Theory of non-Markovian dynamics in resonance fluorescence spectrum

Abstract

We present a detailed theoretical study of non-Markovian dynamics in the fluorescence spectrum of a driven semiconductor quantum dot (QD), embedded in a cavity and coupled to a three-dimensional (3D) acoustic phonon reservoir. In particular, we investigate the effect of pure dephasing on one of the side-peaks of the Mollow-triplet spectrum, expressed in terms of the off-diagonal element of the reduced system operator. The QD is modeled as a two-level system with an excited state representing a single exciton, and ground state represents the absence of an exciton. Coupling to the radiative modes of the cavity is treated within usual Born–Markov approximation, whereas dot-phonon coupling is discussed within non-Markovian regime beyond Born approximation. Using an equation-of-motion technique, the dot-phonon coupling is solved exactly and found that the exact result coincides with that of obtained within Born approximation. Furthermore, a Markov approximation is carried out with respect to the phonon interaction and compared with the non-Markovian lineshape for different values of the phonon bath temperature. We have found that coupling to the phonons vanishes for a resonant pump laser. For a non-resonant pump, we have characterized the effect of dot-laser detuning and temperature of the phonon bath on the lineshape. The sideband undergoes a distinct narrowing and acquires an asymmetric shape with increasing phonon bath temperature. We have explained this behavior using a dressed-state picture of the QD levels.

Appendices

Appendix 1: Detector response function

The detector response function in Eq. (9)

$$\begin{aligned} \bar{I}=\int _{0}^{\infty }d\tau \,\sum _{k}g_{k}g_{k}^{D}e^{-i\omega _{k}\tau }, \end{aligned}$$

(82)

where $g_{k}^{D}=|{\mathcal{\wp }}_{\alpha \beta }|\varepsilon _{k}$. Applying the continuum of modes after replacing the sum via integral $\sum _{k}g_{k}^{2}\rightarrow \int _{0}^{\infty }D(\epsilon )|g(\epsilon )|^{2}d\epsilon$ and using the well-known formula

$$\begin{aligned} \frac{1}{X\pm i0^{+}}={\mathcal{P}}\left( \frac{1}{X}\right) \mp i\pi \delta (X), \end{aligned}$$

(83)

(${\mathcal{P}}$ indicates the principal part) one can write the square of detector response function in terms of principal part and delta function as

$$\begin{aligned} \bar{I}^{2}&=\int _{0}^{\infty }d\epsilon \,D_{c} (\epsilon )|g(\epsilon )|^{2}\bigg [-i{\mathcal{P}} \left( \frac{1}{\epsilon }\right) +\pi \delta (\epsilon )\bigg ]\nonumber \\&\quad \times \int _{0}^{\infty }d\nu \,D(\nu )|g^{D}(\nu )|^{2}\bigg [i{\mathcal{P}} \left( \frac{1}{\nu }\right) +\pi \delta (\nu )\bigg ], \end{aligned}$$

(84)

where $D_{c}(\epsilon )$ and $D(\nu )$ are photonic density of states of cavity and open space, respectively. The photonic density of states of the cavity is described by a Lorentzian density of states given by Eq. (127).

Appendix 2: Schreiffer–Wolff transformation

Hamiltonian $H'$ resulting from the polaron transformation, is

$$\begin{aligned} H'&=\frac{\varOmega _{R}'}{2}\sigma _{3}+\sum _{k}\varDelta _{k}a_{k}^{\dag }a_{k}+\sum _{q}\lambda _{q}b_{q}^{\dag }b_{q}\nonumber \\&+{\mathbf{c.s}}\,\sigma _{3}\,\sum _{k}g_{k}(a_{k}+a_{k}^{\dag })\nonumber \\&+\frac{({\mathbf{c}}^{2}-{\mathbf{s}}^{2})}{2}\sigma _{3}\,\sum _{q}\lambda _{q}(b_{q}+b_{q}^{\dag })\nonumber \\&+\sum _{k}g_{k}\left[ ({\mathbf{c}}^{2}a_{k}-{\mathbf{s}}^{2}a_{k}^{\dagger })\sigma _{+-}+h.c.\right] \nonumber \\&-{\mathbf{c.s}}\,\sum _{q}\lambda _{q}(\sigma _{+-}+\sigma _{-+})(b_{q}+b_{q}^{\dagger })\nonumber \\&-\sum _{q}\frac{\lambda _{q}^{2}}{4\omega _{q}}+{\mathbf{c.s}}\,\sum _{q}\frac{\lambda _{q} ^{2}}{\omega _{q}}(\sigma _{+-}+\sigma _{-+}). \end{aligned}$$

(85)

Last two terms in above Hamiltonian commute with rest of the Hamiltonian and can be ignored within a secular approximation for large $\varOmega _{R}'$. We get rid of the energy exchange process due to phonon interaction using leading-order Schrieffer–Wolff transformation (Schrieffer and Wolff 1966) and start from the full Hamiltonian:

$$\begin{aligned} H'&=H_{1}+V_{2} \end{aligned}$$

(86)

$$\begin{aligned} H_{1}&=H_{S}+H_{R}+H_{P}+H_{dR}+H_{dP}+H_{SR}, \end{aligned}$$

(87)

where individual terms are defined as

$$\begin{aligned} H_{S}&=\frac{\varOmega '_{R}}{2}\sigma _{3}, \end{aligned}$$

(88)

$$\begin{aligned} H_{R}&=\sum _{k}\varDelta _{k}a_{k}^{\dag }a_{k}, \end{aligned}$$

