S & := & \Omega_-^{\dagger} \Omega_+ ~ ~ ~ \mbox{on
$\left[Range(\Omega_+)\right]\ \cap \
\left[Domain(\Omega_-)\right] \ \neq \ \emptyset$ \ .}
\end{eqnarray}

Note that the range of $\Omega_+$ are states in $\cal{H}_{in}$ while the
domain of $\Omega_-^{\dagger}$ are states in $\cal{H}_{out}$. The scattering operator, $S$, is thus well defined when weak asymptotic
completeness holds. Similarly, we define $S^{\dagger} :=
\Omega_+^{\dagger}\Omega_-$ which is also well defined when weak
asymptotic completeness holds. If in addition asymptotic completeness
holds, then

\begin{eqnarray}
S^{\dagger}S & = & \Omega_+^{\dagger} \Omega_-
\Omega_-^{\dagger} \Omega_+ ~ = ~ \Omega_+^{\dagger}(\mathbb{1}
- P_{bound})\Omega_+ ~ = ~ \mathbb{1} - \mathbb{0} ~ ~
\mbox{likewise,} \\
S S^{\dagger} & = & \Omega_-^{\dagger} \Omega_+
\Omega_+^{\dagger} \Omega_- ~ = ~ \Omega_-^{\dagger}(\mathbb{1}
- P_{bound})\Omega_- ~ = ~ \mathbb{1} - \mathbb{0} ~ .
\end{eqnarray}

Thus the scattering operator is unitary when asymptotic completeness
holds. This is regardless of the Moller operators being non-unitary.
The scattering operator $S$ maps from free solutions $\phi_{in}(t)$
to free solutions $\phi_{out}$. Its matrix elements between the basis
of free solutions define the $S-matrix$. Explicitly, let
$\phi^a_{in}(t)$ and $\phi^b_{out}(t)$ denote incoming and out going
free solutions. Then,

\begin{equation} \label{SMatrixDefn1}
\boxed{	S_{ba} :=\langle\phi^b_{out}(t)|S|\phi^a_{in}(t)\rangle
=:\langle\phi^b_{out}(t)|\tilde{\phi}^a_{out}(t)\rangle \ . }
\end{equation}

The $\tilde{\phi}^a_{out}(t)$ state is a state that evolved from a
specific incoming state while $\phi^b_{out}(t)$ is an arbitrary free
state and $S_{ba}$ is the inner product between the two.

Hence, $S_{ba}$ gives the probability amplitude for a free state
$|\phi_{in}(t)\rangle$ to evolve into an outgoing state
$|\phi_{out}^b(t)\rangle$ - in short a transition amplitude.

The $S$ operator defined in terms of the Moller operators is manifestly
time independent. It matrix elements however seem to have a time
dependence through those of the $\phi_{in/out}(t)$. This apparent time
dependence is actually absent and can be seen as follows.

Recall the boxed relations among $\Omega_{\pm}, \Omega_{\pm}^{\dagger},
H$ and $H_0$. From these we get,

\begin{eqnarray*}
SH_0 & = & \Omega_-^{\dagger}\Omega_+H_0 = \Omega_-^{\dagger}H\Omega_+ =
H_0\Omega_-^{\dagger}\Omega_+ = H_0 S. \\
id_tS_{ba} & = & id_t(\langle\phi^b_{out}(t)|S|\phi^a_{in}(t)\rangle) \\
& = & - \langle\phi^b_{out}(t)H_0 S |\phi^a_{in}(t)\rangle + 0 +
\langle\phi^b_{out}(t)iS H_0 |\phi^a_{in}(t)\rangle \\
& = & \langle\phi^b_{out}(t)|[S, H_0]|\phi^a_{in}(t)\rangle ~ = ~ 0.
\end{eqnarray*}

In summary:

* A generic scattering system has $\cal{H}$ and the three
Hamiltonians $H, H_0, H_0'$ specified;

* There exist asymptotic subspaces of $\cal{H}, \cal{H}_{in},
\cal{H}_{out}$ elements of which give full solutions approaching
some free solutions as $t \to \pm\infty$;

* Provided the system admits scattering states, there exist the
Moller operators, $\Omega_{\pm}$ on $\cal{H}$ with ranges in
$\cal{H}_{in/out}$ respectively;

* If weak asymptotic completeness holds, $\exists$ the scattering
operators $S: \cal{H}_{in} \to \cal{H}_{out}$ and its adjoint
$S^{\dagger}$. The scattering operator $S$ is unitary provided
asymptotic completeness holds.

* Once $S_{ba}$ are known, the cross-section can be computed, since
the number of scattered particles is proportional to
$|S_{ba}|^2$, while the flux of the incoming particles is given
by the probability current corresponding to $\phi_{in}(t)$.

* Much of the rigorous analysis goes in addressing the existence and
uniqueness properties of the proposed scattering system. Further
details may be seen in [9, 10].

In practice, rarely a scattering system is specified in
sufficient details to address these issues, but they are at the
very basic definitional level.

Returning to interacting quantum fields, we have the candidate free
states - the states of free quantum fields with known free Hamiltonian
$H_0$. We may specify the interacting system by additional Lorentz
invariant terms in the Lagrangian or specifying the Hamiltonian $H$.
However, we do not know what the Hilbert space of interacting quantum
fields is let alone if it admits scattering states. But we do know what
properties we would like to have for a scattering interpretation. The
task now is to make suitable assumptions and build a computational
recipe to define and compute the $S-$matrix elements.

§.§ Scattering with quantum fields: Heisenberg Picture

The above discussion has been in terms of the states evolving in time
i.e. in Schrodinger picture. For interacting quantum fields we do not
know the state space (except for the free fields). We do have Poincare
covariance  as a guide and this puts the field operators and their
Poincare transforms at the center stage. We are thus led to use the
Heisenberg picture and formulate the recipe in terms of the evolving
operators. As states do not evolve in Heisenberg picture, we postulate
certain properties for the states, postulate Poincare covariance on the
operator evolution and make specific additional assumption of
“asymptotic condition” to incorporate the scattering feature. A
reference for this material is [11].

* We choose the basis states of the interacting quantum fields to be
labeled by mutually commuting, conserved Noether charges whose
existence is guaranteed by the presumed symmetries. Among these
symmetries is included the Poincare group (proper, orthochronous
with/without discrete symmetries). The Poincare symmetry
immediately gives eigenvalues of $P^{\mu}$ as a set of labels.
Our first assumption is regarding the spectrum of $P^{\mu}$.

* There exists a unique vacuum, eigenstate $|0\rangle$ with
$P^{\mu}|0\rangle = 0$;

* Eigenvalues, $p^{\mu}$, of $P^{\mu}$ lie in the forward
light cone i.e.  $p^2 \le 0, p^0 > 0$;

* $\exists$ stable, single particle states with masses
$m_i$: $P^{\mu}|p^{i}\rangle = p^{\mu}_i|p^i\rangle ~ ,
~ p^2_i = - m_i^2 \ .$

* The eigenvalues $0$ and $m_i^2$ of $P^2$ are
discrete in the sense of being isolated values.
The “multi-particle” states have their
total momentum time-like but the corresponding $-p^2$
can take any value above their total rest mass.

Spectrum for a single, massive scalar
field. The two particle states form a continuum
above and including $-(p_1 + p_2)^2 = 4m^2$ hyperboloid.

Note: If there are “internal symmetries” (i.e.
non-space-time symmetries), then it is conceivable that the
zero eigenvalue of $P^{\mu}$ is degenerate, and carries some
representation of the internal symmetry group (usually compact).
Every one of these states must still be singlets of the Poincare
group. But now, these vacua themselves carry some quantum
numbers and quantum fields acting on these will create single
particle states labeled by tensor product of representations of
the internal symmetry group. This is not done and one stipulates
the vacuum to be a singlet under all symmetries.

Note: We have taken the stable particles to have a
non-zero mass. This is a statement about the spectrum of $P^2$
in the full, exact interacting quantum field theory and not at
some approximation of it. This would seem to exclude the
massless representations of the Poincare group for interacting
quantum fields. In practice i.e. in perturbation theory we will
use massless photons and see the accompanying infrared problems.

* An interacting field operator, denoted schematically by
$\Phi(x)$ satisfies an operator equation of the form,
$(\Box - m_0^2)\Phi(x) = J(x)$ where $J(x)$ is an
operator built out of $\Phi$ and possibly other field
operators. We re-write the field equation as $(\Box -
m^2)\Phi(x) = J(x) - (m^2 - m_0^2)\Phi(x) =:
\tilde{J}(x)$

* The field obeys the equal time (anti-)commutation

\[
\left[\Phi(t, \vec{x}), \Phi(t, \vec{y})\right]
= 0 = \left[\Pi(t, \vec{x}), \Pi(t,
\vec{y})\right]  ~ , ~ \left[\Phi(t, \vec{x}),
\Pi(t, \vec{y})\right] = i \delta^3(\vec{x} -
\vec{y}) \ .
\]

If $J(x)$ has no derivatives of fields, the momentum
field is $\Pi(t,\vec{x}) = \dot{\Phi}(t,\vec{x})$. This
stipulates the quantized nature of the field while the
equation of motion stipulates the nature of interaction.

* $\exists$ another set of fields, quantized $\Phi_{in}(x),
\Phi_{out}(x)$, build out of the interacting field
$\Phi(x)$, and satisfying the same set of equal time
(anti-)commutation relations but obeying free field
equations: $(\Box - m^2)\Phi_{in/out} = 0$ where $m$ is
the “physical” mass.

The in and out fields thus have the same mode
decomposition introducing the creation/annihilation
operators which create/destroy stable particles with
physical mass $m$. These fields also transform the same
way as the field $\Phi(x)$ under the symmetries, in
particular, $\Phi_{in}(x+a) = e^{ia\cdot
P}\Phi_{in}(x)e^{-ia\cdot P}$.

* Asymptotic conditions: The exact interacting quantum
field $\Phi(x)$ is linked to the in and out fields by the
asymptotic conditions,

\[
\boxed{\left. \Phi(x)\right|_{t\to -\infty} ~ \longrightarrow
\sqrt{Z}\Phi_{in}(x) ~ ~ , ~ ~ \left. \Phi(x)\right|_{t
\to \infty} ~ \longrightarrow	\sqrt{Z} \Phi_{out}(x)}
\]

Here $Z$ is a possible normalization constant which we will show
to be necessarily different from 1.

