1
$\begingroup$

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 $p$.

$\endgroup$
  • 1
    $\begingroup$ Please include all relevant equations and definitions into the post and make the title into a question reflective of the actual content. $\endgroup$ – ACuriousMind Apr 11 '15 at 1:59
  • 1
    $\begingroup$ 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. $\endgroup$ – Leandro M. Apr 11 '15 at 2:06
  • $\begingroup$ Sorry about my poor phrasing. I've edited it a little bit and hopefully it is clearer now. $\endgroup$ – user110373 Apr 11 '15 at 2:22
  • $\begingroup$ Related: physics.stackexchange.com/q/56070/2451 and physics.stackexchange.com/q/173603/2451 $\endgroup$ – Qmechanic Apr 11 '15 at 6:54
1
$\begingroup$

I did not get my copy of Srednicki out but from what you have written...

Srednicki is referencing the method of steepest descent. Although these notes look to be better than wikipedia. Another page that is directly applicable to the quantum field theory case is here.

In short, exponential integrals may be estimated by the saddle points of the integrand. Using a Minkowski formalism the saddle points are related to how oscillatory the integral is.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.