1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problems with Fourier transformation: jump at discontinuity

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider a particle in an infinite square well of width  = 1. The particle is in a state
    given by:

    Phi=A(1/2-|x-1/2)

    a=1

    b) Find the general form of the expansion coefficients (the Fourier coefficients, right?)
    for expanding the function in terms of the square-well basis set.
    c) On the same graph, plot the wave function and the expansion in terms of square well
    functions (take the sum to at least five terms or so) to verify the notion of expanding
    the wave function in eigenstates of the well.

    2. Relevant equations

    a_n=1/L *Integrate[f[x]*Cos(n*Pi*x/L),{x,-L,L}]
    b_n=1/L *Integrate[f[x]*Sin(n*Pi*x/L),{x,-L,L}]

    f[x]=a_0/2+Sum[a_n*Cos(n*Pi*x/L)+b_n*Sin(n*Pi*x/L),{n,1,Infinity}]

    3. The attempt at a solution
    I ran through the generalized form for a_n and b_n and got the following values:

    a_n=2*(-(1/(2 n^2 \[Pi]^2)) + (-1)^n/(2 n^2 \[Pi]^2)) Cos[2 n \[Pi] x]

    b_n=2*((-1)^n Sin[2 n \[Pi] x])/(2 n \[Pi])

    I got these by splitting the transformation at 1/2 and rewriting 0 to 1/2 as A(1/2-(1/2-x)=x and 1/2 to 1 as A(1/2-(x-1/2)=(1-x). When :I graph my final transform it lines up nicely with the right side of my phi but not the left. I know I'm missing an a_0 but I can't explain why it is so off for tho 0 to 1/2 portion.
     
  2. jcsd
  3. Oct 3, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You might want to take the time to learn how to write the equations in LaTeX. I found your post incredibly hard to read.
    What does a represent?
    Right idea, but you need to think about what the eigenstates for the square well are. It's not the same as the Fourier basis.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Problems with Fourier transformation: jump at discontinuity
  1. Fourier Transform (Replies: 3)

  2. Fourier transforms (Replies: 3)

  3. Fourier Transform (Replies: 1)

Loading...