Note: The asymptotic conditions are not operator
equations but are weak equations i.e. hold for arbitrary matrix
elements in the basis of normalized states.

Below we elaborate the conditions and note a number of properties.

Asymptotic conditions are consistent with Poincare covariance: A
formal solution of the equation satisfied by $\Phi(x)$ is given by,

\begin{equation}\label{FormalSoln}
\Phi(x) ~ = ~ \int d^4y\ G_{ret}(x-y,m)\tilde{J}(y) +
\sqrt{Z}\Phi_{in}(x) .
\end{equation}

As $t\to -\infty$, the first term vanishes (property of retarded Green
function) recovering the asymptotic condition. This is a heuristic
argument since $J(y)$ it built out of $\Phi(x)$ and some iterative
procedure is implicit. The consistency with covariance is seen as,

\begin{eqnarray*}
\sqrt{Z}\Phi_{in}(x-a) & = & \Phi(x-a) - \int d^4y\
G_{ret}(x-a-y,m)\tilde{J}(y) \\
& = & e^{-ia\cdot P}\Phi(x)e^{ia\cdot P} - \int d^4y\
G_{ret}(x-y, m)\tilde{J}(y-a) \\
& = & e^{-ia\cdot P}\left[\Phi(x) - \int d^4y\ G_{ret}(x-y,
m)\tilde{J}(y)\right]e^{ia\cdot P} ~ = ~ \sqrt{Z}e^{-ia\cdot
P}\Phi_{in}(x)e^{ia\cdot P} \ .
\end{eqnarray*}

Note: Only the $G_{ret}$ is used in the formal solution and
not $G_{Feynman}$.

$\Phi_{in}(x)$ creates states with physical mass $m$: The
infinitesimal form of the transformation law for $\Phi_{in}(x)$ implies,
$[P^{\mu}, \Phi_{in}(x)] = -i\partial^{\mu}\Phi_{in}(x)$ and we also
have the equation of motion $(\Box - m^2)\Phi_{in} = 0$.

For any physical state $|k,\alpha\rangle, P^{\mu}|k,\alpha\rangle =
k^{\mu}|k, \alpha\rangle$. Therefore,

\begin{eqnarray*}
-i\partial^{\mu}\langle k, \alpha|\Phi_{in}(x)|0\rangle & =
&\langle k,\alpha|[P^{\mu}, \Phi_{in}(x)]|k,\alpha\rangle =
k^{\mu}\langle k, \alpha|\Phi_{in}(x)|0\rangle \\
\therefore -\Box \langle k, \alpha|\Phi_{in}(x)|0\rangle & = &
k^2 \langle k, \alpha|\Phi_{in}(x)|0\rangle \\
\therefore (-\Box + m^2)\langle k, \alpha|\Phi_{in}(x)|0\rangle
& = 0 = & (k^2 + m^2)\langle k, \alpha|\Phi_{in}(x)|0\rangle ~
\Rightarrow ~ k^2 = -m^2\ .
\end{eqnarray*}

Thus, $\Phi_{in}(x)$ behaves exactly like a free field producing
physical states from the vacuum and we can be taken to have the usual
mode decomposition,

\begin{eqnarray}\label{InversionFormula0}
\Phi_{in}(x) & = & \int \frac{d^3k} {\sqrt{2\omega_{\vec{k}}
(2\pi)^3}} \left[a_{in}(\vec{k}) e^{ik\cdot x} +
a_{in}^{\dagger}(\vec{k})e^{-ik\cdot x}\right] \nonumber \\
\partial_0\Phi_{in}(x) & = &
\int\frac{d^3k}{\sqrt{2\omega_{\vec{k}}(2\pi)^3}}
\left[-i\omega_{\vec{k}} a_{in}(\vec{k})e^{ik\cdot x} +
i\omega_{\vec{k}}a_{in}^{\dagger}e^{-ik\cdot x}\right] ~ ~
\Rightarrow \nonumber \\
a_{in}^{\dagger}(\vec{k}) & = & \frac{-i}
{\sqrt{2\omega_{\vec{k}} (2\pi)^3}} \int d^3x \ e^{ik\cdot x}\
\overleftrightarrow{\partial_0}\ \Phi_{in}(x) ~ ~ , ~ ~ f
\overleftrightarrow{\partial}g := f\partial g - \partial(f) g \
. \\
\mbox{And,~} a_{in}(\vec{k}) & = & \frac{i}
{\sqrt{2\omega_{\vec{k}} (2\pi)^3}} \int d^3x \ e^{-ik\cdot x}\
\overleftrightarrow{\partial_0}\ \Phi_{in}(x) \nonumber
\end{eqnarray}

Note: In writing the inversion formula, we have done the
spatial integration over a $t = $constant hypersurface $\Sigma_{t}$. The
choice of this hypersurface does not matter ($a_{in}^{\dagger}(\vec{k})$
is independent of $t$). This is kept implicit in all the inversion
formulae below.

It follows immediately, $\langle 0|\Phi_{in}(x)|0\rangle = 0$ and,

\begin{eqnarray*}
\langle k,\alpha|\Phi_{in}(x)|0\rangle & = & \langle k,\alpha|
e^{-ix\cdot P}\Phi_{in}(0)e^{ix\cdot P} |0\rangle ~ = ~
e^{-ik\cdot x} \langle k,\alpha|\Phi_{in}(0)|0\rangle \\
\langle k,\alpha|\Phi_{in}(0)|0\rangle & = & \frac{1}
{\sqrt{2\omega_{\vec{k}}(2\pi)^2}} ~ \Rightarrow ~ \boxed{
\langle k,\alpha|\Phi_{in}(x)|0\rangle = \frac{e^{-ik\cdot x}}
{\sqrt{2\omega_{\vec{k}} (2\pi)^3}} .  }
\end{eqnarray*}

Suffice it to say that corresponding expressions exist for $\phi_{out}$.
In particular the heuristic expression take the form,

\[
\Phi(x) = \sqrt{Z} \Phi_{out}(x) + \int d^4y\ G_{adv}(x-y,
\]

How far can we get with Lorentz covariance and the assumptions on
the spectrum?

§.§ Kallen-Lehmann Representation

Consider the vacuum expectation values of the commutator of interacting
field, $i\Delta'(x,x') := \langle 0|[\Phi(x), \Phi(x')]0\rangle$. We had
evaluated this for the free fields while considering the
spin-statistics, $\sim \Delta_+(x-x')$. We evaluate it as,

\begin{eqnarray}
i\Delta(x,x') & = & \sum_n\langle 0|\Phi(x)|n\rangle\langle
n|\Phi(x')|0\rangle - (x \leftrightarrow x') , \\
& = & \sum_n\langle 0|\Phi(0)e^{ix\cdot P}|n\rangle\langle
n|e^{-ix'\cdot P}\Phi(0)|0\rangle - (x \leftrightarrow x') , \\
& = & \sum_n |\langle 0|\Phi(0)|n\rangle|^2\left(e^{i(x-x')\cdot
k_n} - e^{-i(x-x')\cdot k_n}\right)  ~ ~  ~ \mbox{a function of
$(x-x')$.} ~ ~ ~ ~
\end{eqnarray}

The states $|n\rangle$ contain the vacuum, the single particle states on
the isolated hyperboloid $k_n^2 = -m^2$ as well as the multi-particles
continuum. To collect terms with the same total $k^{\mu}_n$, insert $1 =
\int d^4q \delta^4(q - k_n)$ and write,

\begin{eqnarray*}
\Delta'(x-x') & = & - \frac{i}{(2\pi)^3}\int d^4q\left[(2\pi)^3 \sum_n
\delta^4(q - k_n) |\langle 0|\Phi(0)|n\rangle|^2 \left(e^{i(x-x')\cdot
k_n} - e^{-i(x-x') \cdot k_n}\right)\right] \\
& := &  - \frac{i}{(2\pi)^3}\int d^4q \left(e^{i(x-x')\cdot q} -
e^{-i(x-x')\cdot q}\right)\rho(q)  ~ ~ ~ ~ \mbox{where, } \\
\rho(q) & := & \left[(2\pi)^3 \sum_n \delta^4(q - k_n)|\langle
0|\Phi(0)|n\rangle|^2\right] ~ ~ ~ \mbox{(the spectral density).}
\end{eqnarray*}

Claim: The spectral density is a scalar function of $q^2$.

Proof: Under a Lorentz transformation, $U(\Lambda)$, we have
$U(\Lambda)|0\rangle = |0\rangle$ and $U(\Lambda)\Phi(0)U^{-1}(\Lambda)
= \Phi(\Lambda 0) = \Phi(0)$. Hence,

\[
\rho(q) = (2\pi)^3\sum_n \delta^4(q - k_n)|\langle
0|\Phi(0)U(\Lambda)|n\rangle|^2 \ .
\]

\begin{eqnarray*}
\delta^4(k) & \sim & \int d^4x e^{ik\cdot x} = \int d^4(\Lambda
x)e^{ik\cdot(\Lambda x)} = \int d^4x e^{i(\Lambda^{-1}k)\cdot x}
\sim \delta^4(\Lambda^{-1}k) \ . \\
\therefore \rho(q) & = & (2\pi)^3\sum_n\delta^4(\Lambda^{-1}(q -
k_n)) |\langle 0|\Phi(0)U^{-1}(\Lambda)|n\rangle|^2 \ .
\end{eqnarray*}

Let $U(\Lambda)|n\rangle = |m\rangle$. Then $U^{-1}(\Lambda) P^{\mu}
U(\Lambda) = \Lambda^{\mu}_{~\nu} P^{\nu}$.  Acting on $|n\rangle$
implies, $P^{\mu}|m\rangle = k^{\mu}_{m}|m\rangle$ and we already had
$P^{\mu}|n\rangle = k^{\mu}_{n}|n\rangle$. This in turn implies,
$k^{\mu}_m U^{-1}(\Lambda)|m\rangle = \Lambda^{\mu}_{~\nu} k^{\nu}_n
|n\rangle$ or, $k^{\mu}_m = \Lambda^{\mu}_{~\nu} k^{\nu}_n$. Lowering
the Lorentz index gives, $(k_m)_{\mu} = (k_n)_{\nu}
(\Lambda^{-1})^{\nu}_{~\mu}$. Replacing the sum over $n$ by that over
$m$ gives,

\[
\rho(q) = (2\pi)^3\sum_m \delta^4(k_m - \Lambda^{-1}q)|\langle
0|\Phi(0)|m\rangle|^2 = \rho(\Lambda^{-1}q) \ .
\]

Thus, $\rho(q)$ can only depend on $q$ through $q^2$.

