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I Problem -- Fourier transform

  1. Dec 25, 2016 #1
    [##f^*## represents complex conjugate of ##f##. ]

    [##\widetilde{f}(k)## represents fourier transform of the function ##f(x)##.]

    $$\begin{align}
    \int_{-\infty}^{\infty}f^*(x)e^{ikx}\,dx&=\int_{-\infty}^{\infty}f^*(x)\left(e^{-ikx}\right)^*\,dx\\
    &=\int_{-\infty}^{\infty}\left(f(x)e^{-ikx}\right)^*\,dx\\
    &=\left(\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx\right)^*\\
    &=\left[\widetilde{f}(k)\right]^*\\
    \end{align}
    $$
    Now, let
    $$f(x)=u(x)+iv(x)$$
    where ##u(x)## and ##v(x)## are the real and imaginary parts of ##f(x)##.
    Again, we have,
    $$\begin{align}
    \int_{-\infty}^{\infty}f^*(x)e^{ikx}\,dx&=\int_{-\infty}^{\infty}\left(u(x)-iv(x)\right)e^{ikx}\,dx\\
    &=\int_{-\infty}^{\infty}u(x)e^{-i(-k)x}\,dx-i\int_{-\infty}^{\infty}v(x)e^{-i(-k)x}\,dx\\
    &=\widetilde{u}(-k)-i\widetilde{v}(-k)\\
    &=\left[\widetilde{u}(-k)+i\widetilde{v}(-k)\right]^*\\
    &=\left[\widetilde{f}(-k)\right]^* \text{ [using linearity property of fourier transform]}

    \end{align}
    $$
    So, I am getting different results. What is wrong with this calculation.
     
  2. jcsd
  3. Dec 25, 2016 #2

    mfb

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    There is a sign error between (3) and (4).
     
  4. Dec 25, 2016 #3
    I used the defination,
    $$\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx=\widetilde{f}(k)$$
     
  5. Dec 25, 2016 #4

    Svein

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    Well, not quite. The definition is [itex] \widetilde f(k)= \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx[/itex]
     
  6. Dec 25, 2016 #5

    micromass

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    There are actually many non-equivalent definitions. Most just differ in the constant in front.
     
  7. Dec 25, 2016 #6

    mfb

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    Well, it doesn't matter which definition you use, but keep it consistent. Either eikx or e-ikx. Using the former, (3) is the Fourier transform for -k instead of k.
     
  8. Dec 25, 2016 #7
    I did not understand why (3) is the Fourier transform of -k instead of k. Look, I used the defination,
    $$
    \int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx=\widetilde{f}(k)\\
    $$
    Taking complex conjugate on both sides,
    $$
    \left(\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx\right)^*=\left[\widetilde{f}(k)\right]^*\\
    $$
    So (3) and (4) are justified.
     
  9. Dec 26, 2016 #8

    mfb

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    Ah, I misread the signs, sorry.

    There is a mistake between (7) and (8). u and v are real, but their Fourier transformations in general won't be real. You can use ##\widetilde{u}(-k) = \widetilde{u}(k)^*##:

    $$\begin{align}
    \widetilde{u}(-k)-i\widetilde{v}(-k)
    &= \widetilde{u}(k)^* -i \widetilde{v}(k)^*\\
    \end{align}$$

    Not sure how to simplify that.
     
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