Calculation of electron vertex correction in Peskin and Schroeder I am trying two days now to simplify the numerator of the electron vertex correction in the one-loop contribution. My problem is to prove that
$$\bar{u}(p')\left[-\frac{1}{2}\gamma^\mu l^2+(-y \gamma ^{\nu} q_{\nu}+z\gamma ^{\nu} p_{\nu})\gamma^{\mu}((1-y) \gamma ^{\nu} q_{\nu}+z\gamma ^{\nu} p_{\nu})+m^2\gamma^{\mu}-2m((1-2y)q^\mu +2zp^\mu)\right]u(p)$$
$$=\bar{u}(p')\left[\gamma^\mu\left(-\frac{1}{2} l^2+(1-x)(1-y)q^2+(1-2z-z^2)m^2\right) +(p'^\mu+p^\mu)mz(z-1)+q^{\mu}m(z-2)(x-y)\right]u(p)$$
where $q=k'-k=p'-p$, $ x+y+z=1$ $(x,y,z$ are the feynmann parameters$)$,$l=k+yq-zp$, $\bar{u}(p')$ and $u(p)$ are dirac spinors with amd $m$ is the electron mass.
My calculations made me half way. I am curently here
$$=\bar{u}(p')\left[\gamma^\mu\left(-\frac{1}{2} l^2+(1-x)(1-y)q^2+(1-2z-z^2)m^2\right)+2m^2z\gamma^\mu+2p^\mu m z(z-2)-q^\mu m(2y(z-2)+2))\right]u(p)$$
As you can see i am very close but i can't go all the way. Can anyone help me with this proof?
(It's in the page 191 and 192 in Peskin and Schroeder "An Introduction to Quantum Field Theory")
 A: I would like to share my idea:

*

*at first, only $(-y{\not q}+z{\not p}) \gamma^\mu ((1-y) {\not q}+z{\not p})$ need special concern. We rewrite terms before $\gamma^\mu$ in ${\not p'}$ and ${\not q}$ using $p=p'-q $ while kept terms after $\gamma^\mu$ unchanged, we use ${\not p}u(p)=mu(p)$ and $\overline{u}(p'){\not p'}=\overline{u}(p')m$, only terms involves ${\not q}$ and $\gamma^\mu$ remains. With $\overline{u}(p')$ and $u(p)$ implicit,

$(-y{\not q}+z{\not p}) \gamma^\mu ((1-y) {\not q}+z{\not p})$
$=(z{\not p'}-(y+z){\not q}) \gamma^\mu ((1-y) {\not q}+z{\not p})$
$=(mz-(y+z){\not q}) \gamma^\mu ((1-y) {\not q}+mz)$
$=\underline{m^2 z^2\gamma^\mu} + mz(1-y)\gamma^\mu{\not q} -mz(y+z){\not q}\gamma^\mu -(y+z)(1-y){\not q}\gamma^\mu{\not q}$
2a. First term is fine, we swap the last term, as it gives $\gamma^\mu q^2$, remember $\overline{u}(p')q^\mu{\not q}u(p)=\overline{u}(p')q^\mu({\not p'}-{\not p})u(p)=\overline{u}(p')({\not p'}q^\mu-q^\mu{\not p})u(p)=\overline{u}(p')(mq^\mu-q^\mu m)u(p)=0$
$-(y+z)(1-y){\not q}\gamma^\mu{\not q}$
$=-(y+z)(1-y)(2q^\mu-\gamma^\mu{\not q}){\not q}$
$=-2(y+z)(1-y)q^\mu{\not q} + (y+z)(1-y)\gamma^\mu q^2$
$=\underline{(y+z)(1-y)\gamma^\mu q^2}$
2b. swap $\gamma^\mu {\not q}$ gives terms like ${\not q}\gamma^\mu$ which is no good, rewrite back to $p'-p$ then swap gives terms of $m\gamma^\mu$, $p^\mu$ and $p'^\mu$. Remember, only ${\not p'}$ before $\gamma^\mu$ and ${\not p}$ after $\gamma^\mu$ can contract with their corresponding $u$ to give $m$, $\gamma^\mu {\not p'}$ cannot contract with $\overline{u}(p')$ as there is a $\gamma^\mu$ between them.
$mz(1-y)\gamma^\mu{\not q}$
$=mz(1-y)[\gamma^\mu({\not p'}-{\not p})]$
$=mz(1-y)[\gamma^\mu{\not p'}-\gamma^\mu m]$
$=mz(1-y)[2p'^\mu -{\not p'}\gamma^\mu - m\gamma^\mu]$
$=mz(1-y)[2p'^\mu -m\gamma^\mu - m\gamma^\mu]$
$=\underline{-2m^2z(1-y)\gamma^\mu+2mz(1-y)p'^\mu}$
Similarly,
$-mz(y+z){\not q}\gamma^\mu$
$=-mz(y+z)[({\not p'}-{\not p})\gamma^\mu]$
$=-mz(y+z)[m\gamma^\mu-{\not p}\gamma^\mu]$
$=-mz(y+z)[m\gamma^\mu-2p^\mu+\gamma^\mu{\not p}]$
$=-mz(y+z)[m\gamma^\mu-2p^\mu+\gamma^\mu m]$
$=\underline{-2m^2z(y+z)\gamma^\mu + 2mz(y+z)p^\mu}$
Up to here, we actually get everything, the rest is simple algebra grouping terms.


