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I Lorentzian line shape integration

  1. Jun 9, 2017 #1
    Hi,
    I know this might be a bit dum but I'm currently stuck with this integral.
    In this link: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_III/Linienbreite.htm

    I know he's doing the right thing, but I really don't understand the integral of a(omega).
    How come it is E(1/(i(ω-ω0) -γ) - 1/(i(ω+ω0) +γ))?
    Does it even converge?

    Thanks!
     
  2. jcsd
  3. Jun 9, 2017 #2

    BvU

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    Note it's not ##a(\omega)## that is integrated, but ##|a(\omega)|^2##. A Lorentzian is essentially of the form ##1\over 1+x^2## and normalizable.

    Unless you wonder how the (complex) ##a(\omega)## itself is found. That is with the ##E(t)## at the top of the section. Write it out and see.
     
  4. Jun 9, 2017 #3
    Note that:
    (1) There is a typo in the expression for ##\alpha(\omega)## - it should be ##E_{0}## instead of ##E##
    (2) ##E(t)## is not given by the first expression for all time - the excitation only happens at ##t = 0##.
     
  5. Jun 9, 2017 #4
    Thanks for the replies and corrections.

    I've written the integral down, but I mean I don't really understand how the result came out. As it appears to me, at t = inf, it will be 0, and at t = -inf, it will blow up.

    I mean isn't it like this: ∫ exp(-γt + i(ω0 - ω)t)dt = 1/(-γ + i(ω0 - ω)) * exp(-γt + i(ω0 - ω)t) at t = inf and -inf

    Fightfish, do you mean that the integral should start from 0 rather than -inf?
     
  6. Jun 9, 2017 #5
    Or more precisely,
    [tex]
    E(t) = \left [ E_{0} e^{-\gamma t} e^{i \omega_{0} t} + \mathrm{c.c.} \right] \theta (t),
    [/tex]
    where ##\theta(t)## is the Heaviside step function.
     
  7. Jun 9, 2017 #6
    Thank you for the clarification!

    This makes good sense. The negative range of the integral is indeed pretty peculiar.
     
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