Field theory:functional derivative involving Fourier Transform

Question

I have to solve the following functional derivative $$ \frac{\delta}{\delta \Lambda(\mathbf{x})}\log[A-\mathbf{k}^2\Lambda(\mathbf{k})] $$ where $\Lambda(\mathbf{k})$ is the Fourier transform of $\Lambda(\mathbf{x})$, namely $$ \Lambda(\mathbf{k})=\int d\mathbf{x\prime} e^{-i\mathbf{k}\cdot\mathbf{x\prime}}\Lambda(\mathbf{x\prime}) $$

My interpretation is to consider $\Lambda(\mathbf{k})$ as a functional over $\Lambda(\mathbf{x})$ and hence apply the chain rule $$ \frac{\delta}{\delta \Lambda(\mathbf{x})}\log[A-\Lambda(\mathbf{k})]= \int d\mathbf{k{\prime}} \frac{\delta\log[A-\mathbf{k}^2\Lambda(\mathbf{k})]}{\delta \Lambda(\mathbf{k\prime})} \frac{\delta \Lambda(\mathbf{k\prime})}{\delta \Lambda(\mathbf{x})} $$ obtaining $$ \frac{\delta \Lambda(\mathbf{k\prime})}{\delta \Lambda(\mathbf{x})}= e^{-i\mathbf{k\prime}\cdot\mathbf{x}} $$ and $$ \frac{\delta\log[A-\mathbf{k}^2\Lambda(\mathbf{k})]}{\delta \Lambda(\mathbf{k\prime})}= \frac{-\mathbf{k}^2}{A-\mathbf{k}^2\Lambda(\mathbf{k})}\delta(\mathbf{k}-\mathbf{k\prime}) $$

The final result would be $$ \frac{\delta}{\delta \Lambda(\mathbf{x})}\log[A-\mathbf{k}^2\Lambda(\mathbf{k})]= \frac{-\mathbf{k}^2}{A-\mathbf{k}^2\Lambda(\mathbf{k})} \int d\mathbf{k{\prime}} \delta(\mathbf{k}-\mathbf{k\prime}) e^{-i\mathbf{k\prime}\cdot\mathbf{x}}\\ =\frac{-\mathbf{k}^2}{A-\mathbf{k}^2\Lambda(\mathbf{k})} e^{-i\mathbf{k}\cdot\mathbf{x}} $$ but in the paper I'm studying, the final result has a positive exponent $exp[i\mathbf{k}\cdot\mathbf{x}]$. What am I doing wrong?

Answer 1

You aren't doing anything wrong, the paper made a mistake. It probably doesn't affect the result at all, since it is only a complex conjugation difference. But you are working a little too hard. First note:

$$ {\delta \Lambda(k) \over \delta \Lambda(x) } = {\delta\over\delta\Lambda(x)} \int e^{-ikx'} \Lambda(x') dx' = e^{-ikx} $$

you could say by definition. Then

$$ {\delta\over\delta \Lambda x} \log( A - k^2 \Lambda(k)) = {- k^2 e^{-ikx}\over A-k^2\Lambda(k)}$$

There's no need to do formal steps, you can write it down immediately.

Answer 2

The only thing that comes to my mind is that when taking the derivative of $A-\textbf{k}^2\Lambda(\textbf{k})$ with respect to $\Lambda(\textbf{k'})$, the delta function should be of opposite argument, i.e., $\delta(\textbf{k'}-\textbf{k})$. That shouldn't really matter, since it is an even distribution, but the opposite ordering results in $\exp(i\textbf{k}\cdot\textbf{x})$ after the integration. But all this is just a wild guess.