(89)

$$\begin{aligned} H_{P}&=\sum _{q}\lambda _{q}b_{q}^{\dag }b_{q}, \end{aligned}$$

(90)

$$\begin{aligned} H_{dR}&={\mathbf{c.s}}\,\sigma _{3}\,\sum _{k}g_{k}(a_{k}+a_{k}^{\dag }), \end{aligned}$$

(91)

$$\begin{aligned} H_{dP}&=\frac{({\mathbf{c}}^{2}-{\mathbf{s}}^{2})}{2}\,\sigma _{3}\,\sum _{q}\lambda _{q}(b_{q}+b_{q}^{\dag }), \end{aligned}$$

(92)

$$\begin{aligned} H_{SR}&=\sum _{k}g_{k}\left[ ({\mathbf{c}}^{2}a_{k}-{\mathbf{s}}^{2}a_{k}^{\dagger })\sigma _{+-}+h.c.\right] , \end{aligned}$$

(93)

$$\begin{aligned} V_{2}&=-{\mathbf{c.s}}\,\sum _{q}\lambda _{q}(\sigma _{+-}+\sigma _{-+})(b_{q}+b_{q}^{\dagger }). \end{aligned}$$

(94)

We apply a transformation, $\bar{H}=e^{A}He^{-A}$, generated by an anti-Hermitian operator $A=-A^{\dag }$ to eliminate the transition terms to the first order. Using Baker–Campbell–Hausdorff formula and expanding $\bar{H}$ in the powers of A, we obtain

$$\begin{aligned} \bar{H}=H_{1}+V_{2}+[A,H_{1}]+[A,V_{2}]+\frac{1}{2}[A,[A,H]]+... \end{aligned}$$

(95)

In order to get rid of transition term $V_{2}$ to leading order, we set $V_{2}=-[A,H_{1}]$, where A can be written as

$$\begin{aligned} A=\frac{1}{L_{1}}V_{2}, \end{aligned}$$

(96)

and $L_{1}{\mathcal{O}}=[H_{1},{\mathcal{O}}]$. Here A is of order of transition term $V_{2}$. Substituting for A, we obtain Hamiltonian up to the second or higher order in $V_{2}$

$$\begin{aligned} \bar{H}=H_{1}+\frac{1}{2}[A,V_{2}]+... \end{aligned}$$

(97)

Using the definitions of $H_{1}$ and $V_{2}$, we obtain the expression for A as:

$$\begin{aligned} A=-{\mathrm{c.s}}\sum \limits _{q}\frac{\lambda _{q}}{\varOmega _{R}}(b_{q}+b_{q}^{\dag })(\sigma _{+-}-\sigma _{-+}). \end{aligned}$$

(98)

Therefore, the transformed Hamiltonian to the first order in the transition terms due to phonon can be well approximated and written as free and perturbed parts as

$$\begin{aligned} \bar{H}\simeq H_{0}+H_{V}, \end{aligned}$$

(99)

where the free part is

$$\begin{aligned} H_{0}&=H_{S}+H_{R}+H_{P}, \end{aligned}$$

(100)

$$\begin{aligned} H_{S}&=\frac{\varOmega '_{R}}{2}\sigma _{3}, \end{aligned}$$

(101)

$$\begin{aligned} H_{R}&=\sum _{k}\varDelta _{k}a_{k}^{\dag }a_{k}, \end{aligned}$$

(102)

$$\begin{aligned} H_{P}&=\sum _{q}\omega _{q}b_{q}^{\dag }b_{q} \end{aligned}$$

(103)

and perturbed parts is given by

$$\begin{aligned} H_{V}&=H_{dR}+H_{SR}+H_{dP}, \end{aligned}$$

(104)

$$\begin{aligned} H_{dR}&={\mathbf{c.s}}\,\sigma _{3}\,\sum _{k}g_{k}(a_{k}+a_{k}^{\dag }), \end{aligned}$$

(105)

$$\begin{aligned} H_{dP}&=\frac{{\mathbf{c}}^{2}-{\mathbf{s}}^{2}}{2}\,\sigma _{3}\,\sum _{q}\lambda _{q}(b_{q}+b_{q}^{\dag }), \end{aligned}$$

(106)

$$\begin{aligned} H_{SR}&=\sum _{k}g_{k}\left[ ({\mathbf{c}}^{2} a_{k}-{\mathbf{s}}^{2} a_{k}^{\dagger })\sigma _{+-}+h.c.\right] , \end{aligned}$$

(107)

which are Eqs. (18)–(26) in the main text.

