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$^1$ $$\begin{align}\widetilde{\phi}(k)~=~& \int \!d^dx~e^{-ik\cdot x}\phi(x),\cr \phi(x)~=~& \int \!\frac{d^dk}{(2\pi)^d}~e^{ik\cdot x}\widetilde{\phi}(k), \end{align} \tag{1}$$ are the Fourier transform and its inverse in a typical physics convention.

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: $$\begin{align}\delta S~=~&\int \!d^dx~\frac{\delta S}{\delta \phi(x)}\delta\phi(x)\cr ~=~&\int \!\frac{d^dk}{\color{red}{(2\pi)^d}}~\frac{\delta S}{\delta \widetilde{\phi}(\color{red}{-}k)}\delta\widetilde{\phi}(k).\end{align}\tag{2}$$ These conventions are e.g. inspired by the Plancherel theorem/convolution theorem: $$ \int \!d^dx~\phi(x)\psi(x) ~=~\int \!\frac{d^dk}{(2\pi)^d}~\widetilde{\phi}(-k)\widetilde{\psi}(k).\tag{3}$$

3. From (2) it is easy to see that OP's conjecture is correct: $$ FT\left[x\mapsto \frac{\delta S}{\delta \phi(x)}\right](k)~=~ \frac{\delta S}{\delta \widetilde{\phi}(k)}.\tag{4} $$

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 (4) 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.

Answer 3

A more direct way to derive the result is with the aid of the chain rule. So, consider a function $E[f]$, where $f(x)$ is a function of the one dimensional variable $x$. It is related to a spectrum via the Fourier and inverse Fourier transform $$ f(x) = \mathcal{F}^{-1}\{g(k)\} = \int g(k) \exp(ikx) \frac{dk}{2\pi} $$ $$ g(k) = \mathcal{F}\{f(x)\} = \int f(x) \exp(-ikx) dx . $$

Now, we consider the functional derivative with the aid of the chain rule $$ \frac{\delta E[f]}{\delta g(k)} = \int \frac{\delta E[f]}{\delta f(x)} \frac{\delta f(x)}{\delta g(k)} dx . $$ The last functional derivative becomes $$ \frac{\delta f(x)}{\delta g(k)} = \frac{\delta}{\delta g(k)} \int g(k') \exp(ik'x) \frac{dk'}{2\pi} = \int \delta(k-k') \exp(ik'x) \frac{dk'}{2\pi} = \frac{1}{2\pi} \exp(ikx) . $$ After substituting it back, we get $$ \frac{\delta E[f]}{\delta g(k)} = \frac{1}{2\pi} \int \frac{\delta E[f]}{\delta f(x)} \exp(ikx) dx . $$ It looks like a Fourier transform, but the sign in the argument of the exponential is wrong. We can fix this sign by change the sign of either $x$ or $k$. So in the end we have $$ \frac{\delta E[f]}{\delta g(-k)} = \frac{1}{2\pi} \mathcal{F}\left\{ \frac{\delta E[f]}{\delta f(x)}\right\} . $$ We did not need to invoke the Plancherel theorem. We used one-dimensional functions for convenience. The result can be readily generalized to higher dimensions.