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Now that we are properly equipped with our sumtemperature notion, we are ready to use it within the context of the coherent and incoherent scenario. We remind the reader that unless otherwise stated, we will in the following assume that we have a dS <dimensional systemSat hand as well as adM <dimensional machineM with maximal energy gapEmax = EdM1 < ∞. As usual,Sand Mwill be assumed to have an initial state thermal atTR and the underlying Hamiltonian to be non-interacting. Given such a system and machine, we will be in the following interested in investigating what is the lowest temperature, in terms of sumtemperature, that can be achieved on the target system state when one is allowed to apply the coherent and incoherent scenario an unbounded number of times. In short, we will be interested in Λcoh(ρS)andΛinc(ρS).

Our first result in this direction will consist of a bound on the achievable tem-perature ofΛcoh(ρS)andΛinc(ρS), i.e., a bound that holds for both scenarios. Before we turn to the bound itself, we would like to state and prove a preliminary result that will allow simplifying our analysis as well as clarifying the interplay between both scenarios in terms of state attainability. This result states that given a system stateσSdiagonal in the energy eigenbasis, one can reach any sumtemperature via a single application of the coherent scenario that one can reach via a single application of the incoherent scenario. We remind the reader that this relation is a priori not clear.

Indeed, while the incoherent scenario restricts to energy conserving unitaries only – as opposed to arbitrary unitaries for the coherent scenario – the state of the machine in the incoherent scenario is allowed to have a temperature gradient, and as such has a richer structure than in the coherent scenario. The formal result reads as follows.

Lemma 20. Let our system and machine, S and M, be given. LetσSbe a state of S diagonal in the energy eigenbasis. Then, for all v ∈ RdS such that there exists a Λinc with v = D(Λinc(σS)), there also exists aΛcohsuch that v=D(Λcoh(σS)).

Proof. The proof is not hard. It relies mostly on the fact that a hotter thermal state is majorized by a colder one, i.e., for us

τMH2τM2, (12.19)

and that majorization is stable under tensor product, see Corollary 1.2. of [90].

For us that means

ρR,MM =τMR1τMH2τMR1τMR2 = ρM. (12.20) From this we get, using the stability under tensor product again, that

σSρR,HMσSρM, (12.21) which means

λ

σSρR,HM

λ(σSρM). (12.22)

Now let ˜Λinc be a specific application of the incoherent scenario. Let ˜Uinc be the energy conserving unitary such that ˜Λinc(·) = TrM(U˜inc· ⊗ρR,HMinc ). Then by Schur’s Theorem, Theorem1, we have that

Chapter 12. Qudit System 72 Now using Horn’s Theorem, Theorem2, there exists a unitary ˜Usuch that

D

What Lemma20means is that, for a given machine M, we cannot cool a given systemS more with a single application of the incoherent scenario than with one application of the coherent scenario. There are two things one should point out at this stage. First of all, this result does not tell us that we can cool more within the coherent scenario than within the incoherent one. It might indeed well be that for a given machine, both scenarios, within their single application regime, allow reaching the same temperature. Whether a gap exists between both scenarios therefore remains an open question. Second of all, this result really only makes a statement about a single application of each scenario. One could imagine that the incoherent scenario is slow to start off, but that it catches up on the coherent one upon repeated applications and maybe even eventually allows for more cooling. Our next result gets rid of this possibility. It shows that for a given numberk ∈ Nof applications of each scenario, any temperature thatΛkinc(σS)can reach can also be reached byΛcohk (σS). The statement reads as follows.

Lemma 21. Let S and M be given. Let k∈N. LetσSbe a state of S diagonal in the energy eigenbasis. Then for all v∈ RdS for which there exists aΛkincsuch that v = D(Λinck (σS)), there also exists aΛcohk such that v =D(Λcohk (σS)).

Note that we writeΛkcohandΛinck to meankapplications of each scenario but that we allow the specific maps chosen at each step to vary from one step to the other, i.e., Λkcoh=Λ(cohk)Λ(cohk1)◦ · · · ◦Λ(coh1), (12.28) andΛ(cohi) does not have to be equal toΛ(cohj) fori6=j. Similarly forΛkinc.

The statement of Lemma21being a straightforward generalization of Lemma20, one would expect the proof of it to straightforwardly harness the result of Lemma20.

Since the initial stateσS is in both cases demanded to be diagonal and that each scenario does not necessarily deliver diagonal state, one nevertheless have to deal with a few subtleties first. To that end we first state and prove the following technical Lemma.

