Adiabatic Lindbladian Evolution with Small Dissipators

Abstract

We consider a time-dependent small quantum system weakly coupled to an environment, whose effective dynamics we address by means of a Lindblad equation. We assume the Hamiltonian part of the Lindbladian is slowly varying in time and the dissipator part has small amplitude. We study the properties of the evolved state of the small system as the adiabatic parameter and coupling constant both go to zero, in various asymptotic regimes. In particular, we analyse the deviations of the transition probabilities of the small system between the instantaneous eigenspaces of the Hamiltonian with respect to their values in the purely Hamiltonian adiabatic setup, as a function of both parameters.

Appendix: Integration by Parts

We present here a reformulation of the integration by parts argument used in [ASY] to prove the adiabatic theorem of quantum mechanics, suited to our setup.

Let ${{\mathcal {Z}}}$ be a Banach space and assume ${{\mathcal {G}}}:[0,1]\rightarrow {{\mathcal {B}}}({{\mathcal {Z}}})$, ${{\mathcal {K}}}:[0,1]\rightarrow {{\mathcal {B}}}({{\mathcal {Z}}})$ are bounded operator valued $C^\infty$ functions on [0, 1], in the norm sense. Let $\varepsilon >0$, and consider the two-parameter propagators $({{\mathcal {X}}}(t,s))_{1\le s\le t\le 1}$ and $({{\mathcal {Y}}}(t,s))_{1\le s\le t\le 1}$, solution to the equations

$$\begin{aligned} \left\{ \begin{array}{l} \varepsilon \partial _t{{\mathcal {X}}}(t,s)={{\mathcal {G}}}(t){{\mathcal {X}}}(t,s), \\ {{\mathcal {X}}}(s,s)={{\mathbb {I}}}, \ \ 0\le s \le t \le 1, \end{array} \right. \end{aligned}$$

(8.1)

and

$$\begin{aligned} \left\{ \begin{array}{l} \varepsilon \partial _t{{\mathcal {Y}}}(t,s)=({{\mathcal {G}}}(t)+\varepsilon {{\mathcal {K}}}(t)){{\mathcal {Y}}}(t,s), \\ {{\mathcal {Y}}}(s,s)={{\mathbb {I}}}, \ \ 0\le s \le t \le 1. \end{array} \right. \end{aligned}$$

(8.2)

The smooth propagators ${{\mathcal {X}}}(t,s)$ and ${{\mathcal {Y}}}(t,s)$ are determined by the corresponding Dyson series, both depend on $\varepsilon >0$ with norms that diverge as $\varepsilon \rightarrow 0$, a priori . Moreover, they satisfy the integral relation

$$\begin{aligned} {{\mathcal {X}}}(t,r)={{\mathcal {Y}}}(t,r)-\int _r^t {{\mathcal {Y}}}(t,s) {{\mathcal {K}}}(s) {{\mathcal {X}}}(s,r)ds, \ \ \forall 1\ge t\ge r \ge 0. \end{aligned}$$

(8.3)

Assume the existence of gaps in the spectrum of ${{\mathcal {G}}}(t)$, uniformly in $t\in [0,1]$. For $d\in {{\mathbb {N}}}^*$,

$$\begin{aligned} \sigma ({{\mathcal {G}}}(t))=\cup _{1\le j\le d}\, \sigma _j(t)\subset {{\mathbb {C}}}, \ \ \inf _{t\in [0,1], 1\le j\ne k\le d}\mathrm{dist} (\sigma _j(t),\sigma _k(t))\ge G >0. \end{aligned}$$

(8.4)

Consider the corresponding spectral projector

$$\begin{aligned} {{\mathcal {P}}}_j(t)=-\frac{1}{2 \pi \mathrm{i}}\int _{\gamma _j} ({{\mathcal {G}}}(s)-z)^{-1}dz, \end{aligned}$$

(8.5)

where $\gamma _j$ is a simple loop in $\rho ({{\mathcal {G}}}(t))$, the resolvent set of ${{\mathcal {G}}}(t)$, encircling $\sigma _j(t)$ and such that for all $k\ne j$, $\text{ int }\gamma _j\cap \sigma _k(t)=\emptyset$. For ${{\mathcal {B}}}:[0,1]\rightarrow {{\mathcal {B}}}({{\mathcal {Z}}})$, a smooth bounded operator valued function, define for any $t\in [0,1]$

$$\begin{aligned} {{\mathcal {R}}}_j({{\mathcal {B}}})(t)=-\frac{1}{2\pi \mathrm{i}} \oint _{\gamma _j} ({{\mathcal {G}}}(t)-z)^{-1}{{\mathcal {B}}}(t)({{\mathcal {G}}}(t)-z)^{-1}dz, \end{aligned}$$

