Why does Schwartz discard the product of counterterms $\delta_2\delta_m$?

Question

In Quantum Field Theory and the Standard Model by Schwartz the author starts with the QED bare Lagrangian, defines $A_\mu^0 = \sqrt{Z_3}A_\mu$, $\psi^0 = \sqrt{Z_2}\psi$, $m_0=Z_m m_R$ and $e_0 = Z_e e_R$. Further defining $Z_1 = Z_eZ_2\sqrt{Z_3}$ the bare Lagrangian gets written as \begin{eqnarray} {\cal L} &=& -\frac{1}{4}Z_3 F_{\mu\nu}F^{\mu\nu}+iZ_2\bar \psi\gamma^\mu{\partial_\mu}\psi- Z_2Z_m m_R\bar \psi\psi - e_R Z_1 \bar \psi \gamma^\mu A_\mu \psi.\tag{19.8} \end{eqnarray} Now he writes $$Z_i = 1+\delta_i.\tag{19.9+10}$$ The problem is that in the term $Z_2Z_m$ we would also have one term $\delta_2\delta_m$ in eq. (19.12) and this term is discarded by Schwartz.

I don't see how to justify it. I mean, even if you write a series expansion for each $\delta_i$ in powers of $e_R^2$ why would we just keep the leading result?

Being more explicit. Each $\delta_i$ starts at order $e_R^2$. Now say we want a result of order $e_R^4$. Clearly $\delta_2\delta_m$ should contribute because combining the two leading $O(e_R^2)$ contributions will give one $O(e_R^4)$ contribution.

So why one may just drop the $\delta_2\delta_m$ contribution as Schwartz seems to be doing?

Answer 1

Yes, OP is correct: The counterterm in Schwartz eq. (19.12) should be $$Z_2Z_m-1~=~ (1+\delta_2)(1+\delta_m)-1~=~\delta_2+\delta_m+\color{red}{\delta_2\delta_m}.$$ See e.g. Peskin & Schroeder eq. (10.38).

Answer 2

I believe that Schwartz is considering terms in the Lagrangian up to $O(e_R^3)^1$. So all terms of order $O(e_R^4)$, like the mentioned $\delta_2 \delta_m$ are not included. Of course, in general we should also include the $\delta_2 \delta_m$ term, which would vanish once we restrict ourselves to $O(e_R^3)$.

On top of this, if we wanted to include $\delta_2 \delta_m$, we would have to compute the counterterms $\delta_i$ up to $O(e_R^4)^2$. Otherwise, we would have an incorrect expansion of the Lagrangian up to $O(e_R^4)$.

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$^1$ Note that $e_R Z_1 \bar \psi \gamma^\mu A_\mu \psi$ will be of order $O(e_R^3)$.

$^2$ Remember that, as you have mentioned, Schwartz has only considered an expansion of $\delta_i$ up to $O(e_R^2)$.