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I have some schematic notes on computing the effective action and I would like someone to help me fill the gaps.

We start with \begin{equation*} \int{}\mathcal{D}\phi\,e^{-iS[\phi]} \end{equation*} employing the background field method we write \begin{equation*} \phi=\phi_0+\Delta\phi \end{equation*} so we have \begin{equation*} \int{}\mathcal{D}(\Delta\phi)\,e^{-iS[\phi_0+\Delta\phi]} \end{equation*} Taylor expanding around $\phi_0$

$$S[\phi_0+\Delta\phi]=S[\phi_0]+\int{}d^4x_1\,\frac{\delta{}S}{\delta\phi(x_1)}\Delta\phi(x_1)$$ $$+\frac{1}{2}\int{}d^4x_1d^4x_2\frac{\delta^2S}{\delta\phi(x_1)\delta\phi(x_2)}\Delta\phi(x_1)\Delta\phi(x_2)+$$ $$\frac{1}{3!}\int{}d^4x_1d^4x_2d^4x_3\frac{\delta^3S}{\delta\phi(x_1)\delta\phi(x_2)\delta\phi(x_3)}\Delta\phi(x_1)\Delta\phi(x_2)\Delta\phi(x_3)+\ldots$$ since $\phi_0$ satisfies the equations of motion the linear term in $\Delta\phi$ vanishes. Then we have

$$e^{-iS[\phi_0]}\int{}\mathcal{D}(\Delta\phi)e^{-i\frac{1}{2}\int{}d^4x_1d^4x_2\frac{\delta^2S}{\delta\phi(x_1)\delta\phi(x_2)}\Delta\phi(x_1)\Delta\phi(x_2)+\ldots}$$

from here on my notes neglect terms cubic,quartic... in $\Delta\phi$. Can anybody tell me why?.

Also, after this it is written $$e^{-iS[\phi_0]}det(\ldots)$$ where the dots represent (I think) a functional determinant of something. Can anybody tell me what goes inside the determinant, and where this comes from?

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1)They neglect higher powers of $\Delta \phi$ because this effective action desribes dynamics of the fluctuations $\Delta \phi$ above the background fields $\phi_0$. Namely $\Delta \phi$ is small

2)As you know \begin{equation} \int d^n r\, e^{-r_i A_{ij}r_j}=\frac {(2\pi)^{n/2}}{(\mbox{det}\, A)^{1/2}}, \end{equation} where $A$ is a matrix with positive eigenvalues. You can prove this formula by rewriting $A$ in diagonal basis. So $-1/2$ power of determinant.

If your target space is real, then functional integral is just a limit: \begin{equation} \lim_{N\to\infty}\prod_{k=1}^{N} \int d^k \phi\, e^{-\phi_i A_{ij}\phi_j} \end{equation} The only reason you can get first power of determinant is if you are working with fermionic (Grassman) variables. Click here for details. To say something further you need actual form of the action. I suggest to discretize your functional integral and calculate determinant of a matrix (it will be a function of $N$). Limit $N\to \infty$ should give you the answer.

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  • $\begingroup$ If you are insisting on working in coordinate representation and want to know what is the determinant of the operator then you have to think about it as product of all eigenvalues of this operator. But I think that my suggestion above is much simpler. Let me know if it helps. $\endgroup$ – stinglikeabeer Jun 27 '15 at 20:24
  • $\begingroup$ it does help ;) $\endgroup$ – Yossarian Jul 5 '15 at 16:57

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