Green's Function for Harmonic Oscillator

Question

I am trying to prove that $$G(t-t')=\frac i{2w}e^{-iw|t-t'|}$$ satisfies $$\left(\frac{\partial^2}{\partial t^2}+w^2\right)G(t-t')=\delta(t-t').$$ (Problem 7.2 from Quantum Field Theory by Mark Srednicki.)

Using the fact that $\frac{\partial}{\partial t}|t-t'|=sgn(t)$ and $\frac{\partial}{\partial t}sgn(t)=2\delta(t)$, I was able to show that $$\frac{\partial^2}{\partial t^2}G(t-t')=-w^2G(t-t')+e^{-iw|t-t'|}\delta(t).$$ However, I don't see how to get rid of the exponential in front of the delta-function to prove the identity. I would appreciate any help.

Answer 1

The $\delta$ distribution is defined by the result of integrating it against some collection of test functions. Specifically \begin{align} \int dt\,\delta(t - t') f(t) = f(t'). \end{align} If we replace $\delta(t)$ with $g(t) = e^{-i \omega |t|}\delta(t)$, \begin{align} \int dt\,g(t-t') f(t) &= \int dt\, \delta(t-t') e^{-i\omega|t-t'|}f(t)\\ &=e^{-i\omega|t'-t'|}f(t')\\ &=f(t'). \end{align} Since we get the same result integrating against $g$ as against $\delta$, these are equal as distributions.