In chapter 12.1 of Peskin and Schroeder we derive the propagator of a high momenta shell of a $\phi^4$ theory. Following the derivation, we ignore quartic terms as well as mass terms.

The original action is $$ \int{d^dx\left(\frac{1}{2}(\partial_\mu\hat{\phi})^2+\frac{1}{2}m^2\hat{\phi}^2+\lambda\left(\frac{1}{6}\phi^3\hat{\phi}+\frac{1}{4}\phi^2\hat{\phi}^2+\frac{1}{6}\phi\hat{\phi}^3+\frac{1}{4!}\hat{\phi}^4\right)\right)}\tag{12.5} $$ then we only keep the first term giving us $$ \mathcal{S}=\int{d^dx\frac{1}{2}(\partial_\mu\hat{\phi})^2}=\int{d^dx\frac{1}{2}\hat{\phi}\partial^2\hat{\phi}}=\frac{1}{2}\int{\frac{d^dk}{(2\pi)^d}\hat{\phi}^*(k)k^2\hat{\phi}(k)}. $$ First of all, my question is the following. Shouldn't the second $\phi$ be $\phi^*$? I was under the impression that the '$^*$' is used because the fields are Fourier transformed.

Furthermore, how do we get the following propagator (how can we derive it?)? $$ \frac{\int{\mathcal{D}\hat{\phi}}\exp(-\mathcal{S})\hat{\phi}(k)\hat{\phi}(p)}{\int{\mathcal{D}\hat{\phi}\exp(-\mathcal{S})}}=\frac{1}{k^2}(2\pi)^d\delta^{(d)}(k+p)\Theta(k)\tag{12.8} $$ where $\Theta(k)$ is defined in eq. (12.9).

Note: the momentum integrals are down between $b\Lambda \leq k < \Lambda$ for which $\Lambda$ is the momenta cutoff and $b$ a parameter between 0 and 1.

From these I understand the use of the step-function $\Theta$ but I'm not sure how we rederive the propagator. This is essentially the same as the klein-gordon propagator but in Euclidean space and in the limit of low mass ($m^2\ll \Lambda^2$).

  • $\begingroup$ $\hat{\phi}^*$ denotes the complex conjugate of $\hat{\phi}$. $\hat{\phi}$ is used for both the field and its Fourier transformation in this case. If you just plug the definition of $\hat\phi(x)$ in terms of $\hat\phi(k)$ into the integral, you get the relationship $\int{d^dx\frac{1}{2}\hat{\phi}\partial^2\hat{\phi}}=\int{\frac{d^dk}{(2\pi)^d}\hat{\phi}^*(k)k^2\hat{\phi}(k)}$ $\endgroup$ Feb 9, 2022 at 13:45
  • $\begingroup$ Thanks! Can you also explain why we get the complex conjugate? $\endgroup$ Feb 9, 2022 at 13:53

1 Answer 1


In answer to your first question, $$\int{d^dx\frac{1}{2}\hat{\phi}(x)\partial^2\hat{\phi}(x)}=\int d^dx \frac{1}{2}\int d^dk e^{-ikx}\hat\phi(k)\partial^2\left(\int d^dk' e^{-ik'x}\phi(k')\right)$$ $$=\int d^dx \frac{1}{2}\int d^dk e^{-ikx}\hat\phi(k)\left(\int d^dk' i^2k'^2e^{-ik'x}\phi(k')\right)$$ $$=\int d^dx \frac{1}{2}\int d^dk e^{ikx}\hat\phi(-k)\left(\int d^dk' (-1)k'^2e^{-ik'x}\phi(k')\right)$$ $$=-\int d^dkd^dk'\left(\int d^dx e^{i(k'-k)x}\right)\frac{1}{2}\hat\phi(-k) k'^2\phi(k')$$ $$=-\int d^dkd^dk'\frac{1}{(2\pi)^d}\delta^{(d)}\left(k-k'\right)\frac{1}{2}\hat\phi(-k) k'^2\phi(k')$$ $$=-\frac{1}{2}\int \frac{d^dk}{(2\pi)^d}\hat\phi(-k) k^2\phi(k)=-\frac{1}{2}\int \frac{d^dk}{(2\pi)^d}\hat\phi^*(k) k^2\phi(k).$$ There may be a sign error in there somewhere.


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