Influence Phase

The influence phase can be obtained directly from its definition, Eq. (7.173),

$${\cal F}[q(t'),q'(t')] \equiv {\rm Tr_B} \left(\rho_B U_B^{\dagger}[q'] U_B[q] \right) = {\rm T... ... \left(\rho_B \tilde{U}^{\dagger}[q'] e^{iH_0t} e^{-iH_0t} \tilde{U}[q] \right)$$

$$= {\rm Tr_B} \left(\rho_B e^{iC'\hat{p}} e^{iB'\hat{x}}e^{iA'} e^{-iA}e^{-iB\hat{x}}e^{-iC\hat{p}} \right)$$

$$= e^{i(A'-A)} \int dx \langle x\vert \rho_B \vert e^{iC'\hat{p}} e^{i(B'-B)\hat{x}} e^{-iC\hat{p}} \vert x \rangle$$

$$= e^{i(A'-A)} \int dx \langle x\vert \rho_B \vert x +C-C' \rangle e^{i(B'-B)(x+C)},$$

where for a moment we abbreviated $A$, $A'$ etc. for the integrals Eq. (7.181) with $g(t') \equiv f(q(t'))$ in the undashed and $g'(t') \equiv f(q'(t'))$ in the dashed (not the derivative) quantities. We now assume a thermal equilibrium for the density operator $\rho_B$,

$$\rho_B = \frac{e^{-\beta H_0}}{Z},\quad Z={\rm Tr}e^{-\beta H_0 }= \frac{1}{2 \sinh \beta \Omega/2}$$ (181)

$$\langle x \vert \rho_B \vert x' \rangle = \frac{1}{Z}\sqrt{\frac{M\Omega}{2\pi \sinh \Omega \beta}} \exp \l... ... } \left [ \left( x^2+x'^2\right) \cosh \beta \Omega - 2 x x' \right] \right\},$$

where we used the matrix elements of the propagator $\langle x\vert e^{-iH_0 t}\vert x'\rangle$ for $it = \beta$ (Wick rotation of the time $t$). Doing the Gaussian integral yields

$${\cal F}[q(t'),q'(t')] = e^{i(A'-A)} \int dx \langle x\vert \rho_B \vert x +C-C' \rangle e^{i(B'-B)(x+C)}$$

$$= e^{i(A'-A)} \frac{1}{Z}\sqrt{\frac{M\Omega}{2\pi \sinh \Omega \beta}}\int dx e^{i(B'-B)(x+C)}$$

$$\times \exp \left\{ -\frac{M\Omega}{2 \sinh \beta \Omega } \left [ \left( x^2+(x+C-C')^2\right) \cosh \beta \Omega - 2 x (x+C-C') \right] \right\}$$

$$= \left[\mbox{\rm use } \tanh \frac{x}{2}=\frac{\cosh x -1}{\sinh x},\quad \coth x -\frac{1}{2}\tanh \frac{x}{2}=\frac{1}{2}\coth \frac{x}{2}\right]$$

$$= \exp \left\{ i(A'-A)+\frac{i}{2} (B'-B)(C+C') \right \}$$

$$\times \exp \left\{ -\frac{1}{4M\Omega} \coth\frac{\beta\omega}{2}\left[ (B'-B)^2 + M^2\Omega^2(C-C')^2\right] \right\}.$$ (182)

the last step now is to re-insert the definitions of $A,B,C,A',B',C'$. The resulting long expression

$${\cal F}[q(t'),q'(t')]$$ (183)

$$= \exp \left\{ -\frac{1}{4M\Omega}\coth\frac{\beta\omega}{2} \int_0^tdt'\int_0^{t}ds (g'_{t'}-g_{t'})(g'_{s}-g_{s}) \cos \Omega(t'-s) \right\}$$

$$\times \exp\left\{ {-\frac{i}{M\Omega} \int_0^tdt'\int_0^{t'}ds (g'_{t'}g'_s-g_{t'}g_s)\cos \Omega s \sin \Omega t'}\right\}$$

$$\times \exp\left\{ {\frac{i}{2M\Omega} \int_0^tdt'\int_0^{t}ds (g'_{t'}g'_s-g_{t'}g_s)\cos \Omega t' \sin \Omega s}\right\}$$

$$\times \exp\left\{ {\frac{i}{2M\Omega} \int_0^tdt'\int_0^{t}ds (g'_{t'}g_s-g_{t'}g'_s)\cos \Omega t' \sin \Omega s}\right\}$$ (184)

can be further simplified with $\sin \alpha\cos \beta = \frac{1}{2}[\sin (\alpha-\beta) + \sin ( \alpha+\beta)]$ and carefully considering the limits of the integrals and the symmetry of the integrands. Re-installing furthermore $g_{t'}\equiv g(t') \equiv f[q_{t'}]$ (we write the time-arguments as an index to avoid bulky expressions with too many brackets), the result can be written in a compact form, the Feynman-Vernon Influence Functional for the coupling of a single particle to a single harmonic oscillator in thermal equilibrium,

$$H = H_S[q] + H_B[x] + H_{SB}[xq]= H_S[q] + \frac{p^2}{2M}+ \frac{1}{2}M\Omega^2 x^2 + f[q] x$$

$$\langle q \vert \rho(t) \vert q'\rangle = \int dq_0dq_0' \langle q_0 \vert\rho(0)\vert q_0'\rangle \int_{q_... ...p\left[ i \left(S[q] - S[q']\right)\right] \underline{{\cal F}[q_{t'},q'_{t'}]}$$

$${\cal F}[q_{t'},q'_{t'}] = \exp \left\{-\Phi[q_{t'},q'_{t'}] \right\}$$

$$\Phi[q_{t'},q'_{t'}] = \int_{0}^{t}dt'\int_{0}^{t'}ds \left\{ f[q_{t'}] - f[q'_{t'}] \right\} \left\{ L(t'-s) f[q_{s}] - L^*(t'-s) f[q'_{s}]\right\}$$

$$L(\tau) = \frac{1}{2M\Omega} \left( \coth \frac{\beta\Omega}{2}\cos \Omega \tau - i \sin \Omega \tau\right).$$ (185)