Succinct Description and Efficient Simulation of Non-Markovian Open Quantum Systems

Abstract

Non-Markovian open quantum systems represent the most general dynamics when the quantum system is coupled with a bath environment. The quantum dynamics arising from many important applications are non-Markovian. Although for special cases, such as Hamiltonian evolution and Lindblad evolution, quantum simulation algorithms have been extensively studied, efficient quantum simulations for the dynamics of non-Markovian open quantum systems remain underexplored. The most immediate obstacle for studying such systems is the lack of a universal succinct description of their dynamics. In this work, we fulfill the gap of studying such dynamics by (1) providing a succinct representation of the dynamics of non-Markovian open quantum systems with quantifiable error, and (2) developing an efficient quantum algorithm for simulating such dynamics with cost ${\mathcal {O}}(t\, \textrm{polylog}(t/\epsilon ))$ for evolution time t and precision $\epsilon$. Our derivation of the succinct representation is based on stochastic Schrödinger equations, which could lead to new alternatives to deal with open quantum systems as well.

Appendices

Appendix A The Derivation of the Extended Stochastic Dynamics Eq. (35)

As a building block to approximate stationary Gaussian processes in the NMSSE (Eq. (30)), we consider the Ornstein-Uhlenbeck (OU) type of processes [67], expressed as the solution of the following linear SDEs,

$$\begin{aligned} i \dot{\zeta }_k = - d_k \zeta _k + \gamma _k \dot{w}_j(t), \end{aligned}$$

(A1)

for $k=1,2,\ldots , K$, with $K \ge M$. Here $\gamma _k\ge 0$ and $d_k$ is a complex number with positive imaginary part. In addition, each of these independent OU processes has an initial variance $\theta _k^2$,

$$\begin{aligned} {\mathbb {E}}[\zeta _k(0)^\dagger \zeta _k(0)] = \theta _k^2. \end{aligned}$$

If we pick $\gamma _k$, such that $\gamma _k^2= 2\text {Im}(d_k),$ then $\zeta _k(t)$ is a stationary Gaussian process with correlation,

$$\begin{aligned} {\mathbb {E}}[\zeta _j^\dagger (t) \zeta _k(t')]= \theta _j^2 \delta _{j,k} \exp \big ( -id_j^* (t- t') \big ). \end{aligned}$$

(A2)

We now construct an ansatz for approximating the bath correlation function. The case when $M=1$ has been thoroughly investigated in [54]. As a time correlation function, C(t) can be expressed in terms of its power spectrum, denoted here by $\widehat{C}(\omega ),$ as a Fourier integral,

$$\begin{aligned} C(t) = \int _{-\infty }^{\infty } \widehat{C}(\omega ) e^{-i\omega t} d\omega . \end{aligned}$$

(A3)

Known as the power spectrum, $\widehat{C}(\omega )$ is Hermitian and positive semidefinite. Hence it can be diagonalized using the transformation:

$$\begin{aligned} \widehat{C}(\omega ) = \sum _{j} \lambda _j(\omega )^2 {|{Q_j(\omega )}\rangle } {\langle {Q_j(\omega )}|}. \end{aligned}$$

Therefore, a direct numerical approximation of Eq. (A3) will certainly lead to an ansatz like Eq. (5).

Another alternative is to use a contour integral in the upper half plane, and the Cauchy residue theorem, reducing the Fourier integral in Eq. (A3) to a summation over poles [54]. This approach will also lead to the ansatz like Eq. (5).

To ensure the approximation accuracy, in practice, the ansatz in Eq. (5) is often obtained by a least-squares approach. In addition, we will consider the poles $d_k$ within a cut-off frequency $d_\text {max}$, i.e., $\left|{d_k}\right| \le d_\text {max}$.

Next we show that we can find an approximation of $\eta (t)$ that exactly satisfies the relation in Eq. (31), where C(t) is represented by Eq. (5). Specifically, we approximate the noise $\eta (t)$ using the OU processes $\zeta _k(t)$ from Eq. (A1),

$$\begin{aligned} \eta _\beta (t)= \sum _{k=1}^K \langle {Q_k}|\beta \rangle \theta _k \zeta _k(t). \end{aligned}$$

(A4)

In light of Eq. (A2), the approximation in Eq. (A4) corresponds to an approximation of the time correlation function in Eq. (5).

For the case of a single interaction term, where $M=1$ and the matrix C(t) is corresponding to a scalar function, this reduces to the standard approach using a sum of exponentials [54]. But Eq. (5) provides a more general scheme to handle multiple interaction terms.

Next we show how this can simplify the NMSSE in Eq. (30) when the bath correlation functions are expressed as Eq. (5). We first define the operators $T_j$ according to Eq. (36). This simplifies the NMSSE in Eq. (30) to,

$$\begin{aligned} i \partial _t \psi = {\hat{H}}_S \psi - i \lambda ^2\sum _{k=1}^K \int _0^t T^\dagger _k e^{-i ({\hat{H}}_S + d_k^* )\tau } T_k \psi (t-\tau ) d\tau +\lambda \sum _{j=1}^K \zeta _j(t) T_j \psi (t).\quad \end{aligned}$$

(A5)

Here the multiplication by $d_k^*$ represents the operator $d_k^*I_S$, where $I_S$ is the identity matrix.

Eq. (A5) still contains memory. But compared to the original NMSSE in Eq. (30), the correlation C(t) has been broken down to exponential functions, which can be combined with the unitary operator $e^{-i t {\hat{H}}_S}.$ In addition, the noise is expressed as the OU process $\zeta _j$’s, which can be treated using Itô calculus.

