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I was doing fourier transform of lorentzian function

$$ B(k)=\frac{1}{1+\left(\frac{k-k0}{\Delta k}\right)^2} =\frac{\Delta k^2}{\Delta k^2+\left(k-k_0\right)^2} $$

$B(k)$ is neither even nor odd function, so Fourier transform of $B(k)$

$$ \mathscr{F}[B(k)]=\Delta k^2\int_{\infty}^{\infty} \frac{e^{-ik\delta}}{\Delta k^2+\left(k-k_0\right)^2}dk $$

substitution $ k-k_0=\kappa $ , $\qquad k=k_0+\kappa $

$$ \mathscr{F}[B(k)]=\Delta k^2e^{-ik_0\delta}\int_{\infty}^{\infty} \frac{e^{-i\kappa\delta}}{\Delta k^2+\kappa^2}d\kappa = \Delta k^2e^{-ik_0\delta}\int_{\infty}^{\infty} \frac{e^{-i\kappa\delta}}{(\kappa+i\Delta k)(\kappa-i\Delta k)} d\kappa $$

Cauchy's integral formula, I used the formula once for each pole.

When $\delta<0$, the contour is on the upper plane, and it is counter clockwise

$$ \int_{\infty}^{\infty}f(\kappa)e^{-i\kappa\delta}d\kappa=2\pi i \sum Res[f(\kappa)e^{-i\kappa\delta}] $$

$$ \mathscr{F}[B(k)] = 2\pi i \Delta k^2e^{-ik_0\delta} \frac{e^{-\Delta k\delta}}{2i\Delta k} = \pi \Delta k e^{-ik_0\delta}e^{-\Delta k\delta} $$

When $\delta>0$, the contour is on the lower plane, and it is clockwise (which adds minus(-) sign to the integral)

$$ \mathscr{F}[B(k)] = -2\pi i \Delta k^2e^{-ik_0\delta} \frac{e^{\Delta k\delta}}{-2i\Delta k} = \pi \Delta k e^{-ik_0\delta}e^{\Delta k\delta} $$

Together they comprise

$$ \pi \Delta k e^{-\Delta k |\delta|-ik_0\delta} $$

This is some other's work, and there is something I quite don't understand.

How does $\delta>0$ corresponds to upper plane and $\delta<0$ to lower plane?

(besides, did I wrote the correct answer for FT?)

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  • $\begingroup$ You have a sign error in the exponent, it's $e^{-\Delta k \lvert\delta\rvert - i k_0\delta}$. $\endgroup$ Mar 11, 2016 at 16:25
  • $\begingroup$ Thank you for your keen observation $\endgroup$
    – user65452
    Mar 11, 2016 at 16:31
  • $\begingroup$ uh.. I checked again, i think it is still $e^{\Delta k}$ $\endgroup$
    – user65452
    Mar 11, 2016 at 17:04
  • $\begingroup$ No, for $\delta < 0$, when we use the semicircle in the upper half-plane, we must plug $i\Delta k$ into $e^{-i\kappa\delta}$, and $-i\cdot (i\Delta k) = \Delta k$, so we get $e^{\Delta k\delta} = e^{-\Delta k\lvert\delta\rvert}$ then. $\endgroup$ Mar 11, 2016 at 17:11
  • $\begingroup$ oh. you're right. sorry for the confusion $\endgroup$
    – user65452
    Mar 11, 2016 at 20:07

2 Answers 2

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Cauchy's formula requires a closed contour in the complex plane. We typically close the contour with a semicircle in the upper or lower half plane. When $\delta \lt 0$, the integral over the semicircle in the upper half plane vanishes (and that over the lower half plane diverges), so we use the upper half plane. Vice-versa for $\delta \lt 0$.

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Note that for $\delta<0$ the integration over a semi-circular contour $C_R$, with radius $R$ and centered at the origin in the upper-half plane becomes

$$\int_{C_R}\frac{e^{-i\delta z}}{z^2-\Delta k^2}\,dz = \int_0^\pi \frac{e^{-i\delta R e^{i\phi}}}{R^2e^{i2\phi}-\Delta k^2}(iRe^{i\phi})\,d\phi$$

Note that the term $e^{-i\delta R e^{i\phi}}=e^{-i\delta R\cos(\phi)}e^{\delta R\sin(\phi)}$ decays exponentially for $0<\phi <\pi$ as $R\to \infty$ when $\delta<0$. Therefore, as $R\to \infty$ the integral over $C_R$ vanishes.

It is important to point out that this last statement is not obvious since for $\phi \approx 0$ and $\phi \approx \pi$ the exponential is close to zero for fixed $R$. To show that the integral vanishes, we write the following estimates for $R>\Delta k$:

$$\begin{align}\left|\int_0^\pi \frac{e^{-i\delta R e^{i\phi}}}{R^2e^{i2\phi}-\Delta k^2}(iRe^{i\phi})\,d\phi\right|&\le \int_0^\pi \frac{Re^{\delta R\sin(\phi)}}{|R^2e^{i2\phi}-\Delta k^2|}\,d\phi\\\\&\le \frac{2R}{R^2-\Delta k^2}\int_0^{\pi/2} e^{2 \delta R\,\phi/\pi }\,d\phi\\\\&=\frac{\pi(1-e^{-|\delta| R})}{|\delta| (R^2-\Delta k^2)}\end{align}$$ So, it is evident that as $R\to \infty$, the integral over $C_R$ tends to $0$. Note that in the development, we used $\sin(\phi)\ge 2\phi/\pi$ for $0\le \phi\le \pi/2$.

However, when $\delta>0$, the term $e^{-i\delta R e^{i\phi}}=e^{-i\delta R\cos(\phi)}e^{\delta R\sin(\phi)}$ increases exponentially for $0<\phi <\pi$ as $R\to \infty$. Therefore, when $\delta >0$, we close the contour with the semi-circular arc $C'_{R}$, with radius $R$ and centered at the origin in the lower-half plane. On $C'_{R}$, the exponential term decays as $R\to \infty$ and the integration over $C'_{R}$ approaches zero.

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