Interpretation of renormalisation of composite operators

Question

The notion of renormalization is probably one of the most difficult to understand and bizarre properties of the QFT. As for the renormalisation of couplings it seems strange, from the first sight, that the coupling constants in the bare Lagrangian of interacting theory mostly have to be infinite in order to get a finite results (amplitudes and cross-sections) for physical properties.

Renormalised fields and the bare fields are related by multiplicative way for the renormalizable theories: $$ \phi_R = Z_{\phi}^{1/2} \phi_0 $$ The same is true for mass, interaction coupling e.t.c. However, when in comes to the renormalisation of composite operators, renormalisation of an operator involves not a simple multiplication by some factor, but mixing with operators of the same dimension and properties under Lorentz transformations.

I am reading Collins'book Renormalization https://www.cambridge.org/core/books/renormalization/6EA5EEEBB9A02190F7C805856244181B. And in the 6th chapter the renormalisation of an operator is introduced by considering the $\phi^2$ operator for $\phi^3$ theory in 6D.

In order to deduce the expression for the renormalized operator, he is considering the Green's function: $$ \langle 0 | T \phi(x) \phi(y) \phi^2(z) | 0 \rangle $$ Then he considers all one-loop graphs for the theory. Some divergences are eliminated by the counterterms for the mass. But to kill the remaining divergences, he adds new counterterms with the same divergences, as those emerging, when performing integrations in the loops. And the resulting operator is: $$ \frac{1}{2} [\phi^2] = Z_{a} \frac{1}{2} \phi^2 + \mu^{d/2 - 3} Z_b m^2 \phi + \mu^{d/2 - 3} Z_c \phi $$ And the question is - how to correctly interpret the mixing of operators? Speaking roughly, I had a pig, but it turned out to have fish's scale, wings and horns, and it is in fact not a pig, but a strange hybrid, consisting of pig, deer, salmon and eagle. If a want to compute correlation functions with the given composite operator, QFT says:

Correlation function with $\phi^2$ is undefined, maybe you meant $\frac{1}{2}Z_a \phi^2 + \ldots$

Or, in another words, I cannot obtain any sensible result with the $\phi^2$, but there is a combination of $\phi^2$ and other stuff - $m^2 \phi$, $\Box \phi$ - that gives a finite result.

i apologize if the analogies are silly, I wanted to try to give myself some easy explanation.

Answer 1

The fact that renormalization mixes a composite operator with other operators can be understood using lattice QFT, which is manifestly finite from the beginning. Lattice QFT is messy, but for this question, the only important thing is that the lattice provides a non-perturbative cutoff. Renormalization is about changing the scale of the cutoff, and the goal is to understand what this does to a composite operator like $\phi^2$.

Work in euclidean signature for simplicity. In Wilson's picture, renormalization amounts to lowering the cutoff from $\Lambda_H$ to $\Lambda_L$ by integrating out modes with wavenumbers between those two scales. The subscripts $H$ and $L$ stand for high and low. Write $$ \phi(x) = \phi_H(x)+\phi_L(x), \tag{1} $$ where $\phi_{H}(x)$ is the part involving only wavenumbers between the two cutoffs, and $\phi_L(x)$ is the part involving only wavenumbers below $\Lambda_L$. The correlation functions of interest are generated by $$ Z[J,K]\propto \int [d\phi]\ \exp\left(-S[\phi] +\int \phi(x) J(x) +\int \phi^2(x) K(x)\right). \tag{2} $$ The quadratic source term $\int\phi^2 K$ is redundant in equation (2), because we can generate insertions of $\phi^2(x)$ either by taking one derivative with respect to $K(x)$ or by taking two derivatives with respect to $J(x)$.

When we integrate out modes with wavenumbers between $\Lambda_H$ and $\Lambda_L$, we're agreeing to consider only correlation functions of operators that have been smeared enough to avoid resolving anything above the scale $\Lambda_L$. The key point is that smearing does not commute with squaring, so inserting $\phi^2(x)$ in a correlation function and then smearing it is not the same as inserting $\phi_L^2(x)$ in a correlation function: $$ \big(\phi^2(x)\big)_L\neq \phi_L^2(x). \tag{3a} $$ More explicitly, $$ \int dx'\ f(x')\phi^2(x+x')\neq \left(\int dx'\ f(x')\phi(x+x')\right)^2 \tag{3b} $$ where the integral with kernel $f$ is designed to eliminate the high-wavenumber modes. The generating function for correlation functions of sufficiently-smeared operators is obtained from (2) by setting $J_H=0$ and $K_H=0$, which prevents us from inserting any high-resolution operators. This leaves $$ Z[J_L,K_L]\propto \int [d\phi]\ \exp\left(-S[\phi] +\int \phi(x) J_L(x) +\int \phi^2(x) K_L(x)\right). \tag{4} $$ Modes with different wavenumbers are orthogonal to each other, so we have $$ Z[J_L,K_L]\propto \int [d\phi]\ \exp\left(-S[\phi] +\int \phi_L(x) J_L(x) +\int \big(\phi^2(x)\big)_L K_L(x)\right). \tag{5} $$ Thanks to the inequality (3), the quadratic source term $\int \phi^2 K_L$ is no longer redundant.

When we integrate over the high-wavenumber modes $\phi_H$, the $J_L$-source term is not affected, because it doesn't involve $\phi_H$. In contrast, the $K_L$ source term is affected, because the inequality (3) says that the composite operator $\big(\phi^2(x)\big)_L$ does involve $\phi_H$. The details depend on the action $S[\phi]$, but even without working through those details, the inequality (3) already suggests that renormalizing a composite operator will mix it with other operators.