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I review my QFT lecture notes and I am having hard times to figure out the significance of Ward identity in vacuum polarization.

vacuum polarization

In class, we calculated one loop correction stated as $$ i\Pi^{\mu\nu}_2(q)=\frac{(-ie)}{(2\pi)^4}^2\int d^4k Tr\Bigg[\gamma^{\mu}\frac{i(\require{cancel}\cancel p+m)}{k^2-m^2}\gamma^{\nu}\frac{(\cancel k+\cancel q + m)}{(k+q)^2-m^2}\Bigg] $$ Then we follow up the next equation after Wick rotation and Feynman Parametrization $$ i\Pi^{\mu\nu}_2(q)= -4ie^2\int_{0}^{1}dx\int \frac{d^4l}{(2\pi)^4}\Bigg(\frac{\frac{-1}{2}g^{\mu\nu}l^2+g^{\mu\nu}l^2-2x(1-x)q^{\mu}q^\nu+g^{\mu\nu}(m^2+x(1-x)q^2}{[l^2+\Delta^2]^2}\Bigg) $$ which is actually $$ i\Pi^{\mu\nu}_2 \propto g^{\mu\nu}\Pi(q^2) $$ But my lecturer claimed that if our choice on $\Pi_2^{\mu\nu}$ is such that, this will violate Ward identity. So we continued the calculation and with dimensional regularization come up with $$ i\Pi^{\mu\nu}_2 \propto i(q^2g^{\mu\nu}-q^{\mu}q^{\nu})\Pi(q^2) $$ And that is the correct choice of the $\Pi^{\mu\nu}_2$ because does not violate Ward identity. But I do not understand the choice. I know Ward identity claims $q_{\mu}M=0$ but, $q_{\mu}\Pi^{\mu\nu}_2$ also does not equal zero even if I choose the second notation. How can I see it is consistent with Ward?

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Your statement that the integral "is actually" $\propto g^{\mu\nu}\Pi$ is incorrect because you can clearly see the $q^\mu q^\nu$ in the numerator of the integrand.

The correct expression that you have in your final equation is not a choice, it is the result of calculating the loop using dimensional regularization.

Finally, indeed $q_\mu \Pi_2^{\mu\nu}$ is zero because $q_\mu\left( q^2 g^{\mu\nu} - q^\mu q^\nu\right)=q^2 q^\nu - q_\mu q^\mu q^\nu =0$

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  • $\begingroup$ Yes you are right for sure. I already knew Ward indicate $q_{\mu}\Pi^{\mu\nu}_2=0$ but I miscalculate it! Can not believe myself. Thanks anyway. I give a tick. $\endgroup$ – aQuestion Jul 23 '15 at 13:21

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