11.3 Two Qubit Machine

11.3.1 Single Cycle

Given a machine dimensiondMand a systemS, the main challenge of the incoher-ent scenario is to idincoher-entify which specific machines allow cooling systemS. From Section10.2we know that the machine has to have a tensor product structure to be able to coolS. Indeed, if not, the only way it can use the hot thermal bath is to either stay thermal atTRor to entirely thermalize toTH. But we know from Lemma9and Lemma8that this doesn’t lead to any cooling ofS, whatever the degeneracies of

Chapter 11. Qubit System 52 HSM. For machines of dimension 4, i.e.,dM =4, this means that the useful machines have to consist of 2 qubits. Furthermore, the only way of making good use of the hot thermal bath is to leave one qubit atTRand thermalize the other toTH. W.l.o.g. let M1remain atTRandM2be the one thermalizing toTH. We therefore have

ρRSM,H =τSτM1τMH2. (11.62) The next step is to screen through the potential degeneracies ofHSMto see which ones enable to coolS. From Lemma10we know that degenerate subspaces consisting of degeneracies of the machine only are of no use. This gets rid of the degeneracies EM1 =0,EM2 =0, andEM1+EM1 = 0 as potential candidates. Similarly, Lemma11 tells us thatES = 0 is also not a useful degeneracy to have inHSM. This leaves us with 5 potential degeneracies

1. ES=EM1 2. ES=EM2 3. ES=EM1 +EM2 4. EM1 =ES+EM2 5. EM2 =ES+EM1.

One can show explicitly, see Appendix A of [91], that none of these degeneracies lead to cooling except number4

EM1 =ES+EM2. (11.63)

This seems to fully take care of the degeneracy question fordM =4. There is one small subtlety, however. Degenerate subspaces can be defined by more than just one degeneracy condition. To fully conclude the proof that one cannot cool incoherently with machines of dimension 4 unless

EM1 =ES+EM2, (11.64)

we therefore have to make sure that combined subspaces cannot get activated, i.e., cool, although they individually were useless. For the 2 qubit machine this can be done explicitly. For the details we refer to Appendix A of [91]. We therefore have the following result

Theorem 5. Given a qubit target system and a machine of dimension 4, the incoherent scenario can cool the target if and only if the following holds

1. The machine comprises two qubits M1and M2 2. EM1 =ES+EM2

3. Only M2is brought into contact with the hot thermal bath.

Once the machine, as well as which part to heat up, has been identified, the cooling part of the incoherent scenario is fairly straightforward. Indeed, given the hot temperatureTH, one heats upM2toTHand then performs the unitary cooling maximally within the degenerate subspace. Note that the unitary operation is in particular performed at no cost. In the case of the 2 qubit machine this unitary is given by

Chapter 11. Qubit System 53

U= |010i h101|SM+|101i h010|SM+1spanc{|010i,|101i}, (11.65) where spanc{|010i,|101i}denotes the complement of the set span{|010i,|101i}. Doing so we get

rinc(TH) =rSrM1+ [(1−rS)rM1 +rS(1−rM1)] (1−rHM2). (11.66) The associated temperatureTinc(TH)is given by the usual formula,

Tinc(TH) = ES ln

rinc(TH) 1rinc(TH)

. (11.67)

The work cost is calculated from the hot bath. According to Eq.8.4this amounts to calculating the heat drawn from the bath, QH, which is equal to the change of energy ofM2upon heating it. We therefore get

QH =EM2(rM2−rHM2) (11.68) and

∆Finc(TH) =EM2(rM2−rHM2)

1− TR TH

. (11.69)

From this we get the * quantities by taking the limitTH∞, i.e.,

rinc= 1

2(r+rM1) (11.70)

Tinc = ES ln r

S+rM1

2−(rS+rM2)

(11.71)

∆Finc = EM2(rM21

2). (11.72)

Coherent Scenario

We now turn our attention to the coherent scenario. Since we are mostly interested in seeing how it compares to the incoherent scenario, we will impose here the restriction

EM1 =ES+EM2, (11.73)

although this is not needed here for the machine to cool. This will also simplify the analysis, as will become clearer in the following. As in Section11.2.2, we are interested in solving a special instance of the general problem of Eq.10.28. More precisely, we are interested in solving

vminD(ρSM)v·D(HSM), s.t.

