Toward conjecture: Heating is faster than cooling
TToward Conjecture: Warming is Faster than Cooling
Tan Van Vu ∗ and Yoshihiko Hasegawa † Department of Information and Communication Engineering,Graduate School of Information Science and Technology,The University of Tokyo, Tokyo 113-8656, Japan
An asymmetry in thermal relaxation toward equilibrium has been uncovered for Langevin systemsnear stable minima [Phys. Rev. Lett. 125, 110602 (2020)]. It is conjectured that given the samedegree of nonequilibrium of the initial states, relaxation from a lower-temperate state (warming) isalways faster than that from a higher-temperature state (cooling). Here we elucidate this conjecturefor discrete-state systems undergoing Markovian dynamics. We rigorously prove that warming isalways faster than cooling for arbitrary two-state systems, whereas for systems with more thantwo distinct energy levels, the conjecture is, in general, no longer valid. Moreover, for systemswhose energy levels are degenerate to two energy states, we find that there exist critical thresholdsof the energy gap. Depending on the magnitude of this energy gap, warming can be definitelyfaster or slower than cooling, irrespective of the transition rates between states. Our results clarifythe conjecture for discrete-state systems and reveal several hidden features inherent in thermalrelaxation.
Introduction. —A system attached to thermal reser-voirs will relax toward a stationary state. Such ther-mal relaxation is ubiquitous in nature and possesses richfeatures from both dynamic and thermodynamic pointsof view. One of the counterintuitive behaviors is theMpemba effect [1], where cooling down a hot systemis faster than doing so with the same system preparedat a colder temperature. Such nonmonotonic relaxationphenomena have been observed in various systems [2–6] and theoretically analyzed for microscopic dynamics[7–9]. In the same line of this finding, Ref. [10] foundthat cooling the system before heating it could lead toan exponentially fast relaxation. From the perspective ofthermodynamics, the irreversibility of a relaxation pro-cess can be quantified by irreversible entropy production[11]. Notably, it has been shown that irreversible entropyproduction during relaxation is bounded from below byinformation-theoretical [12, 13] and geometrical [14] dis-tances between the initial and final states, resulting instronger inequalities than the conventional second lawof thermodynamics. These relations impose geometricalconstraints on the possible relaxation path. Since ther-mal relaxation is important in condensed matter [15] andheat engines [16], deepening our understanding of relax-ation would benefit research in these areas.Consider the task of preparing a thermal state by let-ting the system relax as coupled to a single reservoir ata fixed temperature. In this case, the relaxation timeis a quantity of interest and can be approximately esti-mated by the convergence rate of the system state to-ward the equilibrium state [17]. Given two identical sys-tems initiated in thermal states, one at a lower tempera-ture and the other at a higher temperature, then a nat-ural question is risen: Which one relaxes faster? Re-cently, considering a pair of thermodynamically equidis-tant temperature quenches, Ref. [18] has unveiled an un-foreseen asymmetry in thermal relaxation for continuous- state Langevin systems. That is, relaxation from a lowertemperature is always faster than that from a higher tem-perature. Roughly speaking, it implies that warming upcold objects is faster than cooling down hot objects. Thisphenomenon has been proved for quenches of dynamicsnear stable minima, and the case of general overdampeddynamics remains to be a conjecture. Taking the aver-age energy into account, one can observe that the en-ergy of a thermal state at a higher temperature is largerthan that at a lower temperature. This allows us to saythat from the energetic point of view, uphill relaxationis faster than downhill relaxation, which is counterintu-itive to an extent. Moreover, relaxation speed cannot becharacterized solely by thermodynamic quantities such asdissipation or frenesy [19]. Therefore, it is nontrivial thatthe free energy emerges as a quantifier