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Energetic instability of passive states in thermodynamics

Carlo Sparaciari carlo.sparaciari.14@ucl.ac.uk Department of Physics and Astronomy, University College London, London WC1E 6BT, United Kingdom    David Jennings david.jennings@physics.ox.ac.uk Department of Physics, University of Oxford, Oxford, OX1 3PU, United Kingdom Department of Physics, Imperial College London, London SW7 2AZ, United Kingdom    Jonathan Oppenheim j.oppenheim@ucl.ac.uk Department of Physics and Astronomy, University College London, London WC1E 6BT, United Kingdom
(July 30, 2025)
Abstract

Passivity is a fundamental concept in thermodynamics that demands a quantum system’s energy cannot be lowered by any reversible, unitary process acting on the system. In the limit of many such systems, passivity leads in turn to the concept of complete passivity, thermal states, and the emergence of a thermodynamic temperature. In contrast, here we need only consider a single system and show that every passive state except the thermal state is unstable under a weaker form of reversibility. More precisely, we show that given a single copy of any athermal quantum state we may extract a maximal amount of energy from the state when we can use a machine that operates in a reversible cycle and whose state is left unchanged. This means that for individual systems the only form of passivity that is stable under general reversible processes is complete passivity, and thus provides a single-shot and more physically motivated identification of thermal states and the emergence of temperature. The machine which extracts work from passive states exploits the fact that one can find a subspace which acts as a virtual hot reservoir, and a subspace which acts as a virtual cold reservoir. We show that an optimal amount of work can be extracted, and that the machine operates at the Carnot efficiency between pairs of virtual reservoirs.

I Introduction

Within thermodynamics, heat engines are devices that operate in a thermal context so as to extract ordered energy in the form of work. The canonical scenario involves an engine that operates cyclically between two temperatures ThotT_{\text{hot}}, TcoldT_{\text{cold}} and performs a quantity of mechanical work. To do so the engine absorbs heat from the hot reservoir, converts some of this energy to mechanical work and releases heat into the cold reservoir in accordance with the Second Law of thermodynamics. The largest possible efficiency, η=1TcoldThot\eta=1-\frac{T_{\text{cold}}}{T_{\text{hot}}}, occurs for the reversible Carnot engine carnot_reflections_1824 ; callen_thermodynamics_1985 and provides a fundamental thermodynamic bound on the amount of ordered energy that can be obtained. Carnot engines, and more in general heat engines, have been extensively studied in the microscopic regime scully_extracting_2003 ; scovil_three-level_1959 ; geusic_quantum_1967 ; alicki_quantum_1979 ; howard_molecular_1997 ; geva_classical_1992 ; hanggi_artificial_2009 ; feldmann_quantum_2006 ; rousselet_directional_1994 ; faucheux_optical_1995 ; linden_how_2010 ; horodecki_fundamental_2013 ; brunner_virtual_2012 ; tajima_optimal_2014 ; ito_optimal_2016 ; verley_unlikely_2014 ; gardas_thermodynamic_2015 ; uzdin_collective_2015 ; uzdin_quantum_2016 ; kosloff_quantum_2016 ; woods_maximum_2015 ; frenzel_quasi-autonomous_2016 ; lekscha_quantum_2016 ; niedenzu_operation_2016 (as well as, of course, in the macroscopic regime).

However, the issue of ordered energy extraction can also be considered in scenarios in which no notion of temperature exists, and can provide a broader notion of equilibrium states. For example, more general equilbrium states can occur in physical realisations when a system has been perturbed and has not had enough time to fully thermalise. They can also arise in the context of non-equilibrium steady states seifert2012stochastic . Given a quantum system in a state ρ\rho one can ask if it is possible to extract energy from it solely by performing a reversible unitary transformation on the system. The largest amount of ordered energy that can be extracted (the “ergotropy”, see Refs. allahverdyan_maximal_2004 ; alicki_entanglement_2013 ; sparaciari_resource_2016 ) depends non-trivially on the quantum state. If no energy can be extracted in this way then ρ\rho is called passive pusz_passive_1978 ; lenard_thermodynamical_1978 ; perarnau-llobet_extractable_2015 ; perarnau-llobet_most_2015 ; hovhannisyan_entanglement_2013 and constitutes a primitive form of equilibrium.

In this work, we consider a scenario that is intermediate between the above two contexts, and is motivated by the fact that a work extraction machine should be considered as a system which is involved in the process. Our core question is whether there exist passive states ρS\rho_{S} for which energy can be extracted if one performs a reversible unitary process over the system SS together with a second quantum system MM, which starts and finishes in the same quantum state ρM\rho_{M}. This second quantum system is the machine, such as the working body in a Carnot cycle, which undergoes a cyclic evolution. This class of processes is reminiscent of the ones taking place inside heat engines, and has been termed a catalytic thermal operation brandao_second_2013 .

Indeed, the system SS described by a passive state represents both the reservoirs, while the additional system MM is the machine which exchanges energy with the reservoirs in a cyclic manner. Due to the fact that microscopic heat engines can nowadays be realised in the laboratory scovil_three-level_1959 ; rousselet_directional_1994 ; faucheux_optical_1995 ; howard_molecular_1997 ; schulman_molecular_1999 ; scully_quantum_2002 , it seems reasonable to extend the analysis on passive states to the case in which this broader class of operations is allowed. Crucially, this analysis has fundamental implications for the notion of passivity. In fact, if energy can be extracted from a passive state with these reversible processes, and no entropy is generated, then it seems that associating passivity of the state is a restricted idealisation, unstable under this simple extension.

The paper is structured as follows. In Sec. II we provide the definition of passive and completely passive states, together with their description in terms of virtual temperatures. Any three levels of a passive state can be thought of as containing a subsystem which acts as a virtual hot reservoir, and one which acts as a virtual cold one. Sec. III provides the description of a protocol involving finite-sized engines which enable us to extract work in a cyclic way from a single passive state. In Sec. IV we prove that in fact any passive state which is not thermal can be activated by such an engine. Finally in Sec. V we show that such an activation can be done optimally via a quasi-static, reversible process with zero generation of entropy. We provide a simple expression for the amount of work which can be extracted from an arbitrary passive state, Eq. (38), and show that our machine can operate at the efficiency of a Carnot engine operating between two virtual heat reservours. We conclude that the only states in the single-shot regime that are stable under general reversible processes are thermal states.

II Passive states

Consider a finite-dimensional quantum system associated with the Hilbert space d\mathcal{H}\equiv\mathbb{C}^{d} (a qudit), with Hamiltonian H=i=0d1Ei|ii|H=\sum_{i=0}^{d-1}E_{i}\ket{i}\bra{i}, and described by the state ρ\rho. We say that the state ρ\rho is passive iff its average energy cannot be lowered by acting on it with unitary operations, that is,

Tr[Hρ]Tr[HUρU],U(),UU=UU=𝕀.\mathrm{Tr}\left[{H\,\rho}\right]\leq\mathrm{Tr}\left[{H\,U\rho\,U^{\dagger}}\right]\ ,\ \forall\,U\in\mathcal{B}(\mathcal{H})\ ,\ UU^{\dagger}=U^{\dagger}U=\mathbb{I}. (1)

This implies that no work can be extracted from the state via a unitary process, since by conservation of energy, lowering the energy of a system would mean that this energy has been transfered to a work storage device.

We can also introduce a more restrictive notion of passivity. Let us consider nn\in\mathbb{N} independent and identically distributed (i.i.d.) copies of our system, with a total Hamiltonian H(n)=i=1dHiH^{(n)}=\sum_{i=1}^{d}H_{i}, where each HiH_{i} is a single-system Hamiltonian acting on a different copy of the system. The state of this global system is described by ρn\rho^{\otimes n}. Then, we say that the state ρ\rho is completely passive if and only if the state ρn\rho^{\otimes n} is passive for all nn\in\mathbb{N}. It can be shown pusz_passive_1978 that the completely passive states of a system with Hamiltonian HH are the ones satisfying the KMS condition kubo_statistical-mechanical_1957 ; martin_theory_1959 ; haag_equilibrium_1967 . Specifically, these states are the ground state and the thermal states with temperature β0\beta\geq 0, that is, τβ=eβH/Z\tau_{\beta}=e^{-\beta H}/Z with Z=Tr[eβH]Z=\mathrm{Tr}\left[{e^{-\beta H}}\right]. Any state which is not of this form, is called athermal.

Refer to caption
Figure 1: (Left) The spectrum of a qutrit passive state ρ=i=02pi|ii|\rho=\sum_{i=0}^{2}p_{i}\ket{i}\bra{i} over the eigenbasis of its Hamiltonian H=i=02Ei|ii|H=\sum_{i=0}^{2}E_{i}\ket{i}\bra{i}. The occupation probabilities are ordered in a decreasing order, from the one associated with the ground state of HH to the one associated with the maximally excited one, as described in Eq. (2). (Right) A passive state can equally be described by virtual temperatures. Indeed, we can associate a virtual temperature to each pair of eigenstates of ρ\rho, as shown in Eq. (3). In the plot, the pair of levels |0\ket{0} and |1\ket{1} is associated with the hot temperature βhot1\beta_{\text{hot}}^{-1}, while the pair |1\ket{1} and |2\ket{2} is associated with the cold temperature βcold1\beta_{\text{cold}}^{-1}.

A characterization of all passive states can be easily obtained. A system in a passive state is such that the ground state has the highest probability of being occupied, and the probability of occupation decreases as the energy associated with the eigenstate of HH increases, Fig. 1, left plot. Specifically, a state ρ\rho is passive iff ρ=f(H)\rho=f(H), where ff is a monotone non-increasing function. Simply put, this means that the state can expressed as

ρ=i=0d1pi|ii|,such thatpipi+1i=0,,d2,\rho=\sum_{i=0}^{d-1}\mathrm{p}_{i}\ket{i}\bra{i}\ ,\ \text{such that}\ \mathrm{p}_{i}\geq\mathrm{p}_{i+1}\ \forall\,i=0,\ldots,d-2, (2)

where {|i}i=0d1\left\{\ket{i}\right\}_{i=0}^{d-1} are the eigenvectors of HH, ordered so that EiEi+1E_{i}\leq E_{i+1} for all ii (for the case of equal energies Ei=Ei+1E_{i}=E_{i+1} we must make an additional stability assumption to ensure that pi=pi+1p_{i}=p_{i+1}).

We can describe the probability distribution of the passive state ρ\rho by using virtual temperatures brunner_virtual_2012 ; skrzypczyk_passivity_2015 . In fact, for any given passive state, we can associate a (non-negative) virtual temperature with each pair of its eigenstates. For example, if we consider the pair (|i,|j)\left(\ket{i},\ket{j}\right), we define the virtual temperature associated with them as the βij10\beta_{ij}^{-1}\geq 0 such that

pipj=:eβij(EiEj),\frac{\mathrm{p}_{i}}{\mathrm{p}_{j}}=:e^{-\beta_{ij}\left(E_{i}-E_{j}\right)}, (3)

where pi\mathrm{p}_{i} is the probability of occupation of the state |i\ket{i}, and EiE_{i} is the energy associated with the state (similarly for jj). Thus, each pair of states can be regarded as an effective thermal state at a specific temperature. When all pairs of states has the same virtual temperature, we have that the passive state is completely passive, that is, it is the thermal state of HH at that temperature.

III The core protocol

We now introduce an engine that extracts work by acting individually on a passive state. The engine is composed by two main elements, namely, a qudit “machine” system with trivial Hamiltonian (H=0H=0), and a particular passive state. It suffices to consider qutrit systems, as a similar construction works more generally. The qutrit is assumed to have a Hamiltonian

HP=i=02Ei|ii|P,H_{P}=\sum_{i=0}^{2}E_{i}\ket{i}\bra{i}_{P}, (4)

and EiEi+1E_{i}\leq E_{i+1}. The qutrit system is described by the state

ρP=i=02pi|ii|P,\rho_{P}=\sum_{i=0}^{2}\mathrm{p}_{i}\ket{i}\bra{i}_{P}, (5)

where pipi+1\mathrm{p}_{i}\geq\mathrm{p}_{i+1} (Fig. 1, left plot).

In the following we assume that the passive state ρP\rho_{P} is described by the virtual temperature Thot=βhot1>0T_{\text{hot}}=\beta_{\text{hot}}^{-1}>0 associated with the pair of eigenstates (|0P,|1P)\left(\ket{0}_{P},\ket{1}_{P}\right), and the virtual temperature Tcold=βcold1>0T_{\text{cold}}=\beta_{\text{cold}}^{-1}>0 associated with the pair (|1P,|2P)\left(\ket{1}_{P},\ket{2}_{P}\right). We assume for simplicity that Thot>TcoldT_{\text{hot}}>T_{\text{cold}}, but a similar analysis applies for Thot<TcoldT_{\text{hot}}<T_{\text{cold}}. In the Supplemental Material, Sec. I, the cycle is presented in full detail. The relation between the probability distribution of ρP\rho_{P} and the temperatures ThotT_{\text{hot}} and TcoldT_{\text{cold}} is given by

p1p0\displaystyle\frac{\mathrm{p}_{1}}{\mathrm{p}_{0}} =:eβhotΔE10,\displaystyle=:e^{-\beta_{\text{hot}}\Delta E_{10}}, (6a)
p2p1\displaystyle\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}} =:eβcoldΔE21,\displaystyle=:e^{-\beta_{\text{cold}}\Delta E_{21}}, (6b)

where ΔE10=E1E00\Delta E_{10}=E_{1}-E_{0}\geq 0, and ΔE21=E2E10\Delta E_{21}=E_{2}-E_{1}\geq 0. Thus, the pair of states (|0P,|1P)\left(\ket{0}_{P},\ket{1}_{P}\right) can be viewed as representing a “hot virtual reservoir”, while the pair of states (|1P,|2P)\left(\ket{1}_{P},\ket{2}_{P}\right) represent a “cold virtual reservoir”. It is worth noting that the other pair of states, (|0P,|2P)\left(\ket{0}_{P},\ket{2}_{P}\right), is associated with a virtual temperature that is intermediate between TcoldT_{\text{cold}} and ThotT_{\text{hot}}, as we can easily verify from Eqs. (6).

The engine extracts work by means of the following cycle. A single system, described by the state ρP\rho_{P}, is put in contact with the machine, described by the state ρM=j=0d1qj|jj|M\rho_{M}=\sum_{j=0}^{d-1}\mathrm{q}_{j}\ket{j}\bra{j}_{M}. Then, we perform mm swaps between the hot virtual reservoir of the passive state and mm different pairs of states of ρM\rho_{M}, followed by nn swaps between the cold virtual reservoir and other nn different pairs of states of ρM\rho_{M}. In order to perform the swaps on different pairs of states, we need the machine to have at least m+nm+n levels, and therefore we fix d=m+nd=m+n. Specifically, we apply the following unitary operation to the global system

Sm,n=S(1,2)(0,m)S(1,2)(m,m+1)S(1,2)(m+1,m+2)S(1,2)(m+n2,m+n1)S(0,1)(m1,m+n1)S(0,1)(m2,m1)S(0,1)(m3,m2)S(0,1)(0,1),S_{m,n}=S_{(1,2)}^{(0,m)}\circ S_{(1,2)}^{(m,m+1)}\circ S_{(1,2)}^{(m+1,m+2)}\circ\ldots\circ S_{(1,2)}^{(m+n-2,m+n-1)}\circ S_{(0,1)}^{(m-1,m+n-1)}\circ S_{(0,1)}^{(m-2,m-1)}\circ S_{(0,1)}^{(m-3,m-2)}\circ\ldots\circ S_{(0,1)}^{(0,1)}, (7)

where the operator S(a,b)(c,d)S_{(a,b)}^{(c,d)} is a swap between system and machine, performed through the permutation |aP|dM|bP|cM\ket{a}_{P}\ket{d}_{M}\leftrightarrow\ket{b}_{P}\ket{c}_{M}. A graphical representation of this global operation is shown in Fig. 2, where each swap is depicted by an arrow acting over the states of the machine. Although in the figure we represent the eigenstates of ρM\rho_{M} in a ladder, they are all associated with the same energy, and therefore the order in which we present them is only functional for visualising the cycle Sm,nS_{m,n}.

