arXiv:2204.00643v1 [quant-ph] 1 Apr 2022

Towards reconciliation of completely positive open system dynamicswith the equilibration postulate

Marcin Lobejko, Marek Winczewski, Gerardo Suarez, Robert Alicki, Micha l HorodeckiInternational Centre for Theory of Quantum Technologies, University of Gdansk, 80-308 Gdansk, Poland

(Dated: April 5, 2022)

Almost every quantum system interacts with a large environment, so the exact quantum me-chanical description of its evolution is impossible. One has to resort to approximate description,usually in the form of a master equation. There are at least two basic requirements for such a de-scription: first, it should preserve the positivity of probabilities; second, it should correctly describethe equilibration process for systems coupled to a single thermal bath. Existing two widespreaddescriptions of evolution fail to satisfy at least one of those conditions. The so-called Davies masterequation, while preserving the positivity of probabilities, fails to describe thermalization properly.On the other hand, the Bloch-Redfield master equation violates the first condition, but it correctlydescribes equilibration, at least for off-diagonal elements for several important scenarios. However,is it possible to have a description of open system dynamics that would share both features? In thispaper, we partially resolve this problem in the weak-coupling limit: (i) We provide a general form ofthe proper thermal equilibrium state (the so-called mean-force state) for an arbitrary open system.(ii) We provide the solution for the steady-state coherences for a whole class of master equations, andin particular, we show that the solution coincides with the mean-force Hamiltonian for the Bloch-Redfield equation. (iii) We consider the cumulant equation, which is explicitly completely positive,and we show that its steady-state coherences are the same as one of the Bloch-Redfield dynamics(and the mean-force state accordingly). (iv) We solve the correction to the diagonal part of thestationary state for a two-level system both for the Bloch-Redfield and cumulant equation, showingthat the solution of the cumulant is very close to the mean-force state, whereas the Bloch-Redfielddiffers significantly.

I. INTRODUCTION

A basic property of the open system interacting with a single thermal reservoir is that it tends to a thermalequilibrium state. Another fundamental property of any system is that evolution should preserve the positivity ofthe measurement outcomes probabilities. Since it is, except for a few particular cases, impossible to describe opensystems dynamics exactly, one applies approximate descriptions in the form of master equations [1, 2]. Clearly, suchapproximate descriptions should satisfy the above two postulates.

A widespread description of the open systems weakly coupled with the bath stems from Born approximation,followed by Markovian approximation. This leads to Bloch-Redfield equation [3, 4]. If one subsequently applies yetanother approximation (called the secular approximation) one obtains the most popular master equation in Gorini-Kossakowski-Linblad-Sudarshan form [5, 6], called Davies equation [7]. The first one is more exact but does notpreserve the positivity of the state. The second one, while preserving positivity and working perfectly in typicalquantum optical regimes, cannot describe subtleties related to nearly degenerated levels. The Born approximationimplies that both equations describe the dynamics up to the second order in the coupling strength λ.

One can now ask to what extent the equilibration postulate is satisfied. Since the equations are up to second orderin λ, the equilibrium state should not be the Gibbs state of the free Hamiltonian anymore, but rather it should be apartial trace of a total (system plus bath) thermal equilibrium state, expanded up to second order in λ. Imaginingthat it is a Gibbs state with respect to some Hamiltonian, one calls the latter “mean-force Hamiltonian”, and thestate is called mean-force Gibbs state. This issue has been a matter of intense research, see [8] for a review.

It is known, in particular, that for some standard coupling to a bosonic bath, the Bloch-Redfield equation satisfies inpart this basic requirement. Namely, the state it equilibrates to has the same off-diagonal elements as the mean-forceGibbs state [9–13]. However, the above is not known for arbitrary coupling, i.e., on the level of generality assumedwhen deriving these master equations. It should be emphasized that the general form of the second-order correctionfor either the mean-force Hamiltonian or the steady-state Hamiltonian for the Bloch-Redfield equation has not beencalculated so far. Regarding diagonal elements, there is an agreement between the steady state and the mean-forceGibbs state only up to zeroth order, and we cannot expect more with second order master equations as argued in[9, 13]. On the other hand, the Davies equation does not satisfy the equilibration postulate even for off-diagonalelements, since its steady-state is a Gibbs state of the bare Hamiltonian, or employing the renormalization proposedin [14], its second-order correction is equal to the so-called Lamb-Stark shift (known also as dynamical correction), ingeneral, different than the mean-force Hamiltonian.

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We are thus in quite an unfortunate situation. The Bloch-Redfield equation, that predicts the correct equilibration(at least for off-diagonal elements) is not completely positive. On the other hand, the completely positive masterequation - the Davies one, fails to describe thermalization much more severely (cf. [13]). Thus, we are quite far awayfrom achieving the two basic postulates:

(i) open systems thermalize,

(ii) probabilities are non-negative,

simultaneously in the description of the evolution of open systems in a weak coupling scenario.There is yet another candidate for dynamical equation that might join the advantages of both the equations above.

This is the “cumulant equation” (also called the refined weak coupling limit) derived in [14] and then independentlydiscovered and further developed in [15, 16] (see [17] for recent development). This is a completely positive evolutionthat is not based on the secular approximation nor Markovian assumption in any form. Since it is just the secularapproximation that makes Davies equation problematic in many respects, one would think that this new equationmay indeed be a solution for the above quite fundamental problem. However, the cumulant equation is believed toreduce to Davies one for the long time limit [14, 15], which would imply that the steady-state must be that of Daviesequation, hence the postulate would not be satisfied.

In this paper, we make a progress towards reconciliation of the two postulates - positivity and thermalization. Tothis end we provide several separate and general results, that are of independent interest. Then the main message,which follows from putting them together, is that the cumulant equation is at the moment quite close to satisfy bothpostulates. We also show where and to what extent it fails to do so.

Let us now present the mentioned results. To this end, let us set up a vocabulary. Namely, in the discussion above,three types of corrections to bare Hamiltonian appear: mean-force correction, steady-state correction and dynamicalcorrection. As explained, the mean-force correction is the following: we imagine that the thermal equilibrium state isthe Gibbs state according to perturbed bare Hamiltonian, and the perturbation is the mean-force correction. Regard-ing steady-state correction, we similarly imagine that the steady-state is the Gibbs state according to a perturbedbare Hamiltonian and that (in general different) perturbation is the steady-state Hamiltonian correction. Finally, thedynamical correction is the well-known Hamiltonian contribution from the environment. The form of the latter isknown for the Davies equation and Bloch-Redfield equation.

So far, we have put more emphasis on the problem of relations between the mean-force and the steady-statecorrections. Of course, ideally, we would like to have all three corrections equal to each other. However, existingresults (see, e.g., [12]) show that, in general, this is not possible, yet it is hard to find a proper physical explanationof such a discrepancy. In our studies, we show how those three corrections are related to each other in a general opensystems framework. Let us also stress that all our results concern second-order corrections - we will thus not explicitlyrepeat this in the following.

Here are the most general results that do not involve any particular master equation: We derive

• general form of mean-force correction for an arbitrary open system, which so far was only known for particularcouplings. This form has been also independently derived by G. Timofeev and A. Trushechkin [18];

• coherences of the steady-state correction for quite a general form of master equation in terms of its Kossakowskimatrix.

In addition, our formulas for the above corrections explicitly show their relations with the dynamical correction. Wenext concentrate on the Bloch-Redfield equation and the cumulant equation: We show that

• for a general coupling, the Bloch-Redfield equation predicts the coherences of a steady-state correction thatcoincide with those of the mean-force (this was known before only for some particular models, see, e.g., [10, 12,19, 20]);

• the generator of cumulant master equation, to lowest order, agrees with Bloch-Redfield equation.

The above results already achieve the goal partially: the last item implies that the cumulant equation has the samesteady-state correction coherences as the Bloch-Redfield one. Since the cumulant equation is completely positive,hence we obtain a reconciliation of complete positivity with thermalization for coherences. This also shows that thecumulant equation does not converge to the Davies equation for long times.

Subsequently, we deal with a much more complicated issue of diagonal elements, we provide:

• general methodology for computing diagonal elements of the steady-state correction.

• formula for those elements in the case of the two-level system, obtaining elegant relation between diagonalelements of the mean-force and those of the steady-state correction.

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Next, we specialize to Bloch-Redfield and cumulant equation:

• We show that diagonal elements of the mean-force and the steady-state correction are quite close to one anotherfor cumulant equation, while for Bloch-Redfield, they significantly differ.

Combined with the above results on coherences, these last results show that the cumulant equation is quite close toreconciling the postulates of positivity and thermalization.

The paper is organized as follows. First, in Section II we provide the notation and concepts that will be used inthe paper, i.e., we define and derive three different corrections to the bare Hamiltonian, namely the dynamical, thesteady-state, and the mean force one. Next, in Section III, we describe the main results regarding the off-diagonalelements of the Hamiltonian. Later in section IV, we introduce the cumulant equation, which is a second-orderdynamical map whose differential form can be used to obtain diagonal corrections up to the second-order, which isdone in Section V. These results are then applied to a two-level system in Section VI.

II. PRELIMINARIA: DYNAMICAL, MEAN-FORCE AND STEADY STATE CORRECTION

We consider a general Hamiltonian of the interacting system S with the thermal reservoir R of the form:

H = H0 +HR + λHI , (1)

HI =∑α

Aα ⊗Rα, (2)

where H0 is a bare Hamiltonian of the system, HR is free Hamiltonian of the bath, Aα, Rα are interaction operators(acting on the system and bath Hilbert spaces, respectively), and λ is a coupling constant. In the following, wedefine a Gibbs state of the thermal reservoir γR = Z−1

R e−βHR at inverse temperature β, where ZR = Tr[e−βHR

].

Additionally, we consider the time-dependent operators A(t) = ei(H0+HR)tRαe−i(H0+HR)t, and we use an abbreviation

〈A〉γR = Tr[AγR]. We assume that that bath operators are centralized, i.e., 〈Rα〉γR = 0.The main object of interest of this article are three different corrections to the bare Hamiltonian of the system H0

in the weak coupling limit (i.e., λ 1), defined according to the Hamiltonians: dynamical Hdyn, steady-state Hst

and the mean-force Hmf. Due to centralization of the bath operators, the leading order of the perturbation calculusis λ2. In accordance, we introduce the second-order corrections:

Hk −H0 = λ2H(2)k + . . . , (3)

which we represent in the basis of jump operators:

H(2)k (t) =

∑ω,ω′

∑α,β

Υ(k)αβ (ω, ω′, t)A†α(ω)Aβ(ω′), (4)

where k indicates to dynamical (dyn), steady-state (st) or mean-force (mf) correction, and the jump operators aregiven by:

Aα(ω) =∑

ε′−ε=ωΠ(ε)AαΠ(ε′). (5)

where Π(ε) is the projector on the subspace with energy ε, such that H0 =∑ε ε Π(ε).

A. Dynamical correction

We start with the dynamical correction H(2)dyn, which is defined according to the generator of the master equation

in the Schrodinger picture of the following general form:

Lt[ρ] = i[ρ,H0 + λ2H(2)dyn(t)] + λ2

∑ω,ω′

∑αβ

Kαβ(ω, ω′, t)

(Aβ(ω′)ρA†α(ω)− 1

2A†α(ω)Aβ(ω′), ρ

)(6)

= i[ρ,H0] + λ2∑ω,ω′

∑αβ

[iΥ

(dyn)αβ (ω, ω′, t)[ρ,A†α(ω)Aβ(ω′)] +Kαβ(ω, ω′, t)

(Aβ(ω′)ρA†α(ω)− 1

2A†α(ω)Aβ(ω′), ρ

)](7)

4

where K is the so-called Kossakowski matrix and for a while we do not determine it: For particular choice of Kand Υ(dyn) we will obtain a given master equation, such as the Bloch-Redfield or Davies-GKSL one. Notice that themaster equation Eq. (6) leads to a completely positive dynamics if the matrix Kαβ(ω, ω′, t) is positive semi-definite.Later, we consider the long-time limit, when t→∞, for which we use the abbreviations:

Υ(dyn)αβ (ω, ω′) ≡ lim

t→∞Υ

(dyn)αβ (ω, ω′, t), Kαβ(ω, ω′) ≡ lim

t→∞Kαβ(ω, ω′, t). (8)

1. Bloch-Redfield and Davies master equations

Let us consider generators of the two most known master equations, i.e., the Bloch-Redfield LRt and the Davisgenerator LDt . For the Bloch-Redfield master equation we have

Kαβ(ω, ω′, t) = Γαβ(ω′, t) + Γ∗βα(ω, t) ≡ γαβ(ω, ω′, t), (9)

Υ(dyn)αβ (ω, ω′, t) =

1

2i[Γαβ(ω′, t)− Γ∗βα(ω, t)] ≡ Sαβ(ω, ω′, t), (10)

where

Γαβ(ω, t) =

∫ t

0

ds eiωs〈Rα(s)Rβ(0)〉γR . (11)

It is well-known that the Bloch-Redfield equation, in general, does not preserve the positivity of the state sinceγαβ(ω, ω′) is not a positive semi-definite matrix. Commonly, this issue is solved by applying the so-called secularapproximation, which leads to the Davies master equation in the GKSL form. In accordance, applying the secularapproximation by putting ω = ω′, and taking a limit t→∞, we obtain the coefficients for the Davies dynamics:

γαβ(ω, ω′, t)sec. approx.−−−−−−−→

t→∞γαβ(ω)δω,ω′ , (12)

Sαβ(ω, ω′, t)sec. approx.−−−−−−−→

t→∞Sαβ(ω)δω,ω′ , (13)

where

γαβ(ω) = limt→∞

γαβ(ω, ω, t) =

∫ +∞

−∞ds eiωs〈Rα(s)Rβ(0)〉γR , (14)

is the Fourier transform of the auto-correlation function, and

Sαβ(ω) = limt→∞

Sαβ(ω, ω, t) = P 1

2π

∫ +∞

−∞dΩ

γαβ(Ω)

ω − Ω, (15)

where P denotes the principal value integral. We also have the following relation:

Γαβ(ω) = limt→∞

Γαβ(ω, t) =1

2γαβ(ω) + iSαβ(ω). (16)

One may believe that the dynamical correction (i.e., the so-called Lamb-Stark shift), together with the bare Hamil-tonian, describes the physical Hamiltonian of the system, i.e., the one to which the system should thermalize in contactwith a single bath. However, there is no simple physical argument why this should be the case. On the contrary,introduced in the next section, the mean-force Hamiltonian should describe a proper local equilibrium based on thesecond law of thermodynamics. As we will see later, those two corrections, in general, are entirely different. What ismore, they have to be different to ensure the proper steady-state of the master equation.

