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Feb 26, 2012 at 1:57 comment added José Figueroa-O'Farrill It's simply a choice which simplifies calculations. Choosing an orthonormal basis for a vector space simplifies any calculation involving an inner product, of which in computing Feynman diagrams there are a-plenty. (Is this really a question for MO?)
Feb 24, 2012 at 21:18 comment added HAJV Of course, it is a choice. Any idea why G-M preferred or required the one he chose?
Feb 24, 2012 at 19:28 history answered H. Arponen CC BY-SA 3.0