Constructing an entangled Unruh Otto engine and its efficiency

Dipankar Barman∗, Bibhas Ranjan Majhi†

Department of Physics, Indian Institute of Technology Guwahati,

Guwahati 781039, Assam, India

November 2, 2021

Abstract

Uniformly accelerated frame mimics a thermal bath whose temperature is proportional to theproper acceleration. Using this phenomenon we give a detailed construction of an Otto cycle betweentwo energy eigenstates of a system, consists of two entangled qubits. In the isochoric stages thethermal bath is being provided via the vacuum fluctuations of the background fields for a monopoleinteraction by accelerating them. We find that Otto cycle is possible when two qubits are acceleratingin the right Rindler wedge (RRW); i.e. in parallel motion, with the initial state of the system is takenas anti-symmetric Bell state. The same is also possible for one qubit is accelerating on the RRWand other one is moving in left Rindler wedge; i.e. in anti-parallel motion, with the initial state is anon-maximally entangled one. Moreover, the first situation can provide the efficiency greater thanthat of the usual single qubit quantum Otto engine, while later one does not. We give the parameterspace of the available quantities in the system for increased efficiency. On the other hand, Otto cycleis not possible if one considers the symmetric initial Bell state or non-maximally entangled state forparallel motion. Moreover, for both initial symmetric and anti-symmetric Bell states we do not findany possibility of the cycle for qubits’ anti-parallel motion.

1 Introduction

Thermodynamics of a quantum system has been a very active area and considerable amount of effort hasbeen devoted to develop a fruitful theoretical formalism in order to explore the quantum thermodynamics.The definition of work, energy and heat of a quantum system is properly addressed in this paradigm.Consider a quantum system, described by a density operator ρ(t), is evolving under a time dependentHamiltonian H0(t). The variation of the expectation value of the energy, ∆〈E〉 from time ti to tf satisfies[1]

∆〈E〉 = 〈Q〉+ 〈W 〉 , (1)

where we identify

∆〈E〉 =

∫ tf

ti

dtd

dt〈E〉 ; (2)

〈Q〉 =

∫ tf

ti

dt Tr(dρ(t)

dtH0

); (3)

and

〈W 〉 =

∫ tf

ti

dt Tr

(ρ(t)

dH0

dt

), (4)

as total energy change, heat transfer and work done on the system, respectively. The above concepts ofdifferent thermodynamic quantities has been successfully implemented to construct quantum version of

∗E-mail: dipankar1998@iitg.ac.in†E-mail: bibhas.majhi@iitg.ac.in

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different classical engines; e.g. Carnot engine, Otto engine, etc. [2, 3, 4, 5, 6]. A Quantum Otto Engine(QOE) can be constructed with a single qubit [3, 4], which has four steps. The system undergoes twoadiabatic processes and two isochoric processes. In adiabatic process, there is work done on or by thesystem and in isochoric process, system has heat exchange with the environment. The efficiency turnsout to be

η0 = 1− ω1

ω2, (5)

with η0 < 1. Here ω1 (ω2) is the energy gap before (after) the adiabatic expansion of the levels of thequbit. Note that, contrary to the classical Otto engine in which efficiency depends on the temperaturesof the thermal baths during the isochoric processes, η0 for QOE depends on the energy gaps between thequantum levels of the qubit system.

On the other hand the combination of relativity and quantum mechanics, best described by quantumfield theory, brings several interesting and important phenomenon, like Hawking [7, 8] and Unruh [9, 10,11] effects, in front of us. These two phenomenon naturally create a thermal environment for a specificclass of observers. Recently, using the concept of Unruh phenomenon [9, 10, 11], a Unruh quantum Ottoengine (UQOE) has been proposed in [12, 13, 14]. This, contrary to QOE, is a completely relativisticset up and therefore the measured time in each stages is denoted by qubit’s proper time (so the role oftime parameters t in Eqs. (2) – (4) is played by proper time τ of qubit’s frame). According to the Unruheffect, a uniformly accelerating observer can see particles in the Minkowski vacuum and this acts as thethermal bath for the accelerating frame. The temperature of the bath which is proportional to the properacceleration can be utilised as heat source for an Otto cycle. So the hot (cold) thermal baths during theisochoric processes can be mimicked by giving uniformly acceleration (deceleration) to the qubit. This isknown as Unruh Quantum Otto Engine (UQOE). The efficiency of this cycle came out to be that givenin (5) (see [12, 13], for details).

Using the idea of UQOE, one of the authors of this paper with Kane very recently proposed aquantum Otto cycle consists of two entangled qubits [15]. The underlying inspiration comes from thefact that Unruh effect is greatly influenced by entanglement between the accelerated observer with anotheraccelerated frame [16, 17, 18, 19] . Moreover the entanglement phenomenon itself is observer dependentquantity [20, 21, 22, 23]. In this regard it may also be noted that the quantum entanglement harvestingbetween two causally disconnected accelerated detectors is possible within a relativistic setup. All theseevents imply a possibility of considerable influence on the efficiency of engines when one considers thecycle between entangled states of two accelerated qubits. The analysis in [15] revealed that the cyclebetween an entangled state and their collective excited state, depending on the relative accelerations ofthe qubits during the isochoric process, may have greater efficiency than a standard quantum Otto cycle.This is named as Entangled Unruh Quantum Otto Engine (EUQOE).

In this paper we will readdress this EUQOE and its nature of efficiency in a much more broaderperspective. Moreover the analysis in [15] is found to have few limitations and is not complete. Let usnow mention them.

• To make a cycle efficient for work output, the amount of work done has to be positive. In additionduring the isochoric processes the heat absorption (rejection) must happen in order to make afruitful engine. We will see later that these are supplemented by condition given by Eq. (11) (seealso below Eq. (25) of [15]). In order to fulfilment of these one must choose the values of theavailable parameters (e.g. accelerations of the qubits, the spacing of energy levels of the qubits, theinteraction time during isochoric phases, etc.), appearing in the system. This identification of theparameter space of all these quantities has not been done earlier. Therefore it is not clear whetherany such parameter space is available in all practical purposes in order to construct a EUQOE.

• The time evolution of the state of system has been done by assuming the satisfaction of commutationbetween different time Hamiltonians. But we know that in general this may not be always true andin that situation time-ordering is necessary.

• In the existing analysis considered that the two qubits are accelerating in the same side of theRindler wedge (right Rindler wedge (RRW)) during the isochoric stages. But we know that accel-erated frame can be constructed in the left Rindler wedge (LRW) as well. Also since the frames,accelerating in these two opposite wedges, can harvest quantum entanglement [24, 25], it wouldbe interesting to incorporate this situation in constructing EUQOE. Such generalization was ab-sent in [15]. Therefore, the question of the possibility of making a EUQOE when one detector isaccelerating in the right and another in the left Rindler wedge remains an option to investigate.

2

Here we aim is to present a complete picture of the EUQOE. All the gaps, mentioned above, will beaddressed here. We scan the parameter space for which making of a EUQOE is possible. We observedthat the necessary conditions are getting satisfied when both detectors are in the right Rindler wedge andinitial state is anti-symmetric Bell state (which is maximally entangled). However there are no continuousrange of values of the parameters for that. The efficiency of the EUQOE get enhanced at some discreteparameter values. Here we will specify those values of the parameters. For all other choices, making ofEUQOE is not possible for parallely moving detectors. We also observe that either maximally symmetricor anti-symmetric both do not lead to any fruitful cycle when the qubits are accelerating in two oppositeRindler wedges. The construction of Otto engine is prohibited by the dissatisfaction of the requiredconditions. However, making of EUQOE is possible with initially non-maximally entangled states fordetectors are in anti-parallel motion. However, the efficiency get suppressed for all the parameter values.

The organization of the paper is as follows. In Section 2 we begin with a brief description on ourmodel. This section also describes the stages of our EUQOE and provides the expression of the efficiencyof the cycle in terms of elements of time evolved density matrix of the two detectors’ system. In the nextSection 3 we provide the expressions of elements of the density matrix for the detectors are acceleratingin same and opposite Rindler wedges. Thereafter, we analyzed for the possibility of making an EUQOEin various possible parameter spaces. We also provide the parameter values for which efficiency of thecycle get enhanced. Finally, we conclude this article in Section 4 with discussion of our results. Fiveappendices are also provided at the end in order to show the explicit steps to evaluate several importantrelations and results.

2 Set Up

We construct a similar system as considered in [15]. It consists of two identical qubits (chosen as two levelUnruh De-Witt detectors [26]), A and B, having energy levels with energy gj = −ω/2 for the groundstate and ej = ω/2 for the exited state. The corresponding eigenkets are |gj〉 and |ej〉 (where j = A,B),respectively. The free Hamiltonian of the detectors can be chosen as [16, 27]

H0 =ω(τ)

2

(dτAdτ

SzA ⊗ 1B +dτBdτ

1A ⊗ SzB), (6)

where Szj = (|ej〉〈ej | − |gj〉〈gj |), ω is the energy gap of the detectors and τA, τB are the proper times ofthe respective detectors. In the above τ is the proper time of our observer which we choose as the detectorA and hence τ = τA. Same Hamiltonian has been chosen earlier in various investigations [16, 27]. Theenergy eigenstates and the corresponding eigenvalues can be obtained for the composite system as

Ee = ω, |e〉 = |eA〉|eB〉;

Es = 0, |s〉 =1√2

(|eA〉|gB〉+ |gA〉|eB〉);

Ea = 0, |a〉 =1√2

(|eA〉|gB〉 − |gA〉|eB〉);

Eg = −ω, |g〉 = |gA〉|gB〉; (7)

Here, as discussed in [15, 19] transition from |g〉 → |e〉 or |e〉 → |g〉 is not possible for the following typeof interaction between the background real scalar field and monopole detectors [28, 29, 24, 19, 25]:

Hint =∑j=A,B

cjχj(τ)mj(τj)φ(xj(τj))dτjdτ

. (8)

In the above cj is coupling constant of interaction. We choose cA = cB = c, as the detectors in our caseare identical. For our present purpose we will use this type of interaction. The monopole operator of j-thdetector, mj at the initial time is defined by

mj(0) = |ej〉〈gj |+ |gj〉〈ej | . (9)

It has been shown in [19] that the transition probability form |a〉 → |e〉 and |a〉 → |g〉 are same, as theyhave same energy spacing. Similarly the probabilities for the transitions |s〉 → |e〉 and |s〉 → |g〉 are also

3

same. Notice that |s〉 and |a〉 are two Bell states appearing in the composite system which are maximallyentangled ones. Moreover, for a general initial state, |D〉 = b1|eAgB〉+ b2|gAeB〉 with ED = 0, transitionprobabilities |D〉 → |e〉 and |D〉 → |g〉 are also same. With the above information here we can try toconstruct the Otto cycle between |s〉 and |e〉 or |g〉, |a〉 and |e〉 or |g〉 (also between |D〉 and |e〉 or |g〉), formaximally and non-maximally (depending on the values of b1 and b2 with b21 + b22 = 1) entangled states.We will check all of these possibilities here.

We choose the detector A as our primary observer. Therefore we set τ = τA and define dτB/dτ = αwhich is constant during detectors’ motions with uniform velocities as well as uniform accelerations (see[17] for relation between τA band τB ; an additional discussion has been given in Appendix A as well). Laterfor notational convenience, we will add a subscript with α: v for constant velocity and ak (k = H, C) foruniform acceleration in heating process (H) and cooling process (C). We have calculated α’s for differentstages of the cycle in Appendix A. The free Hamiltonian (6) can be represented as

H0 =ω

2

(1 + α) 0 0 0

0 (1− α) 0 00 0 (−1 + α) 00 0 0 (−1− α)

= ωhα , (10)

in the basis {|eAeB〉, |eAgB〉, |gAeB〉 |gAgB〉}.

2.1 Stages of cycle

The stages of our EUQOE are already being discussed in [15] which are along the line of those given in[12, 13] for UQOE. For convenience, here again let us briefly describe the stages of the cycle.

• Adiabatic expansion: In step I, the detectors A and B travel at velocities −vA and −vB (vB foranti-parallel motion) for time intervals ∆τ1

A and ∆τ1B , respectively (here superscripts for the stage

number and subscripts for detectors). In this process work is done on the composite system, sothat the energy gap increased from the initial gap ω1 to ω2. However, the population density of thestates remain unchanged.

