I found this function while looking at quantum field theory. Defined by:

$$K(x,y) = \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} e^{i(x k + y q)}\sqrt{ (m^2 + k^2 + q^2) } dk dq$$

where $m$ is a constant.

Without the square root it would look like derivatives of the dirac delta function. But I'm not sure if this one would be more spread out. I'm not even sure how to approximate it.

They are not well defined but need to be regularised in some way, much like the dirac delta function definition:

$$\delta(x)=\int\limits_{-\infty}^{\infty} e^{ixk} dk$$

Another function is:

$$L(x,t) = \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} e^{i(x k)}\frac{\sqrt{ (m^2 + k^2+q^2 ) }}{\sin(t\sqrt{ (m^2 + k^2+q^2 )})+i \varepsilon } dk dq$$

How would I go about plotting these on a 2D plot or even finding an approximation (if one exists!).


1 Answer 1


I will write $x = (x_1,x_2)$ and $z=(k,q)$ to simplify the notations, and $|x| = \sqrt{x_1^2+x_2^2}$, so your function can be written $$ f_m(x) = \int_{\mathbb{R}^2} e^{-i\, x·z} \,\sqrt{m^2+|z|^2}\,\mathrm{d}z $$ and so, rigorously, it can be interpreted as the Fourier transform of the function $z\mapsto \sqrt{m^2+|z|^2}$, which can also be written $(m^2-\Delta)^{1/2}\delta_0$. Even if this is a singular function, it is not localized in one point since $(m^2-\Delta)^{1/2}$ is a nonlocal operator. First, to get the dependence in $m$, we can do the change of variable $z = m\,\xi$ to get (since $\mathrm{d}z = m^2 \,\mathrm{d}\xi$) $$ f_m(x) = m^3\int_{\mathbb{R}^2} e^{-i\, m\,x·\xi} \,\sqrt{1+|\xi|^2}\,\mathrm{d}\xi = m^3 f_1(m\,x). $$

In order to compute $f_1$, which is a singular distribution, one can write it as $$ f_1(x) = (1-\Delta)^{1/2}\delta_0 = (1-\Delta)(1-\Delta)^{-1/2}\delta_0 = (1-\Delta)\, G_1 $$ where $G_1 := (1-\Delta)^{-1/2}\delta_0$ is known as the Bessel potential of order $1$ (see https://en.wikipedia.org/wiki/Bessel_potential), and is just the Fourier transform of $z\mapsto (1+|z|^2)^{-1/2}$. Radial Fourier transform in dimension $2$ can be seen as a Hankel transform (https://en.wikipedia.org/wiki/Hankel_transform) by the formula $$ \hat{g}(x) = (2\pi) F_{g,0}(|x|) $$ and the Hankel transform of $1/\sqrt{1+r^2}$ is $e^{-r}/r$, therefore $$ G_1(x) = 2\pi\, \frac{e^{-|x|}}{|x|} $$ and for $x≠0$ $$ f_1(x) = 2\pi\, (1-\Delta)\frac{e^{-|x|}}{|x|} = 2\pi\, \frac{(|x|+1)\,e^{-|x|}}{|x|^3} $$ so the function you are looking for is (for $x≠0$) $$ f_m(x) = 2\pi\, \frac{(|m\,x|+1)\,e^{-|m\,x|}}{|x|^3} $$

  • $\begingroup$ Wow thanks very much! $\endgroup$
    – zooby
    Jul 22, 2020 at 12:32

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