m^2_{\mu}p\cdot p' - m_{e}^2k\cdot k' + 2 m_e^2m_{\mu}^2]
\end{equation}

where  $q = p + p'$.

To relate to cross-section, we can use the center of mass frame. For the
electron-muon scattering, we may choose

\[
p^{\mu} = (E_e,p\hat{z})\ ,\ k^{\mu} = (E_{\mu}, -p\hat{z}) ~ ~ , ~
~ p^{'\mu} = (E'_e, \vec{p}')\ ,\ k^{'\mu} = (E'_{\mu}, -\vec{p}')
~ , ~ cos(\Theta_{cm}) := \hat{z}\cdot\hat{\vec{p}}' .
\]

Notice that $E_e + E_{\mu} = \sqrt{p^2 + m_e^2} + \sqrt{p^2 + m_{\mu}^2}
= E'_e + E'_{\mu} = \sqrt{(p')^2 + m_e^2} + \sqrt{(p')^2 + m_{\mu}^2}$,
implies that $|p'| = |p|$. The magnitude of the center of mass momentum
and the scattering angle are the only two independent parameters given
the electron and muon masses.

For $e^+e^- \to \mu^+\mu^-$, a good deal of convenience ensues. The
equality of magnitudes of momenta and masses being identical in both
initial and final states, implies that the energies of individual
particles in initial and final state are equal and total energy
conservation implies these energies are equal too. The initial and final
momenta magnitudes are then simply related by the masses. Thus we may

\[
p^{\mu} = (E,p\hat{z})\ ,\ (p')^{\mu} = (E, -p\hat{z}) ~ ~ , ~ ~
k^{\mu} = (E, \vec{k})\ ,\ (k')^{\mu} = (E, -\vec{k}) ~ , ~
cos(\Theta_{cm}) := \hat{z}\cdot\hat{\vec{k}} .
\]

Again only the center of mass energy and the scattering angle are the
only independent parameters given the masses.

It remains to simplify the expression for the cross-sections. We will
give it for the muon production process and leave the $e\mu$ scattering
process as an exercise. We will also neglect the electron mass compared
to the muon mass.

The dot products of momenta, in the center of mass take the form, $q^2 =
(p+p')^2 = - (2E)^2 = -4(k^2 + m^2_e) \approx -4 k^2$. Next, $p\cdot p'
= -E^2 - p^2 \approx -2E^2$ and

\[
p\cdot k = p'\cdot k' = -E^2 + Ekcos(\Theta_{cm})~ ~,~ ~ p\cdot k' =
p\cdot k = -E^2 - Ekcos(\Theta_{cm})\ .
\]

Substituting in the invariant amplitude gives ($m_e = 0$ set),

\begin{eqnarray}
\frac{1}{4}\sum_{spins} |\mathcal{M}|^2_{e^+e^-\to\mu^+\mu^-}  & = &
\frac{8e^4}{(-4E^2)^2}\left\{(E^2 + Ekcos(\Theta_{cm}))^2 + (E^2 -
Ekcos(\Theta_{cm}))^2 - m_{\mu}^2(-2E^2)\right\} \nonumber \\
& = & e^4\left\{ 1 + \frac{k^2 cos^2(\Theta_{cm})}{E^2} +
\frac{m^2_{\mu}}{E^2} \right\} \nonumber \\
& = & e^4\left[ \left(1 + \frac{m^2_{\mu}}{E^2}\right) +  \left(1 -
\frac{m^2_{\mu}}{E^2}\right) cos^2(\Theta_{cm}) \right]
\end{eqnarray}

The differential cross-section is given by (<ref>),

\begin{eqnarray}
\frac{d\sigma}{d\Omega_{cm}} & = &
\frac{1}{64\pi^2}\frac{1}{s}\frac{k'_{cm}}{k_{cm}}\left(\frac{1}{4}\sum_{spins}
|\mathcal{M}|^2\right)  \mbox{\hspace{1.0cm} where, $s = 4E^2,
m_e = 0, m'_1 = m'_2 = m_{\mu}$} ~ ~ \Rightarrow \nonumber \\
& = & \frac{1}{64\pi^2}\frac{1}{s}\sqrt{\frac{s^2 -
4sm_{\mu}^2}{s}}\left[ \left(1 + \frac{m^2_{\mu}}{E^2}\right) +
\left(1 - \frac{m^2_{\mu}}{E^2}\right) cos^2(\Theta_{cm})
\right] \nonumber \\
& = & \left(\frac{e^2}{4\pi}\right)^2 \frac{1}{4\cdot 4E^2}\sqrt{1 -
\frac{m^2_{\mu}}{E^2}}\left[ \left(1 + \frac{m^2_{\mu}}{E^2}\right) +
\left(1 - \frac{m^2_{\mu}}{E^2}\right) cos^2(\Theta_{cm}) \right]
\end{eqnarray}

Putting $\alpha := \Case{e^2}{4\pi}$, the fine structure constant and
$E_{cm} = 2E$, we write the unpolarized, $e^+e^-\to\mu^+\mu^-$
cross-section as,

\begin{eqnarray} \label{MuProdCrossSecns}
\frac{d\sigma}{d\Omega_{cm}}(E,\Theta_{cm}) & = &
\frac{\alpha^2}{4E_{cm}^2}\sqrt{1 - \frac{m^2_{\mu}}{E^2}}\left[ \left(1
+ \frac{m^2_{\mu}}{E^2}\right) + \left(1 - \frac{m^2_{\mu}}{E^2}\right)
cos^2(\Theta_{cm}) \right] \\
\sigma_{total} := \int d\Omega \frac{d\sigma}{d\Omega} & = &
\frac{\pi\alpha^2}{3E_{cm}^2}\sqrt{1 - \frac{m^2_{\mu}}{E^2}} \left(1 +
\frac{m^2_{\mu}}{2E^2}\right)
\end{eqnarray}

Remarks: The square root factor shows that $E \ge m_{\mu}$
must hold for the muon production to proceed. In the ultra-relativistic
limit, $E \gg m_{\mu}$, the differential cross-section has the
characteristic $(1 + cos^2(\Theta))$ angular dependence.

Since the cross-section is a measured quantity and the center of mass
energy is under our control, by comparing the pair production
cross-sections for two different final states, eg $\mu, \tau$, and
taking the ratio, we can obtain bounds on the masses of the heavier

Exercises: Obtain the differential and the total cross-section for
$e\mu \to e\mu$ process. Also for the Moller and the Bhabha processes.

§.§ Compton Scattering:

Recall the total amplitude from the two diagrams (we take the photon
polarizations to be real),

\begin{eqnarray} %HH
i\mathcal{M} & = & \varepsilon_{\nu}(\vec{k}',\lambda')
\bar{u}(p')(ie\gamma^{\nu}) \left[\frac{-i (-(\dsl{p} + \dsl{k}) +
m)}{(p+k)^2 + m^2_e - i\epsilon}\right]
(ie\gamma^{\mu})u(p) \varepsilon_{\mu} (\vec{k},\lambda) \nonumber \\
& & +	\varepsilon_{\nu}(\vec{k},\lambda) \bar{u}(p')(ie\gamma^{\nu})
\left[\frac{-i (-(\dsl{p}' - \dsl{k}) + m)}{(p'-k)^2 + m^2_e -
(ie\gamma^{\mu})u(p) \varepsilon_{\mu} (\vec{k}',\lambda')  \\
\mathcal{M} & = & e^2\varepsilon_{\mu}(\vec{k},\lambda)
\varepsilon_{\nu} (\vec{k}',\lambda') \bar{u}(p')\left[
\frac{\gamma^{\nu}(-(\dsl{p} + \dsl{k}) +
m)\gamma^{\mu}}{(p+k)^2 + m^2_e - i\epsilon} +
\frac{\gamma^{\mu}(-(\dsl{p}' - \dsl{k}) +
m)\gamma^{\nu}}{(p'-k)^2 + m^2_e - i\epsilon}\right]u(p) \nonumber \\
\end{eqnarray}

In the second (crossed diagram) terms, we have used $p - k' = p' -k$
which is more convenient.

Simplification: Since the momenta are on shell, $p^2 = -m^2,
k^2 = 0 = (k')^2$, we get

\[ (p+k)^2 + m^2 = p^2 + m^2 + k^2 + 2p\cdot k = 2p\cdot k ~ ~ , ~ ~
(p'-k)^2 + m^2 = -2p'\cdot k ;
\]

\[
(-\dsl{p} + m)\gamma^{\mu}u(p) = (2p^{\mu} +
\gamma^{\mu}(\dsl{p} + m))u(p) = 2p^{\mu}u(p) ~ ,
\]

\[
\bar{u}(p')\gamma^{\mu}(-\dsl{p}' + m) = \bar{u}(p')(2(p')^{\mu}
+ (\dsl{p}' + m))\gamma^{\mu} = 2(p')^{\mu}\bar{u}(p') ~ .
\]

\begin{eqnarray}
\mathcal{M} & = & e^2\varepsilon_{\mu}(\vec{k},\lambda)
\varepsilon_{\nu} (\vec{k}',\lambda') \bar{u}(p')\left[
\frac{\gamma^{\nu}(2p^{\mu} - \dsl{k}\gamma^{\mu})} {2p\cdot k -
i\epsilon} + \frac{(2(p')^{\mu} + \gamma^{\mu}\dsl{k})\gamma^{\nu})}
{-2p'\cdot k - i\epsilon}\right]u(p) \\
\mathcal{M}^{\dagger} & = & e^2\varepsilon_{\alpha}(\vec{k},\lambda)
\varepsilon_{\beta} (\vec{k}',\lambda') \bar{u}(p)\left[
\frac{(2p^{\alpha} - \gamma^{\alpha}\dsl{k})\gamma^{\beta}}
{2p\cdot k + i\epsilon} + \frac{\gamma^{\beta}(2(p')^{\alpha} +
\dsl{k}\gamma^{\alpha})} {-2p'\cdot k + i\epsilon}\right]u(p')
\end{eqnarray}

In the second equation, we have used

\[
[\bar{u}(p')\Gamma u(p)]^* =
\bar{u}(p)(\gamma^0)\Gamma^{\dagger}\gamma^0u(p') =
\bar{u}(p)\Gamma_{reversed}u(p') .
\]

Here $\Gamma_{reversed}$ has the order of the gamma matrices reversed
due to the $\dagger$ operation.

Summing and averaging over the spins and polarizations the
$|\mathcal{M}|^2$, we get,

\begin{eqnarray}\label{ComptonAmpSquared}
\frac{1}{4}\sum_{\sigma,\sigma',\lambda,\lambda'}|\mathcal{M}|^2 & = &
\frac{e^4}{4}\left[Tr\left\{ \left(   \frac{\gamma^{\nu}(2p^{\mu} -
\dsl{k}\gamma^{\mu})} {2p\cdot k - i\epsilon} +
\frac{(2(p')^{\mu} +
\gamma^{\mu}\dsl{k})\gamma^{\nu}} {-2p'\cdot k -
i\epsilon}\right)(-\dsl{p} + m) \right.\right. \nonumber \\
& & \mbox{\hspace{1.0cm}}\left. \left(\frac{(2p^{\alpha} -
\gamma^{\alpha}\dsl{k})\gamma^{\beta}} {2p\cdot k + i\epsilon} +
\frac{\gamma^{\beta}(2(p')^{\alpha} + \dsl{k}\gamma^{\alpha})}
{-2p'\cdot k + i\epsilon}\right)(-\dsl{p}' + m)\right\}\times
\nonumber\\
& & \mbox{\hspace{1.0cm}} \left. \left\{
\sum_{\lambda}\varepsilon_{\mu}(\vec{k},\lambda)
\varepsilon_{\alpha}(\vec{k},\lambda) \right\} \left\{
\sum_{\lambda'}\varepsilon_{\nu}(\vec{k}',\lambda')
\varepsilon_{\beta}(\vec{k}',\lambda') \right\}\right]
\end{eqnarray}

We have used the completeness relation (<ref>) to
convert the spin sum. We have a trace over strings of $\gamma$ matrices,
but now we also have sums over the photon polarizations.  An explicit
expression can be obtained by introducing an explicit set of 4
orthonormal vectors.

Consider the polarization sum.  Given a $k, k^2 = 0$ we can write
$k^{\mu} = (|\vec{k}|, \vec{k})$.  Introduce another vector $\tilde{k}
:= C(|\vec{k}|, -\vec{k})$ so that $\tilde{k}^2 = 0$. Fix $C$ by
demanding $k\cdot\tilde{k} = -2 \leftrightarrow C = 1/|\vec{k}|^2$. We
have two transverse directions and we take polarizations
$\varepsilon(\vec{k},\lambda)$ along these two directions and mutually
orthonormalized. These are space-like vectors and have no time
component. These 4 vectors, $k, \tilde{k}, \varepsilon(\vec{k},1),
\varepsilon(\vec{k},2)$ are independent and the completeness relation
takes the form,

\[
-\frac{k^{\mu}\tilde{k}^{\nu} + k^{\nu}\tilde{k}^{\mu}}{2} +
\varepsilon^{\mu}(\vec{k},1)\varepsilon^{\nu}(\vec{k},1) +
\varepsilon^{\mu}(\vec{k},2)\varepsilon^{\nu}(\vec{k},2) ~ = ~
\eta^{\mu\nu} \\
\]

Thus we get the sum over photon polarizations as,

\begin{equation}
\sum_{\lambda=1,2} \varepsilon_{\mu}(\vec{k},\lambda)
\varepsilon_{\nu}(\vec{k},\lambda) ~ = ~ \eta_{\mu\nu} +
\frac{k_{\mu}\tilde{k}_{\nu} + k_{\nu}\tilde{k}_{\mu}}{2}
\end{equation}

Claim: The terms containing $k, \tilde{k}$ in the
polarization sum, do not contribute to the $\sum|\mathcal{M}|^2$.

Proof: Observe that we can always express $\mathcal{M} =
\varepsilon_{\mu}(\vec{k},\lambda)M^{\mu}$, by just taking out the
$\varepsilon_{\mu}$. Then $\sum_{\lambda}|\mathcal{M}|^2 =
M^{\mu}(M^*)^{\nu}\{\eta_{\mu\nu} + \Case{1}{2}(k_{\mu}\tilde{k}_{\nu} +
k_{\nu}\tilde{k}_{\mu})\}$. The $k$ terms contracting with $M$ can be
seen by replacing $\varepsilon_{\mu} \to k_{\mu}$. It is more convenient
to write the Fermion propagators using, $\Case{- \dsl{q} + m}{q^2+m^2
-i\epsilon} = \Case{1}{\dsl{q}+m}$ since $(-\dsl{q}+m)(\dsl{q}+m) =
q^2+m^2$. Then,

\begin{eqnarray*}
M|_{\varepsilon_{\mu} = k_{\mu}} & = & e^2\bar{u}(p')
\left[\dsl{\varepsilon}(k') \frac{1}{\dsl{p}+\dsl{k}+m}
\{\dsl{k}+\dsl{p}+m - \dsl{p}-m\} \right. \\
& & \mbox{\hspace{2.0cm}}\left.  + \{\dsl{k}-\dsl{p}'-m + \dsl{p}'+m\}
\frac{1}{\dsl{p}'-\dsl{k}+m} \dsl{\varepsilon}(k') \right]u(p) \\
& = & e^2\left[\bar{u}(p')\dsl{\varepsilon}(k')u(p) -\bar{u}(p')
\dsl{\varepsilon}(k') \frac{1}{\dsl{p}+\dsl{k}+m}(\dsl{p}+m)\})u(p)
\right.\\
& & \mbox{\hspace{1.0cm}}\left.  \bar{u}(p')(-\dsl{\varepsilon}(k'))u(p)
+ \bar{u}(p')(\dsl{p}'+m) \frac{1}{\dsl{p}'-\dsl{k}+m}
\dsl{\varepsilon}(k')u(p)\right]  ~ ~ = 0 ~ !
\end{eqnarray*}

We have used $p -k' = p'-k$ in the second term in the first equation. In
the last equation, the first and the third terms cancel while the second
and the fourth terms vanish by equation of motion. This proves the
claim. Hence, effectively, each of the polarization sums give only the

Note: The total scattering amplitude vanishes when the
photon transverse polarization is replaced by a longitudinal one, is a
general result known as a `Ward identity'. This is a consequence of
gauge invariance. While we do not discuss the general proof here, it
suffices to note that (a) we used the Dirac equation: $(\dsl{p}+m)u(p) =
0 = \bar{u}(p')(\dsl{p}' + m)$; (b) the QED coupling has the form $\sim
A_{\mu}J^{\mu}$ with $J^{\mu} \sim \bar{\Psi}\gamma^{\mu}\Psi \sim
\bar{u}(p')\gamma^{\mu}u(p)$; (c) $\delta A_{\mu} = \partial_{\mu}
\Lambda$ is a gauge transformation ($\leftrightarrow
\varepsilon_{\mu}(k) \to \varepsilon_{\mu}(k) + k_{\mu}\lambda(k)$) and
gauge invariance of the interaction implies $\partial_{\mu}J^{\mu} = 0.$
Thus, the vanishing of the amplitude is related to gauge invariance.

The averaged $|\mathcal{M}|^2$ in equation (<ref>)

\begin{eqnarray}\label{ComptonAmpSquared4}
\sum_{\sigma,\sigma',\lambda,\lambda'}|\mathcal{M}|^2 & = &
e^4\left[Tr\left\{ \left(   \frac{\gamma^{\nu}(2p^{\mu} -
\dsl{k}\gamma^{\mu})} {2p\cdot k - i\epsilon} +
\frac{(2(p')^{\mu} +
\gamma^{\mu}\dsl{k})\gamma^{\nu}} {-2p'\cdot k -
i\epsilon}\right)(-\dsl{p} + m) \right.\right. \nonumber \\
& & \mbox{\hspace{1.0cm}}\left. \left. \left(\frac{(2p_{\mu} -
\gamma_{\mu}\dsl{k})\gamma_{\nu}} {2p\cdot k + i\epsilon} +
\frac{\gamma_{\nu}(2(p')_{\mu} + \dsl{k}\gamma_{\mu})} {-2p'\cdot k +
i\epsilon}\right)(-\dsl{p}' + m)\right\}\right] \\
& = & e^4\left[Tr\left\{ \frac{(2p^{\mu}\gamma^{\nu} -
\gamma^{\nu}\dsl{k}\gamma^{\mu})\dsl{p}(2p_{\mu}\gamma_{\nu} -
\gamma_{\mu}\dsl{k}\gamma_{\nu})\dsl{p}'}{(2p\cdot k)^2}
\right\}_{\circled{1}} ~ ~ + \right. \nonumber \\
& & \hspace{0.75cm} Tr\left\{ \frac{(2(p')^{\mu}\gamma^{\nu} +
\gamma^{\mu}\dsl{k}\gamma^{\nu})\dsl{p} (2p'_{\mu}\gamma_{\nu} +
\gamma_{\nu}\dsl{k} \gamma_{\mu})\dsl{p}'}{(-2p'\cdot k)^2}
\right\}_{\circled{2}} ~ ~ + \nonumber \\
& & \hspace{0.75cm} Tr\left\{ \frac{(2p^{\mu}\gamma^{\nu} -
\gamma^{\nu}\dsl{k}\gamma^{\mu})\dsl{p}(2p'_{\mu}\gamma_{\nu} +
\gamma_{\nu}\dsl{k} \gamma_{\mu})\dsl{p}'}{-4 p\cdot k p'\cdot
k}\right\}_{\circled{3}} ~ ~ + \nonumber \\
& & \hspace{0.75cm} \left. Tr\left\{ \frac{(2(p')^{\mu}\gamma^{\nu} +
\gamma^{\mu}\dsl{k} \gamma^{\nu})\dsl{p} (2p_{\mu}\gamma_{\nu} -
\gamma_{\mu}\dsl{k} \gamma_{\nu})\dsl{p}'} {-4 p\cdot k p'\cdot
k}\right\}_{\circled{4}} \right]  \nonumber \\
& & + m^2 e^4\left[ Tr\left\{ \frac{(2p^{\mu}\gamma^{\nu} - \gamma^{\nu}
\dsl{k}\gamma^{\mu})(2p_{\mu}\gamma_{\nu} -
\gamma_{\mu}\dsl{k}\gamma_{\nu})}{(2p\cdot k)^2} \right\}_{\circled{5}}
~ ~ + \right.  \nonumber \\
& & \hspace{0.75cm} Tr\left\{ \frac{(2(p')^{\mu}\gamma^{\nu} +
\gamma^{\mu}\dsl{k}\gamma^{\nu}) (2p'_{\mu}\gamma_{\nu} +
\gamma_{\nu}\dsl{k} \gamma_{\mu})}{(-2p'\cdot k)^2}
\right\}_{\circled{6}} ~ ~ + \nonumber \\
& & \hspace{0.75cm} Tr\left\{ \frac{(2p^{\mu}\gamma^{\nu} -
\gamma^{\nu}\dsl{k}\gamma^{\mu})(2p'_{\mu}\gamma_{\nu} +
\gamma_{\nu}\dsl{k} \gamma_{\mu})}{-4 p\cdot k p'\cdot
k}\right\}_{\circled{7}} ~ ~ + \nonumber \\
& & \left. \hspace{0.75cm} Tr\left\{ \frac{(2(p')^{\mu}\gamma^{\nu} +
\gamma^{\mu}\dsl{k} \gamma^{\nu}) (2p_{\mu}\gamma_{\nu} -
\gamma_{\mu}\dsl{k} \gamma_{\nu})} {-4 p\cdot k p'\cdot
k}\right\}_{\circled{8}} \right]
\end{eqnarray}

The terms linear on $m$ have odd number of $\gamma$'s and hence vanish
under trace.

It may be checked easily that the $\circled{2}$ is obtained from
$\circled{1}$ by $p \leftrightarrow - p'$ and likewise for $5
\leftrightarrow 6$. Noting the identity
$Tr(\gamma_1\gamma_2\cdots\gamma_n) =
Tr(\gamma_n\gamma_{n-1}\cdots\gamma_2\gamma_1)$, it follows that $3 = 4$
and $7 = 8$.  The first group of 4 terms involves trace of a maximum of
8 gamma matrices while the last 4 terms involve a maximum of 6 gamma's.
These are simplified using various identities among the gamma matrices.

