# Fourier transform for radial wave function of infinitely deep circular well (Bessel function)

1. Oct 5, 2014

### dongsh2

Hi everyone,

do you know how to calculate the Fourier transform for the infinitely deep circular well (confined system)? The radial wave function is given by R=N_m J_m (k r). k=\alpha_{mn}/R. R is the radius of the circular well. R(k R)=0. Thanks.

Another question is that The k in J_{m}(k r) is same as the wave number k in momentum p=\hbar k?

2. Oct 6, 2014

### moriheru

Another question is that The k in J_{m}(k r) is same as the wave number k in momentum p=\hbar
k?

Simly said:yes k is the same,thus the root of 2mE/h(bar)^2
But p= d^2/dx^2+...*-h(bar)^2/2m in three dimensions(I don't know if this was just accidental but anyway?!), what you probably ment was that the Schrodinger eqaution can be rewriten as

d^2/dx^2psi(x)=-k^2, that is in one dimensions with respect to x in three dimensions you would just use the laplacian.

Last edited: Oct 6, 2014
3. Oct 15, 2014

### dongsh2

I think the k in J_{m}(k r) is different from the wave number k in p=\hbar k. The wave number k is a vector. The k in J_{m}(kr) is a scalar, k=\alpha_{mn}/R, in which R is the radius of the circular well.