Next, we demonstrate how to embed the dynamics in Eq. (A5) into an extended, but Markovian, dynamics. The simple observation that motivated the Markovian embedding is that a convolution integral in time can be represented as the solution of a differential equation:

$$\begin{aligned} f(t)= \int _0^t \exp (-a(t-\tau )) g(\tau ) d\tau \quad \Longrightarrow f' = -a f + g, \; f(0)=0. \end{aligned}$$

(A6)

Here f will be regarded as an auxiliary variable, introduced to reduce the memory integral. Compared to computing the integral at every time step using direct quadrature formulas, it is much more efficient to solve the differential equation.

We now show that we can use the same idea to define auxiliary orbitals. Specifically, by inserting Eq. (5) in the SSE in Eq. (30), and by letting, for $k=1,2,\ldots , K,$

$$\begin{aligned} \chi _k^\text { I}(t) = \int _0^t e^{-i ({\hat{H}}_S + d_k^*) \tau } T_k \psi (t-\tau ) d\tau , \end{aligned}$$

we arrive at the equation,

$$\begin{aligned} i\partial _t \chi _k^\text { I}= (H_S + d_k^* ) \chi _k^\text { I}+ i \lambda T_k \psi (t). \end{aligned}$$

(A7)

This simplifies Eq. (A5) to,

$$\begin{aligned} i \partial _t \psi = {\hat{H}}_S \psi - i \lambda ^2 \sum _{j=1}^K T_j^\dagger \chi ^\text { I}_j + \lambda \sum _{j=1}^K T_j \psi (t) \zeta _j(t). \end{aligned}$$

(A8)

To incorporate the noise term, we define,

$$\begin{aligned} \chi ^\text { I\!I}_k(t)= i\psi (t) \zeta _k(t). \end{aligned}$$

(A9)

This reduces Eq. (A8) to,

$$\begin{aligned} i \partial _t \psi = {\hat{H}}_S \psi - i \lambda ^2 \sum _{j=1}^K T_j^\dagger \chi ^\text { I}_j -i \lambda \sum _{j=1}^K T_j \chi _j^\text { I\!I}(t). \end{aligned}$$

(A10)

It remains to derive a closed-from equation for Eq. (A9). Using the Itô’s formula, we obtain,

$$\begin{aligned} i\partial _t \chi ^\text { I\!I}_k = ({\hat{H}}_S - d_k) \chi ^\text { I\!I}_k +i \gamma _k \psi (t) \dot{w}_k + \lambda \zeta _k(t) T_k^\dagger \chi ^\text { I}_k + \lambda \zeta _k(t) T_k \chi ^\text { I\!I}_k. \end{aligned}$$

(A11)

When the coupling parameter $\lambda$ is sufficiently small, one can add or drop ${\mathcal {O}}(\lambda )$ terms, which will contribute to an ${\mathcal {O}}(\lambda ^2)$ error when substituted into Eq. (A10). If such an error is acceptable, by collecting equations, we have extended Schrödinger equations,

$$\begin{aligned} \left\{ \begin{array}{l} i \partial _t \psi =\displaystyle {\hat{H}}_S \psi - i \lambda \sum _{k=1}^K T_k^\dagger \chi ^\text { I}_k - i \lambda \sum _k T_k \chi ^\text { I\!I}_k, \\ \begin{array}{ll} i\partial _t \chi ^\text { I}_k =&{} ({\hat{H}}_S + d^*_k ) \chi ^\text { I}_k + i \lambda T_k \psi (t),\\ i\partial _t \chi ^\text { I\!I}_k = &{} ({\hat{H}}_S - d_k) \chi ^\text { I\!I}_k + i \lambda T_k^\dagger \psi (t) + i \gamma _k \psi (t) \dot{w}_k. \end{array} \qquad k=1,2,\ldots , K.\end{array} \right. \end{aligned}$$

(A12)

Rather than dropping ${\mathcal {O}}(\lambda )$ terms in Eq. (A11), one can continue such a procedure and incorporate the high-order terms. Specifically, noticing the similarity between the ${\mathcal {O}}(\lambda )$ terms in Eq. (A11) with Eq. (A9), we define,

$$\begin{aligned} \chi _k^\text { I\!I\!I}= i \zeta _k(t) \chi _k^\text { I}, \quad \chi _k^\text { I\!V}= i \zeta _k(t) \chi _k^\text { I\!I}. \end{aligned}$$

(A13)

By repeating the above procedure, one can derive similar equations for these auxiliary wave functions. These embedding steps yield the extended Schrödinger equations (ESE) Eq. (35).

The Proof of Lemma 8

Proof

Our proof will mainly target statement (4). The rest of the lemma will become self-evident throughout the proof. In light of the structure of the GQME in Eq. (44), it is enough to consider the case $K=1.$ In this case, $\Gamma$ can be viewed as a $5\times 5$ block matrix. We first write $V_1$ in a block matrix form,

$$\begin{aligned}V_1= \left( \begin{array}{cc} 0 &{} 0 \\ R_1 &{} 0 \end{array} \right) , \end{aligned}$$

where $R_1$ is a $3\times 3$ block matrix with diagonals $\sqrt{2\nu _1}I_S, \sqrt{2\nu _1}I_S$ and $2\sqrt{2\nu _1}I_S$. With direct calculations, we can show that the last term in Eq. (44) can be written as,

$$\begin{aligned}V_1\Gamma V_1^\dagger = \left( \begin{array}{cc} 0 &{} 0 \\ 0 &{} R_1 \Gamma _{0:2,0:2} R_1^\dagger \end{array} \right) . \end{aligned}$$

Here $\Gamma _{0:2,0:2}$ refers to the first $3\times 3$ sub-matrix of $\Gamma$.

