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Functional derivative of effective action

  1. Nov 1, 2015 #1
    1./2. The problem statement, all variables and given/known data

    In my QFT lecture we were introduced to the 1PI effective action ##\Gamma[\varphi]## for a scalar theory (in Euclidean space-time). In one-loop approximation we've found [tex] \Gamma^{(\text{1-loop})}[\varphi] = S[\varphi] + \frac{1}{2} \operatorname{Tr} \log D^{-1}[/tex] where ##D^{-1} = \delta^2 S[\varphi] / \delta \varphi \delta \varphi## denotes the inverse classical propagator.

    Now I'm asked to compute the derivative of ##\Gamma^{(\text{1-loop})}## with respect to ##D##. And this is is point where I got stuck.

    2. Relevant equations

    We had the functional derivative in the exercises and we derived formulas like [tex]\frac{\delta}{\delta f(y)} f(x) = \delta(x-y) \; , \qquad \frac{\delta}{\delta f(y)} \int dx \, f(x)^n = n f(y)^{n-1} \quad (n \in \mathbb{N})\; .[/tex] Furthermore, I know that the functional derivative obeys the product and chain rule.

    3. The attempt at a solution

    Since the action does not depend on ##D##, I have: [tex] \frac{\delta}{\delta D(x,y)} \Gamma^{(\text{1-loop})}[\varphi] = \frac{1}{2} \frac{\delta}{\delta D(x,y)} \operatorname{Tr} \log D^{-1}[/tex] Now I'm not sure how do deal with the trace...I think it is meant in the functional sense, thus [tex]\frac{\delta}{\delta D(x,y)} \Gamma^{(\text{1-loop})}[\varphi] = \frac{1}{2} \frac{\delta}{\delta D(x,y)} \int d^4z \, \operatorname{tr} \log D^{-1}(z,z)[/tex] where the lowercase tr runs over internal indices (colour, flavour...). Now I have no idea how to proceed...there is this trace, a log and furthermore the inverse of ##D## instead ##D## itself. Can someone give me a hint, please?
     
  2. jcsd
  3. Nov 6, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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