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Expansion of the Influence Phase

In order to derive a semiclassical limit from the double path integral, the central idea is to expand the influence phase in powers of the paths $ y_{t'}$. The $ y_{t'}$-paths describe `off-diagonal excursions' from the diagonal paths $ x_{t'}$ in the time-evolution of $ \rho(t)$. We write
$\displaystyle {\cal F}[x_{t'},y_{t'}]$ $\displaystyle =$ $\displaystyle \exp \left\{-\Phi[x_{t'},y_{t'}] \right\}$   Influence Functional  
$\displaystyle -\Phi[x_{t'},y_{t'}]$ $\displaystyle =$ $\displaystyle -\sum_{\alpha=1}^N\int_{0}^{t}dt'\int_{0}^{t'}ds \left\{
f_\alpha[x_{t'}+y_{t'}/2] - f_\alpha[x_{t'}-y_{t'}/2] \right\}$  
  $\displaystyle \times$ $\displaystyle \left\{ S_\alpha(t'-s) f_\alpha[x_{s}+y_{s}/2] - S_\alpha^*(t'-s) f_\alpha[x_{s}-y_{s}/2]\right\}$  
  $\displaystyle =$ $\displaystyle -\sum_{\alpha=1}^N\int_{0}^{t}dt'\int_{0}^{t'}ds
f'_\alpha[x_{t'}]y_{t'}$  
  $\displaystyle \times$ $\displaystyle \left\{\mbox{\rm Re } S_\alpha(t'-s) f'_\alpha[x_{s}]y_{s} + 2i \mbox{\rm Im }
S_\alpha(t'-s) f_\alpha[x_{s}]\right\} + O\left[y_s\right]^3.$ (229)

In the semiclassical approximation, we thus can write the influence as
$\displaystyle {\cal F}_{\rm sc}[x_{t'},y_{t'}]$ $\displaystyle \equiv$ $\displaystyle \exp \left\{-\Phi_{\rm }[x_{t'},y_{t'}] \right\}
= \exp \left\{i\phi_1 - \phi_2 \right\}$ (230)
$\displaystyle i\phi_1$ $\displaystyle \equiv$ $\displaystyle -
i \int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_1[x_s] \underline{\underline{y_{t'}}},\quad
\varphi_1[x_s]\equiv \sum_{\alpha=1}^N 2$Im $\displaystyle S_\alpha(t'-s)f'_\alpha[x_{t'}]f_\alpha[x_{s}]$  
$\displaystyle \phi_2$ $\displaystyle \equiv$ $\displaystyle \int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_2[x_s] \underline{\underline{y_{t'}y_{s}}},\quad
\varphi_2[x_s]\equiv \sum_{\alpha=1}^N$   Re $\displaystyle S_\alpha(t'-s)f'_\alpha[x_{t'}]f'_\alpha[x_{s}].$  

Exercise: Check that for the linear model (coupling linear in $ q$), Eq.(7.212), the influence phase becomes

$\displaystyle \Phi[x_{t'},y_{t'}]$ $\displaystyle =$ $\displaystyle \int_{0}^{t}dt'\int_{0}^{t'}ds
y_{t'}
\left\{ \mbox{\rm Re }L(t'-s) y_{s} + 2i \mbox{\rm Im }L(t'-s) x_{s}\right\}$  
  $\displaystyle +$ $\displaystyle i {\mu}\int_0^t dt'x_{t'}y_{t'}.$ (231)

For the linear model, the semiclassical expansion of the influence phase is therefore exact.

In a similar way, we expand the potential $ V(x\pm y/2)$ in the action in powers of the off-diagonal path $ y$, thus arriving at

$\displaystyle \rho_{\rm sc}(x,y,t)$ $\displaystyle =$ $\displaystyle \int dx_0 dy_0 \rho_0(x_0,y_0) J_{\rm sc}(x,y,t;x_0,y_0)$ (232)
$\displaystyle J_{\rm sc}(x,y,t;x_0,y_0)$ $\displaystyle =$ $\displaystyle \int_{x_0}^x {\cal D}x \int_{y_0}^y{\cal D}y \exp \left[
i\int_{0}^{t}dt'\left(M\dot{x}_{t'}\dot{y}_{t'} - V'(x_{t'})y_{t'} \right)\right]$  
  $\displaystyle \times$ $\displaystyle \exp \left\{
-i \int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_1[x_s] {{y_{t'}}}
-\int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_2[x_s] {{y_{t'}y_{s}}}
\right\}.$  

The first step now is to perform an integration by parts to transform the term $ M\dot{x}_{t'}\dot{y}_{t'}$, and to re-arrange
    $\displaystyle J_{\rm sc}(x,y,t;x_0,y_0) = \int_{x_0}^x {\cal D}x \int_{y_0}^y{\cal D}y \exp \left[iM( \dot{x}_{t}y
- \dot{x}_{0}y_0)\right]$  
  $\displaystyle \times$ $\displaystyle \exp \left[
-i\int_{0}^{t}dt' y_{t'}
\left\{ M\ddot{x}_{t'} + V'(...
...\right\}
-\int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_2[x_s] {{y_{t'}y_{s}}}
\right]$  
$\displaystyle F_B[x_s,t']$ $\displaystyle \equiv$ $\displaystyle \int_{0}^{t'}ds \varphi_1[x_s]
= \int_{0}^{t'}ds\sum_{\alpha=1}^N 2$   Im $\displaystyle S_\alpha(t'-s)f'_\alpha[x_{t'}]f_\alpha[x_{s}]$  
  $\displaystyle =$ $\displaystyle -\sum_{\alpha=1}^N \int_{0}^{t'}ds
\frac{\sin \Omega_\alpha (t'-s)} {M_\alpha\Omega_\alpha} f'_\alpha[x_{t'}]f_\alpha[x_{s}].$ (233)

This is an interesting expression: the term in the brackets $ \{ \}$ looks likely to lead to a classical equation of motion,
$\displaystyle M\ddot{x}_{t'} + V'(x_{t'}) + F_B[x_s,t'] =0,$     (234)

where $ -V'(x_{t'})$ is the force due to the potential $ V(x)$, and $ F_B[x_s,t']$ is a retarded, position-dependent deterministic friction force due to the bath. In addition, however, there is the term quadratic in $ y$ containing the function
$\displaystyle \varphi_2[x_s]$ $\displaystyle \equiv$ $\displaystyle \sum_{\alpha=1}^N$   Re $\displaystyle S_\alpha(t'-s)f'_\alpha[x_{t'}]f'_\alpha[x_{s}]$  
  $\displaystyle =$ $\displaystyle \sum_{\alpha=1}^N \frac{1}{2M_\alpha\Omega_\alpha}
\coth \frac{\beta\Omega_\alpha}{2}
\cos \Omega_\alpha (t'-s) f'_\alpha[x_{t'}]f'_\alpha[x_{s}],$ (235)

which is the only place where the bath temperature $ T=1/\beta$ enters.


next up previous contents index
Next: Completing the Square Up: `Semiclassical' Limit for Damped Previous: `Semiclassical' Limit for Damped   Contents   Index
Tobias Brandes 2004-02-18