# Functional derivative of effective action

Tags:
1. Nov 1, 2015

### Ravendark

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 $$\Gamma^{(\text{1-loop})}[\varphi] = S[\varphi] + \frac{1}{2} \operatorname{Tr} \log D^{-1}$$ 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 $$\frac{\delta}{\delta f(y)} f(x) = \delta(x-y) \; , \qquad \frac{\delta}{\delta f(y)} \int dx \, f(x)^n = n f(y)^{n-1} \quad (n \in \mathbb{N})\; .$$ 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: $$\frac{\delta}{\delta D(x,y)} \Gamma^{(\text{1-loop})}[\varphi] = \frac{1}{2} \frac{\delta}{\delta D(x,y)} \operatorname{Tr} \log D^{-1}$$ Now I'm not sure how do deal with the trace...I think it is meant in the functional sense, thus $$\frac{\delta}{\delta D(x,y)} \Gamma^{(\text{1-loop})}[\varphi] = \frac{1}{2} \frac{\delta}{\delta D(x,y)} \int d^4z \, \operatorname{tr} \log D^{-1}(z,z)$$ 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. Nov 6, 2015