Fourier transform of a functional derivative

Question

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?

Answer 1

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.

Answer 2

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.