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In Srednicki's QFT book, section 9, he introduces the $\varphi^3$ lagrangian:

$$\mathcal{L}= -\frac{1}{2}Z_\varphi(\partial_\mu\varphi)(\partial^\mu\varphi) -\frac{1}{2}Z_mm^2\varphi^2 +\frac{1}{6}Z_g g \varphi^3 +Y \varphi.\tag{9.1} $$

Then for the next several chapters, he says we "expect" $$Z_i~=~1+O(g^2), \qquad i\in\{\varphi,m,g\},\tag{1}$$ and then he goes ahead and assumes this is the case when solving for them in section 14, but he never justifies this assumption. Why is it justified, i.e., why can't they be $1+O(g)$?

I think it has something to do with the lowest-order correction to the propagator being a tree-level diagram with two vertices, not one, which brings in an $O(g^2)$ correction. But why not have the counterterms provide an $O(g)$ correction, so that they dominate the lowest-order correction to the propagator instead of the additional vertices?

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The renormalization factors $Z_{i}$ are produced by loop diagrams. For $Z_{\varphi}$ and $Z_{m}$, this means diagrams with two external legs, since these factors multiply operators with two factors of the field $\varphi$. You cannot possibly write down a loop diagram (or any diagram!) with two external legs with just one three-particle vertex. Since each three-$\varphi$ vertex is associated with a factor of the coupling $g$, you cannot get a correction to the propagator without having at least two vertices and thus an overall $g^{2}$.

The same argument applies to $Z_{g}$. Notice that each three-point vertex you add to a diagram changes the number of lines from even to odd, or vice versa. Thus, a diagram with one factor of $g$ has an odd number of external lines; a diagram with $g^{2}$ has an even number; and you go back to an odd number at order $g^{3}$. This actually ensures that every $Z_{i}$ is a power series in $g^{2}$, not just $g$.

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There's a cute argument based on symmetry, at least at when $Y = 0$ is treated as a perturbation. The $g=Y=0$ theory has a $\mathbb{Z}_2$ symmetry $\phi \to -\phi$. Let $X$ be some observable that's non-zero at tree level, admitting a series expansion $X = X(g)$. Then $X(g)$ must be an even function of $g$. QED!

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  • $\begingroup$ That's a little too cute, as in, it's so brief I don't know what you're getting at. $\endgroup$ – octonion Apr 15 at 3:38
  • $\begingroup$ I think there are a few steps missing in the logic. Are you promoting $g$ to a $\mathbb{Z}_2$ spurion that transforms as $g \to - g$ under $\mathbb{Z}_2$? $\endgroup$ – MannyC Apr 15 at 3:38
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    $\begingroup$ @MannyC yes, in modern parlance you could say that the theory is $\mathbb{Z}_2$ invariant if $g$ and $Y$ transform as spurions. $\endgroup$ – Hans Moleman Apr 15 at 4:02
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OP's eq. (1) is not true in general. It depends crucially on the choice of renormalization conditions$^{\dagger}$, see eqs. (14.7), (14.8) & (16.14) in Ref. 1. E.g. only the product $Z_gg$ enters the Lagrangian density (9.1), so clearly some sort of condition is needed to fix $Z_g$ and $g$ individually. The choices made in Ref. 1 are arguably the simplest choices.

Of course the heart of the $O(g^2)$ argument is already guessed by OP and elaborated in user Buzz's answer.

References:

  1. M. Srednicki, QFT, 2007; Chapters 9 + 14 + 16. A prepublication draft PDF file is available here.

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$^{\dagger}$ The 2 renormalization conditions in eq. (9.2) of Ref. 1. turn out to be not irrelevant for eq. (1).

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