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 $A$,

$$\fbox{$ \begin{array}{rcl} \displaystyle %% e^{-y^2/(2a)} &=& \sq... ...dx_n e^{-\frac{1}{2}{\bf x} A^{-1} {\bf x} + i {\bf y} {\bf x}} \end{array}$\ }$$ (236)

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

We now obtain

$$\exp \left[-\frac{1}{2}\int_{0}^{t}\int_{0}^{t}dt'ds A(t',s) {{y_... ...y} \exp \left[-\frac{\varepsilon^2}{2} \sum_{j,k=0}^{N-1} A_{jk}y_j y_k \right]$$

$$= \lim_{N\to \infty} \left[2\pi \det A \right]^{-N/2} \int d\xi_0..... ...}_{jk}}{\varepsilon^2} \xi_k + i \varepsilon\sum_{j=0}^{N-1} y_j \xi _j \right]$$

$$= \int {\cal D}\xi \exp \left[- \int_{0}^{t}\int_{0}^{t}dt'ds \xi_{t'}A^{-1}(t',s)\xi_{s} + i \int_{0}^{t} dt' y_{t'} \xi_{t'} \right]$$ (237)

$${\cal D}\xi \equiv \lim_{N\to \infty} \left[2\pi \det A \right]^{-N/2} d\xi_0...d\xi_{N-1}$$

$$A(t',s) = \varphi_2[x](t',s), .$$

Here, we have used the fact that the discrete inverse of an operator needs to be divided by $\varepsilon^2$,

$$A^{-1}(t',s) \leftrightarrow \frac{1}{\varepsilon^2} A^{-1}_{jk}.$$ (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:

$$f(x) = \int dx' \delta(x-x') f(x'),\quad f_m = \sum_m \delta_{mn} f_n = {\varepsilon}\sum_m \frac{\delta_{mn}}{\varepsilon} f_n$$

$$\leadsto \delta(x-x') \leftrightarrow \frac{\delta_{mn}}{\varepsilon}$$

$$\int dx' A^{-1}(x,x') A(x',x'') = \delta(x-x'),\quad \varepsilon\sum_{m'} \frac{A^{-1}_{mm'}}{\varepsilon^2} A_{m'm''} = \frac{\delta_{mm''}}{\varepsilon}$$

$$\leadsto A^{-1}(x,x') \leftrightarrow \frac{1}{\varepsilon^2} A^{-1}_{mm'}.$$ (239)

Now using the fact that $\varphi_2$ is symmetric in $t'$ and $s$, we have

$$\int_{0}^{t}dt'\int_{0}^{t'}ds \varphi_2[x_s] {{y_{t'}y_{s}}}= \frac{1}{2}\int_{0}^{t}dt\int_{0}^{t}ds \varphi_2[x_s] {{y_{t'}y_{s}}}$$ (240)

and therefore

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

$$\times \exp \left[ -i\int_{0}^{t}dt' y_{t'} \left\{ M\ddot{x}_{t'} + V'(... ...frac{1}{2}\int_{0}^{t}\int_{0}^{t}dt' ds \varphi_2[x_s] {{y_{t'}y_{s}}} \right]$$

$$= \int_{x_0}^x {\cal D}x \int_{y_0}^y{\cal D}y e^{iM( \dot{x}_{t}y ... ...{1}{2} \int_{0}^{t}\int_{0}^{t}dt'ds \xi_{t'}\varphi_2[x_s]^{-1}\xi_{s} \right]$$

$$\times \exp \left[ -i\int_{0}^{t}dt' y_{t'} \left\{ M\ddot{x}_{t'} + V'(x_{t'}) + F_B[x_s,t'] -\xi_{t'} \right\} \right].$$ (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 ( $\varepsilon=t/N$)

$$\int_{y_0}^y{\cal D}y \exp \left[ -i\int_{0}^{t}dt' y_{t'} b_{t'}... ...}\right)^\frac{N}{2} \exp \left[ -{i\varepsilon}\sum_{j=0}^{N-1} y_jb_j \right]$$

$$= \lim_{N\to \infty} \left(\frac{M}{2\pi i \varepsilon}\right)^\fra... ...a(b_{N-1})}{\varepsilon} e^{-i\varepsilon y_0b_0} \equiv \Delta(y_{t'}-b_{t'}),$$ (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

$$J_{\rm sc}(x,y,t;x_0,y_0) = e^{iM( \dot{x}_{t}y- \dot{x}_{0}y_0)}... ...{1}{2} \int_{0}^{t}\int_{0}^{t}dt'ds \xi_{t'}\varphi_2[x_s]^{-1}\xi_{s} \right]$$

$$\times \Delta( M\ddot{x}_{t'} + V'(x_{t'}) + F_B[x_s,t'] -\xi_{t'} ).$$ (243)