3 i=0

vi =c, (11.74)

with

Chapter 11. Qubit System 54

We first would like to determinercohto know what rangecis allowed to evolve in, and also to know what the maximal cooling on the target is. For this we need to determine what the 4 biggest entries ofD(ρSM)are. Looking at this more closely, and taking our restriction of Eq.11.73into account, we find the following ordering

[D(ρSM)]0>{[D(ρSM)]1,[D(ρSM)]4}> [D(ρSM)]2= [D(ρSM)]5> (11.81)

which gives an associated temperature of Tcoh = ES

EM1TR. (11.86)

A transformation achieving this cooling is the unitaryUswapping the energy eigenstates|011iSM

Chapter 11. Qubit System 55 This is however not the unitary operating at the minimal work cost since the state that one gets after performingUis not passive in the respective subspaces of Lemma4. We will now work out the energetically most efficient transformation, which depends on the ordering ofD(ρSM). As the ordering depends on whether EM2 ≤ESorEM2 > ES, we treat both cases separately.

a.EM2 ≤ES In this case we have

[D(ρSM)]0>[D(ρSM)]1,[D(ρSM)]4 >[D(ρSM)]5= [D(ρSM)]2>

>[D(ρSM)]3,[D(ρSM)]6 >[D(ρSM)]7, (11.88) and so the vectorvR8minimizing Eq.11.74forc=rcohis given by

v= ([D(ρSM)]0,[D(ρSM)]1,[D(ρSM)]4,[D(ρSM)]5,[D(ρSM)]2, (11.89) ,[D(ρSM)]3,[D(ρSM)]6,[D(ρSM)]7). (11.90) One checks that, indeed, the first half ofv,

av = ([D(ρSM)]0,[D(ρSM)]1,[D(ρSM)]4,[D(ρSM)]5), (11.91) is inversely ordered with respect the first half ofD(HSM),

aD(HSM)= (0,EM2,EM1,EM1+EM2), (11.92) and that the second half ofv,

bv = ([D(ρSM)]2,[D(ρSM)]3,[D(ρSM)]6,[D(ρSM)]7), (11.93) is also inversely ordered with respect the second half ofD(HSM),

bD(HSM)= (ES,ES+EM2,ES+EM1,ES+EM1+EM2). (11.94) A unitaryU that achievesvis given by

U =U24U35, (11.95)

with

Uij =|i2i hj2|+|j2i hi2|+1spanc{|i2i,|j2i}, (11.96) wherei2is the 3 digit display ofi=0, . . . , 7 in base 2, e.g., 02 =000. This corresponds to swapping the energy eigenstates|010iSMwith |100iSM as well as|011iSMwith

|101iSM. Physically it corresponds to swapping the population ofSwithM1andU can be written compactly as

U =eiπ2LSM1, (11.97) where

LSMk =i|01i h10|SM

k−i|10i h01|SM

k. (11.98)

The work cost associated toU is

∆Fcoh = (EM1−ES)(rM1 −rS) (11.99)

= EM2(rM1 −rS). (11.100)

Chapter 11. Qubit System 56 b.EM2 >ES In this case we have

[D(ρSM)]0 >[D(ρSM)]4,[D(ρSM)]1> [D(ρSM)]5= [D(ρSM)]2> (11.101)

>[D(ρSM)]6,[D(ρSM)]3 >[D(ρSM)]7, (11.102) and so the vectorv>R8minimizing Eq.11.74forc=rcohis given by

v>= ([D(ρSM)]0,[D(ρSM)]4,[D(ρSM)]1,[D(ρSM)]5,[D(ρSM)]2, (11.103) ,[D(ρSM)]6,[D(ρSM)]3,[D(ρSM)]7). (11.104) As before, av> andbv> are indeed inversely ordered w.r.t. aD(HSM) andbD(HSM)

respectively. A unitaryUthat achievesv>is here given by

U> =U24U35U14U36. (11.105) Physically,U14andU36 together correspond to swapping the populations ofS andM2. Once this is done,U24U35is applied, which corresponds to swapping the population ofSwithM1. One can writeU> compactly as

U> =eiπ2LSM1eiπ2LSM2. (11.106) The work cost associated toU> is

∆Fcoh, > = (EM2−ES)(rM2 −rS) +EM2(rM1−rM2). (11.107) All the analysis of this section until now is a straight exemplification of Lemma4.