B. Mean-force correction

Let us then introduce the mean-force Hamiltonian Hmf, defined according to the marginal Gibbs state of the globalequilibrium, i.e.,

e−βHmf

TrS [e−βHmf ]=

TrR[e−βH ]

TrSR[e−βH ]. (17)

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One should notice that the above equation does not specify uniquely the ground state energy of the mean-forceHamiltonian, i.e., the equation is invariant under the transformation Hmf → Hmf + ∆ for arbitrary real constant ∆.Commonly, this constant is fixed by the convention [21]:

TrSR[e−βH ] = TrS [e−βHmf ] TrR[e−βHR ], (18)

such that we have the following relation

e−β(H0+λ2H

(2)mf +...

)= Z−1

R TrR[e−βH ], (19)

where we expanded the mean-force Hamiltonian into a series. Later, we concentrate only on the second-order correction

H(2)mf . Notice, that the zeroth-order Hamiltonian of the mean-force state (19) is fixed by putting λ = 0, i.e.,

Z−1R TrR[e−βH ]

∣∣∣λ=0

= Z−1R TrR[e−β(H0+HR)] = e−βH0 , (20)

which is given by the bare Hamiltonian H0.We stress that the mean-force is the correct physical Hamiltonian of a weakly coupled system to a single thermal

bath, in a sense that the system should thermalize to its corresponding Gibbs state. Otherwise, if the open systemdynamics in the long time limit tends to a different steady-state, one might still extract work by thermalizing thewhole system to its global equilibrium, which would violate the second law of thermodynamics. In accordance, themean-force Hamiltonian is the reference point for any open system dynamics: The validation of the particular masterequation is that its steady-state should be the Gibbs state with respect to the mean-force Hamiltonian. Let us nowthen introduce the final, steady-state correction.

C. Steady-state correction

In analogy to the mean-force correction, we define the steady-state correction H(2)st as the following Gibbs state:

% = e−β(H0+λ2H(2)st +... ). (21)

(We do not introduce the normalization of % since it is unnecessary for the following discussion.) Based on the generalequilibration postulate, we believe that such a Gibbs state should be the stationary state for the system coupled toa single heat bath (at inverse temperature β). Thus, the master equation with the generator Lt that describes thisprocess, should satisfy the following steady-state condition:

limt→∞

Lt[%] ≡ L∞[%] = 0. (22)

To find the solution for the (second-order) steady-state correction H(2)st , we adapt the perturbative method. First, we

expand the generator of the master equation:

L∞[%] = L(0)∞ [%] + λ2L(2)

∞ [%] + λ4L(4)∞ [%] + . . . , (23)

Note that here we consider a general case, such that for any second-order master equation the generators of the fourthand higher orders identically vanish. In particular, for the master equation given by Eq. (6), we have

L(0)∞ [%] = i[%,H0], (24)

L(2)∞ [%] =

∑ω,ω′

∑αβ

[iΥ

(dyn)αβ (ω, ω′)[ρ,A†α(ω)Aβ(ω′)] +Kαβ(ω, ω′)

(Aβ(ω′)ρA†α(ω)− 1

2A†α(ω)Aβ(ω′), ρ

)], (25)

L(4)∞ [%] = L(6)

∞ [%] = · · · = 0. (26)

Similarly, we expand the postulated stationary-state, i.e.,

% = %0 + λ2%2 + λ4%4 . . . , (27)

where by using the Dyson series for the lowest orders we get:

%0 = e−βH0 , %2 = −e−βH0

∫ β

0

dt etH0H(2)st e

−tH0 . (28)

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Finally, from (22), we obtain the set of equations (for each order of λ):

L(0)∞ [%0] = 0 (29)

L(0)∞ [%2] + L(2)

∞ [%0] = 0, (30)

L(0)∞ [%4] + L(2)

∞ [%2] + L(4)∞ [%0] = 0, (31)

. . .

Our main goal is to solve the following equations to find the solution for the steady-state correction H(2)st . However,

as it was highlighted in [10], the second-order equation (30) only provides the solution for off-diagonal terms of thecorrection, whereas the solution for the diagonal part involves the fourth-order equation (31). In the following (SectionIII and IV), we provide the general solution for off-diagonal elements for the master equation of the form (6) and

the cumulant equation (introduced in Section IV). Moreover, in the Section V, we find the total solution for H(2)st

(involving the diagonal part) for the two-dimensional system via solving the fourth-order equation. We also preparethe fundamentals for methods for the general solution.

At the end of this section, we want to make two remarks. First, notice that the choice %0 = e−βH0 in the definitionof (21) satisfies the zeroth-order equation (29) for the generator (24). However, the solution is not unique and, inparticular, it cannot be specified (in contrast to the mean-force correction) in the limit λ → 0. Nevertheless, thischoice is dictated by the general equilibration postulate, i.e., in the limit of a vanishing coupling constant the systemshould thermalize to the Gibbs state with respect to its bare Hamiltonian. By reason of this ambiguity, later weprovide the necessary condition for a master equation (which is in fact the detailed balance condition) to satisfy thehigher order equation for the steady-state % with %0 = e−βH0 . Secondly, similarly to the mean-force correction, Eq.

(22) determines the diagonal part of the correction H(2)st only up to an arbitrary constant, i.e., if % is the solution, then

also is %e−βδ, where δ is the identity matrix multiplied by an arbitrary real number. This provides us an additionaldegree of freedom that does not affect the physical meaning (i.e., the only differences of energies matter and not isabsolute values).

III. RESULTS: MEAN-FORCE AND (OFF-DIAGONAL) STEADY-STATE CORRECTION

We are ready to state our first main result regarding the mean-force correction.

Theorem 1. The mean-force correction is equal to:

Υ(mf)αβ (ω, ω′) =

1

2π

∫ +∞

−∞dΩ D(ω, ω′,Ω) γαβ(Ω), D(ω, ω′,Ω) =

1

ω′ − Ω− (ω − ω′)(eβ(ω−Ω) − 1)

(ω − Ω)(ω′ − Ω)(eβ(ω−ω′) − 1), (32)

or equivalently, in terms of the Sαβ(ω) function, it takes the form:

Υ(mf)αβ (ω, ω′) =

1

eβω − eβω′(eβωSαβ(ω′)− eβω

′Sαβ(ω) + eβ(ω+ω′) (Sβα(−ω′)− Sβα(−ω))

). (33)

Proof. The sketch of the proof is as follows. To solve Eq. (17) we write the exponents from both sides of the equalityin terms of the Dyson series, which formally can be expressed via the time-ordering operator T as:

e−βHmf = e−β(H0+δHmf) = e−βH0T e−∫ β0dt δHmf(t), (34)

e−βH = e−β(H0+HR+λHI) = e−βH0T e−λ∫ β0dt HI(t), (35)

where we put δHmf = Hmf − H0 and we define a time-dependent operators A(t) = et(H0+HR)Ae−t(H0+HR). Conse-quently, the equality (17) can be rewritten in the form:

TrR

[T(e−∫ β0dt δHmf(t) − e−λ

∫ β0dt HI(t)

)γR

]= 0. (36)

Note that this is exact for arbitrary coupling strength λ. Then, considering the weak-coupling limit (λ 1), weexpand the above equality and obtain within the second-order the following condition for the mean-force correction:∫ β

0

dt H(2)mf (t) = −

∫ β

0

dt

∫ t

0

ds⟨HI(t)HI(s)

⟩γR. (37)

After substituting the representation of the mean-force Hamiltonian given by Eq. (4) and the interaction term (2)

with the definition of jump operators (5), we are able to calculate the coefficients Υ(mf)αβ (ω, ω′) (see the detailed proof

in Appendix B).

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Theorem 1 is the first derivation in the literature of the second-order mean-force Hamiltonian for a general weak-coupling of the form (1) (see also [18]). Remarkably the expression (32) does not exhibit any poles, as it is usual

for a dynamical correction. One observes that the coefficients are symmetric, i.e., Υ(mf)αβ (ω, ω′) = Υ

(mf)αβ (ω′, ω), which

together with S∗αβ(ω) = Sβα(ω), ensures the hermiticity of the Hamiltonian. According to this expression, let us

compare the dynamical Hamiltonian with the mean-force. Using Eq. (16), we write down the dynamical correctionin terms of γαβ(ω) and Sαβ(ω), for the Bloch-Redfield master equation:

Υ(dyn)αβ (ω, ω′) =

1

2(Sαβ(ω) + Sαβ(ω′)) +

i

4(γαβ(ω)− γαβ(ω′)), (38)

whereas for the Davies equation the off-diagonal elements vanish (due to the secular approximation). It is seen thatthe mean-force correction is different than the dynamical one; in particular, the dynamical correction has non-zeroimaginary part in contrast to the real coefficients of the mean-force. In Section VI A we compare numerically both ofthe Hamiltonians for the spin-boson model.

Our second main result is the solution of the Eq. (30). In particular, this provides the formula for off-diagonal

terms of the steady-state correction H(2)st , which later is compared to the mean-force Hamiltonian.

Theorem 2. If the following conditions are satisfied:

(i) the detailed balance relation:

Kαβ(ω, ω) = Kβα(−ω,−ω)eβω, (39)

(ii) and for ω 6= ω′:

Υ(st)αβ (ω, ω′) = Υ

(dyn)αβ (ω, ω′) +

i

eβω − eβω′(Kβα(−ω,−ω′)eβ(ω+ω′) − 1

2Kαβ(ω′, ω)(eβω + eβω

′)

), (40)

then % = e−β(H0+λ2H(2)st +... ) is a solution of the equation: L(0)

∞ [%2] + L(2)∞ [%0] = 0.

Proof. In order to construct the solution of the Eq. (30) we transform the operator equation into the algebraic one,by introducing the representation:

L(k)∞ [ρl] =

∑α,β

∑ω,ω′

g(kl)αβ (ω, ω′)e−βH0Aα(ω)Aβ(ω′), (41)

where l + k = 2. In accordance, the Eq. (30) can be rewritten as:∑α,β

∑ω,ω′

(g(02)αβ (ω, ω′) + g

(20)αβ (ω, ω′))e−βH0Aα(ω)Aβ(ω′) = 0. (42)

such that the solution is given by:

g(02)αβ (ω, ω′) + g

(20)αβ (ω, ω′) = 0 (43)

for each ω, ω′ and α, β. In particular, for the master equation of the form (6), we get

− i(ω′ − ω)Υ(st)αβ (ω, ω′)α(ω′ − ω) + iΥ

(dyn)αβ (ω, ω′)(1− e−β(ω′−ω))

+ eβωKβα(−ω′,−ω)− 1

2Kαβ(ω, ω′)(e−β(ω′−ω) + 1) = 0, (44)

where

α(ω) =

∫ β

0

dt e−tω =

1−e−βω

ω , ω 6= 0

β, ω = 0(45)

From which follows the Theorem 2.

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The non-zero value of the correction (40) predicts the so-called finite “steady-state coherences” of the densitymatrix %. Those coherences represents the off-diagonal elements with respect to the basis of the bare HamiltonianH0, such that according to the Gibbs state of this Hamiltonian, % is out of equilibrium. However, as it was mentionedpreviously, the correct (local) equilibrium of the system should be identified with the mean-force Hamiltonian Hmf

rather than the bare one. Indeed, as it follows from the presented below corollary, the Bloch-Redfield master equationpredicts the same (second-order) steady-state coherences as for the Gibbs state of the mean-force Hamiltonian, whichprovides that the off-diagonal part of its stationary state is equal to the proper equilibrium state.

Corollary 2.1. For the Bloch-Redfield equation, i.e., if Kαβ(ω, ω′) = γαβ(ω, ω′) and Υ(dyn)αβ (ω, ω′) = Sαβ(ω, ω′), we

have

Υ(st)αβ (ω, ω′) = Υ

(mf)αβ (ω, ω′) for ω 6= ω′, (46)

whereas if Kαβ(ω, ω′) = γαβ(ω)δω,ω′ , then

Υ(st)αβ (ω, ω′) = Υ

(dyn)αβ (ω, ω′) for ω 6= ω′. (47)

On the contrary to the Bloch-Redfield master equation, if the secular approximation is done for γαβ(ω, ω′), then

the steady-state coherences are equal to the Gibbs state of the bare Hamiltonian with the dynamical correction H(2)dyn.

However, commonly the secular approximation is also applied for the dynamical part, as it is for the Davies equation(see Eq. (12)), such that the dynamical correction is diagonal and the steady-state coherences vanish.

IV. CUMULANT EQUATION

Let us now introduce the cumulant equation [14] that describes the open system dynamics. Unlike the previousmodels of open system, which are derived as the solution of the von Neumann equation, the cumulant equation isinitially derived in the form of the dynamical map (which does not involve the Markovian approximation):

ρ(t) = eK(2)t ρ(0), (48)

where

K(2)t [ρ] = λ2

∫ t

0

ds∑ω,ω′

∑αβ

(iSαβ(ω, ω′, s)[ρ,A†α(ω)Aβ(ω′)] + γαβ(ω, ω′, s)

(Aβ(ω′)ρA†α(ω)− 1

2A†α(ω)Aβ(ω′), ρ

))(49)

is the generator of the map in the interaction picture, such that ρ(t) = eiH0tρ(t)e−iH0t, and

γαβ(ω, ω′, t) = ei(ω−ω′)tγαβ(ω, ω′, t), (50)

Sαβ(ω, ω′, t) = ei(ω−ω′)tSαβ(ω, ω′, t). (51)

We observe that the super-operator K(2)t is closely related to the generator of the Bloch-Redfield master equation.

Indeed, the Bloch-Redfield generator in the interaction picture is given by:

LRt [ρ] = eiH0tLRt [e−iH0tρeiH0t]e−iH0t − i[ρ,H0] (52)

= λ2∑ω,ω′

∑αβ

(i[ρ, Sαβ(ω, ω′, t)A†α(ω)ρAβ(ω′)] + γαβ(ω, ω′, t)

(Aβ(ω′)ρA†α(ω)− 1

2A†α(ω)Aβ(ω′), ρ

)),(53)

which simply implies that

dK(2)t

dt= LRt . (54)

Remarkably, the time-integrated coefficients appearing in the disspative part of the super-operator K(2)t , i.e.,∫ t

0

ds γαβ(ω, ω′, s) (55)

9

are the elements of a positive semi-definite matrix (see, e.g., [15]). This property implies that the K(2)t super-operator

is of the GKSL form. In this way, the cumulant equation defines one parameter family of CPTP dynamical maps.This feature of the cumulant equation is its advantage over the Bloch-Redfield equation, for which the fundamentalproperty of completely positive evolution is not satisfied.

