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

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\centerline{\bf QED Vertex Correction: Working through the Algebra}
\smallskip
At the one-loop level of QED, the 1PI vertex correction comes from a single
Feynman diagram
$$
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\eqn\Diagram
$$
thus
$$
\!\eqalign{
ie\Gamma^\mu_{\rm 1\,loop}(p',p)\ &
=\int_{\rm reg}\!\!{d^4 k\over(2\pi)^4}\, {-ig^{\nu\lambda}\over k^2+i0}\times
    ie\gamma_\nu\times{i\over \D p'+\D k\ -m+i0}\times ie\gamma^\mu\times
    {i\over \D p+\D k\ -m+i0}\times ie\gamma_\lambda\crr
&=\ e^3\int_{\rm reg}\!\!{d^4 k\over(2\pi)^4}\,{{\cal N}^\mu\over{\cal D}}\cr
}\!\eqn\Origin
$$
where
$$
{\cal N}^\mu\
=\ \gamma^\nu (\D k\, +\D p' +m)\gamma^\mu (\D k\, +\D p +m)\gamma_\nu
\eqn\OrigN
$$
and
$$
{\cal D}\ =\ [k^2+i0]\times[(p+k)^2-m^2+i0]\times[(p'+k)^2-m^2+i0].
\eqn\OrigD
$$
Using Feynman parameter trick, we re-write the denominator as
$$
{1\over\cal D}\
= \mathop{\intop\!\!\!\!\intop\!\!\!\!\intop}\limits^1_0\!\!
dx\,dy\,dz\,\delta(x+y+z-1)\,{2\over\Bigl[
  x((p+k)^2-m^2)\,+\,y((p'+k)^2-m^2)\,+\,z(k^2)\,+\,i0\Bigr]^3}
\eqno\eq
$$
and then expand
$$
x((p+k)^2-m^2)\,+\,y((p'+k)^2-m^2)\,+\,z(k^2)\
=\ \ell^2\ -\ \Delta
\eqno\eq
$$
where
$$
\ell\ =\ k\,+\,xp\,+\,yp'
\eqn\Ldef
$$
and
$$
\Delta\ =\ (xp+yp')^2\ +\ x(m^2-p^2)\ +\ y(m^2-p^{\prime2})\
=\ (1-z)^2m^2\ -\ xyq^2\quad
\langle\!\langle{\rm on\ shell}\rangle\!\rangle
\eqn\Deldef
$$
Therefore
$$
\Gamma^\mu_{\rm 1\,loop}(p',p)\
=\ -2ie^2\mathop{\intop\!\!\!\!\intop\!\!\!\!\intop}\limits^1_0\!\!
dx\,dy\,dz\,\delta(x+y+z-1) \int_{\rm reg}\!\!{d^4\ell\over (2\pi)^4}\,
{{\cal N}^\mu\over \bigl[\ell^2-\Delta+i0\bigr]^3}\,,
\eqn\monster
$$
and now we need to simplify the numerator~\OrigN\ {\sl in the context of this
monstrous integral}.


The first step is obvious: Let us get rid of the $\gamma^\nu$ and $\gamma_\nu$
factors using the $\gamma$ matrix algebra,
{\it eg,} $\gamma^\nu\!\D a\gamma_\nu=-2\D a$, \etc.
However, in order to allow for the dimensional regularization, we need to
re-work the algebra for an arbitrary spacetime dimension $D$ where
$\gamma^\nu\gamma_\nu=D\neq 4$.
Consequently,
$$
\eqalign{
\gamma^\nu\!\D a\gamma_\nu\ &=\ -2\D a\ +\ (4-D)\D a,\cr
\gamma^\nu\!\D a\D b\gamma_\nu\ &=\ 4(ab)\ -\ (4-D)\D a\D b,\cr
\gamma^\nu\!\D a\D b\D c\gamma_\nu\ &=\ -2\D c\D b\D a\ +\ (4-D)\D a\D b\D c,\cr
}\eqn\Algebra
$$
and therefore
$$
{\cal N}^\mu\
=\ -2m^2\gamma^\mu\ +\ 4m(p'+p+2k)^\mu\
-\ 2(\D p+\D k)\gamma^\mu(\D p'+\D k)\
+\ (4-D)(\D p'+\D k-m)\gamma^\mu(\D p+\D k-m).
\eqn\First
$$

