Feynman Rules for Two Different Interacting Fields

Question

I'm currently studying how to deduce Feynman rules for general theories, and I've managed to deduce them for $\phi^3$ and $\phi^4$ theories. Up to this point I've considered the same field for all cases, and deduced the Feynman rules by expanding the interacting term in the correlator and using Wick's theorem to make the contractions.

My question is, if we consider a interacting theory for two different fields, how could we deduce the Feynman rules from Wick's theorem. Consider, for example, the decay of a particle given by the interacting lagrangian term

$$L_\mathrm{int}=-\lambda\Phi\phi^2$$

We see directly it only has vertices with 3 lines. If I expand the exponential that usually comes from this:

$$\exp(-i\lambda\int d^4y\ \Phi\phi^2)=1+(i\lambda)\int d^4y\Phi_y\phi_y^3+(i\lambda)^2\int d^4y_1d^4y_4\ \Phi_{y1}\phi_{y1}^3\Phi_{y2}\phi_{y2}^3+...$$ $$$$

However, I'm not sure which terms I should write for the correlators. For simplicity, let's consider the 2 particle correlator:

$$\langle \Omega |T[\phi_1\phi_2]|\Omega\rangle=\lim_{T\rightarrow \infty}\frac{\langle 0 |T[\phi_1\phi_2 \exp(-i\lambda\int d^4y\ \Phi\phi^2)]|0\rangle}{\langle 0 |\exp(-i\lambda\int d^4y\ \Phi\phi^2)|0\rangle}$$

Let's focus on the numerator, which could either be:

$$N(x_1,x_2)=\langle 0 | \phi_{x1}\phi_{x2}\exp(-i\lambda\int d^4y\ \Phi\phi^2) | 0 \rangle$$

Or could it be:

$$N(x_1,x_2)=\langle 0 | \phi_{x1}\Phi_{x2}\exp(-i\lambda\int d^4y\ \Phi\phi^2) | 0 \rangle$$

Or even it could be:

$$N(x_1,x_2)=\langle 0 | \Phi_{x1}\Phi_{x2}\exp(-i\lambda\int d^4y\ \Phi\phi^2) | 0 \rangle$$

Which one should I consider for the expansion?

Answer 1

All three expressions are correct, but they represent different objects. The first one represents (the numerator of) the correlation function $$ \langle0|\mathrm T\{\color{red}{\phi}(x_1),\color{red}{\phi}(x_2)\}|0\rangle, $$ the second one (the numerator of) the correlation function $$ \langle0|\mathrm T\{\color{red}{\phi}(x_1),\color{blue}{\Phi}(x_2)\}|0\rangle, $$ and the third one (the numerator of) the correlation function $$ \langle0|\mathrm T\{\color{blue}{\Phi}(x_1),\color{blue}\Phi(x_2)\}|0\rangle. $$

These three correlation functions are meaningful. In perturbation theory, and as noted in the OP, these three objects are given by $$ \begin{aligned} \langle0|\mathrm T\{\color{red}{\phi}(x_1),\color{red}{\phi}(x_2)\}|0\rangle&=\langle\hat0|\mathrm T\{\color{red}{\hat\phi}(x_1),\color{red}{\hat\phi}(x_2),\exp\left[ i\int L_{\mathrm{int}}(\color{red}{\hat\phi},\color{blue}{\hat\Phi})\right]\}|0\rangle\\ \langle0|\mathrm T\{\color{red}{\phi}(x_1),\color{blue}{\Phi}(x_2)\}|0\rangle&=\langle\hat0|\mathrm T\{\color{red}{\hat\phi}(x_1),\color{blue}{\hat\Phi}(x_2),\exp\left[ i\int L_{\mathrm{int}}(\color{red}{\hat\phi},\color{blue}{\hat\Phi})\right]\}|\hat0\rangle\\ \langle0|\mathrm T\{\color{blue}{\Phi}(x_1),\color{blue}\Phi(x_2)\}|0\rangle&=\langle\hat0|\mathrm T\{\color{blue}{\hat\Phi}(x_1),\color{blue}{\hat\Phi}(x_2),\exp\left[ i\int L_{\mathrm{int}}(\color{red}{\hat\phi},\color{blue}{\hat\Phi})\right]\}|\hat0\rangle \end{aligned} $$ where a hat represents an interaction picture operator, $\hat\psi:=U\psi U^\dagger$, with $\psi\in\{\color{red}{\phi},\color{blue}\Phi\}$ and $|\hat0\rangle=U|0\rangle$, and the unhatted objects are in the Heisenberg picture (cf. this PSE post). (I am neglecting the denominators to keep the notation as simple as possible; they do not play an important role here).