What it the fourier transform of laplacian and shifted funtion?

Question

I'm looking for the Fourier transform of $\nabla^2f(\vec{r}-\vec{a})$ I can assume that the 3D Fourier transform of $f(\vec{r})$ is $\tilde{f}(\vec{q})$ and the vector $\vec{a}$ is a const vector. Thanks.

Answer 1

Let the Fourier transform, $\tilde{f}(\vec{q})$, of $f(\vec r)$ be given by

$$\tilde{f}(\vec{q})=\int f(\vec r)\,e^{i\vec q\cdot \vec r} \,d^3\vec r$$

Then, the Fourier transform of $\nabla^2f(\vec r-\vec a)$ is given by

$$\begin{align} \int \nabla^2f(\vec r-\vec a)\,e^{i\vec q\cdot \vec r} \,d^3\vec r&=-\int \nabla f(\vec r-\vec a)\cdot \,\nabla (e^{i\vec q\cdot \vec r}) \,d^3\vec r \tag 1\\\\ &=\int f(\vec r-\vec a)\,\nabla^2(e^{i\vec q\cdot \vec r}) \,d^3\vec r \tag 2\\\\ &=\int f(\vec r-\vec a)\,(-|\vec q|^2)\,e^{i\vec q\cdot \vec r}\,d^3\vec r \tag 3\\\\ &=-|\vec q|^2\,e^{i\vec q\cdot \vec a}\,\tilde{f}(\vec{q})\tag 4 \end{align}$$

In arriving at $(1)$, we integrated by parts by making use of the product rule $\nabla \cdot (\phi \vec F)=\phi \nabla \cdot \vec F+\nabla \phi \cdot \vec F$ along with the Divergence Theorem. We also assumed that $r^2 \nabla f$ goes to zero as $r \to \infty$.

In going from $(1)$ to $(2)$, we integrated by parts again.

In going from $(2)$ to $(3)$, we evaluated the Laplacian of the exponential term.

And in going from $(3)$ to $(4)$, we made a simple change of variables and carried out the integral.

Answer 2

Denote the Fourier transformation by $\def\F{\mathscr F}\F$, we have \begin{align*} \F[\Delta f(\cdot -a)](\xi) &= \def\c{\frac 1{(2\pi)^{3/2}}}\c \int_{\def\R{\mathbf R}\R^3} \Delta f(x-a)\exp(-i\def\<#1>{\left<#1\right>}\<x,\xi>)\, dx\\ &=-\c \int_{\R^3} \sum_j \partial_j f(x-a)(-i\xi_j)\exp(-i\<x,\xi>)\, dx\\ &= -\c \sum_{j} \xi_j^2 \int_{\R^3} f(x-a)\exp(-i\<x,\xi>)\, dx\\ &= -\<\xi,\xi> \cdot \c \int_{\R^3} f(x)\exp(-i\<x+a, \xi>)\, dx\\ &= -\<\xi,\xi> \exp(-i\<a,\xi>)\F f(\xi) \end{align*}