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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.

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  • $\begingroup$ If $\mu\neq\nu$ you can change sign to only one coordinate and get an overall minus. $\endgroup$
    – MannyC
    Apr 27, 2020 at 14:10
  • $\begingroup$ @MannyC Do we not transform all components of $l^\mu$ at the same time? If so, could you please explain why not? $\endgroup$
    – dzejkob
    Apr 27, 2020 at 14:18
  • $\begingroup$ 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. $\endgroup$
    – MannyC
    Apr 27, 2020 at 14:19
  • $\begingroup$ 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? $\endgroup$
    – dzejkob
    Apr 27, 2020 at 14:29

1 Answer 1

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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 $\mu \neq \nu$.

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    $\begingroup$ 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. $\endgroup$
    – dzejkob
    Apr 27, 2020 at 14:54

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