of nonequilibriumdegree in equidistant temperature quenches.In this Letter, we put forward our understandingof thermal relaxation by elucidating the conjecture fordiscrete-state systems described by Markov jump pro-cesses. First, we prove that for arbitrary two-state sys-tems, warming is always faster than cooling, which par-tially validates the conjecture. For general systems withat least three distinct energy levels, however, we find thatit is not the case. By analytically constructing counterex-amples, we demonstrate that warming can be faster orslower than cooling, depending on the value of transitionrates. This implies that the conjecture does not univer-sally hold for discrete-state systems. Nevertheless, by re-stricting to a specific class of systems, some universal re-sults pertinent to relaxation asymmetry can be obtained.When the energy levels of the system are degenerate totwo energy states, there exist two critical thresholds ofthe energy gap. Depending on whether this energy gapis larger or smaller than these thresholds, it can be con-cluded with certainty that warming is faster or slowerthan cooling. a r X i v : . [ c ond - m a t . s t a t - m ec h ] F e b Setup. —We consider the thermal relaxation process ofan open system with N states. The system is coupledto a thermal reservoir at the inverse temperature β f = T − f . Due to interaction with the reservoir, stochastictransitions between states are induced. The dynamics ofthe system are governed by the master equation: ∂ t | p ( t ) i = R | p ( t ) i , (1)where | p ( t ) i := [ p ( t ) , . . . , p N ( t )] > denotes the probabil-ity distribution of the system at time t . The matrix R = [ R mn ] ∈ R N × N is time-independent with R mn ≥ n to state m ( = n ),and P m R mn = 0. Without loss of generality, we assumethat E ≥ · · · ≥ E N , where E n denotes the energy ofstate n . We also assume that the system satisfies thedetailed balance condition, R mn e − β f E n = R nm e − β f E m .This is the sufficient condition such that the system al-ways relaxes to the thermal Gibbs state | π f i after a suf-ficiently long time, irrespective of the initial state. Here π fn = e − β f E n /Z β f and Z β f = P Nn =1 e − β f E n . The tran-sition rates can be expressed as R mn = Γ e − β f ( B mn − E n ) for m = n , where B mn = B nm are the barrier coefficientsand Γ is a positive constant.Now, let us formulate the problem. We consider re-laxation that starts from a thermal state | π i i associatedwith the inverse temperature β i = T − i . This can be re-garded as a temperature quench T i → T f at time t = 0.Given a pair of cold and hot temperatures T c and T h sat-isfying T c < T f < T h , we explore the relaxation speeddepending on the direction of the quench, T i = T c ↑ T f (warming) and T i = T h ↓ T f (cooling). The degree ofnonequilibrium (or the free energy) of the initial stateis the same, i.e., D ( | π c i , | π f i ) = D ( | π h i , | π f i ), where D ( | p i , | q i ) := P n p n ln( p n /q n ) is the relative entropy be-tween two distributions | p i and | q i . Our aim is to an-swer the question of which direction leads to a fasterrelaxation. To this end, we first need to quantify therelaxation speed, which is evaluated by the distance be-tween the system state and the thermal state. Analo-gous to Refs. [7, 18], the relative entropy is used to mea-sure the distance between states. In thermal relaxation,the relative entropy is closely related to the free energyand irreversible entropy production [20]. Let | p i ( t ) i bethe distribution corresponding to the initial state | π i i ,then warming is said to be faster (slower) than coolingif D ( | p c ( t ) i , | π f i ) < ( > ) D ( | p h ( t ) i , | π f i ) in the long time.In the following, we explain in details how to evaluatethe relaxation speed.Let 0 = λ > λ > · · · > λ N be the eigenvalues ofthe transition matrix and {| v n i} Nn =1 be the set of corre-sponding eigenvectors, R | v n i = λ n | v n i . (2)It should be noted that all eigenvalues are real numberssince the matrix R satisfies the detailed balance condition [21]. The eigenvectors {| v n i} form a basis for the space R N with | v i = | π f i and | v n i is a traceless vector forall n ≥
