# Fourier transform of a functional derivative

Suppose that $$x(q)$$ is the Fourier transform of the function $$x(r)$$, where $$r$$ is the real-space variable and $$q$$ is the Fourier-space variable. Then, suppose that $$E$$ is an energy functional which can be differentiated either in real-space (with respect to $$x(r)$$) or Fourier-space (with respect to $$x(q)$$).

Representing the Fourier-transform operator as FT[], is the following relation logically valid?

$$\mathrm{FT}\left[\frac{\delta E}{\delta x(r)}\right] = \frac{\delta E}{\delta x(q)}$$

If not, what would be the correct mapping between these functional derivatives?

• I am not sure to understand your notation. Is $x_q$ a function? And is it the Fourier transform of the function $x_r$? If the answers are positive also the answer to your question should be positive (with a suitable interpretation of the formulae). Nov 3, 2022 at 7:07
• @ValterMoretti Yes, that's right. I have edited the question now to use conventional notation, i.e. $x_q = x(q)$ and $x_r = x(r)$. $x(q)$ is the Fourier transform of the function $x(r)$. Nov 3, 2022 at 7:40
• Hi Ferreroire. Is your question inspired by a particular physical context? Nov 3, 2022 at 8:04

1. Note that if a functional $$S[\phi]$$ is local in position space with variable $$x$$, it is typically not local in wavevector space with Fourier transformed variable $$k$$, where$$^1$$ $$\widetilde{\phi}(k)~=~ \int \!d^dx~e^{-ik\cdot x}\phi(k)\tag{1}$$ is the Fourier transform.

2. In physical applications, it is often convenient to define functional derivatives in $$x$$- and $$k$$-space with slightly different conventions for an infinitesimal variation: \begin{align}\delta S~=~&\int \!d^dx~\frac{\delta S}{\delta \phi(x)}\delta\phi(x)\cr ~=~&\int \!\frac{d^dk}{\color{red}{(2\pi)^d}}~\frac{\delta S}{\delta \widetilde{\phi}(\color{red}{-}k)}\delta\widetilde{\phi}(k),\end{align}\tag{2} which is inspired by the Plancherel theorem/convolution theorem.

3. Then it is easy to see that OP's conjecture is correct: $$FT\left[x\mapsto \frac{\delta S}{\delta \phi(x)}\right](k)~=~ \frac{\delta S}{\delta \widetilde{\phi}(k)}.\tag{3}$$

4. If one doesn't like to use a different convention (2) for the functional derivative in $$k$$-space marked in $$\rm \color{red}{red}$$ color, then the formula (3) will be modified accordingly.

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$$^1$$ Here we assume for simplicity a real field $$\phi$$, and leave to the reader to generalize to a complex field.

• I think that relevant theorem here is not the convolution one, but the Plancherel theorem... Nov 3, 2022 at 9:06
• Thanks @Valter Moretti. Yes, the Plancherel theorem is indeed closer. Edited. Nov 3, 2022 at 9:10

Without paying much attention to mathematical hypotheses (however they can be fixed), your idea is correct in view of the following "proof". $$E[\hat{x}(q)]= E\left[ \frac{1}{(2\pi)^{n/2}}\int e^{irq} \hat{x}(q) dr\right]$$ Hence $$\int \frac{\delta E}{\delta \hat{x}(q)} \hat{h}(q) dq = \frac{d}{d\alpha}|_{\alpha=0} E\left[ \frac{1}{(2\pi)^{n/2}}\int e^{irq} (\hat{x}(q) + \alpha \hat{h}(q)) dr \right]$$ $$=\frac{d}{d\alpha}|_{\alpha=0} E\left[ \frac{1}{(2\pi)^{n/2}}\int e^{irq} \hat{x}(q) dr+ \alpha (2\pi)^{-n/2}\int e^{irq} \hat{h}(q) dr \right]$$ $$=\frac{d}{d\alpha}|_{\alpha=0} E\left[x(r)+ \alpha (2\pi)^{-n/2}\int e^{irq} \hat{h}(q) dr \right] = \int \frac{\delta E}{ \delta x(r)} \frac{1}{(2\pi)^{n/2}}\int e^{irq} \hat{h}(q) dr dq$$ Using the Plancherel theorem, the found integral can be recast to
$$\int \left( \frac{1}{(2\pi)^{n/2}}\int e^{-iqr} \frac{\delta E}{ \delta x(r)} dr\right) \: \hat{h}(q) \:dq$$ This identity says that $$\frac{\delta E}{\delta \hat{x}(q)} = \frac{1}{(2\pi)^{n/2}}\int e^{-iqr} \frac{\delta E}{ \delta x(r)} dr$$ which is your thesis.