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In Polyakov's book about gauge fields & strings, in chapter devoted to non-linear sigma model he emphasizes problem with large $N$ expansion of this model. Lagrangian of 2D model is $$\frac{1}{2g^2}(\partial_{\mu}{\bf n})^2$$ with constraint ${\bf n}^2=1$. It is possible to add term into action which explicitly contains constraint by introducing additional field $\lambda(x)$. Then, one can integrate out fields ${\bf n}$ (path integral over these fields is gaussian) and obtain effective action in terms of field $\lambda$. Now it is time to use $N\rightarrow\infty$. It is possible to find saddle point of this effective action and see that it corresponds to $\lambda=m^2$, $m>0$ (up to sign or $i$). Then, we can ivestigate fluctuations near saddle point as $m^2+\alpha(x)$, where $\alpha$ is fluctuation.

After all calculations, we can compute all the correlation functions of initial model in terms of effective action. Near to the end of this chapter, he says that propagator of $\alpha(x)$ field in effective action has bad behavior, $$D(q^2)\rightarrow q^2/\ln(q^2/m^2), \quad q^2\rightarrow\infty,$$ and then says that if we do not impose constraint $n_in^i=1$ everything will be ok. Instead of constraint ${\bf n}^2=1$, it is possible to introduce quartic term into initial action and avoid "bad behavior" problem.

Why constraing $n_in^i=1$ creates problem with propagator behaviour for large momenta? Can somebody clarify this moment?

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  • $\begingroup$ What is the question exactly? The propagator should vanish for large energy, which doesn't seem to be the case. $\endgroup$
    – lcv
    Nov 12 '19 at 11:59
  • $\begingroup$ @lcv , the question is about how to solve this problem and understand how this proble appears $\endgroup$ Nov 12 '19 at 22:13
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If you do not impose the constraint $\boldsymbol n^2=1$ the system is linear, i.e., free, which means that the propagator is just $$ D(q^2)=\frac{1}{q^2+m^2} $$ which has a nice UV behaviour. Recall that non-linearities, i.e., interactions, come from the metric $g_{\mu\nu}(\boldsymbol n)$. If you do not impose the constraint, the manifold is flat, and so $g_{\mu\nu}(\boldsymbol n)=\delta_{\mu\nu}$, which means that the Lagrangian is just $L=\frac12 \boldsymbol n\cdot\partial^2\boldsymbol n$, which is gaussian.

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  • $\begingroup$ Your answer doesn't seem to take into account what Polyakov was saying which is the OP's question. $\endgroup$
    – lcv
    Nov 27 '19 at 3:46
  • $\begingroup$ If replace constraint ${\bf n}^2=1$ by non-linear quartic term in initial lagrangian everything will be ok too (Polyakov uses it). With quartic term theory is also non-linear. I will improve my question and provide details about it. $\endgroup$ Nov 27 '19 at 8:22
  • $\begingroup$ @ArtemAlexandrov I'm not sure I understand your point. If you replace the constraint by a quartic term, all you are doing is implementing the constraint via a Lagrange multiplier. In other words, it is the exact same theory, but expressed differently. How could one way of expressing it lead to a bad propagator, and the other way to a good one? They describe the same physics, they are the exact same theory, but expressed differently. Surely there is something I'm missing... $\endgroup$ Nov 28 '19 at 2:17
  • $\begingroup$ @AccidentalFourierTransform , if I start from free theory with constraint ${\bf n}^2=1$ and explicitly add constraint as Lagrange multiplier into action, I obtain propagator with bad UV behavior, $\sim q^2$. However, if I remove constraint ${\bf n}^2=1$ and write down theory with quartic interaction propagator becomes $1/[\mathrm{const}+1/D(q)]$, where $\mathrm{const}=2N/g$ and quartic interaction is $g({\bf n}-1)^2/4$ $\endgroup$ Nov 29 '19 at 6:53
  • $\begingroup$ @ArtemAlexandrov "... if I start from free theory with constraint ...": sorry, but that does not make much sense. The constraint is an interaction, so you cannot have a free theory and a constraint. It is either free or constrained, but not both! $\endgroup$ Nov 29 '19 at 23:40

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