Appendix 3: Final part matrix element

The final part is given by Eq. (45) can be written within Born approximation after transforming to Laplace domain

$$\begin{aligned} \varPhi (s)\simeq -i{\mathrm{Tr}}_{{\rm R}}{\text{Tr}}_{{\rm P}}L_{V}\,\frac{1}{s+iL_{0}}\,Q\varOmega (0). \end{aligned}$$

(108)

In particular, we want the matrix element, $\varPhi _{+-,+-}(s)$ due to phonon interaction, which can be simplified and written as

$$\begin{aligned} \varPhi _{+-,+-}(s)=-i{\mathrm{Tr}}_{{\rm P}}\,L_{Y}^{+}\,\frac{1}{s+i(\varOmega _{R}'+L_{P})}\,[Q\varOmega (0)]_{+-}, \end{aligned}$$

(109)

where final part of the stationary density matrix can be found (Swain 1981) using GME discussed in Sect. 4

$$\begin{aligned}{}[Q\varOmega (0)]_{+-}=-i\lim _{s\rightarrow 0}\frac{s}{s+i(\varOmega _{R}'+L_{P})}L_{Y}^{+}\rho _{P}(t_0)\,[\rho _{S}(s)\sigma _{ab}]_{+-}. \end{aligned}$$

(110)

The propagators in Eq. (110) has no poles at $s=0^{+}$, where $0^{+}$ is a positive infinitesimal. After performing the limit in the above expression, we substitute for $[Q\varOmega (0)]_{+-}$ in the expression for $\varPhi _{+-,+-}(s)$, to obtain and expression for the final term as

$$\begin{aligned} \varPhi _{+-,+-}(s)=-{\mathrm{Tr}}_{{\rm P}}\,L_{Y}^{+}\,\frac{1}{s+i(\varOmega _{R}'+L_{P})} \frac{1}{0^{+}+i(\varOmega _{R}'+L_{P})}L_{Y}^{+}\rho _{P}(t_0)\varOmega _{+-}(0), \end{aligned}$$

(111)

where $\varOmega _{+-}(0)=[\bar{\rho }_{S}\sigma _{ab}]_{+-}$, see Eqs. (178) and (179). Above expression can be written in terms of final part matrix element in the problem as

$$\begin{aligned} \varPhi _{+-,+-}(s)&=G_{+-,+-}^{P}(s)\varOmega _{+-}(0),\end{aligned}$$

(112)

$$\begin{aligned} G_{+-,+-}^{P}(s)&=-{\mathrm{Tr}}_{{\rm P}}\,L_{Y}^{+}\,\frac{1}{s+i(\varOmega _{R}'+L_{P})} \frac{1}{0^{+}+i(\varOmega _{R}'+L_{P})}L_{Y}^{+}\rho _{P}(t_0). \end{aligned}$$

(113)

Solving above expressions and applying the continuum of modes, we found that final part matrix element will be suppressed by $1/\varOmega _{R}'$ compared to the self-energy matrix element due to phonon interaction. In this limit, contribution from final part can be neglected compared to the contribution from the non-Markovian self-energy.

Appendix 4: Self-energy calculations

The reduced self-energy superoperator $\varSigma _{S}(t)$ in Eq. (44) can be transformed in to Laplace domain and using $L_{V}=L_{dR}+L_{SR}+L_{dP}$, we obtain

$$\begin{aligned} \varSigma _{S}(s)&=-i{\mathrm{Tr}}_{{\rm R}}{\text{Tr}}_{{\rm P}}(L_{dR}+L_{SR}+L_{dP})\frac{1}{s+iL}\nonumber \\&\quad\times (L_{dR}+L_{SR}+L_{dP})\rho _{R}(t_{0})\rho _{P}(t_{0}), \end{aligned}$$

(114)

dropping Q from the exponential in Eq. (44) will not affect the final expression (DiVincenzo and Loss 2005). Cross terms in above expression will vanish because $L_{dR}$ and $L_{SR}$ act on an operator in the radiation mode Hilbert space whereas $L_{dP}$ act on phonon Hilbert space and will not contribute in the final trace. Moreover, the cross terms between $L_{dR}$ and $L_{SR}$ will give rise to off-block diagonal matrix elements in self-energy matrix and will be neglected within secular approximation, see Sect. 4.1 in main text. The self-energy superoperator in Eq. (114) can be decomposed into three different parts as:

$$\begin{aligned} \varSigma _{S}(s)=\varSigma _{S}^{dR}(s)+\varSigma _{S}^{dP}(s)+\varSigma _{S}^{SR}(s), \end{aligned}$$

(115)

where first and second terms give rise to pure dephasing ($T_{2}^{*}$ process) due to radiation and phonon modes, respectively, whereas last term in above expression leads to transition ($T_{1}$ process) due to radiation modes coupling. Free propagator in reduced self-energy expression can be expanded in the powers of interacting Liouvillian $L_{V}$ as (Coish and Loss 2004)

$$\begin{aligned} \frac{1}{s+iL}=\frac{1}{s+iL_{0}}\sum _{k}\bigg (-iL_{V}\frac{1}{s+iL_{0}}\bigg )^{2k}, \end{aligned}$$

(116)

because of the form of couplings in present model only even powers 2k in above expression will survive in the final trace. In order to find the matrix elements of the self-energy superoperator, we write the superoperators in matrix form in dressed-state basis as

$$\begin{aligned}{}[L_{S}]= \left( \begin{array}{cccc} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} \varOmega _{R}' &{} 0\\ 0 &{} 0 &{} 0 &{} -\varOmega _{R}' \end{array} \right) , \end{aligned}$$

(117)

where $[L_{S}]_{\alpha \beta ,\gamma \delta }=Tr\lbrace |\beta \rangle \langle \alpha |S|\gamma \rangle \langle \delta |\rbrace$ and $\lbrace \alpha ,\beta \rbrace \in \lbrace +,-\rbrace$. In the dressed state basis, non-interacting Liouvillian is diagonal and can be inverted to write as $2\times 2$ blocks

$$\begin{aligned} \bigg [\frac{1}{s+iL_{0}}\bigg ]= \left( \begin{array}{cc} G_{\parallel }(s) &{} 0\\ 0 &{} G_{\perp }(s) \end{array} \right) , \end{aligned}$$