Refer to caption
Figure 2: The cycle is represented in a pictorial way over the eigenstates of the dd-dimensional machine (where d=m+nd=m+n). Notice that the machine has a trivial Hamiltonian, and we order the eigenstates only to simplify the visualisation of the cycle. The upward arrow connecting two eigenstates of the machine represents a swap between these two states and the pair (|0P,|1P)\left(\ket{0}_{P},\ket{1}_{P}\right) of the passive state. The downward arrow connecting two eigenstates of the machine represents a swap between this pair and the pair (|1P,|2P)\left(\ket{1}_{P},\ket{2}_{P}\right) of the passive state. We initially perform m1m-1 swaps between (|0P,|1P)\left(\ket{0}_{P},\ket{1}_{P}\right) and {(|jM,|j+1M)}j=0m2\left\{\left(\ket{j}_{M},\ket{j+1}_{M}\right)\right\}_{j=0}^{m-2}, and one swap between (|0P,|1P)\left(\ket{0}_{P},\ket{1}_{P}\right) and (|m1M,|m+n1M)\left(\ket{m-1}_{M},\ket{m+n-1}_{M}\right). Then, we perform n1n-1 swaps between (|1P,|2P)\left(\ket{1}_{P},\ket{2}_{P}\right) and {(|jM,|j+1M)}j=mm+n2\left\{\left(\ket{j}_{M},\ket{j+1}_{M}\right)\right\}_{j=m}^{m+n-2}, and one swap between (|1P,|2P)\left(\ket{1}_{P},\ket{2}_{P}\right) and (|0M,|mM)\left(\ket{0}_{M},\ket{m}_{M}\right). If we consider the arrow representation of swaps, we can see that the cycle is close, and this allows us to recover the local state of the machine MM while extracting work.

In order for the engine to be re-usable, we need the local state of the machine to end up in its initial state. Therefore, we impose the following constraint on the state of the machine,

ρM=!TrP[Sm,n(ρPρM)Sm,n].\rho_{M}\overset{!}{=}\mathrm{Tr}_{P}\left[{S_{m,n}\left(\rho_{P}\otimes\rho_{M}\right)S_{m,n}^{\dagger}}\right]. (8)

Through Eq. (8) we can express the probability distribution of the machine in terms of the passive state ρP\rho_{P} (see the Supplemental Material for further details). In our model we do not explicitly include an additional system (a battery) for storing the energy we extract from the passive state. Instead, we implicitly assume the existence of this work-storage system, and we define the work extracted, ΔW\Delta W, as the difference in average energy between the initial and final state of the main system (as the machine MM has a trivial Hamiltonian, and no interaction terms are present between system and machine). Thus, we have that

ΔW=Tr[HP(ρPρ~P)],\Delta W=\mathrm{Tr}\left[{H_{P}\left(\rho_{P}-\tilde{\rho}_{P}\right)}\right], (9)

where the final state of the system is

ρ~P=TrM[Sm,n(ρPρM)Sm,n].\tilde{\rho}_{P}=\mathrm{Tr}_{M}\left[{S_{m,n}\left(\rho_{P}\otimes\rho_{M}\right)S_{m,n}^{\dagger}}\right]. (10)

It is worth noting that the final state of system and machine will in general develop correlations. These correlations are classical, and without them work would not be extracted during the cycle. However, they do not compromise the re-usability of the machine if applied to another uncorrelated quantum system.

For a given system Hamiltonian HPH_{P} and a given cycle Sm,nS_{m,n} we can investigate the amount of work we extract from the state ρP\rho_{P}. In the Supplemental Material we provide all the necessary steps to evaluate ΔW\Delta W in terms of the probability distribution of ρP\rho_{P}. We can express this quantity as

ΔW=α(mΔE10nΔE21)(eβcoldnΔE21eβhotmΔE10),\Delta W=\alpha\left(m\,\Delta E_{10}-n\,\Delta E_{21}\right)\left(e^{\beta_{\text{cold}}n\Delta E_{21}}-e^{\beta_{\text{hot}}m\Delta E_{10}}\right), (11)

where α\alpha is a positive coefficient depending non-trivially on the probability distribution of ρP\rho_{P}. For the class of passive states we are considering (namely, the one in which βcold>βhot\beta_{\text{cold}}>\beta_{\text{hot}}), we find that work can be extracted (ΔW>0\Delta W>0) iff

  1. 1.

    The Hamiltonian HPH_{P} is such that mΔE10>nΔE21m\,\Delta E_{10}>n\,\Delta E_{21}.

  2. 2.

    The temperatures of the two virtual reservoirs are such that βcold>mΔE10nΔE21βhot\beta_{\text{cold}}>\frac{m\,\Delta E_{10}}{n\,\Delta E_{21}}\beta_{\text{hot}}.

Thus, for a fixed cycle (defined by the parameters mm and nn), and for a fixed Hamiltonian HPH_{P}, we find that work can only be extracted if the virtual temperature TcoldT_{\text{cold}} is lower than ThotT_{\text{hot}} by a multiplicative factor which depends on the energy gaps of the Hamiltonian, see Fig. 3, left plot, for an example. In Sec. IV we show that, for a given Hamiltonian HPH_{P}, work can be extracted by any passive (but not completely passive) state, and we characterise the cycle which allows for this extraction.

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Figure 3: We consider the case in which the machine is a qubit system, and we only perform a single hot and cold swap, that is m,n=1m,n=1. This cycle is analogous to the one studied in Ref. linden_how_2010 . (Left) The work extracted in the cycle is positive if the hot energy gap ΔE10\Delta E_{10} lies inside the range [ΔE10;βcoldβhotΔE10]\left[\Delta E_{10};\frac{\beta_{\text{cold}}}{\beta_{\text{hot}}}\Delta E_{10}\right], as we would expect from conditions 1 and 2. (Right) The efficiency of the engine, as given by Eq. (16).

If we analyse in a more detailed way the cycle, we find that the same amount of energy is gained during each swap between the machine MM and the hot virtual reservoir, that is

qhot=αΔE10(eβcoldnΔE21eβhotmΔE10),q_{\text{hot}}=\alpha\,\Delta E_{10}\left(e^{\beta_{\text{cold}}n\Delta E_{21}}-e^{\beta_{\text{hot}}m\Delta E_{10}}\right), (12)

where α\alpha is the same positive coefficient of Eq. (11). Moreover, the same amount of energy is spent during each swap between the machine MM and the cold virtual reservoir,

qcold=αΔE21(eβcoldnΔE21eβhotmΔE10).q_{\text{cold}}=\alpha\,\Delta E_{21}\left(e^{\beta_{\text{cold}}n\Delta E_{21}}-e^{\beta_{\text{hot}}m\Delta E_{10}}\right). (13)

Knowing the amount of energy exchanged during each swap allows us to evaluate the heat exchanged with the virtual reservoirs. In fact, if we identify the pair of levels (|0P,|1P)\left(\ket{0}_{P},\ket{1}_{P}\right) with the hot virtual reservoir, then the energy exchanged during a swap with these levels can be considered as heat coming from the hot virtual reservoir. In this way, the total heat absorbed by the machine is

Qhot=mqhot,Q_{\text{hot}}=m\,q_{\text{hot}}, (14)

while the total heat provided to the cold virtual reservoir is

Qcold=nqcold.Q_{\text{cold}}=n\,q_{\text{cold}}. (15)

From Eqs. (14) and (15) we obtain that the work extracted can be expressed as ΔW=QhotQcold\Delta W=Q_{\text{hot}}-Q_{\text{cold}}, as in a standard heat engine exchanging energy between two reservoirs. Once QhotQ_{\text{hot}} and QcoldQ_{\text{cold}} are defined, we can evaluate an efficiency of this cycle, that is

η=ΔWQhot=1nΔE21mΔE10.\eta=\frac{\Delta W}{Q_{\text{hot}}}=1-\frac{n\,\Delta E_{21}}{m\,\Delta E_{10}}. (16)

The efficiency of the engine (when the machine is finite-dimensional) is sub-Carnot in the virtual temperatures, see Fig. 3, right plot. In fact, work can only be extracted when conditions 1 and 2 are satisfied, and these conditions implied 0<η<1TcoldThot0<\eta<1-\frac{T_{\text{cold}}}{T_{\text{hot}}}. When we consider the case of an infinite-dimensional machine, we find that by a judicious choice of parameters we may obtain Carnot efficiency.

Once the cycle Sm,nS_{m,n} is ended, the local state of the main system is moved to a less energetic state. By solving Eq. (8), we find that the final state of the main system ρ~P\tilde{\rho}_{P} has the following probability distribution

p0\displaystyle\mathrm{p}^{\prime}_{0} =p0+mΔP,\displaystyle=\mathrm{p}_{0}+m\,\Delta\mathrm{P}, (17a)
p1\displaystyle\mathrm{p}^{\prime}_{1} =p1(m+n)ΔP,\displaystyle=\mathrm{p}_{1}-\left(m+n\right)\,\Delta\mathrm{P}, (17b)
p2\displaystyle\mathrm{p}^{\prime}_{2} =p2+nΔP,\displaystyle=\mathrm{p}_{2}+n\,\Delta\mathrm{P}, (17c)

where the unit of probability ΔP\Delta\mathrm{P} depends on the initial state ρP\rho_{P}, and it is given by

ΔP=α(eβcoldnΔE21eβhotmΔE10),\Delta\mathrm{P}=\alpha\left(e^{\beta_{\text{cold}}n\Delta E_{21}}-e^{\beta_{\text{hot}}m\Delta E_{10}}\right), (18)

with α\alpha the same positive coefficient of Eq. (11). Due to condition 2, the unit ΔP>0\Delta\mathrm{P}>0, so that the probability of occupation of |1P\ket{1}_{P} is reduced in favour of the probabilities p0\mathrm{p}_{0} and p1\mathrm{p}_{1}. Thus, energy is extracted from the passive state when mΔPm\,\Delta\mathrm{P} is moved from p1\mathrm{p}_{1} to p0\mathrm{p}_{0} (during the hot swaps), and part of this energy is used to move the probability nΔPn\,\Delta\mathrm{P} from p1\mathrm{p}_{1} to p2\mathrm{p}_{2} (during the cold swaps).

IV Work extraction from any passive state

We now show that, for a given Hamiltonian HPH_{P}, work can be extracted from any passive but not completely passive state. In particular, we first show this for qutrit passive states, and we then generalise to the qudit case. Work extraction is achieved with the cycle presented in Sec. III, for specific values of the parameters mm and nn. In what follows, we represent the passive state with the probabilities of occupation {p0,p1,p2}\left\{\mathrm{p}_{0},\mathrm{p}_{1},\mathrm{p}_{2}\right\}, as opposed to the previous case in which the virtual temperatures were used. In this way, we can consider all possible scenarios, and we are not limited to the case in which a specific pair of eigenstates has a colder (hotter) virtual temperature than the other pair.

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Figure 4: (Left) The set of qutrit passive states. The red region represents the subset R1R_{1}, while the blue region represents the subset R2R_{2}. The black line is R3R_{3}, that is, the set of completely passive states. (Right) The region R1R_{1} (in red) contains the regions R3,1+R^{+}_{3,1} (green), R5,2+R^{+}_{5,2} (purple), and R11,5+R^{+}_{11,5} (yellow), which cover R1R_{1} better and better as mm and nn grow. In both plots we set M=2M=2 and N=1N=1.

The Hamiltonian of the system HPH_{P} is defined in Eq. (4), where the energy gap between ground and first excited state is ΔE10\Delta E_{10}, and the gap between first and second excited states is ΔE21\Delta E_{21}. We assume that

M,Nsuch thatMΔE10NΔE21=0,\exists\,M,N\in\mathbb{N}\ \text{such that}\ M\,\Delta E_{10}-N\,\Delta E_{21}=0, (19)

that is, we ask the ratio between the two energy gaps to be rational. Notice that, even if the ratio is irrational, we can find a suitable NN and MM such that the condition is approximatively satisfied. Once the relation between energy gaps is defined, we can divide the set of passive states into three different subsets, namely

R1\displaystyle R_{1} ={ρPpassive|(p1p2)N>(p0p1)M},\displaystyle=\left\{\rho_{P}\ \text{passive}\ \Big{|}\ \left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{N}>\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{M}\right\}, (20a)
R2\displaystyle R_{2} ={ρPpassive|(p1p2)N<(p0p1)M},\displaystyle=\left\{\rho_{P}\ \text{passive}\ \Big{|}\ \left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{N}<\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{M}\right\}, (20b)
R3\displaystyle R_{3} ={ρPpassive|(p1p2)N=(p0p1)M}.\displaystyle=\left\{\rho_{P}\ \text{passive}\ \Big{|}\ \left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{N}=\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{M}\right\}. (20c)

The union of these three subsets gives the set of all passive states. In particular, one can verify that the subset R3R_{3} contains all the completely passive states, that is, the thermal states of HPH_{P} at any temperature β10\beta^{-1}\geq 0. Moreover, R1R_{1} corresponds to the set of passive states with βhot\beta_{\text{hot}} associated with the pair of eigenstates |0P\ket{0}_{P} and |1P\ket{1}_{P}, and βcold\beta_{\text{cold}} associated with the pair |1P\ket{1}_{P} and |2P\ket{2}_{P}. The set R2R_{2}, instead, contains the passive states with opposite hot and cold virtual temperatures. Since we are considering qutrit systems, we can represent the set of passive states in a two-dimensional diagram, using their probability distribution. Each point in this diagram represents a passive state. In Fig. 4, left plot, we show the three subsets of Eqs. (20).

In the previous section, we have seen that a cycle defined by the parameters mm and nn can activate a passive state ρP\rho_{P} with Hamiltonian HPH_{P} if conditions 1 and 2 are satisfied. These conditions apply to the case in which the passive state is described by Eqs. 6, with βhot<βcold\beta_{\text{hot}}<\beta_{\text{cold}}. In the present, more general scenario we find that work is extracted by the cycle if and only if the passive state belongs to the following subset

Rm,n+={ρPpassive|\displaystyle R^{+}_{m,n}=\bigg{\{}\rho_{P}\ \text{passive}\ \bigg{|}\quad (p1p2)n>(p0p1)mwhenmΔE10nΔE21>0\displaystyle\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n}>\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\text{when}\ m\,\Delta E_{10}-n\,\Delta E_{21}>0
\displaystyle\vee\quad (p1p2)n<(p0p1)mwhenmΔE10nΔE21<0},\displaystyle\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n}<\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\text{when}\ m\,\Delta E_{10}-n\,\Delta E_{21}<0\bigg{\}}, (21)

where these conditions can be obtained by analysing the general expression of the extracted work, see Supplemental Material. We now show that, by tailoring the value of the parameters mm and nn, we can make Rm,n+R^{+}_{m,n} to (asymptotically) cover either the region R1R_{1} or R2R_{2}. Here, we focus on R1R_{1} solely, since R2R_{2} follows from similar arguments. As a first step, we ask mΔE10nΔE21>0m\,\Delta E_{10}-n\,\Delta E_{21}>0, which implies m>MNnm>\frac{M}{N}n, due to Eq. (19). Then, in order to satisfy this condition, we set m=MNn+1m=\frac{M}{N}n+1, where we ask nn to be large enough for mm to be an integer. The set of passive states activated by the cycle is such that

(p1p2)n>(p0p1)m(p1p2)n>(p0p1)MNn+1(p1p2)N>(p0p1)M+Nn.\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n}>\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\Rightarrow\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n}>\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{\frac{M}{N}n+1}\Rightarrow\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{N}>\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{M+\frac{N}{n}}. (22)

We notice that, since ρP\rho_{P} is passive, p0p1\mathrm{p}_{0}\geq\mathrm{p}_{1}, which implies that

(p0p1)M+Nn(p0p1)M.\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{M+\frac{N}{n}}\geq\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{M}. (23)

Thus, Eq. (22) and (23) together assure that RMNn+1,n+R1R^{+}_{\frac{M}{N}n+1,n}\subset R_{1}. Moreover, if nn\rightarrow\infty, we have that M+NnMM+\frac{N}{n}\rightarrow M, which implies that RMNn+1,n+R1R^{+}_{\frac{M}{N}n+1,n}\rightarrow R_{1}. Thus, we have that, for a given Hamiltonian HPH_{P}, and a given passive state ρPR1\rho_{P}\in R_{1}, there exist a cycle Sm,nS_{m,n} such that ρPRm,n+\rho_{P}\in R^{+}_{m,n}. However, the closer (in trace norm) the state ρP\rho_{P} is to the set of completely passive states (R3R_{3}), the larger the parameters mm and nn have to be, that is, the larger the machine has to be (see Fig. 4, right plot).