Consider $\circled{1}$. We have the traces,

\[
Tr\left\{ 4p^2\gamma^{\nu}\dsl{p}\gamma_{\nu}\dsl{p}'
\right\}
\]

Use: $\boxed{\gamma^{\nu}\dsl{p}\gamma_{\nu} = +2\dsl{p}, ~
\dsl{p}\dsl{p} = -p^2, ~ \dsl{k}\dsl{p} + \dsl{p}\dsl{k} = -2p\cdot k, ~
Tr\dsl{p}\dsl{p}' = -4p\cdot p'}$ and the cyclic property of the trace
to get the above trace as,

\begin{eqnarray}
Tr\{\cdots\}_{\circled{1}} & = & (4p^2)(2)(-4p\cdot p') -
2p^2(2)(-4k\cdot p')\times 2 +
(2)(2)Tr(\dsl{k}\dsl{p}\dsl{k}\dsl{p}') \nonumber \\
& = & 32 m^2(p\cdot p' - k\cdot p') + 32(k\cdot p)(k\cdot p')
\end{eqnarray}

We have used $p^2 = -m^2$ and $k^2 = 0$.

Similarly we get the other traces (without the denominator factors) as,

\begin{eqnarray}
Tr\{\cdots\}_{\circled{3}} & = & 4p\cdot p'Tr(\gamma^{\nu} \dsl{p}
\gamma_{\nu} \dsl{p}') - 2 Tr(\dsl{k}\dsl{p}'\dsl{p}
\gamma_{\nu}\dsl{p}'\gamma^{\nu}) + 2 Tr(\gamma^{\nu} \dsl{p}
\gamma_{\nu} \dsl{k}\dsl{p}\dsl{p}') \nonumber \\
& & ~ ~ ~ ~ - Tr(\gamma^{\nu}\dsl{k}\gamma^{\mu}\dsl{p}\gamma_{\nu}
\dsl{k}\gamma_{\mu}\dsl{p}') \\
& = & -32(p\cdot p')^2 + (k\cdot p - k\cdot p') (16m^2 + 32 p\cdot p') +
0 \\
Tr\{\cdots\}_{\circled{5}} & = & 4p^2 Tr(\gamma^{\nu}\gamma_{\nu}) -2
Tr(\dsl{k}\dsl{p}(-4\mathbb{1})) +8 Tr(\dsl{p}\dsl{k}) + 16Tr(\dsl{k}
\dsl{k}) \nonumber \\
& = & 64 m^2 - 64 k\cdot p \\
Tr\{\cdots\}_{\circled{7}} & = & -64p\cdot p' + 32(k\cdot p - k\cdot p')
- Tr(\gamma^{\nu}\dsl{k}\gamma^{\mu}\gamma_{\nu}\dsl{k}\gamma_{\mu})
\nonumber \\
& = &  -64p\cdot p' + 32(k\cdot p - k\cdot p') + 0
\end{eqnarray}

To simplify the expression further, we need to use: $p\cdot p' = -m^2 +
k\cdot p - k\cdot p'$ which follows from $0 = (k')^2 = (p + k - p')^2$.
For comparison with [13], it is useful to note: $k\cdot
p = k'\cdot p'\ ,\ p\cdot k' = p'\cdot k$.

The final expression is,

\begin{equation}\label{ComptonAmpSquared2}
\frac{1}{4}\sum_{spin/Pol}|\mathcal{M}|^2 = 2e^4\left[\left(
\frac{p\cdot k'}{p\cdot k} + \frac{p\cdot k}{p\cdot k'}\right) +
2m^2\left( \frac{1}{p\cdot k'} - \frac{1}{p\cdot k}\right) +
m^4\left(\frac{1}{p\cdot k} - \frac{1}{p\cdot k'}\right)^2\right]
\end{equation}

Exercise: Check the algebra!

The Compton cross-section is usually presented in the lab frame with the
electron initially at rest, i.e. $p^{\mu} = (m,\vec{0}), k^{\mu} =
(\omega, \omega\hat{z}), (p')^{\mu} = (E',\vec{p}]), (k')^{\mu} =
(\omega', \omega'sin{\theta}, 0, \omega'cos(\theta))$. The final
electron and photon momenta define a plane which is taken to be the
$z-x$ plane with the final photon making an angle $\theta$ to the
$z-$axis. These choices give: $p\cdot k' = -m\omega', p\cdot k = -
m\omega$ and,

\begin{equation}\label{ComptonAmpSquared3}
\sum_{spin/Pol}|\mathcal{M}|^2 = 2e^4\left[ \frac{\omega'}{\omega} +
\frac{\omega}{\omega'} + 2m\left(- \frac{1}{\omega'} +
\frac{1}{\omega}\right) + m^2\left(-\frac{1}{\omega} +
\frac{1}{\omega'}\right)^2\right]
\end{equation}

Next, $(p')^2 = (p + k - k')^2$ implies,

\[
-m^2 = -m^2 + 2p\cdot(k-k') + (k-k')^2 \Rightarrow 0 =
m(\omega-\omega') - \omega\omega'(1-cos(\theta)) ,
\]

leading to the Compton formula for shift in the photon wavelength with
the scattering angle:

\begin{equation}
\mbox{Compton Formula:~ ~} \boxed{\frac{1}{\omega'} -
\frac{1}{\omega} = \frac{1-cos\theta}{m}} ~ \leftrightarrow ~
\omega' = \frac{\omega}{1+\Case{\omega}{m}(1-cos\theta)} \ .
\end{equation}

To get the differential cross-section, we simplify the phase space
integral as,

\begin{eqnarray*}
\int \frac{d^3k'}{2\omega'(2\pi)^3}
\int\frac{d^3p'}{2E'(2\pi)^3}(2\pi)^4 \delta^4(p;+k'-p-k) =
\mbox{\hspace{3.0cm}} & & \\
\int \frac{d\omega'
\end{eqnarray*}

We have used up a $\delta^3$ to get $\vec{p}' = \vec{k}' - \vec{k} -
\vec{p} = (\omega'sin\theta, 0, \omega'cos\theta - \omega)$ which gives
$(E')^2 - (\omega')^2 + \omega^2 - 2\omega'\omega cos\theta + m^2$.

\[
\int d\omega'\frac{\omega'}{4E'(\omega')}\delta(\omega' +
E'(\omega') - m - \omega) =
\frac{\omega'_*}{4E'(\omega'_*)}\frac{1}{|1+\Case{dE'}{d\omega'_*}|}
~ ~,~ ~ \frac{dE'}{d\omega'} = \frac{\omega'-\omega
cos\theta}{E'} \ .
\]

Inserting in the phase space integral,

\begin{eqnarray}
\int_{\Gamma_2} & = & \int \frac{dcos\theta}{2\pi}
\frac{\omega'_*}{4E'_*}\frac{1}{|1 + \case{\omega'-\omega
cos\theta}{E'}|_*}  ~ ~ , ~ ~ \omega'_* + E'(\omega'_*) = m + \omega
\nonumber \\
& = & \frac{1}{8\pi}\int dcos\theta \frac{\omega'^2}{m\omega} \ . ~ ~
\mbox{\hspace{2.0cm} We have,} \\
\frac{d\sigma}{dcos\theta} & = &
\frac{1}{4m\omega}\frac{1}{8\pi}\frac{(\omega')^2}{m\omega}[\sum
|\mathcal{M}|^2 ]   \nonumber \\
& = &
\frac{e^4}{16\pi}\frac{1}{m^2}\left(\frac{\omega'}{\omega}\right)^2
\left[\frac{\omega'}{\omega} + \frac{\omega}{\omega'} + \left\{2m
\frac{-1+cos\theta}{m} + m^2 \frac{(1-cos\theta)^2}{m^2}\right\}\right]
\nonumber
\end{eqnarray}

This give the differential cross-section for the Compton scattering,
known as Klein-Nishina formula,

\begin{equation}
\boxed{\left(\frac{d\sigma}{dcos\theta_{lab}}\right) =
\frac{\pi\alpha^2}{m^2} \left(\frac{\omega'}{\omega}\right)^2
\left[\frac{\omega'}{\omega} + \frac{\omega}{\omega'} -
sin^2\theta_{lab}\right]\ , \ \mbox{with}\
\frac{\omega'}{\omega} = \frac{1}{1 +
\Case{\omega}{m}(1-cos\theta)}  .}
\end{equation}

(a) There has been no approximation at the $\alpha^2$ level computation. It
can now be used for various limits. At $\theta = 0$, forward scattering,
there is no frequency change and the cross-section equaling
$\Case{\pi\alpha^2}{m}$ is independent of the photon energy. For a massive charged particle, one defines its Compton length to be
$\lambda_c := h/(mc) = 2\pi/m$ in the natural units, $\hbar = 1 = c$.
The forward scattering cross-section is then $\alpha^2(\pi
\cross{\lambda}^2)$, where $\cross{\lambda} := \lambda/(2\pi)$.

(b) There are two scales in the problems: $m, \omega$. This gives two
natural limits: (i) $\omega \to 0 \leftrightarrow \omega/m \ll 1$ and
(ii) $\omega \to \infty \leftrightarrow \omega/m \gg 1$.

For a generic $\theta$, consider $\omega \to 0$. Then $\omega/\omega'
\to 1 + (\omega/m)(1-cos\theta)$ and

\begin{equation}\label{ThomsonCrossSection}
\left. \frac{d\sigma}{dcos\theta}\right|_{\omega/m \ll 1} =
\frac{\pi\alpha^2}{m^2}\left[1 + cos^2\theta - 4sin^2(\theta/2)\
\frac{\omega}{m}  + \dots\right]  ~ ,  ~
\sigma_{tot} = \frac{8\pi\alpha^2}{3m^2} + \dots
\end{equation}

The leading term is the Thomson cross-section for classical
radiation scattering off free electrons.

In the opposite limit, $m/\omega \ll 1$, we have

\[
\omega'/\omega = \frac{m}{\omega}\frac{1}{1-cos\theta}\left(1 -
\frac{m}{\omega}\frac{1}{1-cos\theta} + \dots\right) ~ := ~
\epsilon(1-\epsilon \dots)~ ~,~ ~ \epsilon(\theta) :=
\frac{m}{\omega}\frac{1}{1-cos\theta} .
\]

In this limit,

\begin{eqnarray}
\frac{d\sigma}{dcos\theta} & \approx &
\frac{\pi\alpha^2}{m^2}\epsilon(\theta) ~ = ~
\frac{\pi\alpha^2}{m^2}\frac{m}{\omega(1-cos\theta)} , \\
\sigma_{tot}(\delta) := \int_{-1}^{1-\delta}dcos\theta
\frac{d\sigma}{dcos\theta} & \approx &
\frac{\pi\alpha^2}{m\omega}\left[ -\ell n(\delta/2) \right] ~
\to \infty\ , \ \mbox{as $\theta\to 0$}.
\end{eqnarray}

For the $\epsilon(\theta) \ll 1$ condition to hold so that the expansion
is meaningful, $\theta \gg \sqrt{2m/\omega}$ must hold. At $\theta = 0$
we know the exact answer which is finite.

§.§ Electron-Positron Annihilation:

Having evaluated the Compton scattering amplitude, this is actually
simple to evaluate. From the table giving the amplitudes (notice the use
of $p'-k$ and $k-p'$ momenta in the crossed diagrams), we can see that
the annihilation diagrams expressions are obtained by making the
substitutions: $\boxed{p \to p, k \to -k, p' \to -p', k' \to k' \
\mbox{and}\ \bar{u} \to \bar{v} .}$ These substitutions preserve the
momentum conservations appropriate for the processes: $p+k = p'+k' \to
p-k = -p'+k'$. The photon polarizations are insensitive to the `sign' of
momentum and will give $\eta_{\mu\nu}$ as before. One of the spin sums
however changes from $[-\dsl{p}'+m] \to [-(-\dsl{p}'+m)] = [\dsl{p}-m]$
which generates an overall minus sign in the summed, squared
amplitude. This is actually an example of what is known as the crossing symmetry of the $S-$matrix: A scattering amplitude for a
particle with momentum $k$ in the initial state is the same as the
amplitude with the initial particle moved to the final state
anti-particle with momentum $-k$.

Without any explicit calculation we can write down the summed squared
amplitude for the annihilation process as,

\begin{equation}\label{AnnihilationAmpSqured}
\sum_{spin/Pol}|\mathcal{M}|^2 = (-)2e^4\left[-\left( \frac{p\cdot
k'}{p\cdot k} + \frac{p\cdot k}{p\cdot k'}\right) + 2m^2\left(
\frac{1}{p\cdot k'} + \frac{1}{p\cdot k}\right) +
m^4\left(\frac{1}{p\cdot k} + \frac{1}{p\cdot k'}\right)^2\right]
\end{equation}

It is more convenient to express the cross-section in the center of mass

\begin{eqnarray*}
p^{\mu} = (E,p\hat{z}) & , & (p')^{\mu} = (E, -p\hat{z}) ~ , ~ E^2 = p^2
+ m^2 ~ ~,~ ~ s = -(p+p')^2 = 4E^2 \\
k^{\mu} & = & (E, Esin\Theta, 0, Ecos\Theta) ~ , ~ (k')^{\mu} = (E,
-Esin\Theta, 0, -Esin\Theta)
\end{eqnarray*}

This gives,

\begin{eqnarray*}
p\cdot k & = & -E^2 + pEcos\Theta = -E(E - pcos\Theta) ~ ,\\
p\cdot k' & = & -E^2 - pEcos\Theta = -E(E + pcos\Theta) , \\
\frac{|k'|_{cm}}{|k|_{cm}} & = & \sqrt{\frac{s^2 -0 + 0^2}{s^2 -2s(2m^2)
+ 0}} ~ = ~ \frac{1}{1-m^2/E^2} ~ = ~ \frac{E}{\sqrt{E^2 - m^2}} =
\frac{E}{p} \\
\end{eqnarray*}

The differential cross-section takes the form,

\begin{eqnarray}
\frac{d\sigma}{d\Omega_{cm}} & = & \frac{1}{64\pi^2} \frac{1}{4E^2}
\frac{E}{|p|} \left(\sum_{spin/Pol}|\mathcal{M}|^2\right)
\mbox{\hspace{2.0cm}with} \nonumber \\
\sum_{spin/Pol}|\mathcal{M}|^2 & = & (-2e^4)\left[
-\left(\frac{E+pcos\Theta}{E-pcos\Theta} +
\frac{E-pcos\Theta}{E+pcos\Theta}\right) \right. \\
& & \mbox{\hspace{1.5cm}} + \frac{2m^2}{-E}\left( \frac{1}{E+pcos\Theta}
+ \frac{1}{E-pcos\Theta}  \right) \\
& & \left. \mbox{\hspace{3.0cm}} + \frac{m^4}{E^2} \left(
\frac{1}{E+pcos\Theta} + \frac{1}{E-pcos\Theta} \right)^2 \right]
\nonumber \\
& = & 4e^4\left[ \frac{E^2+p^2cos^2\Theta}{E^2-p^2cos^2\Theta} +
\frac{2m^2}{E^2-p^2cos^2\Theta} - \frac{2m^4}{(E^2-p^2cos^2\Theta)^2}
\right]
\end{eqnarray}

The differential cross-section for unpolarized, $e^+e^-$ annihilation
process is then given by,

\begin{equation}
\boxed{	\frac{d\sigma}{d\Omega_{cm}}  = \frac{\alpha^2}{s}\frac{E}{|p|}
\left[ \frac{E^2+p^2cos^2\Theta+2m^2}{m^2+p^2sin^2\Theta} -
\frac{2m^4}{(m^2+p^2sin^2\Theta)^2} \right]  ~ ~ , ~ ~ s = 4E^2\ .}
\end{equation}

In the high energy limit, $E \gg m$, this reduces to,

\begin{equation}
\frac{d\sigma}{d\Omega_{cm}} ~ \to ~ \frac{\alpha^2}{s}\left(
\frac{1+cos^2\Theta}{sin^2\Theta} \right)
\end{equation}

To obtain the total cross-section, we need to integrate over the final
state. Since the two photons a identical, $\Theta$ needs to be
integrated between $0, \pi/2$. The Bose factor of 2 (which we did not
include in the amplitude) cancels the factor of 2 from the angular
integration. We get,

\begin{equation}
\sigma_{tot} ~ = ~ \frac{\alpha^2}{s}(2\pi)\int_{0}^{1}dx
\frac{1+x^2}{1-x^2}
\end{equation}

The integral is clearly divergent from the `forward' direction ($\Theta
= 0$). This is artificial since we set $m = 0$ in the integrand. Split
the integration as $\int_0^{1-\Delta} + \int_{1-\Delta}^1$. The first
term is a finite number and this contribution falls off as $s^{-1}$. For
the second term, we go back to the exact differential cross-section,
approximate the integrand for $1-\Delta \le cos(\Theta) \le 1$. Then,
the factors of $E$ all cancel and second term $\sim \alpha^2 (1 -
\Delta)m^{-2}$ where, $\Delta \ll m^2/p^2$ so that the denominator can
be approximated. Thus, the total cross-section does not diverge
but is bounded by $m^{-2}$. This is an example of bounds satisfied by
total cross-sections at very high energies.

§ NUMERICAL ESTIMATES OF CROSS-SECTIONS AND

We have obtained examples of cross-sections for some of the basic
processes in quantum electrodynamics. We also saw how they relate to
non-relativistic versions of the processes eg potential scattering and
inferred the modification due to relativistic effects as well as
estimate `strengths' of interactions. Before we study the higher order
corrections, it is useful to have a sense of order of magnitudes of
cross-sections and how these numbers are used in applications. We will
focus on the QED processes.

What is the value of the fine structure constant?

This is related to the electric charge, the Planck constant due to
quantum framework and the speed of light due to special relativity. We
first obtain the expression in terms of “engineering units" and then
use the measured values to compute the fine structure constant.

Observe that the Feynman rules are derived from the Lagrangian density
which has properly normalized kinetic (quadratic) terms,
$-\Case{1}{4}F_{\mu\nu}F^{\mu\nu} + i\bar{\Psi} \gamma^{\mu}
\partial_{\mu}\Psi - m\bar{\Psi}\Psi$ and the interaction term
introduces `$e$' as $e\bar{\Psi}\gamma^{\mu}\Psi A_{\mu}$ and thus is
dimensionless. This definition gives the equation of motion,
$\partial_{\mu}F^{\mu\nu} = -e \bar{\Psi}\gamma^{\nu}\Psi
\leftrightarrow \partial_{i}E^i = e\Psi^{\dagger}\Psi$, using the
identifications: $F^{0i} := E^i, \partial_{\mu}F^{\mu\nu} = -J^{\nu},
J^{\mu} = (\rho, J^i)$.  Furthermore, the Hamiltonian and hence the
energy density is $\vec{E}^2/2$.

The MKSA (SI) units has the Gauss law in the form: $\vec{\nabla} \cdot
\vec{E}_{SI} = \rho_{SI}/\epsilon_0$. The field energy density (as
inferred from the Poynting theorem) is $\epsilon_0\vec{E}_{SI}^2/2$.
Comparing the energy densities, we introduce the identification, $E :=
\sqrt{\epsilon_0}E_{SI}$ Then $\rho = \vec{\nabla}\cdot \vec{E} =
\sqrt{\epsilon_0}\vec{\nabla}\cdot \vec{E}_{SI} := \sqrt{\epsilon_0}
\Case{\rho_{SI}}{\epsilon_0}$. This leads to the identification,
$\boxed{e := \Case{e_{SI}}{\sqrt{\epsilon_0}}. }$ Thus, we identify the
variables used in the action with those used in engineering variables by
comparing the equations of motion and the energy densities.

In the natural units, $\hbar = 1 = c$ used in writing the action, the
fine structure constant was defined as $\alpha := e^2/(4\pi)$ which is
dimensionless as noted above. Substituting $e$ in terms of $e_{SI}$ and
introducing $\hbar, c$, we write $\alpha := \Case{e^2_{SI}}
{4\pi\epsilon_0} \hbar^{a}c^b$ and determine the powers $a,b$ by
requiring $\alpha$ to be dimensionless. For this, we need the dimensions
of $e_{SI}$ and $\epsilon_0$! Now we also use the SI expression of the
Lorentz force, $F = e_{SI}E_{SI}$. Hence, dimensionally
(Force)$^2$/(energy density) $\sim e^2_{SI}/\epsilon_0 \sim
M^1L^3T^{-2}$. This and $\alpha$ being dimensionless gives $a = b = -1$
and $\boxed{\alpha = \Case{e^2_{SI}}{4\pi\epsilon_0\hbar c} \simeq
1/137\ ,}$ using the values: $(4\pi\epsilon_0)^{-1} = c^210^{-7}, e_{SI}
= 1.6\times 10^{-19}$ Coulomb, $\hbar \simeq 1.05\times 10^{-34}$
Joules.sec and $c \simeq 3\times 10^8$ m/sec.

An Aside: Another set of units are commonly used in gravitational
physics, the so called geometrized units. These are defined by
setting $G = c = 1$. The choice of $c=1$ gives: 1 sec = 3$\times 10^{8}$
meters. Newton's constant in SI units is, $G \approx 6.67\times 10^{-11}
kg^{-1} meter^3 sec^{-2}$. Hence setting $G = 1$ along with $c=1$, gives
1 kg = $7.4\times 10^{-28}$ meters.

It is conventional in gravitational physics to use the Gauss law in the
form $\vec{\nabla}\cdot\vec{E} = 4\pi\rho$ and the electric field energy
density as $E^2/(8\pi)$. This gives the identifications: $E_{geom} =
\sqrt{4\pi\epsilon_0} E_{SI}\ ,\ q_{geom} = q_{SI}/\sqrt{4\pi\epsilon_0}
$.  As before, the Lorentz force equation gives the dimensions as:
$[q^2_{SI}{\epsilon_0}] = [4\pi q^2_{geom}] = M^1 L^3 T^{-2} = L^2$.

Setting $q_{geom} = \Case{q_{SI}}{\sqrt{4\pi\epsilon_0}} G^{\alpha}
c^{\beta}$ must have dimensions of $L$, we infer $\alpha = 1/2$ and
$\beta = -2$ and $\boxed{q_{geom} = \Case{q_{SI}}{\sqrt{4\pi\epsilon_0}}
G^{1/2}c^{-2} = q_{SI}\sqrt{c^2 \times 10^{-7}} G^{1/2}c^{-2}
\approx (8.6\times 10^{-18}) q_{SI}}$ meters.