Next, we 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, but the linear terms fo not contribute
to the $\int\!d^D\ell$ integral because they are odd with respect to
$\ell\to-\ell$ while everything else in that integral is even.
Consequently, in the context of eq.~\monster, we may neglect the linear
terms, thus
$$
\!\eqalign{
{\cal N}^\mu\ &
\cong\  -2m^2\gamma^\mu\ +\ 4m(p+p'-2xp-2yp')^\mu\cr
&\quad-\ 2\D\ell\gamma^\mu\!\D\ell\
    -\ 2(\D p-x\D p-y\D p')\,\gamma^\mu\,(\D p'-x\D p-y\D p')\cr
&\quad+\ (4-D)\D\ell\gamma^\mu\!\D\ell\
    +\ (4-D)(\D p'-y\D p'-x\D p-m)\,\gamma^\mu\,(\D p-x\D p-y\D p'-m)\cr
&=\ -2m^2\gamma^\mu\ +\ 4mz(p'+p)^\mu\ +\ 4m(x-y)q^\mu\
    -\ (D-2)\D\ell\gamma^\mu\!\D\ell\cr
&\quad-\ 2(z\D p' +(x-1)\D q)\,\gamma^\mu\,(z\D p+(1-y)\D q)\cr
&\quad+\ (4-D)(z\D p' +x\D q -m)\,\gamma^\mu\,(z\D p-y\D q-m)\cr
}\!\eqn\Intermed
$$
where the second equality here follows from  $p'-p=q$ and $x+y+z=1$.

Now, let make use of the external fermions being on-shell.
This means more than just $p^2=p^{\prime2}=m^2$:
Effectively, we sandwich the vertex $ie\Gamma^\mu$ between Dirac spinors
$\bar u(p')$ on the left and $u(p)$ on the right.
The two spinors satisfy the appropriate Dirac equations, and hence any
term in $\Gamma^\mu$ something${}\times\D p$ is equivalent to
same thing${}\times m$ because $\D p u(p)=mu(p)$, and likewise any term
of the form $\D p'\times{}$something is equivalent to $m\times{}$same thing
because $\bar u(p')\D p'=\bar u(p')m$.
Consequently, the terms on the last two lines of eq.~\Intermed\
are equivalent to
$$
\eqalign{
(z\D p' +(x-1)\D q)\,\gamma^\mu\,(z\D p+(1-y)\D q)\ &
\cong\ (zm+(x-1)\D q)\,\gamma^\mu\,(zm+(1-y)\D q)\cr
(z\D p' +x\D q -m)\,\gamma^\mu\,(z\D p-y\D q-m)\ &
\cong\ ((z-1)m+x\D q)\,\gamma^\mu\,((z-1)m-y\D q).\cr
}\eqno\eq
$$
Let us combine these two expressions with respective coefficients $-2$ and
$4-D$ (\cf~eq.~\Intermed) and group similar terms together.
Making use of
$$
\D q\gamma^\mu\ =\ q^\mu\ +\ i\sigma^{\mu\nu} q_\nu\quad
{\rm and}\quad \gamma^\mu\!\D q\ =\ q^\mu\ -\ i\sigma^{\mu\nu}q_\nu\,,
\eqn\sigmalgebra
$$
we obtain
$$
\eqalign{
m^2\gamma^\mu\times &
\Bigl(-2z^2+(4-D)(1-z)^2\Bigr)\cr
{}+\ \D q \gamma^\mu\!\D q\times &
\Bigl( 2(1-x)(1-y)-(4-D)xy\Bigr)\cr
{}+\ m q^\mu\times &
(x-y)\Bigl( -2z-(4-D)(1-z)\Bigr)\cr
{}+\ im\sigma^{\mu\nu}q_\nu\times &
\Bigl(2z(2-x-y)-(4-D)(1-z)(x+y)\Bigr),\cr
}\eqno\eq
$$
and hence
$$
\eqalign{
{\cal N}^\mu\ &
\cong\ -(D-2)\D\ell\gamma^\mu\!\D\ell\ +\ 4mz(p'+p)^\mu\cr
&\quad+\ m^2\gamma^\mu\times \Bigl(-2-2z^2+(4-D)(1-z)^2\Bigr)\cr
&\quad+\ \D q \gamma^\mu\!\D q\times\Bigl( 2(1-x)(1-y)-(4-D)xy\Bigr)\cr
&\quad+\ m q^\mu\times (x-y)\Bigl(4 -2z-(4-D)(1-z)\Bigr)\cr
&\quad+\ im\sigma^{\mu\nu}q_\nu\times \Bigl(2z(2-x-y)-(4-D)(1-z)(x+y)\Bigr).\cr
}\eqn\BigMess
$$
Furthermore, in the context of the Dirac sandwich $\bar u(p')\Gamma^\mu u(p)$
we have
$$
\D q \gamma^\mu\!\D q\ =\ 2q^\mu\D q\ -\ q^2\gamma^\mu\ \cong\ -q^2\gamma^\mu
\eqno\eq
$$
because $\bar u(p')\D qu(p)=0$, and also the Gordon identity
$$
(p'+p)^\mu\ \cong\ 2m\gamma^\mu\ -\ i\sigma^{\mu\nu}q_\mu\,.
\eqno\eq
$$
Therefore, re-grouping terms and making use of $x+y+z=1$, we obtain
$$
\eqalign{
{\cal N}^\mu\ &
\cong\ -(D-2)\D\ell\gamma^\mu\!\D\ell\
   -\ m^2\gamma^\mu\times \Bigl(2(1-4z+z^2)-(4-D)(1-z)^2\Bigr)\cr
&\quad-\ q^2\gamma^\mu\times\Bigl(2(z+xy)-(4-D)xy\Bigr)
   -\ im\sigma^{\mu\nu}q_\nu\times(1-z)\Bigl(2z+(4-D)(1-z)\Bigr)\cr
&\quad+\ m q^\mu\times (x-y)\Bigl(4 -2z-(4-D)(1-z)\Bigr).\cr
}\eqn\Penult
$$