IV.1 Work extraction from a generic qudit passive state

Work extraction from a generic qudit passive state ρP(d)\rho_{P}^{(d)} (for any Hamiltonian HP(d)H_{P}^{(d)}) can be achieved with the cycle introduced in Sec. III, even if this work extraction is not optimal (as it might be when we deal with qutrit state, as we see in Sec. V). Indeed, even if the system has dd levels, we only need to focus our analysis on three of them, and perform the cycle on these levels only. Thus, given the state ρP(d)=i=0d1pi|ii|P\rho_{P}^{(d)}=\sum_{i=0}^{d-1}\mathrm{p}_{i}\ket{i}\bra{i}_{P} and the Hamiltonian HP(d)=i=0d1Ei|ii|PH_{P}^{(d)}=\sum_{i=0}^{d-1}E_{i}\ket{i}\bra{i}_{P}, we can consider the subspace Ak=span{|kP,|k+1P,|k+2P}A_{k}=\text{span}\left\{\ket{k}_{P},\ket{k+1}_{P},\ket{k+2}_{P}\right\}, for a given k[0,d3]k\in[0,d-3]. Thus, we can divide the qudit state and the Hamiltonian in two contributions, one with support over AkA_{k}, the other with support over its complement,

ρP(d)\displaystyle\rho_{P}^{(d)} =(iAkpi)ρP(A)+(1iAkpi)ρP(Ac),\displaystyle=\left(\sum_{i\in A_{k}}\mathrm{p}_{i}\right)\rho_{P}^{(A)}+\left(1-\sum_{i\in A_{k}}\mathrm{p}_{i}\right)\rho_{P}^{(A^{c})}, (24a)
HP(d)\displaystyle H_{P}^{(d)} =HP(A)+HP(Ac),\displaystyle=H_{P}^{(A)}+H_{P}^{(A^{c})}, (24b)

where the normalised quantum states are

ρP(A)\displaystyle\rho_{P}^{(A)} =iAkpijAkpj|ii|P,\displaystyle=\sum_{i\in A_{k}}\frac{\mathrm{p}_{i}}{\sum_{j\in A_{k}}\mathrm{p}_{j}}\ket{i}\bra{i}_{P}, (25a)
ρP(Ac)\displaystyle\rho_{P}^{(A^{c})} =iAkpi1jAkpj|ii|P,\displaystyle=\sum_{i\notin A_{k}}\frac{\mathrm{p}_{i}}{1-\sum_{j\in A_{k}}\mathrm{p}_{j}}\ket{i}\bra{i}_{P}, (25b)

while the Hamiltonian contributions are, respectively, HP(A)=iAkEi|ii|PH_{P}^{(A)}=\sum_{i\in A_{k}}E_{i}\ket{i}\bra{i}_{P} and HP(Ac)=iAkEi|ii|PH_{P}^{(A^{c})}=\sum_{i\notin A_{k}}E_{i}\ket{i}\bra{i}_{P}. In the following, we define λ=iAkpi\lambda=\sum_{i\in A_{k}}\mathrm{p}_{i}, so that ρP(d)=λρP(A)+(1λ)ρP(Ac)\rho_{P}^{(d)}=\lambda\,\rho_{P}^{(A)}+\left(1-\lambda\right)\rho_{P}^{(A^{c})}.

We can now introduce an ancillary system (the machine MM) of dimension m+nm+n, described by the state ρM\rho_{M}, together with the global unitary operator UU,

U=PAc𝕀M+PA𝕀MSm,nPA𝕀M,U=P_{A^{c}}\otimes\mathbb{I}_{M}+P_{A}\otimes\mathbb{I}_{M}\circ S_{m,n}\circ P_{A}\otimes\mathbb{I}_{M}, (26)

where the operator Sm,nS_{m,n}, described in Eq. (7), has support on AkMA_{k}\otimes\mathcal{H}_{M}, and therefore commute with PA𝕀MP_{A}\otimes\mathbb{I}_{M}. If we consider the evolution of the system under this operator, we obtain

ρ~P(d)=TrM[U(ρP(d)ρM)U]=λTrM[Sm,n(ρP(A)ρM)Sm,n]+(1λ)ρP(Ac)=λρ~P(A)+(1λ)ρP(Ac),\tilde{\rho}_{P}^{(d)}=\mathrm{Tr}_{M}\left[{U\left(\rho_{P}^{(d)}\otimes\rho_{M}\right)U^{\dagger}}\right]=\lambda\,\mathrm{Tr}_{M}\left[{S_{m,n}\left(\rho_{P}^{(A)}\otimes\rho_{M}\right)S_{m,n}^{\dagger}}\right]+\left(1-\lambda\right)\rho_{P}^{(A^{c})}=\lambda\,\tilde{\rho}_{P}^{(A)}+\left(1-\lambda\right)\rho_{P}^{(A^{c})}, (27)

and we can easily verify, due to the properties of Sm,nS_{m,n}, that the local state of the machine is left unchanged. The amount of work extracted during this cycle is

ΔW=TrP[HP(d)(ρP(d)ρ~P(d))]=λTrP[HP(A)(ρP(A)ρ~P(A))],\Delta W=\mathrm{Tr}_{P}\left[{H_{P}^{(d)}\left(\rho_{P}^{(d)}-\tilde{\rho}_{P}^{(d)}\right)}\right]=\lambda\,\mathrm{Tr}_{P}\left[{H_{P}^{(A)}\left(\rho_{P}^{(A)}-\tilde{\rho}_{P}^{(A)}\right)}\right], (28)

and the problem reduces to the one analysed at the beginning of this section (that is, to the extraction of work from a qutrit system described by the passive state ρP(A)\rho_{P}^{(A)}, with Hamiltonian HP(A)H_{P}^{(A)}), with the only difference of a multiplicative factor λ(0,1)\lambda\in(0,1) in ΔW\Delta W.

IV.2 Work extraction and kk-activable states

The set of passive states can be divided into a hierarchy of classes, which divides the states according to the number of copies needed to activate them. Here, we say that a state is active if it is not passive, and therefore if we can extract work from it with unitary operations. Any passive but not completely passive state can be activated if we tensor together enough copies of it. In particular, when kk copies of a passive state are active, we call the state kk-activable. We now show that, if work is extracted from a qutrit passive state ρP\rho_{P}, with Hamiltonian HPH_{P}, through the cycle Sm,nS_{m,n}, then the state realised by m+nm+n copies of ρP\rho_{P} is active. It worth noting that, while our cycle only requires an additional system of dimension m+nm+n to extract work from ρP\rho_{P}, in order to activate the same state we would need m+n1m+n-1 copies of it, that is, an ancilla whose size is exponential in n+mn+m.

In the following, we consider a qutrit system, although the same argument applies to qudit systems, for the reasons presented in the previous section. If the passive state ρP\rho_{P} is activated by the cycle Sm,nS_{m,n}, then one of the two conditions in Eq. (IV) has to be satisfied. Let us assume that the conditions satisfied by state and Hamiltonian are

mΔE10\displaystyle m\,\Delta E_{10} >nΔE21,\displaystyle>n\,\Delta E_{21}, (29a)
(p1p2)n\displaystyle\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n} >(p0p1)m,\displaystyle>\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}, (29b)

where the other case follows straightforwardly.

Consider now a system composed by n+mn+m copies of the qutrit system under examination, with Hamiltonian Htot=i=1m+nHP(i)H_{\text{tot}}=\sum_{i=1}^{m+n}H_{P}^{(i)}, where the term HP(i)H_{P}^{(i)} acts over the ii-th copy. The state of this global system is ρPm+n\rho_{P}^{\otimes m+n}. Then, let us focus our attention on two eigenstates of HtotH_{\text{tot}}, namely, |1m+n\ket{1}^{\otimes m+n} and |0m|2n\ket{0}^{\otimes m}\otimes\ket{2}^{\otimes n}. The first eigenstate has an energy of (m+n)E1\left(m+n\right)E_{1}, and its occupation probability is p1m+n\mathrm{p}_{1}^{m+n}. The second eigenstate, instead, has energy mE0+nE2m\,E_{0}+n\,E_{2}, and its occupation probability is p0mp2n\mathrm{p}_{0}^{m}\mathrm{p}_{2}^{n}. It is easy to verify that, if the constraints of Eqs. (29) hold, then the inequalities (m+n)E1>mE0+nE2\left(m+n\right)E_{1}>m\,E_{0}+n\,E_{2} and p1m+n>p0mp2n\mathrm{p}_{1}^{m+n}>\mathrm{p}_{0}^{m}\mathrm{p}_{2}^{n} are satisfied, implying that the state ρPm+n\rho_{P}^{\otimes m+n} is active. Thus, we have shown that if a passive state ρP\rho_{P} can be activated with the cycle Sm,nS_{m,n}, then the state ρPm+n\rho_{P}^{\otimes m+n} is active. However, this result does not tell us whether it is possible to activate the state ρP\rho_{P} by tensoring it with less copies. In the same way, we do not know whether the fact that the state ρPm+n\rho_{P}^{\otimes m+n} is active implies that we can extract work from ρP\rho_{P} with the cycle Sm,nS_{m,n}.

V General Instability of Passive States

We can now establish our central claim: that any athermal passive state is energetically unstable under a reversible process that does not generate entropy. We analyse the evolution of a passive state which sequentially interacts with an infinite-dimensional machine MM, and find that the system moves through a continuous trajectory of passive states towards the set of minimum-energy states, that is, the set of the states jaynes_information_1957 .

We consider a cycle composed of infinitely many hot swaps, mm\rightarrow\infty, and infinite many cold swaps, nn\rightarrow\infty, with the assumption that n=αmn=\alpha\,m, where α\alpha is a parameter taking values in a specific range we will describe shortly. Let us now consider the situation in which the main system is a qutrit with Hamiltonian HPH_{P} given in Eq. (4), described by the passive state ρP\rho_{P} whose probability distribution satisfies the equalities of Eqs. (6). Then, ρP\rho_{P} belongs to the subset R1R_{1} defined in Eq. (20a), and the cycle Sm,nS_{m,n} has to satisfy conditions 1 and 2 in order to extract work from it. These conditions are reflected in the allowed range of the parameter α\alpha, that is

βhotΔE10βcoldΔE21<α<ΔE10ΔE21.\frac{\beta_{\text{hot}}\Delta E_{10}}{\beta_{\text{cold}}\Delta E_{21}}<\alpha<\frac{\Delta E_{10}}{\Delta E_{21}}. (30)

If we set α\alpha equal to a value inside the range specified by the previous equation, and we send mm\rightarrow\infty, we find that (see Supplemental Material for details) the state of the machine as obtained from Eq. (8) is given by a mixture of two “thermal” states, one with effective temperature βhot1\beta_{\text{hot}}^{-1}, the other with effective temperature βcold1\beta_{\text{cold}}^{-1} (note we still have HM=0H_{M}=0 for the machine). These distributions have support in two different subspaces, and their weight depends non-trivially on the energy gaps of HPH_{P} and on the virtual temperatures of ρP\rho_{P}. In fact, we can loosely interpret the state of the machine in terms of a thermal mixture

ρM=λτβhot+(1λ)τβcold,\rho_{M}=\lambda\,\tau_{\beta_{\text{hot}}}+\left(1-\lambda\right)\tau_{\beta_{\text{cold}}}, (31)

where

τβhot=eβhotHhotZhot,withHhot=j=0m1jΔE10|jj|MandZhot=Tr[eβhotHhot],\tau_{\beta_{\text{hot}}}=\frac{e^{-\beta_{\text{hot}}H_{\text{hot}}}}{Z_{\text{hot}}}\ ,\ \text{with}\ H_{\text{hot}}=\sum_{j=0}^{m-1}j\,\Delta E_{10}\ket{j}\bra{j}_{M}\ \text{and}\ Z_{\text{hot}}=\mathrm{Tr}\left[{e^{-\beta_{\text{hot}}H_{\text{hot}}}}\right], (32)

and

τβcold=eβcoldHcoldZcold,withHcold=j=0n3jΔE21|j+mj+m|MandZcold=Tr[eβcoldHcold].\tau_{\beta_{\text{cold}}}=\frac{e^{-\beta_{\text{cold}}H_{\text{cold}}}}{Z_{\text{cold}}}\ ,\ \text{with}\ H_{\text{cold}}=\sum_{j=0}^{n-3}j\,\Delta E_{21}\ket{j+m}\bra{j+m}_{M}\ \text{and}\ Z_{\text{cold}}=\mathrm{Tr}\left[{e^{-\beta_{\text{cold}}H_{\text{cold}}}}\right]. (33)

Notice that in order to define these “thermal states” we have introduced two fictitious Hamiltonians, namely, HhotH_{\text{hot}} and HcoldH_{\text{cold}}. These operators are necessary if we want to consider the distribution of the machine as the mixture of two thermal distributions, but they do not enter in any way in the derivation of the extractable work. Indeed, as we have specified at the beginning of Sec. III, the machine MM can have any Hamiltonian (it does not modify the amount of work we extract during the cycle), and we choose to use a trivial one HM=0H_{M}=0, so that the machine acts as a memory. The weight λ\lambda in the mixture is given by

λ=1eβcoldΔE211eβhotΔE10βcoldΔE21.\lambda=\frac{1-e^{-\beta_{\text{cold}}\Delta E_{21}}}{1-e^{-\beta_{\text{hot}}\Delta E_{10}-\beta_{\text{cold}}\Delta E_{21}}}. (34)

Thus, during the cycle, the passive state ρP\rho_{P} first interacts with the “hot reservoir”, by performing a sequence of swaps between the pair of states |0P\ket{0}_{P} and |1P\ket{1}_{P} and the levels of τβhot\tau_{\beta_{\text{hot}}}. Then, the state interacts with the “cold reservoir”, performing a sequence of swaps between the pair |1P\ket{1}_{P} and |2P\ket{2}_{P} and the levels of τβcold\tau_{\beta_{\text{cold}}}.

In this scenario, we find that the probability distribution of the passive state ρP\rho_{P} is infinitesimally modified, and consequently the work extracted is infinitesimally small. In particular, we find that the unit of probability, defined in Eq. (18), tends to 0 with an exponential scaling, ΔPeβhotmΔE10\Delta\mathrm{P}\propto e^{-\beta_{\text{hot}}m\Delta E_{10}} for mm\rightarrow\infty. Let us consider the probability distribution of the final state of the system ρ~P\tilde{\rho}_{P}. Since the distribution only changes infinitesimally during the cycle, we can recast Eqs. (17) as a set of differential equations (see the Supplemental Material for further details). Thus, we can imagine the situation in which infinite many machines are present, so that we can keep infinitesimally changing the state of the main system. In this case, the evolution of the state ρP\rho_{P} is governed by the following equation

dp1dt=(1+α(p0(t),p1(t)))dp0dt,\frac{\mathrm{d}\mathrm{p}_{1}}{\mathrm{d}t}=-\Big{(}1+\alpha\big{(}p_{0}(t),p_{1}(t)\big{)}\Big{)}\frac{\mathrm{d}\mathrm{p}_{0}}{\mathrm{d}t}, (35)

where the parameter tt provides a continuum label for the sequence of cycles we perform on the passive state. We can then solve this equation for extremal cases for the function α(p0,p1)\alpha(p_{0},p_{1}). When the parameter function α(p0,(t),p1(t))\alpha(p_{0},(t),p_{1}(t)) is equal to one of its limiting values, Eq. (35) assumes a clear meaning. In fact,

  • when α(p0(t),p1(t))=ΔE10ΔE21\alpha(p_{0}(t),p_{1}(t))=\frac{\Delta E_{10}}{\Delta E_{21}}, then the differential equation can be recast as a condition over the average energy of the system, that is,

    Tr[HPρP]=Tr[HPρ~P].\mathrm{Tr}\left[{H_{P}\,\rho_{P}}\right]=\mathrm{Tr}\left[{H_{P}\,\tilde{\rho}_{P}}\right]. (36)

    Then, for α\alpha taking this value, the passive state evolves along a trajectory that conserves the energy of the system.

  • when α(p0(t),p1(t))=βhot(t)ΔE10βcold(t)ΔE21\alpha(p_{0}(t),p_{1}(t))=\frac{\beta_{\text{hot}}(t)\Delta E_{10}}{\beta_{\text{cold}}(t)\Delta E_{21}}, instead, the differential equation can be recast as a condition over the entropy of the system, that is,

    S(ρP)=S(ρ~P),S\left(\rho_{P}\right)=S\left(\tilde{\rho}_{P}\right), (37)

    where S(ρ)=Tr[ρlogρ]S(\rho)=-\mathrm{Tr}\left[{\rho\log\rho}\right] is the Von Neumann entropy. Then, for this α\alpha, the passive state evolves along a trajectory that conserves the entropy of the system.

For α\alpha taking values inside the allowed range, we have that any trajectory between the two presented above is possible, and the set of achievable states is shown in Fig. 5, left plot. It is possible to show that the evolution of the system moves the passive state toward the set of thermal states, which are the stationary states of this dynamic. In Fig. 5, right plot, we show the same set of achievable states, represented this time in the energy-entropy diagram sparaciari_resource_2016 . It is clear that, through this evolution, we can obtain any passive state with a smaller average energy and a bigger entropy than ρP\rho_{P}. In Supplemental Material, we show that these states are also the only ones that we can reach with our engines (and with a broader class of maps, called activation maps).

