# I Lorentzian line shape integration

1. Jun 9, 2017

### Josh1079

Hi,
I know this might be a bit dum but I'm currently stuck with this integral.

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. Jun 9, 2017

### BvU

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.

3. Jun 9, 2017

### Fightfish

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 $t = 0$.

4. Jun 9, 2017

### Josh1079

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?

5. Jun 9, 2017

### Fightfish

Or more precisely,
$$E(t) = \left [ E_{0} e^{-\gamma t} e^{i \omega_{0} t} + \mathrm{c.c.} \right] \theta (t),$$
where $\theta(t)$ is the Heaviside step function.

6. Jun 9, 2017

### Josh1079

Thank you for the clarification!

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