• Heat absorption through isochoric process: After step I, the detectors started accelerating withproper accelerations aAH and aBH , respectively. Due to acceleration, the detectors perceives theMinkowski vacuum as thermal bath as a result of interaction between the detectors and the back-ground quantum fields. The system then absorbs the heat. The detectors accelerate for the timeintervals ∆τ2

A and ∆τ2B , when their velocities changes from −vA to vA and −vB to vB (or vB to

−vB), respectively. As the density matrix evolve under the interaction, it goes from initial state ρ0

to ρ0 + δρH . The expressions of δρH for general ρ0 = |D〉〈D| is explicitly calculated in AppendixB (see final expression in (B.26)).

• Adiabatic Compression: The interaction with the background field in this stage is turned off and thepopulation density of the states remain unchanged. The detectors started moving with constantvelocity, vA and vB (or −vB) for the time intervals of ∆τ3

A and ∆τ3B , respectively. The energy

spacing of the system is allowed to decrease from ω2 to ω1.

• Heat ejection through isochoric process: In the final stage, the detectors started decelerating withproper uniform decelerations aAC and aBC , respectively. As a result, their velocities changes fromvA to −vA and vB to −vB (or −vB to vB), respectively. The temperature of background quantumvacuum is taken to be lower by considering ajC < ajH . This can be done by taking longer interactiontime intervals ∆τ4

A and ∆τ4B compared to those in stage II. Then the system transfers heat to the

environment. The density of state goes from ρ0 + δρH to ρ0 + δρH + δρC . To maintain the cyclicityof the engine, we impose a constraint, δρH + δρC = 0.

2.2 Efficiency

For the cycle to be efficient to work, one has to insure the positivity of work done and heat absorptionby the engine. This is being confirmed by the following impositions [15]:

Tr(δρHhαk) > 0 ; (k = v, aH , aC) . (11)

4

This condition must be satisfied by the EUQOE. The conservation of energy requires [15]

ω2Tr(δρHhαaH )− ω1Tr(δρHhαaC ) = (ω2 − ω1)Tr(δρHhαv ) . (12)

The efficiency of our EUQOE can be obtained as [15]

ηE =(

1− ω1

ω2

) Tr(δρHhαv )

Tr(δρHhαaH )= η0

Tr(δρHhαv )

Tr(δρHhαaH ). (13)

Simultaneous satisfaction of the conditions (11), (12) and (13) guarantees that ηE < 1 (see [15] for allthese details). Note that the factor, appearing on the right hand side with η0, contains the variation ofthe density matrix (δρH) only for the second stage, i.e. in the isochoric heating process. This factor isalways greater than zero due to the positivity of the work done, heat absorption and heat rejection by theengine, and so (ηE/η0) > 0. Now in order to make EUQOE more efficient than the UQOE we must have

(ηE/η0) > 1; i.e.Tr(δρHhαv )

Tr(δρHhαaH)> 1, under the constraint η0 < 1 and ηE < 1. In this case, the efficiency

(ηE) of the engine will be increased due to entanglement phenomenon. However, the efficiency (ηE) willalways be less than one. We mention that the expression for efficiency (13) reduced to a very simpleform in [15] for the initial state as |a〉 and |s〉. We found that such is due to the calculation of traces in(13) without proper time ordering in the perturbative expansion. Here we will improve this analysis byincorporating the time ordering prescription.

The explicit expressions of the trace quantities in (13), incorporating the time ordering, have beencalculated in Appendix C through perturbation technique, described in Appendix B. The first relevantcontribution is coming in the second order. Considering till this order we obtain Tr(δρHhα) as (see Eq.(C.5))

Tr(δρHhα) = [b22PA(ω)− b21PA(−ω)) + b1b2∆PAB}+ α[(b21PB(ω)− b22PB(−ω)) + b1b2∆PAB} . (14)

The expressions of the relevant quantities are given by

Pj(ω2) = c2∫ ∫

dτjdτ′jχj(τj)χj(τ

′j)G

+(x′j , xj)eiω2(τj−τ ′j); (j = A,B), (15)

and

∆PAB = PAB(ω2,−ω2)− PAB(−ω2, ω2) , (16)

where

PAB(ω2,−ω2) = c2∫ ∫

dτAdτ′BχA(τA)χB(τ ′B)eiω2(τA−τ ′B)G+(x′B , xA), (17)

with Pj(−ω2) and PAB(−ω2, ω2) are obtained by replacing ω2 → −ω2 in the integrands of (15) and(17). For maximally entangled initial states, b1 = ±b2 = 1/

√2, (14) get further simplified by introducing

∆Pj = Pj(ω)−Pj(−ω). In the above quantities, G+(x′j , xj) is the positive frequency Wightman function.These integrations are calculated during the stage II of the cycle, where ω2 is energy gap of the engine.We do not need to evaluate the same for stage IV of the cycle due to the cyclicity condition.

3 Calculation of efficiency

In order to find the explicit value of ηE in Eq. (13) one needs to evaluate the integrations (15) and (17).Note that in the second stage, the interaction time is ∆τ2

j for the j-th detector (j = A,B). This time isthe time required for changing the velocity from −vj to vj and can be determined in the following way.

For RRW, the relations between the Rindler proper time and the Minkowski coordinates are

tj =1

ajksinh(ajkτj); xj =

1

ajkcosh(ajkτj) , (18)

where index k can be H (C) denoting heating process (cooling process). Now velocity is given by vj =(dxj/dtj) = (dxj/dτj)/(dtj/dτj) = tanh(ajkτj) and this determines the time interval as (2/ajk) tanh−1(vj)when the detector is accelerating on the RRW. Here we define Tjk/2 = (1/ajk) tanh−1(vj) for that, andthen the integration limits will be from −Tjk/2 to Tjk/2.

5

On the other hand, if the detector is accelerating in the LRW, then the coordinate relations are

tj =1

ajksinh(ajkτj); xj = − 1

ajkcosh(ajkτj) . (19)

Now the velocities are given by vj = − tanh(ajkτj) and hence the interaction time interval is given by(2/ajk) tanh−1(vj). So the integration limits are from −Tjk/2 to Tjk/2.

However, evaluation of the integrations, appearing in ∆Pj and ∆PAB , for finite integration limits cannot be done analytically. But a suitable choice of a switching function χj(τj), which vanishes rapidlybeyond the finite integration limits, can allow to extend the interaction time limits from −∞ to +∞. Inthat case there is a possibility of obtaining an analytical expression. Following the argument of [12] andas suggested there (later used in [13] as well), we choose a Lorentzian like compact symmetric switchingfunction:

χj(τj) =(Tjk/2)2

τ2j + (Tjk/2)2

. (20)

This function is non-vanishing for −Tjk/2 < τj < Tjk/2, and approximately zero beyond this domain.We could choose a Gaussian like switching function [12], however that will not help us to extend theintegration limit from ±Tjk/2 to ±∞. Because at τj → i ×∞, the gaussian exponent (exp(−τ2

j /T 2jk

))diverges and that fails Jordon’s lemma for contour integration.

3.1 Both detectors are moving parallely

The two qubits (the detectors here) are accelerating with constant accelerations (aHA and aHB ) in theRRW during the second stage of the cycle. In this case their trajectories are given by (18) and the relationbetween the proper times between them turns out to be (from (A.2))

τB = αaH τA , (21)

where αaH = aAH/aBH (since this is the heating stage). Using the switching function (20) and extendingthe limits of integration from −∞ to +∞ in (15) and (17), one case perform the integrations analytically.In the evaluation of these, since here our designated observer’s frame is qubit A, all the integrationvariables will be expressed in terms of τA by using (21).

The explicit calculations for PA and PB are presented in Appendix D. The values of these quantitiesare given by

PA(ω2) =(aAHTAH/2)2 e−TAHω2

16 sin2(aAHTAH/2)

+a2AHT 2AH

e− 2πω2aAH

64π2

(Φ

(e− 2πω2aAH , 2, 1 +

aAHTAH2π

)− Φ

(e− 2πω2aAH , 2, 1− aAHTAH

2π

))

+aAHT 2

AHω2 e

− 2πω2aAH

32π

(Φ

(e− 2πω2aAH , 1, 1 +

aAHTAH2π

)− Φ

(e− 2πω2aAH , 1, 1− aAHTAH

2π

)),(22)

PA(−ω2) =ω2TAH

8+ PA(ω2), (23)

PB(ω2) =(aAHTAH/2)2 e−TAHαaHω2

16 sin2(aAHTAH/2)

+a2AHT 2AH

e− 2πω2aBH

64π2

(Φ

(e− 2πω2aBH , 2, 1 +

aAHTAH2π

)− Φ

(e− 2πω2aBH , 2, 1− aAHTAH

2π

))

+aAHT 2

AHαaHω2 e

− 2πω2aBH

32π

(Φ

(e− 2πω2aBH , 1, 1 +

aAHTAH2π

)− Φ

(e− 2πω2aBH , 1, 1− aAHTAH

2π

)),(24)

and

PB(−ω2) =TAHω2αaH

8+ PB(ω2). (25)

6

In the above, Φ is the transcendental Lerch-Hurwitz function, defined as

Φ(z, s, a) =

∞∑k=0

zk

(k + a)s. (26)

Similarly ∆PAB has been calculated in Appendix E.1. We obtain PAB for αaH < 1 case as

∆PAB = −αaHaBHT 3

AHcsch(aAHκ)e−

12 (1−αaH )TAHω2

16 κ (κ2 + T 2AH

)(κ sin(αaHκω2) + κ sin(κω2) + TAH cos(αaHκω2)

− TAH cos(κω2)) , (27)

while that for αaH > 1 case is

∆PAB = −αaHaHBT 3

AHcsch(aAHκ)e−

12 (αaH−1)TAHω2

16κ(κ2 + T 2

AH

) (κ sin(αaHκω2) + κ sin(κω2)− TAH cos(αaHκω2)

+ TAH cos(κω2)) . (28)

Here κ is defined through the relation cosh(aAHκ) = (aAH/aBH + aBH/aAH )/2.For the qualitative analysis of the above results we will need the above expressions in terms of scaled

dimensionless parameters. If we define the dimensionless quantities as aAHTAH (= A), aBHTAH (= B),ω2TAH (= W ) and κ/TAH (= K ), one can check that PA, PB and ∆PAB are all dimensionless quantities.In terms of these one finds

PA(ω2) =(A/2)2e−W

16 sin2(A/2)+

A2e−2πW

A

64π2

(Φ

(e−

2πWA , 2, 1 +

A2π

)− Φ

(e−

2πWA , 2, 1− A

2π

))+

AW e−2πW

A

32π

(Φ

(e−

2πWA , 1, 1 +

A2π

)− Φ

(e−

2πWA , 1, 1− A

2π

)), (29)

PB(ω2) =(A/2)2e−αaHW

16 sin2(A/2)+

A2e−2πW

B

64π2

(Φ

(e−

2πWB , 2, 1 +

A2π

)− Φ

(e−

2πWB , 2, 1− A

2π

))+

AWαaHe− 2πW

B

32π

(Φ

(e−

2πWB , 1, 1 +

A2π

)− Φ

(e−

2πWB , 1, 1− A

2π

)), (30)

PB(−ω2) =AαaH

8+ PB(ω2); PA(−ω2) =

A8

+ PA(ω2), (31)

∆PAB = −αaHB csch(AK )e−12 (1−αaH )W

16 K (K 2 + 1)(K sin(αaHKW ) + K sin(KW ) + cos(αaHKW )− cos(KW )) , (32)

for αaH < 1 and

∆PAB = −αaHB csch(AK )e−12 (αaH−1)W

16 K (K 2 + 1)(K sin(αaHKW ) + K sin(KW )− cos(αaHKW ) + cos(KW )) , (33)

for αaH > 1. Using these quantities, the required expression (14) need to be evaluated to determine theefficiency. Below we will analyse the final expressions for initial state, b1|eAgB〉+ b2|gAeB〉 with differentvalues of b1 and b2 (under the constraint b21 + b22 = 1), separately.

3.1.1 Analysis for maximally entangled symmetrical initial state

Here, we have b1 = b2 = 1/√

2. From (23) and (25) we know that PA(ω)−PA(−ω) and PB(ω)−PB(−ω)are negative quantity. Constructing a Otto cycle requires, Tr(δρHhαv ), Tr(δρHhαaH ) and Tr(δρHhαaC )

need to be positive for some values of A, B, W , αaC . We find that Tr(δρHhαv ) is always negative inthe parameter space (for instance, see Fig. 1). This implies that condition (11) is always violated andhence no EUQOE can be made up with two initially symmetric maximally entangled detectors movingin parallel motion.