We first look at the scenario when $\Gamma (0)$ is block diagonal. The zero blocks in $V_1\Gamma V_1^\dagger$, along with the observation that $H_0$ is block diagonal, imply that the off-diagonals do not change. For the diagonal blocks of $\Gamma (t)$, we first have,

$$\begin{aligned} \begin{aligned} \Gamma _{0,0}(t) =&U_S(t) \rho _S(0) U_S(t)^\dagger ,\\ \Gamma _{1,1}(t)=&\exp (-it(-d_1+d_1^*) ) U_S(t) \Gamma _{1,1}(0) U_S(t)^\dagger . \end{aligned} \end{aligned}$$

(B14)

As a result, these two blocks have norms given respectively by,

$$\begin{aligned} \Vert \Gamma _{0,0}(t) \Vert = \Vert \Gamma _{0,0}(0) \Vert , \quad \Vert \Gamma _{1,1}(t) \Vert = \Vert \Gamma _{1,1}(0) \Vert e^{-2\nu _1 t}. \end{aligned}$$

The next three block diagonals will pick up non-homogeneous terms. For instance, we have,

$$\begin{aligned} \partial _t \Gamma _{2,2}= -i[H_s - d_1, \Gamma _{2,2}] + 2 \nu _1 \Gamma _{0,0}. \end{aligned}$$

Using the variation-of-constant formula, we have,

$$\begin{aligned} \Gamma _{2,2}(t) = e^{-2\nu _1 t} U_S(t) \Gamma _{2,2}(0) U_S(t)^\dagger +2 \nu _1 \int _0^t e^{-2\nu _1 \tau } U_S(\tau ) \Gamma _{0,0}(t-\tau ) U_S(\tau )^\dagger d\tau .\nonumber \\ \end{aligned}$$

(B15)

Also by noticing that $\Vert \Gamma _{1,1}(t)\Vert$ is constant in time, the diagonal block $\Gamma _{3,3}$ can be bounded directly as,

$$\begin{aligned} \Vert \Gamma _{2,2}(t)\Vert \le \Vert \Gamma _{2,2}(0)\Vert \exp (-2\nu _1 t) + \Vert \Gamma _{0,0}(0)\Vert (1 - e^{-2\nu _1 t}). \end{aligned}$$

(B16)

The right hand side remains bounded for all time. Similarly, the next diagonal block can be expressed as,

$$\begin{aligned} \begin{aligned} \Gamma _{3,3}(t)&=\exp (-4\nu _1 t) U_S(t) \Gamma _{3,3}(0) U_S(t)^\dagger \\&+2 \nu _1 \int _0^t \exp (-4\nu _1 t) U_S(\tau ) \Gamma _{1,1}(t-\tau ) U_S(\tau )^\dagger d\tau . \end{aligned} \end{aligned}$$

(B17)

Notice that $\Vert \Gamma _{1,1}(t)\Vert$ is proportional to $\exp (-2\nu _1 t)$. Essentially, what leads to the boundedness of the solution is the fact that there is no secular term, implying that $\Vert \Gamma _{3,3}(t)\Vert$ follows a similar bound as $\Vert \Gamma _{2,2}(t)\Vert _2$. The estimate of $\Vert \Gamma _{4,4}(t)\Vert$ follows the same steps.

We now turn to the off-diagonal blocks. By direct calculations, we have, for $j=0,1$, $k=1, 2, 3, 4$, and $k>j$

$$\begin{aligned} \partial _t \Gamma _{j,k} = -i \big (H_S \Gamma _{j,k}-\Gamma _{j,k}(H_s +d_1) \big ), \end{aligned}$$

which yields,

$$\begin{aligned} \Gamma _{j,k}(t) = \exp (-i d_1^* t) U_S(t)\Gamma _{j,k}(0) U_S^{\dagger }(t) \Longrightarrow \Vert \Gamma _{j,k}(t)\Vert = \Vert \Gamma _{j,k}(0) \Vert \exp (-\nu _1 t).\nonumber \\ \end{aligned}$$

(B18)

For the remaining off-diagonal entries, we will check $\Gamma _{2,3}$ as an example. It follows the equation,

$$\begin{aligned} \partial _t \Gamma _{2,3}= -i \big ((H_s-d_1)\Gamma _{2,3} - \Gamma _{2,3} (H_s + d_1 + d_1^*)\big ) + 2\nu _1 \Gamma _{0,1}. \end{aligned}$$

This implies that,

$$\begin{aligned} \Vert \Gamma _{3,4}(t) \Vert \le \Vert \Gamma _{3,4}(0) \Vert e^{-3\nu _1 t} + \Vert \Gamma _{1,2}(0) \Vert 2 e^{-2\nu _1 t}(1 - e^{-\nu _1 t}). \end{aligned}$$

We also see from these calculations that these off-diagonal blocks will become zero if the initial matrix $\Gamma (0)$ is block diagonal.

