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Sep 18, 2022 at 17:04 comment added R. Rankin Is there any reason a $(\Psi \Psi^*)^{-2}$ term can't be present (renormalizable) perhaps with an additive constant to bound it?
May 29, 2020 at 7:56 comment added Mathieu Krisztian @ACuriousMind : thank you
May 28, 2020 at 21:15 comment added ACuriousMind @MathieuKrisztian 1. It would also not be a scalar. 2. Linear terms can be transformed away anyway.
May 28, 2020 at 20:03 comment added Mathieu Krisztian And why is the $\phi$ term at power one is forbidden ?
May 6, 2015 at 11:00 comment added ACuriousMind @SuperCiocia: If you know the power-counting renormalizability argument, it's obvious from the mass dimension the coupling must have for terms of order higher than 4, if not, then not.
May 5, 2015 at 22:14 comment added SuperCiocia Is there an easy/intuitive way of understanding why the $\mathcal{O}(\phi^5)$ term in non-renormalisable?
May 5, 2015 at 18:06 vote accept SuperCiocia
May 5, 2015 at 16:37 comment added innisfree but the effective potential can look a bit different a la Coleman-Weinberg
May 5, 2015 at 15:13 history answered ACuriousMind CC BY-SA 3.0