Monitored non-adiabatic and coherent-controlled quantum unital Otto heat
engines: First four cumulants

Abdelkader El Makouri<EMAIL_ADDRESS>LPHE-Modeling and
Simulation, Faculty of Sciences, Mohammed V University in Rabat, Morocco.
Abdallah Slaoui  0000-0002-5284-3240 Corresponding author:
<EMAIL_ADDRESS>LPHE-Modeling and Simulation, Faculty of Sciences,
Mohammed V University in Rabat, Morocco.  Centre of Physics and Mathematics,
CPM, Faculty of Sciences, Mohammed V University in Rabat, Rabat, Morocco.
Rachid Ahl Laamara<EMAIL_ADDRESS>LPHE-Modeling and Simulation,
Faculty of Sciences, Mohammed V University in Rabat, Morocco.  Centre of
Physics and Mathematics, CPM, Faculty of Sciences, Mohammed V University in
Rabat, Rabat, Morocco.

###### Abstract

Recently, measurement-based quantum thermal machines have drawn more attention
in the field of quantum thermodynamics. However, the previous results on
quantum Otto heat engines were either limited to special unital and non-unital
channels in the bath stages, or a specific driving protocol at the work
strokes and assuming the cycle being time-reversal symmetric i.e.
$V^{\dagger}=U$ (or $V=U$). In this paper, we consider a single spin-1/2
quantum Otto heat engine, by first replacing one of the heat baths by an
arbitrary unital channel and then we give the exact analytical expression of
the characteristic function from which all the cumulants of heat and work
emerge. We prove that under the effect of monitoring, $\nu_{2}>\nu_{1}$ is a
necessary condition for positive work, either for a symmetric or asymmetric-
driven Otto cycle. Furthermore, going beyond the average we show that the
ratio of the fluctuations of work and heat is lower and upper-bounded when the
system is working as a heat engine. However, differently from the previous
results in the literature, we consider the third and fourth cumulants as well.
It is shown that the ratio of the third (fourth) cumulants of work and heat is
not upper-bounded by unity nor lower-bounded by the third (fourth) power of
the efficiency, as is the case for the ratio of fluctuations. Finally, we
consider applying a specific unital map that plays the role of a heat bath in
a coherently superposed, manner and we show the role of the initial coherence
of the control qubit on efficiency, on the average work and its relative
fluctuations.

$\mathbf{Keywords:}$ Monitored asymmetric quantum Otto heat engines, Unital
qubit channels, Ratio of cumulants and coherently superposed channels.

##  I Introduction

The field of quantum thermodynamics (QT) [1] can simply be defined as a new
research area that is trying to develop and formulate a consistent theory of
thermodynamics valid at the quantum scale. This new (actually it is old) field
has started with the work of Scovil and Schulz-Duboiz on the thermodynamics of
masers [2]. Masers can be considered the first prototype of quantum heat
engines. Nowadays QT gains its progress from quantum resource theory to
thermodynamics [3, 4], fluctuations relations [5, 7, 9, 6, 8], thermodynamic
uncertainty relations [13, 12, 10, 11] and so on. In this research area the
quantum version of thermal machines either heat engines or refrigerators,
occupy a very important place due to the role of their classical versions in
our society. They were recently reported experimentally using different
platforms, e.g. trapped ions [14] and NMR setting [15]. For a general overview
of other platforms, we recommend the recently written paper by Cangemi et al.
[16]. Quantum Otto cycle (QOC) [18, 17, 19] is the most cycle studied in QT.
This is because the heat and work exchanges are done in different strokes
which makes their computation easy. This cycle is composed of two isochoric
and two adiabatic quantum transformations. In previous works such as [17, 18]
the source of heat is a thermal reservoir. However, it was shown that quantum
measurement can as well provide heat to the system as long as its associated
operators do not commute with the measured quantity [20]. This idea of using
measurement as a heat source is inspired by Maxwell-demon and Szilard-engine
[21].

Usually in physics, we assume implicitly that events are causally ordered.
More precisely, either the event A is in the past of B (A$\preceq$B), or B is
in the past of A (B$\preceq$A), or in general the events may be space-like
separated, i.e. there is no signaling between the events A and B. However,
quantum mechanics allows for more general processes which have no definite
causal order, and this is because quantum objects can be in a superposition of
different states. This phenomenon would be useful when one wants to unify two
important theories of the 20 century, i.e., quantum mechanics and general
relativity, as was shown by Lucien Hardy in his works on indefinite causal
order (ICO) and quantum gravity [23, 22]. However, differently from ICO, where
one focuses on the order in which the operations are applied [24, 25], e.g.
$\mathcal{E}$ is before $\mathcal{F}$, $\mathcal{F}$ is before $\mathcal{E}$
or a superposition of the two orders, we can consider as well the case which
quantum channel has been applied on the system e.g. $\mathcal{E}$ or
$\mathcal{F}$ or in general it may be a superposition of the two. The latter
case is called a coherent superposition (CS) of the two channels $\mathcal{E}$
and $\mathcal{F}$ [26]. This is motivated by the fact that it was recently
shown that coherently superposed channels (CSCs) as well provide an advantage
e.g. in quantum communication [27] and quantum metrology [28, 29, 30], and
even more as was shown in [29] they improve the ICO in parameter estimation.
Furthermore, CSCs have been not investigated in other areas such as QT to the
authors’ knowledge except [31, 32].

As we said before, recently measurement-based quantum thermal machines [33,
34, 35, 36, 37, 38, 40, 39] draw more attention in the field of QT. However,
the previous results on quantum heat engines were either limited to special
unital [43, 42, 41, 33, 38] or non-unital channels [44, 45, 40] in the bath
stages, or a specific driving protocol [15] at the work strokes and under the
assumption of the cycle being time-reversal symmetric i.e. $V^{\dagger}=U$, an
exception are the results reported in Refs. [46, 47, 48]. Even more, only
counted works have considered quantities beyond averages, such as variance or
in general higher cumulants, which have not been analyzed so far for quantum
Otto heat engines (QOHEs). In this work, we try to fill these gaps. To do so,
inspired by the QOHE studied in [42, 41], we consider the role of replacing
one of the heat baths of the QOC with an arbitrary unital map on efficiency,
heat, work, their relative fluctuations, skewness and kurtosis. In the second
part of this paper, when considering applying the unital channel in a CS we
limit ourselves to the unital map studied in Refs. [41, 42].

For example, in Ref [49] Saryal et al. have shown that in the adiabatic
regime, the ratio of the $n$-th cumulants of work and heat (for quantum Otto
heat engines with completely thermalizing reservoirs), is equal to the $n$-th
power of the Otto efficiency and it is upper bounded by the $n$-th power of
the Carnot efficiency. Translated into our notation this means that,
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{n}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{n}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}=\left(1-\nu_{1}/\nu_{2}\right)^{n}\leq\left(1-\beta_{h}/\beta_{c}\right)^{n}$.
However, the authors did not give an interpretation of these bounds for $n\geq
3$ nor they have considered what would happen beyond the adiabatic regime. In
Ref [50], Saryal et al. have proved that for autonomous coupled and continuous
thermal machines in the linear response regime, the ratio of the $n$-th
cumulants of the input and output currents are lower bounded by the $n$-th
power of the efficiency and upper bounded by the $n$-th power of the Carnot
efficiency. As a consequence of this interesting result, is that the engine’s
efficiency received a tighter bound than the Carnot bound. Furthermore, in
Ref. [51] the authors have also considered the ratio of the $n$-th cumulants
of an ensemble of non-interacting quantum thermal machines. Suppose we have
$i\in[1,N]$ individual machine, and that the ratio of $n$-th cumulants of
every machine satisfy, $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W_{i}^{n}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{Mi}^{n}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}\leq\left(1-\beta_{h}/\beta_{c}\right)^{n}$.
Beyond the linear response limit, the authors have found that the ratio of the
sum of the $n$-th cumulants of individual machines is bounded by the $n$-th
power of the Carnot efficiency, i.e.
$\sum_{i=1}^{N}\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W_{i}^{n}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\sum_{i=1}^{N}\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{Mi}^{n}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}\leq\left(1-\beta_{h}/\beta_{c}\right)^{n}$.
For an ensemble of non-interacting quantum refrigerators, they have shown
analytically and numerically under certain different assumptions, that the
ratio of the sum of the $n$-th cumulants has a lower bound as well, given by
the $n$-th power of the efficiency. The lower bound is saturated for example
in the tight coupling limit, See [50, 51]. However, in the heat engine regime,
only the upper bound was respected, but not the lower bound.

The main five messages to take from this paper can be summarised as follows:

  1. 1.

First, considering a QOHE, we neither specify the driving protocol nor the
unital channel replacing the thermal bath. Then we give the exact analytical
expression of the characteristic function (CF) from which all the cumulants of
heat and work would emerge. We show that considering arbitrary unital channel
have a positive influence on heat absorbed, work and efficiency. Furthermore,
we prove that under the effect of monitoring, changing the gap of the single
spin-1/2 is a necessary condition for work extraction. This is true for either
symmetric or asymmetric driven Otto cycle. We trace this back to the negative
influence of projective measurement used to assess the fluctuations of
thermodynamic quantities. Further, we show that for the asymmetric Otto cycle
not only do the forward and the backward are needed to be treated on equal
footing for meaningful bounds on relative fluctuations (RFs)[46], but also
help in restoring the fact that the Otto efficiency is still an upper bound.

  2. 2.

We prove that when one of the heat baths in the Otto cycle is replaced by an
arbitrary unital map, then the system cannot work as a refrigerator
independently of the parameters. More precisely, the bath cannot be cooled.
This is forbidden by the second law of thermodynamics in accordance with the
Kelvin-Planck statement that heat only flows spontaneously from a hotter body
to a cooler one.

  3. 3.

We show that the ratio of the fluctuations of $W$ and $Q_{M}$ are lower and
upper bounded when the system is working as a heat engine and for some regions
when it is working as an accelerator. This is proved to be true for the
symmetric and asymmetric Otto cycle. We proved as well, that the square of the
Otto efficiency provides a lower bound on the ratio of fluctuations for the
symmetric Otto cycle. Then, numerically we show that the ratio of the third
and the fourth cumulants can not be lower-bounded by the third (fourth) power
of the efficiency as is the case for the relative fluctuations, nor they are
bounded by unity. The reason behind this is purely quantum due to the driving
as we show later on. Furthermore, the fact that we do not have always,
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{n}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{n}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}\geq\left(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}\right)^{n}$ for
$n=3,4$ suggest that the ratio of third and fourth cumulants can not always
provide a bound on the efficiency as it the case for the second cumulants.

  4. 4.

We analyze in detail the effect of considering an arbitrary unital channel on
the average work and its relative fluctuations. We show that the work average
can be enhanced as well as its reliability (note that enhanced reliability is
equivalent to decreasing relative fluctuations). When the inverse temperature
is negative, we comment on the positive rule of non-adiabaticity.

