# Why do correlation functions involving composite fields require special analysis?

For simplicity I will be considering $$\phi^4$$ theory. To analyze correlation functions of the form $$\langle \phi(x_1)\phi(x_2)\ldots\phi(x_n)$$ with $$x_1 \neq x_2 \neq \cdots \neq x_n \tag{1}$$ we can use perturbation theory and the usual Feynman rules.

Now consider a correlation function where the condition (1) is dropped. That is, some of the fields may be located at the same spacetime location and are said to be composite fields, for example

$$\langle \phi^n(x_1) \phi(x_2)\phi(x_3). \tag{2}$$

My questions about these correlation functions are the following:

1. Why do correlation functions involving composite fields require special analysis and a new set of Feynman rules?
2. Why do they require new renormalizations?
3. How do composite fields introduce new divergences? Are these from the propagators, i.e. we have something like $$D(x_1 - x_1)$$?
4. Why is (2) seen as a 3-point function and not an $$n+2$$-point function?
5. Wouldn't the interaction term $$\phi^4$$ also be considered a composite fields?

I apologize for including multiple questions in one post but I thought most of them are too related to open separate questions. I have already taken a look at the following related question: Aren't $\phi^4$ composite operators? but am still confused regarding the above.

• The main problem is regarding how we choose to define $\phi^2(x)$. The standard definition $\phi^2 (x) = \phi(x) \phi(x)$ does not work. Commented Mar 31 at 23:11
• @Prahar When I tried it I got a divergence coming from the propagator, something like $D(x_1 - x_1)$, and I assume this is why new renormalizations and Feynman rules are needed. Is that the problem you are referring to? Commented Mar 31 at 23:16
• Yes, that's exactly right. Note that this is a problem even in free theories, it has nothing to do with interactions. Interactions just make this problem much worse. Commented Mar 31 at 23:19
• @Prahar I see, thank you very much for your help! Commented Mar 31 at 23:38

As already pointed out by @Prahar, the problem of the definition of a composite operator arises already in the free theory. Considering the Lagrangian of a free real scalar field, $$\mathcal{L}[\varphi]=\frac{1}{2}\left(\partial_\mu \varphi \partial^\mu \varphi - (m^2-i \epsilon) \varphi^2\right), \tag{1} \label{1}$$ the naive expression $$\tilde{Z}[f] =e^{i \tilde{W}[f]}= \int [d \varphi] e^{i \int d^dx \left(\mathcal{L}[\varphi]+f(x) \varphi(x)^2\right)}, \qquad \tilde{Z}[0]=1, \tag{2} \label{2}$$for the generating functional of the squared field requires regularization and renormalization and \eqref{2} has to be replaced by $$Z[f]=e^{iW[f]}=\int [d\varphi] e^{i \int d^dx \left( \mathcal{L}[\varphi]+f(x) \varphi(x)^2+c_1f(x)+c_2 f(x)^2\right)}, \qquad Z[0]=1, \tag{3}$$ where the counterterm $$\Delta \mathcal{L}= c_1 f(x) +c_2 f(x)^2 \tag{4}$$ was introduced. Thus, the generating functional is determined only up to the free constants $$c_{1,2}^{\rm ren}$$. Working out the details of the computation of the generating functional up to order $$f^2$$ in the $$\overline{{\rm MS}}$$ scheme is left as an easy homework exercise.
Translated into operator language, the two-point function $$\langle 0 |\phi(x) \phi(0) |0\rangle= \frac{1}{i} \Delta(x) \tag{5}$$ becomes singular in the coincidence limit $$x \to 0$$ ($$x$$ can be seen as a regulator), $$\Delta(x) = \frac{i}{4\pi^2(-x^2 +i \varepsilon)}+\frac{i m^2}{16 \pi^2} \log\left((-x^2+i \epsilon)m^2 e^{\gamma-1}\right)+\mathcal{O}(x^2 \log x^2), \tag{6} \label{6}$$ making it obvious that the definition of $$\langle 0 | \phi^2(x) |0\rangle$$ requires regularization and renormalization via the counterterm $$c_1 f(x)$$. Analogously, the naive correlation function $$\tilde{\Pi}(x)=\langle 0 |{\rm T}\phi^2(x) \phi^2(0)|0\rangle_{\rm c}$$ exhibits a singularity $$\tilde{\Pi}(x) \sim 1/(x^2)^2$$ (for $$d\to 4$$) such that the integral $$\int d^d x \, \tilde{\Pi}(x) g(x)$$ makes sense only for test functions $$g$$ with $$g(0)=0$$. However, $$\Pi(x)=\tilde{\Pi}(x) - \delta^{d}(x) \int d^d y \, \tilde{\Pi}(y)$$ approaches a well-defined limit for $$d\to 4$$. Moreover, any distribution being defined on all test functions and agreeing with $$\tilde{\Pi}(x)$$ for those with $$g(0)=0$$ differs from $$\Pi(x)$$ only by multiple of $$\delta^{d}(x)$$ (corresponding to the conterterm $$c_2 f(x)^2)$$.