开放系统读书会：Lindblad 主方程的微观推导

2023-02-27

-专业笔记

Model

$$\begin{align} H = H_S + H_B + H_I \end{align}$$

von Neumann equation in the INTERACTION PICTURE

$$\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\rho(t) = - \mathrm{i} [H_I(t), \rho(t)] \end{align}$$

A formal solution

$$\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\rho(t) = - \mathrm{i} [H_I(t), \rho(0)] -\int_0^t \mathrm{d} s[H_I(t), [H_I(s), \rho(s)]] \end{align}$$

Four Approximation

1. Separability

At $t=0$ , there are no correlations between the system

$$\begin{align} \rho(0) \approx \rho_S(0) \otimes \rho_B(0) \end{align}$$

2. Born approximation

Large environment means:

Weak coupling means:

In summary

$$\begin{align} \rho(t) \approx \rho_S(t) \otimes \rho_B \end{align}$$

3. Markov approximation

Short-memory environment

The reservior correlation functions decay faster compared to the system.

4. Secular approximation(Not necessary for all systems)

Only consider the resonance transition.

Not necessary for all master equation.

Derivations

In SCHRODINGER PICTURE

$$\begin{align} H_I^{SP} = \sum_{\alpha} A_{\alpha} \otimes B_{\alpha} = \sum_{\alpha, \omega}A_{\alpha}(\omega) \otimes B_{\alpha} \end{align}$$

where

$$\begin{align} A_{\alpha}(\omega) = \sum_{\varepsilon } |\varepsilon\rangle\langle\varepsilon| A_{\alpha} |\varepsilon + \omega \rangle\langle\varepsilon + \omega| \end{align}$$

So, in the interaction picture

$$\begin{align} H_I(t) = \sum_{\alpha,\omega} e^{- \mathrm{i}\omega t}A_{\alpha}(\omega)\otimes B_{\alpha}(t) = \sum_{\alpha,\omega} e^{ \mathrm{i}\omega t}A^{\dagger}_{\alpha}(\omega)\otimes B^{\dagger}_{\alpha}(t) \end{align}$$

use SEPARABILITY, we get

$$\begin{align} [H_I(t), \rho(0)] =& \sum_{\alpha,\omega}e^{\mathrm{i}\omega t}[A^{\dagger}_{\alpha}(\omega)\otimes B^{\dagger}_{\alpha}(t), \rho_S(0)\otimes \rho_B(0)] \\ =& \sum_{\alpha,\omega}e^{\mathrm{i}\omega t} [A^{\dagger}_{\alpha}(\omega) \rho_S(0) \otimes B^{\dagger}_{\alpha}(t) \rho_B(0) - \rho_S(0) A^{\dagger}_{\alpha}(\omega) \otimes\rho_B(0) B^{\dagger}_{\alpha}(t) ] \\ =& \sum_{\alpha,\omega}e^{\mathrm{i}\omega t} [A^{\dagger}_{\alpha}(\omega), \rho_S(0)] \langle B_\alpha(t) \rangle \\ \end{align}$$

where

$$\begin{align} \langle B_\alpha(t) \rangle \equiv \mathrm{Tr}_B[B^{\dagger}_{\alpha}(t) \rho_B] = 0 \end{align}$$

use $[H_I(t), \rho(0)] = 0$ and BORN APPROXIMATION, we get

$$\begin{align} \mathrm{Tr}_B \left[\frac{\mathrm{d}}{\mathrm{d}t}\rho(t) \right] = \frac{\mathrm{d}}{\mathrm{d}t}\rho_S(t) = -\int_0^t \mathrm{d} s \cdot \mathrm{Tr}_B[H_I(t), [H_I(s), \rho_S(s)\otimes \rho_B]] \end{align}$$

According MARKOV APPROXIMATION, we can approximate $\rho_S(s) \approx \rho_S(t)$ , and extend $t$ to $\infty$ . Substitute $s \to t - s$ ,

$$\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\rho_S(t)= \frac{\mathrm{d}}{\mathrm{d}t}\rho_S(t) = -\int_0^{\infty} \mathrm{d} s \cdot \mathrm{Tr}_B[H_I(t), [H_I(t - s), \rho_S(t)\otimes \rho_B]] \end{align}$$

decompose $H_{I}(t)$ , we get (convolution theorem)

$$\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\rho_S(t) = \sum_{\omega, \omega'} \sum_{\alpha, \beta} e^{\mathrm{i}(\omega' - \omega)t} \Gamma_{\alpha\beta}(\omega)\left[A_{\beta}(\omega)\rho_S(t)A^{\dagger}_{\alpha}(\omega') - A_{\alpha}^{\dagger}(\omega') A_{\beta}(\omega)\rho_s(t)\right] + \mathrm{h.c.} \end{align}$$

where

$$\begin{align} \Gamma_{\alpha\beta}(\omega) = \int_0^{\infty} \mathrm{d}s\cdot e^{\mathrm{i}\omega s} \langle B_{\alpha}^{\dagger}(t) B_{\beta}(t - s)\rangle \end{align}$$

Then SECULAR APPROXIMATION, only keep the resonance term, that is $\omega' = \omega$

$$\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\rho_S(t) = \sum_{\omega} \sum_{\alpha, \beta} \Gamma_{\alpha\beta}(\omega)\left[A_{\beta}(\omega)\rho_S(t)A^{\dagger}_{\alpha}(\omega) - A_{\alpha}^{\dagger}(\omega) A_{\beta}(\omega)\rho_s(t)\right] + \mathrm{h.c.} \end{align}$$

if we define

$$\begin{align} \left\{ \begin{matrix} S_{\alpha\beta}(\omega) \equiv& \frac{1}{2\mathrm{i}} \left[ \Gamma_{\alpha\beta}(\omega) - \Gamma_{\beta\alpha}^{ * }(\omega) \right]& \\ \gamma_{\alpha\beta}(\omega) \equiv& \Gamma_{\alpha\beta}(\omega) - \Gamma_{\beta\alpha}^{ * }(\omega) =& \int_{-\infty}^{\infty} \mathrm{d}s e^{\mathrm{i}\omega s} \langle B^{\dagger}_{\alpha}(s) B_{\beta}(0)\rangle\\ \Gamma_{\alpha\beta}(\omega) =& \frac{1}{2}\gamma_{\alpha\beta}(\omega) + \mathrm{i}S_{\alpha\beta}(\omega)& \end{matrix} \right. \end{align}$$

and (Lamb shift Hamiltonian. It seems like the interaction between vacuum energy fluctuations and the hydrogen electron in different orbitals, ref Atomic and Optical Physics I, 07 Atoms III: Fine Structure)

$$\begin{align} H_{LS} = \sum_{\omega} \sum_{\alpha,\beta} S_{\alpha,\beta}(\omega)A_{\alpha}^{\dagger}(\omega) A_{\beta}(\omega) \end{align}$$

then we get our master equation in the interaction picture

$$\begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\rho_{S}(t) = -\mathrm{i}[H_{LS}, \rho_S(t)] + \mathcal{D}[\rho_S(t)] \end{align}$$

where

$$\begin{align} D(\rho_S) = \sum_{\omega}\sum_{\alpha,\beta} \gamma_{\alpha\beta}(\omega) \left[ A_{\beta}(\omega) \rho_SA_{\alpha}^{\dagger}(\omega) - \frac{1}{2}{A^{\dagger}_{\alpha}(\omega)A_{\beta}(\omega), \rho_S} \right] \end{align}$$

For Schrodinger picture, just shift $H_{LS}\to H_{LS} + H_S$ .

Reference

#物理 #Lindblad Master Equation #Open System