(8.6)

with the same loop $\gamma _j$ as in (8.5). This operator is smooth as well, and satisfies the identity

$$\begin{aligned}{}[{{\mathcal {G}}}(t), {{\mathcal {R}}}_j({{\mathcal {B}}})(t)]=[{{\mathcal {B}}}(t),{{\mathcal {P}}}_j(t)]. \end{aligned}$$

(8.7)

Remark 8.1

If $\sigma _k(t)=\{g_k(t)\}$ for all $1\le k\le d$, $({{\mathcal {G}}}(t)-z)^{-1}=\sum _{1\le k \le d} \frac{{{\mathcal {P}}}_k(t)}{(g_k(t)-z)}$, and

$$\begin{aligned} {{\mathcal {R}}}_j({{\mathcal {B}}})(t)=\sum _{1\le k\le d\atop k\ne j} \frac{{{\mathcal {P}}}_j(t){{\mathcal {B}}}(t){{\mathcal {P}}}_k(t)+{{\mathcal {P}}}_k(t){{\mathcal {B}}}(t){{\mathcal {P}}}_j(t)}{g_k(t)-g_j(t)}. \end{aligned}$$

(8.8)

Lemma 8.2

Suppose ${{\mathcal {K}}}(t)$ is off-diagonal for all $t\in [0,1]$, i.e. s.t. ${{\mathcal {P}}}_j(t){{\mathcal {K}}}(t){{\mathcal {P}}}_j(t)\equiv 0$, $\forall 1\le j\le d$. Then

$$\begin{aligned} {{\mathcal {X}}}(t,r)-{{\mathcal {Y}}}(t,r)&= \frac{1}{2}\sum _{1\le j\le d} \varepsilon \Big ({{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(t){{\mathcal {X}}}(t,r)- {{\mathcal {Y}}}(t,r) {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(r)\Big )\nonumber \\&\quad +\frac{1}{2}\sum _{1\le j\le d} \varepsilon \int _r^t \Big \{ {{\mathcal {Y}}}(t,s){{\mathcal {K}}}(s)[{{\mathcal {G}}}(s), {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)] {{\mathcal {X}}}(s,r)\nonumber \\&\quad -{{\mathcal {Y}}}(t,s) (\partial _s{{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)){{\mathcal {X}}}(s,r) \Big \} ds. \end{aligned}$$

(8.9)

Proof

The operator ${{\mathcal {K}}}$ being off-diagonal and (8.7) give

$$\begin{aligned} {{\mathcal {K}}}(t)=\frac{1}{2}\sum _{1\le j\le d}\big [[{{\mathcal {K}}}(t), {{\mathcal {P}}}_j(t)\big ],{{\mathcal {P}}}_j(t)]=\frac{1}{2}\sum _{1\le j\le d}\big [{{\mathcal {G}}}(t), {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(t)\big ].\qquad \end{aligned}$$

(8.10)

Hence, using (8.3), (8.1) and (8.2),

$$\begin{aligned} {{\mathcal {X}}}(t,r)-{{\mathcal {Y}}}(t,r) =&-\frac{1}{2}\sum _{1\le j\le d}\int _r^t {{\mathcal {Y}}}(t,s) \big [{{\mathcal {G}}}(s), {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)\big ] {{\mathcal {X}}}(s,r)ds \end{aligned}$$

(8.11)

where, for each integral in the summand

$$\begin{aligned} -\int _r^t {{\mathcal {Y}}}(t,s) [&{{\mathcal {G}}}(s), {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)] {{\mathcal {X}}}(s,r)ds\nonumber \\ =&\varepsilon \int _r^t \Big \{(\partial _s{{\mathcal {Y}}}(t,s)) {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s){{\mathcal {X}}}(s,r)\nonumber \\&\qquad \quad \quad + {{\mathcal {Y}}}(t,s){{\mathcal {K}}}(s)\big [{{\mathcal {G}}}(s), {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)\big ] {{\mathcal {X}}}(s,r) \nonumber \\&\qquad \qquad + {{\mathcal {Y}}}(t,s) {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)\partial _s{{\mathcal {X}}}(s,r) \Big \} ds. \end{aligned}$$

