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## 物理代写|热力学代写thermodynamics代考|Factorizable Interaction Hamiltonians

We shall explicitly write the foregoing ME for the factorizable interaction Hamiltonian,
$$H_{\mathrm{I}}(t)=S(t) B,$$
where $H_{\mathrm{I}}$ is the product of operators $S$ and $B$ that act on the system and the bath, respectively. We further assume that $\langle B\rangle_{\mathrm{B}}=0$, so that $\left\langle\mathcal{L}{\mathrm{I}}\right\rangle{\mathrm{B}}=0$ and $\mathcal{P} \mathcal{L}{\mathrm{I}} \mathcal{P}=0$. The ME (11.25) then simplifies to $$\dot{\rho}=-i \mathcal{L}{\mathrm{S}}(t) \rho-\int_0^t d \tau\left\langle\mathcal{L}{\mathrm{I}}(t) \mathcal{U}_0(t, \tau) \mathcal{L}{\mathrm{I}}(\tau)\right\rangle_{\mathrm{B}} \mathcal{U}{\mathrm{S}}^{-1}(t, \tau) \rho .$$ Any operator $A$ satisfies the relation $$\mathcal{U}_0(t, \tau) A=U{\mathrm{S}}(t, \tau) U_{\mathrm{B}}(t-\tau) A U_{\mathrm{S}}^{\dagger}(t, \tau) U_{\mathrm{B}}^{\dagger}(t-\tau),$$
where
$$U_{\mathrm{S}}(t, \tau)=\mathrm{T}{+} e^{-i \int{\mathrm{r}}^t H_{\mathbf{S}}\left(t^{\prime}\right) d t^{\prime}},$$

$$U_{\mathrm{B}}(t)=e^{-i H_{\mathrm{B}} t} .$$
The integrand in (11.27) can therefore be written as follows,
\begin{aligned} I(t, \tau) &=\operatorname{Tr}{\mathrm{B}}\left[S(t) B, \mathcal{U}_0(t, \tau)\left[S(\tau) B, \mathcal{U}{\mathrm{S}}^{-1}(t, \tau) \rho(t) \rho_{\mathrm{B}}\right]\right] \ &=\operatorname{Tr}{\mathrm{B}}\left[S(t) B,\left[\tilde{S}(\tau, t) \tilde{B}(\tau-t), \rho(t) \rho{\mathrm{B}}\right]\right], \end{aligned}
where $\tilde{S}(\tau, t)$ and $\tilde{B}(t)$ are the system and bath operators $S(\tau)$ and $B$ in the interaction representation,
\begin{aligned} \bar{S}(\tau, t) &=U_{\mathrm{S}}^{\dagger}(\tau, t) S(\tau) U_{\mathrm{S}}(\tau, t), \ \tilde{B}(t) &=U_{\mathrm{B}}^{\dagger}(t) B U_{\mathrm{B}}(t) . \end{aligned}
The commutativity of $S$ and $B$, and that of $\rho_{\mathrm{B}}$ and $H_{\mathrm{B}}$, using the cyclic property of the trace, yield:
$$I(t, \tau)=\Phi_T(t-\tau)[S(t), \tilde{S}(\tau, t) \rho(t)]+\text { H.c., }$$
where
$$\Phi_T(t)=\langle\tilde{B}(t) B\rangle_{\mathrm{B}}$$
is the bath autocorrelation function. Here and henceforth we take $\rho_{\mathrm{B}}$ to be the thermal density operator,
$$\rho_{\mathrm{B}}=Z^{-1} e^{-H_{\mathrm{B}} / T},$$
where $Z$ is the unit-trace normalization constant and the temperature $T$ is given, from here on, in energy units, with the Boltzmann constant $k_{\mathrm{B}}=1$.

