- Summary
- In evaluating a single loop diagram in the photon self-energy we get (among other things) a 2nd rank tensor integral (see Schwartz Quantum Field Theory and the Standard Model p. 830 Appendix B.3.4). We want to reduce this integral to a scalar integral and we're told that to do this "it must be proportional to the only tensor around, gµν." Why not some other tensor?

We're trying to reduce the tensor integral ##\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}{\rm{ }}## to a scalar integral (where ##{{\Delta ^2}}## is a scalar). We're told that the tensor integral is proportional to ##{g^{\mu \nu }}##, so we have:

##C{g^{\mu \nu }} = \int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}##. Now we can turn the original integral into a scalar integral:

##C{g^{\mu \nu }}{g_{\mu \nu }} = {g_{\mu \nu }}\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}} = \int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k_\nu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}} = \int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^2}}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}##

##C = \frac{1}{4}\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^2}}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}##

##\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}} = C{g^{\mu \nu }} = {g^{\mu \nu }}\frac{1}{4}\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^2}}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}##

It all works out very nicely, but if all we're after is Lorentz covariance what gives us the right to pick the one tensor (##{g^{\mu\nu}}##) that doesn't change under a Lorentz transformation when any old tensor would give us Lorentz covariance. What does Schwartz mean when he says it's the only one around?

##C{g^{\mu \nu }} = \int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}##. Now we can turn the original integral into a scalar integral:

##C{g^{\mu \nu }}{g_{\mu \nu }} = {g_{\mu \nu }}\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}} = \int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k_\nu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}} = \int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^2}}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}##

##C = \frac{1}{4}\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^2}}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}##

##\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^\mu }{k^\nu }}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}} = C{g^{\mu \nu }} = {g^{\mu \nu }}\frac{1}{4}\int {\frac{{{d^4}k}}{{{{\left( {2\pi } \right)}^4}}}} \frac{{{k^2}}}{{{{\left( {{k^2} - {\Delta ^2}} \right)}^n}}}##

It all works out very nicely, but if all we're after is Lorentz covariance what gives us the right to pick the one tensor (##{g^{\mu\nu}}##) that doesn't change under a Lorentz transformation when any old tensor would give us Lorentz covariance. What does Schwartz mean when he says it's the only one around?