# Superficial degree of divergence by power counting of momenta in QED

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?

• Which page? Which eqs? Oct 4, 2022 at 19:40

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.
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!