Let $\gamma$ be an oriented path from a to b in the complex plane. If A(z) is an operator valued function of z, then define the line integral

$\parbox{9cm}{ \begin{displaymath} \int_{\gamma} A(z)dz \equiv \sum_{i=1}^{N} A(z_{i})(z_{i+1}-z_{i}) \end{displaymath}}$ $\parbox{5cm}{ \epsfig{file=path1.eps,angle=0,width=1.2in} }$

The time ordered product along the path $\gamma$ orders the operators along $\gamma$. In the example

$\parbox{9cm}{ \begin{displaymath} T_{\gamma}(A(z_{3})A(z_{2})A(z_{N})) = A(z_{N})A(z_{3})A(z_{2}) \end{displaymath}}$ $\parbox{5cm}{ \epsfig{file=path2.eps,angle=0,width=1.3in} }$

Prove that

$$\left( T_{\gamma} \exp \frac{1}{i} \int_{\gamma} H(\tau)d\ta... ...mma}\exp \frac{1}{i} \int_{\tilde\gamma} \tilde H(z)dz \right)$$

where $\tilde \gamma$ is the path $\gamma$ traversed in the reverse direction (i.e. if $\gamma$ goes from $a$ to b, then $\tilde \gamma$ goes from $b^{*}$ to $a^{*}$), and $\tilde H (z) =(H(z^{*}))^{\dagger}$