Physics Asked on March 6, 2021
Assume $D$-dimensional Minkowski spacetime with one time dimension and $Dge 2$. Using the mass-$m$ Klein-Gordon propagator for a scalar interaction, we can derive a potential energy$$E=-intfrac{d^{D-1}k}{left(2piright)^{D-1}}frac{exp ikcdot r}{k^2+m^2}$$(see e.g. Quantum Field Theory in a Nutshell Chapter I.4), with $r:=x_1-x_2$ a distance between point unit “charges”. It’s easy to show that if $m=0$ this result is proportional to $ln r$ if $D=3$ and $r^{3-D}$ otherwise, and if $mne 0$ to $r^{3-D}e^{-mr}$ provided $Dne 3$. I’m struggling to evaluate the massive case with $D=3$, viz.$$E=-int_0^{2pi}frac{dtheta}{4pi^2}int_0^inftyfrac{kexp left(ikrcosthetaright)}{k^2+m^2}dk.$$Expanding the exponential as a power series, odd-power terms vanish under the $theta$ integration, so$$E=-int_0^{pi/2}frac{dtheta}{pi^2}int_0^inftyfrac{ucos left(ucosthetaright)}{u^2+left(mrright)^2}du,$$which is real as expected. Defining $fleft( a,,bright):=int_0^inftydfrac{ucos au ,du}{u^2+b^2}$, we have$$E=-int_0^{pi/2}frac{dtheta}{pi^2}fleft(costheta,,mrright).$$However, I’ve been unable to evaluate $f$, let alone this integral based on it.
Make a substitution of integration variables: d$theta$ -> dcos$theta$. Then you can get to a integral only in k which can be evaluated using contour integration and the Cauchy Residue theorem.
Answered by George H. on March 6, 2021
Let us denote the integral as $D(r)$. One of the simplest way to evaluate it is using of Schwinger parametrization, $$frac{1}{k^2+m^2}=frac{1}{2}int_{0}^{infty}dalpha,expleft(-frac{alpha(k^2+m^2)}{2}right)$$ and then find the complete square, $$frac{alpha(m^2+{bf k}^2)}{2}-i{bf k}cdot{bf r}=frac{1}{2}left(sqrt{alpha}{bf k}-frac{i{bf r}}{sqrt{alpha}}right)^2+frac{1}{2}left(frac{{bf r}}{sqrt{alpha}}right)^2+frac{alpha m^2}{2}.$$ Next it is easy to perform Gaussian integration, $$intfrac{d^{D-1}k}{(2pi)^{D-1}}expleft(-frac{1}{2}left(sqrt{alpha}{bf k}-frac{i{bf r}}{sqrt{alpha}}right)^2 right)=(2pialpha)^{-(D-1)/2},$$ with keeping in mind the remainder $e^{-r^2/2alpha}$, so finally $$D({bf r})=frac{1}{2(2pi)^{(D-1)/2}}int_0^{infty}dalpha,alpha^{-(D-1)/2}expleft(-frac{alpha m^2}{2}-frac{{bf r}^2}{2alpha}right).$$ Next, we perform substitution $alpha=lambda/t$, so $$frac{alpha m^2}{2}+frac{{bf r}^2}{2alpha}=frac{lambda m^2}{2t}+frac{r^2t}{2lambda} rightarrow lambda=frac{r}{m},$$ which immediately gives $$D(r)=frac{1}{(2pi)^{d/2}}frac{1}{2}frac{1}{lambda}int_{0}^{infty}dt,left(frac{t}{lambda}right)^{d/2-2}expleft(-frac{mr}{2}left(t+frac{1}{t}right)right),$$ where $d=D-1$. The final result is $$boxed{D(r)=frac{1}{(2pi)^{d/2}}left(frac{m}{r}right)^{d/2-1}K_{d/2-1}(mr)},$$ with $d=D-1$. What about the massless case? It is easy to show that $$boxed{E(r)=-frac{2^{d/2-1}}{(2pi)^{d/2}}frac{1}{r^{d-2}}Gammaleft(frac{d}{2}-1right).}$$
P.S.: I can make a mistake with dimension $D$, however the calculation strategy is clear (for me).
P.S.: It seems that the question about asymptotic behavior is discussed here
Answered by Artem Alexandrov on March 6, 2021
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