Refer to caption
Refer to caption
Figure 5: (Left) The state space of a qutrit system, where the subset of passive states is highlighted. The blue region is R1R_{1}, while the green region is R2R_{2}. The black line is the set of thermal states. The initial state ρP\rho_{P} is represented by the black point in the diagram. If α\alpha takes the value ΔE10ΔE21\frac{\Delta E_{10}}{\Delta E_{21}}, then the system evolves along the red trajectory, and the final state is the thermal state at temperature βmin\beta_{\text{min}}. On the other hand, if α=βhot(t)ΔE10βcold(t)ΔE21\alpha=\frac{\beta_{\text{hot}}(t)\Delta E_{10}}{\beta_{\text{cold}}(t)\Delta E_{21}}, the system evolves along the purple line, and the final state is the thermal state at temperature βmax\beta_{\text{max}}. The yellow region represents the subset of achievable states when the initial state is ρP\rho_{P}. (Right) A partial representation of the state space of a dd-level quantum system in the energy-entropy diagram. In this diagram, quantum states are grouped into equivalence classes defined by their average energy EE and their entropy SS. Each point between the xx-axis (the set of pure states) and the black curve (the set of thermal states) represents one of these equivalence classes. The diagram depends on the Hamiltonian HPH_{P} of the system. Here, we only represent the states with average energy lower than E¯=Tr[HPρmm]\bar{E}=\mathrm{Tr}\left[{H_{P}\rho_{\text{mm}}}\right], where ρmm=𝕀d\rho_{\text{mm}}=\frac{\mathbb{I}}{\mathrm{d}} is the maximally-mixed state, since all passive states are contained in this set. For a given initial state ρP\rho_{P}, the green region contains all the passive states which can be achieved with the process.

It is interesting to consider the limiting values of work extraction that can be achieved following the scheme suggested in this section. In particular, when the system evolves along the energy-preserving trajectory, the final state we obtain is the thermal state of HPH_{P} at temperature βmin1\beta_{\text{min}}^{-1}, that is, τβmin\tau_{\beta_{\text{min}}}, where the temperature is such that Tr[HPρP]=Tr[HPτβmin]\mathrm{Tr}\left[{H_{P}\,\rho_{P}}\right]=\mathrm{Tr}\left[{H_{P}\,\tau_{\beta_{\text{min}}}}\right]. In this case, it is easy to see that the engine is not extracting any work, and its only effect consists in raising the entropy of the system. If we consider the efficiency of this cycle, Eq. (16), we see that η=0\eta=0, as expected. The opposite limit is obtained when the system evolves along the entropy-preserving trajectory. In this case, the final state is τβmax\tau_{\beta_{\text{max}}}, that is, the thermal state of HPH_{P} at temperature βmax1\beta_{\text{max}}^{-1}, such that S(ρP)=S(τβmax)S\left(\rho_{P}\right)=S\left(\tau_{\beta_{\text{max}}}\right). In this case, the work extracted by the cycle is

ΔW=Tr[HP(ρPτβmax)],\displaystyle\Delta W=\mathrm{Tr}\left[{H_{P}\left(\rho_{P}-\tau_{\beta_{\text{max}}}\right)}\right]\,\,, (38)

that is the maximum amount one can extract alicki_entanglement_2013 ; sparaciari_resource_2016 . Significantly, in this case the efficiency is equal to the Carnot one, ηCarnot=1βhotβcold\eta_{\text{Carnot}}=1-\frac{\beta_{\text{hot}}}{\beta_{\text{cold}}}.

VI Conclusion

In the paper we have presented an engine that is able to extract work from any single copy of an athermal passive state. The engine utilises an ancillary system for the work extraction, and the local state of this system is recovered at the end of the cycle. In this way, the cycle can be run multiple times, and each time it acts on a new copy of the passive state. We show that, for any given Hamiltonian, work can be extracted from any passive, but not completely passive state. Moreover, we show that, in order to extract work from passive states close to the set of completely passive states, we need an ancillary system of large dimension. With an infinite dimension machine we can also evolve a passive state smoothly toward the set of thermal states. In particular, optimal work extraction can be obtained in this case, and it is achieved by mapping the initial state into the thermal state with the same entropy.

The present work provides some evidence that a resource theory for thermodynamics with an imperfect thermal reservoir presents non-trivial challenges. Such a resource theory could be realised by providing passive states for free. However, an obvious restriction we should make in this resource theory consists in the fact that we could not provide more than k1k-1 copies of a kk-activatable passive state, otherwise work might be extracted with unitary operations from this free state. Moreover, our results show that, even in the case in which a single passive state is provided, an ancillary system exists such that work can be extracted from the individual passive state. Then, in order to build a sensible resource theory, passive states should be always provided at a work cost, equal to the optimal amount of energy extractable from them when a machine is present.

It might also be interesting to analyse which passive states allow for the extraction of the highest amount of work in our cycle. In particular, this problem has been studied in the case of passive states and unitary evolution perarnau-llobet_most_2015 , and a comparison between that class of states and the class of states which allow for maximal work extraction during our cycle might be interesting.

Acknowledgements: We are grateful to Ben Schumacher and Michael Westmoreland for inspiring discussions. JO thanks the Royal Society and an EPSRC Established Career Fellowship for their support. CS is supported by the EPSRC [grant number EP/L015242/1]. DJ is supported by the Royal Society. We thank the COST Network MP1209 in Quantum Thermodynamics.

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Supplemental Material

Appendix A General cycle for work extraction from passive states

We present in full details the general cycle needed to extract work from a qutrit system described by a passive state. Work extraction is achieved through the interaction between a qudit ancilla (the thermal machine) and the main qutrit system. This qutrit system has Hamiltonian

HP=E0|00|P+E1|11|P+E2|22|P,H_{P}=E_{0}\ket{0}\bra{0}_{P}+E_{1}\ket{1}\bra{1}_{P}+E_{2}\ket{2}\bra{2}_{P}, (39)

and we define the energy gap as ΔE10=E1E0>0\Delta E_{10}=E_{1}-E_{0}>0 and ΔE21=E2E1>0\Delta E_{21}=E_{2}-E_{1}>0. The state of the system is passive, meaning that no energy can be extracted with unitary operations, and we can write it as a classical state

ρP=p0|00|P+p1|11|P+p2|22|P,\rho_{P}=\mathrm{p}_{0}\ket{0}\bra{0}_{P}+\mathrm{p}_{1}\ket{1}\bra{1}_{P}+\mathrm{p}_{2}\ket{2}\bra{2}_{P}, (40)

where p0p1p2\mathrm{p}_{0}\geq\mathrm{p}_{1}\geq\mathrm{p}_{2} (which is a direct consequence of the no-energy-extraction condition).

The machine we introduce is a dd-level system with a trivial Hamiltonian, described by the state

ρM=j=0d1qj|jj|M.\rho_{M}=\sum_{j=0}^{d-1}\mathrm{q}_{j}\ket{j}\bra{j}_{M}. (41)

We operate over system and machine with a unitary operation composed by multiple swaps. In particular, we first perform m1m-1 swaps between the pair of states (|0P,|1P)\left(\ket{0}_{P},\ket{1}_{P}\right) and the pairs {(|jM,|j+1M)}j=0m2\left\{\left(\ket{j}_{M},\ket{j+1}_{M}\right)\right\}_{j=0}^{m-2}, followed by a swap between the same pair of states of the system and the pair (|m1M,|m+n1M)\left(\ket{m-1}_{M},\ket{m+n-1}_{M}\right) of the machine. Then, we perform n1n-1 swaps between the pair (|1P,|2P)\left(\ket{1}_{P},\ket{2}_{P}\right) and the pairs {(|jM,|j+1M)}j=mm+n2\left\{\left(\ket{j}_{M},\ket{j+1}_{M}\right)\right\}_{j=m}^{m+n-2}, followed by a swap between the same system’s states and the pair (|0M,|mM)\left(\ket{0}_{M},\ket{m}_{M}\right). In order to perform this cycle, the dimension of the catalyst has to be at least equal to m+nm+n, and indeed in the following we fix d=m+nd=m+n. The unitary we want to apply is

Sm,n=S(1,2)(0,m)S(1,2)(m,m+1)S(1,2)(m+1,m+2)S(1,2)(m+n2,m+n1)S(0,1)(m1,m+n1)S(0,1)(m2,m1)S(0,1)(m3,m2)S(0,1)(0,1),S_{m,n}=S_{(1,2)}^{(0,m)}\circ S_{(1,2)}^{(m,m+1)}\circ S_{(1,2)}^{(m+1,m+2)}\circ\ldots\circ S_{(1,2)}^{(m+n-2,m+n-1)}\circ S_{(0,1)}^{(m-1,m+n-1)}\circ S_{(0,1)}^{(m-2,m-1)}\circ S_{(0,1)}^{(m-3,m-2)}\circ\ldots\circ S_{(0,1)}^{(0,1)}, (42)

where the operation S(a,b)(c,d)S_{(a,b)}^{(c,d)} is a swap between system and machine, performing the permutation |aP|dM|bP|cM\ket{a}_{P}\ket{d}_{M}\leftrightarrow\ket{b}_{P}\ket{c}_{M}.

For the given unitary evolution we can easily evaluate the final state of the global system. This final state presents classical correlations between system and machine, but in the following we only consider the marginal states for system and machine, which are the sole information we need. In fact, the energy of the global system solely depends on the Hamiltonian HPH_{P} of the system (and therefore only on the local state of the system), as the machine has a trivial Hamiltonian, and we do not have an interaction term HintH_{\text{int}}. Moreover, in order for the machine to be re-usable on a new system, we only need its local initial and final states to be equal, and the correlations with the old systems do not affect the engine. The final state of the system is

ρ~P\displaystyle\tilde{\rho}_{P} =TrM[Sm,n(ρPρM)Sm,n]\displaystyle=\mathrm{Tr}_{M}\left[{S_{m,n}\left(\rho_{P}\otimes\rho_{M}\right)S_{m,n}^{\dagger}}\right]
=(p0+j=1m1(p1qj1p0qj)+(p1qm1p0qm+n1))|00|P\displaystyle=\left(\mathrm{p}_{0}+\sum_{j=1}^{m-1}\left(\mathrm{p}_{1}\mathrm{q}_{j-1}-\mathrm{p}_{0}\mathrm{q}_{j}\right)+\left(\mathrm{p}_{1}\mathrm{q}_{m-1}-\mathrm{p}_{0}\mathrm{q}_{m+n-1}\right)\right)\ket{0}\bra{0}_{P}
+(p1j=1m1(p1qj1p0qj)(p1qm1p0qm+n1)j=m+1m+n1(p1qjp2qj1)(p1qmp2q0))|11|P\displaystyle+\left(\mathrm{p}_{1}-\sum_{j=1}^{m-1}\left(\mathrm{p}_{1}\mathrm{q}_{j-1}-\mathrm{p}_{0}\mathrm{q}_{j}\right)-\left(\mathrm{p}_{1}\mathrm{q}_{m-1}-\mathrm{p}_{0}\mathrm{q}_{m+n-1}\right)-\sum_{j=m+1}^{m+n-1}\left(\mathrm{p}_{1}\mathrm{q}_{j}-\mathrm{p}_{2}\mathrm{q}_{j-1}\right)-\left(\mathrm{p}_{1}\mathrm{q}_{m}-\mathrm{p}_{2}\mathrm{q}_{0}\right)\right)\ket{1}\bra{1}_{P}
+(p2+j=m+1m+n1(p1qjp2qj1)+(p1qmp2q0))|22|P,\displaystyle+\left(\mathrm{p}_{2}+\sum_{j=m+1}^{m+n-1}\left(\mathrm{p}_{1}\mathrm{q}_{j}-\mathrm{p}_{2}\mathrm{q}_{j-1}\right)+\left(\mathrm{p}_{1}\mathrm{q}_{m}-\mathrm{p}_{2}\mathrm{q}_{0}\right)\right)\ket{2}\bra{2}_{P}, (43)

while the final state of the machine is

ρ~M\displaystyle\tilde{\rho}_{M} =TrP[Sm,n(ρPρM)Sm,n]\displaystyle=\mathrm{Tr}_{P}\left[{S_{m,n}\left(\rho_{P}\otimes\rho_{M}\right)S_{m,n}^{\dagger}}\right]
=(p0q0+p0q1+p1qm)|00|M+j=1m2(p1qj1+p0qj+1+p2qj)|jj|M\displaystyle=\left(\mathrm{p}_{0}\mathrm{q}_{0}+\mathrm{p}_{0}\mathrm{q}_{1}+\mathrm{p}_{1}\mathrm{q}_{m}\right)\ket{0}\bra{0}_{M}+\sum_{j=1}^{m-2}\left(\mathrm{p}_{1}\mathrm{q}_{j-1}+\mathrm{p}_{0}\mathrm{q}_{j+1}+\mathrm{p}_{2}\mathrm{q}_{j}\right)\ket{j}\bra{j}_{M}
+(p1qm2+p0qm+n1+p2qm1)|m1m1|M+(p0qm+p2q0+p1qm+1)|mm|M\displaystyle+\left(\mathrm{p}_{1}\mathrm{q}_{m-2}+\mathrm{p}_{0}\mathrm{q}_{m+n-1}+\mathrm{p}_{2}\mathrm{q}_{m-1}\right)\ket{m-1}\bra{m-1}_{M}+\left(\mathrm{p}_{0}\mathrm{q}_{m}+\mathrm{p}_{2}\mathrm{q}_{0}+\mathrm{p}_{1}\mathrm{q}_{m+1}\right)\ket{m}\bra{m}_{M}
+j=m+1m+n2(p0qj+p2qj1+p1qj+1)|jj|M+(p1qm1+p2qm+n2+p2qm+n1)|m+n1m+n1|M.\displaystyle+\sum_{j=m+1}^{m+n-2}\left(\mathrm{p}_{0}\mathrm{q}_{j}+\mathrm{p}_{2}\mathrm{q}_{j-1}+\mathrm{p}_{1}\mathrm{q}_{j+1}\right)\ket{j}\bra{j}_{M}+\left(\mathrm{p}_{1}\mathrm{q}_{m-1}+\mathrm{p}_{2}\mathrm{q}_{m+n-2}+\mathrm{p}_{2}\mathrm{q}_{m+n-1}\right)\ket{m+n-1}\bra{m+n-1}_{M}. (44)

As we stated above, in order for the machine to be re-usable we need its final local state ρ~M\tilde{\rho}_{M} to be equal to the initial one ρM\rho_{M}. Correlations with the system do not invalidate the re-usability, as we always discard the system after the cycle, and we take a new copy to repeat the process. In this way, we can extract work from a reservoir of passive states by acting on them individually. The constraint of an equal initial and final state of the machine provides the following set of equalities,

q0\displaystyle\mathrm{q}_{0} =p0q0+p0q1+p1qm\displaystyle=\mathrm{p}_{0}\mathrm{q}_{0}+\mathrm{p}_{0}\mathrm{q}_{1}+\mathrm{p}_{1}\mathrm{q}_{m} (45a)
qj\displaystyle\mathrm{q}_{j} =p1qj1+p0qj+1+p2qj;j=1,,m2\displaystyle=\mathrm{p}_{1}\mathrm{q}_{j-1}+\mathrm{p}_{0}\mathrm{q}_{j+1}+\mathrm{p}_{2}\mathrm{q}_{j}\quad;\quad j=1,\ldots,m-2 (45b)
qm1\displaystyle\mathrm{q}_{m-1} =p1qm2+p0qm+n1+p2qm1\displaystyle=\mathrm{p}_{1}\mathrm{q}_{m-2}+\mathrm{p}_{0}\mathrm{q}_{m+n-1}+\mathrm{p}_{2}\mathrm{q}_{m-1} (45c)
qm\displaystyle\mathrm{q}_{m} =p0qm+p2q0+p1qm+1\displaystyle=\mathrm{p}_{0}\mathrm{q}_{m}+\mathrm{p}_{2}\mathrm{q}_{0}+\mathrm{p}_{1}\mathrm{q}_{m+1} (45d)
qj\displaystyle\mathrm{q}_{j} =p0qj+p2qj1+p1qj+1;j=m+1,,m+n2\displaystyle=\mathrm{p}_{0}\mathrm{q}_{j}+\mathrm{p}_{2}\mathrm{q}_{j-1}+\mathrm{p}_{1}\mathrm{q}_{j+1}\quad;\quad j=m+1,\ldots,m+n-2 (45e)
qm+n1\displaystyle\mathrm{q}_{m+n-1} =p1qm1+p2qm+n2+p2qm+n1,\displaystyle=\mathrm{p}_{1}\mathrm{q}_{m-1}+\mathrm{p}_{2}\mathrm{q}_{m+n-2}+\mathrm{p}_{2}\mathrm{q}_{m+n-1}, (45f)

which, if solved, allow for the probability distribution of the state of the machine to be expressed in terms of the passive state ρP\rho_{P}.

