There is a specific Gaussian functional integral in the renormaization group analysis for $O(3)$ nonlinear spin model to integrate out the high frequency part of the field $h(x)$ to only leave the field $\theta(x)$, $$\int_{\Lambda{'}<p<\Lambda}\mathcal{D}h(p)\exp\left[-\frac{1}{2}\int(\partial_{\mu}h)^{2}d^{2}x\right]\exp\left[\frac{1}{2}\int{h^{2}}(\partial_{\mu}\theta)^{2}d^{2}x\right],$$ where $\Lambda$ is the cut-off and $\Lambda{'}<p<\Lambda$ is the high-momentum slice. It is claimed in this article, Page 709 that the result is $$N\exp\left[\frac{1}{2}\int\langle{h}^{2}\rangle(\partial_{\mu}\theta)^{2}d^{2}x\right]=N\exp\left[\frac{1}{2}\int(\partial_{\mu}\theta)^{2}d^{2}x\int_{\Lambda{'}}^{\Lambda}\frac{d^{2}p}{(2\pi)^{2}}\frac{1}{p^{2}}\right]$$ but I cannot obtain that. I understand that $\langle{h}^{2}\rangle$ is the inverse kernel of the operator $-(\partial_{\mu})^{2}$. In the common Gaussian integral $\int{d}\mathbf{v}\exp\left(\frac{1}{2}\mathbf{v}^{T}\mathbf{A}\mathbf{v}+\mathbf{j}^{T}\mathbf{v}\right)=N\exp\left(\frac{1}{2}\mathbf{j}^{T}\mathbf{A}^{-1}\mathbf{j}\right)$, $\mathbf{v}$ is coupled to the external source linearly while here $h$ is not but the square $h^{2}$ couples to the field $\theta$. How to figure out the integral?

  • $\begingroup$ Could you write the exact definition of $\langle{h}^{2}\rangle$? $\endgroup$ – DelCrosB Oct 9 '16 at 15:32
  • $\begingroup$ Having done that, what do you know about $\langle h^4\rangle_h$? $\langle h^6\rangle_h$? $\langle \exp (...)\rangle_h$? $\endgroup$ – Cosmas Zachos Oct 9 '16 at 19:20
  • $\begingroup$ @DelCrosB $\langle{h}^{2}\rangle$ is the vacuum expectation value or you can also view is as the free propagator in the same place. I have updated it in the original question. Please check it. $\endgroup$ – Wayne Zheng Oct 10 '16 at 1:23

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