I have a Dyson equation for a Green's function that comes in this form:

$$ G[t,x_f;0,x_i]=G_0[t,x_f;0,x_i]+i\int_\Omega\int_0^t\ dx\ d\tau\ G_0[t,x_f;\tau,x]xG[\tau, x;0, x_i] $$

For convenience, I'd like to Fourier transform it in time. Keeping in mind that $G_0[t,\tau] = G[t,\tau] = 0$ if $\tau>t$ (due to time-ordering), I extend the integral in $\tau$ to $+\infty$, then multiply by $e^{-i\omega t}$ and further integrate in $t$ from 0 to $\infty$:

$$ \int_0^\infty dt\ G[t,x_f;0,x_i]e^{-i\omega t}=\int_0^\infty dt\ G_0[t,x_f;0,x_i]e^{-i\omega t} +i\int_0^\infty\int_\Omega\int_0^\infty dt\ dx\ d\tau\ G_0[t,x_f;\tau,x]e^{-i\omega t}xG[\tau, x;0, x_i] $$

Expressing the Fourier transforms of the functions as $\tilde{G}_0,\tilde{G}$ then I have

$$ \tilde{G}[\omega;x_f,x_i]=\tilde{G}_0[\omega;x_f,x_i] +i\int_\Omega\int_0^\infty dx\ d\tau\ \tilde{G}_0[\omega;x_f,x_i]e^{-i\omega \tau}xG[\tau, x;0, x_i] $$

where I applied an origin shift to $G_0$ inside the double integral, and then I carry out the integral in $\tau$ as well to find:

$$ \tilde{G}[\omega;x_f,x_i]=\tilde{G}_0[\omega;x_f,x_i] +i\int_\Omega dx\ \tilde{G}_0[\omega;x_f,x]x\tilde{G}[\omega; x, x_i] $$

This also makes sense because it's basically the convolution theorem, where the Fourier transform of a convolution becomes the product of the two Fourier transforms. No matter how I look at it, it seems correct, yet when I actually apply these formulas numerically (solving in the time domain and then applying a FFT, and then solving in the frequency domain and comparing) I get two different results, and I'm fairly confident the frequency domain is the wrong one. Is there anything wrong with my reasoning here, or some caveat specific to this kind of equation I am missing? Or should I look for my mistake elsewhere?

  • $\begingroup$ OK, so now I intend to delete my earlier comments (including, later today, this one), so as to tidy up. I can't see a problem with the algebra. The lower limits all need extending to $-\infty$ to make them FTs, but I believe all the integrands vanish appropriately. Maybe the mistake lies in the way you applied the FFT algorithm? You need to explicitly include the range $t<0$ over which $G(t,0)$ is zero. Anyway, if this isn't it, hopefully you or someone else will spot the problem. $\endgroup$ – user197851 Oct 5 '18 at 9:16
  • $\begingroup$ Yes, I didn't bother about the integration limit because of the usual time-ordering condition which should zero everything as necessary. There's the problem of Fourier transforming the Heaviside step function but in theory I have that figured out too, I'm not sure if it could be it though. Anyway please leave this last comment, "the algebra is correct" in itself is already an answer! I think so too, so maybe the mistake really is somewhere else. $\endgroup$ – Okarin Oct 5 '18 at 16:49

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