## 物理代写|热力学代写thermodynamics代考|Master Equation for an Oscillator Bath

The above results hold for an arbitrary bath. Below we consider, for definiteness, the case of a bosonic oscillator bath (Ch. 3) in thermal equilibrium. In this case, the bath Hamiltonian in (11.2) reads
$$H_{\mathrm{B}}=\sum_k \omega_k a_k^{\dagger} a_k,$$
where $k$ labels the bath-mode frequencies $\omega_k$, the bosonic creation and annihilation operators $a_k^{\dagger}$ and $a_k$, respectively. The S-B interaction Hamiltonian (11.26) for a TLS coupled to such a bath reads as in (4.10) with antiresonant terms included or as in (4.12) in the RWA. In the interaction picture [as per (11.31)], the bath operator has then the form
$$\tilde{B}(t)=\sum_k\left(\eta_k a_k e^{-i \omega_k t}+\eta_k^* a_k^{\dagger} e^{i \omega_k t}\right),$$
where $\eta_k$ are the dipolar $k$-mode coupling strengths. The bath autocorrelation function (11.33) is then given by
\begin{aligned} \Phi_T(t) &=\sum_k \sum_{k^{\prime}}\left(\eta_k \eta_{k^{\prime}}^* e^{-i \omega_k t}\left\langle a_k a_{k^{\prime}}^{\dagger}\right\rangle_{\mathrm{B}}+\eta_k^* \eta_{k^{\prime}} e^{i \omega_k t}\left\langle a_k^{\dagger} a_{k^{\prime}}\right\rangle_{\mathrm{B}}\right) \ &=\sum_k\left|\eta_k\right|^2\left{e^{-i \omega_k t}\left[\bar{n}T\left(\omega_k\right)+1\right]+e^{i \omega_k t} \bar{n}_T\left(\omega_k\right)\right}, \end{aligned} where we used the equilibrium bosonic bath properties $\left\langle a_k a{k^{\prime}}\right\rangle_{\mathrm{B}}=\left\langle a_k^{\dagger} a_{k^{\prime}}^{\dagger}\right\rangle_{\mathrm{B}}=0$, $\left\langle a_k^{\dagger} a_{k^{\prime}}\right\rangle_{\mathrm{B}}=\delta_{k k^{\prime}}\left[\bar{n}T\left(\omega_k\right)+1\right],\left\langle a_k^{\dagger} a{k^{\prime}}\right\rangle_{\mathrm{B}}=\delta_{k k^k} \bar{n}_T\left(\omega_k\right), \bar{n}_T(\omega)$ being the average quanta number corresponding to the temperature-dependent Planck distribution at frequency $\omega$,
$$\bar{n}_T(\omega)=\left[\exp \left(\frac{\omega}{T}\right)-1\right]^{-1} .$$

# 热力学代写

## 物理代写|热力学代写thermodynamics代考|Factorizable Interaction Hamiltonians

$$H_{\mathrm{I}}(t)=S(t) B,$$

$$\dot{\rho}=-i \mathcal{L S}(t) \rho-\int_0^t d \tau\left\langle\mathcal{L I}(t) \mathcal{U}0(t, \tau) \mathcal{L I}(\tau)\right\rangle{\mathrm{B}} \mathcal{U} \mathrm{S}^{-1}(t, \tau) \rho .$$

$$\mathcal{U}0(t, \tau) A=U \mathrm{~S}(t, \tau) U{\mathrm{B}}(t-\tau) A U_{\mathrm{S}}^{\dagger}(t, \tau) U_{\mathrm{B}}^{\dagger}(t-\tau),$$

$$\begin{gathered} U_{\mathrm{S}}(t, \tau)=\mathrm{T}+e^{-i \int_{\mathrm{r}}{ }^t H \mathbf{S}\left(t^{\prime}\right) d t^{\prime}}, \ U_{\mathrm{B}}(t)=e^{-i H_{\mathrm{B} t}} . \end{gathered}$$

$$I(t, \tau)=\operatorname{Tr} \mathrm{B}\left[S(t) B, \mathcal{U}0(t, \tau)\left[S(\tau) B, \mathcal{U S}^{-1}(t, \tau) \rho(t) \rho{\mathrm{B}}\right]\right] \quad=\operatorname{Tr} \mathrm{B}[S(t) B,[\tilde{S}(\tau, t) \tilde{B}(\tau-t), \rho(t) \rho \mathrm{B}]],$$

$$\bar{S}(\tau, t)=U_{\mathrm{S}}^{\dagger}(\tau, t) S(\tau) U_{\mathrm{S}}(\tau, t), \tilde{B}(t) \quad=U_{\mathrm{B}}^{\dagger}(t) B U_{\mathrm{B}}(t)$$

$$I(t, \tau)=\Phi_T(t-\tau)[S(t), \tilde{S}(\tau, t) \rho(t)]+\text { H.c., }$$

$$\Phi_T(t)=\langle\tilde{B}(t) B\rangle_{\mathrm{B}}$$

$$\rho_{\mathrm{B}}=Z^{-1} e^{-H_{\mathrm{B}} / T},$$

## 物理代写|热力学代写thermodynamics代考|Master Equation for an Oscillator Bath

$$H_{\mathrm{B}}=\sum_k \omega_k a_k^{\dagger} a_k$$

$$\tilde{B}(t)=\sum_k\left(\eta_k a_k e^{-i \omega_k t}+\eta_k^* a_k^{\dagger} e^{i \omega_k t}\right)$$

$\backslash 1 \mathrm{eft}$ 的分隔符缺失或无法识别
㧴们使用了平衡玻色子浴特性 $\left\langle a_k a k^{\prime}\right\rangle_{\mathrm{B}}=\left\langle a_k^{\dagger} a_{k^{\prime}}^{\dagger}\right\rangle_{\mathrm{B}}=0,\left\langle a_{k^{\dagger}}^{\dagger} a_{k^{\prime}}\right\rangle_{\mathrm{B}}=\delta_{k k^{\prime}}\left[\bar{n} T\left(\omega_k\right)+1\right],\left\langle a_k^{\dagger} a k^{\prime}\right\rangle_{\mathrm{B}}=\delta_{k k^k} \bar{n}_T\left(\omega_k\right), \bar{n}_T(\omega)$ 是对应于频率上与温 度相关的普朗克分布的平均量子数 $\omega$ ，
$$\bar{n}_T(\omega)=\left[\exp \left(\frac{\omega}{T}\right)-1\right]^{-1}$$

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## MATLAB代写

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