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Consider the function

$$ f(\vec{x}) = \int_{\Bbb R^3} {\frac{ e^{-i\,\vec{x}\cdot\vec{k}}}{\sqrt{\vec{k}^2 + m^2}}} d^3 k $$

from Zee's Quantum Field Theory in a Nutshell. He argues like this: “the square root cut starting at $±im$ tells us that the characteristic value of $|\vec{k}|$ in the integral is of order $m$, leading to an exponential decay $\sim e^{−m|\vec{x}|}$”. I cannot understand what he means. It would be nice if someone can expand or suggest a reference for me. Thanks.

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  • $\begingroup$ What he likely means is that - to take a 1D analog - the integral may be expressed by Cauchy's Theorem as $$\int_m^{\infty} dk \, \frac{e^{-k x}}{\sqrt{k^2-m^2}} $$ which may be shown to be dominated by the contribution near $k=m$. $\endgroup$ – Ron Gordon May 11 '15 at 18:50
  • $\begingroup$ I have been thinking about the residue stuff too, but how does it lead to the bound $e^{-m|\vec{x}|}$. And what is the precise definition of "be dominated" ? $\endgroup$ – Tuyet Nhi May 11 '15 at 18:57
  • $\begingroup$ Note that I am keeping this to a comment because I do not wish to get into precise definitions at this point. I am using the language of Steepest Descents. Think about approximating the integral with a sum (avoiding the "pole"), what term or terms make the biggest contribution to the sum? That's what I mean by "dominating." $\endgroup$ – Ron Gordon May 11 '15 at 19:00
  • $\begingroup$ Thanks, that Steepest Descent does make sense; although I still cannot see how to bound the integral by $e^{-m|\vec{x}|}$. $\endgroup$ – Tuyet Nhi May 11 '15 at 19:03
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Letting \begin{align*} \vec{x}&=(0,0,x), \\ \vec{k}&=k(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi), \end{align*} and integrating over $\theta$ and $\phi$ we find \begin{align*} \int_{\Bbb R^3} {\frac{ e^{-i\,\vec{x}\cdot\vec{k}}}{\sqrt{\vec{k}^2 + m^2}}}\, d^3 k &= \frac{4\pi}{x} \int_0^\infty \frac{k\sin k x}{\sqrt{k^2+m^2}}\,dk = \frac{2\pi}{x} \int_{-\infty}^\infty \frac{k\sin k x}{\sqrt{k^2+m^2}}\,dk \\ &= \frac{2\pi}{x} \textrm{Im} \int_{-\infty}^\infty \frac{k e^{i k x}}{\sqrt{k^2+m^2}}\,dk. \end{align*} An integral of exactly this form is worked out in detail here.

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