  5. 5.

Finally, when considering CSCs we consider the unital channel studied in [41].
We show, that when the control qubit is projected in $|-\rangle_{c}$ Fourier
basis has a positive influence on the work extracted and its reliability as
well as efficiency, even though, the latter cannot exceed that of the Otto.
When the control qubit is projected in the $|+\rangle_{c}$ Fourier basis we
see the inverse of these conclusions. Note that efficiency enhancement in the
non-adiabatic regime could be helpful when one is interested in both
efficiency and power as in real-world applications.

Our results put forward thoroughly and deeply the ones reported in Refs. [41,
42], by not only considering average quantities but as well the variance,
skewness and kurtosis which were not considered there. Furthermore, inspired
by the results reported in [52, 42, 41]. we will express all the cumulants of
work and heat in terms of only three transitions probabilities $\delta$,
$\theta$ and $\zeta$ defined below, without specifying the unital channel
providing heat to the system nor the driving protocol used to change the gap
of the system and without the assumption that the cycle is time-reversal
symmetric. In the main text below, we show the difference between our work and
the one in [42, 41]. Furthermore, note that skewness and kurtosis have been
taken into account here since they were not analyzed in the previous works on
QOHEs in literature to the authors’ knowledge. However, note that considering
higher cumulants such as the third and fourth comes with more computational
complexity, let alone higher than the fourth cumulant. For example, here we
could not prove that the ratio of the third (fourth) cumulants is upper
bounded by 1 and lower bounded by the third (fourth) power of the efficiency
since their expressions are too complicated in terms of the parameters
especially those of work. Fortunately, numerically we found that they do not
obey the same bound as is the case for the ratio of fluctuations.

This paper is structured as follows: In Sec.II, we take the conventional QOC
with two heat baths and replace one of them with an arbitrary unital qubit
channel which would play the role of a hot bath. We give the exact analytical
expression of the CF. Further, since we use the two-point measurement to
assess fluctuations of work and heat we show that the latter has a negative
effect on the work positive condition. More precisely, differently from [41]
we show that changing the frequency of the spin is a necessary condition for
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0$. The
efficiency and the first four cumulants of work and heat are analyzed in
detail. In Sec.III, first in Sec.III.1 we define what we mean by CSCs, then we
apply the unital channel considered in [41] in a CS and show the effect of
this on efficiency, work as well as its relative fluctuations. Finally, in
Sec.IV, we give a summary of our results. In the appendix section, we provide
some technical details. Throughout the paper, we set $\hbar=k_{B}=1$.

##  II Monitored quantum unital Otto cycle

Before we dive into the main results of the paper let’s remind the reader of
the unmonitored QOC with either two heat baths or one heat bath and an
arbitrary channel.

###  II.1 Unmonitored (asymmetric) QOC

Figure 1:  (a) A two-level system undergoing the four strokes of the
conventional QOC. $\mathbf{A}\rightarrow\mathbf{B}$ an adiabatic expansion
$\nu_{1}\rightarrow\nu_{2}(>\nu_{1})$, $\mathbf{B}\rightarrow\mathbf{C}$ the
system gets equilibrates with a hot bath, $\mathbf{C}\rightarrow\mathbf{D}$ an
adiabatic compression $\nu_{2}\rightarrow\nu_{1}$ and finally
$\mathbf{D}\rightarrow\mathbf{A}$ gets equilibrate with a cold bath, thus
closing the cycle. In this case, we have two heat baths one at inverse
temperature $\beta_{c}$ and the other at inverse temperature $\beta_{h}$.
$\langle Q_{c}\rangle^{um}$ is the heat exchanged with the cold bath, $\langle
Q_{h}\rangle^{um}$ is the heat exchanged with the hot bath and $\langle
W_{1(2)}\rangle^{um}$ is the work done(extracted) on(from) the system during
the expansion (compression) stroke. From the conservation of energy i.e.
$\langle Q_{c}\rangle^{um}+\langle Q_{h}\rangle^{um}+\langle
W_{1}\rangle^{um}+\langle W_{2}\rangle^{um}=0$, the total work is $\langle
W\rangle^{um}=-(\langle W_{1}\rangle^{um}+\langle W_{2}\rangle^{um})=\langle
Q_{c}\rangle^{um}+\langle Q_{h}\rangle^{um}$. (b) In this work instead of
taking the hot bath to be a completely thermalizing channel, we will change it
by a general quantum channel $\mathcal{E}$ described by
$\sum_{i}K_{i}(.)K_{i}^{\dagger}$. Below when studying and analyzing the first
four cumulants, we limit ourselves to the case when this channel is unital,
i.e. $\mathcal{E}(\mathbb{1})=\mathbb{1}$. $\langle Q_{T}\rangle^{um}$ is the
heat exchanged with the bath at inverse temperature $\beta$ and $\langle
Q_{M}\rangle^{um}$ is the heat exchanged with the quantum channel
$\mathcal{E}$. In the same manner as before the total work is $\langle
W\rangle^{um}=-(\langle W_{1}\rangle^{um}+\langle W_{2}\rangle^{um})=\langle
Q_{M}\rangle^{um}+\langle Q_{T}\rangle^{um}$. We see below, that monitoring
the system between strokes would affect the average quantities $\langle
W\rangle^{um}$ and $\langle Q_{T}\rangle^{um}$ for this we used the subscript
$um$(i.e. unmonitored) to distinguish between the average quantities computed
when there is monitoring. Actually for (a) there is no difference in the
averages when monitoring or not the system since the baths are considered to
be completely thermalizing. This is not the case for (b).

For a quantum system e.g. a two-level system, which we consider in our work,
the steps of the conventional QOC with two thermal baths are given as follows
(see Fig. 1(a)): First, the system starts in thermal equilibrium with a cold
bath at inverse temperature $\beta_{c}$. At the point $\mathbf{A}$, the state
of the system and its Hamiltonian are given as follows:
$\rho_{c}=e^{-\beta_{c}H_{1}}/Z_{c}$ and
$H_{1}=\nu_{1}(|+\rangle_{11}\langle+|-|-\rangle_{11}\langle-|)$.
$Z_{c}(=\mathrm{Tr}(e^{-\beta_{c}H_{1}}))$ is the partition function. Second,
from $\mathbf{A}$ into $\mathbf{B}$, we isolate the system from the cold bath,
and then an adiabatic expansion is applied by changing the gap from $\nu_{1}$
to $\nu_{2}$($>\nu_{1}$). This external driving is described by a unitary
given by $U$. At the end of this process, point $\mathbf{B}$, the state and
the Hamiltonian of the system are given as $\rho_{2}=U\rho_{c}U^{\dagger}$ and
$H_{2}=\nu_{2}(|+\rangle_{22}\langle+|-|-\rangle_{22}\langle-|)$. In this
unitary stroke an amount of work $\langle
W_{1}\rangle^{um}=E_{2}-E_{1}=\mathrm{Tr}\left[\rho_{2}H_{2}\right]-\mathrm{Tr}\left[\rho_{c}H_{1}\right]$
is done on the system. Third, from $\mathbf{B}$ into $\mathbf{C}$, we
thermalize again our system with a hot bath at inverse temperature
$\beta_{h}$. At the end of this thermalization (point $\mathbf{C}$) the state
of the system is $\rho_{h}=e^{-\beta_{h}H_{2}}/Z_{h}$, but the Hamiltonian
stays fixed. In this thermalization stroke an amount of heat $\langle
Q_{h}\rangle^{um}=E_{3}-E_{2}=\mathrm{Tr}\left[\rho_{h}H_{2}\right]-\mathrm{Tr}\left[\rho_{2}H_{2}\right]$
is exchange with the bath. Fourth, from point $\mathbf{C}$ into $\mathbf{D}$,
an adiabatic compression is applied by getting back the gap of the system into
its initial value, i.e. $\nu_{1}$. This process is described by another
unitary given by $V$. At the end of this process the state and the Hamiltonian
of the system are given as follows: $\rho_{3}=V\rho_{h}V^{\dagger}$ and
$H_{1}$. Usually, people assume that $V^{\dagger}=U$(or $V=U$)-this equality
is dependent on the expression of the Hamiltonian-which corresponds to the
cycle being time-reversal symmetric, an exception [47, 46, 48]. In this case,
there is no difference between running the cycle forward or backward. In this
second unitary process an amount of work $\langle
W_{2}\rangle^{um}=E_{4}-E_{3}=\mathrm{Tr}\left[\rho_{3}H_{1}\right]-\mathrm{Tr}\left[\rho_{h}H_{2}\right]$
is extracted from the system. And finally, from point $\mathbf{D}$ into
$\mathbf{A}$, we thermalize our system again with the cold bath, to close the
cycle. In this latter thermalization step the system will exchange an amount
of heat $\langle
Q_{c}\rangle^{um}=E_{1}-E_{4}=\mathrm{Tr}\left[\rho_{c}H_{1}\right]-\mathrm{Tr}\left[\rho_{3}H_{1}\right]$
with the cold bath. $E_{1}=\mathrm{Tr}\left[\rho_{c}H_{1}\right]$,
$E_{2}=\mathrm{Tr}\left[\rho_{2}H_{2}\right]$,
$E_{3}=\mathrm{Tr}\left[\rho_{h}H_{2}\right]$ and
$E_{4}=\mathrm{Tr}\left[\rho_{3}H_{1}\right]$ are the average energies at the
points $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and $\mathbf{D}$,
respectively. The total work is given by, $\langle W\rangle^{um}=-(\langle
W_{1}\rangle^{um}+\langle W_{2}\rangle^{um})$. One of the assumptions that we
can relax is that instead of two heat baths as in figure 1 (a), one can
replace one of them with a measurement channel as has been done e.g. if Refs.
[42, 41, 20, 39, 33]. Furthermore, note that the common thing between these
works is that the quantum channels were unital.

In this work, we go a step further and relax two assumptions. The first is
that we do not assume that $V^{\dagger}=U$ nor do we specify the protocol used
to change the gap $\nu_{1}\leftrightarrow\nu_{2}$, and second instead of
considering a specific unital channel replacing the heat hot bath, we consider
an arbitrary unital qubit channel. When $V^{\dagger}(\neq)=U$ the cycle is
(a)symmetric. Here we take the convention that if the heat is entering the
system it is positive and if it flowing out it is negative. Actually what it
means here for the QOC to be asymmetric is that the degree of non-adiabaticity
at the adiabatic expansion/compression stroke may not be equal. For example,
adiabatic compression can be applied in a quasistatic manner but the expansion
in a non-adiabatic or vice-versa and so on. This shows that an asymmetric-
driven Otto cycle is more general than a symmetric case.