(118)

where parallel block is

$$\begin{aligned}{}[G_{\parallel }(s)]= \left( \begin{array}{cc} \frac{1}{s+i(L_{R}+L_{P})} &{} 0\\ 0 &{} \frac{1}{s+i(L_{R}+L_{P})} \end{array} \right) , \end{aligned}$$

(119)

and perpendicular block is given by

$$\begin{aligned}{}[G_{\perp }(s)]= \left( \begin{array}{cc} \frac{1}{s+i(\varOmega _{R}'+L_{R}+L_{P})} &{} 0\\ 0 &{} \frac{1}{s+i(-\varOmega _{R}'+L_{R}+L_{P})} \end{array} \right) . \end{aligned}$$

(120)

In the similar fashion, we can find other matrices as well

$$\begin{aligned}{}[L_{dR(P)}]= \left( \begin{array}{cccc} L_{X(Y)}^{-} &{} 0 &{} 0 &{} 0\\ 0 &{} L_{X(Y)}^{-} &{} 0 &{} 0\\ 0 &{} 0 &{} L_{X(Y)}^{+} &{} 0\\ 0 &{} 0 &{} 0 &{} -L_{X(Y)}^{+}\\ \end{array} \right) , \end{aligned}$$

(121)

here we have defined new Liouvillian for commutation and anti-commutation relations: $L_{X(Y)}^{\pm }{\mathcal{O}}=[X_{R(P)},{\mathcal{O}}]_{\pm }$, where operators $X_{R}$ and $X_{P}$ are given as

$$\begin{aligned} X_{R}&=\frac{\sqrt{\varOmega _{R}^{2}-\varDelta ^{2}}}{2\varOmega _{R}}\sum _{k}g_{k}(a_{k}+a_{k}^{\dag }), \end{aligned}$$

(122)

$$\begin{aligned} X_{P}&=\frac{\varDelta }{2\varOmega _{R}}\sum _{k}\lambda _{q}(b_{q}+b_{q}^{\dag }). \end{aligned}$$

(123)

Furthermore, we also find the matrix for superoperator $L_{SR}$ in the dressed-state basis

$$\begin{aligned}{}[L_{SR}]= \left( \begin{array}{cccc} 0 &{} 0 &{} -Z_{r}^{\dag } &{} Z_{l}\\ 0 &{} 0 &{} Z_{l}^{\dag } &{} -Z_{r}\\ -Z_{r} &{} Z_{l} &{} 0 &{} 0\\ Z_{l}^{\dag } &{} -Z_{r}^{\dag } &{} 0 &{} 0 \end{array} \right) , \end{aligned}$$

(124)

where we have defined the operators for left and right multiplications as:

$$\begin{aligned} Z_{l}{\mathcal{O}}_{R}&=\sum \limits _{k}g_{k}({\mathbf{c}}^{2}a_{k}-{\mathbf{s}}^{2}a_{k}^{\dag }){\mathcal{O}}_{R} \end{aligned}$$

(125)

$$\begin{aligned} Z_{r}{\mathcal{O}}_{R}&={\mathcal{O}}_{R}\sum \limits _{k}g_{k}({\mathbf{c}}^{2}a_{k}-{\mathbf{s}}^{2}a_{k}^{\dag }), \end{aligned}$$

(126)

and ${\mathcal{O}}_{R}$ is an operator in the radiation mode Hilbert space. Reduced self-energy matrix elements of interest can be calculated according to Eq. (46) in the main text.

1.1 Appendix 4.1: Self-energy for photon interaction

We apply a continuum of modes for the cavity density of states given by a Lorentzian spectrum (Ma et al. 2012),

$$\begin{aligned} D_{c}(\epsilon )|g(\epsilon )|^{2}=\frac{1}{\pi }\frac{g^{2}\varGamma _{c}}{(\epsilon -\varDelta _{c})^{2}+\varGamma _{c}^{2}}, \end{aligned}$$

(127)

where $\varDelta _{c}=\omega _{c}-\omega$ is the detuning of cavity from laser pump frequency, and $\varGamma _{c}$ is cavity bandwidth. Reduced self-energy matrix elements due to radiation mode coupling giving rise to transition are given as

$$\begin{aligned} \varSigma _{+-,++}^{SR}(s)&=\frac{-ig^{2}(\varOmega _{R}+\varDelta )^{2}}{4\varOmega _{R}^{2}}\left( \frac{1}{s+i(\varOmega _{R}'-\varDelta _{c})+\varGamma _{c}}+\frac{1}{s-i(\varOmega _{R}'-\varDelta _{c})+\varGamma _{c}}\right) , \end{aligned}$$

(128)

$$\begin{aligned} \varSigma _{+-,--}^{SR}(s)&=\frac{ig^{2}(\varOmega _{R}-\varDelta )^{2}}{4\varOmega _{R}^{2}}\left( \frac{1}{s+i(\varOmega _{R}'+\varDelta _{c})+\varGamma _{c}}+\frac{1}{s-i(\varOmega _{R}'+\varDelta _{c})+\varGamma _{c}}\right) , \end{aligned}$$

