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Completing the Square

This is a useful trick when dealing with functional integrals. We start from the identity for a real symmetric, positive definite $ n\times n$ matrix $ A$,
$\displaystyle \fbox{$ \begin{array}{rcl} \displaystyle
%% e^{-y^2/(2a)} &=& \sq...
...dx_n
e^{-\frac{1}{2}{\bf x} A^{-1} {\bf x} + i {\bf y} {\bf x}}
\end{array}$\ }$     (236)

Exercise: Prove this identity. Hint: use the standard formula for Gaussian integrals and a linear transformation that diagonalises $ A$.

We now obtain

    $\displaystyle \exp \left[-\frac{1}{2}\int_{0}^{t}\int_{0}^{t}dt'ds A(t',s) {{y_...
...y} \exp \left[-\frac{\varepsilon^2}{2} \sum_{j,k=0}^{N-1} A_{jk}y_j y_k \right]$  
  $\displaystyle =$ $\displaystyle \lim_{N\to \infty} \left[2\pi \det A \right]^{-N/2}
\int d\xi_0.....
...}_{jk}}{\varepsilon^2}
\xi_k + i \varepsilon\sum_{j=0}^{N-1} y_j \xi _j \right]$  
  $\displaystyle =$ $\displaystyle \int {\cal D}\xi \exp \left[- \int_{0}^{t}\int_{0}^{t}dt'ds \xi_{t'}A^{-1}(t',s)\xi_{s}
+ i \int_{0}^{t} dt' y_{t'} \xi_{t'} \right]$ (237)
$\displaystyle {\cal D}\xi$ $\displaystyle \equiv$ $\displaystyle \lim_{N\to \infty} \left[2\pi \det A \right]^{-N/2}
d\xi_0...d\xi_{N-1}$  
$\displaystyle A(t',s)$ $\displaystyle =$ $\displaystyle \varphi_2[x](t',s),$   cf. Eq. (7.239)$\displaystyle .$  

Here, we have used the fact that the discrete inverse of an operator needs to be divided by $ \varepsilon^2$,
$\displaystyle A^{-1}(t',s) \leftrightarrow \frac{1}{\varepsilon^2} A^{-1}_{jk}.$     (238)

This can be derived by considering the discrete equivalent of the delta function and leads to the following translation table between continuous and discrete:
$\displaystyle f(x)$ $\displaystyle =$ $\displaystyle \int dx' \delta(x-x') f(x'),\quad
f_m = \sum_m \delta_{mn} f_n = {\varepsilon}\sum_m \frac{\delta_{mn}}{\varepsilon} f_n$  
  $\displaystyle \leadsto$ $\displaystyle \delta(x-x') \leftrightarrow \frac{\delta_{mn}}{\varepsilon}$  
$\displaystyle \int dx' A^{-1}(x,x') A(x',x'')$ $\displaystyle =$ $\displaystyle \delta(x-x'),\quad
\varepsilon\sum_{m'} \frac{A^{-1}_{mm'}}{\varepsilon^2} A_{m'm''}
= \frac{\delta_{mm''}}{\varepsilon}$  
  $\displaystyle \leadsto$ $\displaystyle A^{-1}(x,x') \leftrightarrow \frac{1}{\varepsilon^2} A^{-1}_{mm'}.$ (239)

Now using the fact that $ \varphi_2$ is symmetric in $ t'$ and $ s$, we have
$\displaystyle \int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_2[x_s] {{y_{t'}y_{s}}}=
\frac{1}{2}\int_{0}^{t}dt\int_{0}^{t}ds \varphi_2[x_s] {{y_{t'}y_{s}}}$     (240)

and therefore
    $\displaystyle J_{\rm sc}(x,y,t;x_0,y_0) = \int_{x_0}^x {\cal D}x \int_{y_0}^y{\cal D}y e^{iM( \dot{x}_{t}y
- \dot{x}_{0}y_0)}$  
  $\displaystyle \times$ $\displaystyle \exp \left[
-i\int_{0}^{t}dt' y_{t'}
\left\{ M\ddot{x}_{t'} + V'(...
...frac{1}{2}\int_{0}^{t}\int_{0}^{t}dt' ds \varphi_2[x_s] {{y_{t'}y_{s}}}
\right]$  
  $\displaystyle =$ $\displaystyle \int_{x_0}^x {\cal D}x \int_{y_0}^y{\cal D}y e^{iM( \dot{x}_{t}y
...
...{1}{2}
\int_{0}^{t}\int_{0}^{t}dt'ds \xi_{t'}\varphi_2[x_s]^{-1}\xi_{s} \right]$  
  $\displaystyle \times$ $\displaystyle \exp \left[
-i\int_{0}^{t}dt' y_{t'}
\left\{ M\ddot{x}_{t'} + V'(x_{t'}) + F_B[x_s,t'] -\xi_{t'} \right\}
\right].$ (241)

Here we have explicitly indicated the dependence of the measure $ {\cal D}\xi[x]$ on the paths $ x_s$, which enters through the determinant of the operator $ \varphi_2[x_s]$. The pathintegral over $ {\cal D}y$ is now very easy: we find ( $ \varepsilon=t/N$)
    $\displaystyle \int_{y_0}^y{\cal D}y
\exp \left[ -i\int_{0}^{t}dt' y_{t'} b_{t'}...
...}\right)^\frac{N}{2}
\exp \left[ -{i\varepsilon}\sum_{j=0}^{N-1} y_jb_j \right]$  
  $\displaystyle =$ $\displaystyle \lim_{N\to \infty} \left(\frac{M}{2\pi i \varepsilon}\right)^\fra...
...a(b_{N-1})}{\varepsilon}
e^{-i\varepsilon y_0b_0}
\equiv \Delta(y_{t'}-b_{t'}),$ (242)

Here, $ \Delta$ indicates the product of delta functions that fixes the $ y_{t'}$ path to the $ b_{t'}$ path, and for $ \varepsilon\to 0$ the $ e^{-i\varepsilon y_0b_0}$ becomes irrelevant. Inserting yields
    $\displaystyle J_{\rm sc}(x,y,t;x_0,y_0) = e^{iM( \dot{x}_{t}y- \dot{x}_{0}y_0)}...
...{1}{2}
\int_{0}^{t}\int_{0}^{t}dt'ds \xi_{t'}\varphi_2[x_s]^{-1}\xi_{s} \right]$  
  $\displaystyle \times$ $\displaystyle \Delta( M\ddot{x}_{t'} + V'(x_{t'}) + F_B[x_s,t'] -\xi_{t'} ).$ (243)


next up previous contents index
Next: Wigner Distribution in `Semi-classical' Up: `Semiclassical' Limit for Damped Previous: Expansion of the Influence   Contents   Index
Tobias Brandes 2004-02-18