# Feynman diagrams for scalar field: which particle are we drawing?

Chapter I.7 of Zee's Quantum Field Theory in a Nutshell is an introduction for Feynman diagrams in the context of a scalar field $\varphi$, with Lagrangian $\mathcal{L} = \frac12[(\partial \varphi)^2-m^2\varphi^2]-\frac{\lambda}{4!}\varphi^4+J\varphi$. In page 50, it says this ([brackets] are mine):

We also know that $J(x)$ corresponds to sources and sinks. Thus, if we expand $Z(J)$ [the path integral of $\exp(i\int d^4x \mathcal{L})$] as a series in $J$, the powers of $J$ would indicate the number of particles involved in a process. (Note that in this nomenclature the scattering process $\varphi + \varphi \to \varphi + \varphi$ counts as a 4-particles process: we count the total number of incoming and outgoing particles.)

This is confusing to me. The way I thought I understood the scalar field is that we have some other field which interacts with $\varphi$ through the $J\varphi$ term, and that the $\varphi^4$ term represents the self interaction of the scalar field. In fact, in page 44 there is a footnote that says (talking about free field theory, that is, without $\varphi^4$):

Thanks to the propagation of $\varphi$, the sources coupled to $\varphi$ interact [...], but the particles associated with $\varphi$ do not interact with each other.

If sources and $\varphi$ are different particles, shouldn't we have two kinds of lines in the diagrams? The simplest diagram for $\varphi + \varphi \to \varphi + \varphi$ is a big X, and there is no trace of whatever particle $\varphi$ couples to. If the lines are $\varphi$'s, how does $J$ enter into the equation?

Maybe looking at the situation the following way helps to clarify: The field $J$ is indeed another field, but it is not dynamical (there is no $\partial_\mu J \partial^\mu J$ term in the Lagrangian). In particular that means, that $J$ does not propagate, so there should not be lines associated to $J$ in the Feynman diagrams. This should answer your final question. $J$ only enters the equation through the one-point vertices of $\varphi$, which I explain now.
The $J\varphi$ interaction means that there is a two-point vertex between the dynamical field $\phi$ and the non-dynamical $J$. This two-point vertex is denoted by a line associated with $\varphi$ out of nothing (a one-point vertex for $\varphi$). These one-point vertices are what is called external particles. If there was no interaction with the external $J$, then there would be no external particles and all diagrams you could draw would be vacuum bubbles. The number of $J\varphi$-vertices therefore determines the number of external particles in the process. This explains the first statement from Zee that you quoted.
The second statement you quoted can be understood in the following way. $J$ cannot self-interact and not propagate. The only way it interacts is through $\varphi$. If $\varphi$ does not self-interact itself, then the only diagrams you can have a straight lines associated with $\varphi$, i.e. propagators of $\varphi$ which effectively lead to in interaction of the background field with itself at the two-endpoints of the $\varphi$ propagator.