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

$${\cal F}[x_{t'},y_{t'}] = \exp \left\{-\Phi[x_{t'},y_{t'}] \right\}$$

$$-\Phi[x_{t'},y_{t'}] = -\sum_{\alpha=1}^N\int_{0}^{t}dt'\int_{0}^{t'}ds \left\{ f_\alpha[x_{t'}+y_{t'}/2] - f_\alpha[x_{t'}-y_{t'}/2] \right\}$$

$$\times \left\{ S_\alpha(t'-s) f_\alpha[x_{s}+y_{s}/2] - S_\alpha^*(t'-s) f_\alpha[x_{s}-y_{s}/2]\right\}$$

$$= -\sum_{\alpha=1}^N\int_{0}^{t}dt'\int_{0}^{t'}ds f'_\alpha[x_{t'}]y_{t'}$$

$$\times \left\{\mbox{\rm Re } S_\alpha(t'-s) f'_\alpha[x_{s}]y_{s} + 2i \mbox{\rm Im } S_\alpha(t'-s) f_\alpha[x_{s}]\right\} + O\left[y_s\right]^3.$$ (229)

In the semiclassical approximation, we thus can write the influence as

$${\cal F}_{\rm sc}[x_{t'},y_{t'}] \equiv \exp \left\{-\Phi_{\rm }[x_{t'},y_{t'}] \right\} = \exp \left\{i\phi_1 - \phi_2 \right\}$$ (230)

$$i\phi_1 \equiv - i \int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_1[x_s] \underline{\underline{y_{t'}}},\quad \varphi_1[x_s]\equiv \sum_{\alpha=1}^N 2 S_\alpha(t'-s)f'_\alpha[x_{t'}]f_\alpha[x_{s}]$$

$$\phi_2 \equiv \int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_2[x_s] \underline{\underline{y_{t'}y_{s}}},\quad \varphi_2[x_s]\equiv \sum_{\alpha=1}^N S_\alpha(t'-s)f'_\alpha[x_{t'}]f'_\alpha[x_{s}].$$

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

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

$$+ i {\mu}\int_0^t dt'x_{t'}y_{t'}.$$ (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

$$\rho_{\rm sc}(x,y,t) = \int dx_0 dy_0 \rho_0(x_0,y_0) J_{\rm sc}(x,y,t;x_0,y_0)$$ (232)

$$J_{\rm sc}(x,y,t;x_0,y_0) = \int_{x_0}^x {\cal D}x \int_{y_0}^y{\cal D}y \exp \left[ i\int_{0}^{t}dt'\left(M\dot{x}_{t'}\dot{y}_{t'} - V'(x_{t'})y_{t'} \right)\right]$$

$$\times \exp \left\{ -i \int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_1[x_s] {{y_{t'}}} -\int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_2[x_s] {{y_{t'}y_{s}}} \right\}.$$

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

$$J_{\rm sc}(x,y,t;x_0,y_0) = \int_{x_0}^x {\cal D}x \int_{y_0}^y{\cal D}y \exp \left[iM( \dot{x}_{t}y - \dot{x}_{0}y_0)\right]$$

$$\times \exp \left[ -i\int_{0}^{t}dt' y_{t'} \left\{ M\ddot{x}_{t'} + V'(... ...\right\} -\int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_2[x_s] {{y_{t'}y_{s}}} \right]$$

$$F_B[x_s,t'] \equiv \int_{0}^{t'}ds \varphi_1[x_s] = \int_{0}^{t'}ds\sum_{\alpha=1}^N 2 S_\alpha(t'-s)f'_\alpha[x_{t'}]f_\alpha[x_{s}]$$

$$= -\sum_{\alpha=1}^N \int_{0}^{t'}ds \frac{\sin \Omega_\alpha (t'-s)} {M_\alpha\Omega_\alpha} f'_\alpha[x_{t'}]f_\alpha[x_{s}].$$ (233)

This is an interesting expression: the term in the brackets $\{ \}$ looks likely to lead to a classical equation of motion,

$$M\ddot{x}_{t'} + V'(x_{t'}) + F_B[x_s,t'] =0,$$ (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

$$\varphi_2[x_s] \equiv \sum_{\alpha=1}^N S_\alpha(t'-s)f'_\alpha[x_{t'}]f'_\alpha[x_{s}]$$

$$= \sum_{\alpha=1}^N \frac{1}{2M_\alpha\Omega_\alpha} \coth \frac{\beta\Omega_\alpha}{2} \cos \Omega_\alpha (t'-s) f'_\alpha[x_{t'}]f'_\alpha[x_{s}],$$ (235)

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