I have been reading chapter 9 in Peskin & Schroeder's QFT book and has been stuck in transition from equation 9.26 to 9.27. Equation 9.26 reads: $$\frac{1}{V^2} \Sigma_{m,l} \exp{[-i(k_m.x_1+k_l .x_2)]}(\prod_{k_{n}>0}\int{d\ Re\phi_{n}\ d\ Im\ \phi_{n}}) * (Re\ \phi_{m}+iIm\ \phi_{m})(Re \ \phi_{l}+iIm\ \phi_{l}) * \exp[\frac{-i}{V}\Sigma_{k^{0}_{n}>0}(m^2-k^2_{n})[(Re\ \phi_{n})^2+(Im\ \phi_{n})^2]] \tag{9.26}$$

and with $k_l=+k_m$, equation vanishes while with $k_l = -k_m$ the equation becomes,

$$\frac{1}{V^2}\Sigma_{m} \exp[-ik_m\ .(x_1-x_2)](\prod_{k^{0}_{n>0}}\frac{-i\pi V}{m^2-k^{2}_n})\ \frac{-iV}{m^2-k^2_{m}-i\epsilon}$$

I can work out the term in parenthesis using Gaussian integrals but I can't seem to wrap my head around how we get the term


Can anybody point me in the right direction please? Apparently $(Re\ \phi_{m})^2$ adds up to the $(Im\ \phi_m)^2$ when $k_l = - k_m$ and we get this contribution. I don't understand how.

  • $\begingroup$ Try to see the discussion of replacement $t\rightarrow (t-i\epsilon)$ after (9.23). It seem that the gaussian integral has only imaginary exp, therefore we have to do analytic continuation. May be it will help. $\endgroup$ Mar 15, 2019 at 18:58
  • $\begingroup$ I mean that $\epsilon$-prescription does not come directly in (9.26)$\rightarrow$(9.27). P&S state that in all formulas after (9.21) one should replace $k^2-m^2\rightarrow k^2-m^2+i\epsilon$ to make integrals convergent $\endgroup$ Mar 15, 2019 at 19:01
  • $\begingroup$ I understand that but what I fail to grasp is how the two parts of $\phi$ add up and become this term. I tried to follow along from the definition of fourier expansion of $\phi$ in 9.21 but I am probably too naive to see it... $\endgroup$
    – Blitz
    Mar 17, 2019 at 2:34

1 Answer 1


Consider the case when $k_l=-k_m$. Peskin and Shroeder equation 9.26 would then read

$$ \frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0}\int d\mathrm{Re}\phi_n\ d\mathrm{Im}\phi_n \right)\times (\mathrm{Re}\phi_m\ \mathrm{Re}\phi_{-m} +i\ \mathrm{Re}\phi_m\ \mathrm{Im}\phi_{-m}+i\ \mathrm{Im}\phi_m\ \mathrm{Re}\phi_{-m}-\mathrm{Im}\phi_m\ \mathrm{Im}\phi_{-m}) \times \exp\left[-\frac{i}{V}\sum_{k_n^0>0}(m^2-k_n^2)[(\mathrm{Re}\phi_n)^2+(\mathrm{Im}\phi_n)^2] \right] .$$

In the first term, using $\mathrm{Re}\phi_{-m}=\mathrm{Re}\phi_m^* = \mathrm{Re}\phi_m$, we get

\begin{align} &\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0}\int d\mathrm{Re}\phi_n\ d\mathrm{Im}\phi_n \right)\times (\mathrm{Re}\phi_m)^2 \times e^{-\frac{i}{V}\sum_{k_n^0>0}(m^2-k_n^2)[(\mathrm{Re}\phi_n)^2+(\mathrm{Im}\phi_n)^2]}\\ &=\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0,n\neq m}\int d\mathrm{Re}\phi_n\ e^{-\frac{i}{V}(m^2-k_n^2)(\mathrm{Re}\phi_n)^2} \right) \times\left(\int_{k_m^0>0} d\mathrm{Re}\phi_m\ (\mathrm{Re}\phi_m)^2\ e^{-\frac{i}{V}(m^2-k_m^2)(\mathrm{Re}\phi_m)^2}\right) \left( \prod_{k_n^0>0}\int d\mathrm{Im}\phi_n\ e^{-\frac{i}{V}(m^2-k_n^2)(\mathrm{Im}\phi_n)^2} \right). \end{align}

Using the identities $ \int d\xi\ e^{-b\xi^2} = \sqrt{\frac{\pi}{b}} $ and $ \int d\xi\ \xi^2e^{-b\xi^2} = \frac{1}{2b}\sqrt{\frac{\pi}{b}} $, the equation simplifies as \begin{align} &\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0,n\neq m}\sqrt{\frac{-i\pi V}{m^2-k_n^2}} \right) \left(\frac{1}{2}\frac{-iV}{m^2-k_m^2-i\epsilon}\sqrt{\frac{-i\pi V}{m^2-k_m^2}}\right) \left( \prod_{k_n^0>0}\sqrt{\frac{-i\pi V}{m^2-k_n^2}} \right)\\ &=\frac{1}{V^2} \sum_m e^{-ik_m.(x_1-x_2)}\left(\prod_{k_n^0>0}\frac{-i\pi V}{m^2-k_n^2} \right)\frac{1}{2}\frac{-iV}{m^2-k_m^2-i\epsilon} . \end{align} The second and the third terms are of the form $\int d\xi\ \xi\ e^{-b\xi^2} =0$. Finally for the fourth term we use $\mathrm{Im}\phi_{-m} = \mathrm{Im}\phi_m^* = -\mathrm{Im}\phi_m $ and then it gives the same contribution as the first term which cancels the factor of $1/2$.


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