# Topological photon mass and QED in (2+1)-D

My question was: how to compute topological photon mass in (2+1) QED and why it comes?

• Are you asking about the mass term of the form $m \, \epsilon_{abc} \, A_a \, \partial_b A_c$ ? – Kostas Nov 20 '19 at 12:05
• @Kostas , yes, I asked about it. I believe that I have understood its appearence and have write the answer – Artem Alexandrov Nov 20 '19 at 12:51

First of all, let me say why photon obtain non-zero mass and how. It comes from one-loop correction to photon propagator, which is given by simple fermionic loop: $$\Pi_{\mu\nu}(k)\propto \int\frac{d^3p}{(2\pi)^3}\frac{\mathrm{Tr}\left[\gamma^{\mu}(\gamma\cdot p+m)\gamma^{\nu}(\gamma\cdot(p+k)+m)\right]}{(p^2-m^2+i\epsilon)((p+k)^2-m^2+i\epsilon)}.$$ Trace plays the crucial role in this expression because in (2+1), we have $$\mathrm{Tr}(\gamma^{\mu}\gamma^{\nu}\gamma^{\lambda})=-2i\epsilon^{\mu\nu\lambda}.$$ Therefore, we can write down polarization tensor as $$\Pi_{\mu\nu}^{S}(k)+\Pi_{\mu\nu}^A(k).$$ One can try to use dimensional regularization to deal with this integral but I conjecture that does not work: dimensional regularization is insensitive to odd divergencies. Then, in my view, the best way to compute $$\Pi_{\mu\nu}(k)$$ is dispersion relation. For $$\Pi^{S}_{\mu\nu}(k)$$, I can quickly obtain the following expression for imaganiary part (I know answer for (3+1) and the only difference is the factor $$\sqrt{k^2/4-m^2}$$), $$\mathrm{Im}\,\,\Pi^{S}\propto\frac{4m^2+k^2}{\sqrt{k^2}},$$ where I denote $$\Pi^S(k)=g^{\mu\nu}\Pi_{\mu\nu}^S/2$$ Then, to regularize the expression for $$\Pi_{\mu\nu}^S(k)$$, we use the following dispersion relation: $$\Pi(t)\propto t^2\int_{4m^2}^{\infty}\frac{ds}{s(s-t)}\mathrm{Im}\,\,\Pi^{S}(s),\quad s=k^2,$$ but this part of polarization tensor is not interesting. The same calculations for assymetric part of vacuum polarization gives very similar answer for $$\Pi^A(k)$$ and the most important fact is $$\boxed{\Pi^A(0)\propto\frac{e^2}{4\pi}\frac{m}{|m|}.}$$ Therefore, technically, photon mass appears due to assymetric contribution to vacuum polarization tensor. Physically, it comes from parity anomaly and I do not find the best words, but roughly speaking it means that mass term are invariant on $$P$$-transformations only in even dimensions and I have emphasized it by denotation $$m/|m|$$.
To be honest, I did not check all the calculations and therefore I use $$\propto$$ instead of $$=$$, but intermediate derivitions are the same.