# Deriving 2D Coulomb potential from Born approximation

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?

----(update)----

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 $$r$$.

• Doesn't the $d=2$ version of the original integral (the one on the left-hand side) satisfy Gauss's law? Commented May 11, 2021 at 13:20
• 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...
– EliC
Commented May 11, 2021 at 13:38
• See Qmechanic's answer to physics.stackexchange.com/q/35197 Commented May 12, 2021 at 0:41
• 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.
– EliC
Commented May 13, 2021 at 3:42
• Yet. I have just realized that we should wondering if the Coulomb potential is ill-defined in 2D.
– EliC
Commented May 13, 2021 at 3:45