By examining the block entries of $\Gamma (t)$, we have shown the boundedness of the solution stated in Lemma 8 for all time. $\square$

Appendix C The Proof of Eq. (9)

Proof

We will prove the asymptotic bound using an expansion of the Eq. (27). More specifically, we write the total density matrix in terms of powers of $\lambda ,$

$$\begin{aligned} \rho (t) = \rho ^{(0)}(t) + \lambda \rho ^{(1)}(t) + \lambda ^2 \rho ^{(2)}(t) + {\mathcal {O}}(\lambda ^3). \end{aligned}$$

(C19)

By taking a partial trace over the bath space, we obtain a similar expansion for $\rho _S:$

$$\begin{aligned} \rho _S(t) = \rho _S^{(0)}(t) + \lambda \rho _S^{(1)}(t) + \lambda ^2 \rho _S^{(2)}(t) + {\mathcal {O}}(\lambda ^3). \end{aligned}$$

(C20)

By inserting Eq. (C19) into Eq. (27) and separate terms of different order, we arrive at

$$\begin{aligned} \begin{aligned} i\partial _t \rho ^{(0)}&= [H_S \otimes I_B + I_S \otimes H_B, \rho ^{(0)} ], \;\; \rho ^{(0)}(0)=\rho (0), \\ i\partial _t \rho ^{(1)}&= [H_S \otimes I_B + I_S \otimes H_B, \rho ^{(1)} ] + \sum _{\alpha =1}^M [ S_\alpha \otimes B_\alpha , \rho ^{(0)}], \;\; \rho ^{(1)}(0) =0, \\ i\partial _t \rho ^{(2)}&= [H_S \otimes I_B + I_S \otimes H_B, \rho ^{(2)} ]+ \sum _{\alpha =1}^M [ S_\alpha \otimes B_\alpha , \rho ^{(1)}], \;\; \rho ^{(2)}(0)=0. \end{aligned} \end{aligned}$$

(C21)

Within this expansion, the dynamics of $\rho ^{(0)}$ contains no coupling. Let

$$\begin{aligned} U(t)= U_S(t) \otimes U_B(t), \quad U_S= \exp \left( -i t H_S\right) , \; U_B(t)= \exp \left( -i t H_B\right) , \end{aligned}$$

be the unitary operators. Then we have,

$$\begin{aligned} \rho ^{(0)}(t) = U(t) \rho ^{(0)}(0) U(t)^\dagger . \end{aligned}$$

(C22)

Since all the operators on the right hand side are in tensor product forms, $\rho ^{(0)}(t)$ remains a tensor product:

$$\begin{aligned} \rho ^{(0)}(t) = \rho _S^{(0)}(t) \otimes \rho _B. \end{aligned}$$

(C23)

Here $\rho _S^{(0)}(t) = U_S(t) \rho _S(0) U_S(t)^\dagger .$ Meanwhile, the matrix $\rho _B$ stays because it commutes with $H_B.$

The term $\rho ^{(1)}$ can be expressed using the variation-of-constant formula,

$$\begin{aligned} \begin{aligned} \rho ^{(1)}(t) =&-i \sum _\alpha \int _0^t U(t-t') S_\alpha \otimes B_\alpha \rho ^{(0)}(t') U(t-t')^\dagger dt' \\ {}&+ i \sum _\alpha \int _0^t U(t-t') \rho ^{(0)}(t') S_\alpha \otimes B_\alpha U(t-t')^\dagger dt'. \end{aligned} \end{aligned}$$

In light of Eq. (C23), we can make the same observation that $\rho ^{(1)}(t)$ consists of terms that are tensor products. By following standard notations [1], i.e.,

$$\begin{aligned} S_\alpha (t)= U_S(t)^\dagger S_\alpha U_S(t), \quad B_\alpha (t)= U_B(t)^\dagger B_\alpha U_B(t), \end{aligned}$$

(C24)

we can simplify $\rho ^{(1)}(t)$ as follows,

$$\begin{aligned} \begin{aligned} \rho ^{(1)}(t) =&-i \sum _\alpha \int _0^t S_\alpha (t'-t)\rho _S^{(0)}(t) \otimes B_\alpha (t'-t) \rho _B dt' \\&+ i \sum _\alpha \int _0^t \rho _S^{(0)}(t) S_\alpha (t'-t) \otimes B_\alpha (t'-t) \rho _B dt'. \end{aligned} \end{aligned}$$

Since $\rho _B$ commutes with $H_B$, it commutes with $U_B.$ Therefore,

$$\begin{aligned} \textrm{tr}(B_\alpha (t'-t) \rho _B)= \textrm{tr}(B_\alpha \rho _B) =0, \end{aligned}$$

which shows that

$$\begin{aligned} \textrm{tr}_B \left( \rho ^{(1)}(t) \right) =0. \end{aligned}$$

Therefore, $\rho _S^{(1)}(t)=0$. The correction to $\rho _S$ comes from $\rho _S^{(2)}(t),$ which is similarly expressed as,

$$\begin{aligned} \begin{aligned} \rho ^{(2)}(t) \!=&-i \sum _\alpha \int _0^t U(t-t') S_\alpha \otimes B_\alpha \rho ^{(1)}(t') U(t-t')^\dagger dt' \\ {}&+ i \sum _\alpha \int _0^t U(t-t') \rho ^{(1)}(t') S_\alpha \otimes B_\alpha U(t-t')^\dagger dt',\\ =&\! -\! \sum _\alpha \!\sum _\beta \! \int _0^t \int _0^{t'} S_\alpha (t'-t) S_\beta (\tau -t) \rho _S^{(0)}(t) \!\otimes \! B_\alpha (t'-t) B_\beta (\tau -t) \rho _B d\tau dt',\\&\!+\! \sum _\alpha \!\sum _\beta \! \int _0^t \int _0^{t'} S_\beta (\tau -t) \rho _S^{(0)}(t) S_\alpha (t'-t) \!\otimes \! B_\beta (\tau -t) \rho _B B_\alpha (t'-t) d\tau dt',\\&\!+\! \sum _\alpha \!\sum _\beta \! \int _0^t \int _0^{t'} S_\alpha (t'-t) \rho _S^{(0)}(t) S_\beta (\tau -t) \!\otimes \! B_\alpha (t'-t) \rho _B B_\beta (\tau -t) d\tau dt',\\&\!-\! \sum _\alpha \!\sum _\beta \! \int _0^t \int _0^{t'} \rho _S^{(0)}(t) S_\beta (\tau -t) S_\alpha (t'-t) \!\otimes \! \rho _B B_\beta (\tau -t) B_\alpha (t'-t) d\tau dt'.\\ \end{aligned} \end{aligned}$$

