Energetic instability of passive states in thermodynamics

0.62 MB·

by  Carlo Sparaciari

Energetic instability of passive states in thermodynamics
Carlo Sparaciari,1,∗ David Jennings,2,3,† and Jonathan Oppenheim1,‡
1Department of Physics and Astronomy, University College London, London WC1E 6BT, United Kingdom
2Department of Physics, University of Oxford, Oxford, OX1 3PU, United Kingdom
3Department of Physics, Imperial College London, London SW7 2AZ, United Kingdom
(Dated: January 9, 2017)
Passivityisafundamentalconceptinthermodynamicsthatdemandsaquantumsystem’senergy
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
systemandshowthateverypassivestateexceptthethermalstateisunstableunderaweakerform
ofreversibility. Moreprecisely,weshowthatgivenasinglecopyofanyathermalquantumstatewe
may extract a maximal amount of energy from the state when we can use a machine that operates
7 in a reversible cycle and whose state is left unchanged. This means that for individual systems the
1 only form of passivity that is stable under general reversible processes is complete passivity, and
0 thus provides a single-shot and more physically motivated identification of thermal states and the
2 emergence of temperature. The machine which extracts work from passive states exploits the fact

n that one can find a subspace which acts as a virtual hot reservoir, and a subspace which acts as
a a virtual cold reservoir. We show that an optimal amount of work can be extracted, and that the
J machine operates at the Carnot efficiency between pairs of virtual reservoirs.

6

]
h
p I. INTRODUCTION
-
t
n
Withinthermodynamics,heatenginesaredevicesthatoperateinathermalcontextsoastoextractorderedenergy
a
in the form of work. The canonical scenario involves an engine that operates cyclically between two temperatures
u
q Thot, Tcold and performs a quantity of mechanical work. To do so the engine absorbs heat from the hot reservoir,
[ convertssomeofthisenergytomechanicalworkandreleasesheatintothecoldreservoirinaccordancewiththeSecond
Law of thermodynamics. The largest possible efficiency, η = 1− Tcold, occurs for the reversible Carnot engine [1, 2]
1 and provides a fundamental thermodynamic bound on the amounTthootf ordered energy that can be obtained. Carnot
v
engines, and more in general heat engines, have been extensively studied in the microscopic regime [3–26] (as well as,
3
0 of course, in the macroscopic regime).
7 However,theissueoforderedenergyextractioncanalsobeconsideredinscenariosinwhichnonotionoftemperature
1 exists,andcanprovideabroadernotionofequilibriumstates. Forexample,moregeneralequilbriumstatescanoccur
0 in physical realisations when a system has been perturbed and has not had enough time to fully thermalise. They
.
1 can also arise in the context of non-equilibrium steady states [27]. Given a quantum system in a state ρ one can ask
0 if it is possible to extract energy from it solely by performing a reversible unitary transformation on the system. The
7 largest amount of ordered energy that can be extracted (the “ergotropy”, see Refs. [28–30]) depends non-trivially
1
on the quantum state. If no energy can be extracted in this way then ρ is called passive [31–35] and constitutes a
:
v primitive form of equilibrium.
i In this work, we consider a scenario that is intermediate between the above two contexts, and is motivated by the
X
fact that a work extraction machine should be considered as a system which is involved in the process. Our core
r
questioniswhetherthereexistpassivestatesρ forwhichenergycanbeextractedifoneperformsareversibleunitary
a S
process over the system S together with a second quantum system M, which starts and finishes in the same quantum
state ρ . This second quantum system is the machine, such as the working body in a Carnot cycle, which undergoes
M
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 [36].
