# Path integral in quantum mechanics

I am confused by the derivation in Srednicki QFT's chapter 6 from (6.8) to (6.9). In (6.8), we have

$$<q'',t''|q',t'>~=~\int DqDp \exp[i\int_{t'}^{t''}dt(p\dot{q}-H(p,q))],\tag{6.8}$$

and (6.9) we have

$$<q'',t''|q',t'>=\int Dq \exp[i\int_{t'}^{t''}dt L(\dot{q},q)].\tag{6.9}$$

It is clear for me that one can work out each infinitesimal integral $$<q_k|\exp[-i\delta t \frac{p^2}{2}]|q_{k-1}>~\sim~ \exp[\frac{i(q_k-q_{k-1})^2\delta t}{2}]$$ to derive the above formula. But I'm confused by the way that is presented in the book. It makes it sound like there is a more general way of computing path integral by finding the stationary point, i.e. given

$$\int Dp \exp[i\int f(p,t)dt]$$

is the result $$\exp[i\int F(t)dt]$$ where $F(t)=f(p(t),t)$ such that $p(t)$ is a stationary point of $f(p,t)$ with respect to $p$.

• Please include all relevant equations and definitions into the post and make the title into a question reflective of the actual content. – ACuriousMind Apr 11 '15 at 1:59
• Yeah, I have no idea what you're asking. Are you wondering why it is that the stationarity condition is fulfilled at all since the integrand is a pure phase? Or is it something else? I also don't understand your second question. – Leandro M. Apr 11 '15 at 2:06
• Sorry about my poor phrasing. I've edited it a little bit and hopefully it is clearer now. – user110373 Apr 11 '15 at 2:22
• – Qmechanic Apr 11 '15 at 6:54