7

(a) (b)

Figure 1: Initial state |s〉 for parallel motion: Plots (a) and (b), showing work done by the system,Tr(δρHhαv ) with respect to dimensionless acceleration of the primary detector (A) and dimensionlessenergy gap (W ) of the composite system. Here we fixed acceleration of the second detector as B = 100.

3.1.2 Analysis for maximally entangled anti-symmetrical initial state

As mentioned earlier, a successful Otto cycle can be built, provided the condition (11) is being satisfied.Therefore, satisfaction of this condition determine the parameter space for the available quantities whichare involved in constructing the cycle. In order to do that, in Figs. 2, 3, 4 and 5, we have plottedTr(δρHhαv ) and Tr(δρHhαaH ) with respect to dimensionless quantities A, W . It should be pointed

out that the quantities Tr(δρHhαk) for k = v, aH , aC are also dimensionless. The 3D plots are showingthat there are positive values of Tr(δρHhαv ) and Tr(δρHhαaH ) for some A and W . However there isno continuous range of values for A, for which the above quantities are positive. These quantities arefluctuating rapidly with respect to the dimensionless acceleration of first detector (A) between somepositive and negative values. However the ratio η/η0 can be greater than one for some values of A andW , where both Tr(δρHhαv ) and Tr(δρHhαaH ) has positive values, suggested by figures 2, 3, 4 and 5.

We also need to ensure that Tr(δρHhαaC ) is positive, which is necessary in order to happening of theheat rejection by the cycle in the cooling process. Otherwise, the efficiency of the cycle will be greaterthan one. This can be checked from (12), which states that for Tr(δρHhαaC )< 0, (ω2 − ω1)Tr(δρHhαv )

will be greater than ω2Tr(δρHhαaH ), and in that case one will have ηE > 1 (see, expression of ηE in(13)). As stated earlier, such is not physically possible. We know that velocity of a detector at the start(end) of the second stage is same as at the end (start) of the fourth stage. Using this fact, the followingrelations can be obtained:

aACTAC = −aAHTAH ; (34)

aBCTBC = −aBHTBH ; (35)

and

αaCTAC = TBC ; (36)

here αaC is given by aAC/aBC . In (34) and (35) as the parameters in the heating process (stage II)are already fixed, we can always freely choose ajC (j = A,B). Note that Eq. (36) can be obtained bytaking ratio between (34) and (35) and then using (21). Therefore, any two parameters can be chosenfreely and and then the third parameter is determined. Here we choose αaC and TAC ( TBC will beautomatically determined) such that Tr(δρHhαaC ) > 0. Now a sufficient condition for satisfaction ofenergy conservation ( from (12)) can be taken as

Wω1TAH

(αaH − 1) = (αaC − 1) . (37)

8

(a)

5.002 5.004 5.006 5.008 5.010

-2.×10-9

-1.×10-9

0

1.×10-9

2.×10-9

3.×10-9

TrδρHh α

v

5.001 5.001

-1.5×10-9

0

2.5×10-9

(b)

(c)

5.002 5.004 5.006 5.008 5.010

-2.×10-9

-1.×10-9

0

1.×10-9

2.×10-9

3.×10-9

TrδρHh α

a H

5.001 5.001

-1.5×10-9

0

2.5×10-9

(d)

Figure 2: Initial state |a〉 and αaH > 1 for parallel motion: Plots showing (a) work done by the system,Tr(δρHhαv ) (c) heat absorbed by the system, Tr(δρHhαaH ) with respect to dimensionless acceleration of

the primary detector (A) and dimensionless energy gap (W ) of the composite system, and (b) Tr(δρHhαv ),(d) Tr(δρHhαaH ) with respect to dimensionless acceleration of the primary detector (A). Here we fixeddimensionless acceleration of the second detector as B = 5.0 and the dimensionless energy gap of thesystem as W = 0.011.

Since the values of W and αaH are already determined by considering positivity of Tr(δρHhαv ) andTr(δρHhαaH ), while αaC in (36) is chosen to be free, the above equation will fix the value of dimensionlessenergy gap ω1TAH for a particular choice of αaC . In this connection remember that ω1 is less than ω2.So we have to choose αaC such that ω1 < ω2 and (37) are satisfied simultaneously. One can notice thatω1 is positive, if we chose αaC > 1 (αaC < 1) for αaH > 1 (αaH < 1). It should be pointed out thatfor αaH = 1, we obtain ω1 = 0 (αaC 6= 1). This is not possible as this implies that at the beginning ofstage I, energy gap of the system is zero. In Table 1, we present some example parameter values whichare satisfying the all the essential conditions and also enhancing the efficiency of the cycle. Values ofTr(δρHhαaC ) for the selected αaC ’s, are also given in the table.

3.1.3 Analysis for non-maximally entangled initial state

The initial state for b1 6= 1/√

2 with |b1|2 + |b2|2 = 1, gives several choices for non-maximally entangledstate where 0 < b1 < 1 (except b1 6= 1/

√2). We have scanned the parameter space for different values of

b1, but found no region where all of the trace quantities (see, (14)) satisfy the condition given by (11).Hence, making of EUQOE is not possible in non-maximally entangled initial state, while the detectorsare in parallel motion.

9

(a)

100.01 100.04 100.07 100.1 100.13

-1.×10-6

0

1.×10-6

2.×10-6

3.×10-6

TrδρHh α

v

100.01 100.015.×10-7

7.×10-7

(b)

(c)

100.01 100.04 100.07 100.1 100.13

-2.×10-6

0

2.×10-6

4.×10-6

6.×10-6

TrδρHh α

a H

100.01 100.01

5.×10-7

7.×10-7

(d)

Figure 3: Initial state |a〉 and αaH > 1 for parallel motion: Plots showing (a) work done by the system,Tr(δρHhαv ) (c) heat absorbed by the system, Tr(δρHhαaH ) with respect to dimensionless acceleration of

the primary detector (A) and dimensionless energy gap (W ) of the composite system and (b) Tr(δρHhαv ),(d) Tr(δρHhαaH ) with respect to dimensionless acceleration of the primary detector (A). Here we fixeddimensionless acceleration of the second detector as B = 100 and the dimensionless energy gap of thesystem as W = 1.

3.2 The detectors are moving anti-parallely:

In this case detector A is moving in RRW while other one (i.e. B) is moving in LRW during the acceleratedphase. In this case trajectories of detector A and detector B are given by (18) and (19), respectively.The relation between the proper times of the detectors is given by (from (A.4))

τB = −αaH τA (38)

Using this relation, PA, PB and PAB are calculated in terms of the proper time of the detector A.Expressions of PA are already presented in (22) and (23). In Appendix D, we obtain expressions of PBwhen the detector B is accelerating in LRW. These are found to be same in value for the detector isaccelerating in RRW. Therefore these are given by (24) and (25). In Appendix E.2, we obtain ∆PAB =PAB(ω,−ω)− PAB(−ω, ω) for both αaH < 1 and αaH > 1 case as

∆PAB = 0 . (39)

3.2.1 Analysis for maximally symmetric and anti-symmetric entangled initial state

For the maximally entangled state i.e., |b1|2 = |b2|2 = 1/2, we have

∆PA = −W8

; ∆PB = −AW8B

= αaH∆PA , (40)

10

(a)

24.96 24.97 24.98 24.99 24.998

-2.×10-7

0

2.×10-7

4.×10-7

6.×10-7

TrδρHh α

v

24.9901 24.9902

-1.5×10-8

0

2.×10-8

(b)

(c)

24.96 24.97 24.98 24.99 24.998

-2.×10-7

0

2.×10-7

4.×10-7

6.×10-7

TrδρHh α

a H

24.9901 24.9902

-1.5×10-8

0

2.×10-8

(d)

Figure 4: Initial state |a〉 and αaH < 1 for parallel motion: Plots showing (a) work done by the system,Tr(δρHhαv ) (c) heat absorbed by the system, Tr(δρHhαaH ) with respect to dimensionless acceleration of

the primary detector (A) and dimensionless energy gap (W ) of the composite system and (b) Tr(δρHhαv ),(d) Tr(δρHhαaH ) with respect to dimensionless acceleration of the primary detector (A). Here we fixeddimensionless acceleration of the second detector as B = 25 and the dimensionless energy gap of thesystem as W = 0.25.

Therefore from (14), we can conclude that work done or heat absorption are same for both symmetricand anti-symmetric initial states. For total work done, α = αv = 1/γ is the inverse Lorentz factor, whichis always positive. The total work done by the system can be calculated from (14), given by

Tr(δHhαv ) = (ω2 − ω1)[∆PA + αv∆PB ] = (ω2 − ω1)(1 + αvαaH )∆PA (41)

and the heat absorbed by the system is

Tr(δHhαaH ) = ω2[∆PA + αaH∆PB ] = ω2(1 + α2aH )∆PA . (42)

Here, we can immediately see that the work done by the system and heat absorbed by the system isalways negative due to (40). Therefore, we can not make an Otto cycle in this scenario.

3.2.2 Analysis for non-maximally entangled initial state

In this case, we use (22), (23), (24), (25) and (39) to evaluate the trace quantities given in (14). Hereα = αv =

√1− v2

rel for work done, where by using (18) and (19), one can find vrel = −2 tanh(A)/(1 +

tanh2(A)). For heating stage, we have α = −αaH = −A/B. For cooling stage, α = −αaC , which can bechosen freely due to similar arguments presented in the subsection 3.1.2. Here we find that all the trace

11

(a)

99.7 99.8 99.9 99.99

-3.×10-6

0

4.×10-6

8.×10-6

TrδρHh α

v

99.9881 99.9881-3.×10-6

0

3.×10-6

(b)

(c)

99.7 99.8 99.9 99.99

-3.×10-6

0

4.×10-6

8.×10-6

TrδρHh α

aH

99.9881 99.9881-3.×10-6

0

3.×10-6

(d)

Figure 5: Initial state |a〉 and αaH < 1 for parallel motion: Plots showing (a) work done by the system,Tr(δρHhαv ) (c) heat absorbed by the system, Tr(δρHhαaH ) with respect to dimensionless acceleration of

the primary detector (A) and dimensionless energy gap (W ) of the composite system and (b) Tr(δρHhαv ),(d) Tr(δρHhαaH ) with respect to dimensionless acceleration of the primary detector (A). Here we fixeddimensionless acceleration of the second detector as B = 100 and the energy gap of the system as W = 0.6.

values (see, (14)) are positive for some b1 < 0.5 (depending on other parameter values). For Example, insub-figures (a), (b), (c) of Fig. 6, we have plotted the traces with respect to A and W , for b1 = 0.4 andB = 20. From sub-figure (d) of Fig. (6), we can see that η/η0 (green line) always less than one. It alsoshows W (ω2TAH ) = 0.2 (orange line) which has been chosen by us, and ω1TAH (blue line), which crossesthe value of W for A ≤ 1. So, with these fixed parameters, one can check that when 1 < A < 20, all thetraces are positive and ω2 > ω1 > 0. Therefore for making an Otto cycle for these particular B, W , b1and αaC values, one must have, 1 < A < 20, although efficiency of the cycle has not enhanced.

4 Conclusion

We have investigated the possibility of construction of a EUQOE with different initial entangled statesbetween two qubits (taken as monopole detectors), which was recently initiated in [15]. The thermal bathhas been mimicked by uniformly accelerating the detectors, either in parallel or in anti-parallel motion.As mentioned in the introduction, there were few limitations and incompleteness in the initial attempt.The present discussion raised above those limitations and provided a complete study of constructing aEUQOE.

We found that making of EUQOE is not possible for all cases. There are only two situations when allnecessary conditions for making an Otto engine is satisfied. Those are: (a) the detectors are in parallel

12

Table 1: Example values of η/η0 for anti-symmetrical initial state for the detectors in parallel motion.

