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Double Path Integrals

Now let us come back to our density operator for our system-bath Hamiltonian,
$\displaystyle H \equiv H_S+H_B+ H_{SB}, \chi(t)= e^{-iHt} \chi(t=0) e^{iHt}.$     (164)

For the moment, let us assume that the system has one degree of freedom $ q$ and the bath the degree of freedom $ x$ (the generalisation to many bath degrees of freedom $ x_i$ is straightforward). We then use a representation of $ \chi(t)$ in spatial coordinates,
$\displaystyle \langle x,q \vert \chi(t) \vert q',x'\rangle$ $\displaystyle =$ $\displaystyle \int dq_0dq_0' dx_0 dx_0' \langle x,q \vert
e^{-iHt}\vert q_0,x_0 \rangle \langle x_0 q_0 \vert\chi(t=0) \vert q'_0,x'_0 \rangle$  
  $\displaystyle \times$ $\displaystyle \langle x'_0 q'_0 \vert e^{iHt} \vert q',x'\rangle.$ (165)

We trace out the bath degree of freedoms to obtain an effective density matrix
$\displaystyle \rho(t) \equiv {\rm Tr_B} \chi(t)$     (166)

of the system,
$\displaystyle \langle q \vert \rho(t) \vert q'\rangle$ $\displaystyle =$ $\displaystyle \int dq_0dq_0' dx_0 dx_0' dx \langle x,q \vert
e^{-iHt}\vert q_0,x_0 \rangle \langle x_0 q_0 \vert\chi(t=0) \vert q'_0,x'_0 \rangle$  
  $\displaystyle \times$ $\displaystyle \langle x'_0 q'_0 \vert e^{iHt} \vert q',x\rangle.$ (167)

Now we realise that the Hamiltonian $ H \equiv H_S+H_B+ H_{SB}$ induces a classical action $ S_{\rm total} \equiv S_S[q]+S_B[x]+S_{SB}[xq]$, where in the following for notational simplicity we omit indices at the three $ S$. We use the path integral representation for the propagator matrix elements,
$\displaystyle \langle q \vert \rho(t) \vert q'\rangle$ $\displaystyle =$ $\displaystyle \int dq_0dq_0' dx_0 dx_0' dx
\int_{q_0}^{q} {\cal{D}}q
\int_{x_0}^{x} {\cal{D}}x
\int_{q_0'}^{q'} {\cal{D}^*}q'
\int_{x_0'}^{x'=x} {\cal{D}^*}x'$  
  $\displaystyle \times$ $\displaystyle \exp\left[ i \left(S[q]+S[x]+S[xq]\right)-i \left(S[q']+S[x']+S[x'q']\right) \right]$  
  $\displaystyle \times$ $\displaystyle \langle x_0 q_0 \vert\chi(t=0) \vert q'_0,x'_0 \rangle$  
  $\displaystyle =$ $\displaystyle \int dq_0dq_0'
\int_{q_0}^{q} {\cal{D}}q
\int_{q_0'}^{q'} {\cal{D}^*}q'
\exp\left[ i \left(S[q] - S[q']\right)\right]$  
  $\displaystyle \times$ $\displaystyle \int dx_0 dx_0' dx
\int_{x_0}^{x} {\cal{D}}x
\int_{x_0'}^{x} {\ca...
...^*}x'
\exp\left[ i \left(S[x]+S[xq]\right)-i \left(S[x']+S[x'q']\right) \right]$  
  $\displaystyle \times$ $\displaystyle \langle x_0 q_0 \vert\chi(t=0) \vert q'_0,x'_0 \rangle$  
  $\displaystyle =$ $\displaystyle [$assume $\displaystyle \chi(t=0) = \rho(0) \otimes \rho_B ]$  
  $\displaystyle =$ $\displaystyle \int dq_0dq_0' \langle q_0 \vert\rho(0)\vert q_0'\rangle
\int_{q_...
...exp\left[ i \left(S[q] - S[q']\right)\right] \underline{{\cal F}[q(t'),q'(t')]}$  
$\displaystyle {\cal F}[q(t'),q'(t')]$ $\displaystyle \equiv$ $\displaystyle \int dx_0 dx_0' dx \langle x_0 \vert\rho_B\vert x_0'\rangle$  
  $\displaystyle \times$ $\displaystyle \int_{x_0}^{x} {\cal{D}}x
\int_{x_0'}^{x} {\cal{D}^*}x'
\exp\left[ i \left(S[x]+S[xq]\right)-i \left(S[x']+S[x'q']\right) \right]$  


next up previous contents index
Next: The Influence Functional Up: Feynman-Vernon Influence Functional Theories Previous: Single Path Integrals   Contents   Index
Tobias Brandes 2004-02-18