Looking at the expressions for the cross-sections and noting that in the
natural units energy $\sim$ length$^{-1}$, we see that the
cross-sections are of the form $\alpha/E^2$. This gets more and more
accurate at ultra relativistic energies where masses can be neglected.
Since the cross-section has dimensions of area while energies are
typically given in MeV/GeV/TeV etc, it is useful to have a conversion
between energy units and length units. As already notes, $c = 1
\Rightarrow 1$ second $ = 3\times 10^8$ meters. $\hbar = 1 \Rightarrow
1$ Joule $ = 10^{34} sec^{-1} \simeq 3.3\times 10^{25} meter^{-1}$.
Next, $\boxed{100 MeV = 1.6\times 10^{-19+8}J \simeq 5\times 10^{14}
meter^{-1}}$ . (This is also expressed as $200 MeV.fermi = 1$). Clearly,
a cross-section at 100 MeV (electron mass being 0.5 Mev) is
approximately equal to $\boxed{\sigma_{100Mev} \simeq \alpha^2/E^2
\simeq 2\times 10^{-33} m^2.}$ The conventional unit for these
cross-sections is 1 barn := 100 fermi$^2$ = $10^{-28}m^2$. Typical
numbers encountered in high energy processes (around 1 GeV) dominated by
strong interactions $\sim 10^{-30}m^2$, electromagnetic interactions
$\sim 10^{-36}m^2$ and weak interactions $\sim 10^{-42}m^2$.

So, scattering experiments will provide us with such numbers. How are
they useful?

In many applications, say passage of a collimated set of particles
through some medium, we have the estimate of the cross-section of
scattering of individual particles comprising the `beam' and the medium.
Intuitively, a cross-section may be viewed as a target disk of area
$\sigma$ which is bombarded by `marbles' which get reflected from the
disk. They are reflected if they hit the disk or pass by if not. Thus,
$\sigma$ gives the likelihood of a scattering interaction. Suppose we
have multiple targets (medium) and multiple beam particles. The
probability of interaction is clearly proportional to the ratio of the
effective area of the target particles and the area exposed to the beam

Thus, probability of scattering = (number of target particles exposed o
the beam)$\times \sigma$/Total area presented by the medium). Let $n$ be
the number of target particles exposed per unit area perpendicular to
the beam. Then the number of particles exposed = n$\times$ total area
and the $\boxed{\mbox{probability of scattering\ } = n\cdot\sigma.}$

Let $\rho$ be the number density of the target particles i.e. number per
unit volume. Let $d$ be the thickness of the medium. Then the areal
density $n = \rho\cdot d$. The probability of scattering is $1 n\sigma =
\rho\cdot d\cdot\sigma$ for $d = \hat{d} = (\rho\cdot\sigma)^{-1}$.  Let
$\langle v\rangle$ be the average speed of the beam particles. Then the
average reaction time :=  $\tau_r := \hat{d}/\langle v\rangle =
1/(\rho\langle v\rangle\sigma)$.

Thus, if we have a confined plasma of say electrons and we bombard it
with a beam of positrons and look for production of $\mu^{\pm}$ pairs,
then we should expect to wait for a time $\tau = 1/(\rho\langle
v\rangle\sigma_{e^+e^-\to\mu^+\mu^-})$. This type of use of microscopic
cross-sections is required in discussing bulk processes eg in
nuclear physics or astrophysics	and are useful in estimating
thermalization times.

Sometimes, we aren't interested in tracking individual processes, any
interaction may suffice. In such cases we need to sum over all
possible final states, $\sigma_{tot} = \sum_f \sigma_{i\to f}$. The
$\sigma_{i\to f} \sim |\mathcal{M}_{i\to f}|^2$. We already have
expressions using phase space integration for fixed number of final
state particles. Now we have to sum over the number of particles and the

\[
\sigma_{tot} = \frac{1}{4E_kE_p}v_{lab}\sum_{f,\alpha}\int
\prod_{i=1}^{N_f} \frac{d^3k'_j}{2\omega_{k'}(2\pi)^3} (2\pi)^4
\delta^4(\Sigma k' - P_{in})|\mathcal{M}|^2\ .
\]

This can be related to the forward scattering amplitude via the
“Optical Theorem”.

We have $S^{\dagger} S = \mathbb{1} \Rightarrow \sum_f\langle
i|S^{\dagger}|f\rangle\langle|S|i\rangle = 1 = \sum_f|S_{fi}|^2$.
Substituting for $S_{fi}$ we get,

\[
|S_{fi}|^2 = \delta_{fi}\delta_{fi} + (2\pi)^4\delta^4(P_f-P_i)
\delta_{fi} [i (\mathcal{M}_{fi} - \mathcal{M}_{fi}^*)] +
(2\pi)^8[\delta^4(P_f-P_i)]^2|\mathcal{M}_{fi}|^2 \ .
\]

Summing over $f$ and using unitarity gives,

\begin{equation*}
(2\pi)^4\sum_f\delta^4(P_f-P_i)\delta_{fi}[i(\mathcal{M}_{fi} -
\mathcal{M}_{fi}^*)] = -
\end{equation*}

\begin{equation} \label{OpticalTheorem}
\boxed{ 2 Im(\mathcal{M}_{ii}) = (2\pi)^4\sum_f\delta^4(P_f-P-i)
|\mathcal{M}_{fi}|^2 . }
\end{equation}

Substituting the rhs in the expression for $\sigma_{tot}$ gives,

\[
\sigma_{tot} = \frac{2Im(\mathcal{M}_{ii})}{4k\sqrt{s}} ~ ~ , ~ ~
\mbox{where, $k = k_{CM}$ and $s = -(\sum P_i)^2$ \ .}
\]

The $\mathcal{M}_{ii}$ is the forward scattering, $i \to i$, amplitude and $\sigma_{tot}$ is the inclusive cross-section for $i \to
\mbox{any} f$ processes. This could also be used to estimate the total

Remark: Let us note that we defined the scattering matrix in
the basis of number states or Fock basis. Further, we chose the
idealization of plane waves for both in and out states. This is
convenient but looses the information about spatial locations of
the scattering particles. This also looses any notion of “impact
parameter” - how far the projectile particle is from the target
particle or equivalently the `orbital angular momentum' of the
projectile about the target particle. The information about the
transverse position is of course contained in the parameters of the
outgoing particles. This should be kept in mind while interpreting some
of the divergences of cross-sections.

§ RADIATIVE CORRECTIONS IN QED

Consider now corrections due to higher powers of interaction
Hamiltonian, also called radiative corrections. Having seen the
correspondence between the correction terms and Feynman diagrams, we can
list the corrections directly in terms of the diagrams. A Feynman
diagram is made up of lines (free propagators) and vertices. In QED,
there are two types of propagators and one types of vertex. So consider
corrections to two point functions and the three point function.

§.§ The fermion propagator: self-energy

\begin{eqnarray}
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=0.75of a,blob] (b){All} ;
%			\vertex [right=0.75of a] (b){\circled{G}} ;
\vertex [right=1.0of b] (c) ;
\diagram* {
(a) -- [fermion] (b) -- [fermion]
\end{feynman}
\end{tikzpicture}
& = &
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=of a] (b) ;
\diagram* {
a -- [fermion] b,
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right =0.75of a] (b) ;
\vertex [right =0.75of b] (c) ;
\vertex [right =0.75of c] (d) ;
\diagram* {
(a) -- [fermion] (b),
(b) -- [fermion] (c),
(c) -- [fermion] (d),
(b) -- [photon,half right,looseness=1.5]
(b) -- [photon,half
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=0.6of a] (b) ;
\vertex [right=0.6of b] (c) ;
\vertex [right=0.6of c] (d) ;
\vertex [right=0.6of d] (e) ;
\vertex [right=0.6of e] (f) ;
\diagram* {
(a) -- [fermion] (b),
(b) -- [fermion] (c),
(c) -- [fermion] (d),
(d) -- [fermion] (e),
(e) -- [fermion] (f),
(b) -- [photon,half right,looseness=1.5]
(c) -- [photon,half left,looseness=1.5]
\end{feynman}
\end{tikzpicture}
~ + ~
\nonumber \\
& &
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=0.75of a] (b) ;
\vertex [right=0.75of b] (c) ;
\vertex [right=0.75of c] (d) ;
\vertex [right=0.75of d] (e) ;
\vertex [right=0.75of e] (f) ;
\diagram* {
(a) -- [fermion] (b),
(b) -- [fermion] (c),
(c) -- [fermion] (d),
(d) -- [fermion] (e),
(e) -- [fermion] (f),
(b) -- [photon,half right,looseness=1.5]
(d) -- [photon,half left,looseness=1.5]
\end{feynman}
\end{tikzpicture}
+ \cdots \\
& = &
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=of a] (b) ;
\diagram* {
(a) -- [fermion] (b),
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=0.75of a,blob] (b){1PI} ;
\vertex [right=1.0of b] (c) ;
\diagram* {
(a) -- [fermion] (b) -- [fermion] (c),
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=0.75of a,blob] (b){1PI} ;
\vertex [right=1.25of b,blob] (c){1PI} ;
\vertex [right=1.0of c] (d){ } ;
\diagram* {
(a) -- [fermion] (b) -- [fermion] (c) --
[fermion] (d),
\end{feynman}
\end{tikzpicture}
~ + ~ \cdots
\end{eqnarray}

Here is a graphical expansion of the exact two point function of the
fermions. In the second line, the diagrams have been grouped into a
series of terms involving 1-particle irreducible (1PI) blobs,
connected by free propagators. The 1PI blob represents the sum of all
diagrams with two external legs and which cannot be disconnected by
removing a single fermion line. The 2nd and the 3rd diagrams in the
first line are 1IP (their extreme propagators being the external lines)
while the last diagram is 1-particle reducible. The second line follows
from first one by inspection.

Let $iS'_F(p)$ denote the lhs - the full 2-point function or full
propagator - and $iS_F(p)$ denote the free propagator. Let the 1PI blob
be denoted by $i\Sigma(p)$. Then the series is:

\begin{eqnarray*}
-iS'_F(p) & = & -iS_F(p) + -iS_F\cdot i\Sigma(p)\cdot -iS_F +
-iS_F\cdot i\Sigma\cdot-iS_F\cdot i\Sigma\cdot -iS_F + \dots \\
& = & -iS_F + -iS_F\cdot i\Sigma(-iS_F - -iS_F\cdot i\Sigma\cdot
-iS_F \dots ) \\
\therefore -iS'_F & = & -iS_F + -iS_F\cdot i\Sigma\cdot -iS'_F
\\
\therefore S_F & = & (1 - S_F\Sigma)S'_F ~ \leftrightarrow ~
\mathbb{1} = S^{-1}_F( - \Sigma)S'_F ~ ~ ~ ~ \mbox{and} ~ ~
S_F^{-1} = \dsl{p} + m , \\
\therefore & & \boxed{S_F'(p) = \frac{\mathbb{1}}{\dsl{p} + m -
\Sigma(p)} ~ , ~ ~ ~ \mbox{$\Sigma(p)$ is called the fermion
self energy.}}
\end{eqnarray*}

The $S'_F(p), \Sigma(p)$ are matrices with the Dirac spinor indices
though we may not always be explicit about it. What can we say about its

We can appeal to Lorentz covariance and decompose $\Sigma(p) =
a(p^2)\dsl{p} + b(p^2)\mathbb{1} + c(p^2)\gamma_5$. With parity
conserving interactions such as qed, we can take $c(p^2) = 0$.
Furthermore, to the leading order, $a(p^2) = 0 = b(p^2)$. Thus the
denominator of the exact propagator is,

\begin{eqnarray*}
[S'_F(p)]^{-1} & = & (1-a(p^2))\dsl{p} + (m-b(p^2))  =  (1-a(p^2))\left[
\dsl{p} + \frac{m-b(p^2)} {1 - a(p^2)} \right] ~ ~ ~ \Rightarrow \\
S'_F(p) & = & \frac{1}{1 - a(p^2)}\left[\frac{-\dsl{p} +
\frac{m-b(p^2)}{1 - a(p^2)}}{p^2 +
\left(\frac{m-b(p^2)}{1-a(p^2)}\right)^2 } \right]
\end{eqnarray*}

Clearly, the denominator vanishes for $p^2 = - m_{ph}^2, m_{ph} :=
\Case{m-b(-m_{ph}^2)}{1 - a(-m{ph}^2)}$. Since the numerator has the
form $-\dsl{p} + m_{ph}$, we can regard the pole to be defined by
$\dsl{p} = -m_{ph}$. Since $a, b$ are functions of $p^2 =
-\dsl{p}\dsl{p}$, we may regard the self energy as a function of
$\dsl{p}$ and define the pole by the condition: $\boxed{\dsl{p} + m -
\Sigma(\dsl{p})|_{\dsl{p} = -m_{ph}} = 0. } $ Clearly, $m_{ph} \neq m$.

The exact propagator can be expanded about its pole as,

\begin{eqnarray*}
\dsl{p} + m - \Sigma(\dsl{p}) & = & 0 + (\dsl{p} + m_{ph})\left[1 -
\frac{d\Sigma(\dsl{p})}{d\dsl{p}}|_{\dsl{p} = -m_{ph}}\right] + o(
(\dsl{p}+m_{ph})^2 ) \\
\therefore S'_F(p) & \simeq & \frac{1}{\dsl{p} + m_{ph}}\left[\frac{1}{1 -
\frac{d\Sigma(\dsl{p})}{d\dsl{p}}|_{\dsl{p}=-m_{ph}}}\right]  ~ := ~
\frac{Z_2}{\dsl{p}+m_{ph}}
\end{eqnarray*}

The shift of the position of the pole in the exact propagator from that
of the free propagator, $\boxed{\delta m := m_{ph} - m = -
\Sigma(\dsl{p} = -m_{ph})}$ is important in an $S-$matrix element which
has the spinors on the external lines. These spinors satisfy the
equation of motion with physical mass (= $m_{ph}$) and not
the mass parameter in the free propagator inferred from the Lagrangian.
In particular this means that in the LSZ formula for $S-$matrix element,
the amputation of external fermion legs is to be done with
$(-i\dsl{\partial} + m_{ph})$. This in turn means that in the Green's
functions with external fermion legs, the external lines must include
the self-energy corrections i.e. should have $S'_F$ propagator instead
of the free propagator $S_F$ (and of course no more self energy
corrections on the external lines). We will return to this point later
again while discussing renormalized perturbation series.

We also see that the residue at the pole in the exact propagator is not
1 and we identify it with the $Z_2$ which we know from the
Kallen-Lehmann representation to be less than 1.

§.§ The photon propagator: photon self-energy

\begin{eqnarray}
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=0.75of a,blob] (b){All} ;
\vertex [right=1.0of b] (c) ;
\diagram* {
(a) -- [photon] (b) -- [photon]
\end{feynman}
\end{tikzpicture}
& = &
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=of a] (b) ;
\diagram* {
(a) -- [photon] (b),
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right =0.75of a] (b) ;
\vertex [right =0.75of b] (c) ;
\vertex [right =0.75of c] (d) ;
\diagram* {
(a) -- [photon] (b),
(b) -- [fermion,half right,looseness=1.5] (c),
(c) -- [photon] (d),
(c) -- [fermion,half right,looseness=1.5]
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (v) ;
\vertex [left=0.4of v] (b);
\vertex [left=0.4of b] (a);
\vertex [right=0.4of v] (c);
\vertex [right=0.4of c] (d);
\vertex [above=0.4of v] (e) ;
\vertex [below=0.4of v] (f) ;
\diagram* {
(a) -- [photon] (b),
(c) -- [photon] (d),
(e) -- [photon] (f),
(b) -- [fermion,half left,looseness=1.5]
(c) -- [fermion,half left,looseness=1.5]
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=0.5of a] (b) ;
\vertex [right=0.5of b] (c) ;
\vertex [right=0.5of c] (d) ;
\vertex [right=0.5of d] (e) ;
\vertex [right=0.5of e] (f) ;
\diagram* {
(a) -- [photon] (b),
(c) -- [photon] (d),
(e) -- [photon] (f),
(b) -- [fermion,half right,looseness=1.5]
(c) -- [fermion,half right,looseness=1.5]
(d) -- [fermion,half right,looseness=1.5]
(e) -- [fermion,half right,looseness=1.5]
\end{feynman}
\end{tikzpicture}
\\
& = &
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=of a] (b) ;
\diagram* {
(a) -- [photon] (b),
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=0.75of a,blob] (b){1PI} ;
\vertex [right=1.0of b] (c) ;
\diagram* {
(a) -- [photon] (b) -- [photon] (c),
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=0.75of a,blob] (b){1PI} ;
\vertex [right=1.25of b,blob] (c){1PI} ;
\vertex [right=1.0of c] (d){ } ;
\diagram* {
(a) -- [photon] (b) -- [photon] (c) --
[photon] (d),
\end{feynman}
\end{tikzpicture}
~ + ~ \cdots
\end{eqnarray}

Here is a graphical representation of the exact photon propagator. As
before, we group together the 1-PI diagrams connected by free
propagators. Let the exact propagator be denoted by $-iD'_{\mu\nu}(q)$
and the free propagator by $-iD_{\mu\nu}(q)$. Let the 1PI blob be
denoted by $i\Pi_{\mu\nu}(q)$. In the Lorentz gauge, the free propagator
is: $D_{\mu\nu}(q) = \eta_{\mu\nu}/(q^2 - i\epsilon)$. While the series
can be formally summed as before, it is more convenient to separate the
tensor indices in the $\Pi_{\mu\nu}(q)$ tensor.

Lorentz covariance implies, $\Pi_{\mu\nu}(q) = \eta_{\mu\nu}A(q^2) +
q_{\mu}q_{\nu}B(q^2)$. The $\Pi_{\mu\nu}$ tensor may be thought of as
$\gamma \to \gamma$ process, although $q^2 \ne 0$. We now appeal to a
Ward identity: $q^{\mu}\Pi_{\mu\nu}(q) = 0$. This is done
separately below. Presently, it implies $A(q^2) = -q^2B(q^2)$ and we
define: $\boxed{\Pi_{\mu\nu}(q) := [\eta_{\mu\nu}q^2 -
q_{\mu}q_{\nu}]\Pi(q^2)}$ . With this notation, the exact 2-point
function takes the form,

\begin{eqnarray*}
-iD'_{\mu\nu}(q) & = & -i \frac{\eta_{\mu\nu}}{q^2-i\epsilon} +
\frac{\eta_{\mu\alpha}}{q^2-i\epsilon}\cdot[i(q^2\eta^{\alpha\beta}
- q^{\alpha}q^{\beta})\Pi(q)]\cdot -i
\frac{\eta_{\beta\nu}}{q^2-i\epsilon} + \\
& & \mbox{\hspace{3.0cm}}
-iD_{\beta\rho}\cdot[i(\dots)^{\rho\sigma}]\cdot -iD_{\sigma\nu}
+ \dots
\end{eqnarray*}

\[
i(q^2\eta^{\alpha\beta} - q^{\alpha}q^{\beta})(-i\eta_{\beta\nu}q^{-2})
= \left(\delta^{\alpha}_{~\nu} - \frac{q^{\alpha} q_{\nu}}{q^2}\right)
=: \Delta^{\alpha}_{~\nu} ~ \Rightarrow ~ \Delta^{\alpha}_{~\nu}
\Delta^{\nu}_{~\beta} = \Delta^{\alpha}_{~\beta} \ .
\]

Then the exact propagator series takes the form,

\begin{eqnarray*}
-iD'_{\mu\nu}(q) & = & -i \frac{\eta_{\mu\nu}}{q^2-i\epsilon} + -i
\frac{\eta_{\mu\alpha}}{q^2-i\epsilon} \Delta^{\alpha}_{~\nu}\Pi(q^2) +
-i \frac{\eta_{\mu\alpha}}{q^2-i\epsilon} \Delta^{\alpha}_{~\beta}
\Delta^{\beta}_{~\nu}\Pi(q^2) + \dots \\
& = & -i \frac{\eta_{\mu\nu}}{q^2-i\epsilon} + -i
\frac{\eta_{\mu\alpha}}{q^2-i\epsilon} \Delta^{\alpha}_{~\nu}(\Pi +
\Pi^2 + \Pi^3 + \dots) \\
& = & \frac{-i}{q^2-i\epsilon} \left[\left\{\eta_{\mu\nu} -
\frac{q_{\mu}q_{\nu}}{q^2} + \frac{q_{\mu}q_{\nu}}{q^2}\right\} +
\left(\eta_{\mu\nu} - \frac{q_{\mu}q_{\nu}}{q^2}\right)(\Pi + \Pi^2 +
\Pi^3 \dots) \right] \\
\therefore D'_{\mu\nu}(q^2) & = & \frac{1}{q^2(1 -
\Pi(q^2))}\left(\eta_{\mu\nu} - \frac{q_{\mu}q_{\nu}}{q^2}\right) +
\frac{1}{q^2}\frac{q_{\mu}q_{\nu}}{q^2} .
\end{eqnarray*}

In any $S-$matrix calculation, the $D'$ propagator will land on a
fermion line and thanks to the Ward identity, the terms proportional to
$q^{\mu}, q^{\nu}$ will vanish. Hence, for $S-$matrix element
computations, we can take $\boxed{D'_{\mu\nu}(q) = \Case{\eta_{\mu\nu}}
{(q^2 - i\epsilon)(1 - \Pi(q^2))}}$. The $\Pi(q^2)$ is called the photon
self energy. To the leading order, the self energy vanishes and hence in
perturbation theory, it can never equal 1 and cause another pole at some
other $q^2$. As long as it is regular at $q^2 = 0$, the exact propagator
continues to have a simple pole at $q^2 = 0$, just like the free
propagator. Hence, the photon wavefunction, $\epsilon_{\mu}(\vec{k})$
continues to be transverse and longitudinal polarizations will decouple.