Invoking the bath correlation function,

$$\begin{aligned} C_{\alpha ,\beta }(t)= \textrm{tr}(B_\alpha (t) B_\beta \rho _B), \end{aligned}$$

(C25)

we arrive at an expansion of $\rho _S(t)$ up to ${{\mathcal {O}}\left( \lambda ^2\right) }$,

$$\begin{aligned} \begin{aligned} \rho _S(t)&= \rho _S^{(0)}(t) - \lambda ^2 \sum _\alpha \sum _\beta \int _0^t \int _0^{t'} S_\alpha (t'-t) S_\beta (\tau -t) \rho _S^{(0)}(t) C_{\alpha ,\beta }(t'-\tau ) d\tau dt'\\&\quad + \lambda ^2 \sum _\alpha \sum _\beta \int _0^t \int _0^{t'} S_\beta (\tau -t) \rho _S^{(0)}(t) S_\alpha (t'-t) C_{\alpha ,\beta }(t'-\tau ) d\tau dt' \\&\quad + \lambda ^2 \sum _\alpha \sum _\beta \int _0^t \int _0^{t'} S_\alpha (t'-t) \rho _S^{(0)}(t) S_\beta (\tau -t) C_{\beta ,\alpha }(t'-\tau )^* d\tau dt' \\&\quad - \lambda ^2 \sum _\alpha \sum _\beta \int _0^t \int _0^{t'} \rho _S^{(0)}(t) S_\beta (\tau -t) S_\alpha (t'-t) C_{\alpha ,\beta }(t'-\tau )^* d\tau dt' \\&\quad + {\mathcal {O}}(\lambda ^3). \end{aligned} \end{aligned}$$

(C26)

Here we have used the property of the bath correlation function: $C_{\alpha ,\beta }(t) = C_{\beta ,\alpha }(-t)^*.$ Now we incorporate the function form of the bath correlation function in Eq. (5). We find that,

$$\begin{aligned} \begin{aligned} \rho _S(t) = \rho _S^{(0)}(t)&- \lambda ^2 \sum _k \int _0^t \int _0^{t'} T_k(t'-t) T_k(\tau -t) \rho _S^{(0)}(t) e^{-i(t'-\tau )d_k t } d\tau dt'\\&+ \lambda ^2 \sum _k \int _0^t \int _0^{t'} T_k(\tau -t) \rho _S^{(0)}(t) T_k(t'-t) e^{-i(t'-\tau )d_k t } d\tau dt' \\&+ \lambda ^2 \sum _k \int _0^t \int _0^{t'} T_k(t'-t) \rho _S^{(0)}(t) T_k(\tau -t) e^{i(t'-\tau )d_k t }d\tau dt' \\&- \lambda ^2 \sum _k \int _0^t \int _0^{t'} \rho _S^{(0)}(t) T_k(\tau -t) T_k(t'-t) e^{i(t'-\tau )d_k t } d\tau dt' \\&+ {\mathcal {O}}(\lambda ^3). \end{aligned} \end{aligned}$$

(C27)

Now we show that the GQME in Eq. (44) has an asymptotic expansion that is consistent with Eq. (C26). Expanding $\Gamma$ as,

$$\begin{aligned} \Gamma (t)= \Gamma ^{(0)}(t) + \lambda \Gamma ^{(1)}(t) + \lambda ^2 \Gamma ^{(2)}(t) + {\mathcal {O}}(\lambda ^3), \end{aligned}$$

and substituting it into Eq. (44), one gets,

$$\begin{aligned} \begin{aligned} \Gamma ^{(0)}(t)&= \exp \left( t{\mathcal {L}}_0 \right) \Gamma (0), \\ \Gamma ^{(1)}(t)&= -i \int _0^t \exp \left( (t-t'){\mathcal {L}}_0 \right) [H_1, \Gamma ^{(0)}(t')] dt',\\ \Gamma ^{(2)}(t)&= -i \int _0^t \exp \left( (t-t'){\mathcal {L}}_0 \right) [H_1, \Gamma ^{(1)}(t')] dt'. \end{aligned} \end{aligned}$$

(C28)

The leading term $\Gamma ^{(0)}(t)$ has been shown to be a block diagonal matrix in the previous section. The first diagonal block is precisely $\rho _S^{(0)}(t)$, which is consistent with the ${{\mathcal {O}}\left( 1\right) }$ term in Eq. (C26). To examine $\Gamma ^{(1)}(t)$, we first notice that the commutator in the integral has the following structure,

$$\begin{aligned} \Xi ^{(0)}(t') = [H_1, \Gamma ^{(0)}(t')]= i \left[ \begin{array}{cccccc} 0 &{} \Xi _{0,1}^{(0)}(t') &{}\Xi _{0,2}^{(0)}(t') &{} 0&{} 0 &{} \cdots \\ \Xi _{1,0}^{(0)}(t') &{} 0 &{}0 &{} 0&{} 0 &{} \cdots \\ \Xi _{2,1}^{(0)}(t') &{} 0 &{}0 &{} \Xi _{3,4}^{(0)}(t') &{} \Xi _{3,5}^{(0)}(t') &{} \cdots \\ 0 &{} 0 &{} \Xi _{4,3}^{(0)}(t') &{} 0 &{}0 &{} \cdots \\ 0 &{} 0 &{} \Xi _{5,3}^{(0)}(t') &{} 0 &{}0 &{} \cdots \\ \vdots &{} \vdots &{} \vdots &{}\vdots &{} \vdots &{} \ddots \\ \end{array}\right] .\nonumber \\ \end{aligned}$$

(C29)

Here we highlighted the leading $5\times 5$ submatrix and the zero blocks within it. This is enough for the purpose of the proof.

