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I would like to better understand the main principles of large-$N$ expansion in quantum field theory. To this end, I decided to consider a simple toy model with lagrangian (from Wikipedia)

$ \mathcal{L} = \frac{1}{2}(\partial_{\mu} \phi_a)^2-\frac{m^2}{2}\phi_a^2 - \frac{\lambda}{8N}(\phi_a \, \phi_a)^2 $

I aimed to renormalize this theory in all orders of perturbation theory in the leading order of $\frac{1}{N}$. The calculation of counterterms in two loops in the leading order of $\frac{1}{N}$ almost coincides with the corresponding calculation in $\phi^4$ theory. In leading order of $1/N$ the counterterms are (using MS-scheme):

1 Loop:

$\Delta \mathcal{L}_{\phi^4}^{1} = -\lambda^2 \mu^{2\epsilon} \frac{1}{32 \pi^2 \epsilon} \frac{(\phi_a \, \phi_a)^2}{8N}$

$\Delta \mathcal{L}_{\phi^2}^{1} = - \frac{\lambda}{32 \pi^2 \epsilon} \frac{m^2 \phi_a^2}{2}$

2 Loops:

$\Delta \mathcal{L}_{\phi^4}^{1} = -\lambda^3 \mu^{2\epsilon} (\frac{1}{32 \pi^2 \epsilon})^2 \frac{(\phi_a \, \phi_a)^2}{8N}$

$\Delta \mathcal{L}_{\phi^2}^{1} = - (\frac{\lambda}{32 \pi^2 \epsilon})^2 \frac{m^2 \phi_a^2}{2}$

I am quite sure (though I haven't proven it properly yet) that in $n$ loops the leading contribution to counterterms comes from a chain of "fish" diagrams for 4-point Green's function and chain of bubbles for 2-point Green's function (I think, it's quite easy to imagine):

$\Delta \mathcal{L}_{\phi^4}^{1} = -\lambda^{n+1} \mu^{2\epsilon} (\frac{1}{32 \pi^2 \epsilon})^n \frac{(\phi_a \, \phi_a)^2}{8N}+O(\frac{1}{N})$

$\Delta \mathcal{L}_{\phi^2}^{1} = - (\frac{\lambda}{32 \pi^2 \epsilon})^n \frac{m^2 \phi_a^2}{2}+O(\frac{1}{N})$

If this speculation is correct, the summation of the perturbation series is quite trivial (we have geometric series). When we do it and then take limit $\epsilon \rightarrow 0$ we will find that

$\Delta \mathcal{L}^{\infty}_{\phi^2} = \frac{m^2 \phi_a^2}{2}$

$\Delta \mathcal{L}^{\infty}_{\phi^4} = \frac{(\phi_a \phi_a)^2}{8N}$

and hence the total lagrangian is simply (in the leading order of $1/N$).

$\mathcal{L} = \frac{1}{2}(\partial_{\mu} \phi_a)^2$

This result seems to me highly suspicious... Did anybody do a similar calculation? I looked all over the Internet and didn't find anything :( I will be very grateful for remarks and links to books or maybe articles where a similar problem is considered.

Golden_Hawk
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user43283
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    I have never done explicit renormalization in a large N expansion, so I can't speak with authority, but the leading-order term in the $\frac{1}{N}$-expansion alone is frequently a bad approximation to the full theory, so "suspicious" results for it are nothing to worry too much about. – ACuriousMind Jul 06 '14 at 20:18
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    @ACuriousMind: But if $N$ is large then it should still be a good approximation? – JeffDror Jul 07 '14 at 07:12
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    Not sure if this is relevant, but if you take your original theory and send $N\rightarrow \infty$, you get a free theory, so then the loop corrections in the original theory in the $N\rightarrow \infty$ limit should reproduce the loop corrections in the free theory, ie, they should all be zero. Presumably to get a nontrivial answer you also need to send the coupling $\lambda\rightarrow \infty$ in such a way that the ratio $\lambda/N$ is fixed (you need to do a similar thing in large $N$ of yang mills, there the analogue of $\lambda/N$ is called the "'t Hooft" coupling") . – Andrew Jul 07 '14 at 14:46
  • @Andrew: It's a good idea, but I'm not quite sure that it works here in this particular problem. As far as I understood you, we should take limit $N \rightarrow +\infty, \quad \lambda = g,N$. In $n$ loops the term, that gives leading contribution, behaves like $\frac{\lambda^n}{N}$ and hence the more order of perturbation theory the more contribution from the corresponding term ($\sim g^n N^{n-1}$). Did I understand you correctly? – user43283 Jul 07 '14 at 19:25
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    As explained here, the RG flow of the O(N) model is characterized by different fixed point, depending on the number of space-time dimensions considered. – Dilaton Jul 08 '14 at 11:03
  • @Dilaton: Thank you very much for interesting links! But I don't think that conclusion made here is correct. I don't sure that $\epsilon$-independent part was calculated correctly (or may be I don't understand how it was done..) What do you think? – user43283 Jul 08 '14 at 18:21
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    To me that step is not obvious either at present, but I am confident that he will explain it as soon as he sees your comment. – Dilaton Jul 08 '14 at 23:38
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    I think that what you effectively do when you send $\epsilon\to0$ at fixed $\mu$, you effectively take the continuum limit (corresponding to $\Lambda\to\infty$ with a sharp cut-off). The O(N) model is known to exist in $d=4$ in the continuum limit only if the theory is free, which is indeed what you get. Did you have a look at Zinn-Justin book ? – Adam Jul 10 '14 at 13:40
  • @Adam: You mean "Quantum Field theory and critical phenomena"? No, I am just a novice in QFT. I'm learning it using Peskin and Shroeder's book. What chapters would you advice to read in Zinn-Justin's book?

    P.S. Sorry for late answer.

     – user43283 Jul 11 '14 at 10:56
  • @user43283: Yes. There's whole chapter on the large N limit, though he uses a sharp cut-off. But the problem you have is generic to the $\phi^4$ model for any N, in $d=4$, so any chapter on that would do I guess. The only difference is that in the large N limit the one loop result is essentially exact (if you resume it properly). My advice would be: read the chapter in ZJ but redo the calculation with dimensional regularization, that should solve your problem. Then post the answer ;-) – Adam Jul 11 '14 at 13:18
  • I would also recommend Zinn-Justin's review of large N theories with Moshe-Moshe (https://arxiv.org/abs/hep-th/0306133). The standard and easier approach to study such a theory is to do a field transformation as discussed in Eq 2.5. After such a variable transformation, the N factors out of the action and in the $N\to\infty$ limit one has to simply study an auxilliary theory (in terms of the new $\rho,\lambda$ variables) in the `classical' limit to solve the full original quantum theory. This is exactly correct as long as one is working in $N\to\infty$ limit. (1/3) – nGlacTOwnS May 07 '19 at 20:52
  • In fact, the saddle point equations in this new $\rho,\lambda$ variables are exactly the same as the Schwinger-Dyson equations that you will obtain from summing over all the planar diagrams of the original theory (planar diagrams are the diagrams that survive the large $N$ limit).

    The perturbation theory around this new classical theory computes the $1/N$ corrections in the original theory. (2/3)

     – nGlacTOwnS May 07 '19 at 20:56
  • @Andrew the correct 't Hooft limit is indeed as it appears in the original question with $\lambda \sim \mathcal O(1)$. The idea being that the vanishing contribution of the vertices in the large $N$ limit is compensated by $N$ flavors of $\phi_a$ being exchanged in the loop diagrams. (3/3) – nGlacTOwnS May 07 '19 at 20:59

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