Figure No. W ω1TAH αaC B A Tr(δρHhaC ) η/η0

1 0.011 0.007333 1.0003 5.0 5.001 2.79519× 10−9 1.004810.0088 1.0005 5.002 1.16748× 10−10 1.421210.009167 1.0006 5.0025 2.13342× 10−10 1.37246

2 1 0.846154 1.00065 100 100.055 5.78869× 10−8 1.158290.871795 1.00078 100.068 6.51390× 10−9 2.671980.882353 1.00085 100.075 4.86188× 10−8 1.34444

3 0.25 0.244444 0.99955 25 24.989 1.60600× 10−9 1.922460.243902 0.99959 24.99 5.85615× 10−9 1.212560.241379 0.99971 24.993 1.07349× 10−8 1.05709

4 0.6 0.591549 0.99929 100 99.93 6.54121× 10−8 1.140570.585366 0.99959 99.96 9.25261× 10−8 1.032600.545455 0.99989 99.99 4.68616× 10−6 1.00009

motion with the initial detector’s state is maximally entangled anti-symmetric Bell state and (b) thedetectors are in anti-parallel motion with detector’s initial state is non-maximally entangled states. Inthe first case, no continuous values of acceleration of first detector (A) is found for which making ofEUQOE is possible. Interestingly, some of the parameter values enhances the efficiency of the enginecompared to that of a single qubit Otto cycle. We have presented a table for parameter values withenhanced efficiency. But for the second case, continuous range of values of acceleration of first detector(A) is found, for some fixed values of other parameters. In the allowed A range, the efficiency firstincreasing and then decreasing with respect to A, however the efficiency never goes beyond η0. Heredepending on αaH < 1 (or αaH > 1), we need to choose αaC < 1 (or αaC > 1). For a particular αaC , weget some range of values of A, which satisfies all the essential conditions for making a EUQOE. For allother cases making of EUQOE is not possible. We summarise these outcomes in Table 2.

Table 2: Summery of possible scenarios of EUQOE.

Motion Initial StatesSatisfying all

criteria?Can η/η0 be

greater than one?Maximally entangled, Symmetric No -

Parallel Maximally entangled, Anti-symmetric Yes Yes

Non-maximally entangled No -

Maximally entangled, Symmetric No -

Anti-parallel Maximally entangled, Anti-symmetric No -

Non-maximally entangled Yes No

Couple of comments are in order. In the original QOE, one provides a real thermal bath and thereforethe existence of this bath is classical in nature. While our EUQOE (also UQOE, proposed in [12, 13]) isa pure quantum mechanical in construction in the sense that the thermal bath itself is provided througha pure quantum (plus relativistic) effect(s). Finally, it may be mentioned that the efficiency here canbe regulated by changing the observer’s acceleration. Such a feature of EUQOE can be important forexperimental verification of the Unruh effect as this phenomenon is the main input in our analysis. Inthis study we provided the values of the parameters to setup an Otto cycle. We hope that with the useof those values of the system parameters one will be able to construct an experimental apparatus whichwill also be able to verify the existence of Unruh phenomenon.

Acknowledgments: DB would like to thank the Indian Institute of Technology Guwahati for supportingthis work through Doctoral Fellowship. The research of BRM is supported by a START-UP RESEARCH

13

(a) (b)

(c)

0 5 10 15 200.0

0.2

0.4

0.6

0.8

ω1AH ω2AH η/η0

(d)

Figure 6: Initially non-maximally entangled state with b1 = 0.4 for anti-parallel motion: Plots showing(a) work done by the system, Tr(δρHhαv ), (b) heat absorbed by the system, Tr(δρHhαaH ), (c) heat

rejected by the system, Tr(δρHhαaC ) (we chosed αaC = 0.05) with respect to dimensionless accelerationof the primary detector (A) and dimensionless energy gap (W ) of the composite system and (d) showsη/η0 (green line), energy gap of the system at the beginning of stage I, ω1TAH (blue line) and expandedenergy gap at the end of stage I, ω2TAH (orange line) with respect to dimensionless acceleration of theprimary detector (A). Here we fixed dimensionless acceleration of the second detector as B = 20 and thedimensionless energy gap of the system as W = 0.2.

GRANT (No. SG/PHY/P/BRM/01) from the Indian Institute of Technology Guwahati, India and bya Core Research Grant (File no. CRG/2020/000616) from Science and Engineering Research Board(SERB), Department of Science & Technology (DST), Government of India.

14

Appendices

A Relation between the proper times of the detectors

Two detectors A and B that are uniformly accelerating in the same Rindler wedge (namely RRW) withaccelerations aA and aB , respectively. Their trajectories in terms of their proper times are determinedby the relation between the Minkowski coordinates and accelerated frame proper time:

tA = 1aA

sinh(aAτA); xA = 1aA

cosh(aAτA) ,

tB = 1aB

sinh(aBτB); xB = 1aB

cosh(aBτB) . (A.1)

On a constant t/x line both the detectors satisfy the following relation

τB = αaτA . (A.2)

This yields the required relation between the proper times of the detectors as (21) with α = αa = (aA/aB).For anti-parallelly accelerating detectors A and B, with accelerations aA and −aB , respectively, then

their trajectories in terms of their proper times are given by

tA = 1aA

sinh(aAτA); xA = 1aA

cosh(aAτA) ,

tB = 1aB

sinh(aBτB); xB = − 1aB

cosh(aBτB) . (A.3)

On a constant t/x line both the detectors satisfy the following relation

τB = −αaτA . (A.4)

In this case, therefore we have τB = ατa where α = −αa = −(aA/aB).If the detectors are moving with constant velocities and have relative velocity vrel. Then we can have

τB = αvτA, where

αv =√

1− v2rel . (A.5)

For the detectors are moving with same constant velocity, then vrel = 0 and hence αv = 1.

B Calculation of the late time density matrix

In this Appendix, we will briefly calculate the change in the density matrix (δρ) of two detectors in stageII or stage IV, where two detectors uniformly accelerate and interacts with the background massless realscalar quantum field through monopole coupling. This calculation is for both heating process (H) andcooling process (C). Therefore we drop the subscripts H or C, used in the main text and keep it as ageneral discussion. Also the energy gap of the combined system is taken as ω (instead of ω1 or ω2 usedin the main text). This interaction is governed by the action given by

Sint =

∫cAχA(τA)mA(τA)φ(xA)dτA +

∫cBχB(τB)mB(τB)φ(xB)dτB . (B.1)

We choose the initial state of the combined system in the asymptotic past as

|in〉 = |0〉|D〉 = |0〉[b1|eAgB〉+ b2|gAeB〉], (B.2)

where |0〉 is the vacuum of the quantum field in tensor product with the entangled state of the detectors.In the above we will choose b1 = (1/

√2) = b2 for |s〉, b1 = (1/

√2), b2 = −(1/

√2) for |a〉 and b1 6=

1/√

2 (0 < b1 < 1), b2 = ±√

1− b21 for non-maximally entangled states in the main analysis of our paper.At the asymptotic future, the evolved quantum state can be given by [30]

|out〉 = TeiSint |in〉 . (B.3)

Here T denotes the time order product of the operators.

15

The reduced density matrix of the detectors can be obtained by tracing out the field part of the finaldensity matrix [29]

ρAB = Trφ|out〉〈out|. (B.4)

The elements of the reduced density matrix are evaluated from the Dyson series expansion of (B.3) uptosecond order in cj inside (B.4):

〈nAnB |ρAB |nAnB〉= Trφ〈nAnB |[1 + iSint − T (SintS

′int/2)][b1|eAgB〉+ b2|gAeB〉]|0〉〈0|[b?1〈eAgB |+ b?2〈gAeB |]

[1− iSint − T (SintS′int)/2]|nAnB〉

= 〈nAnB |[b1|eAgB〉+ b2|gAeB〉][b?1〈eAgB |+ b?2〈gAeB |]|nAnB〉+Trφ〈nAnB |Sint|0〉[b1|eAgB〉+ b2|gAeB〉][b?1〈eAgB |+ b?2〈gAeB |]〈0|Sint|nAnB〉−Trφ〈nAnB |T [SintS

′int/2]|0〉〈0|[b1|eAgB〉+ b2|gAeB〉][b?1〈eAgB |+ b?2〈gAeB |]|nAnB〉

−Trφ〈nAnB |[b1|eAgB〉+ b2|gAeB〉][b?1〈eAgB |+ b?2〈gAeB |]|0〉〈0|T [SintS′int/2]|nAnB〉

= ρ0 + [R(1)nAnB ,nAnB

+R(2)nAnB ,nAnB

+R(2)?nAnB ,nAnB

]︸ ︷︷ ︸O(c2)

+O(c3), (B.5)

where ρ0 is the initial density matrix in the bases of |eAeB〉, |eAgB〉, |gAeB〉 and |gAgB〉, given by

ρ0 =

0 0 0 00 |b1|2 b1b

?2 0

0 b2b?1 |b2|2 0

0 0 0 0

. (B.6)

Terms first order in cj vanishes individually due to the trace operation (〈0M |φ(xi)|0M 〉 = 0) and theterms second order in cj are calculated as follows:

R(1)nAnB ,nAnB

= Trφ〈nAnB |Sint[b1|eAgB〉+ b2|gAeB〉]|0〉〈0|[b?1〈eAgB |+ b?2〈gAeB |]Sint|nAnB〉

= Trφ〈nAnB |(∫

cAχA(τA)mA(τA)φ(xA)dτA +

∫cBχB(τB)mB(τB)φ(xB)dτB

)[b1|eAgB〉+ b2|gAeB〉]|0〉〈0|[b?1〈eAgB |+ b?2〈gAeB |](∫

cAχA(τA)mA(τA)φ(xA)dτA +

∫cBχB(τB)mB(τB)φ(xB)dτB

)|nAnB〉

= Trφ

(∫dτAcAχA(τA)〈nAnB |mA(τA)[b1|eAgB〉+ b2|gAeB〉]φ(xA)|0〉

+

∫dτBcBχB(τB)〈nAnB |mB(τB)[b1|eAgB〉+ b2|gAeB〉]φ(xB)|0〉

)(∫

dτ ′AcAχA(τ ′A)[b?1〈eAgB |+ b?2〈gAeB |]mA(τ ′A)|nAnB〉〈0|φ(x′A)

+

∫dτ ′BcBχB(τ ′B)[b?1〈eAgB |+ b?2〈gAeB |]mB(τ ′B)|nAnB〉〈0|φ(x′B)

)= Trφ

[ ∫dτAcAχA(τA)[b1e

i(gA−eA)τAδnA,gAδnB ,gB + b2ei(eA−gA)τAδnA,eAδnB ,eB ]φ(xA)|0〉

+

∫dτBcBχB(τB)[b1e

i(eB−gB)τBδnA,eAδnB ,eB + b2ei(gB−eB)τBδnA,gAδnB ,gB ]φ(xB)|0〉

][ ∫

dτ ′AcAχA(τ ′A)[b?1ei(eA−gA)τ ′AδnA,gAδnB ,gB + b?2e

i(gA−eA)τ ′AδnA,eAδnB ,eB ]〈0|φ(x′A)

+

∫dτ ′BcBχB(τ ′B)[b?1e

i(gB−eB)τ ′BδnA,eAδnB ,eB + b?2ei(eB−gB)τ ′BδnA,gAδnB ,gB ]〈0|φ(x′B)

]≡ AAnAnB ,nAnB +ABnAnB ,nAnB +BBnAnB ,nAnB +BAnAnB ,nAnB , (B.7)

where ej = ω/2 and gj = −ω/2, are the energy levels of the individual detectors. The positive frequency

16

Wightman function, G+(τA, τB) is obtained as

G+(τA, τB) = Trφ

[φ(xB)|0〉〈0|φ(xA)

]=

∑all field states(n)

〈n|φ(xB)|0〉〈0|φ(xA)|n〉

= 〈0|φ(xA)(∑

n

|n〉〈n|)φ(xB)|0〉

= 〈0|φ(xA)φ(xB)|0〉 (B.8)

where completeness identity of the field states has been used in the last equality. We have

AA =

∫ ∫c2AχA(τA)χA(τ ′A)dτAdτ

′AG

+(x′A, xA)

|b2|2eiω(τA−τ ′A) 0 0 b2b

?1eiω(τA+τ ′A)

0 0 0 00 0 0 0

b1b?2eiω(−τA−τ ′A) 0 0 |b1|2eiω(−τA+τ ′A)

=

|b2|2PA(ω) 0 0 b2b

?1PAA(ω)