2, i.e., h | v n i = 0 since h | R = h | . Here, | i ( | i ) denotes the N -dimensional vector with all zero (one)elements. Therefore, the initial distribution | π i i can beexpressed as a linear combination in terms of {| v n i} as | π i i = | π f i + N X n =2 γ in | v n i , (3)where γ in ’s are real numbers. The analytical form of thedistribution at time t can be written as | p i ( t ) i = | π f i + N X n =2 γ in e λ n t | v n i . (4)In the long-time limit, the probability distribution | p ( t ) i can be approximated to second order as | p i ( t ) i ≈ | π f i + γ i e λ t | v i . (5)The relaxation speed can thus be quantified via the valueof | γ i | [7] (see Supplemental Material [22] for proof).Accordingly, warming is faster (slower) than cooling if | γ c | < ( > ) | γ h | .A closed form of γ i can be obtained [8]. The transitionmatrix can be transformed to a symmetric form as R = F / RF − / , (6)where F = [ F mn ] ∈ R N × N with F mn = e β f E n δ mn . Thematrix R has the same eigenvalues as R , and its eigenvec-tors {| f n i} are related to those of R as | f n i = F / | v n i .Moreover, these eigenvectors are mutually orthogonal, h f m | f n i = h v n | F | v m i δ mn . Multiplying h f | F / to bothsides of Eq. (3), one obtains γ i = h f | F / | π i ih f | f i = h f | π i ih f | f i , (7)where | f i := F / | f i . As can be seen, γ i is proportionalto the inner product between the initial distribution andthe vector | f i . Since the sign of γ in can be absorbedby changing the eigenvectors | v n i → −| v n i , hereafter, weassume γ hn ≤ ≤ n ≤ N . Main results. —Given the above setup, we now presentour main results, which are three theorems regarding therelaxation asymmetry.
Theorem 1.
For two-state systems, warming is alwaysfaster than cooling.
This result positively supports the conjecture. Evenfor two-state systems, it is nontrivial that warming isfaster than cooling. For discrete-state Markovian dynam-ics, the speed of state change is constrained by time-antisymmetric dissipation and time-symmetric frenesy(or dynamical activity) [23]. Relaxation from a thermalstate at a higher temperature has a higher dynamical ac-tivity. In other words, it has more jumps between statesthan relaxation from a thermal state at a lower tempera-ture. Consequently, one may intuitively expect that cool-ing, which has higher dynamical activity, is faster thanwarming. However, the result is contrary to our intuition,suggesting that the dynamical activity alone cannot ac-count for this relaxation asymmetry.
Proof of Theorem 1.
It is enough to prove that | γ c | < | γ h | = − γ h . In the N = 2 case, an arbitrary probabilitydistribution | π i i can be expressed as a point ( x i , − x i )in the two-dimensional space. Since E > E , all thethermal states | π c i , | π f i , and | π h i lie on the segmentwith endpoints (0 ,
1) and (1 / , / h f | π f i = 0 since | f i = F / | π f i and | f i are orthogonal. From the con-ditions h f | π f i = 0 and h f | π h i ≤
0, we can concludethat h f | π c i ≥ γ c ≥
0. Thus, we need only show γ c + γ h < h f | ( π c + π h ) / i <
0. This is equivalentto proving that ( x c + x h ) / > x f since h f | π f i = 0. Wecan rewrite the condition D ( | π c i , | π f i ) = D ( | π h i , | π f i )as S ( | π h i ) − S ( | π c i ) = ( x h − x c ) ln 1 − x f x f , (8)where S ( | p i ) := − P n p n ln p n is the Shannon entropyof the distribution | p i . Note that x c < x h < / g ( x i ) := dS ( | π i i ) /dx i = ln[(1 − x i ) /x i ] is a strictly con-vex function over x i ∈ [0 , / g ( x i ), we obtain1 x h − x c Z x h x c g ( x i ) dx i > g (cid:18) x c + x h (cid:19) , (9) FIG. 1. Geometrical illustration of probability distributions.Two vectors | f i and | π f i are orthogonal, h f | π f i = 0. or equivalently, S ( | π h i ) − S ( | π c i ) > ( x h − x c ) g (cid:18) x c + x h (cid:19) . (10)From Eqs. (8) and (10), we have g ( x f ) > g (cid:18) x c + x h (cid:19) . (11)Since g ( x i ) is a strictly decreasing function, we readilyobtain ( x c + x h ) / > x f , which completes the proof.Next, we consider more general systems that have atleast three distinct energy levels, i.e., there exist threeindexes 1 ≤ i < j < k ≤ N such that E i > E j > E k . Forsuch systems, we obtain the following result. Theorem 2.