To further simplify this expression, let us go back to the
symmetries of the integral~\monster.
The integral over the Feynman parameters, the integral $\int\!d^D\ell$,
and the denominator $[l^2-\Delta]^3$ are all invariant under the parameter
exchange $x\leftrightarrow y$.
In~eq.~\Penult\ for the numerator, the first two lines are invariant
under this symmetry, but the last line changes sign.
Consequently, only the first two lines contribute to the integral~\monster\
while the third line integrates to zero and may be disregarded.

Finally, thanks to the Lorentz invariance of the $\int\!d^D\ell$ integral,
$$
\ell_\lambda\ell_\nu\ \cong\ g_{\lambda\nu}\times{\ell^2\over D}\,,
\eqno\eq
$$
and hence
$$
\D\ell\gamma^\mu\!\D\ell\
=\ \gamma^\lambda\gamma^\mu\gamma^\nu\times\ell_\lambda\ell_\nu\
\cong\ \gamma^\lambda\gamma^\mu\gamma^\nu\times
  g_{\lambda\nu}\, {\ell^2\over D}\
=\ -(D-2)\gamma^\mu\times{\ell^2\over D}\,.
\eqn\LGL
$$
Plugging this formula into eq.~\Penult\ and grouping terms according to
their $\gamma$--matrix structure, we arrive at
$$
{\cal N}^\mu\ =\ {\cal N}_1\times\gamma^\mu\
-\ {\cal N}_2\times{i\sigma^{\mu\nu}q_\nu\over 2m}
\eqn\TwoParts
$$
where
$$
\eqalignno{
{\cal N}_1\ &
\cong\  {(D-2)^2\over D}\,\ell^2\ -\ 2(1-4z+z^2)\,m^2\ -\ 2(z+xy)q^2\cr
&\hskip 7em +\ (4-D)((1-z)^2m^2+xyq^2), &
\eqname\None\cr
{\cal N}_2\ &
\cong\ 4z(1-z)m^2\ +\ 2(4-D)(1-z)^2m^2.&
\eqname\Ntwo\cr
}$$
Note that splitting the numerator according to eq.~\TwoParts\
is particularly convenient for calculating the electron's form factors:
$$
\eqalignno{
\Gamma^\mu_{\rm 1\,loop}\ &
=\ F_1^{\rm 1\,loop}(q^2)\times\gamma^\mu\
    +\ F_2^{\rm 1\,loop}(q^2)\times{i\sigma^{\mu\nu}q_\nu\over2m}\,,&
\eqname\FormFactors\cr
F_1^{\rm 1\,loop}(q^2)\ &
=\ -2ie^2\mathop{\intop\!\!\!\!\intop\!\!\!\!\intop}\limits^1_0\!\!
    dx\,dy\,dz\,\delta(x+y+z-1) \int\!\!{d^D\ell\over (2\pi)^D}\,
    {{\cal N}_1\over \bigl[\ell^2-\Delta+i0\bigr]^3}\,,&
\eqname\Fone\cr
F_2^{\rm 1\,loop}(q^2)\ &
=\ +2ie^2\mathop{\intop\!\!\!\!\intop\!\!\!\!\intop}\limits^1_0\!\!
    dx\,dy\,dz\,\delta(x+y+z-1) \int\!\!{d^D\ell\over (2\pi)^D}\,
    {{\cal N}_2\over \bigl[\ell^2-\Delta+i0\bigr]^3}\,.&
\eqname\Ftwo\cr
}$$


\bye