Since the physical spectrum has $q^2 <  0$ with $q^0 > 0$, we write
$\rho(q) := \theta(q^0) \rho(q^2), \ \rho(q^2) = 0 ~ \mathrm{for} ~ q^2
> 0$.  Substituting in $\Delta'$ gives,

\begin{eqnarray}
\Delta'(x-x') & = & \frac{-i}{(2\pi)^3}\int d^4q \rho(q^2) \theta(q^0)
\left(e^{iq\cdot(x-x')} - e^{-iq\cdot(x-x')}\right) \nonumber\\
& = & \int_0^{\infty} d\sigma^2 \rho(\sigma^2) \left[
\frac{-i}{(2\pi)^3} \int d^4q \delta(q^2 - \sigma^2) \epsilon(q^0)
e^{iq\cdot(x-x')}\right] \ . \nonumber\\
\therefore \Delta'(x-x') & = & \int_0^{\infty}d\sigma^2\ \rho(\sigma^2)
\Delta(x-x', \sigma^2) \ . \label{KallenLehmannScalar}
\end{eqnarray}

Here we have introduced the function $\epsilon(x) = \theta(x) -
\theta(-x)$ to combine the two exponentials and also recognized that the
the square bracket is just $-i[\Phi_{in}(x), \Phi_{in}(x')]$ for $m^2 =
\sigma^2$. The last equation is known as the Kallen-Lehmann
representation or the spectral representation for the commutator
function $\Delta'(x-x')$ of the exact interacting quantum fields. This
has followed from Poincare covariance (separation of the
$x-$dependence), the assumptions regarding the spectrum of the
interacting system ($\rho(q^2>0)=0$) and normalizations used for
$\Phi_{in}$ in identifying the $\Delta(x-x',\sigma^2)$.

We can separate the contribution of the 1-particle states at $\sigma^2 =
m^2$. We obtain the $\langle 0|\Phi(x)|k\rangle$ using the asymptotic

\begin{eqnarray}
\langle 0|\Phi(x)|k\rangle & = & \sqrt{Z}\langle 0|\Phi_{in}(x)|k\rangle
+ \int d^4y\ G_{ret}(x-y, m)\langle 0|\tilde{J}(y)|k\rangle ~ , ~
\mbox{but} \nonumber \\
\langle 0|\tilde{J}(y)|k\rangle & = &\langle 0|(\Box -
m^2)\Phi(x)|k\rangle = (\Box - m^2)\left(\langle 0|\Phi(0)|k\rangle
e^{ik\cdot x}\right) \nonumber \\
& = & (-k^2 - m^2)\left(\langle 0|\Phi(0)|k\rangle e^{ik\cdot x}\right)
~ = ~ 0 . ~ ~ \mbox{The first term then gives,}\nonumber \\
\langle 0|\Phi(x)|k\rangle & = & \sqrt{Z}\langle 0|\Phi_{in}(x)|k\rangle
= \sqrt{Z}\frac{e^{ik\cdot x}}{\sqrt{2\omega_{\vec{k}}(2\pi)^3}}
\Rightarrow |\langle 0|\Phi(x)|k\rangle|^2 =
\frac{Z}{2\omega_{\vec{k}}(2\pi)^3} \ . \nonumber \\
\therefore \rho(q^2)|_{1-particle} & = & (2\pi)^3\int d^3k \delta^4(k -
q) \frac{Z}{2\omega_{\vec{k}}(2\pi)^3} = Z \delta(q^2 - m^2)\theta(q^0)
\\
\therefore & & \boxed{\Delta'(x-x')  =  Z \Delta(x-x',m^2) +
\int_{m^2_1}^{\infty}d\sigma^2 \rho(\sigma^2) \Delta(x-x', \sigma^2)} \
. \label{KallenLehmannSeparated}
\end{eqnarray}

The lower limit on the integral is the smallest invariant mass of the
multi-particle continuum and $m^2_1 > m^2$ is assumed.

We can now derive a bound on $Z$. Observe that

\[
\lim_{t'\to t} (i\partial_t \Delta'(x-x')) = \langle 0|[\dot{\Phi}(x),
\Phi(x')]|0\rangle = -i\delta^3(x-x') = (i\partial_t \Delta(x-x',m^2)) \
\]

We are using the assumption that there are no derivative terms in $J(x)$
or equivalently, $\Pi(x) = \dot{\Phi}(x)$ and noting that the in/out
fields satisfy the same commutation relations.

Taking $-i\partial_t$ on eq.(<ref>) and taking
the limit $t' \to t$, we deduce

\begin{eqnarray}
-i \delta^3(\vec{x} - \vec{x}') & = & Z(-i \delta^3(\vec{x} - \vec{x}'))
+ \int_{m_1^2}^{\infty}d\sigma^2\ \rho(\sigma^2)\left[-i\delta^3(\vec{x}
- \vec{x}',\sigma)\right] \nonumber \\
Or, ~ 1 & = & Z + \int_{m_1^2}^{\infty}d\sigma^2 \ \rho(\sigma^2) ~ ~
\Longrightarrow \boxed{ 0 \le Z < 1 . }
\end{eqnarray}

The inequalities on $Z$ can be understood as follows. Since asymptotic
condition is needed to link $\Phi(x)$ with $\Phi_{in/out}(x)$, $Z \neq
0$. To understand the upper limit, notice that the free fields can only
produce 1-particles states since it is linear in creation/annihilation
operators. The interacting field has no such restriction and can produce
multi-particle states as well. Hence the probability for it to produce
1-particle states is less than that for the free fields i.e.
$|\langle 0|\Phi(x)|k\rangle|^2 < |\langle 0|\Phi_{in/out}(x)
|k\rangle|^2$. This results in $Z < 1$.

In principle the $Z$ factors for the in and out fields could have been
different in the asymptotic conditions. However, following identical
steps as above would lead us to the same relation
(<ref>) implying equality of the $Z$'s.

An important point to note is that $Z < 1$ has a physical reason, quite
independent of any infinities and their renormalization that we will
encounter later on.

Remark: In the Kallen-Lehmann spectral representation we
considered the vacuum expectation value of the commutator of the fully
interacting quantum field presumed to satisfy the usual canonical
commutation relations. Because of this, the interacting quantum field
gets identified with the so called “bare” quantum field in the
renormalized perturbation theory. We could instead use a
“renormalized” quantum field, $\Phi_R(x) := \Phi(x)/\sqrt{Z}$ and
correspondingly define, $\Delta'_R(x-x') := Z^{-1}\Delta'(x-x')$. It
follows that  $\rho_R(q^2) = Z^{-1}\rho(q^2)$ and
$[\Phi_R(x),\dot{\Phi}_R(x')] = iZ^{-1}\delta^3(x-x')$. In terms of the
renormalized quantities, there is no factor of $Z$ in equation
(<ref>). The manipulations leading to the bound
on $Z$ will now give,

\[
\lim_{t'\to t} (i\partial_t \Delta'_R(x-x')) = \langle
0|[\dot{\Phi}_R(x), \Phi_R(x')]|0\rangle = -iZ^{-1}\delta^3(x-x') =
Z^{-1}(i\partial_t \Delta(x-x',m^2)) \ ,
\]

and the eq.(<ref>) will lead to $1 =
Z\left[1+\int_0^{\infty}d\sigma^2\rho_R(\sigma^2)\right]$ implying again
that $Z < 1$. These bounds are true provided the spectral integral is
finite. In perturbation theory, the two point functions
(commutator or the Feynman propagator) of the unrenormalized fields are
Ultra-Violet divergent and the bound cannot be inferred (See section
10.7 of [1]). The Kallen-Lehmann representation
eq.(<ref>) itself does not depend on the
canonical commutation relations for the interacting field and can indeed
be derived for composite fields as well.

§.§ The $S$-matrix and its properties

Having the $\Phi_{in/out}(x)$ and their corresponding mode expansions at
our disposal, we can now define the in/out states as, $|k\rangle_{in} :=
a^{\dagger}_{in}(\vec{k})|0\rangle$ and $|k\rangle_{out} :=
a^{\dagger}_{out}(\vec{k})|0\rangle$. Similarly, multi-particle states
can be defined. We can form wave packets for normalizable states and
take limit of infinite sharpness at the end. We bypass these steps and
work with the momentum eigenstates. Note that these states are time
independent. While we can certainly define the general in/out states
and take their spans, we need to (and do) make a further assumption that
these spans generate the full Hilbert space of the interacting quantum
fields[This automatically precludes bound states in the
interacting system and amounts to postulating asymptotic completeness
property mentioned earlier.]. The sets of in and out states just
constitute two orthonormal bases, conveniently transforming by
representations of the Poincare group. This guarantees existence of a
unitary operator connecting the two orthonormal bases. This is our
scattering operator in this Heisenberg picture formulation. Taking the
basis elements to be generated by monomials of $a^{\dagger}_{in}
(\vec{k}, \sigma, \dots)$ and $a^{\dagger}_{out} (\vec{k},
\sigma, \dots)$, the $S-$matrix is defined as,

\begin{equation}\label{SMatrixDefn2}
S_{\beta\alpha} :=\langle\beta\ out|\alpha\ in\rangle ~ ~ =: ~
~\langle\beta\ in|S|\alpha\ in\rangle\
\end{equation}

Note that $\alpha, \beta$ label basis states which are created by
monomials in $ a^{\dagger}_{in}, a^{\dagger}_{out}$. The
definition of $S$ operator shows that it preserves the labels:
$\langle\beta\ out| =:\langle\beta\ in|S$. We stipulate some conditions
on the $S-$matrix.

Stability of the vacuum and 1-particle states: We expect vacuum
state to suffer no scattering spontaneously except may be acquiring a
phase (uniqueness of vacuum allows a phase since rays are defined unto
phases). We choose the phase to be 1, and stipulate $\boxed{S_{00} =
1}$. Likewise presumed stability of single particle states also
disallows any spontaneous change (nothing to scatter against). Thus we
stipulate that $S_{k', k} = \delta^3(\vec{k}' - \vec{k})$. Scattering
takes place with multi-particle states.

Claim: $~ \Phi_{out}(x) = S^{-1} \Phi_{in}(x) S$ .