0 0 0 00 0 0 0

b1b?2PAA(−ω) 0 0 |b1|2PA(−ω)

, (B.9)

BB =

∫ ∫c2BχB(τB)χB(τ ′B)dτBdτ

′BG

+(x′B , xB)

|b1|2eiω(τB−τ ′B) 0 0 b1b

?2eiω(τB+τ ′B)

0 0 0 00 0 0 0

b2b?1eiω(−τB−τ ′B) 0 0 |b2|2eiω(−τB+τ ′B)

=

|b1|2PB(ω) 0 0 b1b

?2PBB(ω)

0 0 0 00 0 0 0

b2b?1PBB(−ω) 0 0 |b2|2PB(−ω)

, (B.10)

AB =

∫ ∫cAcBχA(τA)χB(τ ′B)dτAdτ

′BG

+(x′B , xA)

b2b

?1eiω(τA−τ ′B) 0 0 |b2|2eiω(τA+τ ′B)

0 0 0 00 0 0 0

|b1|2e−iω(τA+τ ′B) 0 0 b1b?2e−iω(τA−τ ′B)

=

b2b

?1PAB(ω,−ω) 0 0 |b2|2PAB(ω, ω)

0 0 0 00 0 0 0

|b1|2PAB(−ω,−ω) 0 0 b1b?2PAB(−ω, ω)

, (B.11)

BA =

∫ ∫cAcBχA(τ ′A)χB(τB)dτ ′AdτBG

+(x′A, xB)

b1b

?2e−iω(τ ′A−τB) 0 0 |b1|2eiω(τ ′A+τB)

0 0 0 00 0 0 0

|b2|2e−iω(τ ′A+τB) 0 0 b2b?1eiω(τ ′A−τB)

=

b1b

?2P?AB(ω,−ω) 0 0 |b1|2P?AB(−ω,−ω)

0 0 0 00 0 0 0

|b2|2P?AB(ω, ω) 0 0 b2b?1P?AB(−ω, ω)

, (B.12)

where Pj(±ω) is defined by (15). We denote

Pjj(±ω) = c2j

∫ ∫χj(τj)χj(τ

′j)dτjdτ

′jG

+(x′j , xj)e±iω(τj+τ

′j), (B.13)

PAB(±ω,±ω) = cAcB

∫ ∫dτAdτBe

i(±ωτA±ωτ ′B)G+(x′B , xA). (B.14)

17

Using these definitions, we can identify that the matrix BA as hermitian conjugate of matrix AB.

The positive frequency Wightman function G+(x′, x) can be written as G+(τ ′, τ), satisfies the follow-ing properties

G+(−τi,−τj) = G+(τj , τi); G+(τi, τj) = [G+(τj , τi)]?. (B.15)

Using the above properties of the positive frequency Wightman function, we can check that PAB (similarly,Pj) is a real quantity, as

PAB(±ω,±ω) = cAcB

∫ ∫dτAdτBe

i(±ωτA±ωτ ′B)G+(τ ′B , τA)

= cAcB

∫ ∫dτAdτBe

−i(±ωτA±ωτ ′B)G+(−τ ′B ,−τA)

= cAcB

∫ ∫dτAdτBe

−i(±ωτA±ωτ ′B)[G+(τ ′B , τA)]?

= P?AB(±ω,±ω), (B.16)

where after the second equality, we have changed τj → −τj . Then, we have used the property of theWightman function, given in (B.15).

The final expression of R(1) can be obtained by adding (B.9), (B.10), (B.11) and (B.12), as

R(1) =

|b2|2PA(ω) + |b1|2PB(ω) 0 0 b2b

?1PAA(ω) + b1b

?2PBB(ω)

+b2b?1PAB(ω,−ω) + b1b

?2PAB(ω,−ω) +|b2|2PAB(ω, ω) + |b1|2PAB(−ω,−ω)

0 0 0 00 0 0 0

b1b?2PAA(−ω) + b2b

?1PBB(−ω) 0 0 |b1|2PA(−ω) + |b2|2PB(−ω)

+|b1|2PAB(−ω,−ω) + |b2|2PAB(ω, ω) +b1b?2PAB(−ω, ω) + b2b

?1PAB(−ω, ω)

,(B.17)

where we have used the identity (B.16).

18

Next we calculate the following term:

R(2)nAnB ,nAnB

= −Trφ〈nAnB |T (SintS′int/2)|0〉〈0|[b1|eAgB〉+ b2|gAeB〉][b?1〈eAgB |+ b?2〈gAeB |]|nAnB〉

= −〈nAnB |〈0|T [

(∫dτAcAχA(τA)mA(τA)φ(xA) +

∫dτBcBχB(τB)mB(τB)φ(xB)

)(∫dτ ′AcAχA(τ ′A)

mA(τ ′A)φ(x′A) +

∫dτ ′BcBχB(τ ′B)mB(τ ′B)φ(x′B)

)/2]|0〉[b1|eAgB〉+ b2|gAeB〉][b?1〈eAgB |+ b?2〈gAeB |]|nAnB〉

= −〈nAnB |〈0|T [

(c2A

∫ ∫dτAdτ

′AχA(τA)χA(τ ′A)mA(τA)mA(τ ′A)φ(xA)φ(x′A)

+c2B

∫ ∫dτBdτ

′BχB(τB)χB(τ ′B)mB(τB)mB(τ ′B)φ(xB)φ(x′B) + cAcB

∫ ∫dτBdτ

′AχB(τB)χA(τ ′A)

mB(τB)mA(τ ′A)φ(xB)φ(x′A) + cAcB

∫ ∫dτAdτ

′BmA(τA)mB(τ ′B)χA(τA)χB(τ ′B)φ(xA)φ(x′B)

)/2]|0〉

[b1|eAgB〉+ b2|gAeB〉][b?1〈eAgB |+ b?2〈gAeB |]|nAnB〉[Now assuming non-primed proper times are greater than primed proper times, i.e., τ > τ ′, see[30]]

= −〈nAnB |[(c2A

∫ ∫dτAdτ

′AχA(τA)χA(τ ′A)mA(τA)mA(τ ′A)Θ(τA − τ ′A)GW (xA, x

′A)

+c2B

∫ ∫dτBdτ

′BχB(τB)χB(τ ′B)mB(τB)mB(τ ′B)Θ(τB − τ ′B)GW (xB , x

′B)

+ cAcB

∫ ∫dτBdτ

′AχB(τB)χA(τ ′A)mB(τB)mA(τ ′A)Θ(τB − τ ′A)GW (xB , x

′A)

+ cAcB

∫ ∫dτAdτ

′BχA(τA)χB(τ ′B)mA(τA)mB(τ ′B)Θ(τA − τ ′B)GW (xA, x

′B)

)][b1|eAgB〉+ b2|gAeB〉]

[b?1δnA,eAδnB ,gB + b?2δnA,gAδnB ,eB ]

= −〈nAnB |(c2A

∫ ∫dτAdτ

′AχA(τA)χA(τ ′A)mA(τA)mA(τ ′A)[b1|eAgB〉+ b2|gAeB〉]Θ(τA − τ ′A)GW (xA, x

′A)

+c2B

∫ ∫dτBdτ

′BχB(τB)χB(τ ′B)mB(τB)mB(τ ′B)[b1|eAgB〉+ b2|gAeB〉]Θ(τB − τ ′B)GW (xB , x

′B)

+ cAcB

∫ ∫dτBdτ

′AχB(τB)χA(τ ′A)mB(τB)mA(τ ′A)[b1|eAgB〉+ b2|gAeB〉]Θ(τB − τ ′A)GW (xB , x

′A)

+ cAcB

∫ ∫dτAdτ

′BχA(τA)χB(τ ′B)mA(τA)mB(τ ′B)[b1|eAgB〉+ b2|gAeB〉]Θ(τA − τ ′B)GW (xA, x

′B)

)[b?1δnA,eAδnB ,gB + b?2δnA,gAδnB ,eB ]

= −(c2A∫ ∫

dτAdτ′AχA(τA)χA(τ ′A)[b1e

iωτAe−iωτ′AδnA,eAδnB ,gB + b2e

−iωτAeiωτ′AδnA,gAδnB ,eB ]Θ(τA − τ ′A)GW (xA, x

′A)

+c2B∫ ∫

dτBdτ′BχB(τB)χB(τ ′B)[b1e

−iωτBeiωτ′BδnA,eAδnB ,gB + b2e

iωτBe−iωτ′BδnA,gAδnB ,eB ]Θ(τB − τ ′B)GW (xB , x

′B)

+cAcB∫ ∫

dτBdτ′AχB(τB)χA(τ ′A)[b1e

iωτB−iωτ ′AδnA,gAδnB ,eB + b2e−iωτB+iωτ ′AδnA,eAδnB ,gB ]Θ(τB − τ ′A)GW (xB , x

′A)

+cAcB∫ ∫

dτAdτ′BχA(τA)χB(τ ′B)[b1e

iωτ ′B−iωτAδnA,gAδnB ,eB + b2e−iωτ ′B+iωτAδnA,eAδnB ,gB ]Θ(τA − τ ′B)GW (xA, x

′B))

[b?1δnA,eAδnB ,gB + b?2δnA,gAδnB ,eB ]

= −(c2A∫ ∫

dτAdτ′AχA(τA)χA(τ ′A)[b1e

iω(τA−τ ′A)δnA,eAδnB ,gB + b2e−iω(τA−τ ′A)δnA,gAδnB ,eB ]Θ(τA − τ ′A)GW (xA, x

′A)

+c2B

∫ ∫dτBdτ

′BχB(τB)χB(τ ′B)[b1e−iω(τB−τ ′B)δnA,eAδnB ,gB + b2e

iω(τB−τ ′B)δnA,gAδnB ,eB ]Θ(τB − τ ′B)GW (xB , x′B)

+cAcB

∫ ∫dτAdτ

′BχA(τA)χB(τ ′B)[b1e−iω(τA−τ ′B)δnA,gAδnB ,eB + b2e

iω(τA−τ ′B)δnA,eAδnB ,gB ][iGF (xA, x′B)])

[b?1δnA,eAδnB ,gB + b?2δnA,gAδnB ,eB ], (B.18)

where in the third term after the second last equality, we have changed τ ′A → τA and τB → τ ′B . Then we

19

used Θ(τA − τ ′B) + Θ(τ ′B − τA) = 1, to obtain the final expression.

This R(2) can be written in a matrix form, given by

R(2) = −

0 0 0 00 b?1(b1FA(ω) + b1FB(−ω) + b2E(ω)) b?2(b1FA(ω) + b1FB(−ω) + b2E(ω)) 00 b?1(b2FA(−ω) + b2FB(ω) + b1E(−ω)) b?2(b2FA(−ω) + b2FB(ω) + b1E(−ω)) 00 0 0 0

,(B.19)

where we defined

Fj(±ω) = c2j

∫ ∫dτjdτ

′jχj(τj)χj(τ

′j)e±iω(τj−τ ′j)Θ(τj − τ ′j)G+(τj , τ

′j), (B.20)

E(±ω) = cAcB

∫ ∫dτAdτ

′BχA(τA)χB(τ ′B)e±iω(τA−τ ′B)[iGF (τA, τ

′B)]. (B.21)

One can see that Fj(±ω) satisfies the following relations,

Fj(±ω) + F ?j (±ω)

= c2j

∫ ∫dτjdτ

′jχj(τj)χj(τ

′j)[e±iω(τj−τ ′j)Θ(τj − τ ′j)G+(τj , τ

′j) + e∓iω(τj−τ ′j)Θ(τj − τ ′j)G+(τ ′j , τj)

]= c2j

∫ ∫dτjdτ

′jχj(τj)χj(τ

′j)e∓iω(τj−τ ′j)G+(τ ′j , τj)

= Pj(∓ω),(B.22)

where in the first term of the second line we interchanged τj and τ ′j . Also, one finds

Fj(±ω) + F ?j (∓ω)

= c2j

∫ ∫dτjdτ

′jχj(τj)χj(τ

′j)[e±iω(τj−τ ′j)Θ(τj − τ ′j)G+(τj , τ

′j) + e±iω(τj−τ ′j)Θ(τj − τ ′j)G+(τ ′j , τj)

]= c2j

∫ ∫dτjdτ

′jχj(τj)χj(τ

′j)e±iω(τj−τ ′j)Θ(τj − τ ′j)(G+(τj , τ

′j) +G+(τ ′j , τj))