Thus differently from figure 1(a), our quantum unital Otto cycle steps are
given as follows (see figure 1(b)): First the system starts in thermal
equilibrium with a heat bath at inverse temperature $\beta$. At the point
$\mathbf{A}$, the state and the Hamiltonian of the system are, respectively,
$\rho_{1}=e^{-\beta H_{1}}/Z$ and $H_{1}$. After this, from the point
$\mathbf{A}$ into $\mathbf{B}$, a unitary transformation $U$ is applied on the
system, and the state becomes $\rho_{2}=U\rho_{1}U^{\dagger}$ and the
Hamiltonian is $H_{2}$. This is an adiabatic expansion. Then, contrary to
figure 1(a), from point $\mathbf{B}$ into $\mathbf{C}$, an arbitray quantum
unital channel $\mathcal{E}$ defined as
$\mathcal{E}(.)=\sum_{j}K_{j}(.)K_{j}^{\dagger}$, is applied to the system,
and the state of the system becomes
$\rho_{3}=\sum_{j}K_{j}\rho_{2}K_{j}^{\dagger}$. The Hamiltonian of the system
is still $H_{2}$. Then from point $\mathbf{C}$ into $\mathbf{D}$ we apply
another unitary $V$ to the state of the system. In this stroke, the gap of the
system changed back to $\nu_{1}$, and the state of the system becomes
$\rho_{4}=V\rho_{3}V^{\dagger}$ and the Hamiltonian $H_{1}$, but in this case,
we do not assume that $V=U^{\dagger}$. Finally from point $\mathbf{D}$ into
$\mathbf{A}$ we thermalize our system again with the bath at inverse
temperature $\beta$, thus closing the cycle. In analogy to figure 1 (a),
$E_{1}=\mathrm{Tr}\left[\rho_{c}H_{1}\right]$,
$E_{2}=\mathrm{Tr}\left[\rho_{2}H_{2}\right]$,
$E_{3}=\mathrm{Tr}\left[\rho_{3}H_{2}\right]$ and
$E_{4}=\mathrm{Tr}\left[\rho_{4}H_{1}\right]$ are the average energies at the
points $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and $\mathbf{D}$,
respectively, and the heat averages are: $\langle
Q_{M}\rangle^{um}=\mathrm{Tr}\left[\rho_{3}H_{2}\right]-\mathrm{Tr}\left[\rho_{2}H_{2}\right]$
and $\langle
Q_{T}\rangle^{um}=\mathrm{Tr}\left[\rho_{1}H_{1}\right]-\mathrm{Tr}\left[\rho_{4}H_{1}\right]$.

At this point, an important remark is needed. Note that, since between the
steps that we just described there is no measurement, this cycle is called an
unmonitored cycle. This is the reason behind the notation $um$. In this work,
we will consider the monitored version of the cycle described in figure 1(b),
since we are interested in analyzing higher cumulants of work and heat as
well.

###  II.2 The CF of the Otto cycle based on arbitrary qubit channels

Before we give the exact analytical expression of the characteristic function
of the forward for the Otto cycle based on unital qubit channels, let’s
consider first computing it for arbitrary qubit channels. Let’s first clarify
an important issue. When $V^{\dagger}\neq U$, in this case, we have to take
into consideration both the forward as well as backward cycle and treat them
on an equal footing. The steps of the forward cycle are
$\mathbf{A}\rightarrow\mathbf{B}\rightarrow\mathbf{C}\rightarrow\mathbf{D}$
and those of the backward cycle are
$\mathbf{D}\rightarrow\mathbf{C}\rightarrow\mathbf{B}\rightarrow\mathbf{A}$.
Otherwise, one would obtain inconsistent results on the bounds of the RF of
work and heat [46]. Below we show that taking into account the backward cycle
is not only necessary to obtain meaningful bounds on fluctuations [46], but as
well one can restore the fact that Otto efficiency is always the maximum
attainable efficiency.

Applying a projective measurements [9] on the Hamiltonian along the cycle
($\mathbf{A}\rightarrow\mathbf{B}\rightarrow\mathbf{C}\rightarrow\mathbf{D}$),
the stochastic quantities $W_{1}$, $Q_{M}$ and $W_{2}$ are given as follows,

$\begin{split}&W_{1}=\nu_{m}-\nu_{n},\\\ &Q_{M}=\nu_{k}-\nu_{m},\\\
&W_{2}=\nu_{l}-\nu_{k}.\end{split}$ (1)

Here, $\nu_{n}$, $\nu_{m}$, $\nu_{k}$ and $\nu_{l}$ are the measured energy at
the four points $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and $\mathbf{D}$ in
figure 1(b), respectively, and $n$, $m$, $k$ and $l$ are their corresponding
quantum nuumbers.

By adopting the two-point measurement scheme [9, 5], the joint probability
distribution (PD) of the stochastic work $W=-(W_{1}+W_{2})$ and the stochastic
heat $Q_{M}$ exchanged with an arbitrary quantum qubit channel with Kraus
operators $K_{j}$, of the forward cycle, is given as follows,

$\begin{split}P(W,Q_{M})_{F}&=\sum_{n,m,k,l}\frac{e^{-\beta\nu_{n}}}{Z}|_{2}\langle
m|U|n\rangle_{1}|^{2}\sum_{j}|_{2}\langle
k|K_{j}|m\rangle_{2}|^{2}|_{1}\langle
l|V|k\rangle_{2}|^{2}\delta(W+(\nu_{m}-\nu_{n}+\nu_{l}-\nu_{k}))\delta(Q_{M}-(\nu_{k}-\nu_{m})).\end{split}$
(2)

To get the first, second (or in general higher cumulants) of work and heat we
use the CF which follows from the PD, written as

$\chi(\gamma_{W},\gamma_{M})_{F}=\int
P(W,Q_{M})_{F}e^{i\gamma_{W}W}e^{i\gamma_{M}Q_{M}}dWdQ_{M}.$ (3)

This is when $W$ and $Q_{M}$ are continuous variables. Using the expression
(2) then the CF becomes,

$\chi(\gamma_{W},\gamma_{M})_{F}=\sum_{n,m,k,l}\frac{e^{-\beta\nu_{n}}}{Z}|_{2}\langle
m|U|n\rangle_{1}|^{2}\sum_{j}|_{2}\langle
k|K_{j}|m\rangle_{2}|^{2}|_{1}\langle
l|V|k\rangle_{2}|^{2}e^{-i\gamma_{W}(\nu_{m}-\nu_{n}+\nu_{l}-\nu_{k})}e^{i\gamma_{M}(\nu_{k}-\nu_{m})}.$
(4)

Here, $\gamma_{M}$ and $\gamma_{W}$ are the Fourier conjugate of $Q_{M}$ and
W, respectively. The cumulants of heats and work of the forward cycle follow
from $\chi(\gamma_{W},\gamma_{M})_{F}$, by the equation

$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{n}Q_{M}^{m}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}=\frac{\partial^{n}\partial^{m}\rm
ln(\chi(\gamma_{W},\gamma_{M})_{F})}{\partial(i\gamma_{W})^{n}\partial(i\gamma_{M})^{m}}\bigg{\rvert}_{\gamma_{W},\gamma_{M}=0}.$
(5)

If we eliminate the logarithmic function then this expression would give us
the central moments, which are denoted for an arbitrary stochastic variable
$\phi$ as
$\langle\phi^{n}\rangle=\left(\partial^{n}\rm\chi(\gamma_{\phi})_{F}/\partial(i\gamma_{\phi})^{n}\right)\big{\rvert}_{\gamma_{\phi}=0}$.
Actually for the first three derivatives, i.e. $n=1,2$ and $3$, the central
moments and cumulants are the same. However, from the fourth derivative, there
is a difference between central moment and cumulant.
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}$,
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}$,
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi^{3}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}$ and
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi^{4}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}$ are,
repectively, the first(average), the second(variance), the third and the
fourth cumulants of $\phi$, in the forward cycle, where $\phi=W,Q_{M}$. The
normalized quantities $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}$,
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi^{3}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{3/2}$ and
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi^{4}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}$ are,
respectively, RFs, skewness and kurtosis of $\phi$.
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi^{4}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}\phi^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}$ is kurtosis
shifted by -3, which is called excess kurtosis. Here, we just call it
kurtosis. Usually, people use the notation $\langle.\rangle$ to mean the
average. In what follows we use $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}.\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ for $W$ and
$Q_{M}$ to mean the average but for the efficiency we use the ordinary
notation, i.e. $\langle.\rangle$. We hope this notation is not confusing. Till
this point everything is clear. However, one should note that there will be a
very important problem when taking into account the backward. This follows,
from the fact that for general maps the adjoint map cannot be taken to be the
reversal, since it may not give a positive probability, i.e. physical states.
For this reason, below we limit ourselves to unital channels.

Based on Table (1) in appendix A and by defining
$\delta=|_{2}\langle+|U|-\rangle_{1}|^{2}$ to be the probability of the system
transitionning from the state $|-\rangle_{1}$ into the state $|+\rangle_{2}$
in the adiabatic expansion,
$\theta=\sum_{j}|_{2}\langle-|K_{j}|+\rangle_{2}|^{2}$ the probability of the
system in the state $|+\rangle_{2}$ being found in the state $|-\rangle_{2}$
after the unital map has been applied on it and finally
$\zeta=|_{1}\langle+|V|-\rangle_{2}|^{2}$ the transition probability between
the states $|-\rangle_{2}$ and $|+\rangle_{1}$ in the adiabatic compression,
then plugging all this in the equation (4) (see appendix A), one can obtain,

$\begin{split}\chi(\gamma_{W},\gamma_{M})_{F}=&(1-\delta)(1-\zeta)\left(\frac{e^{\beta\nu_{1}}}{Z}(h-\theta)+\frac{e^{-\beta\nu_{1}}}{Z}(1-\theta)\right)+(1-\delta)\zeta\left(\frac{e^{\beta\nu_{1}}}{Z}e^{-2i\gamma_{W}\nu_{1}}(h-\theta)+\frac{e^{-\beta\nu_{1}}}{Z}e^{2i\gamma_{W}\nu_{1}}(1-\theta)\right)\\\
&+(1-\delta)\zeta\left(\frac{e^{\beta\nu_{1}}}{Z}(1-h+\theta)e^{2i\gamma_{W}\nu_{2}}e^{2i\gamma_{M}\nu_{2}}+\frac{e^{-\beta\nu_{1}}}{Z}e^{-2i\gamma_{W}\nu_{2}}e^{-2i\gamma_{M}\nu_{2}}\theta\right)\\\
&+(1-\delta)(1-\zeta)\left(\frac{e^{\beta\nu_{1}}}{Z}(1-h+\theta)e^{2i\gamma_{W}(\nu_{2}-\nu_{1})}e^{2i\gamma_{M}\nu_{2}}+\frac{e^{-\beta\nu_{1}}}{Z}e^{-2i\gamma_{W}(\nu_{2}-\nu_{1})}e^{-2i\gamma_{M}\nu_{2}}\theta\right)\\\
&+\delta(1-\zeta)\left(\frac{e^{\beta\nu_{1}}}{Z}e^{-2i\gamma_{W}\nu_{2}}e^{-2i\gamma_{M}\nu_{2}}\theta+\frac{e^{-\beta\nu_{1}}}{Z}e^{2i\gamma_{W}\nu_{2}}e^{2i\gamma_{M}\nu_{2}}(1-h+\theta)\right)\\\
&+\delta\zeta\left(\frac{e^{\beta\nu_{1}}}{Z}e^{-2i\gamma_{W}(\nu_{1}+\nu_{2})}e^{-2i\gamma_{M}\nu_{2}}\theta+\frac{e^{-\beta\nu_{1}}}{Z}e^{2i\gamma_{W}(\nu_{1}+\nu_{2})}e^{2i\gamma_{M}\nu_{2}}(1-h+\theta)\right)\\\
&+\delta\zeta\left(\frac{e^{\beta\nu_{1}}}{Z}(1-\theta)+\frac{e^{-\beta\nu_{1}}}{Z}(h-\theta)\right)+\delta(1-\zeta)\left(\frac{e^{\beta\nu_{1}}}{Z}e^{-2i\gamma_{W}\nu_{1}}(1-\theta)+\frac{e^{-\beta\nu_{1}}}{Z}e^{2i\gamma_{W}\nu_{1}}(h-\theta)\right),\end{split}$
(6)