(129)

$$\begin{aligned} \varSigma _{+-,+-}^{SR}(s)&=\frac{-ig^{2}}{4\varOmega _{R}^{2}}\bigg (\frac{(\varOmega _{R}+\varDelta )^{2}}{s+i\varDelta _{c}+\varGamma _{c}}+\frac{(\varOmega _{R}-\varDelta )^{2}}{s-i\varDelta _{c}+\varGamma _{c}}\bigg ), \end{aligned}$$

(130)

$$\begin{aligned} \varSigma _{+-,-+}^{SR}(s)&=\frac{-ig^{2}(\varOmega _{R}^{2}-\varDelta ^{2})}{4\varOmega _{R}^{2}}\bigg (\frac{1}{s+i\varDelta _{c}+\varGamma _{c}}+\frac{1}{s-i\varDelta _{c}+\varGamma _{c}}\bigg ), \end{aligned}$$

(131)

and self-energy matrix element responsible for pure dephasing due to cavity coupling can be found as

$$\begin{aligned} \varSigma _{+-,+-}^{dR}(s)=\frac{-ig^{2}(\varOmega _{R}^{2}-\varDelta ^{2})}{2\varOmega _{R}^{2}} \left( \frac{1}{s+i(\varOmega _{R}'-\varDelta _{c})+\varGamma _{c}}+\frac{1}{s+i(\varOmega '_{R}+\varDelta _{c})+\varGamma _{c}}\right) . \end{aligned}$$

(132)

For a large band-width cavity, coupling to its radiative modes to system is treated under Markov approximation and above self-energies are replaced by their $s=-i(\varOmega _{R}'+\varDelta \omega )$ frequency parts, refer main text for details.

1.2 Appendix 4.2: Self-energy for phonon interaction

Similarly, we apply a continuum of modes for 3-d acoustic phonons (Krummheuer et al. 2002) with an exponential cut-off at $\epsilon =\epsilon _{c}$

$$\begin{aligned} \sum _{q}\lambda _{q}^{2}\rightarrow \alpha _{P} \int _{0}^{\infty }d\epsilon |\epsilon |^{3}e^{-|\epsilon |/\epsilon _{c}}, \end{aligned}$$

(133)

we obtain the expression for self-energy in Laplace transform as

$$\begin{aligned} \varSigma _{+-,+-}^{dP}(s)&=\frac{-i\alpha _{P}\varDelta ^{2}}{2\varOmega _{R}^{2}}\int _{0}^{\infty }d\epsilon | \epsilon |^{3}e^{\frac{-|\epsilon |}{\epsilon _{c}}}(2n_{B}(\epsilon )+1)\nonumber \\&\times \left( \frac{1}{s+i(\varOmega _{R}'-\epsilon )}+\frac{1}{s+i(\varOmega _{R}'+\epsilon )}\right) , \end{aligned}$$

(134)

where $\alpha _{P}=$ is the phonon coupling parameter in the units of ${\mathrm{freq.}}^{-2}$ and $n_{B}(\epsilon )$ is Bose function. On further simplification, above self-energy matrix element can be decomposed into real and imaginary parts after setting $s=-i\varDelta _{0}$, where $\varDelta _{0}$ is detuning of the probe from the pump laser frequency, as:

$$\begin{aligned} \varSigma _{+-,+-}^{dP}(s=-i\varDelta _{0})=\varDelta \omega _{P}(\varDelta _{0})-i\varGamma _{P}(\varDelta _{0}), \end{aligned}$$

(135)

where $\varDelta \omega _{P}(\varDelta _{0})={\mathrm{Re}}[\varSigma _{+-,+-}^{dP}(\varDelta _{0})]$ is the frequency-shift and $\varGamma _{P}(\varDelta _{0})=-{\mathrm{Im}}[\varSigma _{+-,+-}^{dP}(\varDelta _{0})]$ is the dephasing due to phonon interaction.

1.3 Appendix 4.3: Self-energy matrix element calculated exactly

In the previous section, we have computed the self-energy matrix element for phonon interaction to the second order in Born approximation. In this section, we will discuss an equation-of-motion method to find the phonon interaction self-energy for all orders in perturbed Liouvillian due to phonons $L_{Y}^{+}$ beyond Born approximation, and show that exact approach recovers the result obtained within Born approximation. Using the general form of superoperators matrices, the expression for self-energy matrix element due to phonons can be written in Laplace domain as:

$$\begin{aligned} \varSigma _{+-,+-}^{P}(s)=-i{\mathrm{Tr}}_{{\rm P}}\,L_{Y}^{+}\,\frac{1}{s+i(\varOmega _{R}'+L_{P}+L_{Y}^{+})}\,L_{Y}^{+}\,\rho _{P}(t_0), \end{aligned}$$

(136)

also in the time domain, we have

$$\begin{aligned} \varSigma _{+-,+-}^{P}(t)=-ie^{-i\varOmega _{R}'t} \underbrace{{\mathrm{Tr}}_{{\rm P}}\,\big [L_{Y}^{+}\,e^{-i(L_{P}+L_{Y}^{+})}\,L_{Y}^{+}\,\rho _{P}(t_0)\big ]}_{{\mathcal{C}}(t)}. \end{aligned}$$

(137)