On the other hand, we formulate the cumulant master equation by taking a time derivative of Eq. (48), i.e.,

d

dtρ(t) =

(d

dteK

(2)t

)[ρ(0)] =

(∫ 1

0

ds esK(2)tdK

(2)t

dte(1−s)K(2)

t

)[ρ(0)] (56)

=

(∫ 1

0

ds esK(2)tdK

(2)t

dte−sK

(2)t

)[ρ(t)] ≡ LCt [ρ(t)]. (57)

Equivalently, the generator of the cumulant master equation can be written as:

LCt =e[K

(2)t ,·] − 1

[K(2)t , ·]

dK(2)t

dt, (58)

such that

d

dtρ(t) =

(d

dtK

(2)t +

1

2[K

(2)t ,

d

dtK

(2)t ] + . . .

)ρ(t), (59)

where ddtK

(2)t = LRt . According to this, we define the generator of the cumulant master equation in the Schrodinger

picture:

LCt [ρ] = i[ρ,H0] + e−iH0tLCt [eiH0tρe−iH0t]eiH0t. (60)

From this we obtain the following theorem:

Theorem 3. The generator of the cumulant master equation is, up to the second-order, equal to the generator of theBloch-Redfield master equation, i.e.,

LCt = LRt +O(λ4) (61)

Proof. Expanding the generator LCt to the second-order and putting Eq. (54), we get

LCt [ρ] = i[ρ,H0] + e−iH0tdK

(2)t

dt[eiH0tρe−iH0t]eiH0t +O(λ4) = LRt [ρ] +O(λ4) (62)

Since the generator of the cumulant master equation, up to the second-order, is equal to the Bloch-Redfield genera-

tor, we see (based on to the Corollary 2.1) that the off-diagonal part of the mean-force Hamiltonian H(2)mf is the solution

for a steady-state of the cumulant, too. In this way, we proved a significant advantage of the cumulant equation overthe Bloch-Redfield since both predict the same proper coherences of the stationary state, yet the dynamics governedby the cumulant is contrary to the Bloch-Redfield equation, completely positive.

V. DIAGONAL ELEMENTS FROM FOURTH-ORDER TERMS

In this section, we try to derive the diagonal part of the correction H(2)st , which according the representation (4) is

given by the coefficients Υ(st)αβ (ω, ω). In principle, this could be done by solving the fourth-order equation (31), i.e.,

L(0)∞ [%4] + L(2)

∞ [%2] + L(4)∞ [%0] = 0. (63)

However, on the contrary to the solution for off-diagonal terms, it is much more complex problem. In the following,we present a general method how to solve this equation and later we present the solution for a two-level system forthe Bloch-Redfield and cumulant master equation.

For simplicity, throughout of this section, we assume that the interaction Hamiltonian is given by

HI = S ⊗R (64)

10

i.e., we replace the sum (2) by a single term (which can be straightforwardly generalized by adding the corresponding

indices). According to this, we also simplify notation such that Υ(k)αβ ≡ Υk. Then, similarly to the second-order (see

Eq. (41)), we start with writing the action of the generators in the basis of jump operators:

L(k)∞ [ρl] =

∑ω1,ω2,ω3,ω4

gkl(ω1, ω2, ω3, ω4)e−βH0A(ω1)A(ω2)A(ω3)A(ω4), (65)

where l + k = 4, such that the fourth-order equation reads then as follows:∑ω1,ω2,ω3,ω4

(g04(ω1, ω2, ω3, ω4) + g22(ω1, ω2, ω3, ω4) + g40(ω1, ω2, ω3, ω4))e−βH0A(ω1)A(ω2)A(ω3)A(ω4) = 0. (66)

Then, we propose the following general proposition.

Proposition 1. Eq. (66) is satisfied if and only if the set of equation:∑(ω1,ω2,ω2,ω4)∈G(|k〉→|k〉)

(g22(ω1, ω2, ω3, ω4) + g40(ω1, ω2, ω3, ω4)

)= 0. (67)

for each k, is satisfied, where G(|k〉 → |k〉) denotes the set of all four-tuples (ω1, ω2, ω2, ω4) of the form:

(ω1, ω2, ω2, ω4) = (εl − εk, εm − εl, εj − εm, εk − εj), (68)

and |k〉 is an eigenstate of the bare Hamilton with energy εk.

Proof. We consider diagonal elements of the Eq. (66), such that we obtain the following set of equations∑ω1,ω2,ω3,ω4

(g04(ω1, ω2, ω3, ω4) + g22(ω1, ω2, ω3, ω4) + g40(ω1, ω2, ω3, ω4))e−βεk 〈k|A(ω1)A(ω2)A(ω3)A(ω4) |k〉 = 0.

(69)

for k = 0, 1, 2, . . . . We see that since for arbitrary ρ we have:∑ω1,ω2,ω3,ω4

g04(ω1, ω2, ω3, ω4) 〈k| e−βH0A(ω1)A(ω2)A(ω3)A(ω4) |k〉 = 〈k| L(0)∞ [ρ] |k〉 = 〈k| [H0, ρ] |k〉 = 0, (70)

so our condition is now just∑ω1,ω2,ω3,ω4

(g22(ω1, ω2, ω3, ω4) + g40(ω1, ω2, ω3, ω4)

)e−βεk 〈k|A(ω1)A(ω2)A(ω3)A(ω4) |k〉 = 0. (71)

Moreover, one observes that 〈k|A(ω1)A(ω2)A(ω3)A(ω4) |k〉 is nonzero only if∑k ωk = 0. Consequently, let us denote

by G(|k〉 → |k〉) the set of all four-tuples (ω1, ω2, ω2, ω4) of the form:

(ω1, ω2, ω2, ω4) = (εl − εk, εm − εl, εj − εm, εk − εj), (72)

form which follows Eq. (67).

The coefficients Υst are encoded within the function g22. In particular, for the master equation of the form (6) and%2 given by Eq. (28), we have:

g22(ω1, ω2, ω3, ω4) = Υst(−ω3, ω4)α(ω3 + ω4)

(iΥdyn(−ω1, ω2) +

1

2K(−ω1, ω2)

)−Υst(−ω1, ω2)α(ω1 + ω2)

(iΥdyn(−ω3, ω4)− 1

2K(−ω3, ω4)

)− e−βω1Υst(−ω2, ω3)α(ω2 + ω3)K(−ω4, ω1) (73)

where α(ω) is given by Eq. (45). Finally, (67) provides the general set of equations for the diagonal elements ofthe steady-state correction in terms of the Υst(ω, ω) coefficients. In the next section, we provide the solution for thetwo-level system for the Bloch-Redfield and cumulant master equation.

11

VI. TWO-LEVEL SYSTEM

Let us consider a two-level system with the bare Hamiltonian H0 = −ω0

2 σz coupled to the thermal bath via a singleinteraction term HI = S ⊗ R, where S = ~r · ~σ, ~r = (x, y, z), and ~σ = (σx, σy, σz) are the Pauli matrices. Then, the

correction H(2)k can be written in the following form:

H(2)k =

(x2 + y2)(Υk(ω0, ω0)−Υk(−ω0,−ω0)) (x− iy)z(Υk(0,−ω0)−Υk(ω0, 0))

(x+ iy)z(Υk(0,−ω0)−Υk(ω0, 0)) −(x2 + y2)(Υk(ω0, ω0)−Υk(−ω0,−ω0))

(74)

For the general master equation (6), the off-diagonal terms can be calculated from Eq. (40). In particular, for theBloch-Redfield or cumulant master equation, if ω 6= ω′ we have generally Υst(ω, ω

′) = Υmf(ω, ω′) (see Corollary 2.1).

In the following, we present the formulas for diagonal elements.

Proposition 2. The diagonal elements of the two-level system steady-state correction are given by:

(i) For a master equation of the form (6), obeying the detailed balance condition (i.e., K(ω, ω) = eβωK(−ω,−ω)):

Υst(ω, ω) = 0 (75)

(ii) For the cumulant master equation (60):

Υst(ω, ω) = Υmf(ω, ω)−Υdyn(ω, ω), (76)

where Υdyn(ω, ω) = S(ω).

Proof. The proof is solely based on Eq. (67), where for the two-level system ω ∈ −ω0, 0, ω0. Moreover, the firstpart of the Proposition 2 (i) is for the master equation of the form (6), which does not involve the fourth and higherorder generators, such that identically g40 = 0. Thus, we have∑

(ω1,ω2,ω2,ω4)∈G(|0〉→|0〉)

g22(ω1, ω2, ω3, ω4) =∑

(ω1,ω2,ω2,ω4)∈G(|1〉→|1〉)

g22(ω1, ω2, ω3, ω4) = 0, (77)

from which we obtain:

β(Υst(−ω,−ω)K(−ω,−ω)− e−βωΥst(ω, ω)K(ω, ω)

)= 0, (78)

where ω = ω0 or −ω0. After applying the detailed balance condition, finally we get

Υst(ω0, ω0) = −Υst(−ω0,−ω0). (79)

Since the solution for the stationary state is invariant under arbitrary shift of the energies (see the comment at theend of the Section II C), we simply put Υst(0, 0) = Υst(ω0, ω0) = Υst(−ω0,−ω0) = 0.

The second part of the Proposition 2 (ii) is much more involving since it contains the fourth-order generator fromthe Eq. (59) (written in the interaction picture). Based on this, the fourth-order generator in the Schrodinger pictureis given by:

L(4)t [ρ] =

1

2e−iH0t

([K

(2)t ,

d

dtK

(2)t

][eiH0tρe−iH0t]

)eiH0t =

1

2

∫ t

0

ds e−iH0t([LRs , LRt

][eiH0tρe−iH0t]

)eiH0t (80)

However, to calculate the coefficient g40 we act on the state %0 (that commutes with H0), and we only consider thediagonal term, such that

〈k| L(4)t [%0] |k〉 =

1

2

∫ t

0

ds 〈k|[LRs , LRt

][%0] |k〉 . (81)

From this one can calculate the expression for a time-dependent coefficient g40. Due to its complicated form, we donot present it explicitly. However, for a two-dimensional system we have a significant simplification since similarly tothe coefficient g22, the sum in the long time limit is just given by:∑

(ω1,ω2,ω2,ω4)∈G(|0〉→|0〉)

g40(ω1, ω2, ω2, ω4) =∑

(ω1,ω2,ω2,ω4)∈G(|1〉→|1〉)

g40(ω1, ω2, ω2, ω4)

=1

2e−βω(1 + eβω)γ(ω)

∫ ∞0

ds (e−βωγ(ω, s)− γ(−ω, s)) (82)

12

(a) β Im[Υk(0,−ω0) − Υk(ω0, 0)]/γ (b) βRe[Υk(0,−ω0) − Υk(ω0, 0)]/γ

(c) β[Υk(ω0, ω0) − Υk(−ω0,−ω0)]/γ

FIG. 1: Comparison of the diagonal and off-diagonal elements of the two-level system Hamiltonian corrections (74)for a spin-boson model with respect to βω0, where β is the inverse temperature of the bath and ω0 is the bare

frequency of the qubit. The orange line represents the mean-force coefficient, the blue one represents the dynamical,and the dashed lines represent the steady-state coefficients of the dynamical equations: green is for the

Bloch-Redfield master equation and black for the cumulant equation. All plots were computed for cut-off frequencyβωc = 50.

where, similarly to the previous case, ω = ω0 or −ω0. Finally, we need to solve

β(Υst(−ω,−ω)K(−ω,−ω)− e−βωΥst(ω, ω)K(ω, ω)

)+

1

2e−βω(1 + eβω)γ(ω)

∫ ∞0

ds (e−βωγ(ω, s)− γ(−ω, s)) = 0

(83)

which leads to the solution (76).

Now, we can combine the results on off-diagonal (Theorem 2) and diagonal (Proposition 2) elements for the two-levelsystem, in the following theorem.

Theorem 4. For the two-level system:

H(2)st = H

(2)mf −D[H

(2)dyn] for the cumulant equation, (84)

H(2)st = H

(2)mf −D[H

(2)mf ] for the Bloch-Redfield equation, (85)

where D[H(2)k ] denotes the diagonal part of the representation (74).

A. Spin-boson model

Let us now apply our results to the well-known spin-boson model. We consider a qubit coupled to a bosonic bathwith the Hamiltonian:

H = −ω0

2σz +

∑k

Ωka†kak + (xσx + yσy + zσz)

∞∑k=1

λk(ak + a†k) (86)

13

where ak and a†k are the bosonic annihilation and creation operators. For this model we have the autocorrelationfunction:

〈R(t)R〉γR =

∫ ∞0

dΩ J(Ω)

(coth[

βΩ

2] cos(Ωt)− i sin(Ωt)

)(87)

where we consider the Ohmic environment, i.e.,

J(Ω) = γΩe−|Ω|ωc , (88)

where γ is a parameter which modifies the strength of the interaction and ωc is a cutoff frequency. According to thegeneral representation (74), we then compute the coefficients Υk(ω0, ω0) − Υk(−ω0,−ω0) representing the diagonalpart, and Υk(0,−ω0)−Υk(ω0, 0) which describes the off-diagonal elements, for dynamical, mean-force and steady-statecorrection. The results are presented in the Fig. 1.

The most important conclusion coming from the numerical simulation is the very good agreement of the diagonalcoefficient for the steady-state of the cumulant equation with the mean-force correction, which even though is shifted,still predicts the same behaviour with respect to growing frequency ω0. On the contrary, the Bloch-Redfield equation,which does not involve the fourth-order generator, predicts no change of the diagonal elements at all. As it waspointed out before, the off-diagonal elements predicted by the cumulant and Bloch-Redfield equation coincide withthe mean-force. For completeness, we also plot the dynamical correction, which as it is seen differs significantly fromother corrections. Especially, on the contrary to others, it predicts the imaginary part of the off-diagonal coefficient.

We see that non-vanishing value of the Υmf(0,−ω0)−Υmf(ω0, 0) predicts the steady-state coherences in the equi-librium state. This was already known in the literature, and in the Appendix E we show that our general formulasvery easily predicts the same value reported in [12], if applied for a particular case of a two-level system. Thus, ourgeneral formulas are verified to reproduce the particular cases known in the literature, as well as provides a simplerframework for their calculation.

VII. CONCLUSIONS

We have shown the significant advantage of the cumulant equation over the Bloch-Redfield or Davies masterequation. Let us point out all of its advantages:

(i) Its derivation from the exact dynamics does not involve the Markovian approximation;

(ii) It is completely positive;

(iii) It predicts the proper steady-state coherences at equilibrium in the long-time limit;

(iv) For the two-level system, the diagonal part of its stationary state (in the long-time limit) is very close toequilibrium.

In comparison, the Bloch-Redfield master equation only satisfies the condition (iii) and the Davies equation only (ii).We thus conclude that cumulant equation is the best candidate to proper description of the thermalization processof the open systems, which simultaneously is completely positive. We believe that involving the fourth-order of the

cumulant super-operator (i.e., the additional K(4)t term in the exponent (48)) should result in exact solution for

diagonal part, too. However, this may break down the positivity of the map, such that the trade-off between longtimes and the short or intermediate ones can appear.