Indeed, the system S described by a passive state represents both the reservoirs, while the additional system M 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 [4, 7, 11, 12, 37, 38], it seems reasonable to extend the analysis on
passivestatestothecaseinwhichthisbroaderclassofoperationsisallowed. Crucially,thisanalysishasfundamental
implications for the notion of passivity. In fact, if energy can be extracted from a passive state with these reversible
∗ [email protected]
† [email protected]
‡ [email protected]
2
processes,andnoentropyisgenerated,thenitseemsthatassociatingpassivityofthestateisarestrictedidealisation,
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
Considerafinite-dimensionalquantumsystemassociatedwiththeHilbertspaceH≡Cd(aqudit),withHamiltonian
H =(cid:80)d−1E |i(cid:105)(cid:104)i|, and described by the state ρ. We say that the state ρ is passive iff its average energy cannot be
i=0 i
lowered by acting on it with unitary operations, that is,
Tr[Hρ]≤Tr(cid:2)HUρU†(cid:3) , ∀U ∈B(H) , UU† =U†U =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 n ∈ N independent and identically
distributed (i.i.d.) copies of our system, with a total Hamiltonian H(n) =(cid:80)d H , where each H is a single-system
i=1 i i
Hamiltonian acting on a different copy of the system. The state of this global system is described by ρ⊗n. Then, we
say that the state ρ is completely passive if and only if the state ρ⊗n is passive for all n ∈ N. It can be shown [31]
thatthecompletelypassivestatesofasystemwithHamiltonianH aretheonessatisfyingtheKMScondition[39–41].
Specifically, these states are the ground state and the thermal states with temperature β ≥ 0, that is, τ = e−βH/Z
β
with Z =Tr(cid:2)e−βH(cid:3). Any state which is not of this form, is called athermal.
FIG.1. (Left)Thespectrumofaqutritpassivestateρ=(cid:80)2 p |i(cid:105)(cid:104)i|overtheeigenbasisofitsHamiltonianH =(cid:80)2 E |i(cid:105)(cid:104)i|.
i=0 i i=0 i
The occupation probabilities are ordered in a decreasing order, from the one associated with the ground state of H to the one
associatedwiththemaximallyexcitedone,asdescribedinEq.(2). (Right)Apassivestatecanequallybedescribedbyvirtual
temperatures. Indeed, we can associate a virtual temperature to each pair of eigenstates of ρ, as shown in Eq. (3). In the plot,
the pair of levels |0(cid:105) and |1(cid:105) is associated with the hot temperature β−1, while the pair |1(cid:105) and |2(cid:105) is associated with the cold
hot
temperature β−1 .
cold
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 H increases, Fig. 1, left plot. Specifically, a state ρ is passive iff ρ=f(H), where f
is a monotone non-increasing function. Simply put, this means that the state can expressed as
d−1
(cid:88)
ρ= p |i(cid:105)(cid:104)i| , such that p ≥p ∀i=0,...,d−2, (2)
i i i+1
i=0
3
where {|i(cid:105)}d−1 are the eigenvectors of H, ordered so that E ≤E for all i (for the case of equal energies E =E
i=0 i i+1 i i+1
we must make an additional stability assumption to ensure that p =p ).
i i+1
We can describe the probability distribution of the passive state ρ by using virtual temperatures [15, 42]. 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(cid:105),|j(cid:105)), we define the virtual temperature associated with them as the β−1 ≥0
ij
such that
p
i =:e−βij(Ei−Ej), (3)
p
j
where p is the probability of occupation of the state |i(cid:105), and E is the energy associated with the state (similarly
i i
for j). 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 H 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 =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
2
(cid:88)
H = E |i(cid:105)(cid:104)i| , (4)
P i P
i=0
and E ≤E . The qutrit system is described by the state
i i+1
2
(cid:88)
ρ = p |i(cid:105)(cid:104)i| , (5)
P i P
i=0
where p ≥p (Fig. 1, left plot).
i i+1
In the following we assume that the passive state ρ is described by the virtual temperature T = β−1 > 0
P hot hot
associated with the pair of eigenstates (|0(cid:105) ,|1(cid:105) ), and the virtual temperature T =β−1 >0 associated with the
P P cold cold
pair (|1(cid:105) ,|2(cid:105) ). We assume for simplicity that T > T , but a similar analysis applies for T < T . In the
P P hot cold hot cold
Supplemental Material, Sec. I, the cycle is presented in full detail. The relation between the probability distribution
of ρ and the temperatures T and T is given by
P hot cold
p
1 =:e−βhot∆E10, (6a)
p
0
p
2 =:e−βcold∆E21, (6b)
p
1
where ∆E = E −E ≥ 0, and ∆E = E −E ≥ 0. Thus, the pair of states (|0(cid:105) ,|1(cid:105) ) can be viewed as
10 1 0 21 2 1 P P
representing a “hot virtual reservoir”, while the pair of states (|1(cid:105) ,|2(cid:105) ) represent a “cold virtual reservoir”. It is
P P
worth noting that the other pair of states, (|0(cid:105) ,|2(cid:105) ), is associated with a virtual temperature that is intermediate
P P
between T and T , as we can easily verify from Eqs. (6).