Proof:   Consider, $\langle \beta\ out | \Phi_{out}(x)
|\alpha\ in\rangle$ and evaluate in two ways.

\begin{eqnarray*}
\langle \beta\ out | \Phi_{out}(x) |\alpha\ in\rangle & = & (\langle
\beta\ in|S)\Phi_{out}(x)| \alpha \ in\rangle .
\end{eqnarray*}

Next, $\langle\beta'\ out| := \langle\beta\ out|\Phi_{out}(x)$ is a
linear combination of the out-basis elements.  Each of this can be
expressed as corresponding in-basis element times $S$. And the in-basis
element is also obtained by action of $\Phi_{in}(x)$. In equation,

\begin{eqnarray*}
\langle\beta'\ out| = \sum_{\gamma}C_{\gamma}\langle\gamma\ out| =
\sum_{\gamma}C_{\gamma}\left(\langle\gamma\ in|S\right) = \hspace{6.0cm}
& & \\
\left(\sum_{\gamma}C_{\gamma}\langle\gamma\ in|\right)S =:
\langle\beta'\ in|S = (\langle\beta\ in|\Phi_{in}(x)) S.
\end{eqnarray*}

Taking inner product of the two expressions with $|\alpha\ in\rangle$

\[
\langle\beta\ out|\Phi_{out}(x)|\alpha\ in\rangle =\langle\beta\
in|S\Phi_{out}(x)|\alpha\ in\rangle = \langle\beta\
in|\Phi_{in}(x))S|\alpha\ in\rangle \ \forall \ \alpha, \beta.
\]

Completeness of the in-basis implies, $\boxed{\Phi_{out}(x) =
S^{-1}\Phi_{in}(x) S\ . }$ proving the claim.

Claim:   Covariance of the in/out fields gives
invariance of the scattering operator.

Proof:   We have,

\begin{eqnarray}
U(\Lambda,a)\Phi_{in}(x)U^{-1}(\Lambda,a) & = & \Phi_{in}(\Lambda x + a)
\nonumber \\
U(\Lambda,a)\Phi_{out}(x)U^{-1}(\Lambda,a) & = & \Phi_{out}(\Lambda x +
~ ~ ~ ~\mbox{and} ~ ~ ~ ~\Phi_{out}(x) = S^{-1}\Phi_{in}(x) S \nonumber
\\
\therefore U (S^{-1}\Phi_{in}(x)S)U^{-1} & = & S^{-1}\Phi_{in}(\Lambda x
+ a)S ~ = ~ S^{-1}U\Phi_{in}(x)U^{-1} S \\
\therefore U^{-1}SUS^{-1}\Phi_{in}(x) & = & \Phi_{in}(x)U^{-1}SUS^{-1} ~
=> ~ \nonumber \\ U^{-1}S U S^{-1} & = & \mathbb{1} ~ or ~ \boxed{S
U(\Lambda,a) = U(\Lambda,a) S} \ .
\end{eqnarray}

Question: Are the Schrodinger picture definitions of the
scattering matrix element (<ref>) and the Heisenberg
picture scattering matrix element defined using $\Phi_{out} =
S^{-1}\Phi_{in}S$, related? If so, how?

We now relate the S-matrix elements to vacuum expectation values of time
ordered products of the interacting fields - the
Lehmann-Symanzik-Zimmermann (LSZ) reduction formulae. This is followed
by covariant perturbation series leading to Feynman rules. We will first
detail these steps for the notationally simpler case of a scalar field
and then summarize the corresponding steps for the Dirac and the Maxwell

§.§ Lehmann-Symanzik-Zimmermann Reduction of $S-$matrix

We have defined the $S-$matrix elements using the in and out states
which form the basis elements created by monomials in
$a^{\dagger}_{in}(\vec{k},\dots)$ and $a^{\dagger}_{out}(\vec{k},\dots)$
acting on the unique vacuum. The in and out fields are not independent
or uncorrelated, they are linked through the interacting field $\Phi(x)$
via the asymptotic condition. The LSZ reduction of the $S-$matrix
elements expresses them in terms of the interacting field explicitly.
For simplicity, let us first consider the case of an interacting
Hermitian scalar field.

§.§.§ LSZ reduction for Klein-Gordon field

Consider a matrix element of the form $\langle\beta\ out|\alpha,k\
in\rangle$ where we have separated the $k$ label in the in-state. The
definition gives,

\begin{eqnarray*}
\langle\beta\ out|\alpha,k\ in\rangle & = &\langle\beta\
out|a^{\dagger}_{in}(\vec{k})|\alpha\ in\rangle = \langle\beta\
out|\left(a^{\dagger}_{in}(\vec{k}) - a^{\dagger}_{out}(\vec{k})
+ a^{\dagger}_{out}(\vec{k}) \right)|\alpha\ in\rangle \ .
\end{eqnarray*}

The $a^{\dagger}_{out}(\vec{k})$ acting on the out-state will remove a
particle with label $\vec{k}$ if it is present in the label set $\beta$,
or will annihilate the out-state. For simplicity, let us assume that
there is no such state in the $\beta$ label. Then we have just added and
subtracted a 0. For the first two term in the brackets , use the
inversion formula (<ref>).

\begin{eqnarray*}
\langle\beta\ out|\alpha,k\ in\rangle & = &\langle\beta\
out|\left[\frac{-i}{\sqrt{2\omega_{\vec{k}}(2\pi)^3}}\int d^3x
e^{ik\cdot x} \overleftrightarrow{\partial_0}\Phi_{in}(x) \right. \\
& & \mbox{\hspace{1.5cm}} \left. +
\frac{+i}{\sqrt{2\omega_{\vec{k}}(2\pi)^3}}\int d^3x e^{ik\cdot x}
\overleftrightarrow{\partial_0}\Phi_{out}(x)\right] |\alpha\ in\rangle .
\\
\left[\dots\right] & = & \left(\lim_{t \to +\infty} - \lim_{t\to
d^3x \frac{e^{ik\cdot x} \overleftrightarrow{\partial_0} \Phi(x)}
{\sqrt{Z}}\right\} \\
\therefore \sqrt{Z}\left[\dots\right] & = & \int d^4x\ \partial_0\left\{
e^{ik\cdot x} \overleftrightarrow{\partial_0} \Phi(x)\right\} \\
& = & \int d^4x\ \left\{ e^{ik\cdot x} {\partial^2_0} \Phi(x) -
\partial^2_0(e^{ik\cdot x})\Phi(x)\right\} ~ , ~ (-\partial_0^2
e^{ik\cdot x} = (-\nabla^2 + m^2)e^{ik\cdot x})\\
& = & \int d^4x\ e^{ik\cdot x}\left\{ {\partial^2_0} \Phi(x) +
(-\nabla^2 + m^2)\Phi(x)\right\} ~ , ~ (\nabla^2\ \mbox{flipped onto\ }
\Phi(x)). \\
\therefore \langle\beta\ out|\alpha,k\ in\rangle & = &
\frac{-i}{\sqrt{Z}}\int \frac{d^4x e^{ik\cdot x}}
{\sqrt{2\omega_{\vec{k}} (2\pi)^3}} (\Box - m^2)\langle\beta\
out|\Phi(x)|\alpha\ in\rangle \ .
\end{eqnarray*}

Now let us separate a particle with label $k'$ from the out-state and
write $\langle\beta\ out| =\langle\gamma,k'\ out|$ the inner product on
the right hand side of the above equation. Consider,

\begin{eqnarray*}
\langle\gamma,k'\ out|\Phi(x)|\alpha\ in\rangle & = & \langle\gamma\
out|a_{out}(\vec{k}')\Phi(x)|\alpha\ in\rangle \\
& = & \langle\gamma\ out|(a_{out}(\vec{k}')\Phi(x) -
\Phi(x)a_{in}(\vec{k}') + \Phi(x)a_{in}(\vec{k}'))|\alpha\ in\rangle
\end{eqnarray*}

As before, for simplicity, assume $\alpha$-label does not contain $k'$
and thus drop the third term in the brackets. Notice that we have added
and subtracted the zero term with $a_{in}(\vec{k}')$ to the right of
$\Phi(x)$ while $a_{out}(\vec{k}')$ is naturally to the left of
$\Phi(x)$. We need to maintain this order as we do not have the
commutation relations between the in/out field and the interacting
field. As before, using the inversion formula (<ref>)
and using the asymptotic condition, we get

\begin{eqnarray*}
\langle\gamma,k'\ out|\Phi(x)|\alpha\ in\rangle & = & \langle\gamma\
out|\left[ \frac{i}{\sqrt{2\omega_{\vec{k}'}(2\pi)^3}}\int
\right.  \right. \\
& & \mbox{\hspace{4.5cm}} \left.\left. -
\Phi(x)\overleftrightarrow{\partial'_0}\Phi(x')_{in}\right\}\right]
|\alpha\ in\rangle \\
\therefore \sqrt{Z}\left[\dots\right] & = & \frac{i}{\sqrt {2\omega_
{\vec{k}'} (2\pi)^3}}\int d^3x'\left\{ \lim_{t'\to \infty}(e^{-ik'\cdot
x'} \overleftrightarrow{\partial'_0}\Phi(x'))\Phi(x) \right. \\
& & \left. \mbox{\hspace{3.0cm}} - \lim_{t'\to -\infty} \Phi(x)
(e^{-ik'\cdot x'} \overleftrightarrow{\partial'_0} \Phi(x')) \right\}
\end{eqnarray*}

It is convenient to define a time ordering instruction. Define a time ordered product as: $~\hfill\boxed{T\left\{A(t_1,\vec{x})
B(t_2,\vec{y})\right\} := \theta(t_1 - t_2) A(t_1,\vec{x})
B(t_2,\vec{y}) + \theta(t_2-t_1) B(t_2,\vec{y}) A(t_1, \vec{x})

Notice that the two limits already have the time ordering incorporated.
Hence we can combine the terms and write,

\begin{eqnarray*}
\sqrt{Z}\left[\dots\right] & = & \frac{i}{\sqrt {2\omega_ {\vec{k}'}
(2\pi)^3}}\int d^3x'\left( \lim_{t'\to \infty} - \lim_{t'\to
-\infty}\right) e^{-ik'\cdot x'} \overleftrightarrow{\partial'_0}
T\left\{\Phi(x')\Phi(x)\right\}  \\
& = & \frac{i}{\sqrt {2\omega_ {\vec{k}'} (2\pi)^3}}\int d4x'
\partial'_0 \left( e^{-ik'\cdot x'} \overleftrightarrow{\partial'_0}
T\left\{\Phi(x')\Phi(x) \right\}\right)
\end{eqnarray*}