By inspecting the solutions that correspond to $\exp \left( t{\mathcal {L}}_0 \right)$, we find that the first diagonal block of $\int _0^t \exp \left( (t-t'){\mathcal {L}}_0 \right) \Xi ^{(0)}(t')$ is zero. Therefore, $\Gamma ^{(1)}$ has no contribution to the density matrix $\rho _S$, i.e., there is no ${{\mathcal {O}}\left( \lambda \right) }$ term. This is consistent with Eq. (C26).

To proceed further, we have to identify the nonzero blocks in Eq. (C29). With direct calculations, we have,

$$\begin{aligned} \begin{aligned} \Xi _{0,4k-3}^{(0)}(t) =&- \Gamma _{1,1}^{(0)}(t) T_j = - \rho _{S}^{(0)}(t) T_j, \\ \Xi _{0,4k-2}^{(0)}(t) =&T_k \Gamma _{4k-2,4k-2}^{(0)}(t). \end{aligned} \end{aligned}$$

From Eq. (C28), we can extract the equation,

$$\begin{aligned} i \partial _t \Gamma _{0,4k-3}^{(1)} = H_S\Gamma _{0,4k-3}^{(1)} - \Gamma _{0,4k-3}^{(1)} (H_S + d_k). \end{aligned}$$

Combining the two equations above, we obtain,

$$\begin{aligned} \begin{aligned} \Gamma _{0,4k-3}^{(1)}(t')&= \int _0^{t'} U_S(t'-t'') \rho _S^{(0)}(t'') T_k U_S(t'-t'')^\dagger e^{i (t'-t'') d_k} dt''\\&= \int _0^{t'} \rho _S^{(0)}(t') T_k(t''-t') e^{i (t'-t'') d_k} dt''. \end{aligned} \end{aligned}$$

Again using the solution properties associated with the superoperator $\exp \left( t{\mathcal {L}}_0 \right)$, we have that the first block of $\Gamma ^{(1)}$ is given by,

$$\begin{aligned} \begin{aligned}&- i \int _0^t U_S(t-t') \Xi ^{(1)}(t') U_S(t-t')^\dagger dt' \\ =&\sum _k \int _0^t U_S(t-t') \big ( [\Gamma _{0,4k-3}(t'), T_k] + [\Gamma _{0,4k-2}(t'), T_k^\dagger ] \big ) U_S(t-t')^\dagger dt'. \end{aligned} \end{aligned}$$

Similar to Eq. (C27), we have also obtained four terms after expanding the commutators. Let us examine the first integral term,

$$\begin{aligned} \begin{aligned}&\sum _k \int _0^t \int _0^{t'} U_S(t-t') \rho _S^{(0)}(t') T_k(t''-t') T_k e^{i (t'-t'') d_k} U_S(t-t')^\dagger ) dt''\\&= \sum _k \int _0^t \int _0^{t'} \rho _S^{(0)}(t') T_k(t''-t) T_k(t'-t) e^{i (t'-t'') d_k} dt'' dt'. \end{aligned} \end{aligned}$$

This is the same as the first last integral in Eq. (C27). The rest of the integrals can be similarly verified. $\square$

Appendix D The Proof of Eq. (10)

Proof

In Lemma 8, we have proved a bound for the case when $\lambda =0.$ Denote the solution by $\Gamma _0(t)$, and the solution operator by $\exp (t{\mathcal {L}}_0)$. Therefore, the GQME in Eq. (44) can be written in a perturbation form,

$$\begin{aligned} \partial _t \Gamma = {\mathcal {L}}_0 \Gamma -i \lambda (H_1 \Gamma - \Gamma H_1^\dagger ). \end{aligned}$$

(D30)

The solution can be recast in an integral form,

$$\begin{aligned} \Gamma (t) = \Gamma _0(t) - i \lambda \int _0^t \exp \left( (t-\tau ) {\mathcal {L}}_0\right) (H_1 \Gamma - \Gamma H_1^\dagger ) d\tau . \end{aligned}$$

(D31)

From Lemma 7, the super-operator $\exp (t{\mathcal {L}}_0)$ is bounded. Therefore, we have,

$$\begin{aligned} \Vert \Gamma (t) - \Gamma _0(t) \Vert \le 2\lambda C \Vert H_1\Vert \int _0^t \Vert \Gamma (t) \Vert dt. \end{aligned}$$

As a result, the bound can be obtained by directly using Gronwall’s inequality.