≡ Qj(±ω) = Q?j (∓ω).(B.23)

E(ω) satisfies the following relation

E(ω) + E?(ω)

= cAcB

∫ ∫dτAdτ

′BχA(τA)χB(τ ′B)eiω(τA−τ ′B)[Θ(τA − τ ′B)G+(τA, τ

′B) + Θ(τ ′B − τA)G+(τ ′B , τA)]

+cAcB

∫ ∫dτAdτ

′BχA(τA)χB(τ ′B)e−iω(τ ′B−τA)[Θ(τ ′B − τA)G+(τA, τ

′B) + Θ(τA − τ ′B)G+(τ ′B , τA)]

= cAcB

∫ ∫dτAdτ

′BχA(τA)χB(τ ′B)eiω(τA−τ ′B)[G+(τA, τ

′B) +G+(τ ′B , τA)]

= P?AB(−ω, ω) + PAB(ω,−ω)

= PAB(−ω, ω) + PAB(ω,−ω), (B.24)

where we interchanged τA and τ ′B in the second line. To arrive in the last equality, we have used thefact that PAB is a real quantity, see (B.16). Now adding (B.19) with it’s hermitian conjugate, and using(B.22), (B.23) and (B.24), we obtain

R(2) +R(2)† = −

0 0 0 00 b21(PA(−ω) + PB(ω))+ b1b2(QA(ω) +QB(−ω)) 0

b1b2(PAB(−ω, ω) + PAB(ω,−ω)) +b22E(ω) + b21E?(−ω)0 b1b2(Q?A(ω) +Q?B(−ω)) b22(PA(ω) + PB(−ω))+ 0

+b22E?(ω) + b21E(−ω) b1b2(PAB(−ω, ω) + PAB(ω,−ω))0 0 0 0

.(B.25)

Finally, by adding (B.17) and (B.25), we obtain the variation of the density matrix (δρ) upto secondorder in ci, as

δρ = R(1) +R(2) +R(2)†. (B.26)

20

C Calculation of Tr(δρHhαk)

To evaluate the trace (14), we need to add the diagonal terms of δρHhαk (k = v, αaH , αaC ). This isgiven by multiplication of matrices given by (B.26) and (10). We consider the coefficients of the initialdetector state, b1, b2 are to be real numbers, satisfying |b1|2 + |b2|2 = 1. The diagonal terms are:

(δρHhαk)eAeB ,eAeB = [b22PA(ω) + b21PB(ω) + 2b1b2PAB(ω,−ω)]1 + αk

2; (C.1)

(δρHhαk)eAgB ,eAgB = −[b21(PA(−ω) + PB(ω)) + b1b2(PAB(ω,−ω) + PAB(−ω, ω))]1− αk

2; (C.2)

(δρHhαk)gAeB ,gAeB = −[b22(PA(ω) + PB(−ω)) + b1b2(PAB(ω,−ω) + PAB(−ω, ω))]−1 + αk

2; (C.3)

(δρHhαk)gAgB ,gAgB = [b21PA(−ω) + b22PB(−ω) + 2b1b2PAB(−ω, ω)]−1− αk

2. (C.4)

Trace of δρHhαk is obtained as,

Tr[δρHhαk ] = {b22PA(ω)− b21PA(−ω) + b1b2[PAB(ω,−ω)− PAB(−ω, ω)]}+αk{b21PB(ω)− b22PB(−ω) + b1b2[PAB(ω,−ω)− PAB(−ω, ω)]}. (C.5)

This leads to (14) for any general initially entangled states.

D Calculation of Pj(±ω)The expression of Pj (j = A,B) is given by (15). The integrations need to be evaluated for qualitativeanalysis of efficiency of the cycle. As the interactions are turned on at time −Tjk/2 and runs upto timeTjk/2, the integration limits should be −Tjk/2 to Tjk/2. These integrations can not be done analyticallydue to their finite integration limits. However, we can extend the integration limits to ±∞ by choosinga suitable switching function. We choose the switching function given by (20), which is non-vanishingfor −Tjk/2 < τj < Tjk/2 and approximately vanishing outside this domain. Pj(ω) can be expressed by(taking cA = cB = 1)

Pj(ω) =

∫ ∞−∞

∫ ∞−∞

dτjdτ′jχj(τj)χj(τ

′j)e

iω(τj−τ ′j)G+(τ ′j , τj), (D.1)

To evaluate this quantity, we choose a coordinate transform

T = τA + τ ′A; σ = τA − τ ′A. (D.2)

The Jacobian of this transformation is 1/2. In the case of evaluating PB , we use (A.2) while detector Bis accelerating in RRW and (A.4) while in LRW to express the integrations in terms of τA and τ ′A. Useof (A.2) and (A.4) implies that

χB(τB)χB(τ ′B) =(TB/2)2

τ2B + (TB/2)2

(TB/2)2

τ ′2B + (TB/2)2=

(±αaTA/2)2

(±αaτA)2 + (±αaTA/2)2

(±αaTA/2)2

(±αaτ ′A)2 + (±αaTA/2)2

= χA(τA)χA(τ ′A) (D.3)

With use of (D.2), we can express the combination of the switching functions as

χj(τj)χj(τ′j) =

(TA/2)2

τ2A + (TA/2)2

(TA/2)2

τ ′2A + (TA/2)2

=T 4A

(T 2 − T 21 )(T 2 − T 2

2 ), (D.4)

where, T1 = σ+ iTA and T2 = −σ+ iTA. The Wightman function in (D.1) for detector A can be writtenas a function of σ, as G+(τ ′A, τA) = [G+(τA, τ

′A)]? = [G+(σ)]?. Then we can write (D.1) for j = A, as

PA(±ω) =1

2

∫ ∞−∞

dσe±iωσ[G+(σ)]?∫ ∞−∞

dTT 4A

(T 2 − T 21 )(T 2 − T 2

2 ), (D.5)

21

σ-iA-σ-iA

σ+iA-σ+iA

CRlower

CRupper

RRe(T)

Im(T)

-R

(a)

Figure 7: Plots showing contours of integral T . For an integrand like f(T ) = e±ikT /[(T 2−T 21 )(T 2−T 2

2 )],we choose upper and lower contours for positive and negative sign of the exponential, respectively. Fork = 0, we are free to choose any contour. Both will give same results.

To evaluate T−integral, we used the upper contour given in (fig:7) (see, [12, 13]). As an integrand likef(T ) = 1/[(T 2 − T 2

1 )(T 2 − T 22 )] vanishes for both T → ±∞ (±i∞), we are free to choose upper or lower

contour. In upper contour, the poles are at T = T1 and T = T2. The integration results,

∫ ∞−∞

dT

(T 2 − T 21 )(T 2 − T 2

2 )= 2πi

(Res(f(T ))

∣∣∣T=T1

+Res(f(T ))∣∣∣T=T2

)= 2πi

(1

2T1(T 21 − T 2

2 )+

1

2T2(T 22 − T 2

1 )

)=

π

2TA(σ2 + T 2A)

. (D.6)

Therefore, (D.5) becomes

PA(±ω) =πT 3

A

4

∫ ∞−∞

dσ e±iωσ [G+(σ)]?

σ2 + T 2A

, (D.7)

• The positive frequency Wightman in Minkowski spacetime is given by [26]

G+(x, x′) = − 1

4π2[(t− t′ − iε)2 − |x− x′|2]. (D.8)

The trajectory of a uniformly accelerating observer moving in RRW and LRW is given in (18) and (19).One finds

(tj − t′j − iε)2 − |xj − x′j |2 =( 1

ajsinh(ajτj)−

1

ajsinh(ajτ

′j)− iε

)2

−∣∣∣± 1

ajcosh(ajτj)∓

1

ajcosh(ajτ

′j)∣∣∣2

= − 2

a2j

+2

a2j

(cosh(ajτj) cosh(ajτ

′j)− sinh(ajτj) sinh(ajτ

′j))− iε

= − 2

a2j

(1− [cosh(aj(τj − τ ′j)− iε]

)= − 2

a2j

(1− cosh

(aj(τj − τ ′j)− iε

))=

4

a2j

sinh2(aj(τj − τ ′j)

2− iε

). (D.9)

22

Therefore the positive frequency Wightman function of a uniformly accelerating detector is given by (seealso in [26])

G+(τj , τ′j) = −

a2j

16π2 sinh2(aj(τj − τ ′j)/2− iε)= − 1

4π2

∞∑n=−∞

1

((τj − τ ′j)− iε− 2πin/aj)2. (D.10)

where we have used (a/2)2

sinh2(ax/2)=∑∞n=−∞

1(x−i2πn/a)2 . Now using (D.10) for j = A with (D.2), we can

write (D.7) as

PA(±ω) = − T3A

16π

∫ ∞−∞

dσ e±iωσ

(σ2 + T 2A)

∞∑n=−∞

1

(σ + iε− 2πin/aA)2. (D.11)

The summation in this expression can be splited into three parts as

PA(±ω) = − T3A

16π

∫ ∞−∞

dσ e±iσω

σ2 + T 2A

(1

(σ + iε)2+

∞∑n=1

1(σ − i2πn

aA

)2 +

∞∑n=1

1(σ + i2πn

aA

)2

)

= PA,0(±ω) + PA,n(±ω) + PA,−n(±ω) . (D.12)

Poles contributing to PA(ω) are at iTA, 2πinaA

and poles contributing to PA(−ω) are at −iε, −iTA, − 2πinaA

.The corresponding residues with factor ±2πi are (ε→ 0)

(2πi)Res(P(I)A (ω))

∣∣∣σ=iTA

=T 2Ae−TAω

16

(1

T 2A

+

∞∑n=1

1

(TA − 2πn/aA)2+

∞∑n=1

1

(TA + 2πn/aA)2

)

=(aATA/2)2e−TAω

16 sin2(aATA/2), [using,

(a/2)2

sin2(ax/2)=

∞∑n=−∞

1

(x− 2πn/a)2]

= (−2πi)Res(P(I)A (−ω))

∣∣∣σ=−iTA

, (D.13)

(2πi)

∞∑n=1

Res(P(I)A,n(ω))

∣∣∣σ= 2πin

aA

=

∞∑n=1

T 3Ae− 2πnω

aA

([T 2A − 4π2n2

a2A]ω − 4πn

aA

)8(T 2A −

4π2n2

a2A

)2

= (−2πi)

∞∑n=1

Res(P(I)A,−n(−ω))

∣∣∣σ=− 2πin

aA

, (D.14)

where

∞∑n=1

T 3Ae− 2πnω

aA

([T 2A− 4π2n2

a2A

]ω− 4πnaA

)8

(T 2A−

4π2n2

a2A

)2 =a2AT

2Ae− 2πωaA

64π2

(Φ

(e− 2πωaA , 2, 1 +

aATA2π

)− Φ

(e− 2πωaA , 2, 1− aATA

2π

))

+aAT 2

Aωe− 2πωaA

32π

(Φ

(e− 2πωaA , 1, 1 +

aATA2π

)− Φ

(e− 2πωaA , 1, 1− aATA

2π

))(D.15)

and

(−2πi)Res(P(I)A,0(−ω))

∣∣∣ω=−iε

=TAω

8. (D.16)

Here the notation, P(I)A means the integrand of the the quantity PA. Same notation also will be used

in the following calculations. Using (D.15) and adding (D.13) and (D.14), we obtain expression of PA(ω),given in (22). Similarly, adding (D.13), (D.14) and (D.16), we obtain the expression of PA(−ω), given in(23).