cold hot
The engine extracts work by means of the following cycle. A single system, described by the state ρ , is put in
P
contact with the machine, described by the state ρ = (cid:80)d−1q |j(cid:105)(cid:104)j| . Then, we perform m swaps between the
M j=0 j M
hot virtual reservoir of the passive state and m different pairs of states of ρ , followed by n swaps between the cold
M
virtualreservoirand other n different pairsof statesof ρ . Inorder toperform theswaps ondifferentpairs ofstates,
M
weneedthemachinetohaveatleastm+nlevels,andthereforewefixd=m+n. Specifically,weapplythefollowing
unitary operation to the global system
S =S(0,m)◦S(m,m+1)◦S(m+1,m+2)◦...◦S(m+n−2,m+n−1)◦S(m−1,m+n−1)◦S(m−2,m−1)◦S(m−3,m−2)◦...◦S(0,1), (7)
m,n (1,2) (1,2) (1,2) (1,2) (0,1) (0,1) (0,1) (0,1)
where the operator S(c,d) is a swap between system and machine, performed through the permutation |a(cid:105) |d(cid:105) ↔
(a,b) P M
|b(cid:105) |c(cid:105) . A graphical representation of this global operation is shown in Fig. 2, where each swap is depicted by an
P M
4
FIG.2. Thecycleisrepresentedinapictorialwayovertheeigenstatesofthed-dimensionalmachine(whered=m+n). Notice
that the machine has a trivial Hamiltonian, and we order the eigenstates only to simplify the visualisation of the cycle. The
(cid:0) (cid:1)
upward arrow connecting two eigenstates of the machine represents a swap between these two states and the pair |0(cid:105) ,|1(cid:105)
P P
of the passive state. The downward arrow connecting two eigenstates of the machine represents a swap between this pair and
the pair (cid:0)|1(cid:105) ,|2(cid:105) (cid:1) of the passive state. We initially perform m−1 swaps between (cid:0)|0(cid:105) ,|1(cid:105) (cid:1) and (cid:8)(cid:0)|j(cid:105) ,|j+1(cid:105) (cid:1)(cid:9)m−2,
P P P P M M j=0
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
and one swap between |0(cid:105) ,|1(cid:105) and |m−1(cid:105) ,|m+n−1(cid:105) . Then, we perform n−1 swaps between |1(cid:105) ,|2(cid:105) and
P P M M P P
(cid:8)(cid:0)|j(cid:105) ,|j+1(cid:105) (cid:1)(cid:9)m+n−2, and one swap between (cid:0)|1(cid:105) ,|2(cid:105) (cid:1) and (cid:0)|0(cid:105) ,|m(cid:105) (cid:1). If we consider the arrow representation of
M M j=m P P M M
swaps,wecanseethatthecycleisclose,andthisallowsustorecoverthelocalstateofthemachineM whileextractingwork.
arrow acting over the states of the machine. Although in the figure we represent the eigenstates of ρ in a ladder,
M
they are all associated with the same energy, and therefore the order in which we present them is only functional for
visualising the cycle S .
m,n
Inorderfortheenginetobere-usable,weneedthelocalstateofthemachinetoendupinitsinitialstate. Therefore,
we impose the following constraint on the state of the machine,
ρ =! Tr (cid:2)S (ρ ⊗ρ )S† (cid:3). (8)
M P m,n P M m,n
Through Eq. (8) we can express the probability distribution of the machine in terms of the passive state ρ (see
P
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, ∆W, as the difference in average energy between the initial
and final state of the main system (as the machine M has a trivial Hamiltonian, and no interaction terms are present
between system and machine). Thus, we have that
∆W =Tr[H (ρ −ρ˜ )], (9)
P P P
where the final state of the system is
ρ˜ =Tr (cid:2)S (ρ ⊗ρ )S† (cid:3). (10)
P M m,n P M m,n
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 H and a given cycle S we can investigate the amount of work we extract
P m,n
from the state ρ . In the Supplemental Material we provide all the necessary steps to evaluate ∆W in terms of the
P
probability distribution of ρ . We can express this quantity as
P
∆W =α(m∆E10−n∆E21)(cid:0)eβcoldn∆E21 −eβhotm∆E10(cid:1), (11)
where α is a positive coefficient depending non-trivially on the probability distribution of ρ . For the class of passive
P
states we are considering (namely, the one in which β >β ), we find that work can be extracted (∆W >0) iff