Moreover, our results were proved on a high level of generality, and thus, contribute to other important areas of openquantum systems. In particular, we derive the second-order formula of the mean-force Hamiltonian for an arbitraryform of the coupling, as well as the solution for steady-state coherences for the whole class of master equations.We also discuss the relationship between different corrections to the bare Hamiltonian. Especially, we compute andcompare all of the corrections quantitatively for a particular spin-boson model.

The most important direction for future studies is to analyze the stationary states out of equilibrium (e.g., systemscoupled to two baths in different temperatures). As it was pointed out in [13], for systems coupled to a single bath, onecan always artificially modify the completely positive GKSL master equation, such that it thermalizes to the properequilibrium. This, however, results in a wrong description of the non-equilibrium situation. Nevertheless, since thecumulant equation is derived according to the Born approximation from the exact dynamics, we believe that it shouldalso predict the proper stationary state out of equilibrium (in the weak-coupling limit).

Few final remarks are here in order. As usual, the obtained corrections are cut-off dependent and often divergewith the growing cut-off frequency. This is actually ubiquitous in the literature of the topic - the problem of cut-off

14

dependence is at the moment swept under the carpet (see however [17] and [22] for discussion). Secondly, we do nottouch on the issue of renormalization: the derivation of the master equation should be based not on bare Hamiltonianbut the renormalized one (as advocated in [14] and [22]). We have not followed this in the present manuscript to keepclear the main message.

ACKNOWLEDGEMENT

We thank Antonio Mandarino for drawing our attention to Ref. [12]. We also acknowledge discussions withparticipants of Open Systems Seminars at ICTQT. This work is supported by Foundation for Polish Science (FNP),IRAP project ICTQT, contract no. 2018/MAB/5, co-financed by EU Smart Growth Operational Programme. MH isalso partially supported by Polish National Science Centre grant OPUS-21 (grant No: 2021/41/B/ST2/03207). MWacknowledges grant PRELUDIUM-20 (grant number: 2021/41/N/ST2/01349) from the National Science Center.

[1] R. Alicki and K. Lendi, Quantum Dynamical Semigroups and Applications, Lecture notes in physics (Springer-Verlag,1987).

[2] H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems (2006).[3] A. G. Redfield, IBM Journal of Research and Development 1, 19 (1957).[4] F. Bloch, Phys. Rev. 105, 1206 (1957).[5] V. Gorini, A. Kossakowski, and E. C. G. Sudarshan, J. Math. Phys. 17, 821 (1976).[6] G. Lindblad, Comm. Math. Phys. 48, 119 (1976).[7] E. B. Davies, Comm. Math. Phys. 39, 91 (1974).[8] A. S. Trushechkin, M. Merkli, J. D. Cresser, and J. Anders, (2021), arXiv:2110.01671 [quant-ph].[9] C. H. Fleming and N. I. Cummings, Phys. Rev. E 83, 031117 (2011).

[10] J. Thingna, J.-S. Wang, and P. Hanggi, The Journal of Chemical Physics 136, 194110 (2012).[11] J. D. Cresser and J. Anders, Phys. Rev. Lett. 127, 250601 (2021).[12] A. Purkayastha, G. Guarnieri, M. T. Mitchison, R. Filip, and J. Goold, Npj Quantum Inf. 6 (2020).[13] D. Tupkary, A. Dhar, M. Kulkarni, and A. Purkayastha, (2022), arXiv:2105.12091 [quant-ph].[14] R. Alicki, Phys. Rev. A 40, 4077 (1989).[15] A. Rivas, Physical Review A 95 (2017).[16] A. Rivas, Entropy 21, 725 (2019).[17] M. Winczewski, A. Mandarino, M. Horodecki, and R. Alicki, (2021), arXiv:2106.05776 [quant-ph].[18] G. Timofeev and A. Trushechkin, (2022), arXiv.[19] T. Mori and S. Miyashita, Journal of the Physical Society of Japan 77, 124005 (2008).[20] Y. Subas ı, C. H. Fleming, J. M. Taylor, and B. L. Hu, Phys. Rev. E 86, 061132 (2012).[21] B. Roux and T. Simonson, Biophysical Chemistry 78, 1 (1999).[22] M. Winczewski and R. Alicki, (2021), arXiv:2112.11962 [quant-ph].[23] M. Hunacek, The Mathematical Gazette 92, 380–382 (2008).

15

Appendix A: Preliminaries

We consider the system and bath Hamiltonian:

H = H0 +HR + λHI (A1)

such that H0 and HR are the free Hamiltonians of the system and the bath, respectively, and HI is the interactionHamiltonian. Except the section B 1 and B 2, where we do not assume any particular form of the interaction term,throughout the paper we consider the following explicit form:

HI =∑α

Aα ⊗Rα. (A2)

We introduce the time-dependent operators:

Aα(t) = eiH0tAαe−iH0t, Rα(t) = eiHRtRαe

−iHRt. (A3)

and jump operators (acting on the system Hilbert space):

Aα(ω) =∑

ε′−ε=ωΠ(ε)AαΠ(ε′) (A4)

where Π(ε) is a projector on subspace with energy ε, such that H0 =∑ε ε Π(ε). These obey the following commutation

relation:

[Aα(ω), H0] = ωAα(ω). (A5)

as well as the relations:

A†α(ω) = Aα(−ω),∑ω

Aα(ω) = Aα. (A6)

From this follows also

Aα(ω)ecH0 = ecωecH0Aα(ω) (A7)

where c is the complex number, such that, in particular, the time-dependent operator is given by:

Aα(t) =∑ω

e−iωtAα(ω). (A8)

Finally, we consider the Bloch-Redfield master equation in the Schrodinger picture:

d

dtρ(t) = Lt[ρ(t)] = i[ρ(t), H0 +

∑ω,ω′

∑α,β

Sαβ(ω, ω′, t)A†α(ω)Aβ(ω′)] (A9)

+∑ω,ω′

∑α,β

γαβ(ω, ω′, t)

(Aβ(ω′)ρ(t)A†α(ω)− 1

2A†α(ω)Aβ(ω′), ρ(t)

)(A10)

where ρ is the system density matrix and Lt is the generator, with

γαβ(ω, ω′, t) = Γαβ(ω′, t) + Γ∗βα(ω, t), (A11)

Sαβ(ω, ω′, t) =1

2i

[Γαβ(ω′, t)− Γ∗βα(ω, t)

], (A12)

Γαβ(ω, t) =

∫ t

0

ds eiωs〈Rα(s)Rβ(0)〉γR . (A13)

In the interaction picture (with respect to the bare Hamiltonian H0), the Bloch-Redfield equation takes the followingform:

d

dtρ(t) = Lt[ρ(t)] = i[ρ(t),

∑ω,ω′

∑α,β

Sαβ(ω, ω′, t)A†α(ω)Aβ(ω′)] (A14)

+∑ω,ω′

∑α,β

γαβ(ω, ω′, t)

(Aβ(ω′)ρ(t)A†α(ω)− 1

2A†α(ω)Aβ(ω′), ρ(t)

)(A15)

16

where ρ(t) = eiH0tρ(t)e−iH0t, and

γαβ(ω, ω′, t) = ei(ω−ω′)tγαβ(ω, ω′, t), (A16)

Sαβ(ω, ω′, t) = ei(ω−ω′)tSαβ(ω, ω′, t). (A17)

Additionally, we also use the abbreviation

Sαβ(ω) = limt→∞

Sαβ(ω, ω, t), γαβ(ω) = limt→∞

γαβ(ω, ω, t) (A18)

From the definition it is seen that γαβ(ω) is the Fourier transform of the autocorrelation function, i.e.,

γαβ(ω) = Γαβ(ω) + Γ∗βα(ω) (A19)

=

∫ ∞0

ds eiωs〈Rα(s)Rβ(0)〉γR +

∫ ∞0

ds e−iωs〈Rβ(s)Rα(0)〉∗γR (A20)

=

∫ ∞0

ds eiωs〈Rα(s)Rβ(0)〉γR +

∫ ∞0

ds e−iωs〈Rα(−s)Rβ(0)〉γR (A21)

=

∫ +∞

−∞ds eiωs〈Rα(s)Rβ(0)〉γR (A22)

from which it follows that γαβ(ω) obeys the detailed balance condition, i.e.,

γαβ(ω) = γβα(−ω)eβω. (A23)

Using the inverse Fourier transform for the autocorrelation function and the Sokhostki-Plemelj identity in the form:∫ ∞0

ds eiωs =1

πδ(ω) + iP 1

ω, (A24)

we shall represent Γαβ(ω) as the principal value integral:

Γαβ(ω) =

∫ ∞0

ds eiωs〈Rα(s)Rβ(0)〉γR =1

2π

∫ +∞

−∞dΩ γαβ(Ω)

∫ ∞0

ds ei(ω−Ω)s (A25)

=1

2γαβ(ω) + P 1

2π

∫ +∞

−∞dΩ

iγαβ(Ω)

ω − Ω(A26)

According to this, and since the γ∗αβ(ω) = γβα(ω), we have

Sαβ(ω) =1

2i

[Γαβ(ω)− Γ∗βα(ω)

]= P 1

2π

∫ +∞

−∞dΩ

γαβ(Ω)

ω − Ω(A27)

Appendix B: Mean-force Hamiltonian

We search for the solution for the mean-force Hamiltonian Hmf from the equation:

e−βHmf =1

ZRTrR[e−βH ], (B1)

where Hmf = H0 + δHmf and ZR = Tr[e−βHR

]. Notice, that we used here a standard convention, i.e.,

Tr[e−βH

]= Tr

[e−βHmf

]Tr[e−βHR

](B2)

that fixes the ground state energy of the mean-force Hamiltonian.

17

1. Dyson series

In the following section, we use an abbreviation A(t) = et(H0+HR)Ae−t(H0+HR). We start with the LHS of the Eq.(B1), which we represent by the formal Dyson form:

e−βHmf = e−βH0eβH0e−βHmf = e−βH0T e−∫ β0dt δHmf(t), (B3)

which gives us the series expansion:

T e−∫ β0dt δHmf(t) = 1−

∫ β

0

dt1 δHmf(t1) +

∫ β

0

dt1

∫ t1

0

dt2 δHmf(t1) δHmf(t2) + . . . (B4)

Similarly, for the RHS, we have

e−βH = e−βH0T e−λ∫ β0dt HI(t), (B5)

such that

T e−λ∫ β0dtHI(t) = 1− λ

∫ β

0

dt1 HI(t1) + λ2

∫ β

0

dt1

∫ t1

0

dt2 HI(t1) HI(t2) + . . . (B6)

Finally, one can write

1

ZRTrR[e−βH ] =

1

ZRTrR[e−βH0T e−λ

∫ β0dt HI(t)] = e−βH0 TrR[T e−λ

∫ β0dt HI(t)γR] (B7)

where γR = e−βHRZR is the Gibbs state of the bath. In analogy, we have:

e−βHmf = e−βH0 TrR[T e−∫ β0dt δHmf(t)γB ] (B8)

such that Eq. (B1) can be written as:

TrR

[T(e−∫ β0dt δHmf(t) − e−λ

∫ β0dt HI(t)

)γR

]= 0 (B9)

or in the series form as:

∞∑k=1

(−1)k∫ β

0

dt1

∫ t1

0

dt2· · ·∫ tk−1

0

dtk

(δHmf(t1) δHmf(t2) . . . δHmf(tk)− λk

⟨HI(t1) HI(t2) . . . HI(tk)

⟩γR

)= 0

(B10)where 〈·〉γR = TrR[ · γR].

2. Weak coupling

Now, let us assume that λ 1, and we expand:

Hmf = H0 + λH(1)mf + λ2H

(2)mf + . . . (B11)

such that δHmf = λH(1)mf + λ2H

(2)mf + . . .

Then, we collect terms in the same order of λ appearing in Eq. (B10), i.e.,

λ :

∫ β

0

dt1H(1)mf (t1) =

∫ β

0

dt1

⟨HI(t1)

⟩γR

(B12)

λ2 : −∫ β

0

dt1H(2)mf (t1) +

∫ β

0

dt1

∫ t1

0

dt2 H(1)mf (t1) H

(1)mf (t2) =

∫ β

0

dt1

∫ t1

0

dt2

⟨HI(t1) HI(t2)

⟩γR

(B13)

λ3 : −∫ β

0

dt1H(3)mf (t1) +

∫ β

0

dt1

∫ t1

0

dt2

(H

(1)mf (t1)H

(2)mf (t2) + H

(2)mf (t1)H

(1)mf (t1)

)−∫ β

0

dt1

∫ t1

0

dt2

∫ t2

0

dt3 H(1)mf (t1) H

(1)mf (t2) H

(1)mf (t3) =

∫ β

0

dt1

∫ t1

0

dt2

∫ t2

0

dt3

⟨HI(t1) HI(t2) HI(t3)

⟩γR

(B14)

. . .

18

In general for the n-th order we have

n∑m=1

(−1)m∑

k1,...,km∈Cmn

∫ β

0

dt1· · ·∫ tm−1

0

dtm H(k1)S (t1) H

(k2)S (t2) . . . H

(km)S (tm)

= (−1)n∫ β

0

dt1· · ·∫ tn−1

0

dtn

⟨HI(t1) . . . HI(tn)

⟩γR

where Ckn is set of the k-th order composition of the number n, e.g., C34 = (2, 1, 1), (1, 2, 1), (1, 1, 2).

3. Derivation of general formulas for corrections to the mean-force Hamiltonian

a. First-order correction

Let us first solve the equation for the first-order correction, i.e.,∫ β

0

dt1H(1)mf (t1) =

∫ β

0

dt1

⟨HI(t1)

⟩γR

=∑α

∫ β

0

dt1Aα(t1)⟨Rα(t1)

⟩γR. (B15)

Furthermore, since 〈Rα(t1)〉γR = 〈Rα〉γR (due to commutation of the Gibbs state γR with the free Hamiltonian HR),we get the solution:

H(1)mf =

∑α

〈Rα〉γR Aα. (B16)

From now on, we assume that bath operators are centralized such that 〈Rα〉γR = 0, which implies H(1)mf = 0.

b. Second-order correction

In this section we provide general formula for second order correction for mean-force Hamiltonian. Remarkablythe expression does not exhibit any poles, as is usual for dynamical correction. Yet we also decompose it into bricksthat are used also to build the dynamical corrections, which do exhibit poles, and require principal value to be welldefined.