On further simplification, one obtains

$$\begin{aligned} {\mathcal{C}}(t)&=2Tr_{P}\,\biggl [[X_{P},X_{P}(t)]_{+}\rho _{P}(0)\biggr ]\nonumber \\&=2\langle X_{P}X_{P}(t)\rangle +2\langle X_{P}(t)X_{P}\rangle , \end{aligned}$$

(138)

where

$$\begin{aligned} X_{P}(t)=e^{-i(L_{P}+L_{Y}^{+})t}X_{P}(0). \end{aligned}$$

(139)

Above expression gives rise to a differential equation

$$\begin{aligned} \dot{X}_{P}(t)=-i(L_{P}+L_{Y}^{+})X_{P}(t), \end{aligned}$$

(140)

which can be written as

$$\begin{aligned} \dot{X}_{P}(t)=-i(L_{Y}^{+}-L_{P})X_{P}(t). \end{aligned}$$

(141)

Introducing: ${\tilde{X}}_{P}(t)=e^{-iL_{P}t}X_{P}(t)$, we obtain an equation of motion for ${\tilde{X}}_{P}(t)$

$$\begin{aligned} \dot{{\tilde{X}}}_{P}(t)=-i[ X_{P}^{0}(t),{\tilde{X}}_{P}(t)]_{+} \end{aligned}$$

(142)

where

$$\begin{aligned} X_{P}^{0}(t)=\frac{\varDelta }{2\varOmega _{R}}\sum _{q}\lambda _{q}(b_{q}e^{i\omega _{q}t}+b_{q}^{\dag }e^{-i\omega _{q}t}). \end{aligned}$$

(143)

Solving for one q, the solution for ${\tilde{X}}_{P}(t)$ takes the form

$$\begin{aligned} {\tilde{X}}_{P,q}(t)=U_{q}(t){\tilde{X}}_{P,q}(0)W_{q}(t) \end{aligned}$$

(144)

where

$$\begin{aligned} {\tilde{X}}_{P}(t)&=\sum _{q}{\tilde{X}}_{P,q}(t) \end{aligned}$$

(145)

$$\begin{aligned} \dot{U}_{q}(t)&=-iX_{P,q}^{0}(t)U_{q}(t) \end{aligned}$$

(146)

$$\begin{aligned} \dot{W}_{q}(t)&=-iW_{q}(t)X_{P,q}^{0}(t). \end{aligned}$$

(147)

Using Eqs. (143) and (146), we have (for one q)

$$\begin{aligned} \dot{U}_{q}(t)=-i\lambda _{q}^{'}e^{-iH_{q}t}(b_{q}+b_{q}^{\dag })e^{iH_{q}t}U_{q}(t) \end{aligned}$$

(148)

where

$$\begin{aligned} \lambda _{q}^{'}&=\frac{\varDelta \lambda _{q}}{2\varOmega _{R}} \end{aligned}$$

(149)

$$\begin{aligned} H_{q}&=\omega _{q}b_{q}^{\dag }b_{q}. \end{aligned}$$

(150)

Introducing

$$\begin{aligned} {\tilde{U}}_{q}(t)=e^{iH_{q}t}U_{q}(t)\quad \Rightarrow U_{q}(t)=e^{-iH_{q}t}{\tilde{U}}_{q}(t) \end{aligned}$$

(151)

and taking the time derivative, one can find an expression

$$\begin{aligned} \dot{{\tilde{U}}}_{q}(t)=i[H_{q}- \lambda _{q}^{'}(b_{q}+b_{q}^{\dag })]{\tilde{U}}_{q}(t). \end{aligned}$$

(152)

Using the shifting operator $S_{q}=e^{\frac{\lambda _{q}^{'}}{\omega _{q}}(b_{q}-b_{q}^{\dag })}$ above expression can be written as

$$\begin{aligned} H_{q}-\lambda _{q}^{'}(b_{q}+b_{q}^{\dag })=S_{q}H_{q}S_{q}^{\dag }- \frac{2\lambda _{q}'^{2}}{\omega _{q}}, \end{aligned}$$

(153)

this implies

$$\begin{aligned} \dot{{\tilde{U}}}_{q}(t)=i\left( S_{q}H_{q}S_{q}^{\dag }- \frac{2\lambda _{q}'^{2}}{\omega _{q}}\right) {\tilde{U}}_{q}(t). \end{aligned}$$

(154)

Multiplying by $S_{q}^{\dag }$ on both sides

$$\begin{aligned} \frac{d(S_{q}^{\dag }{\tilde{U}}_{q}(t))}{dt}=i\left( H_{q}- \frac{2\lambda _{q}'^{2}}{\omega _{q}}\right) S_{q}^{\dag }{\tilde{U}}_{q}(t), \end{aligned}$$

(155)

solving for $S_{q}^{\dag }{\tilde{U}}_{q}(t)$

$$\begin{aligned} S_{q}^{\dag }{\tilde{U}}_{q}(t)=e^{iH_{q}t}S_{q}^{\dag }{\tilde{U}}_{q}(0) e^{-i\varDelta _{P,q}t};\quad \varDelta _{P,q}=\frac{2\lambda _{q}^{'2}}{\omega _{q}}, \end{aligned}$$

(156)

using $U_{q}(t)=e^{-iH_{q}t}{\tilde{U}}_{q}(t)$ and $U_{q}(0)=1$, we obtain

$$\begin{aligned} U_{q}(t)=e^{-iH_{q}t}S_{q}e^{iH_{q}t}S_{q}^{\dag }e^{-i\varDelta _{P,q}t}. \end{aligned}$$