A.1 Work extracted and activable passive states

In our framework, we do not explicitly account for a battery, that is, an additional system with a specific Hamiltonian, able to account for any energy exchange between system and machine. Instead, we implicitly assume the battery to be present, so that any change in the average energy of the system is thought as some energy flowing from (or to) the battery. In particular, if the average energy of the system decreases, then the battery is storing this energy, while when the average energy of the system increases, the battery is providing it. All the energy coming from (or going to) the battery is accounted as work. Under this assumptions, the amount of work we extract during one cycle is given by the changing in the average energy of the system, that is

ΔW=TrP[HP(ρPρ~P)],\Delta W=\mathrm{Tr}_{P}\left[{H_{P}\left(\rho_{P}-\tilde{\rho}_{P}\right)}\right], (46)

where ρP\rho_{P} is the initial passive state, and ρ~P\tilde{\rho}_{P} is the final state, whose probability distribution is {p0,p1,p2}\left\{\mathrm{p}^{\prime}_{0},\mathrm{p}^{\prime}_{1},\mathrm{p}^{\prime}_{2}\right\}. We can express the amount of extracted work in terms of the energy gaps of the Hamiltonian HPH_{P}, as

ΔW=ΔE10(p0p0)ΔE21(p2p2),\Delta W=\Delta E_{10}\left(\mathrm{p}^{\prime}_{0}-\mathrm{p}_{0}\right)-\Delta E_{21}\left(\mathrm{p}^{\prime}_{2}-\mathrm{p}_{2}\right), (47)

where this expression has been obtained by applying the normalisation constraint to the initial and final state of the system.

If we replace the probability distribution of the final state of the system, Eq. (A), into the expression of extracted work, Eq. (47), we obtain that

ΔW=ΔE10(j=1m1(p1qj1p0qj)+(p1qm1p0qm+n1))ΔE21(j=m+1m+n1(p1qjp2qj1)+(p1qmp2q0)).\Delta W=\Delta E_{10}\left(\sum_{j=1}^{m-1}\left(\mathrm{p}_{1}\mathrm{q}_{j-1}-\mathrm{p}_{0}\mathrm{q}_{j}\right)+\left(\mathrm{p}_{1}\mathrm{q}_{m-1}-\mathrm{p}_{0}\mathrm{q}_{m+n-1}\right)\right)-\Delta E_{21}\left(\sum_{j=m+1}^{m+n-1}\left(\mathrm{p}_{1}\mathrm{q}_{j}-\mathrm{p}_{2}\mathrm{q}_{j-1}\right)+\left(\mathrm{p}_{1}\mathrm{q}_{m}-\mathrm{p}_{2}\mathrm{q}_{0}\right)\right). (48)

This expression can be highly simplified if we use the properties of the probability distribution of the machine, Eqs. (45). In particular, from Eq. (45b) we find that

p1qj1p0qj=p1qjp0qj+1;j=1,,m2,\mathrm{p}_{1}\mathrm{q}_{j-1}-\mathrm{p}_{0}\mathrm{q}_{j}=\mathrm{p}_{1}\mathrm{q}_{j}-\mathrm{p}_{0}\mathrm{q}_{j+1}\quad;\quad\forall\,j=1,\ldots,m-2, (49)

while from (45c) we have that

p1qm2p0qm1=p1qm1p0qm+n1.\mathrm{p}_{1}\mathrm{q}_{m-2}-\mathrm{p}_{0}\mathrm{q}_{m-1}=\mathrm{p}_{1}\mathrm{q}_{m-1}-\mathrm{p}_{0}\mathrm{q}_{m+n-1}. (50)

Together, these equations reduce the first bracket of Eq. (48) into a single term,

j=1m1(p1qj1p0qj)+(p1qm1p0qm+n1)=m(p1qm1p0qm+n1).\sum_{j=1}^{m-1}\left(\mathrm{p}_{1}\mathrm{q}_{j-1}-\mathrm{p}_{0}\mathrm{q}_{j}\right)+\left(\mathrm{p}_{1}\mathrm{q}_{m-1}-\mathrm{p}_{0}\mathrm{q}_{m+n-1}\right)=m\left(\mathrm{p}_{1}\mathrm{q}_{m-1}-\mathrm{p}_{0}\mathrm{q}_{m+n-1}\right). (51)

If we consider Eq. (45e), instead, we find that

p1qjp2qj1=p1qj+1p2qj;j=m+1,,m+n2,\mathrm{p}_{1}\mathrm{q}_{j}-\mathrm{p}_{2}\mathrm{q}_{j-1}=\mathrm{p}_{1}\mathrm{q}_{j+1}-\mathrm{p}_{2}\mathrm{q}_{j}\quad;\quad\forall\,j=m+1,\ldots,m+n-2, (52)

while Eq. (45d) implies that

p1qm+1p2qm=p1qmp2q0.\mathrm{p}_{1}\mathrm{q}_{m+1}-\mathrm{p}_{2}\mathrm{q}_{m}=\mathrm{p}_{1}\mathrm{q}_{m}-\mathrm{p}_{2}\mathrm{q}_{0}. (53)

These two equations simplify the second bracket of Eq. (48),

j=m+1m+n1(p1qjp2qj1)+(p1qmp2q0)=n(p1qm+n1p2qm+n2).\sum_{j=m+1}^{m+n-1}\left(\mathrm{p}_{1}\mathrm{q}_{j}-\mathrm{p}_{2}\mathrm{q}_{j-1}\right)+\left(\mathrm{p}_{1}\mathrm{q}_{m}-\mathrm{p}_{2}\mathrm{q}_{0}\right)=n\left(\mathrm{p}_{1}\mathrm{q}_{m+n-1}-\mathrm{p}_{2}\mathrm{q}_{m+n-2}\right). (54)

We can now use Eq. (45f) to show that

p1qm1p0qm+n1=p1qm+n1p2qm+n2,\mathrm{p}_{1}\mathrm{q}_{m-1}-\mathrm{p}_{0}\mathrm{q}_{m+n-1}=\mathrm{p}_{1}\mathrm{q}_{m+n-1}-\mathrm{p}_{2}\mathrm{q}_{m+n-2}, (55)

which allows us to express the work we extract as

ΔW=(mΔE10nΔE21)(p1qm+n1p2qm+n2).\Delta W=\left(m\,\Delta E_{10}-n\,\Delta E_{21}\right)\left(\mathrm{p}_{1}\mathrm{q}_{m+n-1}-\mathrm{p}_{2}\mathrm{q}_{m+n-2}\right). (56)

From the above equation we notice that the work extracted is factorised into an Hamiltonian contribution and another contribution associated with the probability distribution of the passive state. Then, for a given Hamiltonian HPH_{P} such that mΔE10>nΔE21m\,\Delta E_{10}>n\,\Delta E_{21}, we will find that certain passive states allow for work extraction (the ones in which p1qm+n1>p2qm+n2\mathrm{p}_{1}\mathrm{q}_{m+n-1}>\mathrm{p}_{2}\mathrm{q}_{m+n-2}), while others do not. Therefore, for every given Hamiltonian (that is, every ΔE10\Delta E_{10} and ΔE21\Delta E_{21}) and for every given cycle (that is, every nn and mm), we find that the set of passive states is divided into two subsets, the ones which allow for work extraction (we can call them activable states), and the ones which do not. In the following we will express the probability distribution of ρM\rho_{M} in terms of the probability distribution of the passive state, so as to define these two subsets for each Hamiltonian and cycle.

As a first step, we want to express the first m2m-2 elements of the sequence {qj}j=0m1\left\{q_{j}\right\}_{j=0}^{m-1} in terms of last two elements, qm2\mathrm{q}_{m-2} and qm1\mathrm{q}_{m-1}. Moreover, we express the first n2n-2 elements of {qj}j=mm+n1\left\{q_{j}\right\}_{j=m}^{m+n-1} in terms of qm+n2\mathrm{q}_{m+n-2} and qm+n1\mathrm{q}_{m+n-1}. This can be done by utilising the equalities given in Eqs. (45b) and (45e), which we recast in the following way.

qj\displaystyle\mathrm{q}_{j} =(1+p0p1)qj+1p0p1qj+2;j=0,,m3,\displaystyle=\left(1+\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)\mathrm{q}_{j+1}-\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\,\mathrm{q}_{j+2}\quad;\quad\forall\,j=0,\ldots,m-3, (57a)
qj\displaystyle\mathrm{q}_{j} =(1+p1p2)qj+1p1p2qj+2;j=m,,m+n3.\displaystyle=\left(1+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)\mathrm{q}_{j+1}-\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\,\mathrm{q}_{j+2}\quad;\quad\forall\,j=m,\ldots,m+n-3. (57b)

It can be proved (see the technical result 1) that the elements of the sequences can be expressed as

qj\displaystyle q_{j} =T1(m(j+2))qm2p0p1T1(m(j+3))qm1;j=0,,m3,\displaystyle=\mathrm{T}_{1}\left(m-(j+2)\right)\mathrm{q}_{m-2}-\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\,\mathrm{T}_{1}\left(m-(j+3)\right)\mathrm{q}_{m-1}\quad;\quad\forall\,j=0,\ldots,m-3, (58a)
qj\displaystyle q_{j} =T2(m+n(j+2))qm+n2p1p2T2(m+n(j+3))qm+n1;j=m,,m+n3,\displaystyle=\mathrm{T}_{2}\left(m+n-(j+2)\right)\mathrm{q}_{m+n-2}-\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\,\mathrm{T}_{2}\left(m+n-(j+3)\right)\mathrm{q}_{m+n-1}\quad;\quad\forall\,j=m,\ldots,m+n-3, (58b)

where T1(h)=l=0h(p0p1)l\mathrm{T}_{1}(h)=\sum_{l=0}^{h}\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{l} and T2(h)=l=0h(p1p2)l\mathrm{T}_{2}(h)=\sum_{l=0}^{h}\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{l}.

We can now express, using Eqs. (45c) and (45f), the elements qm2\mathrm{q}_{m-2} and qm1\mathrm{q}_{m-1} in terms of qm+n2\mathrm{q}_{m+n-2} and qm+n2\mathrm{q}_{m+n-2}. From Eq. (45c) we obtain that

qm2=T1(2)qm+n1p2p1T1(1)qm+n2.\mathrm{q}_{m-2}=\mathrm{T}_{1}(2)\,\mathrm{q}_{m+n-1}-\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\,\mathrm{T}_{1}(1)\,\mathrm{q}_{m+n-2}. (59)

From Eq. (45f), instead, we get that

qm1=T1(1)qm+n1p2p1T1(0)qm+n2.\mathrm{q}_{m-1}=\mathrm{T}_{1}(1)\,\mathrm{q}_{m+n-1}-\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\,\mathrm{T}_{1}(0)\,\mathrm{q}_{m+n-2}. (60)

Then, we can finally express qm+n2\mathrm{q}_{m+n-2} in terms of qm+n1\mathrm{q}_{m+n-1} through Eq. (45d), and we obtain

qm+n2=D(m,n)qm+n1,\mathrm{q}_{m+n-2}=D(m,n)\,\mathrm{q}_{m+n-1}, (61)

where the coefficient D(m,n)D(m,n) is defined as

D(m,n)=p1p2T1(m)+p1p2T2(n2)T1(m1)+p1p2T2(n1).D(m,n)=\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\frac{\mathrm{T}_{1}(m)+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\mathrm{T}_{2}(n-2)}{\mathrm{T}_{1}(m-1)+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\mathrm{T}_{2}(n-1)}. (62)

Thanks to the above result, we can express the overall probability distribution of ρM\rho_{M} in terms of the occupation probability of the state |m+n1M\ket{m+n-1}_{M}. Thus, we have that

qj\displaystyle\mathrm{q}_{j} =(T1(mj)p2p1D(m,n)T1(m(j+1)))qm+n1;j=0,,m1,\displaystyle=\left(\mathrm{T}_{1}(m-j)-\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\,D(m,n)\,\mathrm{T}_{1}\left(m-(j+1)\right)\right)\mathrm{q}_{m+n-1}\quad;\quad j=0,\ldots,m-1, (63a)
qj\displaystyle\mathrm{q}_{j} =(T2(m+n(j+2))D(m,n)p1p2T2(m+n(j+3)))qm+n1;j=m,,m+n3,\displaystyle=\left(\mathrm{T}_{2}\left(m+n-(j+2)\right)\,D(m,n)-\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\,\mathrm{T}_{2}\left(m+n-(j+3)\right)\right)\mathrm{q}_{m+n-1}\quad;\quad j=m,\ldots,m+n-3, (63b)
qm+n2\displaystyle\mathrm{q}_{m+n-2} =D(m,n)qm+n1,\displaystyle=D(m,n)\,\mathrm{q}_{m+n-1}, (63c)

where it is possible to show that each qj\mathrm{q}_{j}, with j=0,,m+n2j=0,\ldots,m+n-2, is positive if qm+n1\mathrm{q}_{m+n-1} is positive (see the technical result 2). From the normalisation condition it then follows that the sequence {qj}j=0m+n1\left\{q_{j}\right\}_{j=0}^{m+n-1} is a proper probability distribution. Moreover, the normalisation condition allows us to evaluate qm+n1\mathrm{q}_{m+n-1} as a function of the probability distribution of the passive state ρP\rho_{P},

qm+n1=T1(m1)+p1p2T2(n1)(T1(m)+p1p2T2(n2))2+((p1p2)n(p0p1)m)(j=0mT1(j)p1p2j=0n3T2(j)).\mathrm{q}_{m+n-1}=\frac{\mathrm{T}_{1}(m-1)+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\mathrm{T}_{2}(n-1)}{\left(\mathrm{T}_{1}(m)+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\mathrm{T}_{2}(n-2)\right)^{2}+\left(\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n}-\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\right)\left(\sum_{j=0}^{m}\mathrm{T}_{1}(j)-\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\sum_{j=0}^{n-3}\mathrm{T}_{2}(j)\right)}. (64)

From Eq. (64) we can express all the other elements of {qj}j=0m+n1\left\{q_{j}\right\}_{j=0}^{m+n-1} in terms of the probability distribution of ρP\rho_{P}.

We can now further characterise the amount of work extracted during our cycle. In fact, if we apply Eq. (63c) into Eq. (56), we obtain

ΔW=(mΔE10nΔE21)p1((p1p2)n(p0p1)m)T1(m1)+p1p2T2(n1)qm+n1,\Delta W=\left(m\,\Delta E_{10}-n\,\Delta E_{21}\right)\frac{\mathrm{p}_{1}\left(\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n}-\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\right)}{\mathrm{T}_{1}(m-1)+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\mathrm{T}_{2}(n-1)}\,\mathrm{q}_{m+n-1}, (65)

where the sign of ΔW\Delta W depends on the sole terms (mΔE10nΔE21)\left(m\,\Delta E_{10}-n\,\Delta E_{21}\right) and ((p1p2)n(p0p1)m)\left(\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n}-\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\right), since the other factors are always positive. Thus, for each cycle, we can characterise which passive states can be activated by that cycle, that is, which states allow for work extraction during the cycle. The subset of activable states is

Rm,n+={ρPpassive|\displaystyle R^{+}_{m,n}=\bigg{\{}\rho_{P}\ \text{passive}\ \bigg{|}\ (p1p2)n>(p0p1)mwhenmΔE10nΔE21>0\displaystyle\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n}>\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\text{when}\ m\,\Delta E_{10}-n\,\Delta E_{21}>0
\displaystyle\vee\ (p1p2)n<(p0p1)mwhenmΔE10nΔE21<0},\displaystyle\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n}<\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\text{when}\ m\,\Delta E_{10}-n\,\Delta E_{21}<0\bigg{\}}, (66)

where this region clearly depends on the Hamiltonian of the system HPH_{P}, and on the number of swaps performed during the cycle, mm and nn.

