Question on the $1/N$ expansion

Question

My question is from Coleman's Aspect of Symmetry, on the large $N$ approximation.

We will consider the $O(N)$ version of the $\phi^4$ theory. Its Lagrangian density is given by: $$ \mathcal{L}=\dfrac{1}{2}\partial_\mu\phi^a\partial^\mu\phi^2 - \dfrac{1}{2}\mu_0^2\phi^a\phi^a-\dfrac{1}{8}\lambda_0\left(\phi^a\phi^a\right)^2, $$ with $a$ ranging from $1$ to $N$. The tree approximation to the scattering of $a$ mesons to $b\neq a$ mesons is of the order of $\lambda_0$; this is clear.

My misunderstanding comes from these two second order Feynman diagrams:

Coleman argues that the first diagram is of order $\lambda_0^2 N$, and this is clear because it's the second order approximation and you sum over all possible intermediate states to get the $N$.

The second diagram, Coleman continues, is of the order $\lambda_0^2$, without a factor of $N$.

My question is: why? Are we still not summing over all intermediate states to get the factor $N$?

Answer 1

The potential may be written as, $$\frac{\lambda_0}{8}\sum_{a,b}\phi^a\phi^a\phi^b\phi^b$$ Those interaction terms require that four lines going to the vertex are organized into two pairs. Inside each pair both lines are required to have the same index.

In the first diagram two external lines going to each vertex are the same therefore the indices internal pair is not tied to the indices of the external pair and we have to sum over that index. However in the second diagram the indices are different. That fixes the indices of the internal lines to $a$ and $b$

Answer 2

Your last question "Are we still not summing over all intermediate states to get the factor $N$?" is too lazy. The diagram in question has a value that is completely fixed by the Lagrangian $\mathcal{L}$, so there's no need to resort to a vague argument.

Let's be more precise. Every vertex is of form $$ \delta \mathcal{L} \sim -\frac{\lambda_0}{8} \delta_{ab} \delta_{cd} \phi^a \phi^b \phi^c \phi^d $$ and the propagator is of the form $G^{ab} \equiv \langle \phi^a(p) \phi^b(-p) \rangle \propto \delta^{ab}$.

In the left diagram we have two vertices that we're bringing together with two propagators. The way we're contracting indices, it looks like $$ \delta_{bb} \delta_{cd} \times G^{cc'} G^{dd'} \times \delta_{c'd'} \delta_{aa} \propto N \delta_{aa} \delta_{bb}$$ using $\delta_{cd}\delta^{cd} = N$ (no summation on $\delta_{aa}$ or $\delta_{bb}$ above).

For the right diagram, since $a \neq b$ the tensor flow is different. We have $$ \delta_{a c} \delta_{b d} \times G^{cc'} G^{dd'} \times \delta_{ac'} \delta_{bd'} \propto \delta_{aa} \delta_{bb}$$ without any factors of $N$.

Since you're struggling with computing these diagrams, it's probably a good idea to compute some amplitudes in this theory explicitly, including all values of $4\pi$ etcetera.