Proceeding exactly as before we get,

\begin{eqnarray}
\langle\gamma,k'\ out|\alpha,k\ in\rangle & = &
\left(\frac{-i}{\sqrt{Z}}\right)^2\int d^4x'\int d^4x \frac{e^{-ik'\cdot
x}}{\sqrt{2\omega_{\vec{k}}(2\pi)^3}} \nonumber \\
& & \mbox{\hspace{0.0cm}} (\Box_{x'} - m^2)(\Box_x -
m^2)\langle\gamma,k'\ out|T\left\{\Phi(x')\Phi(x)\right\}|\alpha,k\
\end{eqnarray}

The generalization is immediate and we state the master formula for the
$S-$matrix element as,

\begin{eqnarray}
\langle k_1,'\dots,k'_n\ out|k_1,\dots,k_m\ in\rangle & = &
\left(\frac{-i}{\sqrt{Z}} \right)^{m+n}\int d^4x'_1..d^4x'_n\int
d^4x_1..d^4x_m \nonumber \\
& & ~ ~ ~ ~ \prod_{j}^n \frac{e^{-ik'_j\cdot x'_j}}
{\sqrt{2\omega_{\vec{k}'} (2\pi)^3}} \prod_{j}^m \frac{e^{+ik_j\cdot
x_j}} {\sqrt{2\omega_{\vec{k}} (2\pi)^3}} (\Box_{x'_1} - m^2) \nonumber
\\
& & ~ ~ ~ ~ \dots(\Box_{x'_m} - m^2)(\Box_{x_1} - m^2)\dots(\Box_{x_n} -
m^2) \nonumber \\
& & ~ ~ ~ ~ \langle 0|T\left\{ \Phi(x'_1) \dots \Phi(x'_n) \Phi(x_1)
\dots \Phi(x_n) \right\}|0\rangle
\end{eqnarray}

The vacuum expectation value of the time ordered product of $n$ quantum
fields (interacting or free) is called an $n-$point function/Green
function/n-point correlation function.

To get the $S-$matrix elements, the n-point function is operated by the
free field equation expression, multiplied by the in/out mode functions
and integrated over all the space-time points.  Note that the time
ordering arises because the definition of $S-$matrix requires all the
creation operators, $a^{\dagger}_{in}(\vec{k})$, have to be on the right
while the annihilation operators, $a_{out}(\vec{k})$ have to be on the
left. This will also naturally lead to the Feynman Green's function
(free, 2-point function).

Let us note the steps that were followed to get the $S-$matrix elements
in terms of the n-point functions.

* Mode decomposition of the free field and the inversion formulae;

* Equation of motion for the interacting field with a `source' term
on the right hand side;

* Equal time(anti-)commutation relations for both the interacting
and the in/out fields;

* Asymptotic conditions;

* Reduction process;

§.§.§ LSZ reduction for Dirac field

We begin by recalling the mode decomposition.

\begin{eqnarray*}
\Psi(x) & = & \int \frac{d^3k}{\sqrt{2\omega_{\vec{k}}(\pi)^3}}
\left[b(\vec{k},\sigma)u(\vec{k},\sigma)e^{ik\cdot x} +
d^{\dagger}(\vec{k},\sigma)v(\vec{k},\sigma)e^{-ik\cdot x}\right] \\
\Psi^{\dagger}(x) & = & \int \frac{d^3k}
\left[b^{\dagger}(\vec{k},\sigma)u^{\dagger}(\vec{k},\sigma)e^{-ik\cdot
x} + d(\vec{k},\sigma)v^{\dagger}(\vec{k},\sigma)e^{ik\cdot x}\right]
\end{eqnarray*}

We have used the $\Psi^{\dagger}$ field instead of $\bar{\Psi}$ because
the inversion formulae take a more convenient form.

We had defined the $u, v$ spinors as,

\begin{eqnarray*}
(\dsl{k}+m)u(\vec{k},\sigma) = 0 ~ , ~
\bar{u}(\vec{k},\sigma)u(\vec{k},\sigma') ~ = ~ &
\delta_{\sigma,\sigma'} & ~ = ~ - \bar{v}(\vec{k},\sigma)
v(\vec{k},\sigma') ~ , ~ (-\dsl{k}+m)v(\vec{k},\sigma) = 0\\
\sum_{\sigma}u(\vec{k},\sigma)\bar{u}(\vec{k},\sigma) =
\frac{-\dsl{k}+m}{2m} & , &
\sum_{\sigma}v(\vec{k},\sigma)\bar{v}(\vec{k},\sigma) =
\end{eqnarray*}

We need to express these in terms of the adjoints rather than in terms
of the Dirac adjoints. This alternative for can be derived as follows.
We have,

\begin{eqnarray*}
\bar{u}(\vec{k},\sigma)\gamma^{\mu}u(\vec{k},\sigma') & = &
\Lambda^{\mu}_{~\nu} \bar{u}(\hat{k},\sigma) \gamma^{\mu}
u(\hat{k},\sigma') ~  ,  ~ k^{\mu} = \Lambda^{\mu}_{~\nu}
\hat{k}^{\nu} ~ , ~ \hat{k}^{\mu} = (m, \vec{0}) ~. ~ \mbox{For
$\mu = 0 , $}\\
u^{\dagger}(\vec{k},\sigma) u(\vec{k},\sigma') & = &
\Lambda^0_{~\nu}u^{\dagger}(\hat{k},\sigma) \gamma^0
\gamma^{\nu} u(\hat{k},\sigma') ~ = ~
\Lambda^0_{~\nu}u^{\dagger}(\hat{k},\sigma) \gamma^{\nu}
u(\hat{k},\sigma') ~ \because ~ \gamma^0u(\hat{k},\sigma') =
u(\hat{k}, \sigma') \\
& = & \Lambda^0_{~0}u^{\dagger}(\hat{k},
\sigma) \gamma^0 u(\hat{k},\sigma') + \Lambda^0_{~i}u^{\dagger}(
\hat{k},\sigma)\gamma^iu(\hat{k},\sigma') \\
\therefore u^{\dagger}(\vec{k},\sigma)u(\vec{k},\sigma')
& = & \Lambda^0_{~0}\bar{u}(\hat{k},\sigma)u(\hat{k}, \sigma')
+ 0 ~ = ~ \Lambda^0_{~0}\delta_{\sigma,\sigma'} + 0
\end{eqnarray*}

The second term in the last equation is zero because, for $\hat{k}$
spinors $\gamma^0 u(\hat{k},\sigma) = u(\hat{k},\sigma)$. This implies,

\[
u^{\dagger}\gamma^iu = u^{\dagger}\gamma^0\gamma^iu = -
u^{\dagger}\gamma^i\gamma^0u = - u^{\dagger}\gamma^i u \ .
\]

Similarly, $v^{\dagger}(\vec{k},\sigma)v(\vec{k},\sigma') = -
\Lambda^0_{~0}\bar{v}(\hat{k}, \sigma)v(\hat{k}, \sigma') = +
\Lambda^0_{~0}\delta_{\sigma,\sigma'}$. But $\Lambda^0_{~0}$ is defined
through, $k^{\mu} = \Lambda^{\mu}_{~\nu}\hat{k}^{\nu}$. For $\mu = 0$
this gives, $\Lambda^0_{~0} = \Case{\omega_{\vec{k}}}{m}$. Our
orthonormality relations then take the form,

\[
\boxed{u^{\dagger}(\vec{k},\sigma)u(\vec{k},\sigma') =
\frac{\omega_{\vec{k}}}{m}\delta_{\sigma,\sigma'} =
v^{\dagger}(\vec{k},\sigma)v(\vec{k},\sigma') ~ ~ , ~ ~
u^{\dagger}v = 0 = v^{\dagger} u \ . }
\]

We can use these relations to derive the inversion formulae. For
instance, multiplying the mode decomposition for $\Psi(x)$ by
$u^{\dagger}(\vec{k},\sigma)e^{-ik\cdot x}$ and integrating over $d^3x$

\[
b(\vec{k},\sigma) = \frac{2m}{\sqrt{2\omega_{\vec{k}}(2\pi)^3}}
\int d^3x u^{\dagger}(\vec{k},\sigma)e^{-ik\cdot x}\Psi(x)
\]

It is convenient to define

\[
U(\vec{k}, \sigma) := 2m\frac{u(\vec{k}, \sigma)}
{\sqrt{2\omega_{\vec{k}} (2\pi)^3}} ~ ~ , ~ ~ V(\vec{k}, \sigma) :=
2m\frac{v(\vec{k}, \sigma)} {\sqrt{2\omega_{\vec{k}} (2\pi)^3}}
\]

The inversion formulae are,

\begin{eqnarray}
b(\vec{k},\sigma) & = & \int d^3x e^{-ik\cdot x} U^{\dagger}
(\vec{k},\sigma) \Psi(x)  \\
d^{\dagger}(\vec{k},\sigma) & = & \int d^3x e^{+ik\cdot x} V^{\dagger}
(\vec{k},\sigma) \Psi(x)  \\
b^{\dagger}(\vec{k},\sigma) & = & \int d^3x e^{+ik\cdot x}
\Psi^{\dagger}(x) U(\vec{k},\sigma)  \\
d(\vec{k},\sigma) & = & \int d^3x e^{-ik\cdot x} \Psi^{\dagger}(x)
\end{eqnarray}

In the reduction process, we will have for instance,

\begin{eqnarray*}
b^{\dagger}_{in}(\vec{k},\sigma) - b^{\dagger}_{out} (\vec{k},\sigma) &
= & \int d^3x\ e^{ik\cdot x}\left(\Psi^{\dagger}(x)_{in} -
\Psi^{\dagger}_{out}(x) \right) U(\vec{k},\sigma) \\
\mbox{and using asymptotic condition} & \longrightarrow &
\frac{1}{\sqrt{Z}}\int d^4x\ \partial_0\left(e^{ik\cdot
x}\Psi^{\dagger}(x)\right) U(\vec{k},\sigma)
\end{eqnarray*}

\begin{eqnarray*}
\partial_0(e^{ik\cdot x}\Psi^{\dagger}(x))U(\vec{k},\sigma) & = &
e^{ik\cdot x}(\partial_0\Psi^{\dagger} + ik_0\Psi^{\dagger})U ; \\
ik_0\Psi^{\dagger}U & = & i\bar{\Psi}(k_0\gamma^0)U = i\bar{\Psi}
(\dsl{k} - k_i\gamma^i)U = i\bar{\Psi}(-mU) + \bar{\Psi}\gamma^i(-ik_iU)
\\
\therefore e^{ik\cdot x}ik_0\Psi^{\dagger}U & = & e^{ik\cdot x}
(-i\bar{\Psi}mU) + \bar{\Psi} \gamma^i(-\partial_i e^{ik\cdot x}) U ~ =
~ e^{ik\cdot x}\left(-im\bar{\Psi} + \partial_i \bar{\Psi} \gamma^i
\right)U \\
\therefore \partial_0\left(e^{ik\cdot x} \Psi^{\dagger}\right)
U(\vec{k},\sigma) & = & e^{ik\cdot x} \left\{\partial_0\bar{\Psi}
\gamma^0 + \partial_i\bar{\Psi}\gamma^i - im \bar{\Psi}\right\} U ~ = ~
(-ie^{ik\cdot x}) \left\{ i\bar{\Psi}\overleftarrow{\dsl{\partial}} +
\end{eqnarray*}

