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Fourier Transform of e^(ip0x)F(x) to F(p)

  1. May 30, 2014 #1
    1. The problem statement, all variables and given/known data
    f(p) is the Fourier transform of f(x). Show that the Fourier Transform of eipox f(x) is f(p- p0).


    2. Relevant equations
    I'm using these versions of the fourier transform:
    f(x)=1/√(2π)∫eixpf(p)dx
    f(p)=1/√(2π)∫e-ixpf(x)dx

    3. The attempt at a solution

    I have:
    f(p)=1/√(2π)∫eix(po-p)f(x)dx
    which is the same as:
    f(p)=1/√(2π)∫e-ix(p-po)f(x)dx
    but I don't know where to go from here. I think I need to make a substitution using the original transform as I don't need to solve the integral. My other idea is that I have nearly proved it so just need to state the theory as to why this proves it; however, I don't know what that theory would be.
    Any help would be appreciated!
     
    Last edited: May 30, 2014
  2. jcsd
  3. May 30, 2014 #2

    jbunniii

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    You are using ##\underline{f}(p)## to refer to two different integrals which are not equal.

    To avoid confusion, I suggest defining a new function. Let ##g(x) = e^{ip_0x}f(x)##. Then the goal is to show that ##\underline{g}(p) = \underline{f}(p - p_0)##. You have already established that
    $$\underline{g}(p) = \frac{1}{\sqrt{2\pi}}\int e^{-ix(p - p_0)}f(x) dx$$
    Can you express this in terms of ##\underline{f}##?
     
  4. May 31, 2014 #3
    I understand why I should of renamed the functions but I really can't figure out the next step.
    I realise this is the final aim:
    g(p) = 1/√(2π)∫eix(p - p0f(x) dx = f(p-p0)
    But I don't know how f(x) is related to f(p-p0) and the best I seem able to do is get g(p) in terms of g(x), which is useless.
    Sorry I'm being dense!
     
  5. May 31, 2014 #4

    jbunniii

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    You have
    $$\underline{f}(p) = \frac{1}{\sqrt{2\pi}} e^{-ixp} f(x) dx$$
    so what is ##\underline{f}(p - p_0)##? Just replace ##p## with ##p - p_0## in the equation above. What can you conclude?

    Maybe to make it clearer, recognize that I can use any variable instead of ##p## and the equation will still be true. For example,
    $$\underline{f}(y) = \frac{1}{\sqrt{2\pi}} e^{-ixy} f(x) dx$$
    What does this give you when ##y = p - p_0##?
     
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