with, $h=\sum_{j}{}_{2}\langle-|K_{j}K_{j}^{\dagger}|-\rangle_{2}$. To obtain
the CF of the backward cycle $\chi(\gamma_{M},\gamma_{W})_{B}$ it is not
straightforward, since the $K_{j}^{\dagger}$ cannot describe a valid channel,
i.e. a physical channel. Thus computing the CF of the backward, is difficult,
for general qubit maps. Furthermore, below all thermodynamic quantities will
be expressed in terms of $\delta$, $\theta$ and $\zeta$. And below, whenever
we omit the subscript $F$ referring to the forward, it means that the forward
and the backward are the same.

###  II.3 $\chi(\gamma_{M},\gamma_{W})_{F}$ of the Otto cycle based on
arbitrary unital qubit channels

When the map $\mathcal{E}$ is unital, completely positive and trace-
preserving, so is its adjoint $\mathcal{E}^{\dagger}$. Therefore, the adjoint
$\mathcal{E}^{\dagger}$ can be regarded as another time evolution. Thus the
backward of,

$\mathcal{E}(.)=\sum_{j}K_{j}(.)K_{j}^{\dagger},\hskip 28.45274pt{\rm is-
defined-as}\hskip
28.45274pt\mathcal{E^{\dagger}}(.)=\sum_{j}K_{j}^{\dagger}(.)K_{j}.$ (7)

Furthermore, for a two-level system considered here, one should note that
every unital map can be decomposed in terms of Pauli operators [53], for this
reason, unital maps are also called Pauli channels, see [53]. Thus,

$\mathcal{E}(.)=\sum_{i=0}^{3}p_{i}\sigma_{i}(.)\sigma_{i}^{\dagger}.$ (8)

with, $\sum_{i=0}^{3}p_{i}=1$. The Pauli operators in the basis of the
Hamiltonian $H_{2}$ are given as follows:
$\sigma_{0}=|+\rangle_{22}\langle+|+|-\rangle_{22}\langle-|$,
$\sigma_{1}=|+\rangle_{22}\langle-|+|-\rangle_{22}\langle+|$,
$\sigma_{2}=i|-\rangle_{22}\langle+|-i|+\rangle_{22}\langle-|$ and
$\sigma_{3}=|+\rangle_{22}\langle+|-|-\rangle_{22}\langle-|$. In this case, we
have, $0\leq\theta=p_{1}+p_{2}\leq 1$, which shows that the highest possible
value of $\theta$ is 1. Below, we comment on this in detail. The backward does
not mean that we reverse time, since in the lab the reverse process of the
forward as well as is run in the forward. The reader should note that the
adjoint of equation (8) is itself, i.e. $\mathcal{E}=\mathcal{E}^{\dagger}$.

When the map is unital one can easily show that $h=1$ (for unital maps we have
$\sum_{j}K_{j}\mathbb{1}K_{j}^{\dagger}=\mathbb{1}$ and thus $h=1$), in this
case, $\chi(\gamma_{M},\gamma_{W})_{F}$ get simplified into a more compact
form, and we have,

$\begin{split}\chi(\gamma_{W},\gamma_{M})_{F}&=(1-\theta)\left(1+\left(\frac{\rm
2cos((2\gamma_{W}+i\beta)\nu_{1})}{Z}-1\right)(\delta+\zeta-2\delta\zeta)\right)\\\
&+\theta\left(\left(1-\delta\right)\left(\zeta\frac{\rm
2cos(2(\gamma_{W}+\gamma_{M})\nu_{2}-i\beta\nu_{1})}{Z}+(1-\zeta)\frac{\rm
2cos(2(\gamma_{W}(\nu_{2}-\nu_{1})+\gamma_{M}\nu_{2})-i\beta\nu_{1})}{Z}\right)\right.\\\
&+\left.\delta\left((1-\zeta)\frac{2\cos(2(\gamma_{W}+\gamma_{M})\nu_{2}+i\beta\nu_{1})}{Z}+\zeta\frac{\rm
2cos(2((\nu_{1}+\nu_{2})\gamma_{W}+\gamma_{M}\nu_{2})+i\beta\nu_{1})}{Z}\right)\right).\end{split}$
(9)

The backward CF follows from this equation, by the correspondence
$\delta\leftrightarrow\zeta$. Note that $\theta_{F}=\theta_{B}=\theta$ since
the channel is unital. The exact analytical equation (9) is the first main
result of the paper. One can check easily that for $\gamma_{M}=\gamma_{W}=0$,
we have $\chi(\gamma_{W},\gamma_{M})_{F}=1$, which is nothing but probability
conservation.

###  II.4 Exact analytical expressions of the first and second cumulants of
work and heat

From equations (5) and (9), one can derive the next cumulants for the forward
cycle

$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}=2(1-2\delta)\theta\nu_{2}\tanh(\beta\nu_{1}),$
(10) $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}=-4\theta\nu_{2}^{2}(-1+(1-2\delta)^{2}\theta\tanh^{2}{(\beta\nu_{1})}),$
(11) $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}=-2(\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta))\nu_{1}\tanh(\beta\nu_{1}),$
(12) $\begin{split}\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}&=2((1-2\delta)\theta\nu_{2}-(\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta))\nu_{1})\tanh(\beta\nu_{1}),\end{split}$
(13) $\begin{split}\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}&=4(\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))\nu_{1}^{2}+8\theta(-1+\delta+\zeta)\nu_{1}\nu_{2}+4\theta\nu_{2}^{2}\\\
&-4((\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))\nu_{1}+(-1+2\delta)\theta\nu_{2})^{2}\tanh^{2}{(\beta\nu_{1})}.\end{split}$
(14)

Note that when $\theta=0$ all the cumulants of $Q_{M}$ become null since in
this case no energy will be exchanged between the system and the channel. The
third and fourth cumulants are not reported here since they are cumbersome and
not illuminating at all. Further, the average heat
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ follows from the
law of energy conservation.

Before we proceed, some important remarks are needed to be mentioned; first,
one can see clearly that when $\beta=0$ (high-temperature limit) the initial
state will be maximally mixed, and one can show easily that
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}$,
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}$ and
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ are null. The
same thing for backward quantities. This follows from the fact that the state
along the cycle will always stay a maximally mixed state, thus no change in
the average energy will be observed. This is true either for the monitored or
unmonitored Otto cycle. This is because the maximally mixed state is a fixed
point of the cycle. This shows that we have to focus on $\beta\neq 0$.
However, this does not mean necessarily that higher cumulants will be zero,
such as the variance and fourth cumulant. One can show that all odd cumulants
will be null since this is due to the fact the distribution of heat and work
will be symmetric around the 0 value, which is nothing but the average value.
For example when $\beta\rightarrow+\infty$ or $-\infty$, the initial state
will be pure state (for $\beta\rightarrow+\infty(-\infty)$, the pure state
will be the ground state, i.e. $|-\rangle_{1}$(exicted state
$|+\rangle_{1}$)).

In appendix B, we prove that for the considered measurement unital channel in
[41], $\theta$ cannot exceed 1/2. However, in our case, we have
$0\leq\theta\leq 1$ as we saw before. As we show below, this can help in
improving the work extracted, decreasing its relative fluctuations and
enhancing the efficiency. It is already seen from the expression of
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ that when
$\delta<1/2$ and for $\beta>0$, the more we increase $\theta$ the more heat
would be provided by the unital channel. The maximum value is reached when
$\theta=1$.

At this point, we want to emphasize that for $p_{1}+p_{2}=1$ and when the
state is classical i.e. there are no off-diagonal elements, e.g. in the
adiabatic regime, the unital channel Eq. (8), will supply the system with
energy in the form of work and not heat, since in this case, the channel will
not change the entropy of the system. Therefore, even though the channel is a
mixture of $\sigma_{1}$ and $\sigma_{2}$ Pauli operators, the diagonal
elements will only be swapped, thus the entropy of an arbitrary state $\rho$
and $\mathcal{E}(\rho)$ is equal. In the case when $p_{1}=p_{2}=0$,
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}=0$ as expected,
since the Kraus operators (i.e. the identity and the $\sigma_{3}$ Pauli
operator,) commute with the Hamiltonian. When the system state is non-
diagonal, i.e. quantum state for example in the non-adiabatic regime, and when
$p_{1}+p_{2}=1$, the bath can provide the system with energy in the form of
heat, as long as $p_{1}$ and $p_{2}$ in Eq.(8) are both non vanished. Thus
when the unital channel is a mixture of the $\sigma_{1}$ and $\sigma_{2}$
Pauli operators acting on a quantum state, then the channel will provide the
system with heat, since in this case entropy will be changed. However, when
either $p_{1}$ or $p_{2}$ is null in this case the unital channel will be
unitary and thus will not change the entropy of the system. In this latter
case, the energy exchanged would be work and not heat. This analysis shows
that in the adiabatic regime and when $p_{1}+p_{2}=1$ the unital channel will
exchange energy in the form of work (ordered energy) with the system. However,
beyond that regime, it can be either heat or work. See appendix C for more
details.

Equation (14) which characterizes the variance of $W$, is difficult to be
analyzed, e.g. if one wants to know when it becomes vanishing, in contrast to
the variance of $Q_{M}$. This is because it depends on the parameters in a
complicated way. When $\delta=\zeta=0$, we have

$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}=4\theta(\nu_{2}-\nu_{1})^{2}(1-\theta\tanh^{2}(\beta\nu_{1})).$
(15)

In this case, we see that the fluctuations of work vanished in three cases:
when $\nu_{1}=\nu_{2}$ or $\theta=\tanh^{2}(\beta\nu_{1})=1$ or when
$\theta=0$. While the variance of $Q_{M}$ when $\delta=\zeta=0$ vanishes in
two cases: $\theta=0$ or $\theta=\tanh^{2}(\beta\nu_{1})=1$. In this case, the
distributions of $W$ and $Q_{M}$ become degenerate and localized around one
fixed value with all other outcomes having zero probability. Therefore, the
variance would be zero. For example when $\theta=0$ then the only possible
value of $W$ and $Q_{M}$ is 0. The PD becomes
$P(W,Q_{M})_{F}=\delta(W)\delta(Q_{M})$ and the CF is $\chi(W,Q_{M})_{F}=1$.
In this case, the third and fourth cumulants as well become null. Even higher
cumulants than the first four cumulants become zero since the characteristic
function is constant.

