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First order coherence classical optics problem

  1. Sep 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi, im trying to work through exercise 2.1 on page 37 of Microcavities (by alexy kavokin, jeremy baumberg, guillaume malpuech and fabrice laussy)


    the problem is to prove
    [tex]
    | g^{(1)}(\tau) | = | cos( \frac{1}{2}(\omega_1 - \omega_2)\tau) ) |
    [/tex]


    where:

    [tex]
    g^{(1)}(\tau)=\frac{\langle E^{\ast}(t)E(t+\tau)\rangle}{\langle |E(t)|^2 \rangle}
    [/tex]
    and
    [tex]
    E(t)=E_0(t)\exp^{i[k_1z-\omega_1t]}+E_0(t)\exp^{i[k_2z-\omega_2t+\varphi]}
    [/tex]
    where
    [tex]
    \varphi [/tex] varies randomly between measurements


    how do you deal mathmatically with [tex] \varphi [/tex] varying??


    Also more simply above the exercise the text states a simpler apparantly 'trivial' result using the same formula for [tex] g^{(1)}(\tau) [/tex] that the sine wave of
    [tex]
    E(t)=E_0(t)\exp^{i[\omega t - kz + \varphi]}
    [/tex]
    by direct application of the formula for [tex] g^{(1)}(\tau) [/tex] yields:

    [tex]
    g^{(1)}(\tau)= \exp^{i\omega \tau}
    [/tex]

    i cannot show even this 'trivial' application, i find that:
    [tex]
    \langle E^{\ast}(t)E(t+\tau)\rangle = \langle E_0\exp^{-i\omega \tau}\rangle
    [/tex]
    and
    [tex]
    \langle |E(t)|^2 \rangle = \langle |E_0^2 \exp^{2i(\omega t -kx + \varphi)} \rangle
    [/tex]
    How does that evaluate to
    [tex]
    g^{(1)}(\tau)= \exp^{-i\omega \tau}
    [/tex] ??
    If anyone can give me any pointers( or show me the complete workings of this XD ) then it would much appreciated. Thanks for reading my post.
     
    Last edited: Sep 8, 2011
  2. jcsd
  3. Sep 11, 2011 #2
    problem no longer. Just assumed by [tex] /varphi [/tex] varying randomly then all components with [tex] /varphi [/tex] cancel in the averageing.
     
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