# Feynman Rules for Two Different Interacting Fields

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?

• It helps to think of your fields as elements of a scalar multiplet (a 2d column vector). Then what you get is just the normal cubic theory but with mass and interaction terms replaced by matrices. That was a hint. The answer has already been posted by Accidental. – Prof. Legolasov Nov 16 '18 at 21:56
• That's actually an interesting idea, and when I see them like that I indeed get the cubic theory relationships. I'm trying to work also from this approach. – Charlie Nov 17 '18 at 3:09

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).

• To fix: $|0\rangle\to|\hat0\rangle$. – AccidentalFourierTransform Nov 16 '18 at 23:23
• I see, then all are correct but they represent different objects (don't worry about the denominator, I know at the end it just cancels the so-called vaccum bubbles). I developed the three objects, but I'm not sure how to deduce or represent the Feynman rules from them, as I'm contracting two different fields. For example, the correlator for $\phi_{x1},\Phi_{x_2}$, if I take second order (so all fields contract as is non-zero), we have: $-i\lambda\int d^4y_1 d^4y_2\langle 0 |\phi_{x1}\Phi_{x2}\Phi_{y1}\phi_{y1}\phi_{y1}\Phi_{y2}\phi_{y2}\phi_{y2}+...|0\rangle$. – Charlie Nov 17 '18 at 3:13
• The first non-trivial contraction would be (I'm representing contractions with superindices): $\phi_{x1}^a\Phi_{x2}^c\Phi_{y1}^a\phi_{y1}^b\phi_{y1}^b\Phi_{y2}^c\phi_{y2}^d\phi_{y2}^d$. This means that, in Feynman notation, there's a line that joins $x_1$ to $y_1$, and a loop on that very same point; same happens for $x_2$, which connects with $y_2$, and a loop on that very same point. However, how can I differentiate between fields? Should I introduce a different figure for each vertice? – Charlie Nov 17 '18 at 3:14
• I think it is important to use consistent notation. It is better to use $\phi,\Phi$ for Heisenberg fields, and $\hat\phi,\hat\Phi$ for interaction-picture fields. In particular, contractions concern the latter, not the former. That being said, one has $\overline{\hat\phi\hat\phi}=\Delta_\phi$, $\overline{\hat\Phi\hat\Phi}=\Delta_\Phi$, and $\overline{\hat\phi\hat\Phi}=0$. So you only contract $\phi$ fields among themselves, and $\Phi$ fields among themselves. You do not contract $\phi$ fields with $\Phi$ fields. – AccidentalFourierTransform Nov 17 '18 at 3:19
• The standard notation is to use solid lines for $\overline{\hat\phi\hat\phi}=\Delta_\phi$, and dashed (or, say, wiggly) lines for $\overline{\hat\Phi\hat\Phi}=\Delta_\Phi$. In your theory, every vertex has two solid lines and one dashed one, because $L_\mathrm{int}$ has two factors of $\phi$ and one factor of $\Phi$. (Maybe you will find this PSE post useful). – AccidentalFourierTransform Nov 17 '18 at 3:21