Theorem 5. The explicit form of second order correction for mean-force Hamiltonian is the following:

H(2)mf =

∑ω,ω′

Υ(mf)αβ (ω, ω′)A†α(ω)Aβ(ω′), (B17)

where

Υ(mf)αβ (ω, ω′) =

1

2π

∫ +∞

−∞dΩ D(ω, ω′,Ω) γαβ(Ω), D(ω, ω′,Ω) =

1

ω′ − Ω− (ω − ω′)(eβ(ω−Ω) − 1)

(ω − Ω)(ω′ − Ω)(eβ(ω−ω′) − 1)(B18)

The coefficients Υαβ(ω, ω′) can be also expressed in terms of the imaginary part of Γαβ(ω) (see Eq. (A27)) as follows

Υ(mf)αβ (ω, ω′) =

1

eβω − eβω′(eβωSαβ(ω′)− eβω

′Sαβ(ω) + eβ(ω+ω′) (Sβα(−ω′)− Sβα(−ω))

). (B19)

Remark. From (B19) one sees that Υ(mf)αβ is symmetric, i.e., Υ

(mf)αβ (ω, ω′) = Υ

(mf)αβ (ω′, ω). This can be also seen by

writing D(ω, ω′,Ω) in explicitly symmetric form

D(ω, ω′,Ω) =1

2

(eβω − eβω′)(ω + ω′ − 2Ω) + (ω − ω′)(eβω + eβω′ − 2eβ(ω+ω′−Ω))

(eβω − eβω′)(ω − Ω)(ω′ − Ω)(B20)

19

Proof. We shall first prove that Eq. (B19) comes from (B18). We shall use Eq. (A27), i.e.,

Sαβ(ω) =1

2π

∫ +∞

−∞dΩ

γαβ(Ω)

ω − Ω, (B21)

from which we also derive

−Sβα(−ω) =1

2π

∫ +∞

−∞dΩ

γβα(Ω)

ω + Ω=

1

2π

∫ +∞

−∞dΩ

γβα(−Ω)

ω − Ω=

1

2π

∫ +∞

−∞dΩ

γαβ(Ω)e−βΩ

ω − Ω, (B22)

where we used the detailed balance condition (A23). Thus, to express Υ(mf)αβ in terms of Sαβ we have to write D in

terms of 1/(ω − Ω) or 1/(ω′ − Ω). Using

ω − ω′

(ω′ − Ω)(ω − Ω)=

1

ω′ − Ω− 1

ω − Ω(B23)

we thus get

D(ω, ω′,Ω) =1

eβ(ω−ω′) − 1

(eβ(ω−ω′) − 1

ω′ − Ω− eβ(ω−Ω)(

1

ω′ − Ω− 1

ω − Ω) +

1

ω′ − Ω− 1

ω − Ω

)(B24)

=1

eβω − eβω′

(eβω

ω′ − Ω− eβω

′

ω − Ω− eβ(ω+ω′−Ω)(

1

ω′ − Ω− 1

ω − Ω)

)(B25)

Rearranging it a bit, and using (B21) and (B22) we obtain (B19).Let us now prove the expression (B18). We start with second-order equation with centralized bath operators, i.e.,∫ β

0

dtH(2)mf (t) = −

∫ β

0

dt

∫ t

0

ds⟨HI(t)HI(s)

⟩γR

(B26)

Next, we put the representation (B17) and according to the relation (A7), we have

H(2)mf (t) =

∑ω,ω′

Υ(mf)αβ (ω, ω′)etH0A†α(ω)Aβ(ω′)e−tH0 =

∑ω,ω′

Υ(mf)αβ (ω, ω′)et(ω−ω

′)A†α(ω)Aβ(ω′). (B27)

Then, the LHS of Eq. (B26) is equal to:∫ β

0

dt H(2)mf (t) =

∑ω,ω′

A†α(ω)Aβ(ω′)

(Υ

(mf)αβ (ω, ω′)

∫ β

0

dt et(ω−ω′)

)(B28)

whereas the RHS is given by:

−∫ β

0

dt

∫ t

0

ds⟨HI(t)HI(s)

⟩γR

= −∫ β

0

dt

∫ t

0

ds Aα(t)Aβ(s)〈Rα(t) Rβ(s)〉 (B29)

= −∑ω,ω′

∫ β

0

dt

∫ t

0

ds etω−sω′A†α(ω)Aβ(ω′)〈Rα(t− s) Rβ〉

= −∑ω,ω′

A†α(ω)Aβ(ω′)

∫ β

0

dt et(ω−ω′)

∫ t

0

ds esω′〈Rα(s)Rβ〉 (B30)

where in the last line we change a variables s→ t− s. Next, according to Eq. (A19), let us observe that

〈Rα(it)Rβ〉 = 〈Rα(t)Rβ〉 =1

2π

∫ +∞

−∞dΩ e−iΩt γαβ(Ω), (B31)

such that

〈Rα(t)Rβ〉 =1

2π

∫ +∞

−∞dΩ e−Ωt γαβ(Ω). (B32)

20

Finally, the RHS is equal to:

−∫ β

0

dt

∫ t

0

ds⟨HI(t)HI(s)

⟩γR

= −∑ω,ω′

A†α(ω)Aβ(ω′)

(1

2π

∫ +∞

−∞dΩ γαβ(Ω)

∫ β

0

dt et(ω−ω′)

∫ t

0

ds es(ω′−Ω)

)(B33)

Equating LHS=RHS, we get∑ω,ω′

[Υ

(mf)αβ (ω, ω′)

∫ β

0

dt et(ω−ω′) +

1

2π

∫ +∞

−∞dΩ γαβ(Ω)

∫ β

0

dt et(ω−ω′)

∫ t

0

ds es(ω′−Ω)

]A†α(ω)Aβ(ω′) = 0 (B34)

which is solved by

Υ(mf)αβ (ω, ω′) = − 1

2π

∫ +∞

−∞dΩ γαβ(Ω)

∫ β0dt et(ω−ω

′)∫ t

0ds es(ω

′−Ω)∫ β0dt et(ω−ω′)

(B35)

We thus obtain

D(ω, ω′,Ω) = −∫ β

0dt∫ t

0ds et(ω−ω

′)es(ω′−Ω)∫ β

0dt et(ω−ω′)

(B36)

Appendix C: Steady-state correction

1. General method

We look for a solution of the equation:

L[%] = 0, (C1)

where L is the generator of the master equation and % is its stationary state. We expand the generator and steady-statein the series, i.e.,

L[ρ] = L0[ρ] + λ2L2[ρ] + λ4L4[ρ] + . . . (C2)

% = %0 + λ2%2 + λ4%4 + . . . (C3)

such that we have the following set of equations (for each order in λ):

L0[%0] = 0 (C4)

L0[%2] + L2[%0] = 0 (C5)

L0[%4] + L2[%2] + L4[%0] = 0 (C6)

. . . (C7)

We postulate the stationary state (in the Gibbs form):

% = e−β(H0+λ2H(2)st +λ4H

(4)st + . . . ) = %0 + λ2%2 + λ2%4 + . . . (C8)

such that

%0 = e−βH0 (C9)

%2 = −e−βH0

∫ β

0

dt etH0H(2)st e

−tH0 (C10)

%4 = −e−βH0

∫ β

0

dt etH0H(4)st e

−tH0 + e−βH0

∫ β

0

dt1

∫ t1

0

dt2 et1H0H

(2)st e

−t1H0et2H0H(2)st e

−t2H0 (C11)

In the following, we use the summation convention, i.e., the repeating indices are summed up. We start withrepresentation of the second-order correction in the basis of jump operators:

H(2)st = Υ

(st)αβ (ω, ω′)A†α(ω)Aβ(ω′). (C12)

21

In accordance, we have the following expression for %2, i.e.,

%2 = −e−βH0Υ(st)αβ (ω, ω′)

∫ β

0

dt etH0A†α(ω)Aβ(ω′)e−tH0 = −Υ(st)αβ (ω, ω′)α(ω′ − ω)e−βH0A†α(ω)Aβ(ω′), (C13)

where we define:

α(ω) =

∫ β

0

dt e−tω =

1−e−βω

ω , ω 6= 0

β, ω = 0(C14)

In general, we are going to transform the operator equations (C5) and (C6) into the algebraic ones. For this we define:

Lk[%l] = g(kl)αβ (ω1, ω2)e−βH0Aα(ω1)Aβ(ω2) (C15)

for the second-order (such that k + l = 2), and

Lk[%l] = g(kl)αβγδ(ω1, ω2, ω3, ω4)e−βH0Aα(ω1)Aβ(ω2)Aγ(ω3)Aδ(ω4) (C16)

for k + l = 4. In accordance, for the second-order equation (C5), we have(g

(02)αβ (ω1, ω2) + g

(20)αβ (ω1, ω2)

)e−βH0Aα(ω1)Aβ(ω2) = 0 (C17)

whereas for the fourth-order:(g

(04)αβγδ(ω1, ω2, ω3, ω4) + g

(22)αβγδ(ω1, ω2, ω3, ω4) + g

(40)αβγδ(ω1, ω2, ω3, ω4)

)e−βH0Aα(ω1)Aβ(ω2)Aγ(ω3)Aδ(ω4) = 0

(C18)

In the following, we will also use the commutation relations:

[Aα(ω), H0] = ωAα(ω) (C19)

from which we get:

Aα(ω)e−βH0 = e−βωe−βH0Aα(ω). (C20)

The commutation relation (C19) can be further generalize for the product of jump operators, i.e.,

[Aα1(ω1)Aα2

(ω2) . . . Aα2(ω2), H0] = (ω1 + ω2 + · · ·+ ωn)Aα1

(ω1)Aα2(ω2) . . . Aα2

(ω2). (C21)

Notice also that A†α(ω) = Aα(−ω).

2. Second-order equation

In the following, we solve Eq. (C17) for a master equation of the form:

L0[ρ] = i[ρ,H0] (C22)

L2[ρ] =∑α,β

∑ω,ω′

[Υ

(dyn)αβ (ω, ω′)[ρ,A†α(ω)Aβ(ω′)] +Kαβ(ω, ω′)

(Aβ(ω′)ρA†α(ω)− 1

2A†α(ω)Aβ(ω′), ρ

)](C23)

We observe that the zeroth-order equation, i.e., [%0, H0] = 0 is obviously satisfied for a choice %0 = e−βH0 .

Let us then calculate the coefficients g(02)αβ and g

(20)αβ for the second-order equation. We start with:

L0[%2] = i[%2, H0] = −iΥ(st)αβ (ω1, ω2)α(ω1 + ω2)[e−βH0A†α(ω)Aβ(ω2), H0] (C24)

= −i(ω1 + ω2)Υ(st)αβ (−ω1, ω2)α(ω1 + ω2)e−βH0Aα(ω1)Aβ(ω2) (C25)

where we used Eq. (C21), such that

g(02)αβ = −i(ω1 + ω2)Υ

(st)αβ (−ω1, ω2)α(ω1 + ω2) (C26)

22

Next, we shall calculate:

L2[%0] = iΥ(dyn)αβ (ω1, ω2)[%0, A

†α(ω1)Aβ(ω2)] +Kαβ(ω1, ω2)

(Aβ(ω2)%0A

†α(ω1)− 1

2A†α(ω1)Aβ(ω2), %0

)(C27)

= iΥ(dyn)αβ (ω1, ω2)[e−βH0 , A†α(ω1)Aβ(ω2)] +Kαβ(ω1, ω2)

(Aβ(ω2)e−βH0A†α(ω1)− 1

2A†α(ω1)Aβ(ω2), e−βH0

)First, let us rewrite the Hamiltonian part in the form:

iΥ(dyn)αβ (ω1, ω2)[e−βH0 , A†α(ω1)Aβ(ω2)] = iΥ

(dyn)αβ (ω1, ω2)(e−βH0A†α(ω1)Aβ(ω2)−A†α(ω1)Aβ(ω2)e−βH0) (C28)

= iΥ(dyn)αβ (ω1, ω2)(1− e−β(ω2−ω1))e−βH0A†α(ω1)Aβ(ω2) (C29)

and then the dissipative part as follows

Kαβ(ω1, ω2)

(Aβ(ω2)e−βH0A†α(ω1)− 1

2A†α(ω1)Aβ(ω2), e−βH0

)(C30)

= Kαβ(ω1ω2)

(Aβ(ω2)e−βH0A†α(ω1)− 1

2A†α(ω1)Aβ(ω2)e−βH0 − 1

2e−βH0A†α(ω1)Aβ(ω2)

)(C31)

= Kαβ(ω1ω2)

(e−βωe−βH0Aβ(ω2)A†α(ω1)− 1

2e−β(ω−ω2)e−βH0A†α(ω1)Aβ(ω2)− 1

2e−βH0A†α(ω1)Aβ(ω2)

)(C32)

=

(eβω1Kβα(−ω2,−ω1)− 1

2Kαβ(ω1, ω2)(e−β(ω2−ω1) + 1)

)e−βH0A†α(ω1)Aβ(ω2) (C33)

Finally, we get

g(20)αβ (ω1, ω2) = iΥ

(dyn)αβ (−ω1, ω2)(1− e−β(ω1+ω2)) + eβω1Kβα(−ω2,−ω1)− 1

2Kαβ(−ω1, ω2)(e−β(ω1+ω2) + 1)(C34)

Now, we postulate the solution

g(20)αβ (ω1, ω2) + g

(20)αβ (ω1, ω2) = 0 (C35)

for each ω1, ω2 and α, β. First, for ω1 = ω2 ≡ ω, we have

eβωKβα(−ω,−ω)−Kαβ(ω, ω) = 0 (C36)

such that the coefficient Kαβ(ω, ω) has to satisfy the detailed balance condition. Furthermore, for ω1 6= ω2 we get

iΥ(st)αβ (−ω1, ω2)(e−β(ω1+ω2) − 1)− iΥ(dyn)

αβ (−ω1, ω2)(e−β(ω1+ω2) − 1)

+ e−βω1Kβα(−ω2, ω1)− 1

2Kαβ(−ω1, ω2)(eβ(ω1+ω2) + 1) = 0 (C37)

This can be further simplified to:

Υ(st)αβ (ω1, ω2) = Υ

(dyn)αβ (ω1, ω2) +

i

eβω1 − eβω2

(eβ(ω1+ω2)Kβα(−ω2,−ω1)− 1

2Kαβ(ω1, ω2)(eβω1 + eβω2)

). (C38)

a. Solutions for the Bloch-Redfield master equation and for secular approximation

For the Bloch-Redfield master equation, we have:

Υ(dyn)αβ (ω, ω′) =

1

2i(Γαβ(ω′)− Γ∗βα(ω)), (C39)

Kαβ(ω, ω′) = Γαβ(ω′) + Γ∗βα(ω) (C40)

Next, we put:

Γαβ(ω) =1

2γαβ(ω) + iSαβ(ω) (C41)

23

such that

Υ(dyn)αβ (ω, ω′) =

1

2i(1

2γαβ(ω′) + iSαβ(ω′)− 1

2γαβ(ω) + iSαβ(ω)) (C42)