Unlike the fermion self energy which shifts the mass, the photon self
energy does not shift the photon mass from zero thanks to the Ward
identity.  However, like the fermion self energy, the residue at the
pole does shift away from 1, $\boxed{ (Z_3)^{-1} := 1 - \Pi(q^2 = 0).}$

§.§.§ The Ward identity claim: $q^{\mu}\Pi_{\mu\nu}(q) = 0$:

\[
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (v);
\vertex [right=of v] (1);
\vertex [right=2.0of v] (11){\(\gamma_{1}\)};
\vertex [below right=of v] (2);
\vertex [below right=2.0of v] (22){\(\gamma_{2}\)};
\vertex [below left=of v] (i);
\vertex [below left=2.0of v] (ii){\(\gamma_{i}\)};
\vertex [left=of v] (j);
\vertex [left=2.0of v] (jj){\(\gamma_{j}\)};
\vertex [above left=of v] (i+1);
\vertex [above left=2.0of v] (ii+1){\(\gamma_{i+1}\)};
\vertex [above right=of v] (n);
\vertex [above right=2.0of v] (nn){\(\gamma_{n}\)};

\diagram*{
(1) --[photon] (11),
(2) --[photon] (22),
(i) --[photon] (ii),
(j) --[photon] (jj),
(i+1) --[photon] (ii+1),
(n) --[photon] (nn),

\draw[fermion, domain=0:-45] (v) plot
node[label={[label distance=8.0] 15:$p_1$}];

\draw[fermion, domain=-45:-135] (v) plot
node[label={[label distance=5.5] 160:$p_i$}];

\draw[fermion, domain=-135:-180] (v) plot

\draw[fermion, domain=-180:-225] (v) plot
node[label={[label distance=4.0] 190:$p_i+q$}];

\draw[fermion, domain=-225:-315] (v) plot
({1.5*cos(\x)},{1.5*sin(\x)}) ;

\draw[fermion, domain=-315:-360] (v) plot
node[label={[label distance=2.8] 80:$p_n+q$}];

%		\draw (2) arc [start angle=-45,end angle=-90,radius=1.44cm];
\end{feynman}
\end{tikzpicture}
\]

Proof: Recall that $\Pi_{\mu\nu}$ is defined without
external propagators i.e. it is an amputated 2-point function. The
momentum $q$ however, need not be on-shell and hence it does not
correspond to an $S-$matrix element (even if we disregard the
polarization factors). Consider diagrams contributing to the photon self
energy with some fixed number of vertices (or order in the coupling).
The set of these diagrams will be generated by vertex injecting momentum
$q^{\mu}$ at various points on the available fermion lines. The fermion
lines will necessarily be loops as there are no external fermions
contributing to the photon self energy. Consider the subset of diagrams
wherein the $q^{\mu}$ vertex is on some particular loop which
contributes a factor of the form: (the fermion arrow points from $i=1$
to $i=n$ and so do the fermion momenta)

\[
Tr\left[S_F(p_n+q)\gamma^{\mu_n} \dots S_F(p_{i+1}+q) \gamma^{\mu_{i+1}}
\left\{S_F(p_i+q) \dsl{q} S_F(p_i)\right\} \gamma^{\mu_{i}} \dots
\]

The $q^{\mu}$ vertex is inserted between the $i^{th}$ and $(i+1)^{th}$
vertices and adds the propagator $S_F(p_i+q)$. All momenta from
$p_{i+1}$ to $p_n$ are shifted by $q$. We have also contracted with
$q^{\mu}$. The expression between the braces simplifies as
[This identity holds as long as both the fermion propagators
have the same mass, $m_{ph}$ or $m$. This is relevant while dealing with
external fermion lines.],

\[
\frac{1}{\dsl{p}_{i} +\dsl{q }+ m}\dsl{q}\frac{1}{\dsl{p}_{i} +
m}  = \frac{1}{\dsl{p}_{i} + m}- \frac{1}{\dsl{p}_i + \dsl{q} +
m} = S_F(p_i) - S_F(p_i+q) \ .
\]

The trace thus becomes,

\begin{eqnarray*}
Tr\left[S_F(p_n+q)\gamma^{\mu_n} \dots S_F(p_{i+1}+q)
\gamma^{\mu_{i+1}}\left\{S_F(p_i)\right\}\gamma^{\mu_i} \dots
S_F(p_1)\gamma^{\mu_1}\right] \mbox{\hspace{2.0cm}} & & \\
- Tr\left[S_F(p_n+q) \gamma^{\mu_n} \dots S_F(p_{i+1}+q)
\gamma^{\mu_{i+1}} \left\{ S_F(p_i +q)\right\} \gamma^{\mu_{i}} \dots
S_F(p_1) \gamma^{\mu_1}\right]
\end{eqnarray*}

Note that the momenta after $i$ (and $i-1$) have $q$ added to them.
Summing over the diagrams where the $q^{\mu}$ vertex is on this fermion
loop i.e from $i = 1, \dots, n$ , there is a pairwise cancellation. The
first term at $i$ cancels the second term at $i+1$ and so on. We are
left with the second term at $i=1$ and the first term at $i=n$, i.e.

\[
\sum_{i=1}^{n} Tr\left[\dots\right] = Tr\left[S_F(p_n)\gamma^{\mu_n}
\dots S_F(p_1) \gamma^{\mu_{1}}\right] -
Tr\left[S_F(p_{n}+q)\gamma^{\mu_{n}} \dots S_F(p_1+q)
\gamma^{\mu_{1}}\right]
\]

Because we have a fermion loop, there is an integration over say $p_1$.
Provided the integral is finite, we can shift the integration
variable and absorb away the $q$. The integrals of the two traces, then
cancel out and the claim is proved.

Remark: We have not yet introduced loop diagrams. The above
fermion loop, before the $q$ insertion, has $n$ vertices and $n$ momenta
entering at those vertices. The conservation of momenta, taking out the
overall conservation, thus leaves one momentum undetermined and this is
to be integrated.

Remark: As an extension of the result, consider an open
fermion line with incoming momentum $p_0$, outgoing momentum $p_n$ and
insertions on it.

\[
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (0);
\vertex [left=of 0] (1);
\vertex [below=of 1] (11){\(\gamma_1\)};
\vertex [left=of 1] (2);
\vertex [below=of 2] (22){\(\gamma_2\)};
\vertex [left=3.0of 1] (i);
\vertex [below=of i] (ii){\(\gamma_i\)};
\vertex [left=1.0of i] (j);
\vertex [above=of j] (jj){\(\gamma_j\)};
\vertex [left=1.0of j] (i+1);
\vertex [below=of i+1] (ii+1){\(\gamma_{i+1}\)};
\vertex [left=3.0of i+1] (nm1);
\vertex [below=of nm1] (nnm1){\(\gamma_{n-1}\)};
\vertex [left=2.0of nm1] (n);
\vertex [below=of n] (nn){\(\gamma_n\)};
\vertex [left=1.5of n] (n+1);
\diagram*{
(0) --[fermion,edge label=\(p_0\)] (1),
(1) --[fermion,edge label=\(p_1\)] (2),
(2) --[fermion] (i),
(i) --[fermion,edge label=\(p_i\)] (j),
(j) --[fermion,edge label=\(p_i+q\)] (i+1),
(i+1) --[fermion] (nm1),
(nm1) --[fermion,edge label=\(p_{n-1}+q\)] (n),
(n) --[fermion,edge label=\(p_n+q\)] (n+1),
(1) --[photon] (11),
(2) --[photon] (22),
(i) --[photon] (ii),
(j) --[photon] (jj),
(i+1) --[photon] (ii+1),
(nm1) --[photon] (nnm1),
(n) --[photon] (nn),
\end{feynman}
\end{tikzpicture}
\]

For the insertion between the $i, i+1$ vertices, we will have a string
of factors as,

\[
S_F(p_{n}+q)\gamma^{\mu_n}S_F(p_n+q) \dots \gamma^{\mu_{i+1}}
\left\{S_F(p_{i}+q) \dsl{q} S_F(p_{i})\right\} \gamma^{\mu_{i}} \dots
\]

Replacing $S_F(p_i+q)\dsl{q}S_F(p_i) = S_F(p_i) - S_F(p_i+q)$ and
summing over the insertions from $i=0,\dots,n$, we get

\begin{eqnarray*}
\sum_{i=0}^{n} \left[\dots\right] & = &
S_F(p_n)\gamma^{\mu_n}S_F(p_{n-1}) \dots S_F(p_{1}) \gamma^{\mu_{1}}
S(p_0) \\
& & \mbox{\hspace{2.0cm}} - S_F(p_n+q)\gamma^{\mu_n}S_F(p_{n-1}+q) \dots
S_F(p_{1}+q) \gamma^{\mu_{1}} S(p_0+q)
\end{eqnarray*}

To get an $S-$matrix contribution, we have to multiply by
$\bar{u}(p_n+q)[S_F(p_n+q)]^{-1}]$ on the left and by
$[S_F(p_0)]^{-1}u(p_0)$ on the right and take the limits $p_0^2 \to
-m^2, (p_n+q)^2 \to -m^2$. Now we see that each of the two terms has
only one of the poles which can cancel the inverse propagator.  In the
on-shell limit then, each of the terms vanishes, thereby proving that
for $S-$matrix elements with external fermion lines, sum over
insertion of photon lines followed by contraction with the photon
momentum gives zero.

Note: There is one subtlety here. For external lines, we
should be using the exact propagator as per the LSZ rules. The
$S\dsl{q}S$ identity requires both propagators to have the same mass,
the physical mass. This would require all internal fermion lines to use the
exact propagator (and then do not consider the fermion self energy
correction). For fermion loop, this issue does not arise. Notice however
that $S'_F(p)^{-1} = S_F(p)^{-1} -\Sigma(p)$ and $\sigma(p)$ is always
of order $\alpha$ and higher. Thus, in using free inverse propagator
instead of exact inverse propagator, we are dropping a higher order
contribution which we can do in a perturbation theory.

The conclusion is then that $q_{\mu}\mathcal{M}^{\mu}(q,\dots) = 0$
which is a statement of Ward identity.

§.§ The Vertex function: Form factors

The expectation is that the electron-photon coupling will undergo a
change due to higher order corrections. Unlike the 2-point functions
above which were considered for arbitrary momentum, we consider the
3-point function with the fermion momenta “on-shell” (and this means
the $q^2$ is necessarily space-like (prove this)) and hence an internal
line. When viewed as part of an $S-$matrix element, the external fermion
legs are amputated with physical mass inverse propagator and
replaced by the $\bar{u}(p'), u(p)$ spinors respectively. Here is a
graphical view of the possible corrections.

\begin{eqnarray}
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [left=0.15of a] (a'){\(\mu\)} ;
\vertex [right=0.75of a,blob] (b){All} ;
\vertex [above right=1.0of b] (c) ;
\vertex [below right=1.0of b] (d) ;
\diagram* {
(a') -- (a),
(a) -- [photon,momentum=\(q\)] (b),
(b) -- [fermion,edge label=\(p'\)] (c),
(d) -- [fermion,edge label=\(p\)] (b),
\end{feynman}
\end{tikzpicture}
& = &
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=0.75of a] (b) ;
\vertex [above right=1.0of b] (c) ;
\vertex [below right=1.0of b] (d) ;
\diagram* {
(a) -- [photon] (b),
(b) -- [fermion] (c),
(d) -- [fermion] (b),
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right =0.75of a] (b) ;
\vertex [above right =0.75of b] (c1) ;
\vertex [above right =1.5of b] (c2) ;
\vertex [below right =1.5of b] (d2) ;
\vertex [below right =0.75of b] (d1) ;
\diagram* {
(a) -- [photon] (b),
(b) -- [fermion] (c1) -- [fermion] (c2),
(d2) -- [fermion] (d1) -- [fermion] (b),
(d1) -- [photon,half
right,looseness=1.0] (c1),
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right =0.5of a] (b) ;
\vertex [above right =0.5of b] (c1) ;
\vertex [above right =1.0of b] (c2) ;
\vertex [above right =1.5of b] (c3) ;
\vertex [below right =1.5of b] (d3) ;
\vertex [below right =1.0of b] (d2) ;
\vertex [below right =0.5of b] (d1) ;
\diagram* {
(a) -- [photon] (b),
(b) -- [fermion] (c1) -- [fermion] (c2)
-- [fermion] (c3),
(d3) -- [fermion] (d2) -- [fermion] (d1)
-- [fermion] (b),
(d1) -- [photon] (c2),
(d2) -- [photon,half
right,looseness=1.0] (c1),
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right =0.5of a] (b) ;
\vertex [above right =0.5of b] (c1) ;
\vertex [above right =1.0of b] (c2) ;
\vertex [above right =1.5of b] (c3) ;
\vertex [above right =2.0of b] (c4) ;
\vertex [above right =2.5of b] (c5) ;
\vertex [below right =1.5of b] (d3) ;
\vertex [below right =1.0of b] (d2) ;
\vertex [below right =0.5of b] (d1) ;
\diagram* {
(a) -- [photon] (b),
(b) -- [fermion] (c1) -- [fermion] (c2)
-- [fermion] (c3) -- [fermion] (c4) --
[fermion] (c5),
(d3) -- [fermion] (d2) -- [fermion] (d1)
-- [fermion] (b),
(d1) -- [photon] (c2),
(d2) -- [photon] (c4),
(c1) -- [photon,half
left,looseness=1.5] (c3),
\end{feynman}
\end{tikzpicture}
\\
ie\Gamma^{\mu}(p,p') & = &
ie \gamma^{\mu} ~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [left=0.15of a] (a'){\(\mu\)} ;
\vertex [right=0.75of a,blob] (b){1PI} ;
\vertex [above right=1.0of b] (c) ;
\vertex [below right=1.0of b] (d) ;
\diagram* {
(a') -- (a),
(a) -- [photon,momentum=\(q\)] (b),
(b) -- [fermion,edge label=\(p'\)] (c),
(d) -- [fermion,edge label=\(p\)] (b),
\end{feynman}
\end{tikzpicture}
~ + ~
\vcenter{\hbox{
\begin{tikzpicture}
\begin{feynman}
\vertex (a) ;
\vertex [right=0.5of a, blob] (b){1PI} ;
\vertex [right=2.0of a] (c) ;
\vertex [above right=0.75of c, blob] (c1){1PI} ;
\vertex [above right=2.0of c] (c2) ;
\vertex [below right=0.75of c, blob] (d1){1PI} ;
\vertex [below right=2.0of c] (d2) ;
\diagram* {
(a) -- [photon] (b) -- [photon] (c),
(c) -- [fermion] (c1) -- [fermion] (c2),
(d2) -- [fermion] (d1) -- [fermion] (c),
\end{feynman}
\end{tikzpicture}
~ + ~ \cdots
\end{eqnarray}

Notice that some corrections on the fermion lines and the photon line
are just self-energy corrections. The remaining corrections have photon
lines necessarily connecting the two fermion lines and these diagrams
are already 1PI. The last diagram in the second equation is 1PR.

We can simplify the form of the exact vertex function,
$\Gamma^{\mu}(p,p')$ by appealing to Lorentz covariance.

\[
\Gamma^{\mu}(p, p') = \gamma^{\mu}A(q^2) + (p^{\mu} + p^{'\mu})
B(q^2)+ (p^{'\mu} - p^{\mu})C(q^2) \ .
\]

With two independent momenta, we have 3 Lorentz scalars, $p^2, (p')^2,
p\cdot p'$. The first two equal $-m^2_{ph}$ while the third is traded
for $q^2$ for convenience.

Claim: $~ \bar{u}(p')q_{\mu}\Gamma^{\mu}u(p) = 0$.

Proof:   This is the Ward identity argument given above.

The above form for $\Gamma^{\mu}$ then implies,

\begin{eqnarray*}
0 & = & \bar{u}(p')\dsl{q}u(p)A(q^2) + (-m^2_{ph} +
m^2_{ph})\bar{u}(p')B(q^2)iu(p) + q^2\bar{u}(p')C(q^2)u(p) \\ &
= & 0 + 0 + q^2\bar{u}(p')C(q^2)u(p) ~ \Rightarrow ~
\bar{u}(p')C(q^2)u(p)  = 0 ~ .
\end{eqnarray*}

The $A, B$ terms can be rewritten using “Gordon identity”,

\[
\boxed{ \bar{u}(p')\gamma^{\mu}u(p) = \frac{1}{2m_{ph}}\bar{u}(p')\left[
(p+p')^{\mu} - 2i\sigma^{\mu\nu}q_{\nu} \right] u(p) }
\]

This can be established as follows ($m_{ph} \to m$ for convenience):

We have, $u(p) = - \Case{1}{m}\dsl{p}u(p)\ , \ \bar{u}(p') = -
\Case{1}{m} \bar{u}(p')\dsl{p}'$. Therefore

\begin{eqnarray*}
\bar{u}(p')\gamma^{\mu}u(p) & = & -\frac{1}{2m}(
\gamma^{\mu}\dsl{p} + \dsl{p}'\gamma^{\mu} ) ~ ~ \mbox{But,} ~ ~
\gamma^{\mu}\dsl{p}  =  p_{\nu}\gamma^{\mu}\gamma^{\nu} =
p_{\nu}\left(-\eta^{\mu\nu} + \frac{1}{2} [\gamma^{\mu},
\gamma^{\nu}]\right) \\
\therefore \gamma^{\mu}\dsl{p} & = & -p^{\mu} -2i
\Sigma^{\mu\nu}p_{\nu} ~ ~ , ~ ~ \dsl{p}'\gamma^{\mu} = -
(p')^{\mu} - 2i \Sigma^{\nu\mu}p'_{\nu} ~ ~ , ~ ~
\Sigma^{\mu\nu} := \frac{i}{4}[\gamma^{\mu}, \gamma^{\nu}]\ ; \\
\therefore \bar{u}(p')\gamma^{\mu}u(p) & = &
-\frac{1}{2m}\bar{u}(p')\left[-p^{\mu} - (p')^{\mu} -
4i\Sigma^{\mu\nu}(p_{\nu} - p'_{\nu})\right]u(p)  \\
& = & \frac{1}{2m}\bar{u}(p')[(p+p')^{\mu}
-2i\sigma^{\mu\nu}q_{\nu}] u(p) ~ ~ ~ , ~ ~ ~ \sigma^{\mu\nu} :=
2\Sigma^{\mu\nu} ~ , ~ q := p'-p \ .
\end{eqnarray*}

Eliminating the $(p+p')^{\mu}$ from the $B(q^2)$ term, and identifying
$A + 2m_{ph}B =: F_1(q^2)$ and $4\Sigma^{\mu\nu}B(q^2) =:
\Case{\sigma^{\mu\nu}}{2m_{ph}}F^2(q^2)$, we take the general form of
$\Gamma^{\mu}(p,p')$ as,

\[
\boxed{ \Gamma^{\mu}(p,p') = \gamma^{\mu}F_1(q^2) + i
\frac{\sigma^{\mu\nu}q_{\nu}}{2m_{ph}}F_2(q^2) ~ , ~ q := p' - p\ . }
\]

The $F_i(q^2)$ are called form factors. To the lowest order,
$F_1(q^2) = 1, F_2(q^2) = 0$ . These get corrected at higher orders.

§.§ Electric Charge and Anomalous Magnetic Moment

The form factor decomposition helps us identify the electric
charge and the anomalous magnetic moment of electron (and other
charged fermions). This is seen as follows.

Consider the scattering of an electron off a heavy charged particle. We
imagine the scattering to be effected by exchanging a single photon
between the heavy charge and the electron. Schematically the amplitude
can be written as,

\[
i\mathcal{M} =
ie^2\ [\bar{u}(p')\Gamma^{\mu}(p',p)u(p)]\ \frac{1}{q^2}\ [\bar{u}(k')\gamma_{\mu}u(k)]
\]

The first factor is from electron, second is the single photon
propagator and the third factor is from the heavy charge. The $\Gamma$
vertex has only 1PI diagrams and excludes the photon self energy on the
propagator. This represents the modification of the electron response
due to the 1PI corrections, the photon self energy would indicate
modification in the electromagnetic field felt by the electron and hence
is also referred to as vacuum polarization. This is ignored in
this calculation.

In the limit of infinitely heavy fermion, the second
and third factors are replaced by an external classical potential
$A_{\mu}(q)$ and the amplitude is defined as,

\[
i\mathcal{M}\ (2\pi)\ \delta( (p'-p)^0 ) = -ie
\bar{u}(p')\Gamma^{\mu}u(p)\tilde{A}^{cl}_{\mu}(p'-p)
\]

We consider two cases - Coulomb potential and magnetic field, and
consider non-relativistic limit of the scattering.

Coulomb potential: $A^{cl}_{\mu}(x) = (\phi(x), \vec{0})$.
Coulomb field is static and we also assume it to be slowly varying
spatially. This means that we take $\tilde{A}^{cl}_{\mu}(q) =
2\pi\delta(q^0)(\tilde{\phi}(q), \vec{0})$ with the
$\tilde{\phi}(q)$ having support near $q \simeq 0$. We can thus take the
limit $q \to 0$ of the first factor. This in turn means that only the
$F_1$ term contributes. In the non-relativistic limit,
$\bar{u}(p')\gamma^{\mu}u(p) \to \bar{u}(p')\gamma^0u(p) =
2mu^{\dagger}(p)u(p)$ and the amplitude takes the form, $i\mathcal{M}
\to -ieF_1(0)\tilde{\phi}(q)(2mu^{\dagger}(p)u(p))$. Comparing with the
Born approximation for scattering off a potential leads to
the identification, $V(x) = eF_1(0)\phi(\vec{x})$. Thus,
$\boxed{\mbox{electric charge} = eF(0). }$ Since $F_1(0) = 1$ at the
leading order, the radiative corrections to $F_1(q^2)$ should vanish as
$q^2 \to 0$.

Magnetic field: $A_{\mu}^{cl} = (0, \vec{A}^{cl}_i(x))$.
Again this is taken to be time independent and spatially slowly varying.
Thus we consider the limit $q \to 0$. Earlier, we dropped the $F_2$
term, because $F_1$ term was non-zero. Now however $\bar{u}(p')
\gamma^iu(p) \simeq o(p/m)$ in the non-relativistic limit which is
comparable to the $F_2$ term.

