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Pauli-Villars regularization instructs us to do such a replacement: $$\frac{1}{p^2-m^2+i\epsilon} \rightarrow \frac{1}{p^2-m^2+i\epsilon} - \frac{1}{p^2-\Lambda^2+i\epsilon}$$ And then claim: such a replacement implies a smoothly cut off when $p\geq\Lambda$.

I did some modified: $$\frac{1}{p^2-m^2+i\epsilon} - \frac{1}{p^2-\Lambda^2+i\epsilon}=\frac{m^2-\Lambda^2}{(p^2-m^2+i\epsilon)(p^2-\Lambda^2+i\epsilon)}$$ But after this, I still can't see how a "smoothly cut off" is applied when $p\geq\Lambda$.

My question is: shouldn't "cut off" be interpret as: when $p\geq\Lambda$ then the entire expression tends to $0$? And how can I see the cut off?

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When $p^2\gg \Lambda$, as you can see directly from your "modified" expression, the propagator scales as $\sim 1/p^4$ instead of $\sim 1/p^2$. If you go through arguments about the superficial degree of divergence, you'll find that this extra negative weight to the propagator will remove the divergent behavior you would otherwise get.

Meanwhile, for $p^2 \ll \Lambda$, you can essentially ignore the "negative propagator" term, and the propagator has the normal behavior. The regulator is smooth because there is a smooth transition between these behaviors (ie, you can differentiate the propagator with respect to $p^2$ as many times as you want without introducing any singularities, unlike a "hard cutoff" where you would have a theta function involving $p^2$).

However note this is just a regularization trick, not a fundamental solution to the issues of divergences in QFT. Unitarity requires that the propagator falls of no faster than $1/p^2$ for large $p$. In the case of Pauli Villars, the price of regulating the divergences by making the propagator fall off faster, is to introduce a ghost, or a particle with the wrong sign kinetic term (which appears here as the wrong sign propagator). If you tried to take the Pauli Villars theory seriously, you'd find it was completely unstable for energies above $\Lambda$ because of the ghost. However, we only are using the Pauli Villars propagator as regulator and plan to send $\Lambda$ to $\infty$ at the end of the day; we don't really think the ghost particle is there, physically.

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