# 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). Commented 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 $x(r)$. Commented Nov 3, 2022 at 7:40
• Hi Ferreroire. Is your question inspired by a particular physical context? Commented 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$$^1$$ \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.

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

• Where did the $dq$ in the last expression of your 3-line expression come from? Is it a typo? Commented May 20 at 4:01

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.