=1

4i(γαβ(ω′)− γαβ(ω)) +

1

2(Sαβ(ω′) + Sαβ(ω)) (C43)

and

Kαβ(ω, ω′) =1

2γαβ(ω′) + iSαβ(ω′) +

1

2γαβ(ω)− iSαβ(ω) (C44)

=1

2(γαβ(ω′) + γαβ(ω)) + i(Sαβ(ω′)− Sαβ(ω)) (C45)

Let us put above expression into Eq. (C38) and collect all of the terms with Sαβ :

1

2(Sαβ(ω′) + Sαβ(ω))− 1

eβω − eβω′(eβ(ω+ω′)(Sβα(−ω)− Sβα(−ω′))− 1

2(Sαβ(ω′)− Sαβ(ω))(eβω + eβω

′)

)(C46)

=1

eβω − eβω′

[eβω − eβω′

2(Sαβ(ω′) + Sαβ(ω))− eβω + eβω

′

2(Sαβ(ω)− Sαβ(ω′))− eβ(ω+ω′)(Sβα(−ω)− Sβα(−ω′))

](C47)

=1

eβω − eβω′[eβωSαβ(ω′)− eβω

′Sαβ(ω) + eβ(ω+ω′)(Sβα(−ω′)− Sβα(−ω))

](C48)

Next, we collect all of the terms with γαβ , i.e.,

1

4i(γαβ(ω′)− γαβ(ω)) +

i

eβω − eβω′(eβ(ω+ω′) 1

2(γαβ(−ω) + γαβ(−ω′))− 1

4(γαβ(ω′) + γαβ(ω))(eβω + eβω

′)

)(C49)

=i

4

(−γαβ(ω′) + γαβ(ω) +

1

eβω − eβω′(

2(eβω′γαβ(ω) + eβωγαβ(ω′))− (γαβ(ω′) + γαβ(ω))(eβω + eβω

′)))

(C50)

=i

4

(−γαβ(ω′) + γαβ(ω) +

1

eβω − eβω′(

2eβω′γαβ(ω) + 2eβωγαβ(ω′)− γαβ(ω′)(eβω + eβω

′)− γαβ(ω)(eβω + eβω

′)))

(C51)

=i

4

(−γαβ(ω′) + γαβ(ω) +

1

eβω − eβω′(eβω

′γαβ(ω) + eβωγαβ(ω′)− eβω

′γαβ(ω′)− eβωγαβ(ω)

))= 0. (C52)

One sees that only terms Sαβ survives. Moreover, these are exactly equal to the expression for a mean-force Hamil-tonian given by Eq. (B19), such that for the Bloch-Redfield or cumulant master equation we have simply:

Υ(st)αβ (ω, ω′) = Υ

(mf)αβ (ω, ω′), (C53)

for ω 6= ω′.Let us observe that if we apply the so-called secular approximation for γαβ coefficients, i.e.

γαβ(ω, ω′)sec. approx−−−−−−−→ γαβ(ω)δω,ω′ (C54)

then

eβ(ω+ω′)γβα(−ω′,−ω)− 1

2γαβ(ω, ω′)(eβω + eβω

′= δω,ω′

(e2βωγβα(−ω)− eβωγαβ(ω)

)= 0 (C55)

due to the detailed balance condition. Finally, for such master equation, for ω 6= ω′, we have

Υ(st)αβ (ω, ω′) = Υ

(dyn)αβ (ω, ω′). (C56)

3. Fourth-order equation

Now, we are going to solve the fourth-order equation (C18). For simplicity, we assume that the interaction term isgiven by HI = S ⊗R, such that we drop the indices, i.e.,

(g04(ω1, ω2, ω3, ω4) + g22(ω1, ω2, ω3, ω4) + g40(ω1, ω2, ω3, ω4)) e−βH0A(ω1)A(ω2)A(ω3)A(ω4) = 0, (C57)

24

where g(kl)αβγδ ≡ gkl. According to the Proposition 1 in the Section V, the above equation is satisfied if and only if the

following set of equations is satisfied:∑(ω1,ω2,ω2,ω4)∈G(|k〉→|k〉)

(g22(ω1, ω2, ω3, ω4) + g40(ω1, ω2, ω3, ω4)

)= 0. (C58)

where G(|k〉 → |k〉) denotes the set of all four-tuples

(ω1, ω2, ω2, ω4) = (εl − εk, εm − εl, εj − εm, εk − εj). (C59)

a. g22 function

We consider the term:

L2[%2] =∑

ω1,ω2,ω3,ω4

g22(ω1, ω2, ω3, ω4)A(ω1)A(ω2)A(ω3)A(ω4) (C60)

where

%2 = α(ω3 + ω4)Υst(−ω3, ω4)%0A(ω3)A(ω4) (C61)

L2[ρ] = −iΥdyn(−ω1, ω2)[A(ω1)A(ω2), ρ] +K(−ω1, ω2)

(A(ω2)ρA(ω1)− 1

2A(ω1)A(ω2), ρ

)(C62)

Then, we have:

L2[%2] = iΥdyn(−ω1, ω2)Υst(−ω3, ω4)α(ω3 + ω4)A(ω1)A(ω2)%2A(ω3)A(ω4) (C63)

− iΥdyn(−ω1, ω2)Υst(−ω3, ω4)α(ω3 + ω4)%2A(ω3)A(ω4)A(ω1)A(ω2) (C64)

+1

2(Υst(−ω3, ω4)α(ω3 + ω4)K(−ω1, ω2)A(ω1)A(ω2)%2A(ω3)A(ω4)) (C65)

− Υst(−ω3, ω4)α(ω3 + ω4)K(−ω1, ω2)A(ω2)%2A(ω3)A(ω4)A(ω1) (C66)

+1

2(Υst(−ω3, ω4)α(ω3 + ω4)K(−ω1, ω2)%2A(ω3)A(ω4)A(ω1)A(ω2)) (C67)

which we may rewrite as:

L2[%2] =

(−iΥdyn(−ω1, ω2)Υst(−ω3, ω4)α(ω3 + ω4) +

1

2(Υst(−ω3, ω4)α(ω3 + ω4)K(−ω1, ω2))

)e−βH0A(ω3)A(ω4)A(ω1)A(ω2)

+

(iΥdyn(−ω1, ω2)e−β(ω2+ω1) +

1

2

(K(−ω1, ω2)e−β(ω2+ω1)

))Υst(−ω3, ω4)α(ω3 + ω4)e−βH0A(ω1)A(ω2)A(ω3)A(ω4)

− Υst(−ω3, ω4)α(ω3 + ω4)K(−ω1, ω2)e−βω2e−βH0A(ω2)A(ω3)A(ω4)A(ω1). (C68)

Since all ωi’s are mute indices, we change them such that one obtains:

g22(ω1, ω2, ω3, ω4) = Υst(−ω3, ω4)α(ω3 + ω4)

(iΥdyn(−ω1, ω2) +

1

2K(−ω1, ω2)

)−Υst(−ω1, ω2)α(ω1 + ω2)

(iΥdyn(−ω3, ω4)− 1

2K(−ω3, ω4)

)− e−βω1Υst(−ω2, ω3)α(ω2 + ω3)K(−ω4, ω1). (C69)

b. g40 function (cumulant equation)

We consider the fourth-order generator of the cumulant in the Schrodinger picture:

L(4)t [ρ] =

1

2

∫ t

0

ds e−iH0t[LRs , LRt

][eiH0tρe−iH0t]eiH0t. (C70)

Acting on %0 that commutes with H0, this simplifies to:

L(4)t [%0] =

1

2

∫ t

0

ds e−iH0t[LRs , LRt

][%0]eiH0t (C71)

25

We then define

L(4)t [%0] =

∑ω1,ω2,ω3,ω4

g40(ω1, ω2, ω3, ω4, t)e−βH0A(ω1)A(ω2)A(ω3)A(ω4). (C72)

To get an expression for g40, we first compute the action of LRs LRt on %0, i.e.,

LRs LRt [%0] =∑

ω1,ω2,ω3,ω4

f(ω1, ω2, ω3, ω4)e−βH0A(ω1)A(ω2)A(ω3)A(ω4) (C73)

where

f(ω1, ω2, ω3, ω4, t, s) = S(−ω1, ω2, s)S(−ω3, ω4, t)e−β(ω1+ω2) − S(−ω1, ω2, s)S(−ω3, ω4, t)e

−β(ω1+ω2+ω3+ω4) (C74)

+i

2S(−ω1, ω2, s)γ(−ω3, ω4, t)e

−β(ω1+ω2) +i

2S(−ω1, ω2, s)γ(−ω3, ω4, t)e

−β(ω1+ω2+ω3+ω4) (C75)

− iS(−ω1, ω2, s)γ(−ω4, ω3, t)e−β(ω1+ω2+ω3) + S(−ω1, ω2, t)S(−ω3, ω4, s)e

−β(ω1+ω2) (C76)

− S(−ω1, ω2, t)S(−ω3, ω4, s) +i

2S(−ω1, ω2, t)γ(−ω3, ω4, s)e

−β(ω1+ω2) − i

2S(−ω1, ω2, t)γ(−ω3, ω4, s) (C77)

− iS(−ω2, ω3, t)γ(−ω4, ω1, s)e−β(ω1+ω2+ω3) + iS(−ω2, ω3, t)γ(−ω4, ω1, s)e

−βω1 (C78)

− i

2S(−ω3, ω4, s)γ(−ω1, ω2, t)e

−β(ω1+ω2) − i

2S(−ω3, ω4, s)γ(−ω1, ω2, t) + iS(−ω3, ω4, s)γ(−ω2, ω1, t)e

−βω1 (C79)

− i

2S(−ω3, ω4, t)γ(−ω1, ω2, s)e

−β(ω1+ω2) +i

2S(−ω3, ω4, t)γ(−ω1, ω2, s)e

−β(ω1+ω2+ω3+ω4) (C80)

+1

4γ(−ω1, ω2, s)γ(−ω3, ω4, t)e

−β(ω1+ω2) +1

4γ(−ω1, ω2, s)γ(−ω3, ω4, t)e

−β(ω1+ω2+ω3+ω4) (C81)

− 1

2γ(−ω1, ω2, s)γ(−ω4, ω3, t)e

−β(ω1+ω2+ω3) +1

4γ(−ω1, ω2, t)γ(−ω3, ω4, s)e

−β(ω1+ω2) (C82)

+1

4γ(−ω1, ω2, t)γ(−ω3, ω4, s)−

1

2γ(−ω2, ω1, t)γ(−ω3, ω4, s)e

−βω1 (C83)

− 1

2γ(−ω2, ω3, t)γ(−ω4, ω1, s)e

−β(ω1+ω2+ω3) − 1

2γ(−ω2, ω3, t)γ(−ω4, ω1, s)e

−βω1 (C84)

+ γ(−ω3, ω2, t)γ(−ω4, ω1, s)e−β(ω1+ω2) (C85)

Consequently, we have

g40(ω1, ω2, ω3, ω4, t) =1

2ei(ω1+ω2+ω3+ω4)t

∫ t

0

ds (f(ω1, ω2, ω3, ω4, t, s)− f(ω1, ω2, ω3, ω4, s, t)). (C86)

4. Two-level system

Now, we shall specialize to the case of a two-level system. We then have k = 0, 1 and ε1 − ε0 = ω0, such that

G(|0〉 → |0〉) =(0, 0, 0, 0), (ω0,−ω0, 0, 0), (ω0, 0,−ω0, 0), (ω0, 0, 0,−ω0), (C87)

(0, ω0,−ω0, 0), (0, ω0, 0,−ω0), (0, 0, ω0,−ω0), (ω0,−ω0, ω0,−ω0) (C88)

The set S(|1〉 → |1〉) is the same but with changed sign of the qubit frequency ω0 → −ω0. Then, according to Eq.(C69), one can first observe that the coefficient g22 summed over first seven four-tuples vanishes, i.e.,

g22(0, 0, 0, 0) + g22(ω0,−ω0, 0, 0) + g22(ω0, 0,−ω0, 0) + g22(ω0, 0, 0,−ω0)

+ g22(0, ω0,−ω0, 0) + g22(0, ω0, 0,−ω0) + g22(0, 0, ω0,−ω0) = 0 (C89)

whereas for the last one we have

g22(ω0,−ω0, ω0,−ω0) = β(Υst(−ω0,−ω0)K(−ω0,−ω0)− e−βω0Υst(ω0, ω0)K(ω0, ω0)

). (C90)

If additionally K(ω, ω) obeys the detailed balance condition, then

g22(ω0,−ω0, ω0,−ω0) = βe−βω0 (Υst(−ω0,−ω0)−Υst(ω0, ω0))K(ω0, ω0). (C91)

26

a. Second-order master equation

Now, since for arbitrary master equation of the form (6), which is up to second order in λ, we also have g40 = 0.From this we conclude that Eq. (C18) is satisfied if

Υst(ω0, ω0) = Υst(−ω0,−ω0). (C92)

Since for a two-level system, in general we have

〈0|H(2)st |0〉 = Υst(0, 0) + Υst(ω0, ω0), 〈1|H(2)

st |1〉 = Υst(0, 0) + Υst(−ω0,−ω0). (C93)

Thus, applying the condition (C92), we finally get:

〈0|H(2)st |0〉 = 〈1|H(2)

st |1〉 = Υst(ω0, ω0). (C94)

b. Cumulant equation

To solve the Eq. (C18) for the cumulant master equation we need to additionally calculate the term involving thecoefficient g40. Putting the expression (C86), we observe that, similarly to the summation of g22, the sum over firstseven tuples vanishes, such that we obtain a very simple expression

limt→∞

∑(ω1,ω2,ω2,ω4)∈G(|0〉→|0〉)

g40(ω1, ω2, ω2, ω4, t) = g40(ω0,−ω0, ω0,−ω0), (C95)

where

g40(ω0,−ω0, ω0,−ω0) =1

2e−βω0(1 + eβω0)γ(ω0)

∫ ∞0

ds (e−βω0γ(ω0, s)− γ(−ω0, s)). (C96)

Since, the leading order of the cumulant master equation is the Bloch-Redfield generator, we also have

g22(ω0,−ω0, ω0,−ω0) = βe−βω0 (Υst(−ω0,−ω0)−Υst(ω0, ω0)) γ(ω0). (C97)

Finally, we need to solve

g22(ω0,−ω0, ω0,−ω0) + g40(ω0,−ω0, ω0,−ω0) = 0 (C98)

which gives us

Υst(ω0, ω0)−Υst(−ω0,−ω0) =1

2β(1 + eβω0)

∫ ∞0

ds (e−βω0γ(ω0, s)− γ(−ω0, s)) (C99)

