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?

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

1 Answer 1


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.

  • $\begingroup$ Your answer doesn't seem to take into account what Polyakov was saying which is the OP's question. $\endgroup$
    – lcv
    Nov 27, 2019 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, 2019 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, 2019 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, 2019 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, 2019 at 23:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.