3
$\begingroup$

In the introduction part of this renowned paper by S. Coleman and E. Weinberg, the authors write

The quartic self-coupling is required for renormalizability, to cancel the logarithmic divergence that arises in the amplitude for scalar Coulomb scattering.

The quartic term in scalar QED must be included in the Lagrangian (density) if one wishes to write the most general power-counting renormalizable Lagrangian (i.e., all non-negative mass-dimensional couplings) compatible with Lorentz invariance and gauge invariance.

The authors' statement although doesn't seem to contradict this but adds that if one neglects the quartic term, retaining only the quadratic mass term, the theory can become nonrenormalizable even with non-negative mass-dimensional term.

Do they mean massive scalar QED with vanishing quartic coupling is nonrenormalizable?

$\endgroup$
3
$\begingroup$

Do they mean massive scalar QED with vanishing quartic coupling is nonrenormalizable?

Yes, they mean precisely that. In Ref.1 you can find the explicit calculation of the quartic counter-term in scalar QED: $$ Z_\lambda=1+\left(\frac{3e^2}{2\pi^2\lambda}+\frac{5\lambda}{16\pi^2}\right)\frac{1}{\epsilon}+\cdots $$

With this, the quartic term in the Lagrangian reads $$ \lambda Z_\lambda|\phi|^4\overset{\lambda\to0}\longrightarrow\frac{3e^2}{2\pi^2}\frac{1}{\epsilon}|\phi|^4 $$

Therefore, if you were to purposely choose to omit this term, you would miss the $\frac{3e^2}{2\pi^2}\frac{1}{\epsilon}$ piece necessary to make the four-point function finite. You would lose renormalisability.

Note: if it turned out that the limit above vanishes, then you would be allowed to omit the quartic term (at least to one-loop order; it may be necessary to higher orders). This is precisely what happens in a scalar theory (in $d=4$): the most general renormalisable Lagrangian includes the terms $\phi^3$ and $\phi^4$. Unlike before, here you may want to choose to omit either of them without interfering with renormalisability. The reason is clear: if you omit the $\phi^3$ term you get the $\phi\to-\phi$ symmetry (which guarantees that no cubic divergence may arise), while if you omit the $\phi^4$ term the theory becomes super-renormalisable (which guarantees that no quartic divergence may arise).

References.

  1. Srednicki, chapter 65.
| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

I can't answer the question in your OP, but I can answer your title question in the affirmative: Srednicki QFT pgs. 131-132 says

We expect that theories with couplings whose mass dimensions are all positive or zero will be renormalizable. A detailed study of the properties of the momentum integrals in Feynman diagrams is necessary to give a complete proof of this. It turns out to be true without further restrictions for theories that have spin-zero and spin-one-half fields only. Theories with spin-one fields are renormalizable for $d = 4$ if and only if these spin-one fields are associated with a gauge symmetry.

| cite | improve this answer | |
$\endgroup$

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.