"Bad" behavior of propagator in $O(N)$ model

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?

• What is the question exactly? The propagator should vanish for large energy, which doesn't seem to be the case.
– lcv
Commented Nov 12, 2019 at 11:59
• @lcv , the question is about how to solve this problem and understand how this proble appears Commented Nov 12, 2019 at 22:13

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.
• 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. Commented Nov 27, 2019 at 8:22
• @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 $g({\bf n}-1)^2/4$ Commented Nov 29, 2019 at 6:53