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 atT_Ror to entirely thermalize toT_H. But we know from Lemma9and Lemma8that this doesn’t lead to any cooling ofS, whatever the degeneracies of

Chapter 11. Qubit System 52 H_SM. For machines of dimension 4, i.e.,d_M =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 atT_Rand thermalize the other toT_H. W.l.o.g. let M₁remain atT_RandM₂be the one thermalizing toT_H. We therefore have

ρ^R_SM^,H =τ_S⊗τ_M1⊗τ_M^H₂. (11.62) The next step is to screen through the potential degeneracies ofH_SMto 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 E_M1 =0,E_M2 =0, andE_M1+E_M1 = 0 as potential candidates. Similarly, Lemma11 tells us thatE_S = 0 is also not a useful degeneracy to have inH_SM. This leaves us with 5 potential degeneracies

1. E_S=E_M1 2. E_S=E_M2 3. E_S=E_M1 +E_M2 4. E_M1 =E_S+E_M2 5. E_M2 =E_S+E_M1.

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

E_M1 =E_S+E_M2. (11.63)

This seems to fully take care of the degeneracy question ford_M =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

E_M1 =E_S+E_M2, (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. E_M1 =E_S+E_M2

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 temperatureT_H, one heats upM₂toT_Hand 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 span^c{|010i,|101i}denotes the complement of the set span{|010i,|101i}. Doing so we get

r_inc(T_H) =r_Sr_M1+ [(1−r_S)r_M1 +r_S(1−r_M1)] (1−r^H_M2). (11.66) The associated temperatureT_inc(T_H)is given by the usual formula,

T_inc(T_H) = ^ES ln

r_inc(T_H) 1−rinc(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, Q^H, which is equal to the change of energy ofM₂upon heating it. We therefore get

Q^H =E_M2(r_M2−r^H_M2) (11.68) and

∆F_inc(T_H) =E_M2(r_M2−r^H_M2)

1− ^TR T_H

. (11.69)

From this we get the * quantities by taking the limitT_H →_{∞, i.e.,}

r^∗_inc= ¹

2(r+rM₁) (11.70)

T_inc^∗ = ^ES ln _r

S+rM1

2−(rS+rM2)

(11.71)

∆F_inc^∗ = E_M2(r_M2 −¹

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

E_M1 =E_S+E_M2, (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

v≺minD(ρ_SM)v·D(H_SM), s.t.

∑

3 i=0

v_i =c, (11.74)

with

Chapter 11. Qubit System 54

We first would like to determiner^∗_cohto 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(ρ_SM)are. Looking at this more closely, and taking our restriction of Eq.11.73into account, we find the following ordering

[D(ρ_SM)]₀>{[D(ρ_SM)]_1,[D(ρ_SM)]₄}> [D(ρ_SM)]₂= [D(ρ_SM)]₅> _(11.81)

which gives an associated temperature of T_coh^∗ = ^ES

E_M1^{TR. (11.86)}

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

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(ρ_SM). As the ordering depends on whether E_M2 ≤E_SorE_M2 > E_S, we treat both cases separately.

a.E_M2 ≤E_S In this case we have

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

>[D(ρ_SM)]_3,[D(ρ_SM)]₆ >[D(ρ_SM)]_7, ^(11.88) and so the vectorv≤ ∈_R⁸minimizing Eq.11.74forc=r^∗_cohis given by

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

a_v≤ = ([D(ρ_SM)]₀,[D(ρ_SM)]₁,[D(ρ_SM)]₄,[D(ρ_SM)]₅), (11.91) is inversely ordered with respect the first half ofD(H_SM),

a_D(HSM)= (_0,E_M2,E_M1,E_M1+E_M2), (11.92) and that the second half ofv≤,

b_v≤ = ([D(ρ_SM)]₂,[D(ρ_SM)]₃,[D(ρ_SM)]₆,[D(ρ_SM)]₇), (11.93) is also inversely ordered with respect the second half ofD(H_SM),

b_D(HSM)= (E_S,E_S+E_M2,E_S+E_M1,E_S+E_M1+E_M2). (11.94) A unitaryU_≤^∗ that achievesv≤is given by

U_≤^∗ =U₂₄U35, (11.95)

with

U_ij =|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|010i_SMwith |100i_SM as well as|011i_SMwith

|101i_SM. Physically it corresponds to swapping the population ofSwithM₁andU_≤^∗ can be written compactly as

U_≤^∗ =e^−iπ2LSM1, (11.97) where

L_SMk =i|01i h10|_SM

k−i|10i h01|_SM

k. (11.98)

The work cost associated toU_≤^∗ is

∆F_coh^∗ = (E_M1−E_S)(r_M1 −r_S) (11.99)

= E_M2(r_M1 −r_S). (11.100)

Chapter 11. Qubit System 56 b.E_M2 >E_S In this case we have

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

>[D(ρ_SM)]₆,[D(ρ_SM)]₃ >[D(ρ_SM)]₇, (11.102) and so the vectorv> ∈_R⁸minimizing Eq.11.74forc=r^∗_cohis given by

v>= ([D(ρ_SM)]₀,[D(ρ_SM)]₄,[D(ρ_SM)]₁,[D(ρ_SM)]₅,[D(ρ_SM)]₂, (11.103) ,[D(ρ_SM)]₆,[D(ρ_SM)]₃,[D(ρ_SM)]₇). (11.104) As before, av> andbv> are indeed inversely ordered w.r.t. a_D(H_SM) andb_D(H_SM)

respectively. A unitaryU^∗_≥that 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_>^∗ =e^−iπ2LSM1e^−iπ2LSM2. (11.106) The work cost associated toU_>^∗ is

∆F_coh,^∗ > = (E_M2−E_S)(r_M2 −r_S) +E_M2(r_M1−r_M2). (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 ofr_coh ∈ [r_S,r^∗_coh]. We also saw that in the caseE_M2 > E_S, 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 r_coh ∈ [r_S,r^∗_coh]. Here we again distinguish between the caseE_M2 ≤ E_SandE_M2 >E_S. ForE_M2 ≤E_Swe find a result that is a complete analog of the 1 qubit machine result of Section11.2.2.