For systems with more than two distinctenergy levels, warming can be faster or slower than cool-ing, depending on the value of transition rates.
Theorem 2 implies that the conjecture is invalid in thegeneral case. With an appropriate choice of the barriercoefficients, one can obtain a system with | γ c | < | γ h | or | γ c | > | γ h | as desired. This difference between thecontinuous- and discrete-state systems can be explainedin the following manner. For simplicity, we considera single-particle system described by the overdampedLangevin equation. In this continuous-state system, theparticle tends to transit to a close place at any instanttime. Whereas, in discrete-state systems, the particle, inprinciple, can jump to anywhere, as long as the transi-tion rate between these states is positive. This degree offreedom could lead to a complicated relaxation as com-pared to the continuous-state system. The constructionof the transition matrix leading to violation of the con-jecture is somewhat artificial; therefore, there may not bea realistic system that has such a transition matrix. Weanticipate that the conjecture may be valid if appropriateconstraints are imposed on the transition rates. Proof of Theorem 2.
We prove the theorem by analyt-ically constructing a transition rate matrix such thatwarming is slower than cooling. A transition rate ma-trix for the opposite case can also be analogously con-structed. First, one can prove that | π c i , | π h i , and | π f i arelinearly independent (see Supplemental Material [22]).Let {| e i , | e i} be an orthogonal basis of the space S spanned by | π f i and | π h i , and P := | e ih e | + | e ih e | be the projection matrix to the space S . Define | f i := | π c i − P | π c i , then | f i 6 = | i because of the linear inde-pendence of | π c i , | π h i , and | π f i . One can easily verifythat h f | e i = h f | e i = 0, thus h f | π f i = h f | π h i = 0.Moreover, h f | π c i = h f | f i > | f i 6 = | i .Now, we construct a transition rate matrix R that re-sults in | γ c | > | γ h | . Set B mn = E m + E n , one can explic-itly calculate that the matrix R has a single zero eigen-value associated with the eigenvector | f i = F / | π f i ,and the remaining eigenvalues are all − Z β f [8]. Let U = {| x i ∈ R N | h x | f i = 0 } be the subspace that is or-thogonal to | f i . Then, one can construct an orthogonalbasis {| f i , | f i , . . . , | f N i} of U , where | f i := F − / | f i .This is always possible since h f | f i = 0, i.e., | f i ∈ U .Obviously, | f n i ( n ≥
2) is an eigenvector of R with thecorresponding eigenvalue − Z β f . Following the idea inRef. [8], we slightly modify R as R → R + N X n =2 (cid:15) n | f n ih f n |h f n | f n i , (12)where Z β f > (cid:15) > · · · > (cid:15) N ≥ R mn ( m = n ). One can immedi-ately check that R | f i = 0 and R | f n i = ( − Z β f + (cid:15) n ) | f n i for all n ≥
2. The matrix R now has N different eigen-values, and | f i is exactly the eigenvector correspondingto the second largest eigenvalue λ = − Z β f + (cid:15) . Thetransition matrix can be recovered as R = F − / RF / ,and the detailed balance condition is fulfilled due to thesymmetry of R . The relation between | γ c | and | γ h | is nowclarified as | γ c | = h f | π c ih f | f i > h f | π h ih f | f i = | γ h | , (13)which completes the proof.Last, let us consider the remaining case, where theenergy levels are degenerate to two energy states. Inother words, there exists an index 1 ≤ n < N such that E = · · · = E n > E n +1 = · · · = E N . Such degeneratetwo-level systems can be seen in atoms [24] and havebeen used to enhance quantum annealing performance[25]. For convenience, we define the energy gap ∆ E := E n − E n +1 >
0. Interestingly, we find that depending onthe magnitude of this energy gap, warming can be fasteror slower than cooling, regardless of the transition rates.Details are summarized in the following theorem.
