5
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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 $\epsilon^{\mu\nu}F_{\mu\nu}$.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
4
$\begingroup$

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 } $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.