Theorem 6. LetE_M2 ≤ E_S. Let r_coh ∈ [r_S,r^∗_coh]. Letµ = _r^rcoh−rS

M1−rS. Let t = arcsin(√ µ). Then

v=D(Uρ_SMU^†), with U =e^−itLSM1 (11.108) minimizes the optimization problem of Eq.11.74for c=r_cohand has an associated work cost of

∆F_coh = (r_coh−r_S)(E_M1−E_S)_{. (11.109)} Proof idea. The idea of the proof is exactly the same as that of Theorem4, namely to rewrite

v·D(H_SM) (11.110)

such that the majorization conditions as well as the constraint can naturally be expressed. The practical rewriting depends on the ordering ofD(ρ_SM)as well as how E_S,E_M1, andE_M2 relate to one another. In the caseE_M1 = E_S+E_M2 andE_M2 ≤ E_Sthe ordering ofD(ρ_SM)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

Q_ij : permutation matrix exchanging coordinatesiandjonly, (11.115) µ(c) = ^c−r_S

r_M1−r_S. (11.116)

As∑³_i=0[v(c)]_i =c,v(r_coh)is the solution of our problem. For further details, the rewriting ofv(r_coh)as in the statement, as well as the expression of∆F_cohwe 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 ofE_M2 > E_Swe also find a minimizes the optimization problem of Eq.11.74for c=r_cohand has an associated work cost of

∆Fcoh=

((r_coh−r_S)(E_M2 −E_S) ,if r≤r_coh <r_M2 (rM₂−r_S)(E_M2 −E_S) + (r_coh−rM₂)(E_M1−E_S) ,if rM₂ ≤r_coh <rM₁.

(11.119)

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

v·D(H_SM) =−v₀E_M2 + (−v₀−v₁−v₄)(E_M1 − E_M2) +v₇E_M2

+ (v₆+v₃+v₇)(E_M1− E_M2) + (v₅+v₆+v₃+v₇)(2E_M2 − E_M1) +E_M1+E_S− E_M2 +

∑

3 i=₀

v_i(E_M2−E_S).

(11.120) This leads to

min

v≺D(ρSM)

∑³i=0vi=_c

v·D(H_SM)≥ v₁(c)·D(ρ_SM), (11.121)

where

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

µ₁(c) = ^c−r_S

r_M2−r_S. (11.123)

As∑³i=0[v₁(c)]_i = cforc ∈ [r_S,rM₂], v₁(r_coh)is the solution of our problem for r_coh∈ [r_S,r_M2]. Forr_coh∈[r_M2,r_M1]we use the following rewriting

v·D(H_SM) =−v₀E_M2 + (−v₀−v₁)(E_M1− E_M2) + (v₆+v₇)(E_M1 − E_M2) (11.124) +v₇E_M2 + (v₃+v₅+v₆+v₇)E_M2+E_S+

∑

3 i=0

v_i(E_M1−E_S). (11.125) This leads to

min

v≺D(_ρSM)

∑³i=0vi=c

v·D(H_SM)≥ v₂(c)·D(ρ_SM), (11.126)

where

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

µ₂(c) = ^c−r_M2 rM₁−rM₂

. (11.128)

As∑³i=0[v₂(c)]_i = cforc∈ [r_M2,r_M1],v₂(r_coh)is the solution of our problem for r_coh∈ [rM₂,rM₁]. For further details we refer to Appendix C of [91].

What Theorem7says is that forr_coh ∈ [r_S,r_M2], 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=e^−iarcsin(

√2µ)L_SM

2. (11.129)

Ifr_coh ∈(r_M2,r_M1], 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 =e^−iarcsin(

√

2µ−1)L_SM1

e^−iπ2LSM2. (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(ρ_SM)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 T_inc(TH), Eq.11.67, and∆Finc(TH), Eq.11.69, parametrically in the hot bath tempera-tureT_H ∈ [T_R,+_∞]. The coherent curve is generated by plottingT_coh(r_coh), Eq.11.4, and ∆F_coh(r_coh), Eq. 11.163, parametrically in the target ground state population r_coh∈ [r_S,r_coh^∗ ].

For the Figure we selectedE_M2 < E_S. But apart from the coherent curve having a discontinuity in the first derivative atr_coh = r_M2, the behavior of the curves for E_M2 ≥E_Sare 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

T_coh^∗ < T_inc^∗ , (11.131)

∆F_coh^∗ < _∆F_inc^∗ , (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∆F_crit, one can cool more incoherently than coherently.

Chapter 11. Qubit System 60

FIGURE11.1: Parametric plot of the relative temperature of the target qubit _T^T_R as a function of its work cost∆FforE_M2 =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,T_coh^∗ <T_inc^∗ and∆F_coh^∗ <_∆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)