- #1
IanBerkman
- 54
- 1
Dear all,
In my quantum mechanics book it is stated that the Fourier transform of the Coulomb potential
$$\frac{e^2}{4\pi\epsilon_0 r}$$
results in
$$\frac{e^2}{\epsilon_0 q^2}$$
Where ##r## is the distance between the electrons and ##q## is the difference in wave vectors.
What confuses me, is how the Fourier transform of the first term is taken since the integral diverges at r = 0.
I hope anyone can clear this up for me.
Thanks,
IanEDIT: It is already solved, ##r## and ##q## need to be taken as vectors. This thread can be deleted.
In my quantum mechanics book it is stated that the Fourier transform of the Coulomb potential
$$\frac{e^2}{4\pi\epsilon_0 r}$$
results in
$$\frac{e^2}{\epsilon_0 q^2}$$
Where ##r## is the distance between the electrons and ##q## is the difference in wave vectors.
What confuses me, is how the Fourier transform of the first term is taken since the integral diverges at r = 0.
I hope anyone can clear this up for me.
Thanks,
IanEDIT: It is already solved, ##r## and ##q## need to be taken as vectors. This thread can be deleted.