A.2 The final state of the system

Let us consider the final state of the passive system after we have applied the cycle Sm,nS_{m,n}. In Eq. (A) we have shown the probability distribution of ρ~P\tilde{\rho}_{P} as a function of {qi}i=0m+n1\left\{\mathrm{q}_{i}\right\}_{i=0}^{m+n-1}. Thanks to the constraints introduced in Eqs. (45), we can simplify the form of ρ~P\tilde{\rho}_{P}, so that we obtain

p0\displaystyle\mathrm{p}^{\prime}_{0} =p0+mΔP,\displaystyle=\mathrm{p}_{0}+m\,\Delta\mathrm{P}, (67a)
p1\displaystyle\mathrm{p}^{\prime}_{1} =p1(m+n)ΔP,\displaystyle=\mathrm{p}_{1}-\left(m+n\right)\,\Delta\mathrm{P}, (67b)
p2\displaystyle\mathrm{p}^{\prime}_{2} =p2+nΔP.\displaystyle=\mathrm{p}_{2}+n\,\Delta\mathrm{P}. (67c)

We can easily notice that the cycle acts on the passive state by modifying the original probabilities by multiples of

ΔP=p1qm+n1T1(m1)+p1p2T2(n1)((p1p2)n(p0p1)m).\Delta\mathrm{P}=\frac{\mathrm{p}_{1}\,\mathrm{q}_{m+n-1}}{\mathrm{T}_{1}(m-1)+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\mathrm{T}_{2}(n-1)}\,\left(\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n}-\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\right). (68)

The expression of the final state ρ~P\tilde{\rho}_{P} allows us to understand how the cycle operates over the system when work is extracted. In particular, we can consider the evolution of the system in two different situations, linked to the two possible scenarios of Eq. (A.1).

Suppose that HPH_{P} is such that mΔE10>nΔE21m\,\Delta E_{10}>n\,\Delta E_{21}. Then, from the conditions in Rm,n+R^{+}_{m,n}, we can verify that ΔP>0\Delta\mathrm{P}>0, so that the map is depleting the population of the state |1P\ket{1}_{P}, while increasing the populations of both |0P\ket{0}_{P} and |2P\ket{2}_{P} (see Fig. 6, left plot). Work is extracted from the cycle since the energy gained while moving mΔPm\,\Delta\mathrm{P} from p1\mathrm{p}_{1} to p0\mathrm{p}_{0} is bigger than the energy paid to move nΔPn\,\Delta\mathrm{P} from p1\mathrm{p}_{1} to p2\mathrm{p}_{2}. In Sec. C, we show that the entropy of the system has to increase during the transformation. This is achieved since p1\mathrm{p}_{1} gets closer to p2\mathrm{p}_{2} after the cycle.

Refer to caption
Figure 6: (Left) The action of the cycle on the system, when work is extracted and the Hamiltonian HPH_{P} is such that mΔE10>nΔE21m\,\Delta E_{10}>n\,\Delta E_{21}. The probability of occupation of |1P\ket{1}_{P} is reduced, while the other two probabilities of occupation increase. (Right) The action of the cycle on the system, when work is extracted and the Hamiltonian HPH_{P} is such that mΔE10<nΔE21m\,\Delta E_{10}<n\,\Delta E_{21}. The map acts on the system in the opposite way compared to the previous scenario.

Let us consider the case in which HPH_{P} is such that mΔE10<nΔE21m\,\Delta E_{10}<n\,\Delta E_{21}. Then, from the conditions in Rm,n+R^{+}_{m,n}, we can verify that ΔP<0\Delta\mathrm{P}<0, so that the map is depleting the populations of the states |0P\ket{0}_{P} and |2P\ket{2}_{P}, while increasing the populations of |1P\ket{1}_{P} (see Fig. 6, right plot). Work is extracted from the cycle since the energy gained while moving nΔPn\,\Delta\mathrm{P} from p2\mathrm{p}_{2} to p1\mathrm{p}_{1} is bigger than the energy paid to move mΔPm\,\Delta\mathrm{P} from p0\mathrm{p}_{0} to p1\mathrm{p}_{1}. Moreover, the entropy of the system increases since p0\mathrm{p}_{0} gets closer to p1\mathrm{p}_{1} after the cycle.

It is worth noting that the final state ρ~P\tilde{\rho}_{P} can be active. This happen, in the case of mΔE10>nΔE21m\,\Delta E_{10}>n\,\Delta E_{21}, when p1<p2\mathrm{p}^{\prime}_{1}<\mathrm{p}^{\prime}_{2}. In the other case, we obtain a final active state if p0<p1\mathrm{p}^{\prime}_{0}<\mathrm{p}^{\prime}_{1}. In these situations, not only are we able to extract work from the passive state ρP\rho_{P} during the cycle, but we can also perform a local unitary operation (permuting |1P\ket{1}_{P} and |2P\ket{2}_{P} in the first case, and |0P\ket{0}_{P} and |1P\ket{1}_{P} in the second) which allows for additional work extraction. It is also possible for the final state of the system to be passive, and to still lie inside the activable region Rm,n+R^{+}_{m,n}. Due to the correlation created between system and machine, however, this state cannot be used again, at least not with the same machine.

Appendix B Asymptotic behaviour of the machine

We are now interested in the study of the cycle Sm,nS_{m,n} when the size of the machine (as well as the number of hot and cold swaps) tends to infinity. In particular, we are interested in the form of the probability distribution of the machine, the work extracted, and the final state of the passive system. Let us consider the Hamiltonian HPH_{P} of the main (qutrit) system. We know that, for any Hamiltonian HPH_{P}, there exists two integer numbers NN and MM such that MΔE10=NΔE21M\,\Delta E_{10}=N\,\Delta E_{21}. We now consider a passive state ρP\rho_{P} describing this system whose probability distribution satisfies (p1/p2)N>(p0/p1)M\left(\mathrm{p}_{1}/\mathrm{p}_{2}\right)^{N}>\left(\mathrm{p}_{0}/\mathrm{p}_{1}\right)^{M}. Notice that this condition implies that the state is in the subset of passive states denoted by R1R_{1} (see the main text, Fig. 5), or equivalently it implies that the hot virtual temperature is associated with the pair of states |0P\ket{0}_{P} and |1P\ket{1}_{P}. One could analyse the opposite situation as well, but the results we obtain would be analogous, due to the symmetry of the problem with respect to the hot and cold interactions.

In order to perform the asymptotic expansion of the probability distribution of the machine, Eqs. (63), we first want to define how the ratio nm\frac{n}{m} behaves as the number of hot and cold swaps goes to infinity. We set this fraction equal to α\alpha, so that n=αmn=\alpha\,m, and we define a range for this parameter, due to the constraints we set on the passive state. Indeed, if we want to extract work, we need mm and nn to satisfy one of the two conditions in Eq. (A.1), and in particular, since we assume the passive state to be in the region R1R_{1}, we need mΔE10>nΔE21m\,\Delta E_{10}>n\,\Delta E_{21} and (p1/p2)n>(p0/p1)m\left(\mathrm{p}_{1}/\mathrm{p}_{2}\right)^{n}>\left(\mathrm{p}_{0}/\mathrm{p}_{1}\right)^{m}. The two inequalities implies that

logp0p1logp1p2<α<ΔE10ΔE21,\frac{\log\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}}{\log\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}}<\alpha<\frac{\Delta E_{10}}{\Delta E_{21}}, (69)

where it is easy to verify that the lower bound is smaller than the upper one, due to the fact that ρPR1\rho_{P}\in R_{1}.

We can now use the assumptions made on the cycle (that is, on the parameters mm and nn) and on the initial passive state in order to expand the probability distribution of ρM\rho_{M} for m,nm,n\rightarrow\infty. As a first step, let us consider the coefficient D(m,n)D(m,n) presented in Eq. (62). When mm and nn tends to infinity, we find that

D(m,n)1+(p0p1)m(p2p1)n(p0p2)(p1p2)(p0p1)+O((p2p1)n;(p0p1)2m(p2p1)2n),D(m,n)\approx 1+\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{n}\frac{\left(\mathrm{p}_{0}-\mathrm{p}_{2}\right)\left(\mathrm{p}_{1}-\mathrm{p}_{2}\right)}{\left(\mathrm{p}_{0}-\mathrm{p}_{1}\right)}+O\left(\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{n};\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{2m}\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{2n}\right), (70)

where it is easy to verify that the term (p0/p1)m(p2/p1)n0\left(\mathrm{p}_{0}/\mathrm{p}_{1}\right)^{m}\left(\mathrm{p}_{2}/\mathrm{p}_{1}\right)^{n}\rightarrow 0 as m,nm,n\rightarrow\infty, and that both (p2/p1)n\left(\mathrm{p}_{2}/\mathrm{p}_{1}\right)^{n} and (p0/p1)2m(p2/p1)2n\left(\mathrm{p}_{0}/\mathrm{p}_{1}\right)^{2m}\left(\mathrm{p}_{2}/\mathrm{p}_{1}\right)^{2n} tends to 0 faster that this first term. However, we cannot say which one is the fastest without further assumptions, and that is the reason we keep both in the OO.

Once the expansion of D(m,n)D(m,n) is known, we can focus on the probability distribution of the machine. For simplicity, we consider the distribution in Eqs. (63), where qm+n1\mathrm{q}_{m+n-1} is not defined yet; we will define it through the normalisation condition once the asymptotic expansion has been performed. We find that

qj\displaystyle\mathrm{q}_{j} qm+n1(p0p2p0p1+O((p0p1)m(p2p1)n))(p0p1)mj;j=0,,m1,\displaystyle\approx\mathrm{q}_{m+n-1}\left(\frac{\mathrm{p}_{0}-\mathrm{p}_{2}}{\mathrm{p}_{0}-\mathrm{p}_{1}}+O\left(\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{n}\right)\right)\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m-j}\quad;\quad j=0,\ldots,m-1, (71a)
qj\displaystyle\mathrm{q}_{j} qm+n1(p2p1p0p2p0p1+O((p0p1)m(p2p1)n))(p0p1)m(p1p2)mj;j=m,,m+n3,\displaystyle\approx\mathrm{q}_{m+n-1}\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\frac{\mathrm{p}_{0}-\mathrm{p}_{2}}{\mathrm{p}_{0}-\mathrm{p}_{1}}+O\left(\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{n}\right)\right)\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{m-j}\quad;\quad j=m,\ldots,m+n-3, (71b)
qm+n2\displaystyle\mathrm{q}_{m+n-2} qm+n1(1+O((p0p1)m(p2p1)n)).\displaystyle\approx\mathrm{q}_{m+n-1}\left(1+O\left(\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{n}\right)\right). (71c)

We are now able to obtain the value of qm+n1\mathrm{q}_{m+n-1} by imposing the normalisation condition over the asymptotic probability distribution of the machine. We find that

qm+n1((p1p2)(p0p1)2p1(p0p2)2+O((p0p1)m(p2p1)n))(p1p0)m,\mathrm{q}_{m+n-1}\approx\left(\frac{\left(\mathrm{p}_{1}-\mathrm{p}_{2}\right)\left(\mathrm{p}_{0}-\mathrm{p}_{1}\right)^{2}}{\mathrm{p}_{1}\left(\mathrm{p}_{0}-\mathrm{p}_{2}\right)^{2}}+O\left(\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{n}\right)\right)\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{0}}\right)^{m}, (72)

that is, qm+n1\mathrm{q}_{m+n-1} tends to 0 as (p1/p0)m\left(\mathrm{p}_{1}/\mathrm{p}_{0}\right)^{m} for mm\rightarrow\infty. Notice that the same result can be obtained by expanding Eq. (64). If we send mm and nn to infinity, we find that the asymptotic probability distribution of the machine is

qj\displaystyle\mathrm{q}_{j} (p1p2)(p0p1)p1(p0p2)(p0p1)j;j=0,,m1,\displaystyle\approx\frac{\left(\mathrm{p}_{1}-\mathrm{p}_{2}\right)\left(\mathrm{p}_{0}-\mathrm{p}_{1}\right)}{\mathrm{p}_{1}\left(\mathrm{p}_{0}-\mathrm{p}_{2}\right)}\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{-j}\quad;\quad j=0,\ldots,m-1, (73a)
qj\displaystyle\mathrm{q}_{j} p2(p1p2)(p0p1)p12(p0p2)(p1p2)mj;j=m,,m+n3,\displaystyle\approx\frac{\mathrm{p}_{2}\left(\mathrm{p}_{1}-\mathrm{p}_{2}\right)\left(\mathrm{p}_{0}-\mathrm{p}_{1}\right)}{\mathrm{p}_{1}^{2}\left(\mathrm{p}_{0}-\mathrm{p}_{2}\right)}\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{m-j}\quad;\quad j=m,\ldots,m+n-3, (73b)
qm+n2qm+n1\displaystyle\mathrm{q}_{m+n-2}\approx\mathrm{q}_{m+n-1} (p1p2)(p0p1)2p1(p0p2)2(p1p0)m.\displaystyle\approx\frac{\left(\mathrm{p}_{1}-\mathrm{p}_{2}\right)\left(\mathrm{p}_{0}-\mathrm{p}_{1}\right)^{2}}{\mathrm{p}_{1}\left(\mathrm{p}_{0}-\mathrm{p}_{2}\right)^{2}}\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{0}}\right)^{m}. (73c)

We can now investigate how the probability distribution of the main system changes, and evaluate the asymptotic work extracted ΔW\Delta W during on cycle. Let us consider the probability unit ΔP\Delta\mathrm{P}, introduced in Eq. (68). If we set mm and nn to infinity, we have that

ΔP(p1p2)2(p0p1)2p1(p0p2)2(p1p0)m,\Delta\mathrm{P}\approx\frac{\left(\mathrm{p}_{1}-\mathrm{p}_{2}\right)^{2}\left(\mathrm{p}_{0}-\mathrm{p}_{1}\right)^{2}}{\mathrm{p}_{1}\left(\mathrm{p}_{0}-\mathrm{p}_{2}\right)^{2}}\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{0}}\right)^{m}, (74)

that tends to 0 with an exponential scaling. Therefore, the heat engine with infinite-dimensional thermal machine only modifies the passive states by an infinitesimal amount. As a consequence, the work extracted has to be infinitesimal as well. Indeed, by considering Eq. (65) it is easy to show that ΔW\Delta W tends to 0 as m,nm,n\rightarrow\infty, since ΔW\Delta W is proportional to ΔP\Delta\mathrm{P} (modulo a multiplying factor proportional to mm, which tends to infinity more slowly than (p1/p0)m\left(\mathrm{p}_{1}/\mathrm{p}_{0}\right)^{m} tends to 0).

B.1 Final state and work extraction over multiple cycles

In the previous section we have seen that, when the machine is infinitely large, we only modify the passive state infinitesimally. We can then consider the situation in which we are given an infinite number of these machines, and we want to evolve the passive state (and extract work) by sequentially applying our cycle with the help of these machines. In order to study the evolution of the passive state, we can consider its probability distribution after one cycle, see Eqs. (67). These equations can be recast as differential equations, since ΔP0\Delta\mathrm{P}\rightarrow 0 in this scenario. It is easy to verify that the differential equations which govern the evolution of the passive state are

dp0dt\displaystyle\frac{\mathrm{d}\mathrm{p}_{0}}{\mathrm{d}t} =(p1p2)2(p0p1)2p1(p0p2)2,\displaystyle=\frac{\left(\mathrm{p}_{1}-\mathrm{p}_{2}\right)^{2}\left(\mathrm{p}_{0}-\mathrm{p}_{1}\right)^{2}}{\mathrm{p}_{1}\left(\mathrm{p}_{0}-\mathrm{p}_{2}\right)^{2}}, (75a)
dp1dt\displaystyle\frac{\mathrm{d}\mathrm{p}_{1}}{\mathrm{d}t} =(1+α)(p1p2)2(p0p1)2p1(p0p2)2,\displaystyle=-\left(1+\alpha\right)\frac{\left(\mathrm{p}_{1}-\mathrm{p}_{2}\right)^{2}\left(\mathrm{p}_{0}-\mathrm{p}_{1}\right)^{2}}{\mathrm{p}_{1}\left(\mathrm{p}_{0}-\mathrm{p}_{2}\right)^{2}}, (75b)

where α=nm\alpha=\frac{n}{m} takes values in the range given by Eq. (69), and we define

dpidt=limmpipiΔp(m),Δp(m)=m(p1p0)mfori=0,1,\frac{\mathrm{d}\mathrm{p}_{i}}{\mathrm{d}t}=\lim_{m\rightarrow\infty}\frac{\mathrm{p}^{\prime}_{i}-\mathrm{p}_{i}}{\Delta\mathrm{p}(m)}\quad,\quad\Delta\mathrm{p}(m)=m\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{0}}\right)^{m}\quad\text{for}\ i=0,1, (76)

with {p0,p1,1p0p1}\left\{\mathrm{p}_{0},\,\mathrm{p}_{1},1-\mathrm{p}_{0}-\mathrm{p}_{1}\right\} the probability distribution of the state before the cycle, and {p0,p1,1p0p1}\left\{\mathrm{p}^{\prime}_{0},\,\mathrm{p}^{\prime}_{1},1-\mathrm{p}^{\prime}_{0}-\mathrm{p}^{\prime}_{1}\right\} the distribution of the state after the cycle. The continuous parameter tt is here related to the number of cycles we perform on the system. It is worth noting that Eqs. 75 share a common (positive) factor. Therefore we have that, as time goes on, the probability of occupation of |0P\ket{0}_{P} increases, while the one of |1P\ket{1}_{P} decreases (as expected from the discussion in Sec. A.2). Moreover, since α>0\alpha>0, the increase in the former is slower than the decreasing of the latter.

