\input phyzzx

\def\D{\!\not\!}

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\centerline{\bf Note on QED Vertex Correction}
\smallskip
As discussed in class, the one-loop correction to the tree-level
QED vertex $\bar u(p')\gamma^\mu u(p)$ is given by the following
monster of an integral:
$$
\bar u(p')\,\delta_1\Gamma^\mu(p,p')\,u(p)\
=\ -2ie^2 \mathop{\intop\!\!\!\intop\!\!\!\intop}\limits^1_0\!\!
dx\,dy\,dz\,\delta(x+y+z-1) \int\!\!{d^4\ell\over (2\pi)^4}\,
{\bar u(p')\,{\cal N}^\mu\,u(p)\over \bigl[\ell^2-\Delta+i0\bigr]^3}
\eqn\monster
$$
where
$$
\eqalignno{
\ell\ &=\ k\,+\,xp\,+\,yp', & \eqname\Ldef\cr
\Delta\ &=\ (xp+yp')^2\ =\ (1-z)^2m^2\,-\,xyq^2,& \eq\cr
{\cal N}^\mu\ &
=\ \gamma^\nu (\D k\, +\D p' +m)\gamma^\mu (\D k\, +\D p +m)\gamma_\nu\,. &
\eqname\OrigN\cr
}$$
In this note I simplify the numerator~\OrigN\
{\sl in the context of the integral}~\monster.

The first step is obvious:
Let us get rid of the $\gamma^\nu$ and $\gamma_\nu$
factors using $\gamma^\nu\!\D a\gamma_\nu=-2\D a$,
$\gamma^\nu\!\D a\D b\gamma_\nu=4(ab)$ and
$\gamma^\nu\!\D a\D b\D c\gamma_\nu=-2\D c\D b\D a$,
thus
$$
{\cal N}^\mu\ =\ -2m^2\gamma^\mu\
+\ 4m(2k+p+p')^\mu\ -\ 2(\D k+\D p)\gamma^\mu(\D k+\D p') .
\eqn\First
$$
Next, we should re-express the right hand side here in terms of the
Feynman's loop momentum~$\ell$ rather than $k$ using eq.~\Ldef;
expanding the result in powers of $\ell$, we get quadratic, linear
and $\ell$--independent terms.
However, the $\int\!\!d^4\ell$ integral in eq.~\monster\
and the denominator are both even with respect to $\ell\to-\ell$,
so the numerator terms which are linear in $\ell$ would not contribute
to the integral because of their oddness.
Disregarding these terms then gives us 
$$
\eqalign{
{\cal N}^\mu\ &
\cong\ 	-2m^2\gamma^\mu\ +\ 4m(p+p'-2xp-2yp')^\mu\cr
&\qquad-\ 2\D\ell \gamma^\mu\! \D\ell\
	-\ 2(\D p-x\D p-y\D p')\,\gamma^\mu\,(\D p'-x\D p-y\D p')\cr
&=\ -2m^2\gamma^\mu\ +\ 4mz\,(p+p')^\mu\ +\ 4m(x-y)q^\mu\
	-\ 2\D\ell \gamma^\mu\! \D\ell\cr
&\qquad-\ 2\,[z\D p' +(x-1)\D q]\,\gamma^\mu\,[z\D p+(1-y)\D q] \cr
}\eqn\Intermed
$$
where the second equality makes judicious use of $p'-p=q$
and $x+y+z=1$.

