# Why does $k_\mu k_\nu \to k^2 \eta_{\mu\nu}/d$ work in QFT calculations?

When doing calculations of Feynman diagrams in QFT I've seen a trick used that goes something like this $$\int \frac{d^dk}{(2\pi)^d} \frac{k_\mu k_\nu}{f(k^2)}\quad\longrightarrow\quad\int \frac{d^dk}{(2\pi)^d} \frac{ k^2 \eta_{\mu\nu}/d}{f(k^2)}.$$ Sometimes also linear terms in $$k_\mu$$ would get dropped. Unfortunately, I haven't really seen an explanation for this...

I think the part about dropping linear terms has something to do with integrating an uneven function over a symmetric interval. But I really have no clue where the $$k_\mu k_\nu\to k^2 \eta_{\mu\nu}/d$$ comes from..

• – Nihar Karve Feb 18 at 9:43

If one replaces $$k^2=k_0 k^0+k_1 k^1+\cdots$$ by $$k^\mu k^\nu$$ in the numerator of your integral, the only components that give anything but $$0$$ are for $$\mu=\nu$$ and there is(are) sign(s) if you are in Minkowski space-time (depending on your choice of signature). Then, we have something proportional to the Minkowski metric. For the factor $$d$$ you just have to see that integrating over $$k^2$$ gives just $$d$$ times the integral over $$k^a k^a$$ (without summation !). That's all, this is just maths.
The reason is really symmetry. I think considering the problem with $$d=2$$ should be elucidating. Additionally let us take $$f(k)=e^{k^2}$$, in general it should be some polynomial on $$k^2$$ (spherical symm.) such that the integral converges. Let us call the integral $$I_{\mu\nu}$$ and consider w.l.g. $$\mu=1,\nu=2$$, so let us see $$I_{12} = \frac{1}{4\pi^2}\int {\rm d} x \,{\rm d} y\, x y\,e^{-x^2-y^2} = 0$$
The computation also checks out when the indices are the same, that I will leave you to check. Hint: $$k^2$$ overcounts directions... therefore the $$1/d$$.
So at the end of the day what we have is that such replacement is true, provided we have a spherically symmetric situation, that is the function $$f$$ must depend only on the norm square of $$k$$ and the integration must be trough out the whole volume. (And as always in physics... assuming the integral actually converges)