The two differential equations can be reshaped in a single, more helpful one,

dp1dt=(1+α)dp0dt,\frac{\mathrm{d}\mathrm{p}_{1}}{\mathrm{d}t}=-\left(1+\alpha\right)\frac{\mathrm{d}\mathrm{p}_{0}}{\mathrm{d}t}, (77)

and we can investigate the solution of this equation for α\alpha close to its limiting values. As a first step, let us consider the case in which α=ΔE10ΔE121mΔE10ΔE12\alpha=\frac{\Delta E_{10}}{\Delta E_{12}}-\frac{1}{m}\approx\frac{\Delta E_{10}}{\Delta E_{12}}. Then, the solution of Eq. (77) is

p1(t)=(1+ΔE10ΔE12)(p0(t)p0(t=0))+p1(t=0),\mathrm{p}_{1}(t)=-\left(1+\frac{\Delta E_{10}}{\Delta E_{12}}\right)\big{(}\mathrm{p}_{0}(t)-\mathrm{p}_{0}(t=0)\big{)}+\mathrm{p}_{1}(t=0), (78)

where {p0(t),p1(t),1p0(t)p1(t)}\left\{\mathrm{p}_{0}(t),\,\mathrm{p}_{1}(t),1-\mathrm{p}_{0}(t)-\mathrm{p}_{1}(t)\right\} is the probability distribution of the state of the system at time tt, and t=0t=0 is the initial time (when the system is in ρP\rho_{P}). If we rearrange Eq. (78), we see that it is equivalent to the following constraint for the evolved state

Tr[HPρP(t)]=Tr[HPρP]t0,\mathrm{Tr}\left[{H_{P}\,\rho_{P}(t)}\right]=\mathrm{Tr}\left[{H_{P}\,\rho_{P}}\right]\quad\forall\ t\geq 0, (79)

that is, the evolution conserves the energy of the system (equivalently, no work is extracted during the evolution). It is easy to see, for instance by representing the solution of Eq. (78) in a two-dimensional plot of p1\mathrm{p}_{1} versus p0\mathrm{p}_{0}, that the passive state is moving toward the set of thermal states, that are the steady states of this evolution. In fact, when a thermal state is considered, we find that (p1/p2)α=(p0/p1)\left(\mathrm{p}_{1}/\mathrm{p}_{2}\right)^{\alpha}=\left(\mathrm{p}_{0}/\mathrm{p}_{1}\right), which implies ΔP=0\Delta\mathrm{P}=0. Thus, after enough time tt is passed, we find that the initial passive state ρP\rho_{P} has been mapped into the thermal state with inverse temperature βmin\beta_{\text{min}}, where

βmin:Tr[HPτβmin]=Tr[HPρP],τβmin=eβminHPZmin,\beta_{\text{min}}\ :\ \mathrm{Tr}\left[{H_{P}\,\tau_{\beta_{\text{min}}}}\right]=\mathrm{Tr}\left[{H_{P}\,\rho_{P}}\right]\quad,\quad\tau_{\beta_{\text{min}}}=\frac{e^{-\beta_{\text{min}}H_{P}}}{Z_{\text{min}}}, (80)

and ZminZ_{\text{min}} is the partition function of the system at temperature βmin1\beta_{\text{min}}^{-1}.

We can now consider the case in which α=logp0logp1logp1logp2+1mlogp0logp1logp1logp2\alpha=\frac{\log\mathrm{p}_{0}-\log\mathrm{p}_{1}}{\log\mathrm{p}_{1}-\log\mathrm{p}_{2}}+\frac{1}{m}\approx\frac{\log\mathrm{p}_{0}-\log\mathrm{p}_{1}}{\log\mathrm{p}_{1}-\log\mathrm{p}_{2}}, that is, when its value is close to its lower bound. We notice that, in this case, α\alpha itself depends on the probability distribution of the passive state. Then, if we replace α\alpha with its lower bound in Eq. (77) we obtain

logp0dp0dt+logp1dp1dt+logp2dp2dt=0,\log\mathrm{p}_{0}\,\frac{\mathrm{d}\mathrm{p}_{0}}{\mathrm{d}t}+\log\mathrm{p}_{1}\,\frac{\mathrm{d}\mathrm{p}_{1}}{\mathrm{d}t}+\log\mathrm{p}_{2}\,\frac{\mathrm{d}\mathrm{p}_{2}}{\mathrm{d}t}=0, (81)

which, if integrated between time 0 and time tt, gives the following constraint on the entropy of the evolved states

S(ρP(t))=S(ρP)t0,S\left(\rho_{P}(t)\right)=S\left(\rho_{P}\right)\quad\forall\ t\geq 0, (82)

where S(ρ)=Tr[ρlogρ]S(\rho)=-\mathrm{Tr}\left[{\rho\log\rho}\right] is the Von Neumann entropy. Therefore, the evolution of the passive state has to preserve the entropy of the system, and the state is moving toward the set of thermal states. For tt\rightarrow\infty, the system is in the thermal state with inverse temperature βmax\beta_{\text{max}}, where

βmax:S(τβmax)=S(ρP),τβmax=eβmaxHPZmax,\beta_{\text{max}}\ :\ S\left(\tau_{\beta_{\text{max}}}\right)=S\left(\rho_{P}\right)\quad,\quad\tau_{\beta_{\text{max}}}=\frac{e^{-\beta_{\text{max}}H_{P}}}{Z_{\text{max}}}, (83)

and ZmaxZ_{\text{max}} is the partition function of the system at temperature βmax1\beta_{\text{max}}^{-1}.

Thus, when we set α\alpha equal to its limiting values, the evolution of the passive state can either follow a trajectory in which energy is conserved, or in which entropy is conserved. However, all intermediate trajectories can be achieved by imposing a different α\alpha inside the range specified by Eq. (69), and consequently all passive states with lower or equal energy, and greater or equal entropy that ρP\rho_{P} can be reached.

Appendix C Activation maps

Consider a specific family of CPT maps which allow for work extraction from a system described by a passive state. The maps of this family, which we call activation maps, can be represented by unitary operations acting globally on both the main system and an ancilla, such that the local state of the ancillary system is preserved. The cycle of Sec. A is a particular instance of these activation maps, and in the following we study the main properties of this family. Let us consider a system SS with Hamiltonian HSH_{S}, described by the state ρS\rho_{S} (this state does not need to be passive). The energy that we extract from the system when we evolve it with the unitary operator USU_{S} is given by the difference in average energy between the initial and final state,

ΔWS=TrS[HS(ρSUSρSUS)].\Delta W_{S}=\mathrm{Tr}_{S}\left[{H_{S}\left(\rho_{S}-U_{S}\,\rho_{S}\,U_{S}^{\dagger}\right)}\right]. (84)

We assume this energy to be stored in an implicit battery, and we refer to it as work. If the state is passive, then ΔWS0\Delta W_{S}\leq 0, that is, we cannot extract work. If the state is active, we can find some unitary operations that allow for a positive work extraction. In particular, the maximum work we can extract is

ΔWSmax=TrS[HS(ρSρSpass)],\Delta W_{S}^{\text{max}}=\mathrm{Tr}_{S}\left[{H_{S}\left(\rho_{S}-\rho_{S}^{\text{pass}}\right)}\right], (85)

where the state ρSpass\rho_{S}^{\text{pass}} is the passive state obtained from the initial state ρS\rho_{S}. In the literature, ΔWSmax\Delta W_{S}^{\text{max}} is known as ergotropy, see Ref. allahverdyan_maximal_2004 . This quantity is 0 if the initial state is passive, and positive otherwise.

We now add an ancillary system AA with a trivial Hamiltonian, described by the state σA\sigma_{A}, and we consider the family of maps

Λ(ρS)=TrA[USA(ρSσA)USA],\Lambda\left(\rho_{S}\right)=\mathrm{Tr}_{A}\left[{U_{SA}\left(\rho_{S}\otimes\sigma_{A}\right)U_{SA}^{\dagger}}\right], (86)

where the unitary operator USAU_{SA} acts globally over system and ancilla, and we require that the final local state of the ancilla is equal to the initial one, that is,

σA=TrS[USA(ρSσA)USA].\sigma_{A}=\mathrm{Tr}_{S}\left[{U_{SA}\left(\rho_{S}\otimes\sigma_{A}\right)U_{SA}^{\dagger}}\right]. (87)

Notice that the global evolution can create correlations between system and ancilla, and our sole constraint regards the local state of the ancilla. The work extracted during the evolution is given by

ΔWSA=TrS[HS(ρSΛ(ρS))],\Delta W_{SA}=\mathrm{Tr}_{S}\left[{H_{S}\left(\rho_{S}-\Lambda(\rho_{S})\right)}\right], (88)

where the only contribution is given by the energy difference in the system, due to the absence of any interaction term between system and ancilla, and to the fact that the final state of the ancilla is equal to its initial one.

We can now introduce the notion of activation of a quantum state,

Definition 1.

Let us consider a system SS with Hamiltonian HSH_{S}, described by the state ρS\rho_{S}. Then, we say that ρS\rho_{S} can be activated iff there exists an ancillary system AA with trivial Hamiltonian, described by the state σA\sigma_{A}, and an activation map Λ\Lambda as in Eq. (86), satisfying the condition of Eq. (87), such that

ΔWSA>ΔWSmax\Delta W_{SA}>\Delta W^{\mathrm{max}}_{S} (89)

that is, if we can extract more work from ρS\rho_{S} by acting with Λ\Lambda than we can do by acting with any unitary operation.

As we noticed before, an example of activation map is the one used in our passive engine, Eq. (A), where the ancillary system is the machine, and the global unitary operation is Sm,nS_{m,n}.

C.1 General properties of the final state of an activation map

Although the family of maps introduced in the previous section is extremely general, we can still use their definition to derive some properties of the final state Λ(ρS)\Lambda\left(\rho_{S}\right). The first, trivial property consists in the fact that the final state of an activation map has to have a lower energy than the one possessed by a the passified version of the initial state,

TrS[HSρSpass]>TrS[HSΛ(ρS)],\mathrm{Tr}_{S}\left[{H_{S}\,\rho_{S}^{\text{pass}}}\right]>\mathrm{Tr}_{S}\left[{H_{S}\,\Lambda\left(\rho_{S}\right)}\right], (90)

where this condition is obtained by replacing Eqs. (85) and (88) into Def. 1.

A second property regards the entropy of the final state. Due to the invariance of Von Neumann entropy under unitary operations, its sub-additivity, and the constraint on the local state of the machine, Eq. (87), we can show that

S(ρS)S(Λ(ρS)),S(\rho_{S})\leq S(\Lambda\left(\rho_{S}\right)), (91)

that is, the entropy of the system cannot decrease during the evolution through Λ\Lambda, and it increases if correlations create between system and machine.

If we use the two constraints on Λ(ρS)\Lambda\left(\rho_{S}\right) together, we can show that any completely passive state cannot be activated. In this case, in fact, we have that ρS=ρSpass=τβ\rho_{S}=\rho_{S}^{\text{pass}}=\tau_{\beta}, that is, the state under examination is the thermal state of Hamiltonian HSH_{S} for a certain β[0,]\beta\in[0,\infty]. But we know that this state is the one with minimum energy for a given entropy, or, vice versa, the one with maximum entropy for given energy. Then, we cannot find another state Λ(ρS)\Lambda\left(\rho_{S}\right) such that the two conditions of Eqs. (90) and (91) are satisfied at the same time. This implies that any completely passive state cannot be activated, pure ground state and maximally-mixed state included.

We can also consider a generic pure state ρS=|ψψ|\rho_{S}=\ket{\psi}\bra{\psi}. The corresponding passified state is the ground state |0\ket{0}. From Eq. (90) it follows that the final state of Λ\Lambda has to have a lower energy than ρSpass\rho_{S}^{\text{pass}}. But since the passified state we obtain, |0\ket{0}, is by definition the state with minimum energy, we cannot satisfy this condition. Thus, we cannot activate, in the sense of Def. 1, any pure state |ψ\ket{\psi}.

C.2 Asymptotic work extraction from passive states

It was proved by Alicki et al. (Ref. alicki_entanglement_2013 ) that, when an infinite number of copies of a passive state ρS\rho_{S} are considered, the optimal extractable work per single copy is given by

ΔWopt=TrS[HS(ρSτβmax)],\Delta W_{\text{opt}}=\mathrm{Tr}_{S}\left[{H_{S}\left(\rho_{S}-\tau_{\beta_{\text{max}}}\right)}\right], (92)

where τβmax\tau_{\beta_{\text{max}}} is the thermal state with inverse temperature βmax\beta_{\text{max}} such that S(τβmax)=S(ρS)S(\tau_{\beta_{\text{max}}})=S(\rho_{S}). We want to compare the work extracted in the asymptotic limit with the work extracted with a generic activation map Λ\Lambda. This comparison can be easily carried out using the main properties of the final state Λ(ρS)\Lambda(\rho_{S}), see Eqs. (90) and (91), together with the properties of τβmax\tau_{\beta_{\text{max}}}.

For any given final state of the system Λ(ρS)\Lambda(\rho_{S}), there always exists an inverse temperature β^\hat{\beta}, and a thermal state τβ^\tau_{\hat{\beta}} at that temperature, such that S(τβ^)=S(Λ(ρS))S(\tau_{\hat{\beta}})=S\left(\Lambda(\rho_{S})\right). Since the state τβ^\tau_{\hat{\beta}} is thermal, we have that its energy is minimum, that is,

TrS[HSΛ(ρS)]TrS[HSτβ^].\mathrm{Tr}_{S}\left[{H_{S}\,\Lambda(\rho_{S})}\right]\geq\mathrm{Tr}_{S}\left[{H_{S}\tau_{\hat{\beta}}}\right]. (93)

Moreover, from Eq. (91) it follows that the entropy of τβ^\tau_{\hat{\beta}} is greater than the entropy of the state τβmax\tau_{\beta_{\text{max}}}, introduced in the previous paragraph. By considering this entropic condition together with the free energy difference Fβmax(τβ^)Fβmax(τβmax)0F_{\beta_{\text{max}}}(\tau_{\hat{\beta}})-F_{\beta_{\text{max}}}(\tau_{\beta_{\text{max}}})\geq 0, we obtain that the state τβ^\tau_{\hat{\beta}} is more energetic than τβmax\tau_{\beta_{\text{max}}}, that is,

TrS[HSτβ^]TrS[HSτβmax].\mathrm{Tr}_{S}\left[{H_{S}\tau_{\hat{\beta}}}\right]\geq\mathrm{Tr}_{S}\left[{H_{S}\tau_{\beta_{\text{max}}}}\right]. (94)

From the above inequalities we have that

ΔWoptΔWSA=TrS[HSΛ(ρS)]TrS[HSτβmax]Eq. (93)TrS[HSτβ^]TrS[HSτβmax]Eq. (94)0.\Delta W_{\text{opt}}-\Delta W_{SA}=\mathrm{Tr}_{S}\left[{H_{S}\,\Lambda(\rho_{S})}\right]-\mathrm{Tr}_{S}\left[{H_{S}\tau_{\beta_{\text{max}}}}\right]\overset{\text{Eq.~\eqref{ineq_energy_1}}}{\geq}\mathrm{Tr}_{S}\left[{H_{S}\tau_{\hat{\beta}}}\right]-\mathrm{Tr}_{S}\left[{H_{S}\tau_{\beta_{\text{max}}}}\right]\overset{\text{Eq.~\eqref{ineq_energy_2}}}{\geq}0. (95)

Therefore, the energy we extract with the aid of an activation map Λ\Lambda is always equal or lower than the energy (per single copy) that we extract by acting over an infinite number of copies of the passive state with a global unitary operator, ΔWoptΔWSA\Delta W_{\text{opt}}\geq\Delta W_{SA}.

Thus, ΔWopt\Delta W_{\text{opt}} is an upper bound for the work extracted by any activation map Λ\Lambda. In Refs. alicki_entanglement_2013 ; sparaciari_resource_2016 it was shown that this upper bound can be actually achieved by acting over infinite many copies of the system with a global unitary operation. In this paper, instead, we have shown that the extraction of an amount of work equal to ΔWopt\Delta W_{\text{opt}} is also achievable by acting on a single copy of the state. However, one needs to utilise infinite many infinite-dimensional machines to do so, as we showed in Sec. B.

