While reading the book “Modern Consensed Matter Theory”, I came across the following calculation.

$f(k) g(k) = \int d^d r e^{-i kr} f(-i \nabla) g(r)$

I know the convolution theorem for Fourier transform, but this formula still confuses me, can anybody explain how this calculation works?

  • 1
    $\begingroup$ It seems that convolution does not play an important role. It seems that it is the formal expression, so $f(k)g(k)=\int_r e^{-ikr}\sum_nf^{(n)}(0)\frac{(-i\nabla)^n}{n!}g(r),$ then you act on $g(r)$ and find the result $\endgroup$ Mar 14, 2020 at 16:59
  • $\begingroup$ Oh that is clever! So we could see it as transforming g only. Thanks! $\endgroup$
    – Jiahao Fan
    Mar 14, 2020 at 17:03

1 Answer 1


Hint: integrate each and every gradient by parts, $$ \int d^d r ~ e^{-i kr} f(-i \nabla)\tilde g(r) = \int d^d r ~ \tilde g(r) f(i \nabla) e^{-i kr} = \int d^d r ~ f( k) \tilde g(r) e^{-i kr} = f(k)g(k). $$

NB. To see why this is exactly equivalent to the convolution theorem, ignore fussy normalization distractions and work in just one dimension. The important thing to remember is Lagrange's translation operator, $$e^{-y\partial_r} \tilde g(r)= \tilde g(r-y).$$ You then see that $$ f(-i\partial_r)~ \tilde g(r) = \int dy ~\tilde f(y)~ e^{-iy(-i\partial_r)} ~ \tilde g(r)= \int dy ~ \tilde f(y) ~\tilde g(r-y), $$ a convolution.


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