Relation between $1/N$ and perturbative expansions in QFT

Question

I heard many times phrases like "large $N$ is a clever way to organize the diagrammatic expansion" or "each diagram in the large $N$ expansion contains an infinite number of usual perturbative diagrams".

I also know that for each order in $N$, the series in the 't Hooft coupling parameter $t$ has better convergence properties that the perturbative expansion in the usual coupling constant - due to the smaller number of diagrams (their number is growing polynomially, not factorially).

Can somebody explain the relation between these two approaches? For example, it would be great to see which perturbative diagrams are contained in some leading-order $1/N$ diagram, and how/why they are resummed.

Any references appreciated as well.

UPDATE

OK, let me try to formulate what I understood by now, please tell me if smth is wrong. So, assume the Lagrangian looks smth like this: \begin{equation} \mathcal{L} = \dfrac{1}{g} \left( (\partial \Phi)^2 + \Phi^4 \right) = \dfrac{t}{N} \left( (\partial \Phi)^2 + \Phi^4 \right) \end{equation}

Where \begin{equation} g N = t \end{equation}

The expansions for, say, partition function in terms of $g$ and $\{t, 1/N\}$ have form: \begin{equation}\begin{alignedat}{9} \mathcal{Z} &= &&\alpha_0 + \alpha_1 g + \alpha_2 g^2 + \ldots \\ &= N &&(a_{0} + a_{1} t + a_{2} t^2 +\ldots)\\ &+ &&(b_{0} + b_{1} t + b_{2} t^2 +\ldots)\\ &+ \dfrac{1}{N}&&(c_{0} + c_{1} t + c_{2} t^2 +\ldots) + \ldots\\ &= N &&(a_{0} + a_{1} (g N) + a_{2} (g N)^2 +\ldots)\\ &+ &&(b_{0} + b_{1} (g N) + b_{2} (g N)^2 +\ldots)\\ &+ \dfrac{1}{N}&&(c_{0} + c_{1} (g N) + c_{2} (g N)^2 +\ldots) + \ldots\\ \end{alignedat}\end{equation}

Which gives us for the coeficients of the $g$-series: \begin{equation}\begin{alignedat}{9} \alpha_0 &= N &&a_0 + b_0 + \dfrac{1}{N} c_0 + \dfrac{1}{N^2} d_0 + \ldots \\ \alpha_1 &= N &&\left( a_1 + b_1 + \dfrac{1}{N} c_1 + \dfrac{1}{N^2} d_1 + \ldots \right) \\ \alpha_2 &= N^2 &&\left( a_2 + b_2 + \dfrac{1}{N} c_2 + \dfrac{1}{N^2} d_2 + \ldots \right) \\ \end{alignedat}\end{equation}

My understanding is that for some reason at large power of $g$ and $t$ the coefficients behave as follows (I actually have some understanding about the factorial growth, but have not learned about the $1/N$ case yet). \begin{equation}\begin{alignedat}{9} \alpha_k &\propto k! \\ a_k,b_k,c_k &\propto \text{(some power of $k$)} \end{alignedat}\end{equation}

I'd like to know if the last statement is correct, and if it's possible to see it from the previous line.

Answer 1

In any perturbative calculation in QFT, you order your diagrams in whatever your small parameter is. Typically, it is the coupling of some interaction that you expand around, say $\lambda$ in scalar $\lambda\phi^4$ theory. This corresponds to doing expansions in the number of loops and interaction vertices. The more loops/interactions, the lower order the diagram. This is what is meant by organization; you organize your diagrams according to the order of some small parameter.

Now, in some QFTs you don't have a small parameter to play with. t'Hooft realized this when he was studying QCD, so he ingeniously introduced a new parameter, namely the large parameter $N$ in order to try to apply all the standard perturbation theory results. What does this do for us? It turns out that the order at which a diagram occurs is intimately tied to its topology. That is to say, what 2D surface (sphere/plane, torus, 2-hole torus, etc) can I draw my Feynman diagram on such that the lines do not intersect. Hence, in the large N expansion, the leading order diagrams which are kept are the planar diagrams: those which can be drawn without intersections on a plane/sphere. The diagrams that can be drawn on a torus are suppressed by $1/N$ (roughly speaking), and so forth. Look up t'Hooft's double line notation for more details on how this works.

This is a massive simplification, but you still have an infinite number of planar diagrams to sum if you want to calculate the leading order effects. If you also assume your interaction vertex is also small, you can organize in the number of vertices as well. I assume that when you ask about the factorial vs polynomial number of diagrams, you're referring to this idea of $n$-vertex diagrams vs $n$-vertex planar diagrams.

As a side note, perturbative QFT calculations tend to be hard. Hence, people will just "organize" diagrams according to some parameter regardless of whether or not it is small. By magic or by design, this actually gives correct results at times. For example, the large $N$ expansion gives corrections for QCD as $1/3$, which actually gives nice results though is not necessarily one's ideal "small" parameter.