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

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 $H_1$ 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 $\mathcal{P L}{\mathcal{I}} \mathcal{P}=0$. 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
$$U_{\mathrm{S}}(t, \tau)=\mathrm{T}{+} e^{-i \int\tau^t H_{\mathrm{S}}\left(t^{\prime}\right) d t^{\prime}},$$ $$U_{\mathrm{B}}(t)=e^{-i H_{\mathrm{B}} t} .$$
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}

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 $\left\langle a_k a{k^{\prime}}\right\rangle_{\mathrm{B}}=\left\langle a_k^{\dagger} a_{k^{\prime}}^{\dagger}\right\rangle_{\mathrm{B}}=0$, $\left\langle a_k^{\dagger} a_{k^{\prime}}\right\rangle_{\mathrm{B}}=\delta_{k k^{\prime}}\left[\bar{n}T\left(\omega_k\right)+1\right],\left\langle a_k^{\dagger} a{k^{\prime}}\right\rangle_{\mathrm{B}}=\delta_{k k^{\prime}} \bar{n}_T\left(\omega_k\right), \bar{n}_T(\omega)$ being the average quanta number corresponding to the temperature-dependent Planck distribution at frequency $\omega$,
$$\bar{n}_T(\omega)=\left[\exp \left(\frac{\omega}{T}\right)-1\right]^{-1} \text {. }$$
From (11.43) and (11.40), we then obtain the thermal-bath response spectrum,
$$G_T(\omega)=G_0(\omega)\left[1+\bar{n}_T(\omega)\right]+G_0(-\omega) n_T(-\omega),$$
where the bath response spectrum at $T=0$ is
$$G_0(\omega)=\sum_k\left|\eta_k\right|^2 \delta\left(\omega-\omega_k\right) .$$

热力学代考

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

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

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

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

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

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

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

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

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

$\left\langle a_k^{\dagger} a_{k^{\prime}}\right\rangle_{\mathrm{B}}=\delta_{k k^{\prime}}\left[\bar{n} T\left(\omega_k\right)+1\right],\left\langle a_k^{\dagger} a k^{\prime}\right\rangle_{\mathrm{B}}=\delta_{k k^{\prime}} \bar{n}T\left(\omega_k\right), \bar{n}_T(\omega)$ 是与温度相关的 普朗克分布相对应的平均量子数 $\omega$ ， $$\bar{n}_T(\omega)=\left[\exp \left(\frac{\omega}{T}\right)-1\right]^{-1}$$ 以上结果适用于任意浴。为了明确起见，我们在下面考虑处于热平衡状态的玻色子振荡 器浴槽 (第 3 章) 的情况。在这种情况下，(11.2) 中的浴哈密顿量为 $$H{\mathrm{B}}=\sum_k \omega_k a_k^{\dagger} a_k$$

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

$$G_T(\omega)=G_0(\omega)\left[1+\bar{n}_T(\omega)\right]+G_0(-\omega) n_T(-\omega),$$

$$G_0(\omega)=\sum_k\left|\eta_k\right|^2 \delta\left(\omega-\omega_k\right) .$$

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