Let’s now give a quick comparison between the case when we have a completely
thermalizing channel and an arbitrary unital channel. When the unital qubit
map is completely unital it describes the thermalization of a system with a
heat bath at infinite temperature i.e. $T=\infty$(or $\beta=0$). In this case,
the two levels become populated by $1/2$. A heat bath at positive inverse
temperature can make the population of the higher level with at most $1/2$.
However, when considering arbitrary unital channel the higher level can become
more populated than the ground state, especially when the Pauli operators
$\sigma_{1}$ and $\sigma_{2}$ become more dominated than the $\sigma_{0}$ and
$\sigma_{3}$ operators. Let’s show this statement with an example. Suppose the
initial state before the unital channel is given by,

$\rho_{i}=\begin{pmatrix}1-a&0\\\ 0&a\\\ \end{pmatrix}.$ (16)

With $0\leq 1-a\leq 1/2$. After applying the unital channel Eq. (8) on Eq.
(16), and with simple steps of calculations one obtains

$\mathcal{E}(\rho_{i})=\begin{pmatrix}1-(a+(1-2a)(p_{1}+p_{2}))&0\\\
0&a+(1-2a)(p_{1}+p_{2})\\\ \end{pmatrix}.$ (17)

In this case one can show that $1-a\leq 1-(a+(1-2a)(p_{1}+p_{2}))\leq a$. The
upper bound is reached when $p_{1}+p_{2}=1$, but as we said before in this
case the energy difference is work. Therefore we see that the population range
of e.g. the excited level populated with $1-a$ before applying the unital
channel, using an arbitrary unital map can vary from $1-a$ to $a$, thus can
exceed $1/2$. This shows the strength of considering arbitrary unital maps.

###  II.5 Positive work condition and efficiency bound for symmetric and
asymmetric driven Otto cycle: heat engine regime

Here we focus on the analysis of the first cumulant of $W$ and $Q_{M}$. Now,
consider $\nu_{2}=\nu_{1}$ and $\zeta=\delta$, in such a case we have

$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}=2(1-2\delta)\theta\nu_{1}\tanh(\beta\nu_{1}),$
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}=-2(\theta+2\delta(1-2\theta)(1-\delta))\nu_{1}\tanh(\beta\nu_{1}),$

and,

$\begin{split}\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}=4\delta(\delta+\theta-2\delta\theta-1)\nu_{1}\tanh(\beta\nu_{1})\leq
0.\end{split}$

The latter inequality follows from the fact that
$0\leq\delta+\theta-2\delta\theta(=\delta(1-\theta)+\theta(1-\delta))\leq 1$.
For the asymmetric Otto cycle case i.e. $\delta\neq\zeta$ but we still have
$\nu_{2}=\nu_{1}$, we obtain,

$\begin{split}\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}&=-2(\delta+\zeta-2\delta\zeta-2\theta\zeta(1-2\delta))\nu_{1}\tanh(\beta\nu_{1}),\end{split}$

and,

$\begin{split}\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}&=-2(\delta+\zeta-2\delta\zeta-2\theta\delta(1-2\zeta))\nu_{1}\tanh(\beta\nu_{1}).\end{split}$

Numerical analysis showed that these two expressions separately can be either
positive or negative, but taking both of the two quantities into consideration
one obtains,

$\begin{split}\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}&=-4(\delta+\zeta-2\delta\zeta-\theta\zeta+\delta\theta(-1+4\zeta))\nu_{1}\tanh(\beta\nu_{1}).\end{split}$

Simple manipulations further give,

$\begin{split}\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}=-4((\delta+\zeta-2\delta\zeta)(1-\theta)+2\delta\theta\zeta)\nu_{1}\tanh(\beta\nu_{1})(\leq
0).\end{split}$ (18)

The latter inequality follows from the fact that
$(\delta+\zeta-2\delta\zeta)(1-\theta)$ and $2\delta\theta\zeta$ are always
positive or null. Therefore for $\nu_{2}=\nu_{1}$, work cannot be extracted
and the system can only work as either an accelerator ($A$) for $\delta<1/2$
or a heater ($H$) for $\delta>1/2$. Thus the condition $\nu_{2}\neq\nu_{1}$ is
necessary for the system to work as a heat engine, independently of the cycle
being symmetric or asymmetric. This is the main second result of the paper. We
trace this to the the negative effect of the projective measurement on the
average energy
$E_{4}=\mathrm{Tr}\left[V\sum_{j}K_{j}U\rho_{1}U^{\dagger}K_{j}^{\dagger}V^{\dagger}H_{2}\right]$
as we explain further below. This is why the engine in [41] could extract work
even when $\nu_{2}=\nu_{1}$, i.e. even without changing the frequency of the
spin system. Therefore, our results and theirs are not in contradiction with
each other as one may think from the first sight. Note that for the asymmetric
cycle we have taken both the forward and backward work averages for a
meaningful study, since as was shown in [46] for one to obtain meaningful
bounds on fluctuations one needs to take both the forward as well the backward
into account.

Figure 2: (a) Plot of efficiency $\langle\eta\rangle$ as a function of
$\delta$ with $\zeta=\delta$, $\beta=0.5$, $\nu_{1}=1$ and $\nu_{2}=2$. The
horizontal dashed red line is the Otto efficiency which is equal to 0.5. We
plot $\langle\eta\rangle$ only for $\delta\in[0,0.5]$, since beyond this
interval $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ becomes
negative. One should note that differently from Ref. [41] we see that the fact
that $\theta$ can be higher than 1/2 enhances the efficiency. (b) Plot of
extracted work $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ as a function of
$\delta$ with $\zeta=\delta$ with the same value of $\beta$, $\nu_{1}$ and
$\nu_{2}$ as in figure (a). For the considered parameters we see that
considering an arbitrary unital channel could increase the value of $\delta$
above which work becomes negative. Even further, we see that the more we
increase $\theta$ the more the extracted work gets enhanced. However, note
that increasing $\delta$ diminishes the work extracted. This shows that non-
adiabatic transitions are detrimental, but we show these transitions can be
harnessed positively under an important circumstance.

For the system to work as a heat engine ($E$) when $\delta=\zeta$, the
conditions $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0$,
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0$ and
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0$ has to be
staified. For $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ to be positive
we have to ensure $\delta<1/2$. $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ as we show
below, is always negative or equal to zero. Thus for
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0$, we have to
have, $|\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}|<\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$. However, in
terms of the parameters, for $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0$ we have the
next condition on $\nu_{2}$,

$\nu_{2}>\left(\frac{\theta+2\delta(1-2\theta)(1-\delta)}{\theta(1-2\delta)}\right)\nu_{1}(\geq\nu_{1}\
\mathrm{for}\ \delta<1/2).$ (19)

This because
$\theta+2\delta(1-2\theta)(1-\delta)-\theta(1-2\delta)(=2\delta(1-\delta-\theta+2\delta\theta))\geq
0$. For an accelerator the conditions are:
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0$,
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0$ and
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0$. For a heater
the conditions are: $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0$,
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0$ and
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0$. To satisfy
these latter conditions we need to have $\delta>1/2$ with all other parameters
are nonzero.

Now let’s consider the case when $\zeta\neq\delta$. For the forward cycle to
work as a heat engine, the condition $\delta<1/2$ and
$\nu_{2}>\frac{\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta)}{\theta(1-2\delta)}\nu_{1}$
are necessary. For the backward cycle, we have to ensure as well that
$\zeta<1/2$ and
$\nu_{2}>\frac{\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta)}{\theta(1-2\zeta)}\nu_{1}$.
Thus for the forward and backward to work as a heat engine we need to ensure
that $\delta$ and $\zeta<1/2$ and
$\nu_{2}>\mathbf{max}\\{\frac{\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta)}{\theta(1-2\delta)}\nu_{1},\frac{\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta)}{\theta(1-2\zeta)}\nu_{1}\\}$.
Of course we may have other combinations such that the overall cycle is a heat
engine. For example, the forward could be a heat engine and the backward an
accelerator or vice versa and so on. In the latter case, when one imposes the
positivity of work only on the sum i.e.
$\left(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}\right)$ to be
$>0$ we have the next condition on $\nu_{2}$,

$\nu_{2}\geq\left(\frac{\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta)}{\theta(1-\delta-\zeta)}\right)\nu_{1}(>\nu_{1}\
\mathrm{for}\ \delta+\zeta<1).$ (20)

This is when we impose the positivity on only the sum of
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}$ and
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}$ and not on
them individually.

Now let’s look at the efficiency. For $\zeta=\delta$, the efficiency is given
by,

$\langle\eta\rangle=\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}=1-\frac{\nu_{1}}{\nu_{2}}\frac{\theta+2\delta(1-2\theta)(1-\delta)}{(1-2\delta)\theta}.$
(21)

Since
$\theta+2\delta(1-2\theta)(1-\delta)-(1-2\delta)\theta=2\delta(1-(\delta+\theta-2\delta\theta))\geq
0$, it follows that
$-\frac{\nu_{1}}{\nu_{2}}\frac{\theta+2\delta(1-2\theta)(1-\delta)}{(1-2\delta)\theta}\leq-\frac{\nu_{1}}{\nu_{2}}$.
Therefore the efficiency is limited by the one of the Otto, i.e.
$1-\nu_{1}/\nu_{2}$. This latter bound is reached only when the adiabatic
parameter $\delta$ is null. Analogously when $\delta\neq\zeta$ the efficiency
expression [46] is given by,

$\langle\eta\rangle=\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}}=1-\frac{\nu_{1}}{\nu_{2}}\frac{(\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))}{(1-\delta-\zeta)\theta}.$
(22)

Further we have,
$((\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta)))-((1-\delta-\zeta)\theta)=((\delta+\zeta-2\delta\zeta)(1-\theta)+2\delta\theta\zeta)\geq
0$. From the latter inequality one can show that,

$\langle\eta\rangle=\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}}\leq
1-\frac{\nu_{1}}{\nu_{2}}.$ (23)

Therefore for the symmetric and asymmetric Otto cycle, the efficiency is
always less or equal to $1-\nu_{1}/\nu_{2}$. We should mention that for the
asymmetric cycle, if one does not take into consideration the backward, the
efficiency can be greater than that of the Otto, i.e. $1-\nu_{1}/\nu_{2}$.
This means that not only one should treat the backward and the forward cycle
on equal footing for meaningful bounds on RF as was stated in Ref. [46], but
also to restore the Otto efficiency being the upper bound of the efficiency as
our result (Eq.(23)) shows.