Hence, instead of $(-\Box_x + m^2)$ acting on $\Phi(x)$, we will have
$(-2mi)(i\overleftarrow{\dsl{\partial}} + m)$ acting on $\bar{\Psi}(x)$
and $(-2mi)(-i\overrightarrow{\dsl{\partial}} + m)$ acting on $\Psi(x)$.
The wavefunction factors will be:

\begin{eqnarray}
\mbox{Particle in in-state:~} U(\vec{k},\sigma)e^{ik\cdot x} & , &
\mbox{Particle in out-state:~} \bar{U}(\vec{k},\sigma)e^{-ik\cdot x} \\
\mbox{anti-Particle in in-state:~} \bar{V}(\vec{k},\sigma)e^{ik\cdot x}
& , & \mbox{anti-Particle in out-state:~} V(\vec{k},\sigma)e^{-ik\cdot
\end{eqnarray}

For multi-particle case we will again have time ordering instruction. For
fermions, it is defined with an relative minus sign,

$~\hfill\boxed{T\left\{\Psi_{\alpha}(x)\Psi_{\beta}(y)\right\} ~ := ~
\theta(x^0 - y^0)\Psi_{\alpha}(x)\Psi_{\beta}(y) - \theta(y^0 -
x^0)\Psi_{\beta}(y)\Psi_{\alpha}(x) }\hfill~$.

and $T\left\{\Psi_{\alpha}(x)\Psi_{\beta}(y)\right\} =
-T\left\{\Psi_{\beta}(y) \Psi_{\alpha}(x)\right\}$.  Thus for fermions,
changing the ordering inside the time ordering generates a minus sign
for each exchange. We summarize the general matrix element for $m$
fermions going into $n$ fermions as,

\begin{eqnarray}
_{out}\langle .. \vec{k}'_{j'}\sigma'_{j'} .. |..
\vec{k}_j\sigma_j ..  \rangle_{in} & = &
\left(\frac{-i}{\sqrt{Z_{\Psi}}}\right)^{m+n} \prod_{j,j'}\int
d^4x'_{j'} d^4x_j
\nonumber \\
& & \mbox{\hspace{0.5cm}}e^{-i\sum_{j'}k'_{j'}\cdot
x'_{j'}}(-i\dsl{\partial}'_{1} + m)\dots(-i\dsl{\partial}'_{n} +
m)\nonumber \\
& & \mbox{\hspace{1.5cm}}\langle 0|T\left\{
\Psi(x_1')..\Psi(x'_n)\bar{\Psi}(x_1)..\bar{\Psi}(x_m)\right\}
|0\rangle \nonumber \\
& & \mbox{\hspace{2.5cm}}(-i\overleftarrow{\dsl{\partial}_{1}} +
m)\dots(-i\overleftarrow{\dsl{\partial}_{m}} + m)
e^{+i\sum_{j}k_{j}\cdot x_{j}}
\end{eqnarray}

For anti-fermions in the in/out states, the spinors $U, \bar{U}$ are
changed to $V, \bar{V}$ as appropriate.

§.§.§ LSZ reduction for Maxwell field

This is a bosonic field so there is no additional minus sign. Like the
Dirac field, it has two polarizations and thanks to the zero mass, these
are transverse polarizations.  We begin by recalling the mode

\begin{eqnarray*}
A_{\mu}(x) & = & \int\frac{d^3k} {\sqrt{2\omega(2\pi)^3}}
\sum_{\lambda=1,2} \left[ a_{\vec{k},\lambda}
\underline{\varepsilon}_{\mu} (\vec{k},\lambda) e^{ik\cdot x} +
a^*_{\vec{k},\lambda} \underline{\varepsilon}^*_{\mu}
(\vec{k},\lambda)e^{-ik\cdot x}\right] ~ ~ \\
& & \underline{\varepsilon}_0(\vec{k},\lambda) ~ := ~ -\frac{\vec{k}
\cdot \vec{\varepsilon} (\vec{k},\lambda)}{\omega_{\vec{k}}} ~ ~ , ~ ~
\underline{\varepsilon}_i(\vec{k},\lambda) ~ := ~ \left(\delta_i^{~j} -
\frac{k_i\ k^j} {\vec{k}^2}\right) \varepsilon_j(\vec{k},\lambda) ~ ~ ,
~ ~ \omega_{\vec{k}} = |\vec{k}| \nonumber \\
& & \underline{\vec{\varepsilon}} (\vec{k},\lambda) \cdot
\underline{\vec{\varepsilon}}^* (\vec{k},\lambda') ~ = ~
\delta_{\lambda,\lambda'} ~ ~ , ~ ~ \sum_{\lambda}
\underline{\varepsilon}_i(\vec{k},\lambda)
\underline{\varepsilon}^*_j(\vec{k},\lambda)  = ~ \left(\delta_i^{~j} -
\frac{k_i\ k^j} {\vec{k}^2}\right)
\end{eqnarray*}

This gives the inversion formulae as,

\begin{eqnarray*}
a(\vec{k},\lambda) ~ = ~ i\varepsilon_i(\vec{k},\lambda)\int
d^3x e^{-ik\cdot x} \overleftrightarrow{\partial_0}A^i(x) & , &
a^{\dagger}(\vec{k},\lambda) ~ = ~
-i\varepsilon^*_i(\vec{k},\lambda)\int d^3x e^{+ik\cdot x}
\overleftrightarrow{\partial_0}A^i(x)
\end{eqnarray*}

In the transverse/radiation/Coulomb gauge,
$\underline{\varepsilon}_{\mu}(\vec{k}, \lambda) =
\varepsilon_{\mu}(\vec{k}, \lambda) ~ , ~\vec{k}\cdot
\vec{\varepsilon}(\vec{k}, \lambda) = 0$ and the equations of motion are
$\Box A_i(x) = 0$. The commutation relations are:

\[
[A_i(t,\vec{x}), \pi_j(t, \vec{y})] ~ = ~ i\left(\delta_{ij} -
\frac{\partial_i\partial_j}{\nabla^2}\right) ~ := ~ i\int
\frac{d^3k}{(2\pi)^3} e^{ik\cdot(x-y)}\left(\delta_{ij}
\]

The reduction formula goes similar to the scalar. We will have $(-\Box_x
+ m^2) \to -\Box_x$ and the wave function factors would be $\boxed{
\mathcal{E}_{\mu}(\vec{k}, \lambda) := \Case{\varepsilon(\vec{k},
\lambda)}{\sqrt{2\omega_{\vec{k}}(2\pi)^3}} . }$ Then the general
$S-$matrix element for $m \to n$ particles takes the form,

\begin{eqnarray}
_{out}\langle .. \vec{k}'_{j'}\lambda'_{j'} .. |..
\vec{k}_j\lambda ..  \rangle_{in} & = &
\left(\frac{-i}{\sqrt{Z_{A}}}\right)^{m+n} \prod_{j,j'}\int
d^4x'_{j'} d^4x_j \mathcal{E}(\vec{k}_j,\lambda)
\mathcal{E}^*(\vec{k}'_{j'},\lambda'_{j'}) \nonumber \\
& & \mbox{\hspace{0.5cm}}e^{-i\sum_{j'}k'_{j'}\cdot
x'_{j'}}(-\Box'_{1})\dots(-\Box'_{n})\nonumber \\
& & \mbox{\hspace{1.5cm}}\langle 0|T\left\{ A_{\mu'_1}(x_1')..
A_{\mu'_n}(x'_n) A_{\mu_1}(x_1).. A_{\mu_m}(x_m)\right\}
|0\rangle \nonumber \\
& & \mbox{\hspace{2.5cm}}(-\overleftarrow{\Box_1}) \dots
(-\overleftarrow{\Box_{m}}) e^{+i\sum_{j}k_{j}\cdot x_{j}}
\end{eqnarray}

Having related the  $S-$matrix elements to the $n-$point functions of
the interacting fields, out next task is to try see if these can be
expressed in terms of the free fields in some systematic, well defined
way. This is achieved in the so-called covariant perturbation series.

§ COVARIANT PERTURBATION THEORY

In formulating the scattering theory for interacting fields, we
postulated the in/out fields satisfying the free field equations. More
importantly, we also postulated them to satisfy the same equal time
(anti-)commutation relations. With some assumptions of existence, these
suffice to develop a perturbation scheme to compute the $n-$point

Since the interacting and the free (in/out) fields satisfy the same
basic quantum conditions, it is plausible that they are related by some
unitary transformation. For systems with finitely many degrees of
freedom, it is theorem that guarantees existence of canonical
transformations at the classical level and unitary transformations at
the quantum level (thanks to the Stone-von-Neumann theorem). For field
theories, there is no such guarantee and we need to postulate the
required existence. Their utility gives a post-facto justification for
the assumptions. Let,

$~\hfill\boxed{\Phi(t,\vec{x}) = U^{-1}(t)\Phi_{in}(t,\vec{x})U(t) ~ ~ ,
~ ~ \Pi(t,\vec{x}) = U^{-1}(t)\Pi_{in}(t,\vec{x})U(t) \ .}\hfill ~$.

