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Fourier Transform of correlation functions

  1. May 2, 2010 #1
    Why are they useful, what do they denote (physically or otherwise)...
  2. jcsd
  3. May 2, 2010 #2


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    Maybe this will help? http://en.wikipedia.org/wiki/Convolution_theorem" [Broken].
    Last edited by a moderator: May 4, 2017
  4. May 2, 2010 #3
    Yes, the correlation integral acts very much like a convolution integral, only with a plus sign... but I am not able to understand why while calculating correlation functions, or even green's functions, authors tend to caclulate their fourier transforms as well.

    thank you
  5. May 2, 2010 #4
    The main reason is that in a translationally invariant system, Green's functions are diagonal in momentum space. This simplifies all calculations and turns matrix equations (e.g. Dyson equation) into algebraic equations that can be easily solved. The diagonality of correlation functions is one way of saying that "momentum is conserved", although the physical meaning of momentum varies from one model to another.
    Last edited: May 2, 2010
  6. May 2, 2010 #5
    Usually the FT of a correlation function is what's experimentally measurable. Using neutron scattering or photoemission one measures the momentum of the emitted particle, and this directly relates to a moment correlation function or Green's function.
  7. May 2, 2010 #6


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    static susceptibility sum rule for magnetic systems:

    [tex]\chi = \frac{\partial M}{\partial H} = k_B T \int_V d^d\mathbf{r} G(\mathbf{r}) = k_B T \hat{G}(\mathbf{k} = \mathbf{0}) [/tex]

    i.e., The magnetic susceptibility is related to the fourier transform of the correlation function at zero wavevector.
  8. May 3, 2010 #7
    As an example:
    Normally one detects light with a photo detector. A photo detection is converted into electronic signal after some kind of filtering (or whatever is the frequency response of the system). Physically for a square-law detector the Fourier transform (FT) of a correlation provides the frequency spectrum of the signal. That is the simplest of the physical explanation. For example, if two waves interfere in which one of the wave is Doppler shifted the correlation function (or the auto-correlation function to be accurate from the detected signal) will be a sin function while its Fourier will be peaked at a frequency corresponding to the Doppler shift. One reason why correlation function is measured versus Fourier is its large dynamic range not affordable in Fourier. One could measure correlation function all the way from nanoseconds to minutes. Try this time scale with Fourier and see the amount of data needed.....
  9. May 3, 2010 #8
    When we want to study the Goldstone modes of the system, the Fourier transform is more desired because the Goldstone modes vanishes when k->0.
  10. Jul 13, 2010 #9
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