- #1
mdb71
- 5
- 3
Hi all, I have a problem working out the algebra of the following expression in Peskin & Schroeder in a smart way to give the result. It is on page 191, regarding the numerator of the vertex correction function.
We want to get from the LHS to the RHS of the following expression
$$\bar{u}(p')[-\frac{1}{2}l^2\gamma^\mu + (z\not{p}-y\not{q})\gamma^{\mu}(z\not{p}+(1-y)\not{q}) + m^2\gamma^{\mu}-2m((1-2y)q^{\mu}+2zp^{\mu})]u(p)$$ $$ = \bar{u}(p')[\gamma^{\mu}(-\frac{1}{2}l^2+(1-x)(1-y)q^2+(1-2z-z^2)m^2)+(p'^{\mu}+p^{\mu})mz(1-z)+q^{\mu}m(z-2)(x-y)]u(p)$$
using
$$ q \equiv p'-p, \not{p}u(p) = mu(p), \bar{u}(p')\not{p'} = m\bar{u}(p'), \bar{u}(p')\not{q}u(p) = 0,\\
x+y+z = 1$$
Thanks a lot in advance!
We want to get from the LHS to the RHS of the following expression
$$\bar{u}(p')[-\frac{1}{2}l^2\gamma^\mu + (z\not{p}-y\not{q})\gamma^{\mu}(z\not{p}+(1-y)\not{q}) + m^2\gamma^{\mu}-2m((1-2y)q^{\mu}+2zp^{\mu})]u(p)$$ $$ = \bar{u}(p')[\gamma^{\mu}(-\frac{1}{2}l^2+(1-x)(1-y)q^2+(1-2z-z^2)m^2)+(p'^{\mu}+p^{\mu})mz(1-z)+q^{\mu}m(z-2)(x-y)]u(p)$$
using
$$ q \equiv p'-p, \not{p}u(p) = mu(p), \bar{u}(p')\not{p'} = m\bar{u}(p'), \bar{u}(p')\not{q}u(p) = 0,\\
x+y+z = 1$$
Thanks a lot in advance!