The full propagator in QFT is the inverse of this two-point 1PI function defined as $$ \tilde{\Gamma}^{(2)}(p)=p^{2}-m^{2}-\Sigma(p) $$ where $\Sigma(p)$ is the self-energy. I am have calculated the self-energy correction for 1 loop in a modified $\phi^4$ theory.

My question is: how do I find the cross setion from this?

I know that usually it is found from the sum of the connected amputated diagrams but I am not too sure how to proceed in this case given that the normal propagator is 1/ my results. Also how would I extend this result to calculate the self- energy for the 4-point 1PI function?

So in normal $\phi^4 $ theory when studying 2-> 2 scattering up to 1loop I would get this expression $$ \begin{aligned} i \mathcal{M}\left(p_{1} p, \rightarrow p_{3} p_{4}\right)=&-i \lambda-\frac{i \lambda^{2}}{32 \pi^{2}} \int_{0}^{1} \mathrm{~d} x\left[\log \left(\frac{m^{2}-x(1-x) s}{m^{2}-x(1-x) 4 m^{2}}\right)\right.\\ &\left.+\log \left(\frac{m^{2}-x(1-x) t}{m^{2}}\right)+\log \left(\frac{m^{2}-x(1-x) u}{m^{2}}\right)\right] \end{aligned} $$ I want to repeat the same process when instead of the usual quartic vertex I have this $$ \frac{-i g}{3}(2 \pi)^{4} \delta\left(p_{1}+p_{2}+p_{3}+p_{4}\right)f(p_1, p_2, p_3, p_4) $$ where $f$ is some function dependent on the momenta.

Edit: What I am looking for is how to go from the 4-point 1PI function to a cross-section. (I am not sure of what the physical meaning of $\Gamma^{(4)}$ is). I am planning to use dimensional regularization to remove divergences.

  • 1
    $\begingroup$ It depends much on the process and perturbative order you consider. But for a two-point function there is no cross section associated since there is no scattering (which requires at least 2 incoming particles). For a 4-point function on first order in coupling g there is no loop at all, and on 2. order g^2 there is no effect of the self-energy on the cross section. Actually, the main effect of the self-energy is on the mass of the involved particle, and very little if not no effect at all on the cross section. $\endgroup$ Dec 4, 2020 at 15:56
  • $\begingroup$ How do I go from the 4-point function to the cross section keeping up to 1 loop? $\endgroup$ Dec 6, 2020 at 16:59

1 Answer 1


First, you need to renormalize the theory to get rid of the logarithmic divergences. Here's the completed final product, copied verbatim from my professor's lecture notes: $$ \hat{\mathcal{L}} = \frac{1}{2} ( 1 + A ) \partial_\mu \hat{\phi}_\mathrm{P} \partial^\mu \hat{\phi}_\mathrm{P} - \frac{1}{2} ( m_\mathrm{P}^2 - B ) \hat{\phi}^2_\mathrm{P} - \frac{1}{4!} (\lambda_\mathrm{P} - C ) \hat{\phi}^4_\mathrm{P} $$ where the subscript $\mathrm{P}$ stands for "physical", and the physical quantities are related to the quantities in the theory before renormalization by: $$ Z m^2 = m_\mathrm{P}^2 - B $$ $$ Z^2 \lambda = \lambda_\mathrm{P} - C $$ $$ Z = 1 + A $$ The three corrections each renormalizes the field ($A$), the mass ($B$), and the coupling constant ($C$), respectively. $$ \hat{\phi}(\mathbf{x}) = \sqrt{Z} ~ \hat{\phi}_\mathrm{P} $$ $$ \frac{i}{ A \mathbf{p}^2 - m^2 - B } = \frac{ i Z }{\mathbf{p}^2 - m_\mathrm{P}^2} $$ The latter relates $A$, $B$, and $Z$ to the self-energy $\tilde{\Sigma} (\mathbf{p}) $.

Then, you can relate the invariant amplitude $\mathcal{M}$ to the cross section using the following relation: $$ \mathrm{d} \sigma = \frac{1}{2 E_1} \frac{1}{2 E_2} |\mathcal{M}|^2 \mathrm{d} \Pi_\mathrm{LIPS}$$ where $E_1$ and $E_2$ are the initial energies of the two particles associated with the 4-momenta $\mathbf{p}_1$ and $\mathbf{p}_2$, respectively, and $\mathrm{d} \Pi_\mathrm{LIPS}$ being the Lorentz invariant phase space, $$ \mathrm{d} \Pi_\mathrm{LIPS} = (2\pi)^4 \delta^4 (\sum_i p^\mu_i - \sum_f p^\mu_f) \prod_j \frac{ \mathrm{d}^3 \vec{p}_j }{(2\pi)^3} \frac{1}{2 E_{p_j}}$$

  • $\begingroup$ You can see the full derivation in Lectures 12 and 20 here. $\endgroup$
    – wyphan
    Dec 4, 2020 at 20:05

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