In figure (2) we plot the efficiency $\langle\eta\rangle$ (left figure) and
work extracted $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ (right figure)
as a function of the adiabatic parameter $\delta$ for the symmetric Otto cycle
$\zeta=\delta$ and for different values of $\theta$. It is seen that work and
efficiency get enhanced as we increase $\theta$. Further, it is found that
efficiency becomes less perturbed by $\delta$, when we increase $\theta$.

###  II.6 Monitored (our engine) and unmonitored engine (Ref. [41])

Let’s now comment on why our heat engine needs the condition (19) to output
work, in contrast to what has been shown in Ref. [41]. As we said before, in
[41], the authors did not consider monitoring the working medium between the
strokes. However, in our case we consider it. Thus we have that the average
work and heats are not equal. Note that if one considers a completely unital
map or a in general a completely thermalizing map then the two approaches give
the same averages and fluctuations of average energies. Therefore monitoring
or not monitoring the state of the system will not change the story. However,
since the unital map is arbitrary, which may not erase the coherence generated
by the unitary $U$, the monitored and unmonitored heat engines would be
different. Our analysis is in agreement with what has been shown in Ref. [54],
that projective measurement may be detrimental to efficiency, work and its
reliability and power. In Ref. [55], the authors have considered a two-stroke
Otto heat engine, where the hot bath being replaced by a non-selective quantum
measurement. They have shown that the invariant state of the cycle is
dependent on whether the cycle is being monitored or not, and consequently the
work average and its fluctuations are different. More precisely, it was found
that depending on the parameters, the work for the unmonitored cycle can be
either greater or less than that of the monitored cycle. The same thing was
found for work reliability. From our study we conclude that monitoring can
have a negative influence on the performance of the heat engine, and this is
seen above as the engine cannot output work under the condition
$\nu_{1}=\nu_{2}$, either in the symmetric or the asymmetric Otto cycle.

Let’s now show a comparison between the monitored and unmonitored Otto cycle
cases in terms of average energy at the point $\mathbf{D}$ in figure 1(b). One
can easily show that $E_{1}$, $E_{2}$ and $E_{3}$, the average energies at the
points $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ in figure 1(b),
respectively, will not be affected by using projective measurement to assess
the statistics of thermodynamic quantities. Actually the state $\rho_{3}$ of
the unmonitored Otto cycle can be written as follows,

$\begin{split}\rho_{3}&=\sum_{i=0}^{3}p_{i}\sigma_{i}\rho_{2}\sigma_{i}\\\
&=\left(|+\rangle_{22}\langle+|+|-\rangle_{22}\langle-|\right)\left(\sum_{i=0}^{3}p_{i}\sigma_{i}\rho_{2}\sigma_{i}\right)\left(|+\rangle_{22}\langle+|+|-\rangle_{22}\langle-|\right)\\\
&=\rho_{3}^{deph}+\sum_{i=0}^{3}p_{i}\sigma_{i}({}_{2}\langle+|\rho_{2}|-\rangle_{2}|+\rangle{}_{2}{}_{2}\langle-|+{}_{2}\langle-|\rho_{2}|+\rangle{}_{2}|-\rangle_{2}{}_{2}\langle+|)\sigma_{i}.\end{split}$
(24)

Note that we have employed Eq.(8). $\rho_{2}=U\rho_{1}U^{\dagger}$ and
$\rho_{3}^{deph}(={}_{2}\langle+|\rho_{3}|+\rangle_{2}|+\rangle_{22}\langle+|+{}_{2}\langle-|\rho_{3}|-\rangle_{2}|-\rangle_{22}\langle-|)$
is the dephased state $\rho_{3}$. Let’s now forget about the second term.
Similarly to Eq.(24) the state $\rho_{4}$ of the unmonitored Otto cycle is,

$\rho_{4}=\rho_{4}^{deph}+V\sum_{i}\sigma_{i}p_{i}(\langle+_{2}|\rho_{2}|-_{2}\rangle|+\rangle_{22}\langle-|+\langle-_{2}|\rho_{2}|+_{2}\rangle|-\rangle_{22}\langle+|)\sigma_{i}V^{+}.$
(25)

One can show that the second term in Eq.(24) will not contribute to the
average energy $E_{3}$. This can be explained by the fact that the Kraus
operators of an arbitrary unital channel, which are Pauli operators, do not
couple the populations and coherences, thus the coherence created in the first
stroke by $U$ can only affect $E_{4}$ and not $E_{3}$, thus the heat provided
by the unital channel $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ is coherence
independent. What this means is that the difference between the average
energies of the monitored and unmonitored will only be in $E_{4}$.
Furthermore, note that the second term in Eq.(25) will not contribute to the
average energy $E_{4}$ when $\rho_{2}$ is diagonal in the basis of the
Hamiltonian $H_{2}$, since
${}_{2}\langle+|\rho_{2}|-\rangle_{2}={}_{2}\langle-|\rho_{2}|+\rangle_{2}=0$.
In this case, there is no difference between the monitored and unmonitored
heat engines from an energetic point of view. Thus when the arbitrary unital
channel is not completely unital, the unitary $V$ may couple the populations
and coherences such that the latter would contribute to the average work
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ and heat
exchanged with the bath $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$.

In appendix D, for the Landau-Zener model, we give a detailed comparison
between the work extracted and efficiency for the monitored and unmonitored
single qubit.

###  II.7 Negative temperature: Can the system work as a refrigerator in this
case?

Now let’s first prove that $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ is always
negative. We have,
$0\leq\delta+\zeta-2\delta\zeta=\delta(1-\zeta)+\zeta(1-\delta)\leq 1$, since
$0\leq\left((1-\delta)\ \rm and,\ (1-\zeta)\right)\leq 1$. The term,
$(\theta+\zeta-2\theta\zeta+\delta(1-2\zeta)(1-2\theta))$ can be rewritten as
follows:
$(\theta+\zeta-2\theta\zeta+\delta(1-2\zeta)(1-2\theta))=\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta)$.
For $0\leq\theta\leq 1/2$ this term is clearly positive or null, since
$\delta+\zeta-2\delta\zeta$ and $1-2\theta$ are greater than or equal to zero.
To show this for $1/2<\theta\leq 1$ we have,
$(\theta+\zeta-2\theta\zeta+\delta(1-2\zeta)(1-2\theta))=\theta-(2\theta-1)(\delta+\zeta-2\delta\zeta)$.
$(2\theta-1)(\delta+\zeta-2\delta\zeta)$ is $\geq 0$. The biggest value that
$(2\theta-1)(\delta+\zeta-2\delta\zeta)$ can take, independtly of $\theta$, is
when $\delta+\zeta-2\delta\zeta=1$, in this case we have,
$\theta-\max_{0\leq\delta,\zeta\leq
1}(\zeta-2\theta\zeta+\delta(1-2\zeta)(1-2\theta))=\theta-(2\theta-1)=1-\theta$,
which is clearly always $\geq 0$. Thus,
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ is always $\leq
0$ as in Ref. [41]. Therefore, heat will only flow out from the system when it
will be in interaction with the heat bath, thus this bath will play the role
of a heat-cold bath and the unital channel the role of hot heat bath. However,
in contrast to them, our proof is valid for arbitrarily monitored Otto heat
engines based on unital qubit channels, not just the specific measurement
unital channel considered there. This is another main result of our work.

The notion of inverse temperature bath was introduced in Refs. [57, 56].
Above, and in the previous work in Ref. [41] it was shown that
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ is always $\leq
0$ but this is true only under the assumption that the temperature of the heat
reservoir is positive. However, if it is negative then in this case the
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ can be positive,
since the sign of $\tanh(\beta\nu_{1})$ gets flipped. When
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ is negative this
means that the average energy of the system at the point $\mathbf{D}$ is
bigger than the one when the system gets equilibrated with its environment
i.e. point $\mathbf{A}$, see Fig.1(b). However, if the temperature is negative
then in this case $E_{4}$ will be smaller than $E_{1}$ thus heat will flow
from the environment into the system. However, in this case, the system is not
a refrigerator as a heat bath with a negative temperature is hotter than any
heat bath with a positive temperature, thus the bath will act as a hot heat
bath [58]. To summarise the operation modes of the system are:

$\mathrm{for}\ \beta>0:\begin{split}\\\
&\mathrm{E}:\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0,\hskip
28.45274pt\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0,\hskip
28.45274pt\ \rm and\hskip 28.45274pt\
\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0,\\\
&\mathrm{A}:\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0,\hskip
28.45274pt\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0,\hskip
28.45274pt\ \rm and\hskip 28.45274pt\
\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0,\\\
&\mathrm{H}:\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0,\hskip
28.45274pt\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0,\hskip
28.45274pt\ \rm and\hskip 28.45274pt\
\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
<EMAIL_ADDRESS>
and

$\mathrm{for}\ \beta<0:\begin{split}\\\
&\mathrm{A}:\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0,\hskip
28.45274pt\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0,\hskip
28.45274pt\ \rm and\hskip 28.45274pt\
\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0,\\\
&\mathrm{E}:\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0,\hskip
28.45274pt\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0,\hskip
28.45274pt\ \rm and\hskip 28.45274pt\
\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0,\\\
&\mathrm{E^{\prime}}:\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0,\hskip
28.45274pt\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0,\hskip
28.45274pt\ \rm and\hskip 28.45274pt\
\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
<EMAIL_ADDRESS>
For $\beta$ negative, we only have to reverse the sign of work and heats given
in the above equation. Therefore when $\beta\rightarrow-\beta$ a heat engine
becomes an accelerator, an accelerator becomes a heat engine and a heater
becomes a heat engine $E^{\prime}$, but the latter comes with unit efficiency
since the system will absorb heat from both baths and transform it into work,
thus $\langle\eta\rangle=\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\left(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}\right)=1$. This
latter case is possible because of two reasons: the first one is the non-
adiabaticity and the second is the negativity of the temperature of the bath.
Since a heater is not possible in the adiabatic case, i.e. a nonequilibrium
thermal machine. However, note that if the cost of preparing baths at negative
temperatures-nonequilibrium baths-is taken into account, then there will be no
inconsistency with the laws of thermodynamics, see Refs. [59, 60] and the
efficiency would be less than 1. The fact that the efficiency of a two-level
Otto cycle with two heat baths, one of them with a positive temperature and
the other with a negative temperature-being enhanced in the non-adiabatic
regime, which seems counterintuitive, was also noted in Ref. [52].

Thus refrigeration is not allowed even when the temperature is negative.
Furthermore, note that the heater is not possible as well, when $\beta$ is
negative, since a heater demands the next conditions:
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0,\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0\ \rm and\
\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0$, which is not
possible, since when $\beta<0$, $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{T}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ will always be
$\geq 0$. We have demonstrated in equation (18) that a positive work condition
is not possible when $\nu_{2}=\nu_{1}$ and when the inverse temperature is
positive. However, as we see in this section, when the inverse temperature is
negative we may have, $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0$ either for
symmetric or asymmetric Otto cycle.

