Bloch-Wangsness-Redfield theory

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1 Derivation
2 References

Derivation

Let us begin in the Schrödinger picture. The time evolution of the density matrix is given by the Liouville equation (essentially, the Schrödinger equation),

$i \hbar \frac{d \rho(t)}{d t} = [H,\rho(t)]$ .

Consider a system $\mathcal{S}\,$ coupled to an environment $\mathcal{E}\,$ (the heat bath) such that

$H = H_0 + V = H_\mathcal{S} + H_\mathcal{E} + V\,$ ,

and $H_I\,$ describes the interaction between the system and the environment. Define the interaction picture density matrix

$\rho_I(t) \equiv e^{\frac{i}{\hbar} (H_\mathcal{S} + H_\mathcal{E})t} \rho(t) e^{-\frac{i}{\hbar} (H_\mathcal{S} + H_\mathcal{E})t}\,$ ,

and similarly

$V_I(t) \equiv e^{\frac{i}{\hbar} (H_\mathcal{S} + H_\mathcal{E})t} V e^{-\frac{i}{\hbar} (H_\mathcal{S} + H_\mathcal{E})t}\,$ .

Then

$i \hbar \frac{d \rho_I(t)}{d t}\,$	$=e^{\frac{i}{\hbar} H_0 t} \left( [H,\rho(t)] - [H_0,\rho(t)] \right)e^{-\frac{i}{\hbar} H_0t}\,$ ,
	$=e^{\frac{i}{\hbar} H_0 t} [V,\rho(t)] e^{-\frac{i}{\hbar} H_0t}\,$ ,
	$= [V_I(t),\rho_I(t)]\,$ .

This has the formal solution

$\rho_I(t) = \mathcal{T} \exp\left( \frac{1}{i\hbar} \int_0^t\!dt_1\, [V_I(t_1),\,\star\,] \right) \rho_I(0)\,$ .

where $\mathcal{T}\,$ denotes time-ordering. Equivalently,

$\rho_I(t) = \rho_I(0) + \frac{1}{i \hbar} \int_0^t\!dt_1\, [V_I(t_1), \rho_I(t_1) ]\,$ .

Expanded once,

$\rho_I(t) = \rho_I(0) + \frac{1}{i \hbar} \int_0^t\!dt_1\, [V_I(t_1), \rho_I(0) ] + \frac{1}{(i \hbar)^2} \int_0^t\int_0^{t_1}\!dt_1\,dt_2\, \left[V_I(t_1), \left[V_I(t_2), \rho_I(t_2) \right] \right]\,$ .

Let us concentrate on the evolution of the (interaction picture) reduced density matrix $\rho_\mathcal{S} = \operatorname{tr}_\mathcal{E}\,\rho_I\,$ obtained by tracing over the environment. To do this, we will make a number of approximations. Note that

$\rho_\mathcal{S}(t) = \rho_\mathcal{E}(0) + \frac{1}{i \hbar} \int_0^t\!dt_1\, \operatorname{tr}_\mathcal{E}\,[V_I(t_1), \rho_I(0) ] + \frac{1}{(i \hbar)^2} \int_0^t\int_0^{t_1}\!dt_1\,dt_2\, \operatorname{tr}_\mathcal{E}\,\left[V_I(t_1), \left[V_I(t_2), \rho_I(t_2) \right] \right]\,$ .

Simplification: Energy shift

We can now make the simplification that $\left\langle V_I(t) \right\rangle_\mathcal{E} \sim \operatorname{tr}_\mathcal{E}\,V_I(t) \rho_\mathcal{E}(0) = 0\,$ . This is not restrictive, since, if $V\,$ is of the form $V = A_\mathcal{S} \otimes B_\mathcal{E}\,$ , then we can replace $V\,$ with $A_\mathcal{S} \otimes \left( B_\mathcal{E} - \left\langle B_\mathcal{E} \right\rangle_\mathcal{E} \right)\,$ , and simultaneously add $A_\mathcal{S}\left\langle B_\mathcal{E} \right\rangle_\mathcal{E}\,$ to $H_\mathcal{S}\,$ . Since $\rho_\mathcal{E}(0)\,$ has the same form in both pictures, the result holds in the interaction picture also. The same argument can be made if $V = \textstyle\sum_\alpha A_{\mathcal{S},\alpha} \otimes B_{\mathcal{E},\alpha}\,$ . Then

$\rho_\mathcal{S}(t) = \rho_\mathcal{E}(0)+ \frac{1}{(i \hbar)^2} \int_0^t\int_0^{t_1}\!dt_1\,dt_2\, \operatorname{tr}_\mathcal{E}\,\left[V_I(t_1), \left[V_I(t_2), \rho_I(t_2) \right] \right]\,$ .

