Functional derivative of effective action

[Ravendark]

1./2. Homework Statement

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.

Homework 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.

The Attempt at a Solution

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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?

 

[mark726]

Thank you for sharing your question regarding the computation of the derivative of the 1-loop effective action with respect to the inverse classical propagator. I understand that this can be a challenging task, but I will try my best to guide you through the process.

Firstly, as you correctly pointed out, the action does not depend on D, so we can focus on the term \frac{1}{2} \frac{\delta}{\delta D(x,y)} \operatorname{Tr} \log D^{-1}. To make things simpler, let's consider the functional derivative with respect to a general function F(D), instead of the specific point (x,y). This will allow us to write the final result in a more compact form.

Using the chain rule, we can write \frac{\delta}{\delta D(x,y)} \operatorname{Tr} \log D^{-1} = \frac{\delta}{\delta F(D(x,y))} \operatorname{Tr} \log D^{-1} \frac{\delta F(D(x,y))}{\delta D(x,y)}. Now, let's focus on the first term on the right-hand side. Using the properties of the functional derivative that you mentioned, we can write it as \frac{\delta}{\delta F(D(x,y))} \operatorname{Tr} \log D^{-1} = \frac{\delta}{\delta F(D(x,y))} \int d^4z \, \operatorname{tr} \log D^{-1}(z,z) = \int d^4z \, \frac{\delta}{\delta F(D(x,y))} \operatorname{tr} \log D^{-1}(z,z). Now, we can use the fact that the trace is a linear operation and split it into a sum of individual traces. This will allow us to use the functional derivative formula for a product, which you mentioned in your post.

So, we have \frac{\delta}{\delta F(D(x,y))} \operatorname{tr} \log D^{-1}(z,z) = \frac{\delta}{\delta F(D(x,y))} \sum_i \operatorname{tr} \log D^{-1}_i(z,z) = \sum_i \frac{\delta}{\delta F(D(x,y))} \operatorname{tr} \log D^{-1}_i(z,z)