cold hot
1. The Hamiltonian H is such that m∆E >n∆E .
P 10 21
2. The temperatures of the two virtual reservoirs are such that βcold > mn∆∆EE2110βhot.
5
Thus, for a fixed cycle (defined by the parameters m and n), and for a fixed Hamiltonian H , we find that work can
P
only be extracted if the virtual temperature T is lower than T by a multiplicative factor which depends on the
cold hot
energygapsoftheHamiltonian,seeFig.3,leftplot,foranexample. InSec.IVweshowthat,foragivenHamiltonian
H , work can be extracted by any passive (but not completely passive) state, and we characterise the cycle which
P
allows for this extraction.
ΔW
η
η
ΔE Carnot
10
ΔE21 ΔE21ββchooldt ΔE
10
ΔE21 ΔE21ββchooldt
FIG. 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=1. This cycle is analogous to the one studied in Ref. [13]. (Left) The work extracted in the cycle is positive if the hot
(cid:104) (cid:105)
energy gap ∆E lies inside therange ∆E ;βcold∆E , as we would expect from conditions 1 and 2. (Right) Theefficiency
10 10 βhot 10
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 M and the hot virtual reservoir, that is
qhot =α∆E10(cid:0)eβcoldn∆E21 −eβhotm∆E10(cid:1), (12)
where α is the same positive coefficient of Eq. (11). Moreover, the same amount of energy is spent during each swap
between the machine M and the cold virtual reservoir,
qcold =α∆E21(cid:0)eβcoldn∆E21 −eβhotm∆E10(cid:1). (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 (|0(cid:105) ,|1(cid:105) ) with the hot virtual reservoir, then the energy
P P
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
Q =mq , (14)
hot hot
while the total heat provided to the cold virtual reservoir is
Q =nq . (15)
cold cold
From Eqs. (14) and (15) we obtain that the work extracted can be expressed as ∆W =Q −Q , as in a standard
hot cold
heatengineexchangingenergybetweentworeservoirs. OnceQ andQ aredefined,wecanevaluateanefficiency
hot cold
of this cycle, that is
∆W n∆E
η = =1− 21 . (16)
Q m∆E
hot 10
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 <1− Tcold. When we consider the case of an infinite-dimensional machine, we find that by a judicious
choice of parametersTwheotmay obtain Carnot efficiency.
Once the cycle S is ended, the local state of the main system is moved to a less energetic state. By solving
m,n
Eq. (8), we find that the final state of the main system ρ˜ has the following probability distribution
P
p(cid:48) =p +m∆P, (17a)
0 0
p(cid:48) =p −(m+n) ∆P, (17b)
1 1
p(cid:48) =p +n∆P, (17c)
2 2
6
where the unit of probability ∆P depends on the initial state ρ , and it is given by
P
∆P=α(cid:0)eβcoldn∆E21 −eβhotm∆E10(cid:1), (18)
with α the same positive coefficient of Eq. (11). Due to condition 2, the unit ∆P > 0, so that the probability of
occupationof|1(cid:105) isreducedinfavouroftheprobabilitiesp andp . Thus,energyisextractedfromthepassivestate
P 0 1
when m∆P is moved from p to p (during the hot swaps), and part of this energy is used to move the probability
1 0
n∆P from p to p (during the cold swaps).
1 2
IV. WORK EXTRACTION FROM ANY PASSIVE STATE
Wenowshowthat,foragivenHamiltonianH ,workcanbeextractedfromanypassivebutnotcompletelypassive
P
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 m and n. In what
follows, we represent the passive state with the probabilities of occupation {p ,p ,p }, as opposed to the previous
0 1 2
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.
