# Four-gauge-boson vertex in non-Abelian gauge theories

In Peskin & Schroeder's book page 524, the following diagram is calculated for the gauge boson self-energy in order $g^2$:

In dimensional regularization, its contribution is given by

$$-g^2C_2(G)\delta^{ab}\int\frac{d^dp}{(2\pi)^d}\frac{1}{p^2}\cdot g^{\mu\nu}(d-1)$$

At this point, Peskin & Schroeder say that "we could simply discard this diagram". It is clear that the integral over $p$ does not give a pole as $d\rightarrow 4$, but it is also divergent. Why can we simply discard this diagram? (Just because it is not logarithmically divergent?) Many thanks in advance!

The only argument I can find for this is a couple pages earlier, where they say

...which in turn implies that the photon self-energy diagrams have the structure

$$= i(q^2 g^{\mu\nu} - q^\mu q^\nu)\Pi(q^2)$$

The only divergence possible is a logarithmically divergent contribution to $$\Pi(q^2)$$. In non-Abelian gauge theories, (16.57) still holds, so the self-energy again has the Lorentz structure (16.58).

Their point is that the total self-energy cannot have a pole in any dimension, and so if we find a term that does have a pole for any value of $$d$$, as with $$d\to 2$$ in the case of the term you're asking about, we know that something else has to cancel it out.

• Is it right to say that the dimensional regularization can only deal with logarithmical divengence? – soliton May 12 '13 at 9:48
• No, it can deal with power divergences as well. – David Z May 12 '13 at 17:47
• Could you give me an example? – soliton May 13 '13 at 4:49
• Have a look at this. It explains dimensional regularization for general $1/p^{2n}$-type integrals. – David Z May 13 '13 at 5:27