Until now we only considered the first cumulants of work and heat. Let’s now
take a step further towards higher cumulants such as variance, skewness and
kurtosis which are defined as measures of fluctuations, asymmetry and
tailedness and peakedness of a probability distribution, respectively.

###  II.8 A lower and upper bound on the ratio of the fluctuations of $W$ and
$Q_{M}$

In this section, we prove rigorously that the ratio of the fluctuations of $W$
and $Q_{M}$ is lower bounded by the square of the efficiency and upper bounded
by 1. We show when these bounds are respected either for the symmetric or the
asymmetric Otto cycle. Furthermore, for the symmetric Otto cycle, we prove
that the square of the Otto efficiency gives a tighter lower bound on the
ratio of fluctuations than the one derived from the square of the efficiency
of the engine.

####  II.8.1 Symmetric Otto cycle

The authors of Ref. [46] have shown that for a symmetric and asymmetric Otto
cycle based on two heat baths, the ratio
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ is lower and
upper bounded when the system is working as a heat engine. Since our Otto
cycle is different from theirs we want to know if this is still true here as
well. Let’s consider $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}-\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}$. We now only
consider the case $\zeta=\delta$. After messy but straightforward steps of
calculations which are not important to be included here, we obtain,

$\begin{split}\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}}-\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}}&=\frac{2(1-\delta)\delta\nu_{1}(-(\theta+2\delta(1-\delta)(1-2\theta))\nu_{1}+2\theta(1-2\delta)\nu_{2})\coth^{2}(\beta\nu_{1})}{\theta(1-2\delta)^{2}((\theta+2\delta(1-\delta)(1-2\theta))\nu_{1}-\theta(1-2\delta)\nu_{2})^{2}}\\\
&=\frac{(1-\delta)\delta\nu_{1}(\langle\langle
W\rangle\rangle+2(1-2\delta)\nu_{2}\tanh(\beta\nu_{1}))\coth^{3}(\beta\nu_{1})}{\theta(1-2\delta)^{2}((\theta+2\delta(1-\delta)(1-2\theta))\nu_{1}-\theta(1-2\delta)\nu_{2})^{2}}.\end{split}$
(26)

This equation can be either positive, equal to zero or negative. The
denominator is always $\geq 0$. For equation (26) to be $\geq 0$ we have to
ensure that,

$\nu_{2}\geq\frac{(\theta+2\delta(1-\delta)(1-2\theta))}{2(1-2\delta)\theta}\nu_{1}.$
(27)

When this latter condition is met we have,

$\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}}\geq\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
<EMAIL_ADDRESS>(28)

Furthermore, note that the lower bound
$\frac{(\theta+2\delta(1-\delta)(1-2\theta))}{2(1-2\delta)\theta}\nu_{1}$ on
$\nu_{2}$ is half of the necessary condition needed for the system to work as
a heat engine. Thus this condition does not ensure that the system is working
as a heat engine. This shows that
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}\geq\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}$ can still be
valid even when the system is not working as a heat engine, such as an
accelerator as we found numerically, to be shown later on. However, when the
system is working as a heater, then we have,

$\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}}<\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
<EMAIL_ADDRESS>(29)

This is because for the system to work as a heater we need $\delta>1/2$, and
under this constraint, one can show that
$2(-(\theta+2\delta(1-\delta)(1-2\theta))\nu_{1}+2\theta(1-2\delta)\nu_{2})\left(=\left(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}+2(1-2\delta)\nu_{2}\tanh(\beta\nu_{1})\right)/\tanh(\beta\nu_{1})\right)<0$,
since $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<0$. Therefore, we
conclude that $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}\geq\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}$ is always
respected (violated) when the system as a heat engine (heater), but it can or
cannot be violated when it is an accelerator. The violation depends on the
parameters. Straightforwardly from Eq.(28) it follows that,

$\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}\geq\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
<EMAIL_ADDRESS>(30)

This equation tells us that when equation (27) is satisfied, the ratio of the
fluctuations of $W$ and $Q_{M}$ is always greater or equal to the square of
efficiency. In the heat engine region, the lower bound is reached in the
adiabatic limit. Now let’s see if the ratio of the relative fluctuations of
$W$ and $Q_{M}$ has any upper bound as in Ref. [46]. In the latter reference,
it was shown that for a symmetric and asymmetric QOHE with two completely
thermalizing baths that $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<1$. Let’s now
investigate under which condition
$1-\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ is $\geq 0$. To
do this, this inequality can be modified into checking if
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}-\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}\geq 0$. We have

$\begin{split}\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}-\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}&=4\nu_{1}(2\delta^{2}(-1+2\theta)\nu_{1}+\theta(\nu_{1}-2\nu_{2})+2\delta(\nu_{1}-2\theta\nu_{1}+2\theta\nu_{2}))(-1+(\theta+2(-1+\delta)\delta(-1+2\theta))\tanh^{2}(\beta\nu_{1}))\\\
&=4\nu_{1}(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}+(\theta+2(-1+\delta)\delta(-1+2\theta))\nu_{1}\tanh(\beta\nu_{1}))(1-(\theta+2(-1+\delta)\delta(-1+2\theta))\tanh^{2}(\beta\nu_{1}))\coth(\beta\nu_{1}).\end{split}$
(31)

The term
$(-1+(\theta+2(-1+\delta)\delta(-1+2\theta))\tanh^{2}{(\beta\nu_{1}}))$ is
$\leq 0$, since
$(\theta+2(-1+\delta)\delta(-1+2\theta))\tanh^{2}{(\beta\nu_{1}})$ is $\leq$
1, thus for $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}-\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}$ to be $\geq 0$
one has to ensure that,
$(2\delta^{2}(-1+2\theta)\nu_{1}+\theta(\nu_{1}-2\nu_{2})+2\delta(\nu_{1}-2\theta\nu_{1}+2\theta\nu_{2}))$
is $\leq 0$ which is equivalent into,
$\nu_{2}\geq\frac{(\theta+2\delta(1-\delta)(1-2\theta))}{2(1-2\delta)\theta}\nu_{1}$,
which is nothing, but the condition given in Eq. (27). Therefore, under the
condition (27), we have

$\langle\eta\rangle^{2}=\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}}\leq\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}\leq 1.$ (32)

This equation is the first main result of this section. It shows that the
ratio of the fluctuations of $W$ and $Q_{M}$ is lower and upper-bounded.
Therefore, similarly to the previous analysis on the lower bound, we have
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}<1$ is always
respected when the system as a heat engine, since
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}-\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}>0$ from the heat
engine conditions, but it can or cannot be violated when it is an accelerator.
The violation depends on the parameters, and to be precise whenever equation
(27) is violated then $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\rangle/\langle
Q_{M}^{2}\rangle\geq 1$. Under the conditions of a heat engine ($\delta<1/2$
and
$\nu_{2}>((\theta+2\delta(1-2\theta)(1-\delta))/(\theta(1-2\delta)))\nu_{1}$),
equation (31) is always positive, which means that the upper bound in equation
(32) is not achieved, it is reached only when the system is working as an
accelerator depending on the values of the parameters.

Let’s now give an interpretation to equation (32). In the adiabatic regime,
the ratio of the fluctuations of $W$ and $Q_{M}$ saturates to its lower bound
which is the square of the Otto efficiency. However, when $0<\delta<1/2$ then
the more we increase $\delta$, but not exceeding 1/2, the more the ratio of
fluctuations increases until the upper bound 1 is violated in the accelerator
regime depending on the parameters. In the heat engine region, it means that
the fluctuations of $W$ are always less than those of $Q_{M}$, but the lower
bound on the ratio of the fluctuations cannot be arbitrary. However, when the
system is working as an accelerator then the fluctuations of $W$ can be either
less, greater or equal to those of $Q_{M}$.

Note that equation can be interpreted also as a witness of the operational
regime. More precisley when the ratio of fluctuations does not respect Eq.
(32), then this indicates that the system now starts working as a useless
machine, such as a heater or an accelerator.

Actually one can show that the square of the Otto efficiency provide a tighter
lower bound on the ratio of fluctuations of $W$ and $Q_{M}$ than the square of
the efficiency of the engine. After messy but strauthforward algebra one can
find that,

$\begin{split}\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}\nu_{2}^{2}-\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}\left(\nu_{2}-\nu_{1}\right)^{2}&=8\delta\nu_{1}\nu_{2}^{2}\left((1-\delta)\nu_{1}+2\delta\theta\nu_{1}+2\theta(\nu_{2}-\nu_{1})\right).\\\
&+16\delta\nu_{1}\nu_{2}^{2}\left(1-\delta-\theta+2\delta\theta\right)\left((1-2\delta)\theta\nu_{2}-(1-\delta)(\delta+\theta-2\delta\theta)\nu_{1}\right)\tanh^{2}\left(\beta\nu_{1}\right).\end{split}$
(33)

For $\nu_{2}\geq\nu_{1}$, one can easily see that
$8\delta\nu_{1}\nu_{2}^{2}\left((1-\delta)\nu_{1}+2\delta\theta\nu_{1}+2\theta(\nu_{2}-\nu_{1})\right)\geq
0$. On the other hand for
$(1-2\delta)\theta\nu_{2}-(1-\delta)(\delta+\theta-2\delta\theta)\nu_{1}$ to
be greater or equal to zero, this translates into

$(1-2\delta)\theta\nu_{2}\geq(1-\delta)(\delta+\theta-2\delta\theta)\nu_{1}.$
(34)

For a positive work we already have proved that we need
$(1-2\delta)\theta\nu_{2}\geq(\theta+2\delta(1-2\theta)(1-\delta))\nu_{1}$,
see Eq. (19). We have,

$(\theta+2\delta(1-2\theta)(1-\delta))\nu_{1}-(1-\delta)(\delta+\theta-2\delta\theta)\nu_{1}=\delta(1-\delta-\theta+2\delta\theta)\nu_{1}\geq
0.$ (35)

Since the necessary condition for positive work is always greater or equal to
the one necessary (Eq. 34) for the term in the second line in Eq. (33) to be
$\geq 0$, this means that even when the positive work condition is not
satisfied still the second tem in Eq. (33) can be $\geq 0$. Therefore under
the condition Eq. (34) we have,

$\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}\geq\left(1-\nu_{1}/\nu_{2}\right)^{2}.$
(36)

We already have proved that efficiency is limited by the Otto efficiency, see
Eq. (20). Therefore we have,

$\langle\eta\rangle^{2}=\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}}\leq\left(1-\nu_{1}/\nu_{2}\right)^{2}\leq\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
<EMAIL_ADDRESS>(37)

This is another cnetral result of this section. Unfortunately, below, we could
not generalize this result to the asymmetric Otto cyclc. From Eq. (38) we
have,

$\langle\eta\rangle\leq\left(1-\nu_{1}/\nu_{2}\right)\leq\sqrt{\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
<EMAIL_ADDRESS>(38)

This means that the square root of the ratio of fluctuations can also provide
an upper bound on the efficiency, however, it is not tighter than the Otto
efficiency. In Ref. [51, 50], the authors have found for autonomous continuous
thermal machines, that the square root of the ratio of the second cumulants
provide a tighter bound on the efficiency than the Carnot efficiency. A result
that we see does not hold here for driven discrete quantum heat engines.