Recall from the NR limit of the Dirac spinors, $u^T := (u_1^T, u_2^T),
\ (\dsl{p} + m)u(p) = 0 \Rightarrow u_2(p) \simeq - \Case{\vec{p} \cdot
\vec{\sigma}}{2m}u_1$ and we have the normalization $u^{\dagger}u(p) =
2m$. We also had,

\[
\bar{u}(p')\gamma^i u(p)  \simeq u_1^{\dagger}(p')\sigma^iu_2(p)
+ u_2^{\dagger}(p')\sigma^iu_1(p) \simeq -\frac{1}{2m}
u_1^{\dagger}(p')\{ \sigma^i\vec{p}\cdot\vec{\sigma} +
\vec{p}'\cdot\vec{\sigma}\sigma^i \}u_1(p) .
\]

Using $\sigma^i\sigma^j = \delta^{ij} + i\epsilon^{ij}_{~k}\sigma^k$, we

\[
\bar{u}(p')\vec{\gamma}\cdot\vec{A}^{cl}u(p) = -\frac{1}{2m}
u_1^{\dagger}(p')\left( \vec{p}\cdot \vec{A}^{cl} + \vec{p}'\cdot\vec{A}^{cl}
-iA_i^{cl}\epsilon^{ij}_{~k}q_j\sigma^k \right)u_1(p)
\]

Noting that $q^0 = 0$, the $F_2$ term has $\sigma^{ij}q_j$. Going over
to the two component spinors, we get

\begin{eqnarray*}
\frac{i}{2m} \bar{u}(p')\sigma^{ij}u(p)q_j & \approx &
\frac{i}{2m}
\epsilon^{ij}_{~k}[u_1^{\dagger}(p')\sigma^ku_1(p)]
\end{eqnarray*}

The contribution of the $u_2$ spinors is negligible in the NR limit.

Combining the terms we get in the NR limit,

\[
\bar{u}(p')\Gamma^iu(p)\tilde{A}_i^{cl}(q) \approx
u_1^{\dagger}(p')\left[\frac{(\vec{p}+\vec{p}')\cdot\vec{A}^{cl}}{2m}F_1(0) -
\frac{i}{2m} \epsilon^{ij}_{~k}q_jA^{cl}_i\sigma^k(F_1(0) +
\]

In the non-relativistic Hamiltonian, $(p - eA)^2/(2m)$ we have the
$e(p\cdot  + A\cdot p)$ terms which are recovered. The remaining terms
however are new. Focusing only on those, we write the amplitude as,

\[
i\mathcal{M} = -ieu_1^{\dagger}(p')\left[-\frac{1}{2m} (F_1(0) +
F_2(0))\sigma^kB_k\right]u_1(p) ~ ~ , ~ ~ B_k :=
-i\epsilon_k^{~ij}q_i\tilde{A}_j^{cl}(q) \ .
\]

This amplitude is interpreted as the Born approximation to a potential
scattering with potential, $V(x) = -\langle\vec{\mu}\rangle\cdot
\vec{B}(\vec{x})$, with the effective magnetic moment,

\[
\boxed{\langle\vec{\mu}\rangle = \frac{e}{m_{ph}}[F_1(0) +
F_2(0)]u_1^{\dagger}(p) \frac{\vec{\sigma}}{2} u_1(p) := g
\frac{e}{2m_{ph}}\vec{S} ~ , ~ g := 2[F_1(0) + F_2(0)] = 2[1 +
\]

$g$ is called the Lande's g-factor, its value being $2$ is
prediction of Dirac equation while $F_2(0) \neq 0$ is a prediction of
QED. It is called the anomalous magnetic moment.

Without actually calculating the higher order corrections, we noted what
they mean for 2-point functions (self-energies). We used Lorentz
covariance to get a form factor decomposition and interpreted them by
appealing to the NR limit. One of the immediate prediction was the
anomalous magnetic moment.

We also note that to the leading order (see diagrams), we have $Z^{-1}_2
= 1, Z^{-1}_3 = 1, F_1(q^2) = 1, F_2(q^2) = 0$. This is the reason that
at the tree level calculations that we did, did not have any factors of
$Z$'s. This will change when we do explicit evaluations of the leading
corrections in the next section.

§ RADIATIVE CORRECTIONS AT 1-LOOP: DIVERGENCES

We now calculate the leading radiative corrections to the self energies
and the QED vertex function i.e. the 1-loop contributions to
$\Sigma(\dsl{p}), \Pi(q^2), F_1(q^2), F_2(q^2)$. Here are the Feynman
diagrams and the corresponding invariant amplitudes.

Fermion self energy

[right=1.5of v1] (v2);
[right=1.5of v2] (v3);
[right=1.5of v3] (v4);

(v1) – [fermion,edge label=\(p\)] (v2),
(v2) – [fermion,edge label=\(k\)] (v3),
(v3) – [fermion,edge label=\(p\)] (v4),
(v2) – [photon,half left,
looseness=1.5, momentum=\(p-k\)] (v3),

\[ i\Sigma(p) = (ie)^2\int \frac{d^4k}{(2\pi)^4}
\gamma^{\mu} \frac{-i (-\dsl{k}+m)}{k^2+m^2-i\epsilon} \gamma^{\nu}
\frac{-i\eta_{\mu\nu}}{(p-k)^2+\mu^2-i\epsilon} \]

Photon self energy (vacuum polarization)

[right=1.5of v1] (v2);
[right=1.5of v2] (v3);
[right=1.5of v3] (v4);

(v1) – [photon,momentum=\(q\)] (v2),
(v2) – [fermion,half
left,looseness=1.5, edge label=\(k+q\)] (v3),
(v3) – [photon,momentum=\(q\)] (v4),
(v3) – [fermion,half left,
looseness=1.5, edge label=\(k\)] (v2),

\[
i\Pi^{\mu\nu}(q) = (ie)^2 (-1)\int \frac{d^4k}{(2\pi)^4}
\]

QED Vertex function, $\delta\Gamma^{\mu} := \Gamma^{\mu} -
\gamma^{\mu}$

(v1) \(\mu\);
[right=1.5of v1] (v2);
[above right=1.5of v2] (v3);
[below right=1.5of v2] (v4);
[above right=1.0of v3] (v5);
[below right=1.0of v4] (v6);

(v1) – [photon,momentum'=\(q\)] (v2),
(v2) – [fermion, edge label=\(p'-k\)] (v3),
(v4) – [photon,momentum'=\(k\)] (v3),
(v4) – [fermion, edge label=\(p-k\)] (v2),
(v3) – [fermion,edge label=\(p'\)] (v5),
(v6) – [fermion,edge label=\(p\)] (v4),

\begin{eqnarray*}
\delta\Gamma^{\mu}(p,p') & = & (ie)^2\int
\frac{d^4k}{(2\pi)^4}
\frac{-i\eta_{\alpha\beta}}{(k)^2+\mu^2-i\epsilon}\times\\
& & \left[\gamma^{\alpha}
\frac{-i(-(\dsl{p}'-\dsl{k})+m)}
{(p'-k)^2+m^2-i\epsilon} \gamma^{\mu}
\frac{-i(-(\dsl{p}-\dsl{k})+m)} {(p-k)^2+m^2-i\epsilon}
\gamma^{\beta}\right]
\end{eqnarray*}

In the photon self energy expression, the $(-1)$ is due to the fermion
loop. The photon propagator is added a small mass $\mu^2$ in
anticipation. In the vertex correction, the expression is understood to
be sandwiched between $\bar{u}(p')$ and $u(p)$.

There are some new points to be noted.

$\bullet$ There is a momentum integration over $k$, unrelated to the
external momenta. This $k$ can be space-like, time-like or light-like.
In any frame, we may consider the region where $k^{\mu} \to \infty$,
then: (i) the electron self-energy $\sim \int d^4k k^{\alpha}/k^4 \sim
\int dk$ which is naively linearly divergent; (ii) the photon
self-energy $\sim \int d^4k/k^2 \sim \int kdk$ which is naively
quadratically divergent and (iii) the vertex correction $\sim d^4k
k^2/k^6 \sim \int dk/k $ which is naively logarithmically

$\bullet$ There are also other regions of integration space which give
divergent contributions, but these will be visible after the integrand
is put in a convenient form.

$\bullet$ The $i\epsilon$ in the denominator is quite significant now.
It implies that the 4-dimensional integration must be defined with the
$k^0$ integration being done first and the spatial integrations done

$\bullet$ The naively divergent integrals need to be regularised
i.e. a prescription for the integration must be supplied which will
manifest the divergence in an explicit form. The divergent contribution
so obtained, must be subtracted to obtain finite answers. This is
the process of renormalization.

$\bullet$ The integrand has a numerator which has tensor/spinor indices
while the denominator is a product of scalar factors. One combines the
denominator factors into a single factor form using the so called Feynman parameters. This is followed by the momentum integration with a
regularization and a finite part is identified. The integration over the
Feynman parameters is done last to get the answer. We explain these
steps now.

The Feynman/Schwinger trick:

\[
\mbox{Claim:\hspace{0.5cm}} \frac{1}{A_1\dots A_n} = (n-1)!
\int_0^{\infty}dx_1\dots dx_n \frac{\delta(\Sigma_i x_i -
1)}{(\Sigma_i x_iA_i)^n}  .
\]

Proof: Observe that $A^{-1} = \int_0^{\infty}d\alpha e^{-\alpha A}~, ~
Re(A) > 0$. Therefore,

\begin{eqnarray*}
\frac{1}{A_1\dots A_n} & = & \int_0^{\infty}d\alpha_1\dots
d\alpha_n e^{-\Sigma_i \alpha_iA_i}\cdot 1 ~ ~ , ~ ~ 1 =
\int_0^{\infty}dt \delta(t - \Sigma_i\alpha_i) ~ ~ \mbox{and} ~
~ \alpha_i \to t x_i ,\\
& = & \int_0^{\infty}dt \int_0^{\infty} dx_1\dots dx_n \ t^n
e^{-t\Sigma_i x_iA_i} \ \frac{1}{t} \ \delta(\Sigma_i x_i - 1)
\\
& = & \int_0^{1} dx_1\dots dx_n \delta(\Sigma_i x_i -
1)\int_0^{\infty}dt \ t^{n-1}e^{-t(\Sigma_i x_iA_i)} \\
& = &  (n-1)!  \int_0^{\infty}dx_1\dots dx_n
\frac{\delta(\Sigma_i x_i - 1)}{(\Sigma_i x_iA_i)^n} ~ ~
\because ~ \int_0^{\infty}dt \ t^{n-1}e^{-t \Lambda} =
\Lambda^{-n}\Gamma(n) \ .
\end{eqnarray*}

Note: Since each of the denominator factors are of the form
$A_i = (p_i-k)^2 + m_i^2 - i\epsilon$, we see that

\begin{eqnarray*}
\Sigma_i x_i A_i & = & k^2 -2k\cdot(\Sigma_i x_ip_i) + \Sigma_i
x_i(p_i^2 + m_i^2) - i\epsilon \\
& = & (k - \Sigma_i x_ip_i)^2 + M^2 - i\epsilon ~ ~ , ~ M^2(x_i,
p_i,m_i) := \Sigma_i x_i(p_i^2+m_i^2) - (\Sigma_i x_ip_i)^2 \ .
\end{eqnarray*}

The obvious step is to shift the integration variable, $k \to k +
\Sigma_i x_i p_i$ which simplifies the denominator - it has the same
form of a single propagator with the same $-i\epsilon$.

Let us begin with the fermion self energy.

§.§ Isolation of Divergence: Fermion Self Energy

\[
i\Sigma(p) = (ie)^2\int \frac{d^4k}{(2\pi)^4} \gamma^{\mu} \frac{-i
(-\dsl{k}+m)}{k^2+m^2-i\epsilon} \gamma^{\nu} \frac{-i\eta_{\mu\nu}}
\]

The Feynman trick gives,

\[
\frac{1}{(k^2+m^2-i\epsilon)( (p-k)^2+\mu^2-i\epsilon)}  = \int_0^1 dx
\frac{1}{[(k-xp)^2+M^2-i\epsilon]^2} ~ ,
\]

with, $ M^2 := x(1-x)p^2 + (1-x^2)m^2 +x\mu^2 $. Shifting $k \to k +
xp$, and using $\gamma^{\mu}\gamma_{\mu} = -4 \mathbb{1},
\gamma^{\mu}\dsl{k}\gamma_{\mu} = 2\dsl{k}$, gives

\[
i\Sigma = e^2\int_0^1\int \frac{d^4k}{(2\pi)^4} \frac{-4m
-2(\dsl{k}+x\dsl{)p}}{[k^2 + M^2(x,p) -i\epsilon]^2}
\]

Recalling that the $k^0$ integration is to be done first and that the
poles are at $k^0 = \pm \sqrt{\vec{k}^2+M^2} \mp i\epsilon$, we rotate
the contour anti-clockwise without crossing any singularity. This is
equivalent to putting $k^0 = i \underline{k}^0$ and $k^2 \to
(\underline{k}^0)^2+\vec{k}^2 = $Euclidean vector norm. This sends $\int
d^4k \to i\int d^4 \underline{k}$. For Euclidean integrals $\int d^4
\underline{k} \Case{\underline{k}^{\alpha}\gamma_{\alpha}}{(
\underline{k}^2 + M^2)^2} = 0 $, since the integrand is odd under $k \to
-k$. We are left with,

\[
i\Sigma = -ie^2\int_0^1dx \int \frac{d^4
\underline{k}}{(2\pi)^4} \frac{4m +
2x\dsl{p}}{(\underline{k}^2+M^2)^2} ~ ~ , ~ ~ M^2 := x(1-x)p^2 +
(1-x^2)m^2 +x\mu^2
\]

The $k-$integral is logarithmically divergent and needs to be regulated.
There are several ways of doing this. One is the Pauli-Villars
regularization which subtracts from the photon propagator, another
identical piece with arbitrarily large mass squared, $\Lambda^2$, i.e.

\[
\frac{1}{(p-k)^2+\mu^2} \to \frac{1}{(p-k)^2+\mu^2} -
\frac{1}{(p-k)^2+\Lambda^2} .
\]

For large $\underline{k}$, both terms go as $\underline{k}^{-2}$ and
cancel each other leaving a finite answer. Thus $i\Sigma \to
i\Sigma_{reg} := i\Sigma(\Lambda=\infty) - i\Sigma(\Lambda)$. The same
combining of denominators will produce identical terms with $M^2(\mu^2)
\to M^2(\Lambda^2)$. The momentum integration becomes,

\[
\int \frac{d^4\underline{k}}{(2\pi)^4}\left[\frac{1}
{(\underline{k}^2+M_{\mu}^2)^2} - \frac{1}
{(\underline{k}^2+M_{\Lambda}^2)^2}\right] = \int
\frac{d\Omega_3}{(2\pi)^4}\int_0^{\infty}dk |k|^3\left[\dots\right]
\]

The angular integration gives: $\Omega_{n-1} = \Case{2\pi^{n/2}}
{\Gamma(n/2)} = 2\pi^2$ for $n = 4$ and the integral becomes,

\begin{eqnarray*}
\int \frac{d^4 \underline{k}}{(2\pi)^2}\left[ \dots - \dots\right] & = &
\frac{1}{8\pi^2} \int_0^{\infty} \frac{dy}{2}
\left[\frac{y}{(y+M_{\mu}^2)^2} -\frac{y}{(y+M_{\Lambda}^2)^2}\right] ~
~ , ~ ~ k^2 =: y~ \mbox{substituted.} \\
& = & \frac{1}{16\pi^2} \int_0^{\infty} dy \left[\frac{1}{y+M_{\mu}^2} -
\frac{1}{y+M^2_{\Lambda}} - \frac{M^2_{\mu}}{(y+M_{\mu}^2)^2} +
\frac{M^2_{\Lambda}}{(y+M^2_{\Lambda})^2}\right] \\
& = & \frac{1}{16\pi^2}\left[ ln\left( \frac{y+M_{\mu}^2}
{y+M^2_{\Lambda}} \right)_0^{\infty} + \left( \frac{M_{\mu}^2}
{y+M^2_{\mu}} - \frac{M_{\Lambda}^2}{y+M^2_{\Lambda}}
\right)_{0}^{\infty} \right] \\
\therefore \int \frac{d^4 \underline{k}}{(2\pi)^2}\left[ \dots -
\dots\right] & = &
\frac{1}{16\pi^2}ln\left(\frac{M_{\Lambda}^2}{M_{\mu}^2}\right) \ .
\end{eqnarray*}

In the limit $\Lambda \to \infty, M^2_{\Lambda} \to x\Lambda^2$ and we

\begin{equation}\label{FermionSelfEnergyCorrection}
\boxed{ i\Sigma(p) = -i \frac{e^2}{8\pi^2}\int_0^1dx \ (2m+x\dsl{p})
ln\left[ \frac{x\Lambda^2}{x(1-x)p^2 + (1-x)m^2 + x\mu^2}\right] }
\end{equation}

Recall that the mass shift as well as the $Z_2$ are obtained from the
self energy as, $\delta m := m_{ph} - m = -\Sigma(\dsl{p} = -m_{ph}),
Z_2^{-1} = 1 - \Case{\Sigma(\dsl{p})}{d\dsl{p}}|_{\dsl{p} = -m_{ph}}$.
Since the self energy already has an explicit factor of $e^2$, in the
integrand, we can use $\dsl{p} = - m_{ph} \approx -m$. The derivative of
$\Sigma$ is obtained as,

\begin{eqnarray*}
\left. \frac{d\Sigma}{d\dsl{p}}\right|_{\dsl{p} = -m_{ph}} & = &
-\frac{\alpha}{2\pi}\int_0^1dx\ \left[ x
ln\left(\frac{x\Lambda^2}{-x(1-x)m^2 + (1-x)m^2 +
x\mu^2}\right) + \right. \\
& & \left. (2m - x\cdot m) \frac{-1}{-x(1-x)m^2 + (1-x)m^2 +
x\mu^2} (-2x(1-x)(-m)) \right]
\end{eqnarray*}

Simplification leads to,

\begin{equation}
\boxed{
\left. \frac{d\Sigma}{d\dsl{p}}\right|_{\dsl{p} = -m_{ph}} =
\frac{\alpha}{2\pi}\int_0^1dx\ \left[ -x
ln\left(\frac{x\Lambda^2}{(1-x)^2m^2+x\mu^2}\right) +
2(2 - x)\frac{x(1-x)m^2}{(1-x)^2m^2+x\mu^2} \right]
\end{equation}

\begin{equation}
\boxed{
\delta m = -\Sigma(\dsl{p} = -m_{ph}) = \frac{\alpha}{2\pi}
\int_0^1dx\
\end{equation}

The integrands are well behaved at both end points so the integrals are
finite and the leading contribution is given by the $\Lambda^2$
dependent part alone and for dimensional reasons we divide by the
fermion mass. Thus, the divergent contributions as $\Lambda \to
\infty$ take the form,

\begin{eqnarray*}
\delta m & \rightsquigarrow & \frac{\alpha}{2\pi}\int_0^1dx
(2-x)ln(\Lambda^2/m^2) = \frac{3\alpha}{4\pi}ln(\Lambda^2/m^2)
\\
Z_2^{-1} & \rightsquigarrow & 1 -
\frac{\alpha}{2\pi}\int_0^1(-xln(\Lambda^2/m^2)) = 1 +
\frac{\alpha}{4\pi}ln(\Lambda^2/m^2) \Rightarrow Z_2 \approx 1 -
\frac{\alpha}{4\pi}ln(\Lambda^2/m^2)
\end{eqnarray*}

The logarithmic divergence is thus manifested in terms of
$ln(\Lambda^2/m^2)$. There are of course finite pieces, but until we
take care of the divergent parts the finite parts are irrelevant.

§.§ Isolation of Divergence: Photon Self-Energy

\[
i\Pi^{\mu\nu}(q) = -e^2 \int \frac{d^4k}{(2\pi)^4} \frac{Tr[
\gamma^{\mu}(-\dsl{k}+m)\gamma^{\nu}(-\dsl{k}
-\dsl{q}+m)]}{(k^2+m^2-i\epsilon)( (k+q)^2+m^2-i\epsilon)}
\]

Using the trace formulae for the Dirac matrices (<ref>), we

\begin{eqnarray*}
\mbox{Numerator:} & = &
Tr[\gamma^{\mu}\dsl{k}\gamma^{\nu}(\dsl{k}+\dsl{q}) +
m^2\gamma^{\mu}\gamma^{\nu}] \\
& = & 4[ k^{\mu}(k+q)^{\nu} - \eta^{\mu\nu}k\cdot(k+q) +
(k+q)^{\mu}k^{\nu} -m^2\eta^{\mu\nu}] \\
& = & 4[ 2k^{\mu}k^{\nu} - \eta^{\mu\nu}(k^2+m^2) + (k^{\mu}k^{\nu} +
k^{\nu}q^{\mu} - \eta^{\mu\nu}k\cdot q)] \\
\mbox{Denominator:} & = & \frac{1}{k^2+m^2-i\epsilon}\frac{1}{(k+q)^2 +
m^2 -i \epsilon)}  \\
& = & \int_0^1 dx\ \frac{1}{[(k+xq)^2 + x(1-x)q^2 +m^2]^2}  ~ ~ ~ ~
\mbox{shift~} k \to k -xq,
\end{eqnarray*}

The shift generates additional terms linear in $k$ in the numerator.
Under integration, these terms drop out and we are left with,

\begin{eqnarray*}
i\Pi^{\mu\nu}(q) & = & (-4e^2)\int \frac{d^4k}{(2\pi)^4}
\frac{\left\{2k^{\mu}k^{\nu} + 2x^2q^{\mu}q^{\nu} -
\eta^{\mu\nu}(k^2+x^2q^2+m^2) -x(2q^{\mu}q^{\nu} -
\eta^{\mu\nu}q^2) \right\}}{(k^2 + M^2 -i\epsilon)^2}
\end{eqnarray*}

where, $\boxed{M^2 = x(1-x)q^2 + m^2}$. Doing Wick rotation as before,

\begin{eqnarray*}
i\Pi^{\mu\nu}(q) & = & (-4e^2)i\int \frac{d^4k}{(2\pi)^4}
\frac{\left\{2\underline{k}^{\mu}\underline{k}^{\nu} - 2x(1-x)
q^{\mu}q^{\nu} + x(1-x)\eta^{\mu\nu}q^2 -
\eta^{\mu\nu}(\underline{k}^2+m^2) \right\}}{(\underline{k}^2 +
M^2 -i\epsilon)^2}
\end{eqnarray*}

The tensorial integral,

\[
\int \frac{d^4 \underline{k}}{(2\pi)^4}
\frac{\underline{k}^{\mu}\underline{k}^{\nu}}{(\underline{k}^2 +
M^2)^2}  = A \delta^{\mu\nu} ~ ~ , ~ ~ \mbox{with}~ ~ A =
\frac{1}{4}\int \frac{d^4 \underline{k}}{(2\pi)^4}
\frac{\underline{k}^2}{(\underline{k}^2 + M^2)^2}
\]

There are several problems to be faced now. First, the integral is
quadratically divergent and shift in the momentum variable is
unjustified.  Second, we have now $\delta^{\mu\nu}$ and $\eta^{\mu\nu}$
in the numerator, confusing Lorentz covariance. Third, naively we see a
quadratically divergent piece with coefficient $\eta^{\mu\nu}$ but only
logarithmically divergent one with coefficient $q^{\mu}q^{\nu}$ casting
doubts on possibility of gauge invariance. We could try the
Pauli-Villars regulation, but we will introduce a different one, the
dimensional regularization. The Pauli-Villars may be seen in
[13]. The dimensional regularization has the main
virtue of keeping the Ward identity satisfied, even for non-abelian
gauge theories.

