Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Oct 5, 2014 #1
    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. jcsd
  3. Oct 6, 2014 #2
    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
  4. Oct 15, 2014 #3
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fourier transform for radial wave function of infinitely deep circular well (Bessel function)
Loading...