Simple act of taking derivatives, I suppose

[vizart]

I am having bit of a problem proving Eq. (0.2):

(0.1) \text{ } G_{\omega}(t-t^\prime)=\theta(t-t^\prime) \frac{e^{-i\omega(t-t^\prime)}}{2\omega}+\theta(t^\prime-t) \frac{e^{i\omega(t-t^\prime)}}{2\omega}
(0.2) \text { }\left (-\partial^2_{t}-\omega^2 \right ) G_\omega (t-t^\prime)=i\delta(t-t^\prime)

The problem is dealing with the first and second order derivatives of the theta function in Eq. (0.1); they don't match the right hand side of (0.2).

MODs: It's not a homework problem.

 

[mathman]

I am totally unfamiliar with these equations. However when I see a delta function in one equation and a complex exponential in the other I think of Fourier transforms.

 

[cpt_carrot]

The derivative of \Theta(t) is \delta(t). You can then use this result, along with the fact that the second derivative of \Theta(t) is peaked at t=0 (so that these terms cancel) to find \partial^2_t G and substitute to find the answer.

If you want any more details please show what you have worked out so far.

 

[vizart]

The derivative of \Theta(t) is \delta(t). You can then use this result, along with the fact that the second derivative of \Theta(t) is peaked at t=0 (so that these terms cancel) to find \partial^2_t G and substitute to find the answer.

If you want any more details please show what you have worked out so far.

Thanks for your response. I knew about the derivative of the delta function, but the problem is when you take first and second derivatives of the theta function, you get one term with delta function and cos\left (\omega(t-t^\prime)\right ) (the other derivative is carried over the exp function) and for the second derivative, the derivative of delta function multiplied by sine of the same variable (all with some additional constant coefficients). It's these cosine and sine functions that I couldn't get them equal to one and zero respectively (modulo those presumably manageable coefficients).

 

[cpt_carrot]

Thanks for your response. I knew about the derivative of the delta function, but the problem is when you take first and second derivatives of the theta function, you get one term with delta function and cos\left (\omega(t-t^\prime)\right ) (the other derivative is carried over the exp function) and for the second derivative, the derivative of delta function multiplied by sine of the same variable (all with some additional constant coefficients). It's these cosine and sine functions that I couldn't get them equal to one and zero respectively (modulo those presumably manageable coefficients).

The sin and cos terms are multiplied by delta functions, right?

So you can take them to be at t=t' and so they give the required values.

 

[vizart]

The sin and cos terms are multiplied by delta functions, right?

So you can take them to be at t=t' and so they give the required values.

Alright, I see that the cosine term that is multiplied by a delta function could be taken as one, but the sine term is multiplied by the derivative of delta function and I can't see how you could set it equal to zero.

 

[cpt_carrot]

The derivative of a delta function has the property
\int_{-\infty}^\infty\delta'(t)f(t)dt=-f'(0)

In your equation you should have a term like
\delta'(\tau)(e^{i\omega\tau}+e^{-i\omega\tau})
which at \tau=0 has a derivative which goes away so this term gives no contribution.

Hope that clears things up.

 

[vizart]

The derivative of a delta function has the property
\int_{-\infty}^\infty\delta'(t)f(t)dt=-f'(0)

In your equation you should have a term like
\delta'(\tau)(e^{i\omega\tau}+e^{-i\omega\tau})
which at \tau=0 has a derivative which goes away so this term gives no contribution.

Hope that clears things up.

Thanks for the explanations CPT, but due to opposite variables in the theta function, when you take the first derivative, a relative negative sign between exponential functions appears which means it's more like \delta'(\tau)(e^{i\omega\tau}-e^{-i\omega\tau}) which as I mentioned is proportional to a sine function. Plus, even if it was a cosine function you should take this as an identity, which means if I multiply \delta'(\tau)\cos(\omega\tau) with another function f(\tau) in a more general context, then that integral identity would not help to make the whole expression vanish taking f'(\tau) \neq 0 -integration implied.

 

[cpt_carrot]

I agree with all of that, and am now stumped, sorry

If you work it out, let me know!