The fields satisfy the equations of motion in the form,

\begin{eqnarray}\label{HamEqns}
\partial_t \Phi_{in}(x) = i[H_{in}(\Phi_{in},\Pi_{in}), \Phi_{in}(x)] ~
& , & ~ \partial_t \Pi_{in}(x) = i[H_{in}(\Phi_{in},\Pi_{in}),
\Pi_{in}(x)] \\
\partial_t \Phi(x) = i[H(\Phi,\Pi), \Phi(x)] ~ & , & ~ \partial_t \Pi(x)
= i[H(\Phi,\Pi), \Pi(x)]
\end{eqnarray}

These enable us to derive an equation for $U(t)$.

\begin{eqnarray*}
\partial_t\Phi_{in} & = & \partial_t[U(t)\Phi U^{-1}(t)] =
\partial_tU\cdot U^{-1}\Phi_{in} + U(\partial_t\Phi)U^{-1} +
\Phi_{in}\cdot U\partial_tU^{-1} \\
& = & [\partial_tU\cdot U^{-1}, \Phi_{in}] + iU[H(\Phi, \Pi), \Phi]
U^{-1} ~ ~ \mbox{But,} \\
U [H(\Phi, \Pi), \Phi] U^{-1} & = & [UH(\Phi,\Pi]U^{-1}, U\Phi U^{-1}] ~
= ~ [H(\Phi_{in}, \Pi_{in}), \Phi_{in}] \\
\therefore \partial_t\Phi_{in} & = & [ \partial_tU\cdot U^{-1} +
i\left\{ H(\Phi_{in},\Pi_{in}) - H_{in}(\Phi_{in}, \Pi_{in}) +
H_{in}(\Phi_{in}, \Pi_{in})\right\}, \Phi_{in} ] \\
\partial_t\Phi_{in} & = & \partial_t\Phi_{in} + [\partial_tU\cdot U^{-1}
+ iH_I(\Phi_{in}, \Pi_{in}), \Phi_{in}] ~ ~ ~ ~ \mbox{Similarly,} \\
\partial_t\Pi_{in} & = & \partial_t\Pi_{in} + [\partial_tU\cdot U^{-1} +
iH_I(\Phi_{in}, \Pi_{in}), \Pi_{in}] ~ .
\end{eqnarray*}

Since $\partial_tU\cdot U^{-1} + iH_I(\Phi_{in},\Pi_{in})$ commutes with
both $\Phi_{in}, \Pi_{in}$, it must be a multiple of identity, say
$E_0\mathbb{1}$. This gives the equation determining the $U(t)$ as,

\begin{equation}\label{UEqn}
\boxed{i\partial_tU(t) ~ = ~ H'_I(\Phi_{in},\Pi_{in})U(t)} ~ , ~
\boxed{H'_I(\Phi_{in}, \Pi_{in}) := H(\Phi_{in},\Pi_{in}) -
H_{in}(\Phi_{in}, \Pi_{in}) + E_0(t)}
\end{equation}

To solve the equation, it is convenient to define the combination
$U(t,t') := U(t)U^{-1}(t')$. It follows that this combination too
satisfies the same first order equation with the initial condition,
$U(t,t) = \mathbb{1}$.  The integral form of this equation is:

\[
U(t,t') ~ = ~ \mathbb{1} - i\int_{t'}^{t}dt'' H'_I(t'')U(t'',t')
\ .
\]

This is solved by iteration.

Define: $U_0(t,t') = \mathbb{1} , $ and for $n \ge 1,\ U_n(t,t') :=
\mathbb{1} - i\int_{t'}^{t}dt_n H'(t_n)U_{n-1}(t_n, t')$. Then $U(t,t')
= \sum_n U_n(t, t')$ is the formal solution of the integral equation.

It is trivial to verify the formal solution. It is formal because no
conditions are imposed for convergence of the series. The first few
terms are,

\begin{eqnarray}
U_0(t,t') & = & \mathbb{1} \nonumber \\
U_1(t,t') & = & \mathbb{1} - i\int_{t'}^tdt_1H'(t_1)\mathbb{1} \nonumber
\\
U_2(t,t') & = & \mathbb{1} - \int_{t'}^tdt_2 H'(t_2)U_1(t_2,t')
\nonumber \\
& = & \mathbb{1} +(-i)\int_{t'}^tdt_2H'(t_2) +
(-i)^2\int_{t'}^tdt_2\int_{t'}^{t_2}dt_1H'(t_2)H'(t_1) \nonumber \\
U_3(t,t') & = & \mathbb{1} + (-i)\int_{t'}^tdt_3
H'(t_3)\left\{\mathbb{1} +(-i)\int_{t'}^tdt_2H'(t_2) \right. \nonumber
\\
& & \mbox{\hspace{4.5cm}} \left. +
(-i)^2\int_{t'}^tdt_2\int_{t'}^{t_2}dt_1H'(t_2)H'(t_1) \right\}
\nonumber \\
& = & \mathbb{1} + (-i)\int_{t'}^tdt_3 H'(t_3) + (-i)^2 \int_{t'}^t dt_3
\int_{t'}^{t_3}dt_2H'(t_3)H'(t_2)\nonumber \\
& & \mbox{\hspace{3.8cm}} + (-i)^3\int_{t'}^tdt_3 \int_{t'}^{t_3} dt_2
\int_{t'}^{t_2} dt_1 H'(t_3)H'(t_2)H'(t_1) ~ ~ ~ ~  \nonumber \\
\therefore U_n(t,t') & = & \mathbb{1} + \sum_{k=1}^n
(-i)^k\int_{t'}^tdt_k \int_{t'}^{t_k}dt_{k-1}\dots\int_{t'}^{t_2}dt_1
H'(t_k)H'(t_{k-1})\dots H'(t_1)
\end{eqnarray}

Consider the term $\int_{t'}^tdt_n\int_{t'}^{t_n}dt_{n-1}$. By
interchanging the order f integration we can write it as ($H'_n :=

\begin{eqnarray*}
\int_{t'}^tdt_n\int_{t'}^{t_n}dt_{n-1}H'_{n}H'_{n-1} & = &
\int_{t'}^tdt_{n-1}\int_{t_{n-1}}^{t}dt_{n} H'_{n}H'_{n-1} ~ = ~
\int_{t'}^tdt_{n}\int_{t_{n}}^{t}dt_{n-1} H'_{n-1}H'_{n} \\
\therefore LHS = \frac{1}{2}(LHS + RHS) & = & \frac{1}{2}\left[
\int_{t'}^tdt_n\int_{t'}^{t_n}dt_{n-1}H'_{n}H'_{n-1} +
\int_{t'}^tdt_{n}\int_{t_{n}}^{t}dt_{n-1} H'_{n-1}H'_{n} \right] \\
& = & \frac{1}{2}\left[
\int_{t'}^tdt_n\int_{t'}^{t_n}dt_{n-1}T\left\{H'_{n}H'_{n-1}\right\}
\right. \\
& & \mbox{\hspace{3.0cm}} \left. +
\int_{t'}^tdt_{n}\int_{t_{n}}^{t}dt_{n-1} T\left\{H'_{n}H'_{n-1}\right\}
\right] \\
& = & \frac{1}{2}\int_{t'}^tdt_n \int_{t'}^{t} dt_{n-1} T
\left\{H'_{n}H'_{n-1}\right\}
\end{eqnarray*}

Similarly, we can symmetrize the higher order terms to replace each
product by their time ordered products and then combine the integrals to
have the same limits of integrations. The $n^{th}$ order term then takes
the form: $\Case{1}{n!} \int_{t'}^tdt_n \int_{t'}^tdt_{n-1} \dots
\int_{t'}^tdt_1 T\{H'(t_n)H'(t_{n-1})\dots H'(t_1)\}$. We write the
iterated formal solution of the (<ref>) as,

\begin{eqnarray} \label{USoln}
U(t,t') & := & T\left(exp\left\{-i\int_{t'}^tdt' H'_I(\Phi_{in}(t'),
\Pi_{in}(t')) \right\}\right) \\
& = & \mathbb{1} + \sum_{n=1}^{\infty}\frac{(-i)^n}{n!}
\int_{t'}^tdt_1\dots\int_{t'}^tdt_n\ T\left\{H'_I(t_1)\dots
\end{eqnarray}

The solution is thus obtained entirely in terms of the `in' fields. Note
that the solution gives $U(t,t')$ and not $U(t)$. While the
equations satisfied by $U(t)$ and $U(t,t')$ are the same, there is no
“initial” condition provided for $U(t)$ and in a sense it is
ill-defined. There is no canonical way to deduce/define  $U(t)$ from
$U(t,t')$. Fortunately, $U(t,t')$ suffice.

Note: From the definition it follows the useful composition
property that $U(t,t') = U(t,t'')U(t'',t')$, regardless of any
inequalities for $t''$. This may also be verified directly.

Let us see an implication of the existence of $U(t)$.  Consider an
$n-$point function for a scalar, $G(x_1,\dots,x_n) =\langle 0| T\left\{
\Phi(x_1)\dots \Phi(x_n)\right\}|0\rangle $. Insert in this $\Phi(x) =
U^{-1}(t)\Phi_{in}(x)U(t)$ which gives,

\begin{eqnarray*}
\Phi(x_1)\Phi(x_2)\dots \Phi(x_n) & = & U^{-1}(t_1) \Phi_{in}(x_1)
U(t_1)\cdot U^{-1}(t_2) \Phi_{in}(x_2) U(t_2)..U^{-1}(t_n)
\Phi(x_n)U(t_n) \\
& = & U^{-1}(t) \left[U(t,t_1)\Phi_{in}(x_1) U(t_1, t_2)
\Phi_{in}(x_2)\dots \right. \\
& & \left. \mbox{\hspace{1.0cm}} U(t_{n-1}, t_n) \Phi_{in}(x_n)
U(t_n,-t) \right]U(-t) ~ ~ , ~ ~ U(t,t') := U(t)U^{-1}(t')
\end{eqnarray*}

Observe that in the limit $t \to \infty$, the factor $U^{-1}(t)$ goes to
the extreme left and $U(t)$ to the extreme right. All other space-time
arguments being finite, we can take these factors outside the time
ordering symbol. The product of the fields $\Phi(x)$'s is already under
time ordering which then allows us to arrange all the factors within
$T\{\dots\}$ to be conveniently grouped. In particular, all the $U(t_i,
t_j)$ can be combined using their composition property noted above.

With the limit of $t \to \infty$ implicit, we can write the $n-$point
function as,

\begin{equation}\label{GwithU}
G(x_1,\dots,x_n) ~ = ~\langle 0|U^{-1}(t)T\left\{\Phi_{in}(x_1) \dots
\Phi_{in}(x_n)\ exp\left(-i\int_{-t}^{t}dt' H'_{I}(t')\right)\right\}
U(t)|0\rangle \ .
\end{equation}

Next, we evaluate the $U(-t)|0\rangle$ in the limit $t\to \infty$ and
its adjoint.

