# Is the off-diagonal part of this rank-2 tensor integrand odd?

Peskin and Schroeder in Introduction to Quantum Field Theory consider the following tensor integral (Eq. 6.46): $$\int \frac{\mathrm{d}^4l}{(2\pi)^4} \frac{l^\mu l^\nu}{D^n} = \int \frac{\mathrm{d}^4 l}{(2\pi)^4} \frac{\frac{1}{4}g^{\mu\nu}l^2}{D^n}$$ where the denominator function $$D$$ is even, i.e. it depends on the mangitude $$l^2$$. They explain that to obtain this relation, we first notice that the integral on the left-hand side vanishes by symmetry unless $$\mu=\nu$$. Then, Lorentz invariance implies its tensor structure needs to be proportional to $$g^{\mu\nu}$$. The coefficient can be found by contracting both sides with $$g_{\mu\nu}.$$

My question is about the first of these steps: why is this integrand odd unless $$\mu=\nu$$? It's clear that if the numerator was $$l^\mu$$, then under the transformation $$l^\mu \rightarrow -l^\mu$$, we obtain an odd integrand. In this case however, do we not have $$l^\mu l^\nu \rightarrow (-)^2l^\mu l^\nu = l^\mu l^\nu$$, making the integrand even? I understand how the next steps imply the final result, it's just this antisymmetry statement I can't grasp.

I checked two other QFT textbooks (Schwartz, Eq. B.51 and Srednicki, Eq. 14.53) and they do not refer to the antisymmetry - they just argue that the integral must be proportional to $$g^{\mu\nu}$$ as $$D$$ is a Lorentz scalar and we integrate over $$l$$, so this is the only tensor available.

• If $\mu\neq\nu$ you can change sign to only one coordinate and get an overall minus. Commented Apr 27, 2020 at 14:10
• @MannyC Do we not transform all components of $l^\mu$ at the same time? If so, could you please explain why not? Commented Apr 27, 2020 at 14:18
• You can make any change of variables you want. $l^\mu = (a,b,c,d) = (a,-b',c,d)$. An integrals in $\mathbb{R}^4$ is just four integral one after the other. Commented Apr 27, 2020 at 14:19
• Sorry, I don't understand why you can do that. I thought you'd have to change all the coordinates like $l^\mu = (a,b,c,d) \rightarrow (-a, -b, -c, -d)$. My logic came from thinking about e.g. wavefunctions in 3D spherical polar coordinates. To check their symmetry, I would do $\textbf{r} \rightarrow -\textbf{r}$, which is equivalent to $(x,y,z) \rightarrow (-x,-y,-z)$. So why are we allowed to switch only one of the coordinates here? Commented Apr 27, 2020 at 14:29

Recall that $$\int \mathrm{d}^4l\,f(l^0,l^1,l^2,l^3) \equiv \int_{-\infty}^\infty \mathrm{d}l^0\int_{-\infty}^\infty \mathrm{d}l^1\int_{-\infty}^\infty \mathrm{d}l^2\int_{-\infty}^\infty \mathrm{d}l^3\,f(l^0,l^1,l^2,l^3)\,.$$ Every integral satisfies the usual rules for changes of variables and has no idea that Lorentz symmetry is there. So we can say $$l^1 = - (l')^1$$, and turn the integral into $$\int_{-\infty}^\infty \mathrm{d}l^0\int_{-\infty}^\infty \mathrm{d}(l')^1\int_{-\infty}^\infty \mathrm{d}l^2\int_{-\infty}^\infty \mathrm{d}l^3\,f(l^0,-(l')^1,l^2,l^3)\,.$$ Doing this change of variables is not a Lorentz transormation nor a parity transformation nor anything meaningful from the spacetime point of view. It's like a no-op. We are changing the name of a dummy index.
With this change of variables the numerator in the op changes sign whenever $$\mu=1,\nu\neq1$$ or viceversa. A similar reasoning can be done for $$0,2,3$$. Thus we see why Peskin says that it has to be zero when $$\mu \neq \nu$$.
• Thanks, that makes sense. I guess I was thinking about is as a parity transformation rather than just a simple redefinition of co-ordinates, which can be applied to e.g. $l^1$ only. Commented Apr 27, 2020 at 14:54