Suppose I have the following Lagrangian density:

$$ \mathcal L = -\frac{1}{2} \sum_{i = 1}^N \left [ \partial_\mu \phi_i\partial^\mu \phi^i +m^2 \phi_i^2\right ] + \frac{g}{2N}\sum_{i=1}^N \sum_{j=1}^N \phi_i^2 \phi_j^2 $$

where the $\phi$'s are real scalar fields. Then we can see that, as $N \to +\infty$, the coupling constant $\frac{g}{N} \to 0$ and we just get a bunch of free scalar fields. However, I'm having trouble with the Feynman diagrams.

Consider the one-loop correction to the propagator of the field $\phi_i$. There's one diagram, which is:

enter image description here

It has one interaction vertex, so it goes as $g/N$. However, there are $N$ such diagrams, since there are $N$ possible fields $\phi_k$ for the loop. So it seems it all cancels out and you would get that the one-loop correction doesn't depend on $N$. But this is false because when $N\to+\infty$ the theory becomes non-interacting and there should be no corrections to the propagator.


The coefficient in front of the interaction terms is a bit arbitrary, in the sense that in principle, you could have written anything, for example $\frac{g}{N^2}$. The way to figure out the 'natural' numerical factor is by considering the renormalized coupling constant at tree level. $i g_\text{physical} = i (\frac{g}{2N}) \times \text{Sum of all X tree level diagrams}+\text{Loop corrections}$. Usually, we pick the convention that at tree level, $g_{physical}=g$. Picking another convention is fine, but it means that $g$ will scale as N increases, as in your case. If I'm not mistaking, for this example, I think that the sum of all X (diagrams that look like an X at tree level) should scale as $N^2$, times factors for interchanging legs and whatnot. So a more "natural" coupling constant would go as $\frac{g}{N^2}$ times combinatoric numbers, thus solving your problem.


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