Green function for forced harmonic oscillator

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Homework Help Overview

The problem involves finding the Green function for a forced harmonic oscillator, specifically deriving \( G(t) \) from \( G(\omega) \) using Fourier transforms. The original poster presents the expression for \( G(\omega) \) and attempts to compute \( G(t) \) through residue calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the necessity of using calculus of residues and the implications of causality in the integration process. Questions arise regarding the reality of the resulting function, particularly the presence of the imaginary unit \( i \) in the expression for \( G(t) \). There is also a discussion about the expected behavior of the system and the form of the solution.

Discussion Status

The discussion is ongoing, with participants questioning the correctness of the original poster's result and exploring the implications of the imaginary component in the solution. Some participants suggest that the sine function is appropriate for the system's response, while others clarify the representation of the sine function in terms of exponential functions.

Contextual Notes

There is an emphasis on ensuring that the Green function remains real, as well as considerations regarding the initial conditions and the physical interpretation of the system's response.

Judas503
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Homework Statement


The problem requires to solve the integration to find ## G(t) ## after ##G(\omega)## is found via Fourier transform. We have G(\omega)= \frac{1}{2\pi}\frac{1}{\omega _{0}^2 - \omega ^2}

Homework Equations


As mentioned previously, the question asks to find ##G(t)##

The Attempt at a Solution


It is obvious that calculus of residues is required. To account for causality (## G(t<0)=0 ##), the poles at ## \omega=\pm \omega_{0} ## are shifted to the lower half plane by ## i\epsilon ## and integrated along the contour in the lower half plane. Then,
G(t)=-\frac{1}{2\pi}\int_{-\infty}^{+\infty}\frac{e^{-i\omega t}}{(\omega +i\epsilon)^2 -\omega _{0} ^2}d\omega
After calculating the residues at ## \omega =\pm \omega _{0} - i\epsilon ##, I found
G(t)=\frac{i}{2\omega_{0}}\sin \omega_{0}t

Is my answer correct?
 
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The presence of ##i## suggests it's not. G(t) should be real, shouldn't it?

The sine makes sense. The system is at rest and then you impart an impulse, causing it to oscillate. It has to be sine because it starts from x=0.
 
It looks like the i and the 2 might both get absorbed into the ##\sin \omega_0 t =\frac{e^{i \omega_0 t }-e^{-i \omega_0 t }}{2i}## term.
 
Yes, sorry! I forgot to put the "i" in the exponential form of sine. That should clear the problem.
 

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