It allowed us to nevertheless gather some important intuition about the general case ofrcoh ∈ [rS,rcoh]. We also saw that in the caseEM2 > ES, one could adopt a better strategy than swapping the populations ofSand M1directly by swapping the popu-lations ofSwith that ofM2first. The achieved temperature on the target is the same but the work cost differs between both strategies.

Next we turn our attention to the case, where rcoh ∈ [rS,rcoh]. Here we again distinguish between the caseEM2 ≤ ESandEM2 >ES. ForEM2 ≤ESwe find a result that is a complete analog of the 1 qubit machine result of Section11.2.2.

Theorem 6. LetEM2 ≤ ES. Let rcoh ∈ [rS,rcoh]. Letµ = rrcohrS

M1rS. Let t = arcsin(√ µ). Then

v=D(UρSMU), with U =eitLSM1 (11.108) minimizes the optimization problem of Eq.11.74for c=rcohand has an associated work cost of

∆Fcoh = (rcoh−rS)(EM1−ES). (11.109) Proof idea. The idea of the proof is exactly the same as that of Theorem4, namely to rewrite

v·D(HSM) (11.110)

such that the majorization conditions as well as the constraint can naturally be expressed. The practical rewriting depends on the ordering ofD(ρSM)as well as how ES,EM1, andEM2 relate to one another. In the caseEM1 = ES+EM2 andEM2 ≤ ESthe ordering ofD(ρSM)is fixed, and given by Eq.11.88, and the useful rewriting is the

Chapter 11. Qubit System 57 Using again that the minimum of the sum is greater than the sum of the minima we get

Qij : permutation matrix exchanging coordinatesiandjonly, (11.115) µ(c) = c−rS

rM1−rS. (11.116)

As∑3i=0[v(c)]i =c,v(rcoh)is the solution of our problem. For further details, the rewriting ofv(rcoh)as in the statement, as well as the expression of∆Fcohwe refer to Appendix C of [91].

The solution of Theorem6corresponds to partially performing the swapping ofS andM1instead of performing the full swap. In the case ofEM2 > ESwe also find a minimizes the optimization problem of Eq.11.74for c=rcohand has an associated work cost of

∆Fcoh=

((rcoh−rS)(EM2 −ES) ,if r≤rcoh <rM2 (rM2−rS)(EM2 −ES) + (rcoh−rM2)(EM1−ES) ,if rM2 ≤rcoh <rM1.

(11.119)

Chapter 11. Qubit System 58 Proof idea. In this case we have two practical rewritings ofv·D(HSM)depending on the value ofrcoh. The first one is practical forrcoh∈ [rS,rM2]and is the following

v·D(HSM) =−v0EM2 + (−v0−v1−v4)(EM1 − EM2) +v7EM2

+ (v6+v3+v7)(EM1− EM2) + (v5+v6+v3+v7)(2EM2 − EM1) +EM1+ES− EM2 +

3 i=0

vi(EM2−ES).

(11.120) This leads to

min

vD(ρSM)

3i=0vi=c

v·D(HSM)≥ v1(c)·D(ρSM), (11.121)

where

v1(c) =T14(µ1(c))T36(µ1(c))D(ρSM), (11.122) with

µ1(c) = c−rS

rM2−rS. (11.123)

As∑3i=0[v1(c)]i = cforc ∈ [rS,rM2], v1(rcoh)is the solution of our problem for rcoh∈ [rS,rM2]. Forrcoh∈[rM2,rM1]we use the following rewriting

v·D(HSM) =−v0EM2 + (−v0−v1)(EM1− EM2) + (v6+v7)(EM1 − EM2) (11.124) +v7EM2 + (v3+v5+v6+v7)EM2+ES+

3 i=0

vi(EM1−ES). (11.125) This leads to

min

vD(ρSM)

3i=0vi=c

v·D(HSM)≥ v2(c)·D(ρSM), (11.126)

where

v2(c) =T24(µ2(c))T35(µ2(c))T14(1)T36(1)D(ρSM), (11.127) with

µ2(c) = c−rM2 rM1−rM2

. (11.128)

As∑3i=0[v2(c)]i = cforc∈ [rM2,rM1],v2(rcoh)is the solution of our problem for rcoh∈ [rM2,rM1]. For further details we refer to Appendix C of [91].