(8.12)

Thanks to the smoothness of all operators in the integrand, we have

$$\begin{aligned} (\partial _s{{\mathcal {Y}}}(t,s) )&{{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s){{\mathcal {X}}}(s,r)+{{\mathcal {Y}}}(t,s) {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)\partial _s{{\mathcal {X}}}(s,r)\nonumber \\&= \partial _s({{\mathcal {Y}}}(t,s) {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s){{\mathcal {X}}}(s,r)) -{{\mathcal {Y}}}(t,s) (\partial _s{{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)){{\mathcal {X}}}(s,r), \end{aligned}$$

(8.13)

which yields the sought for identity. $\square$

As a corollary of Lemma (8.2), if one of the propagators $({{\mathcal {X}}}(t,s))_{0\le s\le t\le 1}$ or $({{\mathcal {Y}}}(t, s))_{0\le s\le t\le 1}$ is uniformly bounded in $\varepsilon$, so is the other, and their difference goes to zero with $\varepsilon$:

Corollary 8.3

Assume $\exists \ \varepsilon _1>0, C_1<\infty$ such that $\sup _{0<\varepsilon \le \varepsilon _1\atop 0\le s\le t\le 1}\Vert {{\mathcal {X}}}(t,s)\Vert \le C_1$. Then, $\exists \ \varepsilon _2>0, C_2<\infty$ such that $\sup _{0<\varepsilon \le \varepsilon _2 \atop 0\le s\le t\le 1}\Vert {{\mathcal {Y}}}(t,s)\Vert \le C_2$. The same statement holds for ${{\mathcal {X}}}$ and ${{\mathcal {Y}}}$ exchanged. Moreover, $\exists \ C_3<\infty$ such that for all $\varepsilon <\varepsilon _2$,

$$\begin{aligned} \sup _{0\le s\le t\le 1}\Vert {{\mathcal {X}}}(t,s)-{{\mathcal {Y}}}(t,s)\Vert \le C_3 \varepsilon , \end{aligned}$$

(8.14)

and, whenever ${{\mathcal {K}}}(0)=0$, there exists $C_4<\infty$ so that for all $t\in [0,1]$

$$\begin{aligned} \Vert {{\mathcal {X}}}(t,0)-{{\mathcal {Y}}}(t,0)\Vert \le C_4 t \varepsilon . \end{aligned}$$

(8.15)

Remark 8.4

If both ${{\mathcal {X}}}(t,s)$ and ${{\mathcal {Y}}}(t,s)$ are a priori uniformly bounded, estimate (8.14) holds for all $\varepsilon$.

Proof

Set

$$\begin{aligned} 2C_0=\max&\Big ( \sum _{1\le j\le d}\sup _{0\le s\le 1}\Vert {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)\Vert , \sum _{1\le j\le d}\sup _{0\le s\le 1}\Vert \partial _s {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)\Vert , \nonumber \\&\sum _{1\le j\le d}\sup _{0\le s\le 1}\Vert {{\mathcal {K}}}(s)\big [{{\mathcal {G}}}(s), {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)\big ]\Vert \Big ) \end{aligned}$$

(8.16)

and consider $\varepsilon <\varepsilon _1$. Lemma (8.2) yields the bound

$$\begin{aligned} \Vert {{\mathcal {Y}}}(t,r)\Vert&\le C_1+\varepsilon C_0 (\Vert {{\mathcal {Y}}}(t,r)\Vert +C_1)+\varepsilon 2C_0C_1\sup _{0\le s \le t\le 1} \Vert {{\mathcal {Y}}}(t,s)\Vert \end{aligned}$$

(8.17)

so that, taking the supremum over $0\le r\le t\le 1$ and for $\varepsilon < \min (\varepsilon _1, 1/(2C_0(1+2C_1))) := \varepsilon _2$, we get in turn

$$\begin{aligned} \sup _{0\le r \le t\le 1}\Vert {{\mathcal {Y}}}(t,r)\Vert&\le \frac{1}{(1-\varepsilon C_0(1+2C_1))}(C_1(1+\varepsilon C_0)) \le C_1\frac{3+4C_1}{1+2C_1}\equiv C_2. \end{aligned}$$

(8.18)

