In an example of operator product expansion applied to $\phi^4$ theory of the book QFT an integrated approach, where the Eulerian Lagrangian is $$\mathcal{L}=\frac{1}{2}\left(\partial_{\mu} \phi\right)^{2}+t a^{-2} \phi^{2}+u a^{D-4} \phi^{4}+h a^{-(1+D / 2)} \phi\tag{15.127}$$ where $$m^{2}=t a^{-2}, \quad \lambda=u a^{D-4}, \quad J=h a^{-(1+D / 2)}\tag{15.128}$$ defines the dimensionless mass, coupling constant and symmetry breaking field in terms of the lattice spacing $a$.

The author then defines the normal ordered composite operator as $$: \phi^{2}:=\phi^{2}-\left\langle\phi^{2}\right\rangle, \quad: \phi^{4}:=\phi^{4}-3\left\langle\phi^{2}\right\rangle \phi^{2}, \quad \text { etc. }\tag{15.129}$$ with $\phi_{n}\equiv: \phi^{n}:$, a calculation of OPF proceeds as follow: $$\begin{aligned} \lim _{y \rightarrow x} \phi_{1}(x) \phi_{1}(y) &=\lim _{y \rightarrow x}: \phi(x):: \phi(y): \\ &=\lim _{y \rightarrow x}\left[\frac{1}{|x-y|^{D-2}}+: \phi^{2}((x+y) / 2):+\cdots\right] \end{aligned}\tag{15.130}$$

Although I think I roughly "understand" (which can be wrong) that we ought to treat $\phi$ to different powers as different field operators in this formalism, and that normal ordering ensures whatever field being normal ordered is not contracted in the application of Wick's theorem, I still don't quite get how this scheme actually works. In particular I think in Fradkin's (the author of the book) lecture recording available at the course webpage, he commented that at the fixed point (scale invariant point), the Lagraingian is $$\mathcal L^*=:\frac{1}{2}\left(\partial_{\mu} \phi\right)^{2}:$$ this is confusing, as it suggests that we should replace the fields in the Lagrangian with the normal ordered fields, how would this be something legitimate to do? And why these normal ordered fields come into play after all? (I can't help but feeling my explanation that the normal ordered field is not contracted by Wick's theorem is only partial).

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    $\begingroup$ I don't really get why the Lagrangian needs to be normal ordered in the first place. The Lagrangian density is a classical field (as in Feynmann path integral), not a quantum field. I think only in Hamiltonian (canonical formalism) you need to normal order operators, as this subtracts the infinite zero-point energy. Normal ordering is used to subtract UV divergences in an OPE to define a new regular quantum field. The fact that you don't need to wick contract inside normal ordering is just because the contraction is already built into the definition. $\endgroup$
    – Yijian Zou
    Jun 23, 2022 at 11:06
  • $\begingroup$ When you're at the fixed point, local operators span the Hilbert space. So this is why you might want to consider operators containing multiple powers of $\phi$ and derivatives. One of them will happen to coincide with the Lagrangian. $\endgroup$ Jun 27, 2022 at 16:00


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