The idea is to think of the perturbation theory to be a member of a
class of similar theories  formulated in general $n$ dimensions. One
chooses a value of $n$ where the integrals are well defined ($n < 4$)
and analytically continues $n \to 4$. This isolates the original
divergence in a particular form (poles) providing the needed
regularization. Since the mass/length dimensions also depend on
space-time dimensions we need the coupling constants to be
dimensionfull. To maintain space-time covariance, we need to regard the
external momenta and the $\gamma$ matrices etc also to be
$n-$dimensional. We set $\epsilon := 4 -n$ and consider the limit
$\epsilon \to 0$. Since the integrals are over Euclidean momenta due to
the Wick rotation, the external momenta and the metric tensor need to
continue back to Minkowski signature at the end of the calculations.

In this regularization scheme, the following rules are adopted.

\begin{eqnarray}
\int \frac{d^nk}{(2\pi)^n}f(|k|^2) & = & \int
\frac{d\Omega_{n-1}}{(2\pi)^n}\int dk|k|^{n-1}f(|k|^2) ~ ~ , ~ ~
\boxed{\int d\Omega_{n-1} = \frac{2\pi^{n/2}}{\Gamma(n/2)}}
\label{DimensionalAngularIntgral}\\
\therefore \int \frac{d^nk}{(2\pi)^n}f(|k|^2) & = &
\left[\frac{2\pi^{n/2}}{\Gamma(n/2)}\right]\frac{1}{(2\pi)^n}\int
dk k^{n-1}f(k^2) \\
\int_0^{\infty}dk \frac{k^{n-1}}{(k^2 + M^2)^{\alpha}} & = &
\frac{1}{2}\int_0^{\infty}dx \frac{x^{\Case{n}{2}
-1}}{(x+M^2)^{\alpha}}  ~ ~ , ~ ~ \mbox{put}~ y =
\frac{M^2}{x+M^2} \nonumber \\
& = & \frac{M^{n-2\alpha}}{2}\int_0^1dy
y^{\alpha-1-\Case{n}{2}}(1-y)^{\Case{n}{2}-1}  =
\frac{M^{n-2\alpha}}{2}\beta(\alpha-n/2, n/2) \nonumber
\end{eqnarray}

\begin{equation} \label{DimensionalRadialIntgral}
\boxed{\therefore \int_0^{\infty}dk \frac{k^{n-1}}{(k^2 + M^2)^{\alpha}}
= \frac{M^{n-2\alpha}}{2}\frac{\Gamma(\alpha - n/2) \Gamma(n/2)}
{\Gamma(\alpha)} }
\end{equation}

\begin{equation} \label{DimensionalMomIntgral}
\boxed{ \int\frac{d^nk}{(2\pi)^n}\frac{1}{(k^2+M^2)^{\alpha}} ~ = ~
\frac{1}{(4\pi)^{n/2}}\frac{\Gamma(\alpha-n/2)}{\Gamma(\alpha)}
(M^2)^{n/2 - \alpha} }
\end{equation}

The $\Gamma$ function has isolated poles at $\alpha - n/2 = 0, -1,
-2,\dots$ i.e.  at $n = 2(\alpha + m), m = 0,1,\dots .$ The Gamma
function has the expansion $\Gamma(\epsilon/2) = 2/\epsilon -\gamma +
o(\epsilon)$ where, $\gamma = 0.57721 \dots$ is the Euler-Mascheroni Constant.

Since finite parts can be affected we need to keep $\epsilon$ in various
place too eg $\{\gamma^{\mu}, \gamma^{\nu}\} =
-2\delta^{\mu\nu}\mathbb{1}, Tr(\mathbb{1}) = n = 4-\epsilon,
\gamma^{\mu}\gamma^{\nu}\gamma_{\mu} = (2-\epsilon)\gamma^{\nu}$. These
terms will give contributions from the $1/\epsilon$ terms. The coupling
constant dimensions work as follows. $[\Psi] = (n-1)/2, [A_{\mu}] =
(n-2)/2 \Rightarrow [e] = n - (n-1) - (n/2 -1) = \epsilon/2$. Therefore
we write $e \to e\mu_0^{\epsilon/2}$ where $\mu_0$ is an arbitrary mass
scale which should disappear from physical quantities. The dimensions
are in mass units. The tensorial integral will now follow from
$k^{\mu}k^{\nu} \to \Case{1}{n}\delta^{\mu\nu}k^2$. We use these in
evaluating the $\Pi^{\mu\nu}(q)$.

\begin{eqnarray*}
\Pi^{\mu\nu}(\underline{q}) & = & (-ne^2\mu_0^{\epsilon})\int_0^1dx
\left[ \left\{ -2x(1-x)\underline{q}^{\mu}\underline{q}^{\nu} +
\delta^{\mu\nu}( x(1-x)\underline{q}^2 -m^2)\right\}\times \right. \\
& & \left. \int \frac{d^n \underline{k}}{(2\pi)^n} \frac{1}
{(\underline{k}^2 + M^2)^2} + \delta^{\mu\nu}(2/n-1) \int \frac{d^n
\underline{k}}{(2\pi)^n}\frac{\underline{k}^2 + M^2 -
M^2}{(\underline{k}^2 + M^2)^2}\right] ~ , \mbox{where,} \\
& & \mbox{\hspace{4.0cm}} \boxed{M^2 = x(1-x)q^2 + m^2}\\
& = & (-ne^2\mu_0^{\epsilon}\int_0^1 dx\left[
\delta^{\mu\nu}\left(\frac{2}{n}-1\right)\underbrace{\int
\frac{d^n
\underline{k}}{(2\pi)^n}\frac{1}{\underline{k}^2 +
M^2}}_{\circled{1}} + \underbrace{\int \frac{d^n
\underline{k}}{(2\pi)^n}\frac{\underline{k}^2}{(\underline{k}^2
+ M^2)^2}}_{\circled{2}}\times \right. \\
& & \left. \mbox{\hspace{1.0cm}} \left\{
-2x(1-x)\underline{q}^{\mu}\underline{q}^{\nu} +
\delta^{\mu\nu}\left(-m^2 - x(1-x)\underline{q}^2
-\left(\frac{2}{n}-1\right)M^2\right)\right\} \right] \\
\circled{1} & = & \frac{1}{(4\pi)^{n/2}} \frac{\Gamma(1-n/2)}{\Gamma(1)}
(M^2)^{n/2-1} ~ = ~ \frac{M^2}{(4\pi)^2} \left(\frac{M^2}{4\pi}\right)
^{-\epsilon/2} \frac{\Gamma(\epsilon/2)}{(\epsilon/2-1)} \\
\circled{2} & = & \frac{1}{(4\pi)^{n/2}} \frac{\Gamma(2-n/2)}{\Gamma(1)}
(M^2)^{n/2-2} ~ = ~ \frac{1}{(4\pi)^2} \left(\frac{M^2}{4\pi}\right)
^{-\epsilon/2} \Gamma(\epsilon/2) \\
\therefore \circled{1} & = &
\left(\frac{M^2}{\epsilon/2-1}\right)\circled{2}  ~ ~ ~ , ~ ~ ~
\frac{\epsilon}{2} -1 = 1-\frac{n}{2} = \frac{n}{2}\left(\frac{2}{n} -
1\right) \\
\Pi^{\mu\nu}(q) & = & (-ne^2\mu_0^{\epsilon})\int_0^1dx \left[
\delta^{\mu\nu} \left( \frac{2}{n}-1 \right)
\frac{M^2}{\frac{n}{2}(\frac{2}{n}-1)} \right. \\
& & \mbox{\hspace{0.5cm}} \left. + \delta^{\mu\nu}\left(-m^2 -
\left(\frac{2}{n}-1\right)M^2 + x(1-x)q^2\right)
-2x(1-x)q^{\mu}q^{\nu}\right]\times\circled{2} \\
But, [\dots] & = & \delta^{\mu\nu}\left\{\cancel{\frac{2}{n}M^2}
\underbrace{-m^2 -\left(\cancel{\frac{2}{n}M^2}-M^21\right) +
x(1-x)q^2}_{2x(1-x)q^2}\right\} -2x(1-x)q^{\mu}q^{\nu} \\
&  = & 2x(1-x)(q^2\delta^{\mu\nu} - q^{\mu}q^{\nu}) ~ ~ \Rightarrow ~ ~
\boxed{\Pi^{\mu\nu}(\underline{q}) = (\underline{q}^2\delta^{\mu\nu} -
\underline{q}^{\mu}\underline{q}^{\nu})\Pi(\underline{q}^2) ~
\mbox{with,}  } \\
\Pi(q^2) & := & -ne^2\mu_0^{\epsilon}\int_0^1
(2x)(1-x) \\
& = & -\frac{2(4-\epsilon)e^2}{(4\pi)^2}\int_0^1dx\
\left\{ln\left(\frac{4\pi\mu_0^2}{M^2}\right) \frac{\epsilon}{2}+1\right\}
\end{eqnarray*}

Using $a^{\epsilon/2} = 1 + \Case{\epsilon}{2}ln(a) + o(\epsilon^2)$, we

\begin{equation}
\boxed{ \Pi(\underline{q}^2) = -\frac{2	\alpha}{\pi}\int_0^1dx\
x(1-x)\left(\frac{2}{\epsilon}-\gamma\right)\left(1 +
\frac{\epsilon}{2}ln \left(\frac{4\pi\mu_0^2}{M^2}\right)
\left(1-\frac{\epsilon}{4}\right)\right) }
\end{equation}

Notice that for $\underline{q}^2$, $M^2 = m^2$ and

\begin{equation}
\boxed{\Pi(0) = -\frac{2\alpha}{\pi}\frac{2}{\epsilon}\int_0^1dx(x-x^2)
+ \mbox{constant}\ = -\frac{2\alpha}{3\pi}\frac{1}{\epsilon},} ~
\Rightarrow ~ \boxed{Z_3 = 1-\frac{2\alpha}{3\pi\epsilon}\ }
\end{equation}

We see the divergence in $\Pi(0), Z_3$ and there is of course no mass
shift. We return to the interpretation of $\Pi(q^2)$ later.

Remark: The $\epsilon^{-1}$ pole in dimensional
regularization corresponds to logarithmic divergence in Pauli-Villars.
this is indicated by comparing the regulated integral
$\int[(k^2+M^2)^{-2} - (k^2\Lambda^2)^{-2}]$ in Pauli-Villars, which
goes over to $\Case{1}{4\pi^2}ln(x\Lambda^2/M^2)$. In the dimensional
regularization, our integral $\circled{2}$ goes as
$\Case{1}{4\pi^2}\Case{2}{\epsilon}$. Hence, $\epsilon^{-1}
\leftrightarrow ln(\Lambda/M)$. Thanks to the dimensional regularization
maintaining the Ward identity, the $q^2\delta^{\mu\nu} - q^{\mu}q^{\nu}$
got pulled out and the superficially quadratically divergent integral go
converted to a logarithmically divergent one.

§.§ Isolation of Divergence: Vertex function

We compute $\delta\Gamma^{\mu} = \Gamma^{\mu} - \gamma^{\mu}$, with $p^2
= (p')^2 = -m^2$ and sandwiching by $\bar{u}(p'), u(p)$ is implicit.

\begin{equation*}
\delta\Gamma^{\mu}(p,p')  =  -ie^2\int \frac{d^4k}{(2\pi)^4}
\frac{\eta_{\alpha\beta}}{k^2+\mu^2-i\epsilon} \left[\gamma^{\alpha}
\frac{(-(\dsl{p}'-\dsl{k})+m)} {(p'-k)^2+m^2-i\epsilon}
\gamma^{\mu} \frac{(-(\dsl{p}-\dsl{k})+m)}
{(p-k)^2+m^2-i\epsilon} \gamma^{\beta}\right]
\end{equation*}

We have three denominators and the Feynman trick leads to,

\begin{eqnarray*}
\frac{1}{k^2+\mu^2-i\epsilon} \frac{1}{(p-k)^2+m^2-i\epsilon}
\frac{1}{(p'-k)^2+m^2-i\epsilon} \hspace{3.0cm} & & \\
= \int_0^1dx\int_0^1dy\int_0^1dz \frac{2}{\left[(k-xp-yp')^2
+M^2-i\epsilon\right]^3} & & \mbox{where,} \\
M^2 := x(p^2+m^2)+y((p')^2+m^2) +z\mu^2 - (xp+yp')^2 & & \\
\therefore M^2 = 0+0++z\mu^2 + m^2(x+y)^2 + q^2xy \hspace{2.3cm} & &
\end{eqnarray*}

Since $q^2 > 0$ (space-like) for this diagram, $\boxed{M^2 = (x+y)^2m^2
+ xyq^2+z\mu^2}$ is manifestly positive. As usual shifting the momentum
$k \to k+xp+xp'$, doing the Wick rotation and dropping the terms linear
in $k$ in the numerator, we get,

\begin{equation*}
\delta\Gamma^{\mu}(p,p') =
\frac{d^4\underline{k}}{(2\pi)^4}
\frac{2\times\mbox{Nr}}{[\underline{k}^2+M^2]^3}
\end{equation*}

\begin{eqnarray*}
Nr & = & \gamma^{\alpha} \left\{(\dsl{p}'-\dsl{k}-x\dsl{p}-y\dsl{p}')
\gamma^{\mu}
(\dsl{p}'-\dsl{k}-x\dsl{p}-y\dsl{p}')\right\}\gamma_{\alpha} +
m^2\gamma^{\alpha}\gamma^{\mu}\gamma_{\alpha} \\
& & -m\gamma^{\alpha}\left\{\gamma^{\mu} (\dsl{p}-x\dsl{p}-y\dsl{p}')+
(\dsl{p}'-x\dsl{p}-y\dsl{p}') \gamma^{\mu}\right\}\gamma_{\alpha} + \
\mbox{linear in $k$}
\end{eqnarray*}

Now we need to use the identities,

\begin{eqnarray}
\gamma^{\alpha}\gamma^{\mu}\gamma_{\alpha} & = &
-2\eta^{\alpha\mu}\gamma_{\alpha} - \gamma^{\mu}(-4) =
2\gamma^{\mu} \\
\gamma^{\alpha}\gamma^{\rho}\gamma^{\sigma}\gamma_{\alpha} & = &
4\eta^{\rho\sigma} \\
\gamma^{\alpha}\gamma^{\rho}\gamma^{\mu}\gamma^{\sigma}\gamma_{\alpha}
& = & 2\gamma^{\sigma}\gamma^{\mu}\gamma^{\rho} \\
\therefore
\gamma^{\alpha}\dsl{k}\gamma^{\mu}\dsl{k}\gamma^{\alpha} & = & 2
\dsl{k}\gamma^{\mu}\dsl{k} = -4k^{\mu}\dsl{k} + 2\gamma^{\mu}k^2
~ ~ \because ~ ~ \dsl{k}\dsl{k} = -k^2 \\
\gamma^{\alpha}\{\dots\}\gamma^{\mu}\{\dots\}\gamma_{\alpha} & =
& 2\{(1-x)\dsl{p}-y\dsl{p}'\}\gamma^{\mu}\{(1-y)\dsl{p}' -
x\dsl{p}\} \\
\gamma^{\alpha}\{\gamma^{\mu}(\dots) + (\dots)\gamma^{\mu}\}
\gamma_{\alpha} & = & 4\left\{(1-x)p^{\mu}-y(p')^{\mu} +
(1-y)(p')^{\mu}-xp^{\mu}\right\} \\ \nonumber
\end{eqnarray}

Using these, the numerator takes the form,

\begin{eqnarray*}
\bar{u}(p')(Nr)u(p) & = & \bar{u}(p')\left([-4k^{\mu}\dsl{k} +
2k^2\gamma^{\mu}]_{\circled{1}} + 2[\left\{(1-x)
\dsl{p}-y\dsl{p}' \right\} \gamma^{\mu} \left\{(1-y)\dsl{p}'-x\dsl{p}
\right\}]_{\circled{2}} \right. \\ & & \hspace{2.0cm}\left.
+2[m^2\gamma^{\mu}]_{\circled{3}} - 4m[(1-2x)p^{\mu} +
\end{eqnarray*}

Observe that $\dsl{p}u(p) = -mu(p), \bar{u}{p'}\dsl{p}' =
-m\bar{u}{p'}$. In $\circled{1}, \int d^4k k^{\mu}\dsl{k} \propto
\gamma^{\mu}$ (after Pauli-Villars). We simplify $\circled{2}$ as,

\begin{eqnarray*}
\frac{1}{2}\bar{u}(p')\circled{2}u(p) & = &
\bar{u}(p')[(1-x)(1-y)\dsl{p} \gamma^{\mu}\dsl{p}'
+xym^2\gamma^{\mu}+x(1-x)m\dsl{p}\gamma^{\mu} +
\dsl{p}\gamma^{\mu} & \to & -2p^{\mu} - (-m)\gamma^{\mu} ~ , ~
\gamma^{\mu}\dsl{p}' \to -2(p')^{\mu} + m\gamma^{\mu}  ~ ~
\mbox{between the spinors}.
\end{eqnarray*}

Thus we see that, between the spinors, all the terms in the numerator
can be put in the form of $(p+p')^{\mu}[\dots] + \gamma^{\mu}[\dots]$
and therefore the integral can be arranged as contributing to the form
factors $F_1, F_2$. The intermediate steps, including the Pauli-Villars
regularization, are left as an exercise. The result is (from

\begin{eqnarray}
F_1(q^2) - 1 & = & \frac{\alpha}{2\pi}\int_0^1dx\ dy\ dz
\delta(1-x-y-z)\left[ln\left(\frac{z\Lambda^2}{M^2}\right) +
\right. \nonumber \\
& & \hspace{2.0cm}\left. \frac{1}{M^2}\left( (1-x)(1-y)q^2 +
(1-4z+z^2)m^2\right)\right] \\
F_2(q^2) & = & \frac{\alpha}{2\pi}\int_0^1dx\ dy\
dz\delta(1-x-y-z)\left[\frac{2m^2z(1-z)}{M^2}\right], ~ ~ ~
\mbox{with} \\
& & \hspace{2.0cm} \boxed{M^2 := q^2xy + m^2(1-z)^2 +z\mu^2} \nonumber
\end{eqnarray}

(i) In the $\delta F_1$, there is a UV divergence ($\Lambda \to \infty$) coming
from the large $k$ region, even for $q^2 = 0$. This violates the
$\delta F_1(0) = 0$ condition and thus changes the electric charge. The
condition can be naturally restored by defining $\boxed{\delta
F_{1,ren}(q^2):= \delta F_1(q^2) - \delta F_1(0). }$ This is a renormalization prescription.

(ii) The $\delta F_1$ also has an “infrared divergence” ($\mu^2 \to
0$) at $q^2 = 0$ coming from the $M^{-2}(0) = m^{-2}(1-z)^{-2}$ from the
$z\to 1$ region.

(iii) The $F_2(q^2)$ has no divergences. Its value at $q^2=0$ gives the
anomalous magnetic moment and can be evaluated explicitly as,

\begin{eqnarray}\label{AnomalousMagMoment}
F_2(0) &  = &  \frac{\alpha}{2\pi}\int_0^1dx\ dy\
dz\delta(1-x-y-z)\frac{2z}{1-z} ~ = ~
\frac{\alpha}{\pi}\int_0^1dz\int_0^{1-z}\frac{z}{1-z}  ~ = ~
\frac{\alpha}{2\pi} \nonumber \\
& \Rightarrow & \boxed{\frac{g-2}{2} = \frac{\alpha}{2\pi}}  ~ ~
\hspace{2.5cm} \mbox{One of the triumphs of QED!}
\end{eqnarray}

The renormalized $F_1(q^2)$ is obtained as,

\begin{eqnarray}\label{Renormalized-F1}
F_{1,ren}(q^2) & = & 1 + \frac{\alpha}{2\pi}\int_0^1dx\ dy\
dz\delta(1-x-y-z) \left[ ln\left( \frac{m^2(1-z)^2} {m^2(1-z)^2+q^2xy}
\right) \right.  \nonumber \\
& & \left. \hspace{1.5cm} +\frac{m^2(1-4z+z^2) +
q^2(1-x)(1-y)}{m^2(1-z)^2+q^2xy+\mu^2z} -
\frac{m^2(1-4z+z^2)}{m^2(1-z)^2+\mu^2z}\right]
\end{eqnarray}

We have to face the divergences now. All three radiative corrections
have UV divergence while the vertex function has IR divergence as well.
For dealing with the IR divergence, we need to look at another process
called “Bremsstrahlung” (breaking radiation in German).

