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?

  • 1
    $\begingroup$ Are you using the same convention for the sign of the exponent in the Fourier transform? $\endgroup$ – DJBunk Oct 2 '12 at 12:50
  • $\begingroup$ Yes I am sure I'm using the same definition... from the details in the paper it seems they obtain a different delta function $\delta(k+k\prime)$ when computing the derivative of log in k... but it does not make any sense! :) $\endgroup$ – Juan Sebastian Totero Oct 2 '12 at 12:59
  • $\begingroup$ Wait, maybe I got it... it seems I am performing a direct Fourier Transform of the $\delta(k)$, and this should produces me the complex coniugate of its anti-transform, am I right? $\endgroup$ – Juan Sebastian Totero Oct 2 '12 at 13:03
  • 2
    $\begingroup$ Remark: it is not good to denote a function ($\Lambda$ in your case) and its Fourier image with the same letter since generally they are different functions of their arguments. $\endgroup$ – Vladimir Kalitvianski Oct 2 '12 at 18:48
  • 1
    $\begingroup$ Qmechanic : It's a private report, so it's not published ATM :) $\endgroup$ – Juan Sebastian Totero Oct 3 '12 at 8:26

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.

  • $\begingroup$ Yes I know you can do it by glance, but I did every step just to be sure something pedantic was not happening here :). The next step in my work here was to integrate the above derivative, notice it is just an inverse fourier transform (due to the supposed positive exponent), and then use the convolution theorem to move back to the real space. Actually, this is not a real problem because I found a much faster solution that avoids me this calculation and produces me the right result in zero time :) Thx, I'll set your answer as right, given it's a good example of a direct calculation :) $\endgroup$ – Juan Sebastian Totero Nov 5 '12 at 11:01
  • $\begingroup$ @JuanSebastianTotero: It is just doing a symmetric Fourier transform, where you don't have to worry about complex conjugations. $\endgroup$ – Ron Maimon Nov 5 '12 at 13:51

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.

  • $\begingroup$ Yes it could make a great difference in this case... but applying the definition of functional derivative, the delta should be $/delta(k-k\prime)$ (the first one is the silent variable, the second one is the actual variable of the resulting derivative)... how do you obtain the inverted delta $\endgroup$ – Juan Sebastian Totero Oct 2 '12 at 15:55
  • $\begingroup$ As I said, this is a very vague argument which just shows the only thing I found. Theoretically, you can switch from $\delta(k-k')$ to $\delta(k'-k)$ since it is an even function, but then the integral should give the same result. This difference in integration is what bothers me most, now. $\endgroup$ – Ondřej Černotík Oct 2 '12 at 16:25

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.