Why does the vacuum polarization in 2D massless Fermion QED,

$$ i\Pi^{\mu\nu}(q) = i(\eta^{\mu\nu}-\frac{q^\mu q^\nu}{q^2})\frac{e^2}{\pi}, $$

have the structure of a photon mass term, as is claimed on Peskin chapter 19 page 653?


Because QED in $D=2$ is a confining theory and as such it develops mass gap. The coulomb potential in $D=2$ is linear with the distance of the charges. It is one of the few exactly solvable confining QFT theories.

Perhaps, I should add that by gauge invariance one can always fix $A_x=0$ while for the other component, $A_t$, the equations of motion give just a constraint, $\partial^2_x A_t\propto j_0$. There is thus no propagating mode associated with the photon field in $D=2$. Solving the constraints for $A_t$ and plugging it back in to the action you generate a mass term for the boson field that describes the fermion fields (and currents) via the so called bosonization (schematically, the correlation functions of scalar fields $\phi$ are logs, their exponential can give the correlation functions of other fields such as the fermions). It is exactly such a mass term that give mass to the ''meson'' state.

Another way to see it, is through the chiral anomaly $\partial_\mu J^\mu_5=\frac{e}{2\pi} \epsilon^{\mu\nu}F_{\mu\nu}$ which, via the equation of motion for $A_\mu$, implies $(\partial_\mu \partial^\mu+e^2/\pi)\epsilon^{\mu\nu} F_{\mu\nu}=0$. This equation says that there is a pole at $p^2=e^2/\pi$ associated withe the pseudoscalar operator $\epsilon^{\mu\nu}F_{\mu\nu}$.

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    $\begingroup$ Hi, thanks for your comment. It indeed feels better now. What I am still wondering about is: in the context that P&S make their claim, it seems as if they expect one to be able to just stare at propagators (which is the vacuum polarization) and "see" that it has a mass. Is there a way to see it like that? $\endgroup$ – PPR Jul 6 '14 at 14:47
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    $\begingroup$ Just look at eq.7.73 and 7.75 in P&S, and I should become apparent. $\endgroup$ – TwoBs Jul 6 '14 at 16:01

The way to see to this a la Peskin and Schroeder is by staring at the renormalized propagator in equation 7.75, \begin{equation} P _{ \mu \nu } \equiv \frac{ - i g _{ \mu \nu } }{ q ^2 ( 1 - \Pi ( q ^2 ) ) } \end{equation} In $ 4 $ dimensions, $ \Pi ( q ^2 ) $ doesn't contain a pole (as expected) and so, \begin{equation} P _{ \mu \nu } = \frac{ - i g _{ \mu \nu } }{ q ^2 (\# ) } \end{equation} In $ 2 $ dimensions, the propagator strangely does gain a pole, $ \Pi ( q ^2 ) = \frac{1}{ q ^2 }\frac{ e ^2 }{ \pi } $. This gives, \begin{equation} P _{ \mu \nu } = \frac{ - i g _{ \mu \nu } }{ q ^2 - \frac{ e ^2 }{ \pi } } \end{equation} giving a photon mass, $ m_\gamma=e / \sqrt{ \pi } $.


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