In Mandl & Shaw QFT the equation for superficial degree of divergence $K$ for QED in 4D is stated to be

$$ K = 4d - f_i - 2b_i, $$

where $d$ is the number of internal momenta not fixed by energy-momentum conservation at the vertices, $f_i$ and $b_i$ are the number of internal fermion and photon lines. This is then used to derive the result

$$ K = 4 - \frac{3}{2}f_e - b_e, $$

where $f_e$ and $b_e$ are the number of external fermion and photon lines. The first of these equations is said to be derived from counting powers of momentum variables of integration, but I don't see how this works and there are no further explanations in the book.

Where does this come from? Is it very important to understand things like primitive divergences to understand renormalizability well?

  • $\begingroup$ Which page? Which eqs? $\endgroup$
    – Qmechanic
    Oct 4, 2022 at 19:40

2 Answers 2


As quick sketch to see where the formula comes from just schematically write out the integral: since $d$ momenta are not fixed we integrate over them \begin{align*} \int (\mathrm{d}^4k)^d \left(\frac{1}{k^2}\right)^{b_i} \left(\frac{1}{k}\right)^{f_i} = \int dk\ k^{4d-1}\ k^{-2b_i-f_i} \propto k^{4d-2b_i - f_i} \Big|_{-\infty}^\infty \end{align*} The result is finite if the exponent is less than zero, hence it's called the superficial degree of divergence. The actual degree can however be lowered by certain symmetries and cancellations etc.


I just expand Wihtedeka's answer which is already correct...
A general propagator, for large momentum (we are interested in divergences for $p \to \infty$), looks like $\frac{1}{p^\alpha}$. In the case of fermions $\alpha=1$, in the case of photons $\alpha=2$. Then for every loop we have an integral (Feynman rule if you wish). The total momentum power of the integral is given by the superficial degree of divergence (4 positive powers for each integration or loop minus the propagators contribution). Note that a degree of divergence 0 is not enough, in general, for the integral to converge!
You may want to take a look here https://universe-review.ca/R15-12-QFT12.htm at the table and the relative description.
The second formula you wrote nicely shows you that the degree of divergence only depends on the number of external legs. For example this means that in QED all diagrams with more than 2 external fermions and 1 boson are convergent. Hence you have only a finite number of divergent amplitudes! In this case one says that the theory is renormalizable because the divergences can be absorbed in a finite number of counterterms.
You may want to read Timo Weigand's lecture notes for more details.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.