####  II.8.2 Asymmetric Otto cycle

Using the Mathematica software, one can show that for the case when
$\zeta\neq\delta$ we have

$\begin{split}&\frac{2(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})}{(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})^{2}}-\frac{2(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})}{(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})^{2}}\\\
&=\frac{\nu_{1}(-(\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))\nu_{1}+2\theta(1-\delta-\zeta)\nu_{2})}{\theta(-1+\delta+\zeta)^{2}}\\\
&\times\frac{-(\theta(\delta-\zeta)^{2}(\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))+(\delta^{2}\theta+\delta(-1+2\zeta-2\theta\zeta)+\zeta(-1+\theta\zeta))\coth^{2}(\beta\nu_{1}))}{((\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))\nu_{1}+\theta(-1+\delta+\zeta)\nu_{2})^{2}}.\end{split}$
(39)

Note that we used the properly symmetrized expression of the relative
fluctuations introduced in [46]. One can show that
$-(\theta(\delta-\zeta)^{2}(\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))+(\delta^{2}\theta+\delta(-1+2\zeta-2\theta\zeta)+\zeta(-1+\theta\zeta))\coth^{2}(\beta\nu_{1}))$
is $\geq 0$. Let’s now prove this latter inequality.

We have,
$\theta(\delta-\zeta)^{2}(\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))\geq
0$ and
$-(\delta^{2}\theta+\delta(-1+2\zeta-2\theta\zeta)+\zeta(-1+\theta\zeta))\coth^{2}(\beta\nu_{1})=-(\theta(\delta-\zeta)^{2}-(\delta+\zeta-2\delta\zeta))\coth^{2}(\beta\nu_{1})$.
One can show that $\theta(\delta-\zeta)^{2}-(\delta+\zeta-2\delta\zeta)\leq
0$. This latter inequality follows from a simple maximazation oevr $\theta$,
since both $(\delta-\zeta)^{2}$ and $(\delta+\zeta-2\delta\zeta)$ are $\geq
0$. From this one can conclude that
$-(\delta^{2}\theta+\delta(-1+2\zeta-2\theta\zeta)+\zeta(-1+\theta\zeta))\coth^{2}(\beta\nu_{1})=-(\theta(\delta-\zeta)^{2}-(\delta+\zeta-2\delta\zeta))\coth^{2}(\beta\nu_{1})\geq
0$. The terms $\theta(-1+\delta+\zeta)^{2}$ and
$((\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))\nu_{1}+\theta(-1+\delta+\zeta)\nu_{2})^{2}$
are $\geq 0$.

Now let’s define the parameter $A$ to be given as follows,

$A=-(\delta^{2}\theta+\delta(-1+2\zeta-2\theta\zeta)+\zeta(-1+\theta\zeta))\coth^{2}(\beta\nu_{1})-\theta(\delta-\zeta)^{2}(\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta)).$
(40)

In such a case our task for proving that $A\geq 0$, reduce to checking that,

$-(\delta^{2}\theta+\delta(-1+2\zeta-2\theta\zeta)+\zeta(-1+\theta\zeta))\coth^{2}(\beta\nu_{1})\geq\theta(\delta-\zeta)^{2}(\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta)).$
(41)

Actually to prove the latter inequality it is sufficient to prove it for the
lowest value of $\coth^{2}(\beta\nu_{1})$ which is one. In this case (i.e.
$\coth^{2}(\beta\nu_{1})=1$) Eq. (40) reduces into,

$A=-(\theta(\delta-\zeta)^{2}-(\delta+\zeta-2\delta\zeta))-\theta(\delta-\zeta)^{2}(\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta)).$
(42)

From this, after a simple algebra, we have

$A=(\delta+\zeta-2\delta\zeta)-\theta(\delta-\zeta)^{2}(1+\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta)).$
(43)

One can show that $(\delta+\zeta-2\delta\zeta)$ and
$\theta(\delta-\zeta)^{2}(1+\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta)$ are
always positive or equal to zero. Further the highest value of
$\theta(\delta-\zeta)^{2}(1+\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta)$ for
arbitrary $\delta$ and $\zeta$ is when $\theta=1$. This letter statement can
be checked easily by using the first and the second derivative with respect to
$\theta$. In this case Eq. (43) becomes

$\begin{split}A&=\delta(1-\delta)+\zeta(1-\zeta)-(\delta-\zeta)^{2}(1-\delta-\zeta+2\delta\zeta).\end{split}$
(44)

After simple steps of calculations, one can find that,

$\begin{split}A=(1-2\delta)(\zeta^{3}+\delta(1-\zeta^{2}))+(1-2\zeta)(\delta^{3}+\zeta(1-\delta^{2}))+2\delta\zeta.\end{split}$
(45)

This equation is more simple to be analyzed by looking at it. Taking into
account the fact that for a heat engine to be possible in the forward as well
as in the backward cycle, $\delta$ and $\zeta$ need to be less than 1/2.
Imposing this condition is only necessary but not sufficient as we have seen
before. Since this leaves the possibility of an accelerator as well. From this
constraint, it is straightforward to show that $A\geq 0$. Therefore for
$2(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})/(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})^{2}\geq
2(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})/(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})^{2}$ one
need to ensure that,

$\nu_{2}\geq\frac{\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta)}{2(1-\delta-\zeta)\theta}\nu_{1}.$
(46)

Actually this lower bound
$\frac{\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta)}{2(1-\delta-\zeta)\theta}\nu_{1}$
on $\nu_{2}$ is half of the necessary condition for
$(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})\geq 0$, see
equation (20). Analogously to the symmetric Otto cycle, let’s now look at
$(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})-(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})$. The
explicit expression of the latter difference is given as,

$\begin{split}&(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})-(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})=\\\
&=-8\nu_{1}((\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))\nu_{1}+2\theta(-1+\delta+\zeta)\nu_{2})(1-(\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))\tanh^{2}(\beta\nu_{1}))\\\
&=4\nu_{1}(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}+2(\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta))\nu_{1}\tanh(\beta\nu_{1}))\\\
&\times(1-(\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta))\tanh^{2}(\beta\nu_{1}))\coth(\beta\nu_{1}).\end{split}$
(47)

Since
$(1-(\theta+\zeta-2\theta\zeta+\delta(-1+2\theta)(-1+2\zeta))\tanh^{2}(\beta\nu_{1}))\geq
0$, one can show that the condition (46) is necessary in this case as well for
$(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})-(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B})$ to be $\geq
0$. Therefore, when equation (46) is satisfied we have

$\langle\eta\rangle^{2}=\left(\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}}\right)^{2}\leq\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}}\leq 1.$ (48)

This shows that the ratio of the fluctuations of $W$ and $Q_{M}$ is still
lower and upper-bounded even in the asymmetric case. Equation (48) is a
generalization of equation (32) to the asymmetric driven Otto cycle.
Furthermore, note that the upper bound value 1, is not achieved when the
system is working as a heat engine. This follows from imposing the heat engine
conditions on the forward and the backward. These conditions are:
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F(B)}>0$ and
$\delta$($\zeta$)$<1/2$. In that case one can easily show that
$(1-(\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta))\tanh^{2}(\beta\nu_{1}))\coth(\beta\nu_{1})$
and $4\nu_{1}(\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}+2(\theta+(1-2\theta)(\delta+\zeta-2\delta\zeta))\nu_{1}\tanh(\beta\nu_{1}))$
are $>0$. From Eq. (48), and analogously to Eq. (38) we have,

$\langle\eta\rangle\leq\sqrt{\left(\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{F}+\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}_{B}}\right)}<1,$
(49)

which is less than the upper bound 1 allowed by the law of energy
conservation. Even though we could not prove it, we believe that the square
root of the ratio of fluctuations of the asymmetric Otto cycle does not
provide a tighter bound on the efficiency of the engine than the Otto
efficiency.

###  II.9 The ratio of the third and fourth cumulants of $W$ and $Q_{M}$

We have proved before, that the efficiency of the asymmetric Otto cycle is
limited by the one of the Otto if we take into account both the forward and
backward and treat them on an equal footing. Then we have proved the ratio of
fluctuations is lower and upper bounded. In Ref [50] Saryal et al. and in Ref
[51] Gerry et al. have proved that for continuous thermal machines, the ratio
of the $n$-th cumulants has a lower and upper bound, where the lower bound is
defined by the $n$-th power of the efficiency and the upper bound by the
$n$-th power of the Carnot efficiency. The lower bound was shown to be
saturated in the tight coupling limit. Therefore, analogously to the bounds on
the ratio of the second cumulants for quantum unital Otto heat engines, let’s
ask if the ratio of the third (fourth) cumulants can be lower bounded by third
(power) power of the efficiency and upper bounded by 1, as it is the case for
autonomous continuous thermal machines?

The standardized third and fourth cumulants are respectively, the skewness and
kurtosis. Skewness is a measure of the asymmetry of a distribution. When it is
negative (positive) it means that the left tail (right tail) of a distribution
is longer than the one on the right (left) side. For kurtosis (or excess
kurtosis) it is a measure of the tailedness and peakedness of a distribution.
These two quantities have to do with the shape of the distribution. In the
quasistatic limit, the ratio of the third and fourth cumulants of heat and
work are given as follows,

$\frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{3}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{3}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}=\left(1-\frac{\nu_{1}}{\nu_{2}}\right)^{3}\\\
\ \rm{and},\ \\\ \frac{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{4}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}{\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{4}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}}=\left(1-\frac{\nu_{1}}{\nu_{2}}\right)^{4}.$
(50)

This is still valid even for higher cumulants in the adiabatic regime.
Furthermore, what these two relations tell us, is that the sign of the third
and fourth cumulants of heat and work are the same in the quasistatic limit,
and that the third (fourth) cumulant of $Q_{M}$ is greater than that of $W$.
However, beyond this regime, their signs can be different, which makes this
regime more interesting to be investigated. We will return to this point in a
moment.

In figure (3) we plot the skewness difference
$\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{3}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{3/2}-\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{3}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{3/2}$ and
kurtosis difference $\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{4}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}Q_{M}^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}-\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{4}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}/\mathopen{\hbox{\set@color${\langle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\langle}$}}W^{2}\mathclose{\hbox{\set@color${\rangle}$}\mkern
2.0mu\kern-3.49998pt\leavevmode\hbox{\set@color${\rangle}$}}^{2}$ as a