PB in RRW: Using (A.2), (D.3) and (D.1), we can write PB as

PB(±ω) = α2a

∫ ∞−∞

∫ ∞−∞

dτAdτ′AχA(τA)χA(τ ′A)e±iωαa(τA−τ ′A)G+

B(αaτ′A, αaτA) . (D.17)

23

The Wightman function for detector B can be written as (from (D.10))

G+B(αaτA, αaτ

′A) = G+(τB , τ

′B) = − a2

B

16π2 sinh2(aB(τB − τ ′B)/2− iε)

= − (aA/αa)2

16π2 sinh2(aA(τA − τ ′A)/2− iε)

= − 1

4π2α2a

∞∑n=−∞

1

(σ − iε− 2πin/aA)2. (D.18)

Using the coordinate transformation given in (D.2) with (D.4), (D.6) and (D.18), we can write (D.17) as

PB(±ω) = − T4A

8π2

∫ ∞−∞

∫ ∞−∞

dσ dT e±iωαaσ

(T 2 − T 21 )(T 2 − T 2

2 )

∞∑n=−∞

1

(σ + iε− 2πinaA

)2

= − T3A

16π

∫ ∞−∞

dσ e±iωαaσ

σ2 + T 2A

∞∑n=−∞

1

(σ + iε− 2πinaA

)2

= PA(±ωαa) . (D.19)

The last equality comes form comparing expression of PB(±ω) after the second equality with expressionof PA(±ω), given in (D.11). The only difference is presence of αa with ω in (D.19). Therefore adding(D.13) and (D.14) with replacing ω with ωαa and using αa = aA/aB , gives expression of PB(ω) in (24).Similary, adding (D.13), (D.14) and (D.16) with replacing ω with ωαa and using αa = aA/aB , givesexpression of PB(−ω) in (25).

PB in LRW: Using (A.4), (D.3) and (D.1), we can write PB as

PB(±ω) =

∫ ∞−∞

∫ ∞−∞

dτBdτ′BχB(τB)χB(τ ′B)e±iω(τB−τ ′B)G+(τ ′B , τB)

= α2a

∫ ∞−∞

∫ ∞−∞

dτAdτ′AχA(τA)χA(τ ′A)e∓iωαa(τA−τ ′A)G+(−αaτ ′A,−αaτA)

= α2a

∫ ∞−∞

∫ ∞−∞

dτAdτ′AχA(τA)χA(τ ′A)e±iωαa(τA−τ ′A)G+(αaτ

′A, αaτA) , (D.20)

where in the last equality, we changed τ → −τ and we know that G+(αaτ′A, αaτA) = [G+(αaτA, αaτ

′A)]?.

This expression of PB is same as given in (D.17). Thus

P(LRW )B (±ω) = P(RRW )

B (±ω) = PB(±ω) . (D.21)

.

E Calculation of PAB(ω,−ω)− PAB(−ω, ω)Using (A.2), (A.4) and (D.4), we obtain the combination of the switching functions for both detectors,as

χA(τA)χB(τ ′B) = χA(τA)χA(τ ′A) =T 4A

(T 2 − T 21 ) (T 2 − T 2

2 ). (E.1)

E.1 Acceleration in Same Direction

Using the coordinate transformation (D.2) with (A.2), we can write

ω(τA − τ ′B) =ω

2(σ + T − αaT + αaσ) =

ω T

2(1− αa) +

ωσ

2(1 + αa). (E.2)

24

Using the trajectories given in (A.1), denominator of the green function can be evaluated as

(tA − t′B − iε)2 − |xA − xB |2

=( 1

aAsinh(aAτA)− 1

aBsinh(aBτ

′B)− iε

)2

−( 1

aAcosh(aAτA)− 1

aBcosh(aBτ

′B))2

= − 1

a2A

− 1

a2B

+2

aAaB(cosh(aAτA) cosh(aBτ

′B)− sinh(aAτA) sinh(aBτ

′B))− iε

= − 2

aAaB

1

2

(aAaB

+aBaA

)+

2

aAaBcosh(aAσ − iε)

=2

aAaB[cosh(aAσ − iε)− cosh(aAκ)]

=4

aAaBsinh

(aA(σ − iε− κ)

2

)sinh

(aA(σ − iε+ κ)

2

), (E.3)

here in the third equality, we used (A.2) and (D.2), and in the fourth equality we defined

cosh(aAκ) =1

2

(aAaB

+aBaA

). (E.4)

The positive frequency Wightman function becomes

G+(τA, τ′B) = G+(σ) = − aAaB

16π2[sinh(aA(σ−iε−κ)2 ) sinh(aA(σ−iε+κ)

2 )]= [G+(τ ′B , τA)]? , (E.5)

Putting (E.2) and (E.5) in (17), we obtain

PAB(±ω,∓ω) = −aAaB αaT 4A

32π2

∫ ∞−∞

dσ e±iω2 (1+αa)σ

sinh(aA(σ+iε−κ)2 ) sinh(aA(σ+iε+κ)

2 )

∫ ∞−∞

dT e±iω2 (1−αa)T

(T 2 − T 21 )(T 2 − T 2

2 ). (E.6)

We are interested to evaluate the quantities PAB(±ω) when aA 6= aB . For both the cases αa > 1 andαa < 1, we need to evaluate the T−integrals separately due to their pole structures in the complexT−plane.

For αa < 1:PAB(ω,−ω): The complex T−integration for PAB(ω,−ω) has the contributing poles at T1 = σ+ iTA andT2 = −σ + iTA in the upper half plane (see fig:(7)). The integration yields

πe−ωTA(1−αa)/2

4 σ TA

(e−i(1−αa)ωσ/2

−iTA + σ+ei(1−αa)ωσ/2

iTA + σ

). (E.7)

Putting this into (E.6), we will get,

PAB(ω,−ω) = −aAaB αaT 3Ae−ωTA(1−αa)/2

128π

∫ ∞−∞

dσ

(σ − iε)[sinh(aA(σ+iε−κ)2 ) sinh(aA(σ+iε+κ)

2 )]

×

(eiωσ

σ + iTA+

eiωσαa

σ − iTA

)

= −aAaB αaT 3Ae−ωTA(1−αa)/2

128π[PAB,1(ω,−ω) + PAB,2(ω,−ω)] . (E.8)

The pole at σ = 0 on the real axis, will be shifted to upper half of the complex σ−plane. The contributingpoles are +iε, + iTA, ± κ − iε + 2πin

aA(n = 1, 2, · · · ,∞) in the upper half plane (see subfigure (a) in

figure (8)). The corresponding residues are

(2πi)Res(P(I)AB(ω,−ω))

∣∣∣σ=iε

= 0 , (E.9)

(2πi)Res(P(I)AB,2(ω,−ω))

∣∣∣σ=iTA

=2π e−ωαaTA

TA sinh(aA(−κ+iTA)

2

)sinh

(aA(κ+iTA)

2

) , (E.10)

25

(2πi)Res(P(I)AB(ω,−ω))

∣∣∣σ=κ−iε+ i2πn

aA

=2πi

(κ+ 2πniaA

)aA2 sinh(aAκ)

[eiω[κ+ 2πni

aA]

κ+ 2πinaA

+ iTA+

eiαaω[κ+ 2πni

aA]

κ+ 2πinaA− iTA

](E.11)

and

(2πi)Res(P(I)AB(ω,−ω))

∣∣∣σ=−κ−iε+ i2πn

aA

=−2πi

(−κ+ 2πniaA

)aA2 sinh(aAκ)

[eiω[−κ+ 2πni

aA]

−κ+ 2πinaA

+ iTA+

eiαaω[−κ+ 2πni

aA]

−κ+ 2πinaA− iTA

],

(E.12)where we have used, cosh(inπ) = (−1)n and sinh(±aAκ+ inπ) = ± sinh(aAκ)(−1)n.

iϵ

i

-iM0

M1

M2

M3

N0

N1

N2

N3

CR

R-R

(a)

iϵ

i

-i

M1

M0

M-1

M-2

M-3

N1

N0

N-1

N-2

N-3

CR

R-R

(b)

Figure 8: Plots showing contours of integral (a) PAB(ω,−ω), (b) PAB(−ω, ω), where contributing polesare in upper and lower half of the complex-σ plane , respectively. The poles Mn and Nn are defined asκ− iε+ i2πn

aAand −κ− iε+ i2πn

aA, respectively.

PAB(−ω, ω): From (E.6), we have

PAB(−ω, ω) = −aAaB αaT 4A

32π2

∫ ∞−∞

dσ e−iωσ2 (1+αa)

sinh(aA(σ+iε−κ)2 ) sinh(aA(σ+iε+κ)

2 )

∫ ∞−∞

dT e−iωT2 (1−αa)

(T 2 − T 21 )(T 2 − T 2

2 ). (E.13)

The complex T−integration has the contributing poles at T = −T1 = −σ − iTA and T = −T2 = σ − iTAin the lower half plane (see fig:(7)). The integration yields

πe−ωTA(1−αa)/2

4σTA

(e−

iωσ2 (1−αa)

σ − iTA+eiωσ2 (1−αa)

σ + iTA

). (E.14)

Putting this into (E.13), we obtain

PAB(−ω, ω) = −aAaB αaT 3Ae−ωTA(1−αa)/2

128π

∫ ∞−∞

dσ

(σ − iε)[sinh(aA(σ+iε−κ)2 ) sinh(aA(σ+iε+κ)

2 )]

×

(e−iωσ

σ − iTA+e−iωσαa

σ + iTA

)

= −aAaB αaT 3Ae−ωTA(1−αa)/2

128π[PAB,1(−ω, ω) + PAB,2(−ω, ω)] . (E.15)

The contributing poles are −iTA, ± κ + iε − 2πinaA

(n = 0, 1, · · · ,∞) in the lower half plane (seesubfigure (b) in figure (8)). The corresponding residues are

(−2πi)Res(P(I)AB,2(−ω, ω))

∣∣∣σ=−iTA

=2π e−ωαaTA

TA sinh(aA(κ−iTA)

2

)sinh

(aA(−κ−iTA)

2

) , (E.16)

(−2πi)Res(P(I)AB(−ω, ω))

∣∣∣σ=κ−iε− i2πnaA

=−2πi

(k − 2πinaA

)aA2 sinh(aAκ)

[e−iω[k− 2πin

aA]

k − 2πinaA− iTA

+e−iωαa[k− 2πin

aA]

k − 2πinaA

+ iTA

],

(E.17)

26

and

(−2πi)Res(P(I)AB(−ω, ω))

∣∣∣σ=−κ−iε− i2πnaA

=−2πi

(k + 2πinaA

)aA2 sinh(aAκ)

[eiω[k+ 2πin

aA]

−k − 2πinaA− iTA

+eiωαa[k+ 2πin

aA]

−k − 2πinaA

+ iTA

],

(E.18)where we have used, cosh(−inπ) = (−1)n and sinh(±aAκ− inπ) = ± sinh(aAκ)(−1)n.

In PAB(ω,−ω)−PAB(−ω, ω), residue given in (E.10) will cancel out with residue (E.16), and for alln > 0 residues given in (E.11) and (E.12) cancel out with residues in (E.18) and (E.17), respectively. Thedifference, PAB(ω,−ω)−PAB(−ω, ω) can be obtained by adding (E.17) and(E.18) for n = 0 (multiplied

with the factor−aAaB αaT 3

Ae−ωTA(1−αa)/2

128π ), given in (27).

For αa > 1:PAB(ω,−ω): From (E.6), we have

PAB(ω,−ω) = −aAaB αaT 4A

32π2

∫ ∞−∞

dσ eiωσ2 (1+αa)

sinh(aA(σ+iε−κ)2 ) sinh(aA(σ+iε+κ)

2 )

∫ ∞−∞

dT e−iωT2 (αa−1)

(T 2 − T 21 )(T 2 − T 2

2 ).(E.19)

The complex T−integration has the contributing poles at T = −T1 = −σ − iTA and T = −T2 = σ − iTAin the lower half plane (see fig:(7)). The integration yields

πe−ωTA(αa−1)/2

4σTA

(e−

iωσ2 (αa−1)

σ − iTA+eiωσ2 (αa−1)

σ + iTA

). (E.20)

Putting this into (E.19), we will get,

PAB(ω,−ω) = −aAaB αaT 3Ae−ωTA(αa−1)/2

128π

∫ ∞−∞

dσ

(σ − iε)[sinh(aA(σ+iε−κ)2 ) sinh(aA(σ+iε+κ)

2 )]

×

(eiωσ

σ − iTA+

eiωσαa

σ + iTA

)

= −aAaB αaT 3Ae−ωTA(αa−1)/2

128π[PAB,1(ω,−ω) + PAB,2(ω,−ω)] . (E.21)

The contributing poles are iε, iTA, ±κ−iε+ 2πinaA

(n = 1, 2, · · · ,∞) in the upper half plane (see subfigure(a) in figure (8)). The corresponding residues are

(2πi)Res(P(I)AB(ω,−ω))

∣∣∣σ=iε

= 0 , (E.22)

(2πi)Res(P(I)AB,1(ω,−ω))

∣∣∣σ=iTA

=2π e−ωTA

TA sinh(aA(−κ+iTA)

2

)sinh

(aA(κ+iTA)

2

) , (E.23)

(2πi)Res(P(I)AB(ω,−ω))

∣∣∣σ=κ−iε+ i2πn

aA

=2πi

(κ+ 2πniaA

)aA2 sinh(aAκ)

[eiω[κ+ 2πni

aA]

κ+ 2πinaA− iTA

+eiαaω[κ+ 2πni

aA]

κ+ 2πinaA

+ iTA

],

(E.24)and

(2πi)Res(P(I)AB(ω,−ω))

∣∣∣σ=−κ−iε+ i2πn

aA

=−2πi

(−κ+ 2πniaA

)aA2 sinh(aAκ)

[eiω[−κ+ 2πni

aA]

−κ+ 2πinaA− iTA

+eiαaω[−κ+ 2πni

aA]

−κ+ 2πinaA

+ iTA

],

(E.25)where we have used, cosh(inπ) = (−1)n and sinh(±aAκ+ inπ) = ± sinh(aAκ)(−1)n.

