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The Wiener Khinchin Theorem for chaotic light

  1. Sep 10, 2015 #1
    1. The problem statement, all variables and given/known data

    It's problem 4:

    https://scontent-sea1-1.xx.fbcdn.net/hphotos-xpa1/v/t1.0-9/12004675_10206509414950788_2644752353357758096_n.jpg?oh=e6292fae7cdc34b881c7ac31a506e315&oe=56680268
    2. Relevant equations

    The Wiener Khinchin theorem gives that the normalized spectral power density (I assume this is what my professor means with "normalized spectral distribution function") is found from: $$F(\vec{r},\omega) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} g^{(1)}(\vec{r},t) e^{i \omega \tau} d \tau$$ where $$g^{(1)}(\vec{r},t) = \frac{<E^*(\vec{r},t)E(\vec{r},t+\tau)>}{<E^*(\vec{r},t)E(\vec{r},t)>}$$.The <> brackets denote taking the mean of the electrical field ##E## over a period ##T##which is much larger than the coherence time.

    3. The attempt at a solution
    Problem 4:
    Hey all. I'm a bit confused by the wording of this problem. What exactly is my professor asking for? The Wiener Khinchin theorem is BASED on using the fourier transform of the E field to get to the spectral power density... But he is saying now that this fourier transform and the W-K theorem give different results??
     
  2. jcsd
  3. Sep 11, 2015 #2
    Sorry was a bit unclear in the OP; the <> brackets denote taking means/autocorrelations of expressions that include the electrical field ##E##.
     
  4. Sep 11, 2015 #3
    I think you might understand the issue if you compare the fourier transform of a plane wave, which is a delta function, with the autocorrelation of a plane wave (i.e. sinc funtion).
     
  5. Sep 11, 2015 #4
    What do you mean? The autocorrelation of a plane wave ## e^{i \omega t}## is another plane wave. $$<E^*(\vec{r},t) E(\vec{r},t+\tau)> = <e^{i \omega \tau}> = e^{i \omega \tau}$$
     
  6. Sep 12, 2015 #5
    Dear Wminus,
    I appologize for my nonsence. I think that if you consider a plane wave with additive noise, the fourier transform will contain all the frequency components of the noise but the fourier transform of the autocorrelation function will not.
     
  7. Sep 12, 2015 #6
    No problem. And yeah I ended up using this kind of argument in my answer. I used the collision broadening of spectral lines as my example.
     
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