(157)

Similarly, for $W_{q}(t)$

$$\begin{aligned} W_{q}(t)=S_{q}e^{-iH_{q}t}S_{q}^{\dag }e^{iH_{q}t}e^{i\varDelta _{P,q}t}. \end{aligned}$$

(158)

Substituting for $U_{q}(t)$ and $W_{q}(t)$, also using $U_{q}(0)=W_{q}(0)=1$, we obtain an expression for $X_{P,q}(t)$

$$\begin{aligned} X_{P,q}(t)=S_{q}e^{iH_{q}t}S_{q}^{\dag }X_{P,q}(0)S_{q}e^{-iH_{q}t}S_{q}^{\dag }. \end{aligned}$$

(159)

Recalling Eq. (142):

$$\begin{aligned} {\mathcal{C}}(t)=2\underbrace{\langle X_{P}X_{P}(t)\rangle }_{{\mathcal{C}}_{1}(t)}+2\underbrace{\langle X_{P}(t)X_{P}\rangle }_{{\mathcal{C}}_{2}(t)}. \end{aligned}$$

and substituting for $X_{P,q}(t)$, we have

$$\begin{aligned} {\mathcal{C}}_{1}(t)&={\mathrm {Tr}}_{P}\biggl [\sum _{q',q'''}X_{P,q'}(0) \biggl (\prod _{q''}S_{q''}e^{iH_{q''}t}S_{q''}^{\dag }{\tilde{X}}_{P,q'''}(0)\nonumber \\&\times S_{q''}e^{-iH_{q''}t}S_{q''}^{\dag }\biggr )\biggl ( \prod _{q}\rho _{P,q}(0)\biggr )\biggr ]. \end{aligned}$$

(160)

Above expression can be solved for $q=q'=q''=q'''$, because other modes do not contribution in the final trace and do not conserve the particle number for the different modes. On further simplification and doing some algebraic manipulation, one can obtain a simplified expression as

$$\begin{aligned} {\mathcal{C}}_{1}(t)=\sum _{q'=q'''}\prod _{q=q'=q''}{\mathrm {Tr}}_{P}\biggl [X_{P,q'}(0)\biggl (S_{q''}e^{iH_{q''}t}S_{q''}^{\dag }{\tilde{X}}_ {P,q'''}(0)S_{q''}e^{-iH_{q''}t}S_{q''}^{\dag }\biggr )\rho _{P,q}(0)\biggr ] \end{aligned}$$

(161)

and similarly for ${\mathcal{C}}_{2}(t)$

$$\begin{aligned} {\mathcal{C}}_{2}(t)=\sum _{q'=q'''}\prod _{q=q'=q''}{\mathrm {Tr}}_{P}\biggl [\biggl (S_{q''}e^{iH_{q''}t}S_{q''}^{\dag }{\tilde{X}}_{P,q'''}(0)S_{q''}e^ {-iH_{q''}t}S_{q''}^{\dag }\biggr )X_{P,q'}(0)\rho _{P,q}(0)\biggr ]. \end{aligned}$$

(162)

Considering the factor in parenthesis which is common in both expressions, we have

$$\begin{aligned} X_{P}(t)=\sum _{q=q'}\prod _{q}\biggl (S_{q}\overbrace{e^{iH_{q}t}\underbrace{S_{q}^{\dag }{\tilde{X}}_{P,q'}(0)S_{q}}_ {Z_{q}}e^{-iH_{q}t}}^{Y_{q}(t)}S_{q}^{\dag }\biggr ), \end{aligned}$$

(163)

where we have defined:

$$\begin{aligned} Y_{q}(t)=e^{iH_{q}t}Z_{q}e^{-iH_{q}t} \quad {\mathrm {and}} \qquad Z_{q}=S_{q}^{\dag }{\tilde{X}}_{P,q}(0)S_{q}. \end{aligned}$$

Using Baker–Campbell–Hausdorff formula:

$$\begin{aligned} Z_{q}&=S_{q}^{\dag }{\tilde{X}}_{P,q}(0)S_{q}\nonumber \\&=\lambda _{q}'(b_{q}+b_{q}^{\dag })-\frac{2\lambda _{q}'^{2}}{\omega _{q}}, \end{aligned}$$

(164)

similarly for $Y_{q}(t)$,

$$\begin{aligned} Y_{q}(t)&=e^{iH_{q}t}Z_{q}e^{-iH_{q}t}\nonumber \\&=\lambda _{q}'(b_{q}e^{-i\omega _{q}t}+b_{q}^{\dag }e^{i\omega _{q}t})- \frac{2\lambda _{q}'^{2}}{\omega _{q}}. \end{aligned}$$

(165)

Substituting for $Y_{q}(t)$ and $Z_{q}$ in the expression for $X_{P}(t)$, and then substituting for $X_{P}(t)$ in the expressions for ${\mathcal{C}}_{1}(t)$ and ${\mathcal{C}}_{2}(t)$, we perform the final trace to obtain the following expressions

$$\begin{aligned} {\mathcal{C}}_{1}(t)&=\frac{\varDelta }{4\varOmega _{R}}\sum _{q}\lambda _{q}^{2}(n_{q}\,e^{i\omega _{q}t}+(n_{q}+1)e^{-i\omega _{q}t}) \end{aligned}$$

