# Large-N of non-linear sigma model

Consider the partition function of non-linear sigma model is: $$Z=\int\mathcal{D}[\textbf n]\mathcal{D}[\lambda]\exp(-S)\\S=\frac{1}{f} \int d^dx \int d \tau [(\partial_\tau \textbf{n})^2+c^2(\nabla_x \textbf n)^2+i \lambda (\textbf n^2-1)]$$

I have two questions when I want to generalize it to large $$N$$:

1. I know that the coefficient $$f$$ need to be scaling as $$1/N$$, thus, the total pre-factor $$1/f$$ is $$N/f_0$$, where $$f_0$$ is fixed. But, I cannot understand the motivation and reason of it because it seems like we just put it for our goal, i.e. the whole pre-factor $$N$$.

2. After integral out the $$\textbf{n}$$ field, we can obtain one term of the effective action: $$\frac{N}{2} Tr \ln(-\partial_\tau^2-c^2\nabla_x+i\lambda)$$

I can understand that the $$N$$ comes from the $$N$$ independent components, but why $$\frac{1}{2}$$? Also, why does the coupling $$f$$ not enter in the logarithm？

• The $f$-dependence of the logarithm multiplies everything, so $\ln(g/f) = \ln(g) - \ln(f)$ and it just contributes an overall constant shift to the action. Alternatively, just rescale $\mathbf{n} \rightarrow \sqrt{f} \mathbf{n}$ before integrating out the $\mathbf{n}$ fields, which will also contribute a multiplicative constant to $Z$ (equivalent to a shift in $S$). Commented Oct 29, 2019 at 16:34
• The one half comes from the fact that the field you are integrating out (it's a Gaussian integral) is real.
– lcv
Commented Oct 29, 2019 at 19:15

The half comes from the fract that the gaussian integral over the bosonic fields gives you the reciprocal of the square root of the Fredholm determinant: In $$A$$ is an $$n$$-by-$$n$$ matrix
$$\int d^nx\, e^{ -x^TAx}= \pi^{n/2} [{\rm det}(A)]^{-1/2}.$$