Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Lorentzian line shape integration
  1. Line integral (Replies: 1)

  2. Line integral (Replies: 3)

  3. Line Integral (Replies: 11)

  4. Line integrals (Replies: 4)

Loading...