(166)

$$\begin{aligned} {\mathcal{C}}_{2}(t)&=\frac{\varDelta }{4\varOmega _{R}}\sum _{q}\lambda _{q}^{2}(n_{q}\,e^{-i\omega _{q}t}+(n_{q}+1)e^{i\omega _{q}t}). \end{aligned}$$

(167)

After substituting the expressions for ${\mathcal{C}}_{1}(t)$ and ${\mathcal{C}}_{2}(t)$ in the expression for reduced self-energy matrix element and performing the final trace, we obtain

$$\begin{aligned} \varSigma _{+-,+-}^{P}(t)=\frac{-i\varDelta ^{2}\,e^{-i\varOmega _{R}'t}}{2\varOmega _{R}^{2}} \sum _{q}\lambda _{q}^{2}(2n_{q}+1)\,\cos (\omega _{q}t). \end{aligned}$$

(168)

Applying the continuum of modes and going back to Laplace domain,

$$\begin{aligned} \varSigma _{+-,+-}^{P}(s)&=\frac{-i\alpha _{P}\varDelta ^{2}}{2\varOmega _{R}^{2}} \int _{0}^{\infty }d\epsilon | \epsilon |^{3}e^{\frac{-|\epsilon |}{\epsilon _{c}}}(2n_{B}(\epsilon )+1)\nonumber \\&\times \left( \frac{1}{s+i(\varOmega _{R}'-\epsilon )}+\frac{1}{s+i(\varOmega _{R}'+\epsilon )}\right) , \end{aligned}$$

(169)

above expression recovers the result obtained for self-energy matrix element within Born approximation, which is Eq. (51) in the main text.

Appendix 5: Initial condition

In this section, we will discuss the stationary density matrix and find the initial condition for the operator $\varOmega (t)$ given in Eq. (33). The stationary density matrix $\bar{\rho }$ accounts for conditions that accumulate between the system and interactions in the time interval $t\in [t_{0} ,0]$. In this time interval, $[t_{0}, 0]$, the stationary density matrix is given by its value at $t=0$ and we can replace $\rho _{S}(t')\rightarrow \rho _{S}(t)$ in Eq. (42), to obtain

$$\begin{aligned} \dot{\rho }_{S}(t)=-iL_{S}\rho _{S}(t)-i\rho _{S}(t)\int _{t_{0}}^{t}dt'\varSigma _{S}(t-t'). \end{aligned}$$

(170)

Changing integration variable: $\tau =t-t'$ to obtain

$$\begin{aligned} \dot{\rho }_{S}(t)=-iL_{S}\rho _{S}(t)- i\rho _{S}(t)\int _{0}^{t-t_{0}}d\tau \varSigma _{S}(\tau ), \end{aligned}$$

(171)

one can extend the upper limit of above integration to infinity by setting $t_{0}\rightarrow -\infty$, and solving the differential equation to obtain

$$\begin{aligned} \rho _{S}(0)=e^{i[L_{S}+\varSigma _{S}(s=0)]t_0}\rho _{S}(t_0), \end{aligned}$$

(172)

and taking Laplace transform on both sides, we obtain

$$\begin{aligned} \frac{\rho _{S}(0)}{s}=\frac{1}{s+iL_{S}+i\varSigma _{S}(s=0)}\rho _{S}(t_{0}), \end{aligned}$$

(173)

here Laplace transform is defined as $f(s)=\int _{0}^{\infty }e^{-st}f(t)dt$. We choose an initial condition when exciton is in excited state $|a\rangle$ given by $\rho _{S}(t_{0})=|a\rangle \langle a|$, and evolves in the presence of pump laser. After performing a secular approximation and using the definition of stationary limit

$$\begin{aligned} \bar{\rho }_{S}=\lim \limits _{s\rightarrow 0^{+}}s\left( \frac{\rho _{S}(0)}{s}\right) , \end{aligned}$$

(174)

one can find all the elements of stationary density matrix operator as:

$$\begin{aligned} \bar{\rho }_{++}=\frac{-\varSigma _{++,--}(s=0)}{\varSigma _{z}(s=0)}, \end{aligned}$$

(175)

where $\varSigma _{z}=\varSigma _{++,++}-\varSigma _{++,--}$. Similarly,

$$\begin{aligned} \bar{\rho }_{--}&=\frac{\varSigma _{++,++}(s=0)}{\varSigma _{z}(s=0)}\end{aligned}$$

(176)

$$\begin{aligned} \bar{\rho }_{+-}&=\bar{\rho }_{-+}=0. \end{aligned}$$

(177)

Recalling, the initial condition for operator $\varOmega (t)$ given by Eq. (33):

$$\begin{aligned} \varOmega (0)=\bar{\rho }\sigma _{ab}, \end{aligned}$$

(178)

and matrix element of interest can be extracted after performing the trace over phonon and photon modes, and substituting for stationary density matrix element $\bar{\rho }_{++}$,

$$\begin{aligned} \varOmega _{+-}(0)=\frac{-{\mathbf{c}}^{2}\varSigma _{++,--}(s=0)}{\varSigma _{z}(s=0)}. \end{aligned}$$

(179)

Dynamics of the operator $\varOmega (t)$ is evaluated using Hamiltonian given by Eq. (18), with an initial condition given by above expression which is Eq. (69) in the main text.