*Put everything into numerator, with $u(p)$, $\overline{u}'(p')$ implicit. Remember $x+y+z=1$
numerator
$=-\frac{1}{2}\gamma^\mu l^2$
$+(m^2 z^2\gamma^\mu + (y+z)(1-y)\gamma^\mu q^2 -2m^2z(1-y)\gamma^\mu+2mz(1-y)p'^\mu -2m^2z(y+z)\gamma^\mu + 2mz(y+z)p^\mu)$
$+m^2 \gamma^\mu -2m((1-2y)q^\mu + 2zp^\mu) $
$=\gamma^\mu(-\frac{1}{2} l^2 +(1-x)(1-y) q^2 + m^2 z^2-2m^2z(1-y) -2m^2z(y+z)+m^2)$
$+2mz(1-y)p'^\mu + (2mz(y+z)-4mz)p^\mu -2m(1-2y)q^\mu$
$=\gamma^\mu(-\frac{1}{2} l^2 +(1-x)(1-y) q^2 + m^2(z^2-2z+2zy-2zy-2z^2+1)$
$+2mz(1-y)p'^\mu + 2mz(y+z-2)p^\mu -2m(1-2y)q^\mu$
$=\gamma^\mu (-\frac{1}{2}l^2+ (1-x)(1-y) q^2 + (1-2z-z^2)m^2)$
$+2mz(1-y)p'^\mu - 2mz(1+x)p^\mu -2m(1-2y)q^\mu$
We got the $\gamma^\mu$ term, and
Observing $ap'^\mu+bp^\mu+cq^\mu=\frac{a+b}{2}(p'^\mu+p^\mu)+\frac{a-b}{2}(p'^\mu-p^\mu)+cq^\mu=\frac{a+b}{2}(p'^\mu+p^\mu)+(\frac{a-b}{2}+c)q^\mu$
We further get:
$+2mz(1-y)p'^\mu - 2mz(1+x)p^\mu -2m(1-2y)q^\mu$
$=\frac{1}{2}(2mz(1-y)-2mz(1+x))(p'^\mu+p^\mu)+(\frac{1}{2}(2mz(1-y)+2mz(1+x))-2m(1-2y))q^\mu$
$=mz(-x-y)(p'^\mu+p^\mu)+m(z(1-y)+z(1+x)-2(x-y+z))q^\mu$
$=mz(z-1)(p'^\mu+p^\mu)+m(z(x-y)+2z-2(x-y)-2z)q^\mu$
$=mz(z-1)(p'^\mu+p^\mu)+m(z-2)(x-y)q^\mu$
and we get everything