C.3 Ancilla as part of a bigger thermal bath

Consider the case in which the ancilla utilised in Λ\Lambda is just a subsystem of an infinite thermal reservoir at temperature β1\beta^{-1}. In this situation, we have to explicitly define an Hamiltonian HAH_{A} (where we have the freedom to rigidly translate the spectrum of this Hamiltonian), so that the state of the ancilla σA\sigma_{A} coincides with the thermal state τβ(A)=eβHA/ZA\tau_{\beta}^{(A)}=e^{-\beta H_{A}}/Z_{A}.

As we have seen, the map Λ\Lambda lowers the energy of the system and builds correlations between system and ancilla, while preserving the local state of the ancillary system. If we consider the ancilla as part of the infinite bath, then we see that Λ\Lambda extracts work from the passive state while no heat is exchanged with the bath (as the local state of the ancilla is unchanged). In the following we show that the energy extracted during this transformation is always lower than the difference in free energy between the initial state ρS\rho_{S} and the thermal state τβ(S)=eβHS/ZS\tau_{\beta}^{(S)}=e^{-\beta H_{S}}/Z_{S}. Even in the case in which Λ\Lambda maps ρS\rho_{S} into τβ(S)\tau_{\beta}^{(S)}, the work extracted is not optimal, as part of this work is locked inside the correlations between system and ancilla. In order to extract the remaining work from the correlations, and thus to perform optimal work extraction, we have to exploit the infinite thermal reservoir, exchanging an amount of heat proportional to the difference in entropy between τβ(S)\tau_{\beta}^{(S)} and ρS\rho_{S}. It is worth noting that, although this second operation allows us to extract an higher amount of work than the one obtained with the sole Λ\Lambda, we do not consider it as an allowed operation in our framework, as it requires an additional ancillary system (the bath) with infinite dimension.

During the first operation we map the initial state ρS\rho_{S} into the final one Λ(ρS)\Lambda(\rho_{S}). This final state might or might not be a thermal state of HSH_{S}, and the sole constraints we have are given by Eqs. (90) and (91) (energy has to decrease while entropy has to increase). The work we extract is the energy difference between the initial and final state, as we show in Eq. (88),

ΔW1=TrS[HS(ρSΛ(ρS))],\Delta W_{1}=\mathrm{Tr}_{S}\left[{H_{S}\left(\rho_{S}-\Lambda(\rho_{S})\right)}\right], (96)

which is positive by definition, since we assume Λ\Lambda to be an activation map, see Def. 1. The final state of system and ancilla is ρ~SA=USA(ρSτβ(A))USA\tilde{\rho}_{SA}=U_{SA}\left(\rho_{S}\otimes\tau_{\beta}^{(A)}\right)U_{SA}^{\dagger}, and correlations are present, quantified by the mutual information

I(S~:A~)=S(Λ(ρS))+S(τβ(A))S(ρ~SA).I\left(\tilde{S}:\tilde{A}\right)=S(\Lambda(\rho_{S}))+S(\tau_{\beta}^{(A)})-S(\tilde{\rho}_{SA}). (97)

The heat Q1Q_{1} exchanged during this transformation is equal to 0, as the local state of the bath does not change.

We now use the power of the infinite thermal reservoir to extract the last part of work from the state ρ~SA\tilde{\rho}_{SA}, by mapping it into τβ(S)τβ(A)\tau_{\beta}^{(S)}\otimes\tau_{\beta}^{(A)}. In this case, work is given by the free energy difference between the two states, that is

ΔW2=Fβ(ρ~SA)Fβ(τβ(S)τβ(A))=1β(D(Λ(ρS)||τβ(S))+I(S~:A~)),\Delta W_{2}=F_{\beta}\left(\tilde{\rho}_{SA}\right)-F_{\beta}\left(\tau_{\beta}^{(S)}\otimes\tau_{\beta}^{(A)}\right)=\frac{1}{\beta}\left(D\left(\Lambda(\rho_{S})||\tau_{\beta}^{(S)}\right)+I\left(\tilde{S}:\tilde{A}\right)\right), (98)

where D(Λ(ρS)||τβ(S))=β(Fβ(Λ(ρS))Fβ(τβ(S)))D\left(\Lambda(\rho_{S})||\tau_{\beta}^{(S)}\right)=\beta\left(F_{\beta}\left(\Lambda(\rho_{S})\right)-F_{\beta}\left(\tau_{\beta}^{(S)}\right)\right) is the relative entropy between Λ(ρS)\Lambda(\rho_{S}) and τβ(S)\tau_{\beta}^{(S)}. Since both the relative entropy and the mutual information are non-negative quantities, we have that work is indeed extracted during this second process. The heat exchanged in this second transformation is equal to the entropy difference (modulo the multiplicative constant β1\beta^{-1}) between the final and initial state

Q2=1β(S(τβ(S)τβ(A))S(ρ~SA))=1β(S(τβ(S))S(ρS)),Q_{2}=\frac{1}{\beta}\left(S\left(\tau_{\beta}^{(S)}\otimes\tau_{\beta}^{(A)}\right)-S\left(\tilde{\rho}_{SA}\right)\right)=\frac{1}{\beta}\left(S\left(\tau_{\beta}^{(S)}\right)-S\left(\rho_{S}\right)\right), (99)

where the last equality follows from the invariance under unitary operations of the Von Neumann entropy.

If we now consider the two transformations as a single one, we see that the total work extracted is

ΔWtot=ΔW1+ΔW2=Fβ(ρS)Fβ(τβ(S)),\Delta W_{\text{tot}}=\Delta W_{1}+\Delta W_{2}=F_{\beta}\left(\rho_{S}\right)-F_{\beta}\left(\tau_{\beta}^{(S)}\right), (100)

that is, ΔWtot\Delta W_{\text{tot}} is optimal, and the heat exchanged is Q2Q_{2}, equal to the entropy difference between τβ(S)\tau_{\beta}^{(S)} and ρS\rho_{S}.

An interesting scenario occurs when Λ\Lambda maps the initial state into τβ(S)\tau_{\beta}^{(S)}. In this case, we see that the work we obtain in the second transformation (the one involving the whole thermal bath) is proportional to the sole mutual information, so that work is exclusively extracted from the correlations between system and catalyst. The amount of work in this case (see also Ref. perarnau-llobet_extractable_2015 , Sec. VI B) is

ΔW2corr=1βI(S~:A~)=1β(S(τβ(S))S(ρS)),\Delta W_{2}^{\text{corr}}=\frac{1}{\beta}I\left(\tilde{S}:\tilde{A}\right)=\frac{1}{\beta}\left(S(\tau_{\beta}^{(S)})-S(\rho_{S})\right), (101)

where the quantity is still non-negative, since Λ\Lambda can map ρS\rho_{S} into τβ(S)\tau_{\beta}^{(S)} only if S(ρS)S(τβ(S))S(\rho_{S})\leq S(\tau_{\beta}^{(S)}), see Sec. C.2.

Appendix D Technical results

In this section we show some of the technical results we have used to analyse the generic cycle on passive states.

Technical Result 1.

Consider the sequence of real numbers {xj}ab\left\{x_{j}\right\}_{a}^{b}, those elements are linked by the following set of equations,

xj=(1+λ)xj+1λxj+2;j=a,,b2,x_{j}=\left(1+\lambda\right)x_{j+1}-\lambda\,x_{j+2}\quad;\quad j=a,\ldots,b-2,

where λ\lambda\in\mathbb{R} and a,ba,b\in\mathbb{N}, ab2a\leq b-2. Then, the elements of this sequence can be expressed in terms of xb1x_{b-1} and xbx_{b} as

xj=T(b(j+1),λ)xb1λT(b(j+2),λ)xb;j=a,,b2,x_{j}=\mathrm{T}(b-(j+1),\lambda)\,x_{b-1}-\lambda\,\mathrm{T}(b-(j+2),\lambda)\,x_{b}\quad;\quad j=a,\ldots,b-2,

where T(h,λ)=l=0hλl=1λh+11λ\mathrm{T}(h,\lambda)=\sum_{l=0}^{h}\lambda^{l}=\frac{1-\lambda^{h+1}}{1-\lambda}.

Proof.

If we insert the solution into the set of equations, we find

T(b(j+1),λ)xb1λT(b(j+2),λ)xb\displaystyle\mathrm{T}(b-(j+1),\lambda)\,x_{b-1}-\lambda\,\mathrm{T}(b-(j+2),\lambda)\,x_{b} =(1+λ)T(b(j+2),λ)xb1λ(1+λ)T(b(j+3),λ)xb\displaystyle=(1+\lambda)\mathrm{T}(b-(j+2),\lambda)\,x_{b-1}-\lambda(1+\lambda)\,\mathrm{T}(b-(j+3),\lambda)\,x_{b}
λT(b(j+3),λ)xb1+λ2T(b(j+4),λ)xb\displaystyle-\lambda\,\mathrm{T}(b-(j+3),\lambda)\,x_{b-1}+\lambda^{2}\,\mathrm{T}(b-(j+4),\lambda)\,x_{b}

for jj taking values from aa to b2b-2. We can re-organise the above equation, and we find that it is satisfied iff

T(b(j+1),λ)\displaystyle\mathrm{T}(b-(j+1),\lambda) =(1+λ)T(b(j+2),λ)λT(b(j+3),λ);j=a,,b2,\displaystyle=(1+\lambda)\,\mathrm{T}(b-(j+2),\lambda)-\lambda\,\mathrm{T}(b-(j+3),\lambda)\quad;\quad j=a,\ldots,b-2, (102a)
T(0,λ)\displaystyle\mathrm{T}(0,\lambda) =(1+λ)T(1,λ)λT(2,λ).\displaystyle=(1+\lambda)\,\mathrm{T}(-1,\lambda)-\lambda\,\mathrm{T}(-2,\lambda). (102b)

These two equalities easily follow from the definition of T(h,λ)\mathrm{T}(h,\lambda), as it can be check by replacing this coefficient with its explicit form in both Eq. (102a) and (102b). ∎

Technical Result 2.

The probability distribution of the state ρM\rho_{M} is positive and normalised.

Proof.

Let us consider the probabilities qj\mathrm{q}_{j} for j=0,,m1j=0,\ldots,m-1, as given in Eq. (63a). If we replace jj with j=mjj^{\prime}=m-j, then the main coefficient in the equation becomes

T1(j)p2p1D(m,n)T1(j1)\displaystyle\mathrm{T}_{1}(j^{\prime})-\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\,D(m,n)\,\mathrm{T}_{1}(j^{\prime}-1) =T1(j)T1(m1)T1(j1)T1(m)T1(m1)+p1p2T2(n1)\displaystyle=\frac{\mathrm{T}_{1}(j^{\prime})\,\mathrm{T}_{1}(m-1)-\mathrm{T}_{1}(j^{\prime}-1)\,\mathrm{T}_{1}(m)}{\mathrm{T}_{1}(m-1)+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\mathrm{T}_{2}(n-1)}
+p1p2T1(j)T2(n1)T1(j1)T2(n2)T1(m1)+p1p2T2(n1).\displaystyle+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\frac{\mathrm{T}_{1}(j^{\prime})\,\mathrm{T}_{2}(n-1)-\mathrm{T}_{1}(j^{\prime}-1)\,\mathrm{T}_{2}(n-2)}{\mathrm{T}_{1}(m-1)+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\mathrm{T}_{2}(n-1)}.

It is clear that the denominator is positive, as T1(h)\mathrm{T}_{1}(h) and T2(h)\mathrm{T}_{2}(h) are positive for all hh\in\mathbb{Z}. We need to show that the nominator is positive as well. The nominator of the first term can be reduced to

T1(j)T1(m1)T1(j1)T1(m)=T1(m1)T1(j1)=l=jm1(p0p1)l0,\mathrm{T}_{1}(j^{\prime})\,\mathrm{T}_{1}(m-1)-\mathrm{T}_{1}(j^{\prime}-1)\,\mathrm{T}_{1}(m)=\mathrm{T}_{1}(m-1)-\mathrm{T}_{1}(j^{\prime}-1)=\sum_{l=j^{\prime}}^{m-1}\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{l}\geq 0,

where the last equality follows from the fact that j=1,,mj^{\prime}=1,\ldots,m. The nominator of the second term can be expressed as

T1(j)T2(n1)T1(j1)T2(n2)=T1(j1)(p1p2)n1+T2(n2)(p0p1)j+(p0p1)j(p1p2)n1>0.\mathrm{T}_{1}(j^{\prime})\,\mathrm{T}_{2}(n-1)-\mathrm{T}_{1}(j^{\prime}-1)\,\mathrm{T}_{2}(n-2)=\mathrm{T}_{1}(j^{\prime}-1)\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n-1}+\mathrm{T}_{2}(n-2)\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{j^{\prime}}+\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{j^{\prime}}\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{n-1}>0.

Thus, the probabilities {qj}j=0m1\left\{q_{j}\right\}_{j=0}^{m-1} are positive when qm+n1\mathrm{q}_{m+n-1} is positive.

We can now focus on the probabilities qj\mathrm{q}_{j} for j=m,,m+n3j=m,\ldots,m+n-3, as given in Eq. (63b). By replacing jj with j=m+n(j+2)j^{\prime}=m+n-(j+2) we obtain that the main coefficient in the equation becomes

T2(j)D(m,n)p1p2T2(j1)\displaystyle\mathrm{T}_{2}(j^{\prime})\,D(m,n)-\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\,\mathrm{T}_{2}(j^{\prime}-1) =(p1p2)T2(j)T1(m)T2(j1)T1(m1)T1(m1)+p1p2T2(n1)\displaystyle=\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)\,\frac{\mathrm{T}_{2}(j^{\prime})\,\mathrm{T}_{1}(m)-\mathrm{T}_{2}(j^{\prime}-1)\,\mathrm{T}_{1}(m-1)}{\mathrm{T}_{1}(m-1)+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\mathrm{T}_{2}(n-1)}
+(p1p2)2T2(j)T2(n2)T2(j1)T2(n1)T1(m1)+p1p2T2(n1).\displaystyle+\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{2}\frac{\mathrm{T}_{2}(j^{\prime})\,\mathrm{T}_{2}(n-2)-\mathrm{T}_{2}(j^{\prime}-1)\,\mathrm{T}_{2}(n-1)}{\mathrm{T}_{1}(m-1)+\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\mathrm{T}_{2}(n-1)}.

As before, the denominator is positive, as T1(h)\mathrm{T}_{1}(h) and T2(h)\mathrm{T}_{2}(h) are both positive h\forall\,h\in\mathbb{Z}. The nominator of the first term can be reduced to

T2(j)T1(m)T2(j1)T1(m1)=T2(j1)(p0p1)m+T1(m1)(p1p2)j+(p1p2)j(p0p1)m>0.\mathrm{T}_{2}(j^{\prime})\,\mathrm{T}_{1}(m)-\mathrm{T}_{2}(j^{\prime}-1)\,\mathrm{T}_{1}(m-1)=\mathrm{T}_{2}(j^{\prime}-1)\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}+\mathrm{T}_{1}(m-1)\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{j^{\prime}}+\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{j^{\prime}}\left(\frac{\mathrm{p}_{0}}{\mathrm{p}_{1}}\right)^{m}>0.

The nominator of the second term can be expressed as

T2(j)T2(n2)T2(j1)T2(n1)=T2(n2)T2(j1)=l=jn2(p1p2)l0,\mathrm{T}_{2}(j^{\prime})\,\mathrm{T}_{2}(n-2)-\mathrm{T}_{2}(j^{\prime}-1)\,\mathrm{T}_{2}(n-1)=\mathrm{T}_{2}(n-2)-\mathrm{T}_{2}(j^{\prime}-1)=\sum_{l=j^{\prime}}^{n-2}\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}\right)^{l}\geq 0,

where the last equality follows from the fact that j=1,,n2j^{\prime}=1,\ldots,n-2. Thus, the probabilities {qj}j=mm+n3\left\{q_{j}\right\}_{j=m}^{m+n-3} are positive when qm+n1>0\mathrm{q}_{m+n-1}>0.

In Eq. (63c), we showed that qm+n2\mathrm{q}_{m+n-2} is related to qm+n1\mathrm{q}_{m+n-1} by the multiplicative coefficient D(m,n)D(m,n), which can be easily shown to be positive for any integer m,n1m,n\geq 1. Finally, the normalisation condition force qm+n1>0\mathrm{q}_{m+n-1}>0, and implies the probability distribution of ρM\rho_{M} to be positive and normalised. ∎