Momentum space representation of the hydrogen atom Schroedinger equation The Schroedinger equation for the hydrogen atom is
$$\Big(-\frac{\hbar^2}{2m} \Delta - \frac{e^2}{r} \large)\Psi(\mathbf{r}) = E \; \Psi(\mathbf{r}).$$
I have found that the momentum representation of the above equation reads
$$\frac{\mathbf{p}^2}{2m}\Psi(\mathbf{p}) -\frac{e^2}{2\pi^2 h} \int \frac{\Psi(\mathbf{p'})\mathrm{d}^3\mathbf{p'}}{|\mathbf{p}-\mathbf{p'}|^2} = E \; \Psi(\mathbf{p}).$$
How to derive it? The Fourier transforms of the Laplace operator part and the right hand side are clear. But how about the potential part?
 A: The problem here is the mistaken impression that the momentum representation is connected to the position representation by a Fourier Transform. Contrary to what is said in textbooks, this is only true for Cartesian Coordinates. For curvilinear coordinates (e.g. r, theta, phi) a different transformation is needed. Check out the article: The Hydrogen Atom in the Momentum Representation, John R. Lombardi, Phys. Rev. A, 22, 797 (1980), in which the correct transform leads to a momentum space where all the momentum variables are chosen to be properly conjugate to the three position space coordinates.
A: You take the Fourier transform of Schrödinger equation in position space. Transformation of the term with potential can be calculated using convolution theorem (Fourier transform of a product is convolution of Fourier transforms - details are easy to find online and the proof is rather straightforward). Fourier transform of $V$ itself can be easily calculated using three facts (I write up to constant factors which can be easily checked):


*

*Fourier transform of Laplace operator acting on function is Fourier transform of that function times $-k^2$ (here $k$ is wavector).

*Laplace operator on $V$ is delta.

*Fourier transform of delta is a constant.