Assumption 1: Born approximation

We can always write (in any picture)

$\rho(0) = \rho_\mathcal{S}(0) \otimes \rho_\mathcal{E}(0) + \rho_\mathrm{correlation}(0)\,$ ,

which may be taken as a definition of $\rho_\mathrm{correlation}\,$ . Let us assume that the interaction is turned on at time $t=0\,$ , and that prior to that the system and environment are not correlated. We will assume that the coupling between the system and the environment is weak, so that

$\rho(t) = \rho_\mathcal{S}(t) \otimes \rho_\mathcal{E}(t)\,$ ,

for timescales over which perturbation theory remains valid. Furthermore, we will assume that the correlation time $\tau_\mathcal{E}\,$ (and thus the relaxation time) of the environment is sufficiently small that $\rho_\mathcal{E}(t) \approx \rho_\mathcal{E}(0)\,$ if $t \gg \tau_\mathcal{E}\,$ .

Note that if the environment has thermalized, then it has a thermal density matrix

$\rho_\mathcal{E}(0) = \tfrac{1}{Z_\mathcal{E}}\textstyle\sum_n e^{-\frac{E_{n_\mathcal{E}}}{k_B T} } \left|n_\mathcal{E}\right\rangle \left\langle n_\mathcal{E} \right|\,$ ,

which is a stationary state, i.e., $[\rho_\mathcal{E}(0),H_\mathcal{E}] = 0\,$ , so that $\rho_\mathcal{E}(0)\,$ has the same form in both the interaction picture and Schrödinger picture.

Formally, then, to second order accuracy,

$\rho_\mathcal{S}(t) = e^{M(t)} \rho_\mathcal{S}(0)\,$ ,

where

$M(t) = \frac{1}{(i \hbar)^2} \int_0^t\int_0^{t_1}\!dt_1\,dt_2\, \operatorname{tr}_\mathcal{E}\,\left[V_I(t_1), \left[V_I(t_2),\,\star \,\otimes \rho_\mathcal{E}(0) \right] \right]\,$ .

Differentiating, we get

$\frac{d}{dt} \rho_\mathcal{S}(t) = \frac{1}{(i \hbar)^2} \int_0^t\! ds\, \operatorname{tr}_\mathcal{E}\,\left[V_I(t), \left[V_I(s), \rho_\mathcal{S}(s) \otimes \rho_\mathcal{E}(0) \right] \right]\,$ .

This should properly be considered a difference equation, since we have assumed that $t \gg \tau_\mathcal{E}\,$ .

Assumption 3: Markov approximation

We will also assume that we are working over timescales that are shorter than the gross timescale over which the system evolves. Thus we can replace $\rho_\mathcal{S}(s)\,$ with $\rho_\mathcal{S}(t)\,$ .

We then get the Redfield equation:

$\frac{d}{dt} \rho_\mathcal{S}(t) = \frac{1}{(i \hbar)^2} \int_0^t\! ds\, \operatorname{tr}_\mathcal{E}\,\left[V_I(t), \left[V_I(s), \rho_\mathcal{S}(t) \otimes \rho_\mathcal{E}(0) \right] \right]\,$ .

We can change variables from $s\to t - s\,$ , so that the limits of integration go from $t\,$ to $0\,$ . Then

$\frac{d}{dt} \rho_\mathcal{S}(t) = \frac{1}{(i \hbar)^2} \int_0^t\! ds\, \operatorname{tr}_\mathcal{E}\,\left[V_I(t), \left[V_I(t-s), \rho_\mathcal{S}(t) \otimes \rho_\mathcal{E}(0) \right] \right]\,$ .

Consider the integral

$\int_0^{t_1}\!dt_2 \operatorname{tr}_\mathcal{E}\, V_I(t_1) V_I(t_2) \rho_I(0) = \int_0^{t_1}\!dt_2 \operatorname{tr}_\mathcal{E} \,V_I(t_1-t_2) V_I(0) \rho_I(0)\,$ .

The correlation time of the thermal bath $\mathcal{E}\,$ is assumed to be very short, so that the correlation function $\left\langle V_I(t_1-t_2)V_I(0) \right\rangle_\mathcal{E}\,$ differs only significantly from zero when $t_1 \approx t_2\,$ . We can therefore extend the limit of integration to $\infty\,$ :

$\int_0^{t_1}\!dt_2 \operatorname{tr}_\mathcal{E}\, V_I(t_1) V_I(t_2) \rho_I(0) \approx \int_0^{\infty}\!dt_2 \operatorname{tr}_\mathcal{E}\, V_I(t_1) V_I(t_2) \rho_I(0)\,$ .

$\frac{d}{dt} \rho_\mathcal{S}(t) = \frac{1}{(i \hbar)^2} \int_0^\infty\! ds\, \operatorname{tr}_\mathcal{E}\,\left[V_I(t), \left[V_I(t-s), \rho_\mathcal{S}(t) \otimes \rho_\mathcal{E}(0) \right] \right]\,$ .