Talk:Schwinger parametrization

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easy?

what exactly is this making easy, or more likely easier? --MarSch 12:23, 18 December 2006 (UTC)[reply]

More

Re(Easy): If A(p) is quadratic in momentum p, then the resulting momentum integral is just a gaussian.

This article should also talk about the use of the Schwinger trick to get representations of propagators/greens functions in a classical background via the heat kernel representation.

Also perhaps about how the Schwinger parametrization leads to propertime regularisation... styler 03:49, 13 March 2007 (UTC)[reply]

Join with Feynman parametrization

Maybe it is worth to join this article into the article about Feynman parametrization, as all these parametrizations mentioned there are derivable from each other - and the Schwinger parametrization is even named and used in the derivation. This joined article then of course should carry a more general name, e.g. loop integral parametrizations. Stefan Groote (talk) 05:43, 21 September 2022 (UTC)[reply]