=1

2β

∫ ∞0

ds (γ(ω0, s) + e−βω0γ(ω0, s)− γ(−ω0, s)− eβω0γ(−ω0, s)) (C100)

=1

2β

∫ ∞0

ds (γ(ω0, s)− eβω0γ(−ω0, s))−1

2β

∫ ∞0

ds (γ(−ω0, s)− e−βω0γ(ω0, s)) (C101)

Finally, without loss of generality, we can put

Υst(ω, ω) =1

2β

∫ ∞0

ds (γ(ω, s)− eβωγ(−ω, s)). (C102)

At the end, let us compare it with the mean-force coefficient, i.e.,

Υmf(ω, ω) =1

2π

∫ +∞

−∞dΩ D(ω, ω,Ω)γ(Ω) (C103)

where

D(ω, ω,Ω) = − 1

β

∫ β

0

dt

∫ t

0

ds es(ω−Ω) =1− eβ(ω−Ω) + β(ω − Ω)

β(ω − Ω)2. (C104)

27

In order to compare both expression, we propose another representation for the mean-force coefficient. Using thedetailed balance condition, we see that∫ +∞

−∞dΩ f(Ω)γ(Ω) =

∫ +∞

−∞dΩ f(Ω)eβΩγ(−Ω) =

∫ +∞

−∞dΩ f(−Ω)e−βΩγ(Ω) (C105)

Consequently, we can write

Υmf(ω, ω) =1

2π

∫ +∞

−∞dΩ D(ω, ω,Ω)γ(Ω) =

∫ +∞

−∞dΩ

1 + β(ω − Ω)− eβ(ω−Ω)

β(ω − Ω)2γ(Ω) (C106)

=1

2π

∫ +∞

−∞dΩ

(1 + β(ω − Ω)

β(ω − Ω)2− eβ(ω−Ω)

β(ω − Ω)2

)γ(Ω) (C107)

=1

2πβ

∫ +∞

−∞dΩ

(1

(ω − Ω)2− eβω

(ω + Ω)2+

β

ω − Ω

)γ(Ω) (C108)

=1

2πβ

∫ +∞

−∞dΩ

(1

(ω − Ω)2− eβω

(ω + Ω)2

)γ(Ω) +

1

2π

∫ +∞

−∞dΩ

γ(Ω)

ω − Ω(C109)

Finally, let us observe that ∫ ∞0

ds γ(ω, s) =1

π

∫ +∞

−∞dΩ

γ(Ω)

(ω − Ω)2, (C110)

and

S(ω) =1

2π

∫ +∞

−∞dΩ

γ(Ω)

ω − Ω, (C111)

such that according to the previous choice for Υst(ω, ω), we have simply

Υmf(ω, ω) = Υst(ω, ω) + S(ω). (C112)

Appendix D: Cumulant equation

Consider a system interacting with a thermal reservoir whose Hamiltonian is given by:

H = H0 +HR + λHI (D1)

Let us also consider the Born Approximation such that ρ(0) = ρS(0) ⊗ ρR where ρR is a stationary state of theenvironment. In the interaction picture the reduced state at time t is:

ρS(t) = TrR(U(t, t0)ρS(t0)⊗ ρR(t0)U†(t, t0)

)(D2)

One may expand the evolution operator in the interaction picture U(t, 0) = T e−i∫ t0HI(t′)dt′ and rearrange terms (of

the same power of HI) to obtain:

ρS(t) = ρS(0)−λ2 T2

∫ t

0

dt1

∫ t

0

dt2 TrR ([HI(t1), [HI(t2), ρS(0)⊗ ρR]])︸ ︷︷ ︸K

(2)t

+O(λ3) (D3)

The terms O(H3I ) can be neglected for weak coupling or short times. We already considered the initial state of the

bath to be thermal ρB(0) = ρβ = e−βHB/Tre−βHB

and the bath operators to be centralized. Let us know focus

on the second term, let us apply time-ordering explicitly so that:

K(2)t = −λ

2

2

∫ t

0

dt1

∫ t

0

dt2θ(t1 − t2) TrR ([HI(t1), [HI(t2), ρS(0)⊗ ρR]])

− λ2

2

∫ t

0

dt1

∫ t

0

dt2θ(t2 − t1) TrR ([HI(t2), [HI(t1), ρS(0)⊗ ρR]]) (D4)

28

Let us know expand the double commutators:

K(2)t = −λ

2

2

∫ t

0

dt1

∫ t

0

dt2θ(t1 − t2) TrR

[HI(t1)HI(t2)ρS(0)ρR −HI(t1)ρS(0)ρRHI(t2)−HI(t2)ρS(0)ρRHI(t1)

+ ρS(0)ρRHI(t1)HI(t2)]− λ2

2

∫ t

0

dt1

∫ t

0

dt2θ(t2 − t1) TrR

[HI(t2)HI(t1)ρS(0)ρR −HI(t2)ρS(0)ρRHI(t1)

− HI(t1)ρS(0)ρRHI(t2) + ρS(0)ρRHI(t2)HI(t1)]

From here it can be seen that we have three kind of terms, namely H2I ρ,HIρHI , ρH

2I . Let us consider each of those

independently

HIρHI :λ2

2

∫ t

0

dt1

∫ t

0

dt2(θ(t1 − t2) + θ(t2 − t1)

)TrR

[HI(t1)ρS(0)ρRHI(t2) +HI(t2)ρS(0)ρRHI(t1)

](D5)

=λ2

2

∫ t

0

dt1

∫ t

0

dt2 TrR

[HI(t1)ρS(0)ρRHI(t2) +HI(t2)ρS(0)ρRHI(t1)

](D6)

= λ2

∫ t

0

dt1

∫ t

0

dt2 TrR

[HI(t1)ρS(0)ρRHI(t2)

](D7)

where in the last step we used a change of variables on the second term, such that t1 ↔ t2. Next, we consider theother two missing terms

H2I ρ : −λ

2

2

∫ t

0

dt1

∫ t

0

dt2

(θ(t1 − t2) TrR

[HI(t1)HI(t2)ρS(0)ρR

]+ θ(t2 − t1) TrR

[HI(t2)HI(t1)ρS(0)ρR

])(D8)

= −λ2

2

∫ t

0

dt1

∫ t

0

dt2

(θ(t1 − t2) TrR

[[HI(t1), HI(t2)]ρS(0)ρR

]+ TrR

[HI(t2)HI(t1)ρS(0)ρR

])(D9)

ρH2I : −λ

2

2

∫ t

0

dt1

∫ t

0

dt2

(θ(t1 − t2) TrR

[ρS(0)ρRHI(t2)HI(t1)

]+ θ(t2 − t1) TrR

[HI(t1)HI(t2)ρS(0)ρR

])(D10)

= −λ2

2

∫ t

0

dt1

∫ t

0

dt2

(θ(t1 − t2) TrR

[ρS(0)ρR[HI(t2), HI(t1)]

]+ TrR

[ρS(0)ρRHI(t1)HI(t2)

])(D11)

In both cases the step taken from one line to the other was summing a zero so that the terms could

be recast in that form, they were ± 12

∫ t0dt1∫ t

0dt2θ(t1 − t2) TrR

[HI(t2)HI(t1)ρS(0)ρR

]in the first case and

± 12

∫ t0dt1∫ t

0dt2θ(t1 − t2) TrR

[ρS(0)ρRHI(t1)HI(t2)

]in the second one. Regrouping all terms we have

K(2)t = λ2

∫ t

0

dt1

∫ t

0

dt2

(TrR

[HI(t1)ρS(0)ρRHI(t2)

]− 1

2

(TrR

[ρS(0)ρRHI(t1)HI(t2)

]+ TrR

[HI(t2)HI(t1)ρS(0)ρR

]))

− λ2

2

∫ t

0

dt1

∫ t

0

dt2θ(t1 − t2)(

TrR

[[HI(t1), HI(t2)]ρS(0)ρR

]− TrR

[ρS(0)ρR[HI(t1), HI(t2)]

])(D12)

= λ2

∫ t

0

dt1

∫ t

0

dt2

(TrR

[HI(t1)ρS(0)ρRHI(t2)

]− 1

2

(TrR

[ρS(0)ρRHI(t1)HI(t2)

]+ TrR

[HI(t2)HI(t1)ρS(0)ρR

]))− iλ2[Λ(t), ρS(0)] (D13)

where:

Λ(t) =1

2i

∫ t

0

dt1

∫ t

0

dt2θ(t1 − t2) TrR

[[HI(t1), HI(t2)]ρR

](D14)

=1

2i

∫ t

0

dt1

∫ t

0

dt2sgn(t1 − t2) TrR

[HI(t1)HI(t2)ρR

](D15)

29

where we used θ(x) = 1+sgn(x)2 . Now, if we expand the interaction Hamiltonian in the interaction picture

HI =∑w,k e

iwtAk(w)Bk =∑w,k e

−iwtA†k(w)Bk

Λ(t) =1

2i

∑w,w′

∑αβ

∫ t

0

dt1

∫ t

0

dt2sgn(t1 − t2)ei(wt1−w′t2)A†α(w)Aβ(w′)〈Bα(t1)Bβ(t2)〉R

=∑w,w′

∑αβ

Ξ(w,w′, t)A†α(w)Aβ(w′) (D16)

So we obtain:

K(2)t [ρS(0)] = −i

∑w,w′

∑αβ

Ξ(w,w′, t)[A†α(w)Aβ(w′), ρS(0)] + ξαβ(w,w′, t)(Aβ(w′)ρS(0)A†α(w)− 1

2A†α(w)Aβ(w′), ρS(0)

),

where

ξαβ(w,w′, t) =

∫ t

0

dt1

∫ t

0

dt2ei(wt1−w′t2)〈Rα(t1)Rβ(t2)〉. (D17)

We may rewrite this in terms of previously obtained quantities as:

ξαβ(w,w′, t) =

∫ t

0

ds

∫ t

0

dωei(ws−w′ω)〈Rα(s)Rβ(w)〉 (D18)

=

∫ t

0

ds

∫ t

s

dωei(ws−w′ω)〈Rα(s)Rβ(w)〉+

∫ t

0

ds

∫ s

0

dωei(ws−w′ω)〈Rα(s)Rβ(w)〉 (D19)

=

∫ t

0

dω

∫ ω

0

dsei(ws−w′ω)〈Rα(s)Rβ(w)〉+

∫ t

0

ds

∫ s

0

dωei(ws−w′ω)〈Rα(s)Rβ(w)〉 (D20)

=

∫ t

0

ds

∫ s

0

dωei(wω−w′s)〈Rα(w)Rβ(s)〉+

∫ t

0

ds

∫ s

0

dωei(ws−w′ω)〈Rα(s)Rβ(w)〉 (D21)

=

∫ t

0

ds

∫ s

0

dωei(wω−w′s)〈Rα(w − s)Rβ〉+

∫ t

0

ds

∫ s

0

dωei(ws−w′ω)〈Rα(s− w)Rβ〉 (D22)

=

∫ t

0

ds

∫ s

0

dξei((w−w′)s−wξ)〈Rα(−ξ)Rβ〉+

∫ t

0

ds

∫ s

0

dξei((w−w′)s+ξw′)〈Rα(ξ)Rβ〉 (D23)

=

∫ t

0

dsei(w−w′)s(Γ∗βα(w, s) + Γαβ(w′, s)) (D24)

=

∫ t

0

dsei(w−w′)sγαβ(w,w′, s) =

∫ t

0

dsγαβ(w,w′, s) (D25)

Now, we can notice that the derivative of such coefficient corresponds to:

d

dtξαβ(w,w′, t) = ei(w−w

′)tγ(w,w′, t) = γ(w,w′, t) (D26)

Furthermore from [22] we know that:

d

dtΞαβ(w,w′, t) =

ei(w′−w)t

2i(Γαβ(w′, t)− Γ∗βα(w, t)) = ei(w

′−w)tSαβ(w,w′, t) = Sαβ(w,w′, t) (D27)

One may then rewrite K(2)t as:

K(2)t [ρ] = λ2

∫ t

0

ds∑ω,ω′

∑αβ

(iSαβ(ω, ω′, s)[ρ,A†α(ω)Aβ(ω′)] + γαβ(ω, ω′, s)

(Aβ(ω′)ρA†α(ω)− 1

2A†α(ω)Aβ(ω′), ρ

))(D28)

30

Appendix E: Comparison with dynamical Hamiltonian and the steady state - qubit case

In this section we consider the particular case of a qubit coupled to a bosonic bath given by

H =ω0

2σz +

∑k

Ωka†kak + S

∞∑k=1

λk(ak + a†k) (E1)

where we take S to be a general interaction operator in the pauli basis:

S = xσx + yσy + zσz (E2)

This form of Hamiltonian with y = 0 has been studied previously in [12]. Where it was reported that such Hamiltonianhad steady-state coherences, they used a similar expansion to the general results presented in this article. In thissection, we see that the general framework presented agrees with their results. Using equation (B17), and thisinteraction the second-order correction to the Hamiltonian takes the form:

H(2)k =

z2Υ(0, 0) + (x2 + y2)Υ(ω, ω) (x− iy)z(Υ(0,−ω)−Υ(ω, 0))

(x+ iy)z(Υ(0,−ω)−Υ(ω, 0)) z2Υ(0, 0) + (x2 + y2)Υ(−ω,−ω)

(E3)

where k indicates to dynamical (dyn), steady-state (st) or mean-force (mf) correction. We can rewrite this correc-tion as a linear combination of the Pauli Matrices such that:

H(2)k = A1 +Bσx + Cσy +Dσz (E4)

A = z2Υ(0, 0) +x2 + y2

2(Υ(w,w) + Υ(−w,−w)) (E5)

B = xz(Υ(0,−w)−Υ(w, 0)) (E6)

C = yz(Υ(0,−w)−Υ(w, 0)) (E7)

D =x2 + y2

2(Υ(w,w)−Υ(−w,−w)) (E8)

We can see how the different approaches differ qualitatively by looking at the structure of the different Υ(w,w′)given by each approach. It is important to remark that any approach that performs the secular approximation willhave both B and C equal to zero, meaning the correction will be diagonal and as such won’t be able to describethe off-diagonal elements of the steady states accordingly. While nonsecular approaches such as the Bloch-Redfieldequation, will have non-diagonal corrections, leading to a more appropriate description of the off-diagonal elementsof the correction as well as steady state coherences. Let us for a moment recall the structure of the Bloch-Redfieldcoefficients which are given by (C43), simply substituting the appropriate frequencies for the qubit leads to:

Υ(0,−w)−Υ(w, 0) =S(−w)− S(w)

2+ i

γ(0)− (γ(w) + γ(−w))

4(E9)

and

Υ(w,w) + Υ(−w,−w) = S(w) + S(−w) (E10)

Υ(w,w)−Υ(−w,−w) = S(w)− S(−w) (E11)