FIG. 4. (Left) The set of qutrit passive states. The red region represents the subset R , while the blue region represents the
1
subset R . The black line is R , that is, the set of completely passive states. (Right) The region R (in red) contains the
2 3 1
regions R+ (green), R+ (purple), and R+ (yellow), which cover R better and better as m and n grow. In both plots we
3,1 5,2 11,5 1
set M =2 and N =1.
The Hamiltonian of the system H is defined in Eq. (4), where the energy gap between ground and first excited
P
state is ∆E , and the gap between first and second excited states is ∆E . We assume that
10 21
∃M,N ∈N such that M∆E −N∆E =0, (19)
10 21
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 N and M 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
(cid:40) (cid:12) (cid:18)p (cid:19)N (cid:18)p (cid:19)M(cid:41)
R = ρ passive (cid:12) 1 > 0 , (20a)
1 P (cid:12) p p
2 1
(cid:40) (cid:12) (cid:18)p (cid:19)N (cid:18)p (cid:19)M(cid:41)
R = ρ passive (cid:12) 1 < 0 , (20b)
2 P (cid:12) p p
2 1
(cid:40) (cid:12) (cid:18)p (cid:19)N (cid:18)p (cid:19)M(cid:41)
R = ρ passive (cid:12) 1 = 0 . (20c)
3 P (cid:12) p p
2 1
7
The union of these three subsets gives the set of all passive states. In particular, one can verify that the subset R
3
contains all the completely passive states, that is, the thermal states of H at any temperature β−1 ≥0. Moreover,
P
R corresponds to the set of passive states with β associated with the pair of eigenstates |0(cid:105) and |1(cid:105) , and β
1 hot P P cold
associated with the pair |1(cid:105) and |2(cid:105) . The set R , instead, contains the passive states with opposite hot and cold
P P 2
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 m and n can activate a passive state
ρ with Hamiltonian H if conditions 1 and 2 are satisfied. These conditions apply to the case in which the passive
P P
state is described by Eqs. 6, with β < β . In the present, more general scenario we find that work is extracted
hot cold
by the cycle if and only if the passive state belongs to the following subset
Rm+,n =(cid:26)ρP passive (cid:12)(cid:12)(cid:12)(cid:12) (cid:18)pp1(cid:19)n >(cid:18)pp0(cid:19)mwhen m∆E10−n∆E21 >0
2 1
(cid:18)p (cid:19)n (cid:18)p (cid:19)m (cid:27)
∨ 1 < 0 when m∆E −n∆E <0 , (21)
p p 10 21
2 1
where these conditions can be obtained by analysing the general expression of the extracted work, see Supplemental
Material. Wenowshowthat,bytailoringthevalueoftheparametersmandn,wecanmakeR+ to(asymptotically)
m,n
cover either the region R or R . Here, we focus on R solely, since R follows from similar arguments. As a first
1 2 1 2
step, we ask m∆E −n∆E >0, which implies m> Mn, due to Eq. (19). Then, in order to satisfy this condition,
10 21 N
we set m= Mn+1, where we ask n to be large enough for m to be an integer. The set of passive states activated by
N
the cycle is such that
(cid:18)p1(cid:19)n >(cid:18)p0(cid:19)m ⇒(cid:18)p1(cid:19)n >(cid:18)p0(cid:19)MNn+1 ⇒(cid:18)p1(cid:19)N >(cid:18)p0(cid:19)M+Nn . (22)
p p p p p p
2 1 2 1 2 1
We notice that, since ρ is passive, p ≥p , which implies that
P 0 1
(cid:18)p0(cid:19)M+Nn ≥(cid:18)p0(cid:19)M. (23)
p p
1 1
Thus, Eq. (22) and (23) together assure that R+ ⊂R . Moreover, if n→∞, we have that M+N →M, which
Mn+1,n 1 n
N
implies that R+ → R . Thus, we have that, for a given Hamiltonian H , and a given passive state ρ ∈ R ,
Mn+1,n 1 P P 1
N
there exist a cycle S such that ρ ∈ R+ . However, the closer (in trace norm) the state ρ is to the set of
m,n P m,n P
completely passive states (R ), the larger the parameters m and n have to be, that is, the larger the machine has to
3
be (see Fig. 4, right plot).
