Consider the Lagrangian of $\phi^4$ theory in 4-dimensions $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4\tag{1}$$ For a term in the Lagrangian of the form $C_{m,n}\phi^n(\partial_\mu\phi)^m$, the corresponding coefficient in 4-dimensions scales as $$C^\prime_{n,m}=b^{n+m-4}C_{n,m}\tag{2}.$$ For a reference, see Eq. 12.27, page 402 of Peskin and Schroeder.

The Lagrangian $\mathcal{L}$ in Eqn.(1), contains all relevant and marginal operators except $$\phi^2(\partial_\mu\phi)^2, \phi(\partial_\mu\phi)^3,\phi^3(\partial_\mu\phi),(\partial_\mu\phi)^4,\phi^3.\tag{3}$$ Even if we exclude the $\phi^3$ terms by demanding a symmetry under $\phi\to-\phi$, or by demanding the Hamiltonian to be bounded from below, one is still left with the other 4 possibilities.

If $\phi^4$ theory is regarded as a low-energy effective theory, I do not understand why the first 4 terms in (3) are they not considered? Is it just for simplicity?

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    $\begingroup$ all the terms in $(3)$ are irrelevant (except for the last one, $\phi^3$). $\endgroup$ – AccidentalFourierTransform May 29 '17 at 19:02

your power-counting is not correct. In $d=4$ a gradient $\partial_{\mu}$ counts like a field $\phi$. So for instance you should have $C_{m,n}'=b^{n+2m-4}C_{m,n}$

  • $\begingroup$ I used Peskin's formula 12.27 for $d=4$. Is that wrong there? $\endgroup$ – SRS May 29 '17 at 19:03
  • $\begingroup$ you misread the formula which has a number of fields $N$ and of derivatives $M$. In your monomials $N=n+m$ and $M=m$. $\endgroup$ – Abdelmalek Abdesselam May 29 '17 at 19:13
  • $\begingroup$ Okay. Now I understand. :-) $\endgroup$ – SRS May 29 '17 at 19:17

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