Since $\Gamma _0(t)$ is block diagonal, the above inequality also shows that the off-diagonal blocks of $\Gamma (t)$ are of order $\lambda .$ Finally, the bounds for the trace can be verified from Eqs. (B14), (B15) and (B17) in the proof of Lemma 8. $\square$

Appendix E The Proof of Eq. (11)

Proof

From Eq. (D30), we may take the trace.

$$\begin{aligned} \textrm{tr}(\Gamma (t) ) - \textrm{tr}(\Gamma ^{(0)}(t) ) = - i \lambda \int _0^t \textrm{tr}( \Sigma (t,\tau ) ) d\tau , \end{aligned}$$

(E32)

where,

$$\begin{aligned} \Sigma (t,\tau ) =\exp (t{\mathcal {L}}_0) \Xi , \end{aligned}$$

with $\Xi$ from Eq. (52). By the definition of $\exp (t{\mathcal {L}}_0)$, $\Sigma (t,\tau )$ is the solution of the equation,

$$\begin{aligned} \partial _t \Sigma = {\mathcal {L}}_0 \Sigma , \quad \Sigma (\tau , \tau )= (H_1 \Gamma - \Gamma H_1^\dagger ). \end{aligned}$$

Using the property in Eq. (59) of the super-operator $\exp (t{\mathcal {L}}_0)$ in Eq. (10), we obtain the bound,

$$\begin{aligned} \mid \textrm{tr}\big ( \Sigma (t,\tau ) \big )\mid \le 3K \mid \textrm{tr}\big ( \Sigma _{0,0} (\tau , \tau ) \big )\mid + \sum _{k>0} \mid \textrm{tr}\big ( \Sigma _{k,k} (\tau , \tau ) \big )\mid . \end{aligned}$$

We now invoke the estimate in Lemma 7. The trace of each diagonal block of $\Sigma (\tau , \tau )$ is bounded by the off-diagonal blocks of $\Gamma (t)$, which is of order $\lambda$. Namely, there exists a constant, such that,

$$\begin{aligned} \mid \textrm{tr}\big ( \Sigma (\tau ,\tau ) \big )\mid \le C \lambda . \end{aligned}$$

Collecting terms, we have,

$$\begin{aligned} \mid \textrm{tr}(\Gamma (t) ) - \textrm{tr}(\Gamma ^{(0)}(t) )\mid \le Ct \lambda ^2. \end{aligned}$$

Alternatively, one can start with Eq. (D30), and apply the formula again to replace the $\Gamma (t)$ in the integral:

$$\begin{aligned} \begin{aligned} \Gamma (t) - \Gamma ^{(0)}(t) =&- i \lambda \int _0^t \exp \left( (t-\tau ) {\mathcal {L}}_0\right) (H_1 \Gamma ^{(0)}(\tau ) - \Gamma ^{(0)}(\tau ) H_1^\dagger ) d\tau \\&- \lambda ^2 \int _0^t \int _0^\tau \exp \left( (t-\tau ) {\mathcal {L}}_0\right) (H_1 \Sigma (\tau ,s) - \Sigma (\tau ,s) H_1^\dagger ) ds d\tau + {\mathcal {O}}(\lambda ^3). \end{aligned} \end{aligned}$$

Here $\Sigma (\tau ,s) = \exp \left( (\tau -s) {\mathcal {L}}_0\right) \Xi _0$, with

$$\begin{aligned} \Xi ^{(0)}=H_1 \Gamma _0(s) - \Gamma _0(s) H_1^\dagger . \end{aligned}$$

(E33)

For the ${\mathcal {O}}(\lambda )$ term, we notice that $\Gamma {(0)}$ is block diagonal, and so in light of Lemma 7, the matrix $\Xi ^{(0)}$ has zero trace. This implies that,

$$\begin{aligned} \textrm{tr}\big ( \Gamma (t) ) - \Gamma ^{(0)}(t) \big ) = {\mathcal {O}}(\lambda ^2). \end{aligned}$$

For the ${\mathcal {O}}(\lambda ^2)$ term, from Eq. (55), we have, the trace of the first block of $H_1 \Sigma (\tau ,s) - \Sigma (\tau ,s) H_1^\dagger$ is given by,

$$\begin{aligned} \sum _{k=1}^K \textrm{tr}\Big ( (\Sigma _{0,4k-3} - \Sigma _{4k-2,0}) T_k \Big )+ \sum _{k=1}^K \textrm{tr}\Big ( T_k^\dagger ( \Sigma _{0,4k-2} - \Sigma _{4k-3,0}) \Big ). \end{aligned}$$

(E34)

Meanwhile, from the proof of Lemma 8, we see that the superoperator does not change the off-diagonal blocks in the first row and column. Thus, $\Sigma _{0,j}(\tau , s)= \Sigma _{0,j}(s, s) = \Xi _{0,j}^{(0)}(s)$.

With direct matrix multiplications, we find that,

$$\begin{aligned} \Xi _{4k-3,0}= -T_k \Gamma _{4k-3,4k-3} - \Gamma _{0,0} T_k^\dagger , \quad \Xi _{4k-2,0} = T_k \Gamma _{4k-2,4k-2}. \end{aligned}$$

From the proof of Lemma 8, we also have $\Gamma _{4k-3,4k-3}(t)=0.$ Combining these steps, we find that the trace in Eq. (E34) is zero. Therefore, we have

$$\begin{aligned} \textrm{tr}(\rho _S(t)) = 1 + {\mathcal {O}}(\lambda ^3). \end{aligned}$$

