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Second order polarization

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi

    I am trying to solve problem 2.1 in these notes (it is on the very first page): http://qis.ucalgary.ca/quantech/673/notes/chapter_two.pdf [Broken]. The problem tells me to show that the Fourier transform of

    [tex]
    P^{(2)} (t) = \varepsilon _0 \frac{1}{{(2\pi )^2 }}\int_{ - \infty }^\infty {\int_{ - \infty }^\infty {\chi ^{(2)} (t_1 ,t_2 )E(t - t_1 )E(t - t_2 )dt_1 dt_2 } }
    [/tex]

    is given by

    [tex]
    P^{(2)}(\omega ) = \varepsilon _0 \int_{ - \infty }^\infty {\int_{ - \infty }^\infty {\chi ^{(2)} (\omega _1 ,\omega _2 )\delta (\omega - \omega _1 - \omega _2 )E(\omega _1 )E(\omega _2 )d\omega _1 d\omega _2 } }
    [/tex]


    3. The attempt at a solution
    OK, so I know that the FT of a convolution yields a product and that the FT of a delta-function is a constant. I am not sure whether to

    1) tack on a factor of exp(-iωt) and integrate over t on both sides OR
    2) write every factor in terms of its FT, and then compare terms in the end

    I choose option 2. Here goes

    [tex]
    \int_{ - \infty }^\infty {P(\omega )e^{ - i\omega t} } = \varepsilon _0 \frac{1}{{(2\pi )^2 }}\int_{ - \infty }^\infty {\int_{ - \infty }^\infty {\chi ^{(2)} (\omega _1 ,\omega _2 )e^{ - i\omega _1 t_1 - i\omega _2 t_2 } E(\omega _1 )e^{ - i\omega _1 (t - t_1 )} E(\omega _2 )e^{ - i\omega _2 (t - t_2 )} d\omega _1 d\omega _2 dt_1 dt_2 } }
    [/tex]

    But this makes t1 and t2 go out in the exponentials, which is no good. But if I go back to option 1, I can't see how I would ever end up with a delta-function in ω-ω12.

    I would really appreciate a hint.

    Best,
    Niles.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 10, 2011 #2
    No one has a hint? In principle this seems straightforward, but I simply cannot crack this nut.
     
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