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

  • $\begingroup$ Are you asking about the mass term of the form $m \, \epsilon_{abc} \, A_a \, \partial_b A_c$ ? $\endgroup$
    – Kphysics
    Nov 20, 2019 at 12:05
  • $\begingroup$ @Kostas , yes, I asked about it. I believe that I have understood its appearence and have write the answer $\endgroup$ Nov 20, 2019 at 12:51

1 Answer 1


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 $m/|m|$.

Furthermore, Coleman & Hill showed that higher order corrections are zero for all possible matter fields and couplings to electromagnetic field. In addition, the same effect arises in every odd dimensional QED theory.

Finally, photon obtains mass without any topological terms in initial lagrangian and one can roughly say that it is an analogue of chiral anomaly existing in even dimensions.

To be honest, I did not check all the calculations and therefore I use $\propto$ instead of $=$, but intermediate derivitions are the same.

This answer is based on the following papers:

  1. Scharf, G., et al. "Causal approach to (2+ 1)-dimensional QED." Annals of Physics 231.1 (1994): 185-208, link
  2. Coleman, Sidney, and Brian Hill. "No more corrections to the topological mass term in QED3." Physics Letters B 159.2-3 (1985): 184-188, link

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