At this point, we can use the fact that the vertex correction
$\Gamma^\mu$ --- and hence the numerator ${\cal N}^\mu$ is always
sandwiched between $\bar u(p')$ and $u(p)$.
Consequently, on the last line of eq.~\Intermed\ we may replace
the $\D p'$ in the first factor with $m$ since $\bar u(p')\D p'=
\bar u(p') m$; likewise, in the last factor, we replace
the $\D p$ with $m$ since $\D p u(p)=m u(p)$.
Thus, the last line of eq.~\Intermed\ becomes
$$
\eqalign{
-2\,[z\D p' +(x-1)\D q\,]\,\gamma^\mu\,[z\D p+(1-y)\D q\,]\hskip -4em\cr
&\cong\ -2\,[mz -(1-x)\D q\,]\,\gamma^\mu\,[mz+(1-y)\D q\,]\cr
&=\ -2z^2m^2\,\gamma^\mu\ +\ 2(1-x)(1-y)\,\D q\gamma^\mu\!\!\D q\cr
&\qquad+\ 2z(1-x)m\D q\gamma^\mu\ -\ 2z(1-y)m\,\gamma^\mu\!\!\D q\cr
&\cong\ -2\,[z^2m^2+(1-x)(1-y)q^2]\,\gamma^\mu\cr
&\qquad+\ 2z(y-x)mq^\mu\ +\ 2z(x+y-2)m\times i\sigma^{\mu\nu}\!q_\nu\cr
}\eqn\LastLine
$$
where the last equality (or rather equivalence in the context of eq.~\monster)
follows from
$$
\bar u(p')\D q\, u(p)\ =\ 0\quad\Rightarrow\
\bar u(p')\bigl[\D q \gamma^\mu\! \D q] u(p)\ =\ -q^2\times \bar u'\gamma^\mu u
%\eqno\eq
$$
as well as
$$
\gamma^\mu\D\! q\ =\ q^\mu\,+\,i\sigma^{\mu\nu}q_\nu\,,\qquad
\D q\gamma^\mu\ =\ q^\mu\,-\,i\sigma^{\mu\nu}q_\nu\,.
%\eqno\eq
$$
%
Substituting eq.~\LastLine\ into eq.~\Intermed\ and making use
of the Gordon identity, we arrive at
$$
\eqalign{
{\cal N}^\mu\ \cong\ {}&
 -2\D\ell \gamma^\mu\! \D\ell\
    -\ 2\left[(1-4z+z^2)m^2+(1-x)(1-y)q^2\right]\times\gamma^\mu\cr
&-\ 4z(1-z)m^2\times{i\sigma^{\mu\nu}\!q_\nu\over 2m}\
    +\ 2(2-z)(x-y)\,m q^\mu .\cr
}\eqn\Penult
$$
Note that the last term here is odd under interchange
$x\leftrightarrow y$ of the two Feynman parameters.
Since everything else in the integral~\monster\ is symmetric
with respect to this interchange, the last term in eq.~\Penult\
integrates to zero and may be safely disregarded.

Finally, I would like to make use of the Lorentz symmetry of the
integral~\monster\ over the loop momentum~$\ell$.
Because of this symmetry, the integral of a quadratic term
$\ell^\alpha\ell^\beta$ should be proportional to the Lorentz
invariant $g^{\alpha\beta}$, hence in four spacetime dimensions,
$$
\ell^\alpha\ell^\beta\ \cong\ \coeff14\ell^2 g^{\alpha\beta}
\eqno\eq
$$
and therefore
$$
\D\ell\gamma^\mu\!\D\ell\
=\ \ell^\alpha\ell^\beta\times \gamma_\alpha\gamma^\mu\gamma_\beta\
\cong\ \coeff14\ell^2\times \gamma_\alpha\gamma^\mu\gamma^\alpha\
=\ -\half\ell^2\,\gamma^\mu .
\eqn\ELL
$$
Applying this equivalence to the first term in eq.~\Penult\
we arrive at
$$
{\cal N}^\mu\ \cong\ 
[\ell^2-2(1-4z+z^2)m^2-2(1-x)(1-y)q^2]\times\gamma^\mu\
-\ 4z(1-z)m^2\times{i\sigma^{\mu\nu}q_\nu\over 2m}
\eqn\Ultimate
$$
and at this point, I give up
--- it does not get any simpler than this.

\bye
