# Renormalization massless phi4 theory in Peskin.

by bifolco84
Tags: phi-4, renormalization
 P: 6 In Peskin- Schroeder, pag 412: "In massless phi4 theory, the one-loop propagator correction is completely canceled by mass counterterm." So, do massless theory provides mass counterterm? How is it generated? Maybe from a bare mass...don't have any clue. I'm confused because it seems that Peskin uses such a counterterm to eliminate the propagator ultraviolet divergencies (at 1 loop) but every kind of reference in a mass counterterm disappear in the following pages. So...no clue!
 P: 720 If you don't allow yourself a mass counterterm, I think the theory is not renormalizable, because it's impossible to cancel the divergence in the one-loop correction to the propagator. Having introduced the counterterm, you should be able to adjust it so that the pole in the full propagator occurs at zero momentum, so that the physical mass of the particle is zero. So to get a physical mass of zero in phi-fourth theory you need a nonzero mass term in the bare Lagrangian. Hopefully somewhat can correct me if I have this wrong.
 Sci Advisor Thanks P: 1,741 The tadpole diagram in massless $\phi^4$ theoy is 0 in dimensional regularization. So from there no trouble comes. Also in BPHZ, this term immediately cancels completely since it's momentum independent. As in any massless theory, mass is generated by the anomalous breaking of conformal symmetry. You start with a theory, which has no dimensionful parameters, but as soon as you calculate loop corrections, you encounter UV and IR divergences. The IR divergences must be tamed by appropriate soft-particle resummations. For the UV divergences you have to renormalize. Due to the infrared singularities, you cannot renormalize the divergent vertex function (1PI diagrams with four external legs) at all four momenta set to 0 as in the massive case, because you would hit IR singularities, and the counter terms at this original BPHZ prescription become ill defined. For instance take the one-loop four vertex ("fish diagram") of order $\lambda^2$. If $s$ is the center-of mass momentum squared running through the diagram, this vertex function has a branch cut along the whole positive real axis in the complex $s$ plane (in the massive case the cut starts in the time-like and runs along $s>(2m)^2$ due to the two-particle threshold). Thus you must not subtract the diagram at $s=0$ but at some scale $s_{\Lambda}=-\Lambda^2<0$. In this way you introduce a momentum scale, and the conformal invariance of the quantized theory is broken. Your renormalized coupling parameter becomes a function of this momentum scale, $\Lambda$, and thus you produce mass terms at higher orders, and you have to introduce a mass-counter term to get rid of the corresponding UV divergences.

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