It is known that one could recover electrostatic Coulomb potential from QED as a nonrelativistic Born approximation. The 3D case could be found in the standard textbook P&S Intro to QFT. This method indeed could (or should can) be generalized to arbitrary spatial dimension $d$. Theories in different dimension share free propagator in common form $$ G_{00}(q)=\frac{i}{q^2+i\epsilon} $$ due to the fundamental quadratic principle in physics. In the nonrelativistic limit, $$ G_{00}(q)\sim\frac{i}{-|\mathbf q|^2+i\epsilon}+\mathcal O(|\mathbf q|^4). $$ The Coulomb potential is its spaitial Fourier transformation $$ \int\mathrm d^d\mathbf q\ \frac{ie^{i\mathbf q\cdot\mathbf r}}{-|\mathbf q|^2+i\epsilon}=\int\mathrm dq\mathrm d\Omega\ \frac{iq^{d-1}e^{iqr\cos\theta}}{-q^2+i\epsilon}=\frac{1}{r^{d-2}}\int\mathrm du\mathrm d\Omega\ \frac{iu^{d-1}e^{iu\cos\theta}}{-u^2+i\epsilon'} $$ Assuming the integral has been properly regularized, it is a constant independent of $r$ so that Coulomb potential $$V(r)=\frac C{r^{d-2}}.$$

The problem arises in the case $d=2$. In this case $V(r)$ is a constant, which violates the result from applying Gauss's law in 2D $$V(r)=C\ln(\frac rL).$$

So what is wrong here?


Inspired by the post mentioned in @Chiral Anormal's comment, I found the integral could be expressed with help of Bessel functions $$ \int\mathrm dq\mathrm d\theta\frac{qe^{iqr\cos\theta}}{-q^2+i\epsilon}=2\pi\int\mathrm d q\ \frac{qJ_0(qr)}{-q^2+i\epsilon}=-2\pi K_0(-i^{3/2}r\sqrt{\epsilon}) $$

As is commonly done in QFT the divergence was arranged by $\epsilon$ and asymptotically $$ -2\pi K_0(-i^{3/2}r\sqrt{\epsilon})\sim 2\pi \ln(r\sqrt{\epsilon})+\mathcal O(1),\quad \epsilon\rightarrow 0 $$ as expected. The "characteristic length" $L$ in logarithmic plays a role as the regulator was argued by @Luboš Motl in Coulomb potential in 2D. It partially solves my question. However, still, I am troubled about the consistency of the QED propagator in 1+2d spacetime and the gap in my argument, i.e., how the last integral depends on $r$.

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    $\begingroup$ Doesn't the $d=2$ version of the original integral (the one on the left-hand side) satisfy Gauss's law? $\endgroup$ Commented May 11, 2021 at 13:20
  • $\begingroup$ Emmm...I am not sure if I have fully understood the point. The integral on the LHS do satisfy the Poisson equation $\nabla^2 V(r)=\delta(r)$ with $d=2$, as I think it should be. However the RHS seems not. That do trouble me a lot... $\endgroup$
    – EliC
    Commented May 11, 2021 at 13:38
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    $\begingroup$ See Qmechanic's answer to physics.stackexchange.com/q/35197 $\endgroup$ Commented May 12, 2021 at 0:41
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    $\begingroup$ Thanks a lot! I have read the answer and it takes a genius transformation in calculating the propagator in a general form. But for me it is not... emmm the final answer. Things are still not properly well-defined. I mean if one takes a priori $d=2$ and $m=0$ the integral is not regularized. In the answer it just let the known result evolve (or extrapolate) to an indetermined case and see the tendance. I have updated my question and make a further claim about it. $\endgroup$
    – EliC
    Commented May 13, 2021 at 3:42
  • $\begingroup$ Yet. I have just realized that we should wondering if the Coulomb potential is ill-defined in 2D. $\endgroup$
    – EliC
    Commented May 13, 2021 at 3:45


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