Then, inserting this estimate into (8.9), one gets, uniformly in $0\le s\le t\le 1$,

$$\begin{aligned} \Vert {{\mathcal {X}}}(t,s)-{{\mathcal {Y}}}(t,s)\Vert \le C_3 \varepsilon , \end{aligned}$$

(8.19)

with $C_3=C_0(C_1+C_2+2C_1C_2)$. Finally, for the initial time $s=0$, the integrated contribution in (8.9) reduces to $\frac{1}{2}\sum _{1\le j\le d} \varepsilon {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(t){{\mathcal {X}}}(t,0)$ when ${{\mathcal {K}}}(0)=0$, and since either ${{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(t)=0$ or

$$\begin{aligned} {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(t)=\int _0^t\partial _s {{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(s)ds, \end{aligned}$$

(8.20)

we have in any case

$$\begin{aligned} \Big \Vert \frac{1}{2}\sum _{1\le j\le d}{{\mathcal {R}}}_j([{{\mathcal {K}}}, {{\mathcal {P}}}_j])(t)\Big \Vert \le tC_0. \end{aligned}$$

(8.21)

The integral term in (8.9) is of order $\varepsilon t$, so that the bound (8.15) holds with $C_4=C_0C_1(1+2C_2)$.

The fact that ${{\mathcal {X}}}$ and ${{\mathcal {Y}}}$ can be exchanged in all arguments above follows from the structure of the RHS of (8.9). $\square$

Proof of Lemma 3.2:

We briefly prove estimate (3.7). Here the Banach space is ${{\mathcal {Z}}}={{\mathcal {H}}}$, the propagators are ${{\mathcal {X}}}=U$, ${{\mathcal {Y}}}=V$, and the generators are constructed with ${{\mathcal {G}}}=-iH$ and ${{\mathcal {K}}}=K=\sum _{1\le j\le d} P'_jP_j$.

Using $P(t)P'(t)P(t)\equiv 0$ for any smooth projector P(t), one has the identities $P_j(t)K(t) P_j(t)\equiv 0$, for all $1\le j\le d$, and actually both U and V are bounded a priori, since they are unitary. Moreover, $K(0)=0$, under Reg. Hence Lemma 3.2 derives from Corollary 8.3. $\square$

Proof of Lemma 3.12:

As a second application, we derive here estimate (3.37). We need to show that

$$\begin{aligned} P_n(t)\int _0^t{{\mathcal {V}}}^0(t,s)\circ {{\mathcal {L}}}^1_{s}\circ {{\mathcal {V}}}^0(s,0) (\rho _j)ds\, P_m(t)={{{\mathcal {O}}}}(\varepsilon ). \end{aligned}$$

(8.22)

We first note that by Lemma 3.7, ${{\mathcal {V}}}^0(s,0) (\rho _j)={\tilde{\rho }}_j(s)$, where $\partial _s {\tilde{\rho }}_j(s)= [K(s), {\tilde{\rho }}_j(s)]$ is continuous in trace norm and $\varepsilon$-independent. Moreover,

$$\begin{aligned} P_n(t){{\mathcal {V}}}^0(t,s)(\cdot )P_m(t)=P_n(t)\big ({{\mathcal {V}}}^0(t,0)\circ {{\mathcal {V}}}^0(0,s)(P_n(s) \cdot P_m(s))\big )P_m(t), \end{aligned}$$

(8.23)

thanks to the intertwining property (7.19). Using the definition of ${{\mathcal {L}}}_s^1$ , we have

$$\begin{aligned}&P_n(s) {{\mathcal {L}}}^1_{s}\circ {{\mathcal {V}}}^0(s,0) (\rho _j) P_m(s)=\sum _{l\in I}P_n(s)(\Gamma _l(s) {\tilde{\rho }}_j(s)\Gamma _l^*(s)P_m(s) \nonumber \\&\quad -\delta _{mj}\frac{1}{2}P_n(s)\Gamma _l^*(s)\Gamma _l(s) {\tilde{\rho }}_j(s)P_m(s) - \frac{1}{2} \delta _{nj}P_n(s) {\tilde{\rho }}_j(s)\Gamma _l^*(s)\Gamma _l(s) P_m(s), \end{aligned}$$

