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?
Answer
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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