FT is a functional integrating over a domain of r. Thus the unit of the solution changes with the unit of the differential dr. However, FT has a 'calibration factor' depending on a particular definition and it can also change the unit. The important point, the definition of FT is consistent and after an inverse FT you get the same solution.   

Hi Jan,

thanks for your answer - precise to the point as always :)

My particular problem involves a plasma density profile, which is in fact Gaussian in a sperical geometry. I want to incorporate this profile into a certain dispersion relation that is "living" in the k-space. So, I tried just to make the FT of my Gaussian and put it into the dispersion relation. The maths is straightforward and I checked the result with different methods. However, if I put this n(k) into my dispersion relation the units on both sides do not match, which is strange in my opinion...

Hi Jan and Johannes,

First, for setting the subject straight, the original density  n  is better to write in a slightly modified  form

D(r) =  M  (A / \pi)^{3/2}  exp(- A r²)    

which implies that  M is the total  mass spread over the 3-dimensional  space. Then its Fourier transform  calculated by the following formula

F(k)  :=  \int_{R3}   D(r)  exp[ - i (k;r) ]   dr

gives us    F(k)  =  M  exp[ - k2 /4A],  which is of dimension  M, which is the total mass. Consequently,   F(k)  is not dimensionless!

Note 1. In particular,  from the probability point of view, the lack of dimension of  F(k) is perfectly OK  since the total mass of probability equls M=1  and is dimensionless.

Note 2. In the proposed formula of  density, the quantity   "  M (A / \pi)^{3/2} " is the density at the origin.

Regards, Joachim

Dear Joachim, thanks a lot for your answer - the whole business start to get clearer for me :)