27

PAB(−ω, ω): From (E.6), we have

PAB(−ω, ω) = −aAaB αaT 4A

32π2

∫ ∞−∞

dσ e−iωσ2 (1+αa)

sinh(aA(σ+iε−κ)2 ) sinh(aA(σ+iε+κ)

2 )

∫ ∞−∞

dT eiωT2 (αa−1)

(T 2 − T 21 )(T 2 − T 2

2 ).(E.26)

The complex T−integration has the contributing poles at T = T1 = σ + iTA and T = T2 = −σ + iTA inthe upper half plane (see fig:(7)). The integration yields

πe−ωTA(αa−1)/2

4σTA

(e−i(αa−1)ωσ/2

σ − iTA+ei(αa−1)ωσ/2

σ + iTA

). (E.27)

Putting this into (E.26), we obtain

PAB(−ω, ω) = −aAaB αaT 3Ae−ωTA(αa−1)/2

128π

∫ ∞−∞

dσ

(σ − iε)[sinh(aA(σ+iε−κ)2 ) sinh(aA(σ+iε+κ)

2 )]

×

(e−iωσ

σ + iTA+e−iωσαa

σ − iTA

)

= −aAaB αaT 3Ae−ωTA(αa−1)/2

128π[PAB,1(−ω, ω) + PAB,2(−ω, ω)] . (E.28)

The contributing poles are −iTA, ±κ− iε− 2πinaA

(n = 0, 1, · · · ,∞) in the lower half plane (see subfigure(b) in figure (8)). The corresponding residues are

(−2πi)Res(P(I)AB,1(−ω, ω))

∣∣∣σ=−iTA

=2π e−ωTA

TA sinh(aA(κ−iTA)

2

)sinh

(aA(−κ−iTA)

2

) , (E.29)

(−2πi)Res(P(I)AB(−ω, ω))

∣∣∣σ=κ−iε− i2πnaA

=−2πi

(k − 2πinaA

)aA2 sinh(aAκ)

[e−iω[k− 2πin

aA]

k − 2πinaA

+ iTA+e−iωαa[k− 2πin

aA]

k − 2πinaA− iTA

],

(E.30)and

(−2πi)Res(P(I)AB(−ω, ω))

∣∣∣σ=−κ−iε− i2πnaA

=2πi

(−k − 2πinaA

)aA2 sinh(aAκ)

[eiω[k+ 2πin

aA]

−k − 2πinaA

+ iTA+

eiωαa[k+ 2πin

aA]

−k − 2πinaA− iTA

],

(E.31)where we have used, cosh(−inπ) = (−1)n and sinh(±aAκ− inπ) = ± sinh(aAκ)(−1)n.

In PAB(ω,−ω) − PAB(−ω, ω), residue given in (E.23) cancel out with residue in (E.29) and for alln > 0 residues given in (E.24) and (E.25) cancel out with residues in (E.31) and (E.30), respectively. Thedifference, PAB(ω,−ω)−PAB(−ω, ω) can be obtained by adding (E.31) and (E.30) for n = 0 (multiplied

with the factor−aAaB αaT 3

Ae−ωTA(1−αa)/2

128π ), given in (28).

E.2 Acceleration in Opposite Direction

If the detector A and B are accelerating anti-parallely, their trajectories are given by (A.3). Therefore,denominator of the green function can be calculated as

(tA − t′B − iε)2 − |xA − x′B |2

=( 1

aAsinh(aAτA)− 1

aBsinh(aBτ

′B)− iε

)2

−( 1

aAcosh(aAτA) +

1

aBcosh(aBτ

′B))2

= − 1

a2A

− 1

a2B

− 2

aAaB(cosh(aAτA + aBτ

′B) + iε)

= − 2

aAaB(cosh(aAσ + iε) + cosh(aAκ))

= − 4

aAaBcosh

(aA(σ + iε+ κ)

2

)cosh

(aA(σ + iε− κ)

2

), (E.32)

28

where in the third equality, we have used (A.4), (D.2) and (E.4). The corresponding Wightman functionis

G+(τA, τ′B) = G+(σ) = +

aAaB

16π2[cosh(aA(σ+iε−κ)2 ) cosh(aA(σ+iε+κ)

2 )]= [G+(τ ′B , τA)]? . (E.33)

The Wightman function has poles at

σ = ±κ− iε± iπ(2n+ 1)

aA; n = 0, 1, 2, · · ·∞ (E.34)

As there is no poles on the real axis, the absence or presence of the ‘iε’ in the green function, will noteffect the complex integral of σ. We will drop the ‘iε’ from the Wightman function. We also can checkthat

ω(τA − τB) = ω(τA + αaτ′A) =

ωσ

2(1− αa) +

ωT

2(1 + αa) . (E.35)

iϵ

i

-iM-1

M0

M1

M2

N-1

N0

N1

N2

CR

R-R

(a)

iϵ

i

-i

M0

M-1

M-2

M-3

N0

N-1

N-2

N-3

CR

R-R

(b)

Figure 9: Plots showing contours of integral (a) PAB(ω,−ω), (b) PAB(−ω, ω), where contributing polesare in upper and lower half of the complex-σ plane, respectively. The poles Mn and Nn are defined as

κ+ iπ(2n+1)aA

and −κ+ iπ(2n+1)aA

, respectively.

Using (E.33), (E.35) and (E.1) in (17), we obtain expression of PAB , as

PAB(±ω,∓ω) = −aAaB αaT 4A

32π2

∫ ∞−∞

dσ e±iω2 (1−αa)σ

cosh(aA(σ−κ)2 ) cosh(aA(σ+κ)

2 )

∫ ∞−∞

dT e±iω2 (1+αa)T

(T 2 − T 21 )(T 2 − T 2

2 ). (E.36)

PAB(ω,−ω): From (E.36), the complex T−integration has the contributing poles at T = T1 = σ+ iTAand T = T2 = −σ + iTA in the upper half plane (see fig:(7)). The integration yields

πe−12ωTA(αa+1)

4σTA

(ei2 (αa+1)ωσ

σ + iTA+e−

i2 (α+1)ωσ

σ − iTA

). (E.37)

Putting this into (E.36)

PAB(ω,−ω) = −aAaB αaT 3A e−(1+αa)TAω/2

128π

×∫ ∞−∞

dσ

(σ − iε)[cosh(aA(σ−κ)2 ) cosh(aA(σ+κ)

2 )]

( eiωσ

σ + iTA+e−iαaωσ

σ − iTA

)= −aAaB αaT 3

A e−(1+αa)TAω/2

128π[PAB,1(ω,−ω) + PAB,2(ω,−ω)] . (E.38)

The contributing poles are +iε, ± κ ± iπ(2n+1)aA

(n = 0, 1, · · · ,∞) for the upper and lower half planes.The residues of the first term in the upper half plane (see subfigure (a) of figure (9)) are obtained as

Res(P(I)AB,1(ω,−ω))

∣∣∣σ=iε

=2π

TA cosh2(aAκ

2

) (E.39)

29

and

(2πi)Res(P(I)AB,1(ω,−ω))

∣∣∣σ=κ+

πi(2n+1)aA

+ (2πi)Res(P(I)AB,1(ω,−ω))

∣∣∣σ=−κ+

πi(2n+1)aA

=4πi csch(aAκ)

aA sinh2( (2n+1)iπ2 )

(eiω[k+

(2n+1)iπaA

](k + (2n+1)iπ

aA)(k + (2n+1)iπ

aA+ iTA)

− eiω[−k+

(2n+1)iπaA

](k − (2n+1)iπ

aA)(k − (2n+1)iπ

aA− iTA)

),

(E.40)where we have used cosh(±ak ± (2n + 1) iπ2 ) = sinh(±ak) sinh(±(2n + 1) iπ2 ) and for the second term inthe lower half plane (see subfigure (b) of figure (9)), we obtain

(−2πi)Res(P(I)AB,2(ω,−ω))

∣∣∣σ=κ−πi(2n+1)

aA

+ (−2πi)Res(P(I)AB,2(ω,−ω))

∣∣∣σ=−κ−πi(2n+1)

aA

=−4πi csch(aAκ)

aA sinh2(− (2n+1)iπ2 )

(e−iωαa

[k− (2n+1)iπ

aA

](k − (2n+1)iπ

aA)(k − (2n+1)iπ

aA− iTA)

− e−iωαa

[−k− (2n+1)iπ

aA

](k + (2n+1)iπ

aA)(k + (2n+1)iπ

aA+ iTA)

).

(E.41)PAB(−ω, ω): From (E.36), the complex T−integration has the contributing poles at T = −T1 =

−σ − iTA and T = −T2 = σ − iTA in the lower half plane (see fig:(7)). The integration yields

πe−12 (αa+1)TAω

4σTA

(ei2 (αa+1)σω

σ + iTA+e−

i2 (αa+1)σω

σ − iTA

). (E.42)

Putting this into (E.36), we obtain

PAB(−ω, ω) = −aAaB αaT 3A e−(1+αa)TAω/2

128π

×∫ ∞−∞

dσ

(σ − iε)[cosh(aA(σ−κ)2 ) cosh(aA(σ+κ)

2 )]

( eiαaωσ

σ + iTA+

e−iωσ

σ − iTA

)= −aAaB αaT 3

A e−(1+αa)TAω/2

128π[PAB,1(−ω, ω) + PAB,2(−ω, ω)] . (E.43)

The contributing poles are +iε, ± κ± iπ(2n+1)aA

(n = 0, 1, · · · ,∞) for the upper half plane and lower halfplane. The corresponding residues for the first term in the upper half plane (see subfigure (a) of figure(9)), are

Res(P(I)AB,1(−ω, ω))

∣∣∣σ=iε

=2π

TA cosh2(aAκ

2

) (E.44)

and

(2πi)Res(P(I)AB,1(−ω, ω))

∣∣∣σ=κ+

πi(2n+1)aA

+ (2πi)Res(P(I)AB,1(−ω, ω))

∣∣∣σ=−κ+

πi(2n+1)aA

=4πi csch(aAκ)

aA sinh2( (2n+1)iπ2 )

(eiωαa

[k+

(2n+1)iπaA

](k + (2n+1)iπ

aA)(k + (2n+1)iπ

aA+ iTA)

− eiωαa

[−k+

(2n+1)iπaA

](k − (2n+1)iπ

aA)(k − (2n+1)iπ

aA− iTA)

).

(E.45)For the second term in the lower half plane (see subfigure (b) of figure (9)), we obtain

(−2πi)Res(P(I)AB,2(−ω, ω))

∣∣∣σ=κ−πi(2n+1)

aA

+ (−2πi)Res(P(I)AB,2(−ω, ω))

∣∣∣σ=−κ−πi(2n+1)

aA

=−4πi csch(aAκ)

aA sinh2(− (2n+1)iπ2 )

(e−iω[k− (2n+1)iπ

aA

](k − (2n+1)iπ

aA)(k − (2n+1)iπ

aA− iTA)

− e−iω[−k− (2n+1)iπ

aA

](k + (2n+1)iπ

aA)(k + (2n+1)iπ

aA+ iTA)

).

(E.46)So, in ∆PAB , residues in (E.39), (E.40), and (E.41) cancel out with residues in (E.44), (E.46) and

(E.45), respectively. Hence, ∆PAB vanishes for anti-parallel acceleration of the detectors.

30

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