Let us now compare this coefficient with the one obtained with the mean force approach. We will only be consideringthe off-diagonal of the correction:

K(ω) = Υ(0,−ω)−Υ(ω, 0) =1

2π

∫ +∞

−∞dΩγ(Ω)C(ω,Ω) (E12)

where

C(ω,Ω) =

(ω2(1− e−βΩ) coth[βω2 ] + (1 + e−βΩ)Ωω

Ω(Ω2 − ω2)

)(E13)

Additionally, our coefficients satisfy detailed balance conditions such that:

γ(−Ω) = γ(Ω)e−βΩ, C(ω,−Ω) = C(ω,Ω)eβΩ (E14)

31

Using those we see that γ(−Ω)C(ω,−Ω) = γ(Ω)C(ω,Ω) and

K(ω) =1

π

∫ +∞

0

dΩ γ(Ω)C(ω,Ω) (E15)

Let us now separate γ(Ω) into its symmetric and anti-symmetric parts:

γs(Ω) = 12 (γ(Ω) + γ(−Ω)) = 1

2 (1 + e−βΩ)γ(Ω) (E16)

γa(Ω) = 12 (γ(Ω)− γ(−Ω)) = 1

2 (1− e−βΩ)γ(Ω) (E17)

Then we may write:

K(ω) =2

π

∫ +∞

0

dΩ

(ω2γa(Ω) coth[βω2 ] + γs(Ω)Ωω

Ω(Ω2 − ω2)

)(E18)

As mentioned before this system with y = 0 had been previously considered in [12], let us now compare our resultsto those previously available in the literature. Their effective Hamiltonian is given by:

HS =

λ2f21 Υ(0, 0)− 1

2 (ω − 2λ2f22 Υ(ω, ω)) λ2f1f2K(ω)

λ2f1f2K(ω) λ2f21 Υ(0, 0) + 1

2 (ω − 2λ2f22 Υ(ω, ω))

(E19)

The couplings in this notation are x = f2, z = f1 and y = 0. They also use ω′ = ω − 2λ2f22 Υ(ω, ω) and E0 =

λ2f21 Υ(0, 0) such that:

HS =

E0 − ω′

2 λ2f1f2K(ω)

λ2f1f2K(ω) E0 + ω′

2

(E20)

Then we may find that

〈σx〉 =Tr[σxe−βHS

]Tr[e−βHS ]

= −x tanh[√x2 + z2β]√

x2 + z2(E21)

where x = λ2f1f2K(ω) and z = ω′

2 .

〈σx〉 = −2x

ω′tanh[

βω

2] +O(λ3) (E22)

In [12] the authors also put ω′ = ω, such that

〈σx〉 = −4λ2f1f2

πω

∫ +∞

0

dΩ

(γs(Ω)ω tanh[βω2 ]

Ω2 − ω2+

ω2γa(Ω)

Ω(Ω2 − ω2)

)(E23)

(E24)

Now let us compare γa,s(Ω) with the correlation function for a bosonic bath:

f(t) =

∫ ∞0

dΩ J(Ω)

(coth[

βΩ

2] cos(Ωt)− i sin(Ωt)

)(E25)

f(t) =1

2

∫ +∞

−∞dΩ e−iΩtγ(Ω) =

1

2

∫ +∞

−∞dΩ γ(Ω) (cos(Ωt)− i sin(Ωt)) (E26)

=1

2

∫ +∞

0

dΩ (γs(Ω) + γa(Ω)) (cos(Ωt)− i sin(Ωt)) +1

2

∫ +∞

0

dΩ (γs(Ω)− γa(Ω)) (cos(Ωt) + i sin(Ωt)) (E27)

=1

π

∫ +∞

0

dΩ (γs(Ω) cos(Ωt)− iγa(Ω) sin(Ωt)) =1

π

∫ +∞

0

dΩ γa(Ω)

(γs(Ω)

γa(Ω)cos(Ωt)− i sin(Ωt)

)(E28)

=1

π

∫ +∞

0

dΩ γa(Ω)

(1 + e−βΩ

1− e−βΩcos(Ωt)− i sin(Ωt)

)=

1

π

∫ +∞

0

dΩγa(Ω)

(coth[

βΩ

2] cos(Ωt)− i sin(Ωt)

)(E29)

32

According to this, we have the following relations:

γa(Ω) = πJ(Ω) = πωa(Ω), γs(Ω) = πJ(Ω) coth[βΩ

2] = πωs(Ω) (E30)

and the final result becomes:

〈σx〉 = −4λ2f1f2

ω

∫ +∞

0

dΩ

(ωs(Ω)ω tanh[βω2 ]

Ω2 − ω2− ω2ωa(Ω)

Ω(Ω2 − ω2)

)(E31)

on the other hand they have

〈σx〉 =2λ2f1f2

ω[∆s(ω) tanh[

βω

2] + ∆a(ω)−∆a(0)] (E32)

where

∆s(ω) =∫∞

0dΩ ωs(Ω)

(1

Ω+ω −1

Ω−ω

)= −2

∫∞0dΩωs(Ω)ω

Ω2−ω2 (E33)

∆a(ω) =∫∞

0dΩ ωa(Ω)

(1

Ω+ω + 1Ω−ω

)= 2

∫∞0dΩωa(Ω)Ω

Ω2−ω2 (E34)

∆a(ω)−∆a(0) = 2∫∞

0dΩ ωa(Ω)Ω2−(Ω2−ω2)

Ω(Ω2−ω2) = 2∫∞

0dΩ ωa(Ω)ω2

Ω(Ω2−ω2) (E35)

such that

〈σx〉 = −4λ2f1f2

ω

∫ ∞0

dΩ

(ωs(Ω)ω tanh[βω2 ]

Ω2 − ω2− ω2ωa(Ω)

Ω(Ω2 − ω2)

)(E36)

Appendix F: Bloch-Redfield master equation (derivation)

We shall derive the Bloch-Redfield master equation in terms of γ (A16) and S (A17) coefficients starting from thevon Neumann equation:

LRt [ρ(t)] = −∫ t

0

ds TrR[HI(t), [HI(s), ρ(t)⊗ γR]], (F1)

which is derived according to the Born-Markov approximation. We expand commutators and put an explicit form ofthe interaction Hamiltonian (A2):

LRt [ρ(t)] = −∫ t

0

ds TrR[HI(t), [HI(s), ρ(t)⊗ γR]] =

∫ t

0

dsTrR ([HI(s)ρ(t)⊗ γR, HI(t)]− [ρ(t)⊗ γR HI(s), HI(t)])

=

∫ t

0

dsTrR[HI(s)ρ(t)⊗ γRHI(t)− ρ(t)⊗ γRHI(s)HI(t)−HI(t)HI(s)ρ(t)⊗ γR +HI(t)ρ(t)⊗ γRHI(s)

]=∑α,β

∫ t

0

ds[Aα(s)ρ(t)Aβ(t)〈Rβ(t)Rα(s)〉γR +Aβ(t)ρ(t)Aα(s)〈Rα(s)Rβ(t)〉γR

−ρ(t)Aα(s)Aβ(t)〈Rα(s)Rβ(t)〉γR −Aβ(t)Aα(s)ρ(t)〈Rβ(t)Rα(s)〉γR]

=∑α,β

∫ t

0

ds [〈Rα(t)Rβ(s)〉γR (Aβ(s)ρ(t)Aα(t)−Aα(t)Aβ(s)ρ(t))] + h.c.

After introducing the jump operators (A4), we get

LRt [ρ(t)] =∑ω,ω′

∑α,β

∫ t

0

ds e−i(ω′s+ωt)〈Rα(t)Rβ(s)〉γR (Aβ(ω′)ρ(t)Aα(ω)−Aα(ω)Aβ(ω′)ρ(t)) + h.c. (F2)

=∑ω,ω′

∑α,β

∫ t

0

ds ei(ωt−ω′s)〈Rα(t)Rβ(s)〉γR

(Aβ(ω′)ρ(t)A†α(ω)−A†α(ω)Aβ(ω′)ρ(t)

)+ h.c. (F3)

=∑ω,ω′

∑α,β

Γαβ(ω, ω′, t)(Aβ(ω′)ρ(t)A†α(ω)−A†α(ω)Aβ(ω′)ρ(t)

)+ h.c. (F4)

33

where we put the definition:

Γαβ(ω, ω′, t) ≡∫ t

0

ds ei(ωt−ω′s)〈Rα(t)Rβ(s)〉γR . (F5)

This can be further simplified to the form:

Γαβ(ω, ω′, t) = ei(ω−ω′)t

∫ t

0

ds eiω′s〈Rα(s)Rβ(0)〉γR ≡ ei(ω−ω

′)t Γαβ(ω′, t) (F6)

where we changed the variables in the integrand s→ t−s and use the property 〈Rα(t)Rβ(s)〉γR = 〈Rα(t−s)Rβ(0)〉γR .Next, we rewritten the hermitian conjugate part in the form:∑

ω,ω′

∑α,β

Γ∗αβ(ω, ω′, t)(Aβ(ω′)ρ(t)A†α(ω)−A†α(ω)Aβ(ω′)ρ(t)

)†(F7)

=∑ω,ω′

∑α,β

Γ∗αβ(ω, ω′, t)(Aα(ω)ρ(t)A†β(ω′)− ρ(t)A†β(ω′)Aα(ω)

)(F8)

=∑ω,ω′

∑α,β

Γ∗βα(ω′, ω, t)(Aβ(ω′)ρ(t)A†α(ω)− ρ(t)A†α(ω)Aβ(ω′)

)(F9)

Finally, we get

LRt [ρ(t)] =∑ω,ω′

∑α,β

(Γαβ(ω, ω′, t) + Γ∗βα(ω′, ω, t))Aβ(ω′)ρ(t)A†α(ω) (F10)

− 1

2

∑ω,ω′

∑α,β

(Γαβ(ω, ω′, t)A†α(ω)Aβ(ω′)ρ(t) + Γ∗βα(ω′, ω, t)ρ(t)A†α(ω)Aβ(ω′)

)(F11)

− 1

2

∑ω,ω′

∑α,β

(Γαβ(ω, ω′, t)A†α(ω)Aβ(ω′)ρ(t) + Γ∗βα(ω′, ω, t)ρ(t)A†α(ω)Aβ(ω′)

)(F12)

− 1

2

∑ω,ω′

∑α,β

(Γ∗βα(ω′, ω, t))A†α(ω)Aβ(ω′)ρ(t) + Γαβ(ω, ω′, t)ρ(t)A†α(ω)Aβ(ω′)

)(F13)

+1

2

∑ω,ω′

∑α,β

(Γ∗βα(ω′, ω, t))A†α(ω)Aβ(ω′)ρ(t) + Γαβ(ω, ω′, t)ρ(t)A†α(ω)Aβ(ω′)

)(F14)

where the last two lines sum up to zero. After rearranging terms and putting the definition (A16) and (A17), wefinally obtain the master equation in the form:

LRt [ρ(t)] =∑ω,ω′

∑α,β

[iSαβ(ω, ω′, t)[ρ(t), A†α(ω)Aβ(ω′)] + γαβ(ω, ω′, t)

(Aβ(ω′)ρ(t)A†α(ω)− 1

2A†α(ω)Aβ(ω′), ρ(t)

)].

(F15)

Appendix G: The cumulant equation in the Schrodinger pircture

The cumulant equation is originally derived in the interaction picture. In order to transform the cumulant equationinto the Schrodinger picture we start with a simple observation.

ρ(t) = ei[H0,·]tρ(t). (G1)

The super-operator in the r.h.s. of the equation above has its unique inverse, and ρ(0) = ρ(0), therefore:

ρ(t) = e−i[H0,·]teK(2)t ρ(0) = eK

(2)t ρ(0). (G2)

The r.h.s. of the equation above defines the Schrodinger picture cumulant eqaution super-operator K(2)t :

eK(2)t = e−i[H0,·]teK

(2)t . (G3)

34

The explicit form of K(2)t can be found with the aid of the Baker–Campbell–Hausdorff (BCH) formula.

eXeY = exp

X + Y +

1

2[X,Y ] +

1

12[X, [X,Y ]]− 1

12[Y, [X,Y ]] + · · ·

. (G4)

We observe that in a generic case the super-operator K(2)t in not of the GKSL form. This follows from the presence

of multi-commutator terms in the formula (G4). These terms do not vanish, as [K(2)t , H0] is not central. Therefore,

eK(2)t is an example of a one-parameter family of CPTP dynamical maps that are not of the GKSL form.

Appendix H: The cumulant equation in the differential form

We start this Section with the following Lemma on the properties of the derivative of an exponential map.

Lemma 1. The derivative of the exponential map is given by

d

dteX(t) =

(e[X(t),·] − 1

[X(t), ·]dX(t)

dt

)eX(t). (H1)

Proof. The proof of the above relation is identical to the proof of Theorem 5 in reference [23] up to small modifications.

Using Lemma 1 we instantly obtain the cumulant equation in the differential form:

d

dtρ(t) =

(e[K

(2)t ,·] − 1

[K(2)t , ·]

dK(2)t

dt

)ρ(t) = LCt ρ(t). (H2)

This result can also be obtained with integration of equation (G1). When truncated to the leading order, the above

formula reproduces the Bloch-Redfield master equation sincedK

(2)t

dt = LRt .Equation (H2) can be readily transformed into the Schrodinger picture. This is done with iterative application of

the e±iH0t operators to the jump operators Ai(ω) inside K(2)t super-operator.

d

dtρ(t) =

(−i[H0, ·] + e−i[H0,·]t e

[K(2)t ,·] − 1

[K(2)t , ·]

LRt ei[H0,·]t

)ρ(t) (H3)

=

(−i[H0, ·] +

e[K(2)t ,·] − 1

[K(2)t , ·]

LRt

)ρ(t) = LCt ρ(t), (H4)

where

K(2)t [ρ]

= λ2

∫ t

0

ds∑ω,ω′

∑αβ

ei(ω−ω′)(s−t)

(iSαβ(ω, ω′, s)[ρ,A†α(ω)Aβ(ω′)] + γαβ(ω, ω′, s)

(Aβ(ω′)ρA†α(ω)− 1

2A†α(ω)Aβ(ω′), ρ

)),

(H5)

LRt = LRt + i[H0, ·]. (H6)

Moreover, we observe that:

LCt = LRt +O(λ4). (H7)

Equation (H4) can be compared with the differential form of the Schrodinger picture cumulant equation obtainedwith Lemma 1 and the super-operator in equation (G3).

d

dtρ(t) =

(e[K

(2)t ,·] − 1

[K(2)t , ·]

dK(2)t

dt

)ρ(t) = LCt ρ(t). (H8)

The above formula has only a formal meaning, as the K(2)t super-operator does not possess a closed form formula.

We present it only for the curiosity of the reader.