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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?
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.