Theorem 3. If β h ∆ E ≥ ln[ n/ ( N − n )] , then warming isfaster than cooling. Conversely, if β c ∆ E ≤ ln[ n/ ( N − n )] ,then warming is slower than cooling. Theorem 3 indicates that there are two critical thresh-olds of the energy gap. As the energy gap is above orbelow these thresholds, a universal conclusion on asym-metry in thermal relaxation can be made. It is also highlynontrivial that the value of the energy gap affects therelaxation speeds of warming and cooling in this way.When ∆ E is large, the jump from energy state E n +1 to E n is less likely to occur as compared to the oppo-site jump. Thus, one may expect that warming takes alonger time than cooling. However, contrary to our intu-ition, warming is faster than cooling as ∆ E increases. Inaddition, as long as n ≤ N/
2, warming is always fasterthan cooling, regardless of the value of ∆ E . This im-plies that the number of high-energy states also plays adeterminant role in relaxation speed. Proof of Theorem 3.
We employ the same strategy as inthe proof of Theorem 1. We prove the former case first,i.e., β h ∆ E ≥ ln[ n/ ( N − n )] leads to a faster warming.It can be observed that all points lying on the segment ‘ with endpoints | π c i and | π h i are thermal states. Ob-viously, | π c i , | π f i , and | π h i are linearly dependent, i.e.,there exists a real number a ∈ (0 ,
1) such that | π f i = a | π c i + (1 − a ) | π h i . Since h f | π f i = 0 and h f | π h i ≤ h f | π c i ≥ γ c ≥ h f | p i for | p i ∈ ‘ . Thus,we need only show γ c + γ h < h f | ( π c + π h ) / i < x c + x h ) / > x f ,where x i = π in completely characterizes the thermal state | π i i . The condition D ( | π c i , | π f i ) = D ( | π h i , | π f i ) can berewritten as S ( | π h i ) − S ( | π c i ) = ( x h − x c ) n ln 1 − nx f ( N − n ) x f . (14)Note that x c < x h ≤ / (2 n ) since β h ∆ E ≥ ln[ n/ ( N − n )],and g ( x i ) := dS ( | π i i ) /dx i = n ln[(1 − nx i ) / ( N − n ) x i ] isa strictly convex function over x i ∈ [0 , / (2 n )]. Applyingthe Hermite–Hadamard inequality for g ( x i ) and followingthe same steps as in Eqs. (9) and (10), one readily obtains( x c + x h ) / > x f , which proves the former case.When β c ∆ E ≤ ln[ n/ ( N − n )], one can derive that1 / (2 n ) ≤ x c < x h , and g ( x i ) is a strictly concave functionover x i ∈ [1 / (2 n ) , g ( x i ), we have g ( x f ) = 1 x h − x c Z x h x c g ( x i ) dx i < g (cid:18) x c + x h (cid:19) , (15)or x f > ( x c + x h ) /
2, which implies | γ h | < | γ c | . Conclusion. —In this Letter, we elucidated the con-jecture regarding the relaxation asymmetry for discrete-state systems described by Markov jump processes. Weproved that the conjecture holds for any two-state sys-tems but is invalid for general systems with more thantwo distinct energy levels. For systems with two degen-erate energy levels, we obtained some universal resultsindicating that the asymmetry in relaxation depends onthe energy gap and the number of high-energy states.While the conjecture is shown to be invalid in the gen-eral discrete-state case, it is possible that under someconstraints on transition rates, the conjecture is valid.Such a question requires further investigation and willbe addressed in future work. It is also interesting tostudy the relaxation asymmetry for non-Markovian sys-tems [26] and open quantum systems [27, 28] where co-herence emerges in the initial state.We thank Keiji Saito for many insightful discussions.This work was supported by Ministry of Education, Cul-ture, Sports, Science and Technology (MEXT) KAK-ENHI Grant No. JP19K12153.