A. Work extraction from a generic qudit passive state
Work extraction from a generic qudit passive state ρ(d) (for any Hamiltonian H(d)) can be achieved with the
P P
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 d levels, we only need to focus our analysis on three of
them, and perform the cycle on these levels only. Thus, given the state ρ(d) = (cid:80)d−1p |i(cid:105)(cid:104)i| and the Hamiltonian
P i=0 i P
H(d) =(cid:80)d−1E |i(cid:105)(cid:104)i| , we can consider the subspace A =span{|k(cid:105) ,|k+1(cid:105) ,|k+2(cid:105) }, for a given k ∈[0,d−3].
P i=0 i P k P P P
Thus, we can divide the qudit state and the Hamiltonian in two contributions, one with support over A , the other
k
with support over its complement,
(cid:32) (cid:33) (cid:32) (cid:33)
ρ(d) = (cid:88) p ρ(A)+ 1− (cid:88) p ρ(Ac), (24a)
P i P i P
i∈Ak i∈Ak
H(d) =H(A)+H(Ac), (24b)
P P P
8
where the normalised quantum states are
ρ(A) = (cid:88) pi |i(cid:105)(cid:104)i| , (25a)
P (cid:80) p P
i∈Ak j∈Ak j
ρ(Ac) = (cid:88) pi |i(cid:105)(cid:104)i| , (25b)
P 1−(cid:80) p P
i∈/Ak j∈Ak j
while the Hamiltonian contributions are, respectively, H(A) =(cid:80) E |i(cid:105)(cid:104)i| and H(Ac) =(cid:80) E |i(cid:105)(cid:104)i| . In the
P i∈Ak i P P i∈/Ak i P
following, we define λ=(cid:80) p , so that ρ(d) =λρ(A)+(1−λ)ρ(Ac).
i∈Ak i P P P
Wecannowintroduceanancillarysystem(themachineM)ofdimensionm+n,describedbythestateρ ,together
M
with the global unitary operator U,
U =P ⊗I +P ⊗I ◦S ◦P ⊗I , (26)
Ac M A M m,n A M
where the operator S , described in Eq. (7), has support on A ⊗H , and therefore commute with P ⊗I . If
m,n k M A M
we consider the evolution of the system under this operator, we obtain
ρ˜(d) =Tr (cid:104)U(cid:16)ρ(d)⊗ρ (cid:17)U†(cid:105)=λTr (cid:104)S (cid:16)ρ(A)⊗ρ (cid:17)S† (cid:105)+(1−λ)ρ(Ac) =λρ˜(A)+(1−λ)ρ(Ac), (27)
P M P M M m,n P M m,n P P P
and we can easily verify, due to the properties of S , that the local state of the machine is left unchanged. The
m,n
amount of work extracted during this cycle is
(cid:104) (cid:16) (cid:17)(cid:105) (cid:104) (cid:16) (cid:17)(cid:105)
∆W =Tr H(d) ρ(d)−ρ˜(d) =λTr H(A) ρ(A)−ρ˜(A) , (28)
P P P P P P P P
andtheproblemreducestotheoneanalysedatthebeginningofthissection(thatis,totheextractionofworkfroma
qutritsystemdescribedbythepassivestateρ(A),withHamiltonianH(A)),withtheonlydifferenceofamultiplicative
P P
factor λ∈(0,1) in ∆W.
B. Work extraction and k-activable states
Thesetofpassivestatescanbedividedintoahierarchyofclasses,whichdividesthestatesaccordingtothenumber
ofcopiesneededtoactivatethem. Here,wesaythatastateisactive ifitisnotpassive,andthereforeifwecanextract
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 k copies of a passive state are active, we call the state k-activable.
Wenowshowthat, ifworkisextractedfromaqutritpassivestateρ , withHamiltonianH , throughthecycleS ,
P P m,n
thenthestaterealisedbym+ncopiesofρ isactive. Itworthnotingthat,whileourcycleonlyrequiresanadditional
P
system of dimension m+n to extract work from ρ , in order to activate the same state we would need m+n−1
P
copies of it, that is, an ancilla whose size is exponential in n+m.
