Another Look at Influence Functionals for General Baths

(This sub-section is partly due to private communications from W. Zwerger). Feynman and Vernon realised that the coupling of a system $S$ to any bath $B$ can be mapped onto the coupling to an equivalent oscillator bath, if the coupling is weak and second order perturbation theory can be applied: let us have another look at the operator form of the influence functional, Eq. (7.174),

$${\cal F}[q(t'),q'(t')] \equiv {\rm Tr_B} \left(\rho_B U_B^{\dagger}[q'] U_B[q] \right)$$

$$H_{B}(t) = H_0+ V(t), \quad H'_{B}(t)= H_0+ V'(t),$$ (211)

where $U_B[q]$ is the time-evolution operator for $H_{B}(t)$ and $U_B^{\dagger}[q']$ the (backwards in time) evolution operator for $H'_{B}(t)$. Here, $H_{B}(t)$ and $H'_{B}(t)$ refer to different paths $q$ and $q'$.

Example: For a Fermi bath, we could have

$$H_0= \sum_k \varepsilon_k c^{\dagger}_k c_k,\quad V(t)\equiv V[q_t] = \sum_{kk'} M_{kk'} \exp ({i(k-k')q_{t}}) c^{\dagger}_{k'} c_k.$$ (212)

where $c^{\dagger}_k$ creates a Fermion with quantum number $k$.

We again introduce the interaction picture and write

$$U_B[q] = e^{-iH_0t} \left\{ 1 + i \int_{0}^{t}dt' \tilde{V}(t') - \int_{0}^{t}\int_{0}^{t'}dt'ds \tilde{V}(t')\tilde{V}(s)+...\right\}$$

$$U^{\dagger}_B[q'] = \left\{ 1 - i \int_{0}^{t}dt' \tilde{V'}(t') - \int_{0}^{t}\int_{0}^{t'}dt'ds \tilde{V'}(s)\tilde{V'}(t')+...\right\}e^{iH_0t}$$ (213)

The product of the two time-evolution operators therefore becomes

$$U^{\dagger}_B[q'] U_B[q] = 1 - i \int_{0}^{t}dt' \left\{ \tilde{V'}(t') - \tilde{V}(t') \right\} + \int_{0}^{t}dt' \tilde{V'}(t') \int_{0}^{t}ds \tilde{V}(s)$$

$$- \int_{0}^{t}\int_{0}^{t'}dt'ds \left\{ \tilde{V'}(s)\tilde{V'}(t') + \tilde{V}(t')\tilde{V}(s)\right\} +...$$

$$= 1 - i \int_{0}^{t}dt' \left\{ \tilde{V'}(t') - \tilde{V}(t') \rig... ...t}dt'ds \left\{ \tilde{V'}(t')\tilde{V}(s) +\tilde{V'}(s)\tilde{V}(t') \right\}$$

$$- \int_{0}^{t}\int_{0}^{t'}dt'ds \left\{ \tilde{V'}(s)\tilde{V'}(t') + \tilde{V}(t')\tilde{V}(s)\right\} +...$$

$$= 1 - i \int_{0}^{t}dt' \left\{ \tilde{V'}(t') - \tilde{V}(t') \rig... ...'}dt'ds \left\{ \tilde{V'}(t')\tilde{V}(s) +\tilde{V'}(s)\tilde{V}(t') \right\}$$

$$- \int_{0}^{t}\int_{0}^{t'}dt'ds \left\{ \tilde{V'}(s)\tilde{V'}(t') + \tilde{V}(t')\tilde{V}(s)\right\} +...$$

$$= 1 - i \int_{0}^{t}dt' \left\{ \tilde{V'}(t') - \tilde{V}(t') \rig... ...'}dt'ds \left\{\left[ \tilde{V'}(t')-\tilde{V}(t')\right] \tilde{V}(s) \right\}$$

$$- \int_{0}^{t}\int_{0}^{t'}dt'ds \left\{\tilde{V'}(s) \left[ \tilde{V'}(t') - \tilde{V}(t')\right]\right\}+...$$ (214)

In order to be a little bit more definite, a useful parametrisation of the interaction operators might be

$$\hat{V}(t)\equiv \sum_{\alpha\beta}g_{\alpha\beta}(t) \hat{X}_{\a... ... \hat{V'}(t)\equiv \sum_{\alpha\beta}g'_{\alpha\beta}(t) \hat{X}_{\alpha\beta},$$ (215)

with bath operators $\hat{X}_{\alpha\beta}$. Note that this comprises the cases considered so far (harmonic oscillator, Fermi bath). Taking the trace over $\rho_B$, we obtain

$${\cal F}[q(t'),q'(t')] \equiv {\rm Tr_B} \left(\rho_B U_B^{\dagger}[q'] U_B[q] \right)\equiv \langle U_B^{\dagger}[q'] U_B[q] \rangle_0$$

$$= 1 - i\sum_{\alpha\beta} \int_{0}^{t}dt' \left\{ g'_{\alpha\beta}(t')-g_{\alpha\beta}(t')\right\} \langle \tilde{X}_{\alpha\beta}(t') \rangle_0$$

$$+ \sum_{\alpha\beta\gamma\delta} \int_{0}^{t}\int_{0}^{t'}dt'ds \le... ...ta}(s) \langle \tilde{X}_{\alpha\beta}(t') \tilde{X}_{\gamma\delta}(s)\rangle_0$$

$$- g'_{\gamma\delta}(s) \langle \tilde{X}_{\gamma\delta}(s) \tilde{X}_{\alpha\beta}(t') \rangle_0 \Big] +...$$ (216)

Introducing the correlation tensor

$$L_{\alpha\beta\gamma\delta}(t',s)\equiv \langle \tilde{X}_{\alpha\beta}(t') \tilde{X}_{\gamma\delta}(s)\rangle_0,$$ (217)

this can be written as

$${\cal F}[q(t'),q'(t')] = 1 - i\sum_{\alpha\beta} \int_{0}^{t}dt' ... ...}(t')-g_{\alpha\beta}(t')\right\} \langle \tilde{X}_{\alpha\beta}(t') \rangle_0$$

$$+ \sum_{\alpha\beta\gamma\delta} \int_{0}^{t}\int_{0}^{t'}dt'ds \le... ...amma\delta}(t',s) - g'_{\gamma\delta}(s) L_{\gamma\delta\alpha\beta}(s,t')\Big]$$

$$+ ...$$ (218)