§.§ Bremsstrahlung Cross-section to $o(\alpha)$

We know from the classical electrodynamics that accelerated charge
radiates. A quantum mechanical view of such a radiation is depicted in
the diagrams below.

\[
\parbox[c]{3.0cm}{
\begin{tikzpicture}[scale=0.5]
\begin{feynman}
\vertex (a);
\vertex [right=of a] (c) ;
\vertex at ($(a)!0.5!(c)$) (b) ;
\vertex [right=of c] (d) ;
\vertex [below=0.75of c] (e) ;
\vertex [left=of e] (f);
\vertex [right=of e] (g);
\vertex [above right=0.75of b] (h) ;

\diagram* {
(a) -- [fermion] (b) -- [fermion] (c) --
[fermion] (d) ,
(f) -- [fermion] (e) -- [fermion] (g) ,
(c) -- [photon] (e) ,
(b) -- [photon,momentum=\(k\)] (h) ,
\end{feynman}
\end{tikzpicture}
\hspace{0.5cm} + \hspace{0.5cm}
\parbox[c]{3.0cm}{
\begin{tikzpicture}
\begin{feynman}
\vertex (a);
\vertex [right=of a] (b) ;
\vertex [right=of b] (d) ;
\vertex at ($(b)!0.5!(d)$) (c) ;
% shifts the vertex closer	\vertex [below=0.75 of b] (e) ;
\vertex [below=0.75of b] (e) ;
\vertex [left=of e] (f);
\vertex [right=of e] (g);
\vertex [above right=0.75of c] (h) ;

\diagram* {
(a) -- [fermion] (b) -- [fermion] (c) --
[fermion] (d) ,
(f) -- [fermion] (e) -- [fermion] (g) ,
(b) -- [photon] (e) ,
(c) -- [photon,momentum=\(k\)] (h) ,
\end{feynman}
\end{tikzpicture}
\hspace{0.5cm} + \hspace{0.5cm} \mbox{emission of any number of
\]

The diagrams depict a scattering process which causes the top electron
to `accelerate' and `emit' a (on-shell) photon. This is the QFT view of
`radiation from an accelerated charge'. The bottom charged line
represents some heavy fermion/boson/source, while the connecting wavy
line denotes mediation of the interaction. The first diagram suggests
photon emission before the kick while the second diagram suggests
photon emission after the kick. As per the rules of the
relativistic QFT, both contributions are to be added and mod-squared to
get to the cross-section. To the leading order - single photon emission
- the cross-section is order $\alpha$. This cross-section also has a an
IR divergence as $k \to 0$. An on-shell photon with arbitrarily small
momentum is called a `soft photon'.

To isolate the divergence, focus on the following diagrams:

\[
\parbox[c]{3.0cm}{
\begin{tikzpicture}
\begin{feynman}
\vertex (a);
\vertex [right=0.75 of a, blob] (b) {};
\vertex [above right=of b] (c) ;
\vertex [below right=of b] (d) ;
\vertex at ($(b)!0.5!(d)$) (d') ;
\vertex [above right=0.75 of d'] (e) ;

\diagram* { (a) -- [scalar] (b) ,
(b) -- [fermion,edge label=\(p'\)]
(c) , (b) -- [anti fermion, edge
label'=\(p-k\)] (d') -- [anti
fermion,edge label'=\(p\)] (d) , (d') --
[boson,edge label'=\(k\)] (e), };
\end{feynman}
\end{tikzpicture}
\hspace{1.0cm} + \hspace{1.0cm}
\parbox[c]{3.0cm}{
\begin{tikzpicture}
\begin{feynman}
\vertex (a);
\vertex [right=0.75 of a, blob] (b) {};
\vertex [above right=of b] (c) ;
\vertex at ($(b)!0.5!(c)$) (c') ;
\vertex [below right=of b] (d) ;
\vertex [right=0.75 of c'] (e) ;

\diagram* { (a) -- [scalar] (b) ,
(b) -- [fermion,edge label=\(p'+k\)]
(c') -- [fermion, edge label=\(p'\)] (c),
(b) -- [anti fermion, edge label'=\(p\)]
(c') -- [boson,edge label'=\(k\)] (e),
\end{feynman}
\end{tikzpicture}
\]

We are computing the amplitude for the process $e \to e + \gamma$. Let
$\mathcal{M}_0(k',l')$ denote the part of the amplitude of electron
interacting with the external source, it is depicted as the blob. Here
$k', l'$ denote the out-going and in-coming momenta respectively. The
full amplitude is then given by (polarizations
$\varepsilon(\vec{k},\lambda)$ taken to be real),

\begin{eqnarray*}
i\mathcal{M}(p',p,k) & = & (ie) \bar{u}(p')\left[ \mathcal{M}_0(p',p-k)
\frac{-i(-(\dsl{p}-\dsl{k}) + m)}{(p-k)^2+m^2-\epsilon} \gamma^{\mu}
\varepsilon_{\mu} \right. \\
& & \hspace{3.0cm} \left. \varepsilon_{\mu}\gamma^{\mu}
\frac{-i(-(\dsl{p}'+\dsl{k}) + m)}{(p'+k)^2+m^2 -i\epsilon}
\mathcal{M}_0(p'+k,p)\right]u(p)
\end{eqnarray*}

To compare with the classical radiation formula, consider the limit
wherein the external source interaction approximates as,
$\mathcal{M}_0(p',p-k) \approx \mathcal{M}_0(p'+k,p) \approx
\mathcal{M}_0(p',p)$. Next, since the photon is soft, we neglect the
$\dsl{k}$ in the fermion numerators. The denominators simplify as,
$(p-k)^2+m^2 = -m^2 -2p	\cdot k + 0 = m^2 = -2p\cdot k$ and likewise,
$(p'+k)^2+m^2 = +2p'\cdot k$. Furthermore, $(-\dsl{p} +
m)\gamma^{\mu}u(p) \to 2p^{\mu}u(p)$, and
$\bar{u}(p')\gamma^{\mu}(-\dsl{p}'+m) \to \bar{u}(p')(p')^{\mu}$. Thus
the amplitude takes the form,

\[
i\mathcal{M}(p',p,k) = e \underbrace{\left[\bar{u}(p')
\mathcal{M}_0(p',p) u(p)\right]}_{\mbox{elastic scattering
amplitude}} \times \underbrace{\left(-\frac{\varepsilon\cdot
p}{p\cdot k} + \frac{\varepsilon\cdot
p'}{p'\cdot k} \right)}_{\mbox{extra $k$
dependent factor}}
\]

The total, unpolarized cross-section is obtained as,

\begin{equation}
d\sigma(p\to p'+k) = d\sigma(p\to p')\int \frac{d^3k}{2|k|(2\pi)^3}
\sum_{\lambda=1,2} e^2 \left| \frac{\varepsilon\cdot p'}{p'\cdot k} -
\frac{\varepsilon\cdot p}{p'\cdot k} \right|^2
\end{equation}

The integration is over soft photons i.e. $|\vec{k}| < k_{max}$. Notice
that The integral is dimensionless.

The integrand is the probability density of radiating a photon of
momentum $k$ within $d^3k$ and any transverse polarization, accompanying
the electron scattering with $p\to p'$ (“acceleration kick”, $p'\neq
p+k$) while the integral is the total probability of emission of a
soft photon with energy up to $k_{max}$.

Multiplying the integrand by the photon energy, $|\vec{k}|$ gives the
expectation value of the energy carried away by soft photons up to
energy $k_{max}$,

\[
\mathcal{E}_{soft}(k_{max}) := \langle E\rangle=
\int\frac{d^3k}{(2\pi)^3} \sum_{\lambda} \frac{e^2}{2}
\left|\vec{\varepsilon}\cdot \left(\frac{\vec{p}'}{p'\cdot k} -
\frac{\vec{p}}{p\cdot k}\right)\right|^2
\]

The polarization sum in the cross-section can be done as follows. Noting
that $k\cdot(\Case{p'}{p'\cdot k} - \Case{p}{p\cdot k}) = 0$, we can
drop the $k\tilde{k}$ terms in the completeness relation for the
polarizations, effectively taking $\Sigma_{\lambda} \varepsilon_{\mu}
\varepsilon_{\nu} = \eta_{\mu\nu}$. Hence,

\[
\sum_{\lambda}\left[\left(\frac{p'}{p'\cdot k} - \frac{p}{p\cdot
\left[\left(\frac{p'}{p'\cdot k} - \frac{p}{p\cdot
k}\right)^{\nu}\varepsilon_{\nu}\right] =
\left(\frac{p'}{p'\cdot k} - \frac{p}{p\cdot
k}\right)\cdot\left(\frac{p'}{p'\cdot k} - \frac{p}{p\cdot
k}\right) ,
\]

and the expectation value of energy carried by soft photons is given by,

\begin{equation} \label{ProbRadn}
\boxed{ \mathcal{E}_{soft}(k_{max}) =
\int\frac{d^3k}{(2\pi)^3}\sum_{\lambda}\frac{e^2}{2}
\left[-\frac{m^2}{(p'\cdot k)^2} - \frac{m^2}{(p\cdot k)^2} -
2\frac{p\cdot p'}{(p\cdot k)(p'\cdot k)}\right] }
\end{equation}

Note that the integral represents average radiated energy inferred from
the quantum mechanical probability distribution for emission of a
single soft photon with $|\vec{k}| < k_{max}$.

An Aside: Incidentally, integrand in the r.h.s. is also the
classical  formula for energy carried away by the $k^{th}$ Fourier
mode of the electromagnetic field. To see this, consider a classical
trajectory $z^{\mu}(\tau) = p^{\mu}/m$ for $\tau < 0$ and equals
$(p')^{\mu}/m$ for $\tau > 0$. The corresponding current is,

\begin{eqnarray*}
J^{\mu}(x) & = &
\int_0^{\infty}d\tau\frac{(p')^{\mu}}{m}\delta^4(x - z(\tau)) +
\int_{-\infty}^0d\tau\frac{p^{\mu}}{m}\delta^4(x - z(\tau)) ~ ~
\Rightarrow \\
A^{\mu}(\vec{k}) & = & -
\frac{e}{|\vec{k}|}\left(\frac{(p')^{\mu}}{p'\cdot k} -
\frac{p^{\mu}}{p\cdot k} \right) ~ ~ ~ ~  \mbox{(using the
Retarded Green function.)}
\end{eqnarray*}

The energy of the above radiation field is the precisely as given by the
integrand in the r.h.s. of eq.(<ref>) [13].
The integral now represents the total energy carried by Fourier modes
with momenta $|\vec{k}| < k_{max}$.

The integral, $\mathcal{E}_{soft}(k_{max})$,  thus has two
different interpretations.

Returning to the evaluation of the integral in eq.(<ref>), let
us choose a frame in which  the initial and final energies of the
electron are the same: $E' = E \Rightarrow (p')^0 = p^0$ and we write $p
= E(1, \vec{v}),\ p' = E(1,\vec{v}'),\ k = (k := |k|, \vec{k})$. This is
equivalent to $p^2 = -m^2$ and $m^2/E^2 = 1-\vec{v}^2$.  With this
choice, $p\cdot k = Ek(-1+\hat{k}\cdot \vec{v}),\ p'\cdot k =
Ek(-1+\hat{k} \cdot \vec{v}')$ and $p\cdot p' =
E^2(-1+\vec{v}\cdot\vec{v}')$. The $k^2$ from the $d^3k$ cancels and the
integrated emitted energy takes the form,

\begin{eqnarray*}
\mathcal{E}_{soft}(k_{max}) & = & -\frac{e^2}{2}\int dk\int
\frac{d\Omega}{8\pi^3}\left[
\frac{m_0^2}{E^2}\left(\frac{1}{(1-\hat{k}\cdot \vec{v})^2}    +
\frac{1}{(1-\hat{k}\cdot \vec{v}')^2} \right) -
\frac{2(1-\vec{v}\cdot \vec{v}')}{(1-\hat{k}\cdot
\vec{v})(1-\hat{k}\cdot \vec{v}')} \right]  \\
I(\vec{v}, \vec{v}') & := & \int\frac{d\Omega_{\hat{k}}}{4\pi}\left[
\frac{2(1-\vec{v}\cdot \vec{v}')}{(1-\hat{k}\cdot
\vec{v})(1-\hat{k}\cdot \vec{v}')} -
\frac{m^2}{E^2}\left(\frac{1}{(1-\hat{k}\cdot \vec{v})^2}    +
\frac{1}{(1-\hat{k}\cdot \vec{v}')^2} \right) \right] ~ ~ \Rightarrow \\
\mathcal{E}_{soft}(k_{max}) & = & \frac{e^2}{4\pi^2}\int dk I(\vec{v},
\vec{v}') ~ ~ \approx ~ ~ \frac{\alpha}{\pi}k_{max} I(\vec{v}, \vec{v}')
\end{eqnarray*}

The integral is actually divergent at the upper end. It has been cut off
at $k_{max}$ which is provided by the inverse of the duration over
which the electron receives the kick. This is given by $k_{max} \sim
|\vec{p} - \vec{p}'| := |\vec{q}|$. This is a physical cutoff.

The angular integration receives the maximum contribution from when $
\hat{k}$ is parallel to $\vec{v}$ or $\vec{v}'$ depending upon if it is
the initial or the final state bremsstrahlung. Given the directions $v
\hat{v},\ \hat{v}'$, the $\hat{k}$ varies between these two directions
(smaller angle). Thus, to pick up contribution when $\hat{k}$ is
parallel to $\vec{v}$, we define $cos\theta$ by $\hat{k}\cdot\vec{v} =
vcos\theta$ which varies between $|\vec{v}|cos\theta =
\vec{v}\cdot\vec{v}'$ and $cos\theta = 1$. Similarly for contribution
from nearly parallel to $\vec{v}'$, we define $cos\theta$ by
$\hat{k}\cdot\vec{v}' = |\vec{v}'|cos\theta$ which varies from
$|\vec{v}'|cos\theta = \vec{v}\cdot\vec{v}'$ to $\cos\theta = 1$.

Additionally, in the extreme relativistic limit where we can neglect the
$m^2/E^2$ terms, we can approximate
$(1-\hat{k}\cdot\vec{v})(1-\hat{k}\cdot\vec{v}') \approx
(1-|\vec{v}|cos\theta)(1-\vec{v}\cdot\vec{v}')\ or\ \approx
(1-|\vec{v}'|cos\theta)(1-\vec{v}\cdot\vec{v}')$.  The angular integral
then approximates as,

\begin{eqnarray*}
I(\vec{v}, \vec{v}') & \approx & \int^1_{|\vec{v}|cos\theta =
\vec{v}\cdot\vec{v}'} dcos\theta \frac{1-\vec{v}\cdot
\vec{v}'}{(1-vcos\theta)(1-\vec{v}\cdot\vec{v}')}  +
\int^1_{|\vec{v}'|cos\theta = \vec{v}\cdot\vec{v}'} dcos\theta
\frac{1-\vec{v}\cdot
\vec{v}'}{(1-v'cos\theta)(1-\vec{v}\cdot\vec{v}')} \\
& \simeq &
ln\left(\frac{1-\vec{v}\cdot\vec{v}'}{1-|\vec{v}|}\right) +
\approx ln\left( \frac{(p\cdot p')^2}{E^2(E-p)(E-p')} \right) ~
~ \\
& \simeq  & 2 ln \frac{q^2}{m^2}   ~ ~ \mbox{where} ~ ~ q :=
\end{eqnarray*}

We have used: $(1-\vec{v}\cdot\vec{v}') = -p\cdot p'/E^2\ ,
(1-|\vec{v}|) = (E-|\vec{p}|)/E, \ (E-|\vec{p}|)(E - |\vec{p}'|) \approx
(E -|\vec{p}|)^2$ and $E(E-|\vec{p}|) \approx E^2 - \vec{p}^2 = m^2$ in
getting to the last equation.

Thus the radiated energy in the soft modes is given by,

\begin{equation}
\mathcal{E}_{soft}(k_{max}) ~ = ~
\frac{\alpha}{\pi}\int_0^{k_{max}} dk I(\vec{v}, \vec{v}') ~
\xrightarrow[E \gg m]{} ~ \frac{2\alpha}{\pi}\int_0^{k_{max}}dk\
ln(q^2/m^2) \ .
\end{equation}

Interpreting the above energy classically, what would be the total
number of photons emitted, $N_{\gamma}$? Notice that $I(\vec{v},
\vec{v}')dk$ is the contribution to the radiated energy from the Fourier
modes with energy, $|\vec{k}|$. The equivalent number of photons would
be $\Case{I(\vec{v}, \vec{v}')dk}{|\vec{k}|}$ and the total number
$N_{\gamma}$ is obtained by integrating over the energies:

\[
N_{\gamma} =
\frac{\alpha}{\pi}\int_0^{k_{max}}\frac{dk}{k}I(\vec{v},
\vec{v}') \approx
\frac{2\alpha}{\pi}\int_0^{k_{max}}\frac{dk}{k}\ ln(q^2/m^2)
\]

This is clearly divergent from the lower limit.

But this is the same expression for the total quantum mechanical
probability for emission of a soft photon up to energy $k_{max}$ and thus
is divergent.

This is the IR divergence of the QED cross-section for
the bremsstrahlung process.

If the photon is given a small mass $\mu$, then the $\int dk/k$ will
give $ln((|q|\simeq k_{max})/\mu)$ while $I(\vec{v}, \vec{v}')$ gives
the $ln(q^2/m^2)$.  Hence,

\begin{equation}
\boxed{ d\sigma(p\to p'+k) = d\sigma(p\to p')\cdot
\frac{\alpha}{\pi}\underbrace{\left[ln(q^2/\mu^2)\cdot
ln(q^2/m^2)\right]}_{\mbox{Sudakov double log}} }
\end{equation}

§ TREATMENT OF DIVERGENCES:

At the first non-trivial attempt at computing radiative corrections, we
encounter divergences of the UV type (from large loop momentum) and of
IR type (from small loop momentum in massless propagators). How do we
understand the physical origin of these? How do we adjust the
computational procedure so as to make unambiguous predictions to
be confronted with observations? Let us recapitulate what we have got.

$\bullet$ Quite generally, using only Poincare covariance and assumption
about the possible spectrum of a theory with mass gap, the
Kallen-Lehmann representation gave us,

\begin{eqnarray*}
\Delta'_F(p) & = & Z\Delta_0(p, m_{ph}) +
\int_{m^2_{th}}^{\infty} d\sigma^2 \rho(\sigma^2) \Delta_0(p,
\sigma)  ~ ~ , ~ ~ 0 < Z < 1 \\
& & 	\Delta_0(p,\sigma) ~ := ~ \frac{-i}{p^2 +
\sigma^2-i\epsilon} \hspace{1.5cm} \mbox{(Free propagator of
mass $\sigma$)} \\
S'_F(p) & = & Z_2 S_0(p,m_{ph}) + \int_{m^2_{th}}^{\infty}
d\sigma^2 \frac{-i\{\rho_1(\sigma^2)\dsl{p} +
\rho_2(\sigma^2)\}} {p^2+\sigma^2-i\epsilon} \\
& & 	S_0(p, \sigma) ~ := ~ -i\frac{(-\dsl{p}+
\sigma)}{p^2+\sigma^2-i\epsilon} \hspace{1.0cm} \mbox{(Free
propagator of mass $\sigma$)} \\
(D'_F)_{\mu\nu}(q) & = & \left(\eta_{\mu\nu} -
\frac{q_{\mu}q_{\nu}}{q^2}\right)\left\{
\frac{Z_3}{q^2-i\epsilon}  + \int_{0}^{\infty} d\sigma^2
\frac{\Pi(\sigma^2)} {q^2+\sigma^2-i\epsilon}\right\}
\end{eqnarray*}

The $Z$'s are the field (or wavefunction) renormalization constants.
They are determined as the residues at the (isolated) pole at the physical mass, $m_{ph}$. This is without any perturbation series.

In perturbation series though we obtained (for fermion and photon),

\[
S'_F(p) = \frac{1}{\dsl{p}+m-\Sigma(\dsl{p})} ~ ~ ~ \& ~ ~ ~
D'_F(q) = \left(\eta_{\mu\nu} -
\frac{q_{\mu}q_{\nu}}{q^2}\right)\left[\frac{1}{q^2}\
\frac{1}{1-\Pi(q^2)}\right]
\]

This is consistent with the general expectation that the physical masses
are determined by the poles in the exact propagator while the
corresponding residues determine the $Z$'s. Thus,

\[
\delta m := m_{ph} - m = -\Sigma(\dsl{p}=-m_{ph}) ~ , ~ Z_2^{-1}
= 1 - \frac{d\Sigma(\dsl{p})}{d\dsl{p}}|_{\dsl{p} = -m_{ph}} ~ ,
~ Z_3^{-1} = 1 - \Pi(0).
\]

and of course the photon physical mass is zero.