Inthefollowing, weconsideraqutritsystem, althoughthesameargumentappliestoquditsystems, forthereasons
presentedintheprevioussection. Ifthepassivestateρ isactivatedbythecycleS ,thenoneofthetwoconditions
P m,n
in Eq. (21) has to be satisfied. Let us assume that the conditions satisfied by state and Hamiltonian are
m∆E >n∆E , (29a)
10 21
(cid:18)p (cid:19)n (cid:18)p (cid:19)m
1 > 0 , (29b)
p p
2 1
where the other case follows straightforwardly.
Consider now a system composed by n + m copies of the qutrit system under examination, with Hamiltonian
H = (cid:80)m+nH(i), where the term H(i) acts over the i-th copy. The state of this global system is ρ⊗m+n. Then,
tot i=1 P P P
let us focus our attention on two eigenstates of H , namely, |1(cid:105)⊗m+n and |0(cid:105)⊗m ⊗ |2(cid:105)⊗n. The first eigenstate
tot
has an energy of (m+n)E , and its occupation probability is pm+n. The second eigenstate, instead, has energy
1 1
mE +nE , and its occupation probability is pmpn. It is easy to verify that, if the constraints of Eqs. (29) hold,
0 2 0 2
then the inequalities (m+n)E > mE +nE and pm+n > pmpn are satisfied, implying that the state ρ⊗m+n is
1 0 2 1 0 2 P
active. Thus, we have shown that if a passive state ρ can be activated with the cycle S , then the state ρ⊗m+n
P m,n P
is active. However, this result does not tell us whether it is possible to activate the state ρ by tensoring it with less
P
copies. In the same way, we do not know whether the fact that the state ρ⊗m+n is active implies that we can extract
P
work from ρ with the cycle S .
P m,n
9
V. GENERAL INSTABILITY OF PASSIVE STATES
Wecannowestablishourcentralclaim: thatanyathermalpassivestateisenergeticallyunstableunderareversible
process that does not generate entropy. We analyse the evolution of a passive state which sequentially interacts with
an infinite-dimensional machine M, 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 [43].
We consider a cycle composed of infinitely many hot swaps, m → ∞, and infinite many cold swaps, n → ∞, with
the assumption that n = αm, where α 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 H given in Eq. (4), described
P
by the passive state ρ whose probability distribution satisfies the equalities of Eqs. (6). Then, ρ belongs to the
P P
subset R defined in Eq. (20a), and the cycle S has to satisfy conditions 1 and 2 in order to extract work from it.
1 m,n
These conditions are reflected in the allowed range of the parameter α, that is
β ∆E ∆E
hot 10 <α< 10. (30)
β ∆E ∆E
cold 21 21
If we set α equal to a value inside the range specified by the previous equation, and we send m→∞, we find that
(seeSupplementalMaterialfordetails)thestateofthemachineasobtainedfromEq.(8)isgivenbyamixtureoftwo
“thermal” states, one with effective temperature β−1, the other with effective temperature β−1 (note we still have
hot cold
H = 0 for the machine). These distributions have support in two different subspaces, and their weight depends
M
non-trivially on the energy gaps of H and on the virtual temperatures of ρ . In fact, we can loosely interpret the
P P
state of the machine in terms of a thermal mixture
ρ =λτ +(1−λ)τ , (31)
M βhot βcold
where
τβhot = e−βZhotHhot , with Hhot =m(cid:88)−1j∆E10|j(cid:105)(cid:104)j|M and Zhot =Tr(cid:2)e−βhotHhot(cid:3), (32)
hot
j=0
and
τβcold = e−βZcoldHcold , with Hcold =n(cid:88)−3j∆E21|j+m(cid:105)(cid:104)j+m|M and Zcold =Tr(cid:2)e−βcoldHcold(cid:3). (33)
cold
j=0
Notice that in order to define these “thermal states” we have introduced two fictitious Hamiltonians, namely, H
hot
andH . Theseoperatorsarenecessaryifwewanttoconsiderthedistributionofthemachineasthemixtureoftwo
cold
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 M 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 H =0, so that the machine acts as a memory.
