Fourier Transform of Poisson Equation While trying to solve the Poisson Equation by using Green's Function I have to Fourier transform the equation i.e
$$-\nabla^{2}\phi(r)=\rho(r).$$ 
In the book after Fourier transform, the solution is written as $$k^{2}\phi(k)=\rho(k).$$ 
I can understand the expression in the right hand side, but why there is $k^{2}$ in the left hand side?
 A: i think $\nabla^{2}\phi(r)=\rho(r)$ is the same as $$ \frac{d^2}{dr^2} \phi(r) = \rho(r) $$.  the convention i am used to in electrical engineering is to use large case letters for the Fourier Transform and use $\omega$ for angular frequency instead of "$k$".  in that notation, Fourier transforming both sides is $$ (i \omega)^2 \Phi(\omega) = \mathrm{P}(\omega) $$
since $i^2 = -1$, i am curious if you're missing a minus sign with $k^2$.
A: It's easiest to see if you start with the definition of the inverse fourier transform
$$ f(\mathbf x) = \int d\mathbf k \, \hat f(\mathbf k) e^{i \mathbf k \cdot \mathbf x} $$
and take a laplacian of both sides
$$ \nabla^2 f(\mathbf x) = \nabla^2 \int d\mathbf k \, \hat f(\mathbf k) e^{i \mathbf k \cdot \mathbf x} = \int d\mathbf k\, \hat f(\mathbf k) \nabla^2 e^{i \mathbf k \cdot \mathbf x} = \int d\mathbf k \, [ -k^2 \hat f(\mathbf k) ] e^{i \mathbf k \cdot \mathbf x}$$
So we say
$$ \mathcal{F} \left[ \nabla^2 f(\mathbf x) \right] = -k^2 \hat f(\mathbf k)$$
Note: depending on your field and book, your mileage may vary when it comes to factors of $2\pi$ or signs