$\bullet$ The vertex function for on shell fermion masses takes the

\[
\gamma^{\mu}(p,p') := \gamma^{\mu}F_1(q^2) +
i\sigma^{\mu\nu}q_{\nu}F_2(q^2) ~ , ~ \sigma_{\mu\nu} :=
\frac{i}{2}[\gamma{\mu},\gamma_{n}] ~ , ~ \dsl{p} = -m_{ph} =
\dsl{p}'\ .
\]

This decomposition is understood to be sandwiched between $\bar{u}(p')$
and $u(p)$. Since $m_{ph} = m + \delta m = m + o(\alpha)$, for the first
order corrections, we can take $m_{ph} = m$. Here are the expressions we

\begin{eqnarray}
(I) \hspace{1.3cm} \Sigma(p) & = &
\frac{\alpha}{2\pi}\int_0^1dx\ (2m+x\dsl{p}) ln
\frac{x\Lambda^2}{x(1-x)p^2+(1-x)m^2+x\mu^2-i\epsilon}\\
\delta m & = & \frac{\alpha}{2\pi}\int_0^1dx\ (2-x)ln
\frac{x\Lambda^2}{(1-x)^2m^2 + x\mu^{2}} , \\
Z_2 & = & 1 + \frac{\alpha}{4\pi}ln(\Lambda^2/m^2) ~ ~ \\
(II) \hspace{1.1cm} \Pi(q^2) & = & -\frac{2\alpha}{\pi}\int_0^1dx\
x(1-x)\left\{\frac{2}{\epsilon} - \gamma -
ln\left(\frac{m^2+q^2x(1-x)}{\mu_0^2}\right)\right\} \\
& = & -\frac{2\alpha}{3\pi}\frac{1}{\epsilon} + o(\epsilon^0) ~
\Rightarrow ~ Z_3 = 1-\frac{2\alpha}{3\pi}\frac{1}{\epsilon} \\
(III) \hspace{0.9cm} F_1(q^2) & = & 1 + \frac{\alpha}{2\pi}\int dx\ dy\
dz\ \delta(1-x-y-z)\left\{ln \left(\frac{z\Lambda^2}{M^2}\right) \right.
\nonumber  \\
& & \left. \hspace{2.0cm} \frac{-(1-x)(1-y)q^2 +
(1-4z+z^2)m^2}{M^2}\right\} \\
F_2(q^2) & = & \frac{\alpha}{2\pi}\int dx\ dy\ dz\delta(1-x-y-z)\left\{
\frac{2m^2z(1-z)}{M^2} \right\} ~ , \\
& & \hspace{3cm} M^2 := q^2xy + m^2(1-z)^2 + \mu^2z \\
\therefore F_1(0) & = & 1 + \frac{\alpha}{2\pi}\int dx\ dy\ dz
\delta(1-x-y-z) \left\{ln\left( \frac{z\Lambda^2}{(1-z)^2m^2} \right)
\right. \nonumber \\ & & \left. \hspace{4.0cm} +
\frac{1-4z+z^2}{(1-z)^2}\right\} \\
F_2(0) & = & \frac{\alpha}{2\pi}\int dx\ dy\ dz\delta(1-x-y-z)
\left(\frac{z}{1-z}\right) = \frac{\alpha}{2\pi}  ~ =: ~ \frac{g-2}{2}
\\
(IV) \hspace{0.5cm} d\sigma_{p\to p'+k} & = & d\sigma_{p\to
p'}\int\frac{d^3k}{2k(2\pi)^2} \sum_{\lambda} e^2\left( \frac{
\vec{\varepsilon}\cdot \vec{p}'}{\vec{p}'\cdot \vec{k}} - \frac{
\vec{\varepsilon}\cdot \vec{p}}{\vec{p}\cdot \vec{k}}\right)^2 \\
& = & d\sigma_{p\to p'}\left[\frac{e^2}{4\pi^2}\int\frac{dk}{k}I(\vec{v},
\vec{v}')\right] ~ ~ \mbox{where,} \\
I(\vec{v}, \vec{v}') & = & \int\frac{d\Omega}{4\pi}\left\{
\frac{2(1-\vec{v}\cdot\vec{v}')}{(1-\hat{k}\cdot\vec{v})
(1-\hat{k}\cdot\vec{v}')} - \frac{m^2/E^2}{(1-\hat{k}\cdot\vec{v})^2} -
\frac{m^2/E^2}{(1-\hat{k}\cdot\vec{v}')^2}\right\} \hspace{0.8cm} \\
& & \mbox{where} ~ p^{\mu} = (E, E\vec{v})\ , \ (p')^{\mu} = (E,
E\vec{v}')\ , \ k^{\mu}(|\vec{k}| , \vec{k}) ~ \mbox{is used.} \\
& & \mbox{For $|\vec{v}| \approx |\vec{v}'| \approx 1\ $},
I(\vec{v},\vec{v}') \approx 2ln(\vec{q}^2/m^2), \ q := p' - p ~
\Rightarrow \nonumber \\
d\sigma_{p\to p'+k} & = & d\sigma_{p\to
p'}\left[\frac{\alpha}{\pi}ln(q^2/\mu^2)\ ln(q^2/m^2)\right]
\end{eqnarray}

§.§ Treatment of the IR divergences

Let us consider the IR problem first. Since $F_1$ has both the UV and
the IR divergence, we will separate these by using the UV renormalized
$F_{1,ren}$ which was defined through, $\delta F_{1,ren}(q^2) := \delta
F_1(q^2) - \delta F_1(0)$. This guarantees that the renormalized $F_1$
satisfies the condition $F_1(0) = 1$. Using the expressions above, we
see that,

\begin{eqnarray*}
\delta F_{1,ren}(q^2) & = & \frac{\alpha}{2\pi}\int dx\ dy\ dz\
\delta(1-x-y-z) \left\{\frac{-(1-x)(1-y)q^2 +
(1-4z+z^2)m^2}{M^2(q^2)} \right. \\
& & \left. \hspace{4.0cm} - \frac{(1-4z+z^2)m^2}{(1-z)^2m^2+z
\mu^2}\right\}
\end{eqnarray*}

The IR divergence comes from $z\to 1 \leftrightarrow x\sim y\sim 0
\leftrightarrow M(q^2,x,y)$ would vanish but for the photon mass
$\mu^2$. To isolate the divergence, it suffice to take $x=y=0,\ z=1$ in
the numerator. Doing the $x$ integration using the delta function gives
$x = 1-y-z > 0 \Rightarrow y < 1-z$ and leads to,

\[
\delta F_{1,ren} \approx \frac{\alpha}{2\pi}\int_0^1dz \int_0^{1-z}dy
\left\{ \frac{-2m^2-q^2}{(1-z)^2m^2+y(1-y-z)q^2+\mu^2}
\frac{2m^2}{(1-z)^2m^2+\mu^2} \right\}
\]

Substituting $y := (1-z)u, v:= 1-z$, the leading contribution in the
limit $\mu \to 0$ is expressed as [13],

\begin{eqnarray}
& & \boxed{F_{1,ren}(q^2) \approx 1 -\frac{\alpha}{2\pi}f_{IR}(q^2)\
ln\left( \frac{m^2 \ or\ q^2}{\mu^2} \right) } ~ ~ ~ \mbox{where,}\\
& & \boxed{f_{IR}(q^2) \ = \ \int_0^1du
\left[\frac{m^2+q^2/2}{m^2+u(1-u)q^2}\right] - 1 ~ }
\end{eqnarray}

Since $F_1$ is the coefficient of the $\gamma^{\mu}$ term, we can
replace $e \to eF_{1,ren}$ in the electron scattering off a classical potential. The cross-section is then given by,

\[
\left(\frac{d\sigma}{d\sigma}\right) \simeq
\left(\frac{d\sigma}{d\sigma}\right)_{tree}\left[1-\frac{\alpha}{\pi}
f_{IR}(q^2) ln\left( \frac{m^2 \ or\ q^2}{\mu^2} \right)\right] .
\]

Notice that $e\to eF_1 \Rightarrow e^2\to e^2F_1^2 \Rightarrow
\alpha/(2\pi) \to \alpha/(\pi)$.

Not only is this divergent as $\mu \to 0$, for non-zero $\mu$ it is
actually negative implying negative cross-section! In the limit of
$q^2 \to \infty$ ($q^2$ is space-like and hence positive), $f_{IR}(q^2)
\to ln(q^2/m^2) \Rightarrow F_{1,ren}(q^2) \simeq 1 -
\Case{\alpha}{2\pi}ln(q^2/m^2)\,ln(q^2/\mu^2)$ and hence,

\begin{equation}\label{CrossIR}
\boxed{ d\sigma(p\to p') \simeq d\sigma_{tree}(p\to
p')\left[1-\frac{\alpha}{\pi}ln(q^2/m^2)\ ln(q^2/\mu^2)\right] ~
~ , ~ ~ q^2 \to \infty\ , \ \mu^2\to 0. }
\end{equation}

The Bremsstrahlung cross-section on the other hand is,

\begin{equation}\label{BremCross}
\boxed{ d\sigma(p\to p'+k) = d\sigma(p\to
ln\frac{q^2}{\mu^2}\right] = d\sigma(p\to
ln\frac{q^2}{\mu^2}\right]. }
\end{equation}

Both the cross-sections above are IR divergent and both suffer from
ambiguity from contamination of soft photons.

We already noted that detectors are unable to distinguish a charged
particle accompanied by soft photons below detector sensitivity. What is
the appropriate theoretically computed quantity which reflects this
limitation of the detection process?

When an experimenter reports a detection of a scattered electron, he/she
is actually giving an estimate of the probability that an electron,
$e(p')$ is detected and a photon is not detected. This
probability is the probability for a process with no emitted photon plus
the probability that there are accompanying soft photons with energies
below the detection threshold, $\varepsilon_{th}$. In equation,

\begin{equation} \label{MeasuredCross}
\boxed{ (d\sigma)_{measured} = d\sigma\ (\ p\to p'\ ) + d\sigma\
(\ p\to p'+k\ ;\  |\vec{k}| < \varepsilon_{th}\ )\  . }
\end{equation}

But this is precisely the sum of the two cross-sections given in
eqns.(<ref>,<ref>). We see that the leading
contribution as $q^2 \to \infty, \mu^2\to 0$ is exactly canceled out in
the observed cross-section!

Note: For a general $q^2$, the sum of the two cross-sections
is given by [13],

\[
(d\sigma)_{measured} = d\sigma_{tree}(\ p\to p'\ )\left\{ 1 -
\frac{\alpha}{\pi} f_{IR}(q^2) ln\left( \frac{q^2\ or\ m^2\
}{\mu^{2}}\right) + \frac{\alpha}{2\pi}\ I(\vec{v}, \vec{v}')\ ln(
\varepsilon_{th}/\mu^2 ) \right\}
\]

It turns out that without the limit $q^2\to \infty$, it still holds that
$I(\vec{v}, \vec{v}') \to 2 f_{IR}$ [13]. For general
$q^2$, the coefficient of $f_{IR}$ is $ln(q^2)$ or $ln(m^2)$. For large
$q^2$ we can of course drop  $m^2$. Furthermore, experimentally it is
easier to track the behavior of the cross-section as a function of
$q^2$, so taking $q^2 \gg m^2$, we write the unambiguous and measurable
prediction as,

\begin{equation}\label{FinalMeasured}
\boxed{ \left(\frac{d\sigma}{d\Omega}\right)_{measured} =
\left(\frac{d\sigma}{d\Omega}\right)_{tree}(p\to p')\left[1 -
\frac{\alpha}{\pi}ln\frac{q^2}{m^2}\ ln\frac{q^2}{\varepsilon_{th}^2} +
o(\alpha^2)\right]_{q^2\gg m^2} }
\end{equation}

Appreciate that we had parametrised the IR divergence in terms of the
photon mass $\mu^2$; we used the renormalized $F_{1,ren}(q^2)$ to
separate the IR from the UV and finally, identified the quantity which
is actually reported by experiments which is the sum of the two

This has been a demonstration at the 1-loop. It is non-trivial result of
a great deal of work, that the basic mechanism of cancellation works to
all orders in $\alpha$. There are other types of divergences analogous
to the IR divergences, the so called “mass singularities” which do not
cancel but can be factorised in a convenient form and then eliminated
from the observed cross-sections. See the book by G. Sterman

UV divergences is a different ball game and needs a different procedure.

§.§ Treatment of the UV divergences

All the three radiative corrections in, $\Sigma, \Pi$ and $F_1$ have the
UV divergences parametrised in terms of the Pauli-Villars cut-off
$\Lambda$ or the dimensional regularization $\epsilon^{-1}$ pole. Note
that the bremsstrahlung process does not have a UV divergence as there
is physical cut-off for the energy of the “soft photons”. We need to
pay attention to the $Z$ factors now.

Recall that the $S-$matrix element definition, via the LSZ reduction
procedure, had a factor of $\Case{1}{\sqrt{Z}}$ for each external line.
There was also an amputation of the external line, effected by the
equation of motion operator eg $\Box-m^2$ acting on the Feynman
propagator $\Delta_F$ for the external line. The $\sqrt{Z}$ entered from
the asymptotic condition relating the interacting field to the in/out
fields. The in/out fields satisfy their respective field equations with
physical masses and also have the normalization factors with
$\omega_k$ which also contain the physical masses. In the momentum
space, these factors associated with external lines are of the form
$\Case{1}{\sqrt{Z}}(p^2+m^2)\Delta'_{F}(p)$ where the $\Delta_F$
contains the self-energy giving it the form $\Delta'_F(p) =
(p^2+m^2-\Pi(p^2))^{-1}$ (for scalars).

In perturbative computation of the self-energy however, the $m$ is the
mass parameter in the $\cal{L}$ which is not the physical mass. In fact
from the Kallen-Lehmann representation, we know that the physical mass
is determined from $p^2+m^2-\Pi(p^2)|_{p^2=-m_{ph}^2} = 0
\leftrightarrow -m^2_{ph}+m^2 = \Pi(-m^2_{ph})$. Expanding $\Pi(p^2)$
about $-m^2_{ph}$ gives us,

\begin{eqnarray*}
\Pi(p^2) & = & \pi(-m^2_{ph}) + (p^2+m^2_{ph})\Pi'(-m^2_{ph}) +
\dots  ~ ~ \Rightarrow \\
(\Delta'_{F})^{-1}(p) & = & p^2+m^2-\Pi(-m^2_{ph}) -
(p^2+M^2_{ph})\Pi'(-m^2_{ph}) + \dots \\
& = & (p^2 + m^2_{ph})\left\{1 - \Pi'(-m^2_{ph})+\dots\right\} ~
= ~ (p^2 + m^2_{ph})\left\{Z_{\Phi}^{-1}+\dots\right\} \\
\therefore \frac{1}{\sqrt{Z_{\Phi}}}(p^2+m^2_{ph})\Delta'_F(p) &
= & \sqrt{Z_{\Phi}}\left\{1 + o(\ (p^2+m^2_{ph})^2\ ) \right\}
\end{eqnarray*}

Since we evaluate the $S-$matrix elements on shell, all the higher order
terms vanish. The net result is that:

In an $S-$matrix element, for each external line
introduce a factor of $\sqrt{Z_{field}}\times\ $ appropriately
normalized wavefunction and now do not include self-energy
corrections on the external lines.

Consider the electron scattering off external potential. The
corresponding invariant amplitude is given by $e(\sqrt{Z_2})^2
\bar{u}(p') \Gamma^{\mu}(p',p) u(p)$ with $\Gamma^{\mu} = \gamma^{\mu}
F_1 + (\dots)F_2$. The $F_2$ has no divergences while $Z_2, F_1$ both
are divergent. Thus we write, $Z_2 = 1 + \delta Z_2, F_1 = 1 + \delta
F_1$ with the $\delta$'s representing the $o(\alpha)$ divergent
corrections.  Thus we write,

\begin{eqnarray*}
\mathcal{M}_{e\to e}(p',p) & = & e(1+\delta Z_2)\left\{
\bar{u}(p')\gamma^{\mu}u(p)\ (1+\delta F_1(q^2)) +
\bar{u}(p')\frac{i\sigma^{\mu\nu}q_{\nu}}{2m_{ph}}u(p) \ \delta
F_2(q^2)\right\} \\
& = & e\left[\bar{u}(p')\gamma^{\mu}u(p)\right]\left\{1+\delta
F_1+\delta Z_2\right\} + \bar{u}(p')
\frac{i\sigma^{\mu\nu}q_{\nu}}{2m_{ph}}u(p) \ \delta F_2(q^2) ~ ~
\mbox{But,}\\
\delta Z_2 & = & \frac{\alpha}{2\pi}\int_0^1dz \left[-zln
\frac{z\Lambda^2}{M^2} + 2(2-z)\frac{z(1-z)m^2}{M^2}\right] ~ , ~ M^2 =
(1-z)^2m^2 + z\mu^2\\
\delta F_1(0) & = & \frac{\alpha}{2\pi}\int_0^1dz(1-z)\left[ln
\frac{z\Lambda^2}{M^2} + (1+z^2-4z)\frac{m^2}{M^2}\right] \\
\therefore \delta F_1(0)+\delta Z_2 & = & \frac{\alpha}{2\pi}\int_0^1dz
\left[(1-2z)ln \frac{z\Lambda^2}{M^2} + \frac{m^2}{M^2}\times \right. \\
& & \left. \hspace{4.0cm} \left\{(4-2z)(1-z)z +
\end{eqnarray*}

Integrate the first  term by parts,

\begin{eqnarray*}
\int_0^1dz(1-2z)ln \frac{z\Lambda^2}{M^2} & = & (z-2z^2)\left.
ln\frac{z\Lambda^2}{M^2}\right|_0^1 -
\int_0^1dz(z-z^2)\left\{\frac{1}{z} - \frac{2(1-z)(-1)m^2 +
\mu^2}{M^2}\right\} \\
& = & -\int_0^1dz \frac{1-z}{M^2}\{ M^2 + 2z(1-z)m^2 - z\mu^2\}\\
& = & -\int_0^1dz\frac{1-z}{M^2}\{m^2(1-z)(1+z)\} = -\int_0^1dz
\frac{m^2(1-z)^2(1+z)}{m^2\left((1-z)^2 +z\frac{\mu^{2}}{m^2}\right) }\\
%& = & \mbox{non-singular as $z\to 1$ and thus finite.} \\
\therefore \delta F_1(0) + \delta Z_2 & = & \frac{\alpha}{2\pi}
\int_0^1dz\left[-\frac{m^2(1-z)(1-z^2)}{M^2} + \frac{m^2}{M^2}(1-z)\{ 4z
- 2z^2 + 1+z^2-4z\}\right] \\
& = & \frac{\alpha}{2\pi}\int_0^1dz\frac{m^2}{M^2}(1-z)(1-z^2)\{-1 + 1\}
= 0 ~ \mbox{(!)}
\end{eqnarray*}

Thus not only has the $ln\Lambda^2$ divergence canceled, we have got
$\delta Z_2 = - \delta F_1(0)$ and thus the entire correction takes the

\[
\mathcal{M}_{e\to e}(p',p) = e\bar{u}(p')\gamma^{\mu}u(p)\left\{1 +
\underbrace{\delta F_1(q^2) - \delta F_1(0)}_{\delta F_{1,ren}}\right\}
+i\bar{u}(p')\frac{i\sigma^{\mu\nu}q_{\nu}}{2m_{ph}}u(p)\delta F_2(q^2)
\]

We recover the ad hoc prescription of using $\delta F_{1,ren}$
introduced earlier while discussing the IR divergences. Note that the
subtracted term was crucial for the IR divergence cancellation. It was
of course  physically expected since $\delta F_1(q^2=0) = 0$ should
actually hold to all orders. Note that the UV divergence in the vertex
function, $\delta F_1(q^2)$, is canceled against the divergence in
$\delta Z_2$ coming from electron self-energy.

Now we are left with the divergence in $Z_3$ from the $\Pi(q^2=0)$ for
the photon self-energy. We still have the electron mass-shift $\delta m$
which is divergent.

§.§.§ 1-Loop Renormalization: Charge Screening and Lamb shift

Consider charged particle scattering by exchanging a photon, necessarily
off-shell (space-like in fact). Replace the free propagator
$(D_F)_{\mu\nu}(q^2)$ by the exact propagator which includes the photon
self-energy, $(D'_F)_{\mu\nu}(q^2)$. The exact propagator has additional
terms $q_{\mu}q_{\nu}$. But thanks to the Ward identity, these do not
contribute to $S-$matrix elements and effectively, $(D'_F)_{\mu\nu}(q^2)
= -i\eta_{\mu\nu}[q^2(1-\Pi(q^2))]^{-1}$. This has a pole at $q^2=0$
with residue $Z_3 = [1-\Pi(0)]^{-1}$. The replacement thus gives a
factor of $Z_3$. The photon internal line connects two vertices
contributing $e^2$. Thus, if we identify $\boxed{e^2Z_3 =:
e^2_{ph}}$ , then all divergences coming from the photon self-energy
can be neatly absorbed in the physical, measured charge which is
finite.  For the mass parameter too, we identified $m^2_{ph} = m^2 +
\delta m$, which absorbs the divergence in $\delta m$ into the
Lagrangian parameter $m$. The identification $e_{ph} := e\sqrt{Z_3}$ is
called charge renormalization while $m^2_{ph} := m^2 + \delta
m$ is called mass renormalization. For external photon line, we
will just have $\sqrt{Z_3}$ factor and no self-energy corrections and of
course no mass shift.

Thus the exact photon propagator may be replaced by the free propagator
by simultaneously replacing $\alpha$ by $\alpha_{eff}$ defined below.

\begin{eqnarray*}
\alpha (D'_F)_{\mu\nu}(q^2) & = & (D_F)_{\mu\nu}(q^2)
\frac{\alpha}{1-\Pi(q^2)} = \frac{\eta_{\mu\nu}} {q^2-i\epsilon}
\left[ \frac{\alpha_{ph}(1-\Pi(0))}{1-\Pi(q^2)} \right] :=
\alpha_{eff}(q^2)\left[ \frac{\eta_{\mu\nu}} {q^2-i\epsilon}
\right] \\
& & \mbox{with,} \hspace{1.0cm} \boxed{\alpha_{eff}(q^2) =
\frac{\alpha_{ph}}{1-(\Pi(q^2) - \Pi(0))}}
\end{eqnarray*}

To appreciate the implication of the above procedure, recall the
discussion of the effective potential inferred from the tree level
scattering for both the Yukawa and the QED coupling. Let us use the same
formula (for QED), but include the self-energy correction for the photon
propagator. The inferred potential is,

\begin{eqnarray}
V(\vec{x}) & = & \int\frac{d^3q}{(2\pi)^3}e^{i\vec{q}\cdot
\vec{x}} \frac{-e^2_{ph}}{\vec{q}^2(1-\Pi(q^2)+\Pi(0))} \\
\Pi(q^2) & = & -\frac{2\alpha}{\pi}\int_0^1dx\ x(1-x)\left\{
\frac{2}{\epsilon} -\gamma -
ln\left(\frac{m^2+q^2x(1-x)}{\mu_0^2}\right)\right\} \\
\Pi(q^2)-\Pi(0) & = & -\frac{2\alpha}{\pi}\int_0^1dx\ x(1-x)\left\{
ln\left(\frac{m^2}{m^2+q^2x(1-x)}\right)\right\} \\
& \xrightarrow[]{\vec{q}^2 \ll m^2} & \frac{2\alpha}{\pi}\int_0^1dx\
x^2(1-x)^2\frac{q^2}{m^2} = \frac{\alpha}{15\pi}\frac{q^2}{m^2} ~ ~
\mbox{(NR limit)}\\
\therefore V(\vec{x}) & \approx & \int\frac{d^3q}{(2\pi)^3}
e^{i\vec{q}\cdot \vec{x}} \frac{-e^2_{ph}} {\vec{q}^2 (1-
\frac{\alpha}{15\pi} \frac{q^2}{m^2})} \\