Recent content by hamad12a

it's on page 51, Peskin & Schroeder's book.

P&S had calculated this expression almost explicitly, except that I didn't find a way to exchange the $$\nu \lambda$$ indices, but I'm sure the below identity is used, $$ \begin{aligned}\left(\overline{u}_{1 L} \overline{\sigma}^{\mu} \sigma^{\nu} \overline{\sigma}^{\lambda} u_{2...

the mistake was in applying the formulae a^3-b^3 which equals to : a3 – b3 = (a – b)(a2 + ab + b2) i wrote 2ab instead of ab only without myltiplying it by 2 so, again the roots are: -1/2+sqrt(3)/2*i and -1/2-sqrt(3)/2 by using the D=b^2-4ac for the quadratic one and a=b for the first term...

Homework Statement in a given activity: solve for z in C the equation: z^3=1 Homework Equations prove that the roots are 1, i, and i^2 The Attempt at a Solution using z^3-1=0 <=> Z^3-1^3 == a^3-b^3=(a-b)(a^2+2ab+b^2) it's clear the solution are 1 and i^2=-1 but i didn't find "i" as a solution...