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Tan Van Vu ∗ and Yoshihiko Hasegawa † Department of Information and Communication Engineering,Graduate School of Information Science and Technology,The University of Tokyo, Tokyo 113-8656, Japan
This supplemental material describes the details of calculations introduced in the main text. The equations andfigure numbers are prefixed with S [e.g., Eq. (S1) or Fig. S1]. Numbers without this prefix [e.g., Eq. (1) or Fig. 1]refer to items in the main text.
Quantification of relaxation speed
For convenience, we define D ( | p i ) := D ( | p i , | π f i ). Given | π i i = | π f i + P Nn =2 γ in | v n i and | γ c | < | γ h | , we will showthat D ( | p c ( t ) i ) < D ( | p h ( t ) i ) as t → ∞ . Note that | p i ( t ) i = | π f i + P Nn =2 γ in e λ n t | v n i . In the long-time limit, the term P Nn =2 γ in e λ n t | v n i vanishes. Since D ( | p i , | p + dp i ) = N X k =1 dp k p k + O (∆ ) , (S1)where ∆ = P Nk =1 | dp k | , we can approximate D ( | p i ( t ) i ) = N X k =1 (cid:16)P Nn =2 γ in e λ n t v nk (cid:17) p ik + O (∆ ) ( p ik ) − =( π fk ) − + O (∆) = N X k =1 (cid:16)P Nn =2 γ in e λ n t v nk (cid:17) π fk + O (∆ ) . (S2)Subsequently, we have D ( | p h ( t ) i ) − D ( | p c ( t ) i ) = (cid:2) ( γ h ) − ( γ c ) (cid:3) N X k =1 v k π fk e λ t + X m,n ≥ { m,n } > a mn e ( λ m + λ n ) t + O (∆ ) , (S3)where a mn are constants. The first term in the right-hand side is positive since | γ c | < | γ h | . The remaining terms maybe negative; however, they are negligible compared to the first term in the long-time limit. Therefore, D ( | p c ( t ) i ) < D ( | p h ( t ) i ) as t → ∞ . Proof of linear independence
To prove the linear independence of | π c i , | π f i , and | π h i , it is sufficient to show that the determinant of the followingmatrix is always negative. X = e − β c E e − β c E e − β c E e − β f E e − β f E e − β f E e − β h E e − β h E e − β h E . (S4)Here, E > E > E and β c > β f > β h . Without loss of generality, one can assume that E = 0 and β h = 0. In thiscase, the determinant can be calculated as | X | = (1 − e − β c E )(1 − e − β f E ) − (1 − e − β c E )(1 − e − β f E ) . (S5)Therefore, | X | < − e − β c E − e − β f E < − e − β c E − e − β f E . (S6) a r X i v : . [ c ond - m a t . s t a t - m ec h ] F e b Set h ( x ) := (1 − e − β c x ) / (1 − e − β f x ), Eq. (S6) is equivalent to h ( E ) < h ( E ). We need only prove that h ( x ) is astrictly decreasing function over x >
0. Taking the derivative of h ( x ) with respect to x , one obtains dh ( x ) dx = e ( β f − β c ) x [ β c ( e β f x − − β f ( e β c x − e β f x − . (S7)Since β c ( e β f x − − β f ( e β c x −
1) = ∞ X n =1 x n n ! ( β c β nf − β f β nc )= ∞ X n =1 x n n ! β c β f ( β n − f − β n − c ) < ∵ < β f < β c ) , we have dh ( x ) /dx