(8.24)

where all terms are independent of $\varepsilon$. Hence, to get the result, we are lead to show that for a smooth trace class operator $[0,1]\ni s\rightarrow F(s)$, such that $\partial _sF(s)\in {{\mathcal {T}}}({{\mathcal {H}}})$, independent of $\varepsilon$, and $n\ne m$,

$$\begin{aligned} \int _0^tV^0(0,s) P_n(s)F(s)P_m(s)V^0(s,0)ds\, ={{{\mathcal {O}}}}(\varepsilon ). \end{aligned}$$

(8.25)

We have thanks to (8.7) with ${{\mathcal {G}}}=H$

$$\begin{aligned} P_n(s)F(s)P_m(s)=[P_n(s)F(s)P_m(s), P_m(s)]=[H(s),{{\mathcal {R}}}_m(P_nFP_m)(s)]\qquad \end{aligned}$$

(8.26)

so that, by a slight variation of Lemma 8.2

$$\begin{aligned}&\int _0^tV^0 (0,s) [H(s),{{\mathcal {R}}}_m(P_nFP_m)(s)]V^0(s,0)ds\nonumber \\&\quad =-\mathrm{i}\varepsilon V^0(0,s) {{\mathcal {R}}}_m(P_nFP_m)(s)]V^0(s,0)|_0^t +\mathrm{i}\varepsilon \int _0^tV^0(0,s) \Big \{ \partial _s{{\mathcal {R}}}_m(P_nFP_m)(s)\nonumber \\&\qquad -K(s){{\mathcal {R}}}_m(P_nFP_m)(s)+{{\mathcal {R}}}_m(P_nFP_m)(s)K(s)\Big \}V^0(s,0)ds. \end{aligned}$$

(8.27)

As F(s) and its derivative are trace class, the expression above is ${{{\mathcal {O}}}}(\varepsilon )$ in trace norm. $\square$

Let us finally note that, making use of the projectors appearing (8.23), we can further integrate by parts the last integral term, provided $\partial _sF(s)$ is continuously differentiable in trace norm, in which case

$$\begin{aligned}&\int _0^tV^0(0,s) P_n(s)F(s)P_m(s)V^0(s,0)ds\nonumber \\&\quad = \mathrm{i}\varepsilon \big ({{\mathcal {R}}}_m(P_nFP_m)(0) -V^0(0,t) {{\mathcal {R}}}_m(P_nFP_m)(t)V^0(t,0)\big )+{{{\mathcal {O}}}}(\varepsilon ^2). \end{aligned}$$

(8.28)

Proof of Lemma 3.13:

For any $\rho \in {{\mathcal {T}}}({{\mathcal {H}}})$, we need to consider

$$\begin{aligned} \int _0^t\int _0^{s_1}\dots \int _0^{s_{n-1}} {{\mathcal {V}}}^0(t,s_1)\circ {{\mathcal {L}}}^1_{s_1} \circ {{\mathcal {V}}}^0(s_1,s_2) \circ {{\mathcal {L}}}^1_{s_2}\dots {{\mathcal {L}}}^1_{s_n} \circ {{\mathcal {V}}}^0(s_n,0) (\rho )ds_{n}\dots ds_{2}ds_{1}. \end{aligned}$$

(8.29)

Noting with (3.15) that for any $0\le s,t\le 1$ ${{\mathcal {V}}}^0(t,s)(\rho )={{\mathcal {V}}}^0(t,0)\circ {{\mathcal {V}}}^0(0,s)(\rho ),$ we can write (8.29) as ${{\mathcal {V}}}^0(t,0)(I_n(t))$, where $I_n(t)\in {{\mathcal {T}}}({{\mathcal {H}}})$ is defined inductively by

$$\begin{aligned} I_n(t)&=\int _0^t {{\mathcal {V}}}^0(0,s_1)\circ {{\mathcal {L}}}^1_{s_1} \circ {{\mathcal {V}}}^0(s_1,0)(I_{n-1}(s_1))ds_1,\nonumber \\ I_1(t)&=\int _0^t {{\mathcal {V}}}^0(0,s_n) \circ {{\mathcal {L}}}^1_{s_n} \circ {{\mathcal {V}}}^0(s_n,0)(\rho )ds_n. \end{aligned}$$

(8.30)

Lemma 3.12 shows the existence of $C_1<\infty$, such that for all $0\le t\le 1$ and $\varepsilon$ small enough,

$$\begin{aligned} \Vert I_1(t)-{{\mathcal {P}}}_0(0)(I_1(t))\Vert _1\le \varepsilon C_1 \Vert \rho \Vert _1. \end{aligned}$$