What Theorem7says is that forrcoh ∈ [rS,rM2], the best strategy is to partially swap the populations ofSand M2. Indeed, in that caseµ ∈ [0,12]and sog(µ) =0 and f(µ) =arcsin(p2µ), so that

U=eiarcsin(

)LSM

2. (11.129)

Ifrcoh ∈(rM2,rM1], Theorem7tells us that the best strategy is to fully swap the populations ofSandM2and then subsequently partially swap the populations ofS andM1. Indeed, in that caseµ∈(12, 1]so that f(µ) = π2 andg(µ) =arcsin(p2µ−1).

Chapter 11. Qubit System 59 We therefore have

U =eiarcsin(

1)LSM1

eiπ2LSM2. (11.130) Note that it is a priori not evident at all, at least to us, that out of all the 8×8 unitaries that one could potentially choose, that the ones performing the partial swaps of the end result are the most efficient in terms of energy expenditure. The proof has to deal with all these possibilities at first and makes sure that no other transformation performs better. To do so it uses a lot of the fine-tuned information about the problem, i.e., the specific ordering ofD(ρSM)as well as the relationship of the different energy levels of the joint systemSM. Given the end results and the fact that they are so intuitive suggests, however, that this fine-tuned information might not, after all, be that crucial. It would be interesting to find a proof that is based on a more general principle. This might allow generalizing the result to more complicated machines more easily as well.

Coherent vs. Incoherent

Now that we have analyzed both scenarios in detail for the two qubit machine cool-ing a qubit target, we would like to see how they compare both in terms of coolcool-ing performance and in terms of cooling for a given amount of injected work cost. This comparison is summarized in Figure11.1, where the amount of cooling vs. the as-sociated work cost is mapped out. The incoherent curve is generated by plotting Tinc(TH), Eq.11.67, and∆Finc(TH), Eq.11.69, parametrically in the hot bath tempera-tureTH ∈ [TR,+]. The coherent curve is generated by plottingTcoh(rcoh), Eq.11.4, and ∆Fcoh(rcoh), Eq. 11.163, parametrically in the target ground state population rcoh∈ [rS,rcoh ].

For the Figure we selectedEM2 < ES. But apart from the coherent curve having a discontinuity in the first derivative atrcoh = rM2, the behavior of the curves for EM2 ≥ESare qualitatively the same.

There are a few interesting observations that can be made from this comparison.

First of all, looking at the end points of the curve, we see that one can coherently cool more than incoherently and that one does so at a lower work cost. This is true in general as one can easily analytically prove that

Tcoh < Tinc , (11.131)

∆Fcoh < Finc , (11.132) always holds, see [91]. The coherent scenario therefore always performs better than the incoherent scenario for maximal cooling in the single cycle regime. However, the coherent scenario is not universally superior to the incoherent one. Indeed, for sufficiently low work cost, the incoherent scenario always outperforms the coherent one. This is suggested by Figure11.1and can be easily proven in general by com-puting the initial slope of each curve. The incoherent scenario starts with an infinite slope while the coherent one with a finite one. The incoherent curve therefore always lies below the coherent one at the beginning. The incoherent and coherent curves must eventually cross since the endpoint of the coherent curve lies below that of the incoherent one. There therefore exists a critical work cost∆Fcritsuch that if one only has at its disposal some work cost∆F to invest that is smaller than∆Fcrit, one can cool more incoherently than coherently.

Chapter 11. Qubit System 60

FIGURE11.1: Parametric plot of the relative temperature of the target qubit TTR as a function of its work cost∆FforEM2 =0.4 andTR =1.

The red solid curve corresponds to the incoherent scenario, the blue dashed, to the coherent scenario. When the cooling is maximal (i.e., the work cost is unrestricted), the coherent scenario always outperforms the incoherent one,Tcoh <Tinc and∆Fcoh <∆Finc . However, below a critical work cost∆Fcrit, the incoherent scenario always outperforms

the coherent one..

Dans le document Optimal Manipulation of Correlations and Temperature in Quantum Thermodynamics (Page 62-71)