0

$D$-dim Screened Poisson equation or Klein-Gordon equation in Euclidean spacetime is $$\left(-\sum_{i=1}^D \partial_i\partial_i +m^2\right)u(x)= f(x) $$ The Green's function is $$\int_{-\infty}^{\infty}\frac{d^D k}{(2\pi)^D}\frac{e^{ik\cdot x}}{k^2+m^2}$$

1.How to prove for $D>2$: $$\int_{-\infty}^{\infty}\frac{d^D k}{(2\pi)^D}\frac{e^{ik\cdot x}}{k^2+m^2}=\frac{1}{(2\pi)^{D/2}}\left(\frac{m}{|x|}\right)^{D/2-1}K_{D/2-1}(m|x|)$$ with $k\cdot x= \sum_{i=1}^D k^i x^i$ and $k\cdot k= \sum_{i=1}^D k^i k^i$, $|x|=\sqrt{\sum_{i=1}^{D}(x^i)^2}$. $K_\alpha(x)$ is modified Bessel function of the second kind.

2.What's the above integration when $D=2$ and $D=1$?

3.I can't understand why such a integration can be convergent for $D\ge2$. It seems that when $D=2$ this integration should be logarithmical divergent. $D>2$ should be divergent as power $D-2$. Why this argument is wrong?

4.How to prove the analytic form of above function when $m\rightarrow0$ for $D>2$ is $$\frac{2^{D/2-2}\Gamma(D/2-1)}{(2\pi)^{D/2}x^{D-2}}$$

5.What's the limit when $D=2$ and $D=1$?

Qmechanic
  • 12,298
maplemaple
  • 1,211
  • Too many questions, most of which aren't questions but rather requests to solve the problem for you. I'll answer part 3 since it's an actual question. The reason it's wrong is because there is an $e^{ik\cdot x}$ in the integral that causes oscillations that allow the integral to converge. Your argument implies that for $D\ge 2$ it is not absolutely integrable and more importantly that the function should diverge as $x\to 0.$ – spaceisdarkgreen Oct 15 '17 at 21:19
  • @spaceisdarkgreen Yes. The integrability of function with $e^{ikx}$ always make me confused. I don't learn this part in my calculus course. Is there some textbook or literature which treat them thoroughly? – maplemaple Oct 15 '17 at 21:26
  • Well it's a Fourier transform so there should be resources that handle transform integrals in more than one dimension (I don't know anything comprehensive off the top of my head.) $D=1$ is easy here (just a contour integral) and $D=3$ is elementary as well. The way I learned to do it for general $D$ was to write $$\frac{1}{k^2+m^2} = \int_0^\infty e^{-t(k^2+m^2)}dt $$ and then change order of integration to do the $k$ integral first, which is just a Gaussian integral. Then massage the rest into one of the integral representations of Bessel K. – spaceisdarkgreen Oct 16 '17 at 01:35
  • 1
    This is the free field propagator for quantum field theory so it's done in a lot of QFT books. See eqn 5.4.3 here http://www.wiese.itp.unibe.ch/lectures/fieldtheory.pdf – spaceisdarkgreen Oct 16 '17 at 01:44
  • 1
    This is called the https://en.wikipedia.org/wiki/Bessel_potential. This is done in dimension $1$ here https://math.stackexchange.com/questions/4220006/computing-the-fourier-transform-of-exponential-decay-in-mathbbr2/4220125#4220125 with the method indicated by spaceisdarkgreen. You can find the full derivation in the book of Stein Singular integrals and differentiability properties of functions (1970) . – LL 3.14 Dec 31 '22 at 17:50

0 Answers0