(8.31)

Let us show by induction that for for each n, there exists $C_n<\infty$ so that for $\varepsilon$ small enough

$$\begin{aligned} \Vert I_n(t)-{{\mathcal {P}}}_0(0)(I_n(t))\Vert _1\le \varepsilon C_n \Vert \rho \Vert _1. \end{aligned}$$

(8.32)

Assuming the result for $n\ge 1$, we consider the step $n+1$. We get

$$\begin{aligned}&I_{n+1}(t)-{{\mathcal {P}}}_0(0)(I_{n+1}(t))=\sum _{1\le j\ne k\le d}P_j(0)I_{n+1}(t)P_k(0)\nonumber \\&\quad = \sum _{1\le j\ne k\le d}\int _0^t P_j(0)\big \{{{\mathcal {V}}}^0(0,s) \circ {{\mathcal {L}}}^1_{s} \circ {{\mathcal {V}}}^0(s,0)(I_n(s))\big \}P_k(0)ds\nonumber \\&\quad = \sum _{1\le j\ne k\le d}\int _0^t\!\! P_j(0)\big \{{{\mathcal {V}}}^0(0,s) \circ {{\mathcal {L}}}^1_{s} \circ {{\mathcal {V}}}^0(s,0)\circ {{\mathcal {P}}}_0(0)(I_n(s)))\big \}P_k(0)ds +{{{\mathcal {O}}}}_n(\Vert \rho \Vert _1 \varepsilon ), \end{aligned}$$

(8.33)

by the induction hypothesis and recalling the operator ${{\mathcal {V}}}^0(t,s)$ is isometric and ${{\mathcal {L}}}_s^1$ is uniformly bounded. Then we observe that for any $j\ne k$, and any $[0,1]\ni s\mapsto A(s)\in {{\mathcal {T}}}({{\mathcal {H}}})$, $C^1$ in trace norm, see Lemmas 3.4, 3.5 and (3.23)

$$\begin{aligned}&\int _0^t P_j(0)\big \{{{\mathcal {V}}}^0(0,s) (A(s))\big \}P_k(0)ds\nonumber \\&\quad =\int _0^t e^{\frac{\mathrm{i}}{\varepsilon }\int _{0}^s(e_j-e_k)(u)du}P_j(0)\big \{{{\mathcal {W}}}^0(0,s) (A(s))\big \}P_k(0)ds\nonumber \\&\quad =\frac{-\mathrm{i}\varepsilon }{e_j(s)-e_k(s)}e^{\frac{\mathrm{i}}{\varepsilon }\int _{0}^s(e_j-e_k)(u)du}P_j(0)\big \{{{\mathcal {W}}}^0(0,s) (A(s))\big \}P_k(0)\Big |_0^t\nonumber \\&\qquad +\int _0^t\mathrm{i}\varepsilon e^{\frac{\mathrm{i}}{\varepsilon }\int _{0}^s(e_j-e_k)(u)du}P_j(0)\partial _s\Big (\frac{{{\mathcal {W}}}^0(0,s) (A(s))}{e_j(s)-e_k(s)}\Big )P_k(0)ds. \end{aligned}$$

(8.34)

The trace norm of the RHS is bounded above by $\varepsilon c (\sup _{0\le s\le 1}\Vert A(s)\Vert _1+\sup _{0\le s\le 1}\Vert \partial _s A(s)\Vert _1)$, where c is a constant independent of $\varepsilon$. The integral term of the RHS of (8.33) has the form (8.34) with

$$\begin{aligned} A(s)= {{\mathcal {L}}}^1_{s} \circ {{\mathcal {V}}}^0(s,0)\circ {{\mathcal {P}}}_0(0)(I_n(s))= {{\mathcal {L}}}^1_{s} \circ {{\mathcal {W}}}^0(s,0)\circ {{\mathcal {P}}}_0(0)(I_n(s)), \end{aligned}$$

(8.35)

where ${{\mathcal {W}}}^0(s,0)$ and ${{\mathcal {L}}}_s^1$ are smooth, independent of $\varepsilon$ and bounded on ${{\mathcal {T}}}({{\mathcal {H}}})$, while $I_n(s)$ and $\partial _t I_n(s)$ are continuous and bounded in trace norm by a constant (uniform in $\varepsilon$) time $\Vert \rho \Vert _1$, see (8.30), which ends the proof. $\square$