Lemma 22. Letσ1andσ2be diagonal system states such thatσ1σ2. Then for all v∈RdS for which there exists a Λinc such that v = D(Λinc(σ1)), there exists a Λcoh such that v=D(Λcoh(σ2)).

Chapter 12. Qudit System 73 Proof. Using the stability of majorization under tensor product, Corollary 1.2. of [90], we have that

σ1ρMσ2ρM. (12.29)

Since majorization is transitive one can straightforwardly replace all the instances ofσ1ρM in the proof of Lemma20byσ2ρM, which proves our result.

Now we are ready to prove Lemma21.

Proof of Lemma21. The proof is by induction over k. k = 1 is the statement of Lemma 20. Now suppose that the statement is true for k, we will show that it holds fork+1. Letv∈RdSfor which there exists aΛkinc+1 =Λ(inck+1)Λ(inck)◦ · · · ◦Λ(inc1), with associated unitariesUinc(i), such that

v=D(Λkinc+1(σS)). (12.30) We look atΛinck =Λ(inck)◦ · · · ◦Λ(inc1)kinc(σS)may not be diagonal but by Lemma6 we know that it is diagonalizable via a local energy conserving unitary, call it ˜Uinc. Then is diagonal. As before, there is aΛkcohsuch that

Λcohk (σS) =U˜Λ˜cohk (σS)U˜. (12.37) Furthermore, by Schur’s Theorem (Theorem1),

D

Chapter 12. Qudit System 74 So by Lemma22, there exists aΛcohsuch that

v=D

Λinck+1(σS)=D(Λinc(σ1)) =D(Λcoh(σ2)) =D

Λkcoh+1(σS), (12.42) whereΛkcoh+1=ΛcohΛcohk .

With these preliminary results, we have gathered some good intuition about how both scenarios compare to one another in terms of state attainability and are ready to move to the main result of this section. The result states that in the limit of unbounded, as well as infinite if well-defined, applications of each scenario, all the states that one can ever reach are sumhotter than the following state

σS = terms, the statement reads as follows.

Theorem 11(Universal bound). Let S be a system. Let M be a machine with maximal energy gapEmax, such thatρSσS. Then for all k ∈N Proof Idea. The proof idea is the following. First of all, since the incoherent scenario cannot perform better than the coherent scenario, Lemma21, one really only needs to prove the result for the coherent scenario. For the coherent scenario one can show, see Section A of the Supplementary material of [92], that ifρSσSthen also Λcoh(ρS) ≺ σS for any choice ofΛcoh. To prove this, one first only considersΛcoh, the coherent operations cooling the most, i.e., the Λcoh that has the associatedU reorderingρSMsuch that the greatest eigenvalues ofρSM contribute to the ground state and so on. One showsΛcoh(ρS)≺ σS and then uses that for any other choice

Chapter 12. Qudit System 75 For any other sequence Λkcoh(ρS) for which all partial sums converge, the same argument holds and

Λcoh(ρS)≺σS. (12.48)

For a qubit target system the bound can be expressed as a temperature and tells us that all the states that can be reached with both scenarios have a temperature higher or equal than

T= ES

EmaxTR, (12.49)

provided that the initial state has a temperature higher or equal thanT, i.e., provided T ≤ TR. The qubit bound was first derived in [93] and also appears in [94]. To relate this result to the machines that we considered in chapter11, for the one qubit machine,Emax=EM, and for the two qubit machine,Emax= EM1+EM2.

Note also that the stateσScan be viewed as a thermal state at inverse temperature βRof the modified target Hamiltonian

S =

dS1 k

=0

kEmax|ki hk|S. (12.50) That is

σS = e

βRH˜S

Tr

eβRH˜S. (12.51)

We will, however, try to avoid this notation since writingσS as such might give the wrong impression that what we are doing is modifying the Hamiltonian of the system fromHSto ˜HSinstead of actually cooling the system. Indeed, changing the Hamiltonian of a system is equivalent to completely changing it and might therefore be considered as cheating. As a comparison, if upon given a warm beer to cool one is given back a cold glass of water, most of us would not see this procedure as cooling a beer.

Finally, note that a particular feature of this bound is that it does not depend on all the intricacies of the machine or even on its dimension. It only cares about its maximal energy gap,Emax. This simplifies a lot the analysis as one can very efficiently determine what the lowest achievable temperature on the target system is given some machineM.

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