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

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.

• What exactly do you mean by the two approaches? Do you mean the differences in which diagrams are kept at each order in the large N vs non-large N cases? Nov 2, 2016 at 20:09
• This as well - e.g., what perturbative diagrams correspond to some $1/N$ diagram and vice versa. I've never seen it clearly explained so far. I would also like to understand how we manage to get rid of the factorial divergence. Nov 2, 2016 at 21:43

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.
• But don't we have theories where we can use both? For example, assume one has a toy model $H = \dfrac{N}{g^2}\left((\partial \Phi)^2 + \Phi^4\right)$ with $\Phi$ being $N\times N$ matrix field. Can't we have expansions both 1) in $g^2$ 2) in $1/N$ and then in $t = N g^2$. If so, how are they related? Nov 2, 2016 at 22:24
• Right, so this is what I'm noting in my third paragraph. If you do a $1/N$ expansion followed by a $t$ expansion, you get $n$-vertex diagrams that are also planar. This is contrasted with the simpler $g$ expansion, where you get all $n$-vertex diagrams (namely you add the non-planar diagrams) Nov 2, 2016 at 22:41
• @mavzolej I do not have a proof, but I feel that the last few lines you wrote are roughly correct. However, I don't think its obvious from the previous lines. For example, the $b_k$ diagrams will be much more numerous than the $a_k$ diagrams, and $c_k$ diagrams more so than $b_k$. This is not captured in your equations. Fundamentally the number of diagrams has to be counted, so it becomes a combinatorial graph theory question on an arbitrary genus surface. I cannot prove it to you, but my gut feeling is that the number of diagrams is polynomial in $k$. Nov 5, 2016 at 3:42