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I am attempting to verify that $G$ is a Green function for the oscillator equation of motion such that:

$\bigg(\frac{\partial^2}{\partial t^2} + \omega^2 \bigg)G(t-t')=\delta (t-t')$

The Green function is

$G(t-t') = \frac{i}{2\omega}exp(-i\omega |t-t'| ) $

If we set $t'=0$ we carry out the differentiation and get as far as:

$\frac{\partial^2 G}{\partial t^2}= -\frac{1}{2}i\omega e^{-i \omega |t| }sgn^2(t) + e^{-i \omega |t|} \delta (t) $

The worked solution then goes to

$\frac{\partial^2 G}{\partial t^2}=-\omega^2 G(t) + \delta(t)$

I can see where the square of the signum function would go to $1$ but what happened to the exponential factor which was multiplying the delta function?

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You should put an Heaviside function multiplying your solution. In your case is $$ G(t-t')=\theta(t-t')\frac{\sin[\omega(t-t')]}{\omega}. $$ Then, $$ \frac{dG(t)}{dt}=\theta(t)\cos(\omega t) $$ and $$ \frac{d^2G(t)}{dt^2}=\delta(t)-\omega^2\theta(t)\frac{\sin(\omega t)}{\omega} $$ and you are done. I have used the fact that $$ \frac{d\theta(t)}{dt}=\delta(t) $$ and everything should be intended in the sense of distributions.

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  • $\begingroup$ Is there a reason why you don't get a delta function from the product rule when you differentiate G the first time, also why is it $\delta$ in the second derivative and not $delta$ times the cosine, is this something to do with distributions? $\endgroup$ – Tom Jul 26 '18 at 9:46
  • $\begingroup$ Because it is multiplied by a 0 being the $\sin(\omega t)$ 0 for $t=0$. You should intend this in the sense of distributions and so, there is an integration involved. $\endgroup$ – Jon Jul 26 '18 at 9:58

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