$\square$

Appendix F The Proof of Lemma 15

Proof

We use an intermediate superoperator ${\mathcal {I}}+\delta {\mathcal {K}}$. Assume the Hilbert space ${\mathcal {K}}$ is acting on has dimension N. Consider a Hilbert space of arbitrary dimension $N'$. For any operator Q acting on ${\mathbb {C}}^{N}\otimes {\mathbb {C}}^{N'}$ with $\left\Vert {Q}\right\Vert _1 = 1$, we have

$$\begin{aligned} \begin{aligned}&\left\Vert {({\mathcal {M}}_{\delta }\otimes {\mathcal {I}}_{N'} - ({\mathcal {I}}_{N} + \delta {\mathcal {K}})\otimes {\mathcal {I}}_{N'})(Q)}\right\Vert _1 \\&= \left\Vert {\sum _{j=0}^m(A_j\otimes I)Q(A_j\otimes I)^{\dag } - (Q + \delta ({\mathcal {K}}\otimes {\mathcal {I}}_{N'})(Q)}\right\Vert _1 \\&= \left\Vert {\delta ^2(H\otimes I)Q(H^{\dag }\otimes I)}\right\Vert _1 \\&\le \left\Vert {H\otimes I}\right\Vert ^2\\&\le (\delta \Lambda )^2. \end{aligned} \end{aligned}$$

(F35)

Now, we have

$$\begin{aligned} \left\Vert {({\mathcal {M}}_{\delta } - ({\mathcal {I}}_{N} + \delta {\mathcal {K}})}\right\Vert _{\diamond } \le (\delta \Lambda )^2. \end{aligned}$$

(F36)

To bound the distance between ${\mathcal {I}}+\delta {\mathcal {K}}$ and $e^{\delta {\mathcal {K}}}$, we assume $0 \le \delta \left\Vert {{\mathcal {K}}}\right\Vert _{\diamond } \le 1$. Consider any X such that $\left\Vert {X}\right\Vert _1 \le 1$, we have

$$\begin{aligned} \left\Vert {(e^{\delta {\mathcal {K}}} - ({\mathcal {I}}+\delta {\mathcal {K}}))(X)}\right\Vert _1&= \left\Vert {\sum _{s=2}^{\infty }\frac{\delta ^s}{s!}{\mathcal {K}}^s(X)}\right\Vert _1 \le \sum _{s=2}^{\infty }\frac{\delta ^s}{s!}\left\Vert {{\mathcal {K}}^s(X)}\right\Vert _1 \end{aligned}$$

(F37)

$$\begin{aligned}&\le \sum _{s=2}^{\infty }\frac{\delta ^s}{s!}\left\Vert {{\mathcal {K}}(X)}\right\Vert _1^s \le (\delta \left\Vert {{\mathcal {K}}(X)}\right\Vert _1)^2 \le (\delta \left\Vert {{\mathcal {K}}}\right\Vert _1)^2, \end{aligned}$$

(F38)

where the penultimate inequality follows from the fact that $e^z - (1+z) \le z^2$ when $0 \le z \le 1$.

Now, we extend this bound to the diamond norm. Note that, for two Hilbert spaces ${\mathbb {C}}^{N}$ and ${\mathbb {C}}^{N'}$,

$$\begin{aligned} (e^{\delta {\mathcal {K}}} - ({\mathcal {I}}_{N}+\delta {\mathcal {K}}))\otimes {\mathcal {I}}_{N'} = e^{\delta ({\mathcal {K}}\otimes {\mathcal {I}}_{N'})} - ({\mathcal {I}}_{N \times N'} + \delta ({\mathcal {K}} \otimes {\mathcal {I}}_{N'})). \end{aligned}$$

(F39)

When $N = N'$, we have that $\left\Vert {{\mathcal {K}}\otimes {\mathcal {I}}_{N'}}\right\Vert _1 = \left\Vert {{\mathcal {K}}}\right\Vert _{\diamond }$. This implies that

$$\begin{aligned} \left\Vert {(e^{\delta {\mathcal {K}}} - ({\mathcal {I}}_{N'}+\delta {\mathcal {K}})}\right\Vert _{\diamond }&= \left\Vert {\left( e^{\delta {\mathcal {K}}} - ({\mathcal {I}}_{N}+\delta {\mathcal {K}})\right) \otimes {\mathcal {I}}_{N'}}\right\Vert _1 \end{aligned}$$

(F40)

$$\begin{aligned}&= \left\Vert {e^{\delta ({\mathcal {K}}\otimes {\mathcal {I}}_{N'})} - ({\mathcal {I}}_{N\times N'}+\delta ({\mathcal {K}}\otimes {\mathcal {I}}_{N'}))}\right\Vert _1 \end{aligned}$$

(F41)

$$\begin{aligned}&\le (\delta \left\Vert {{\mathcal {K}}\otimes {\mathcal {I}}_{N'}}\right\Vert _1)^2 \end{aligned}$$

(F42)

$$\begin{aligned}&\le (\delta \left\Vert {{\mathcal {K}}}\right\Vert _{\diamond })^2. \end{aligned}$$

(F43)

To see the relationship between $\left\Vert {{\mathcal {L}}}\right\Vert _{\diamond }$ and $\left\Vert {{\mathcal {K}}}\right\Vert _1$, first observe that $\left\Vert {{\mathcal {K}}}\right\Vert _1 \le 2\Lambda$. Then using the fact the $\left\Vert {M\otimes I}\right\Vert = \left\Vert {M}\right\Vert$ for all M, it follows that $\left\Vert {{\mathcal {K}}}\right\Vert _{\diamond } \le 2\Lambda$. Together with Eq. (F43), we have

$$\begin{aligned} \left\Vert {(e^{\delta {\mathcal {K}}} - ({\mathcal {I}}_{N'}+\delta {\mathcal {K}})}\right\Vert _{\diamond } \le (2\delta \Lambda )^2. \end{aligned}$$

(F44)

This result in the lemma now directly follows from Eqs. (F36) and (F44). $\square$