M
The weight λ in the mixture is given by
1−e−βcold∆E21
λ= . (34)
1−e−βhot∆E10−βcold∆E21
Thus,duringthecycle,thepassivestateρ firstinteractswiththe“hotreservoir”,byperformingasequenceofswaps
P
between the pair of states |0(cid:105) and |1(cid:105) and the levels of τ . Then, the state interacts with the “cold reservoir”,
P P βhot
performing a sequence of swaps between the pair |1(cid:105) and |2(cid:105) and the levels of τ .
P P βcold
In this scenario, we find that the probability distribution of the passive state ρ is infinitesimally modified, and
P
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, ∆P ∝ e−βhotm∆E10 for m → ∞. Let us consider the probability
distribution of the final state of the system ρ˜ . Since the distribution only changes infinitesimally during the cycle,
P
we can recastEqs. (17)as a setof differential equations(see the SupplementalMaterial 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 ρ is governed by the following equation
P
dp1 =−(cid:16)1+α(cid:0)p (t),p (t)(cid:1)(cid:17)dp0, (35)
dt 0 1 dt
where the parameter t 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(p ,p ). When the parameter function α(p ,(t),p (t))
0 1 0 1
is equal to one of its limiting values, Eq. (35) assumes a clear meaning. In fact,
10
• when α(p0(t),p1(t))= ∆∆EE1201, then the differential equation can be recast as a condition over the average energy
of the system, that is,
Tr[H ρ ]=Tr[H ρ˜ ]. (36)
P P P P
Then, for α taking this value, the passive state evolves along a trajectory that conserves the energy of the
system.
• whenα(p (t),p (t))= βhot(t)∆E10,instead,thedifferentialequationcanberecastasaconditionovertheentropy
0 1 βcold(t)∆E21
of the system, that is,
S(ρ )=S(ρ˜ ), (37)
P P
where S(ρ) = −Tr[ρlogρ] is the Von Neumann entropy. Then, for this α, the passive state evolves along a
trajectory that conserves the entropy of the system.
Forαtakingvaluesinsidetheallowedrange,wehavethatanytrajectorybetweenthetwopresentedaboveispossible,
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 [30]. It is
clearthat,throughthisevolution,wecanobtainanypassivestatewithasmalleraverageenergyandabiggerentropy
than ρ . In Supplemental Material, we show that these states are also the only ones that we can reach with our
P
engines (and with a broader class of maps, called activation maps).
FIG.5. (Left)Thestatespaceofaqutritsystem,wherethesubsetofpassivestatesishighlighted. TheblueregionisR ,while
1
thegreenregionisR . Theblacklineisthesetofthermalstates. Theinitialstateρ isrepresentedbytheblackpointinthe
2 P
diagram. Ifαtakesthevalue ∆E10,thenthesystemevolvesalongtheredtrajectory,andthefinalstateisthethermalstateat
∆E21
temperature β . On the other hand, if α= βhot(t)∆E10, the system evolves along the purple line, and the final state is the
min βcold(t)∆E21
thermal state at temperature β . The yellow region represents the subset of achievable states when the initial state is ρ .
max P
(Right)Apartialrepresentationofthestatespaceofad-levelquantumsystemintheenergy-entropydiagram. Inthisdiagram,
quantumstatesaregroupedintoequivalenceclassesdefinedbytheiraverageenergyE andtheirentropyS. Eachpointbetween
the x-axis (the set of pure states) and the black curve (the set of thermal states) represents one of these equivalence classes.
ThediagramdependsontheHamiltonianH ofthesystem. Here,weonlyrepresentthestateswithaverageenergylowerthan
P
E¯ =Tr[H ρ ], where ρ = I is the maximally-mixed state, since all passive states are contained in this set. For a given
P mm mm d
initial state ρ , the green region contains all the passive states which can be achieved with the process.
P
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 H at temperature β−1, that is, τ , where the temperature is such that
P min βmin
Tr[H ρ ]=Tr[H τ ]. Inthiscase,itiseasytoseethattheengineisnotextractinganywork,anditsonlyeffect
P P P βmin
consists in raising the entropy of the system. If we consider the efficiency of this cycle, Eq. (16), we see that η = 0,
as expected. The opposite limit is obtained when the system evolves along the entropy-preserving trajectory. In this