The superficial degree of divergence for a diagram is defined as the power of $k$ in the nominator minus the power of $k$ in the denominator. It is written to be equal to $4\times$ (number of loops)$ - $(number of internal Fermion lines)$ - 2 \times $(number of internal Boson lines).

Why is it 4 and not 3? When you do a 4-dimensional integral in spherical coordinates, you get a factor of $k^3$ from your volume element.

In particular, when you calculate the QED vertex, for instance, you get:

$$ \int_{\mathbb{R}^4} \frac{d^4k}{(2\pi)^4} \frac{\gamma^\rho (\not\!{p_1}-\not\!{k}+m)\gamma^\mu(-\not\!{p_2}-\not\!{k}+m)\gamma_\rho}{k^2\left[\left(p_1-k\right)^2-m^2\right]\left[\left(p_2+k\right)^2-m^2\right]} \overset{k\to\infty}{\propto} \int_{\mathbb{R}^4} \frac{d^4k\, k^2}{k^6} \overset{\text{spherical}}{\underset{\text{coordinates}}{\propto}} \int_0^\infty\frac{dk}{k} $$

Thus I would say that the superficial degree of divergence of this diagram is $3\times1-2-2=-1$ and not $0$.

What am I missing?

  • 3
    $\begingroup$ The degree of divergence of your last integral is 0 and not -1, it still diverges (albeit logarithmically)!! Every integration does, in fact, introduce a 4 and not a 3. You still have to integrate over $k_0$, even in spherical coordinates. $\endgroup$
    – Danu
    Commented Feb 10, 2014 at 12:39
  • $\begingroup$ But the "degree of divergence" is defined as the power of k in the nominator minus the power of k in the denominator. In the last term on the most right hand side, the power of k in the nominator is 0 and the power of k in the denominator is 1, so the "degree of divergence" is -1. $\endgroup$
    – PPR
    Commented Feb 10, 2014 at 12:42
  • 6
    $\begingroup$ But $dk$ still has units. Thus when counting the degree of divergence, you must include it. In some sense it is just a definition. A degree of divergence of $0$ is defined to be expected to diverge logarithmically. $\endgroup$
    – JeffDror
    Commented Feb 10, 2014 at 12:43
  • 1
    $\begingroup$ nominator? $\endgroup$
    – Qmechanic
    Commented May 12, 2014 at 18:18
  • 1
    $\begingroup$ You get $k^3$, yes, but $\int_0^R k^3 = R^4/4$. It's the degree of divergence of the integral that matters, not of the integrand. $\endgroup$ Commented Sep 19, 2014 at 1:55

1 Answer 1


No matter which coordinate system you choose, you must integrate over four coordinates, or $D$ when working in an arbitrary number of dimensions. This $d^Dk$ has units $[L]^{-D}$, where $[L]$ are units of length. The integral $$\int \frac{d^Dk}{k^n}$$ has units $[L]^{n-D}$ and is expected to diverge (as $k\rightarrow \infty$) whenever $D\geq n$. The limiting case $D=n$ might be a little confusing, but you can see that this diverges too. Take, for example, $D=n=3$: $$\int \frac{d^3k}{k^3}\propto \int \frac{dk}{k}=\log k\Biggr|_{0}^{\infty}=\text{divergent}$$ Here, it is said that the integral diverges logarithmically, for obvious reasons.


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