# Aren't $\phi^4$ composite operators?

I have this trouble with terminology. I wonder why authors introduce the concept of composite operators after they've already talked about eg phi four theory, it phi cubed. Aren't these operators already composite ones? What's the difference? In particular when considering renormalization, say QED, the procedure goes like introducing factors of $Z^{1/2}$ for each field and so on. So why is the renormalization procedure in QED+Yukawa so different that one needs to consider how composite operators get renormalized? What's the big difference, isn't say $m\bar{\psi}\psi$ of QED already composite?

• What definition of composite operator (which, in my experience, is just that of a local operator, i.e. some product of fields) are you using? Who are the authors that introduce "composite operators" not already in the simplest applications of Wick's theorem, and that do not consider $\phi^4$ to be a composite one? – ACuriousMind Dec 7 '14 at 1:00
• Perhaps phi 4 was a bad example. But what's the difference between Yukawa term and the m-term of QED in this regard? Cheng is the author, but I could be wrong about them not regarding some oleraots as not composite because I've just skimmed through the book, so forgive my ignorance. – Your Majesty Dec 7 '14 at 14:21
• Well, since a local operator is "a product of fields", $\bar\psi\psi$ qualifies as well. My initial comments still stands with the replacement $\phi^4 \mapsto \bar\psi\psi$. – ACuriousMind Dec 7 '14 at 15:38
• I must take a deeper look at the book, I'll come back to this. – Your Majesty Dec 7 '14 at 15:40

Consider a correlator like $$\frac{1}{Z}\int \phi(x)^4 \phi(y_1)\ldots\phi(y_n)\ e^{-\int \phi(z)^4 dz}\ d\mu(\phi)$$ where $d\mu$ is the perturbed free field measure. Both $\phi^4$'s are composite fields but very different ones. The $\phi(z)^4$ in the exponential is a composite field for the free field theory $d\mu$ whereas the $\phi(x)^4$ outside is a composite field for the full interacting theory. In fact, and this is Nelson's proof, it is possible to rigorously construct this interacting theory in 2d (in finite volume) by defining $\phi(z)^4$ as a $d\mu$ composite field (this is just the trivial Wick power renormalization) then sticking it in the exponential and integrating $d\mu(\phi)$.
Basic renormalization theory tells you how to get correlators $$\frac{1}{Z}\int \phi(x_1) \phi(x_2) \phi(x_3) \phi(x_4) \phi(y_1)\ldots\phi(y_n)\ e^{-\int \phi(z)^4 dz}\ d\mu(\phi)$$ AFTER taking the limit $\Lambda\rightarrow\infty$ of removing the UV cut off, for non-coinciding points $x_1,\ldots,y_n$. To get to the above correlator with $\phi(x)^4$ you still have to follow this by the analysis of the very singular limit $x_1,x_2,x_3,x_4\rightarrow x$. This needs more advanced renormalization theory, typically in the following chapters of QFT textbooks about composite fields and the OPE. The process I described is in fact the construction of the composite field $\phi^4$ in the interacting theory using the OPE. An alternative construction (in fact the more standard one) is to take the limit $x_1,x_2,x_3,x_4\rightarrow x$ first and then take $\Lambda\rightarrow\infty$. Both procedures should give the same result.