Claim: $U(-t)|0\rangle \to \lambda_-|0\rangle$ as $t\to
\infty$.

Proof: Consider an in-state containing at least one particle
of momentum $k, |\alpha,k\rangle_{in}$. Then,

\begin{eqnarray*}
_{in}\langle\alpha,k|U(-t)|0\rangle & = &
_{in}\langle\alpha|a_{in}(\vec{k})U(-t)|0\rangle \\
& = & i\int_{\Sigma_{t'}}
\frac{d^3x}{\sqrt{2\omega_{\vec{k}}(2\pi)^3}}e^{-ik\cdot
x}\overleftrightarrow{\partial'_{0}}_{in}\langle \alpha|
\Phi_{in}(-t',\vec{x})U(-t)|0\rangle
\end{eqnarray*}

We have used the inversion formula for $a_{in}$ by choosing an arbitrary
$\Sigma_{t'}$ hyper-surface and $a_{in}$ is of course independent of

In the integrand above, substitute $\Phi_{in}(t',\vec{x}) = U(t')
\Phi(t',\vec{x}) U^{-1}(t')$ and evaluate the time derivative.

\begin{eqnarray*}
\mbox{Integrand} & = & e^{-ik\cdot x}\langle\alpha| \left[\dot{U}(-t')
U^{-1}(-t') \Phi_{in}(-t',\vec{x}) U(-t) + \Phi_{in}(-t',\vec{x}) U(-t')
\partial_{t'}U^{-1}(t') U(-t) \right. \\
& & \left. + U(-t')\partial'_0\Phi(-t')U^{-1}(-t')U(-t) -
U(-t')\Phi(-t')U^{-1}(-t')U(-t)(\partial'_0(-ik\cdot x))\right]|0\rangle
\end{eqnarray*}

Now take choose $t' = t$. Then, the first two terms on the r.h.s.
combine as,

\[
[\dot{U}(-t)U^{-1}(-t), \Phi_{in}(-t, \vec{x})]U(-t) ~ = ~
-i[H'_I(\Phi_{in}, \Pi_{in}), \Phi_{in}(-t, \vec{x})]U(-t) ~ = ~ 0 \ .
\]

In the usual theories, the interaction Hamiltonian has no $\Pi$
dependence and hence at equal time, commutes with the field giving 0.

In the last two terms, the $U^{-1}(-t')U(-t)$ factor becomes
$\mathbb{1}$ for $t' = t$. These terms combine to give, $_{in}
\langle\alpha| U(-t) \left(e^{-ik\cdot x} \overleftrightarrow
{\partial_0} \Phi(-t) \right)|0\rangle$.  Taking the  limit $t \to
\infty$ and invoking the asymptotic condition, converts $\Phi(x)$ into
$\sqrt{Z}\Phi_{in}(x)$.  The spatial integration converts the expression
into $\sqrt{Z}_{in} \langle\alpha| U(-t) a_{in}(\vec{k})|0\rangle = 0$.

Thus we conclude that $_{in}\langle\alpha,k|U(-t)|0\rangle \to 0$ as $t
\to \infty$ for all states containing at least one particle.
$U(-t)|0\rangle$ must therefore be proportional to $|0\rangle$ itself
thereby proving the claim.

By similar reasoning, we obtain $U(t)|0\rangle \to \lambda_+|0\rangle$
as $t \to \infty$.

\begin{equation} \label{GNorm}
\lambda_+^*\lambda_- = \left[\langle 0|T exp \left(-i
\int_{-\infty}^{\infty}dt' H'_I(t') \right) |0\rangle \right]^{-1}
\end{equation}

The proof is simple,

\begin{eqnarray*}
\lambda_+^*\lambda_- & = & \lim_{t\to \infty}\langle
0|U^{-1}(t)|0\rangle\langle 0|U(-t)|0\rangle \\
& \approx & \langle 0|U^{-1}(t)|U(-t)|0\rangle ~ ~ \mbox{since
intermediate states do not contribute by above result}\\
& \approx & \langle 0|U(-t)U^{-1}(t)|0\rangle ~ = ~ \langle
0|U(-t,t)|0\rangle \\
& \approx & \langle 0|(U(t,-t))^{-1} |0\rangle ~ = ~ (\langle
0|U(t,-t)|0\rangle)^{-1} \ .
\end{eqnarray*}

The final expression follows by substituting the formal solution in the
limit $t \to \infty$.

Using the equations (<ref>, <ref>), we express the $n-$
point function as,

\begin{equation}\label{GEqn}
\boxed{ G(x_1,\dots,x_n) ~ = ~ \frac{\langle
-i\int_{-\infty}^{\infty}dt' H_I(t')\right\}|0\rangle}{\langle
0|T\left\{exp -i\int_{-\infty}^{\infty}dt' H_I(t')\right\}|0\rangle} }
\end{equation}

Notice that the multiple of identity $E_0(t)\mathbb{1}$ has been
canceled between the numerator and the denominator before taking the
limit $t\to \infty$.

Suffice it to say that analogous expressions exist for other fields as
well. However we will not write them explicitly.

The reduction formulae give $S-$matrix elements in terms of the
$n-$point functions and the $n-$point functions have a perturbative
expansion in terms of the free fields alone. These two together are the
master formulae for computations.

A further simplification is provided by the Wick's Theorem.

§.§ Normal ordering and Wick's theorem

Recall that the mode expansion of the free quantum fields can be
grouped as $\Phi(x) = \Phi_+(x) + \Phi_-(x) $ with  $\Phi_{\pm}(x)$
denoting the sum over positive/negative  frequency modes. The positive
frequency part is made up of annihilation operators while the negative
frequency part is made up of creation operators alone. For a Hermitian
quantum field, the negative frequency part is the Hermitian adjoint of
the positive frequency part. A product of fields at different points
will mix these parts and the idea is to bring all annihilation operators
to the right and the creation operators to the left. The vacuum
expectation value of so ordered groups will of course be zero. In this
process, several commutators of fields at different space-time points
are generated, but they all are c-numbers. This leads to a

We define normal ordered products of two free fields as:

\begin{eqnarray}
\mbox{For bosons~ ~}	:\Phi_-(x)\Phi_+(y): ~ = ~ \Phi_-(x)\Phi_+(y) ~
~ &,&~ ~ :a^{\dagger} a: ~ := ~ a^{\dagger} a \\
:\Phi_+(x)\Phi_-(y): ~ = ~ \Phi_-(y)\Phi_+(x) ~ ~ &,&~ ~ :a a^{\dagger}:
~ := ~ a^{\dagger} a \\
\mbox{For fermions~ ~}	:\Psi_-(x)\Psi_+(y): ~ = ~ \Psi_-(x)\Psi_+(y) ~
~ &,&~ ~ :b^{\dagger} b: ~ := ~ b^{\dagger} b \\
:\Psi_+(x)\Psi_-(y): ~ = ~ - \Psi_-(y)\Psi_+(x) ~ ~ &,&~ ~ :b
b^{\dagger}: ~ := ~ - b^{\dagger} b
\end{eqnarray}

For product of two or more positive (or negative) frequency fields, the
normal ordering does not change order. It is immediately obvious that
$\langle 0| : Operator : |0\rangle = 0$.

Note that the normal ordering are meaningful only for
products of free fields (in/out). Hence the $in/out$ suffixes are
suppressed for these.

To get a general form for several bosonic as well as fermionic free
fields, notice that for a quantum field $Q(x) = Q_+(x)+ Q_-(x)$ a
product of $n$ fields generates terms with the $k$ number of $Q_+$'s
interspersed with $(n-k)$ number of $Q_-$'s, with the order of the
space-time points maintained, with $k = 0,\dots,n$. Under normal ordering
each of these terms will shift the $Q_+$'s to the right and $Q_-$'s to
the left, generating a permutation of the space-time points, say $p$. If
$Q$ is a fermionic field, the term with permutation $p$ will get a
factor of $\sigma_p = sgn(p)$ while a bosonic field will have $\sigma_p
= 1$. Note that for any given $k$, there will be several terms eg
$Q_1Q_2, Q_3Q_7,\dots$ etc. Each will generate its own permutation under
normal ordering. With this understood, we can write the general
expression for normal ordered product of $n-$fields as,

$~\hfill \boxed{
:Q(x_1)\dots Q(x_n): ~ := ~ \sum_{A,B} \sigma_p
\prod_{i\in A} Q_-(x_{p(i)}) \prod_{j\in B}
Q_+(x_{p(j)})\ .
}\hfill ~$

The $A,B$ are two groups of space-time points corresponding to the $k,
(n-k)$ mentioned above and the sum refers to various possible groupings
of the space-time points within each class.

Exercise: Verify for $n = 4$.

Consider now the relation between time ordered products and normal
ordered products. For a single field, both orderings are trivial and
their vacuum expectation value vanishes: $\langle 0|T\{\phi_{in}(x)\}|0
\rangle = 0 =\langle 0|:\phi(x):|0\rangle$.

For time ordered product of two fields, for any particular instances of
time they appear in a particular order which can be explicitly put in
the normal ordered form, eg. $a(\vec{k})a^{\dagger}(\vec{k}') =
``\mathbb{1}" + a^{\dagger}(\vec{k}')a(\vec{k}) =
:a(\vec{k})a^{\dagger}(\vec{k}'): + ``\mathbb{1}"$. The last term is a
c-number (multiple of the identity operator). Similarly, $\Phi(x)\Phi(y)
= :\Phi(x)\Phi(y): +\ c-$number.  And the c-number is trivial to
evaluate by just taking the vev (vacuum expectation value): $c-$number =
$\langle 0|\Phi(x)\Phi(y)|0\rangle$. Noting that $T\{\Phi(x)\Phi(y)\} =
\theta(x^0 - y^0)\Phi(x)\Phi(y) + \theta(y^0-x^0)\Phi(y)\Phi(x)$, we
get, $\boxed{T\{\Phi(x)\Phi(y)\} = :T\{\Phi(x)\Phi(y)\}: + \langle
0|T\{\Phi(x)\Phi(y)\}|0\rangle}$. Notice that normal ordering and time
ordering commute: $:T\{A(x)B(y)\}: = T\{:A(x)B(y):\}$, for all
non-trivial $A(x), B(y)$ operators i.e. for operators which are
polynomials in the basic field operators of degree greater that zero.
The generalization of this to product of arbitrary number of field is
the Wick's theorem. We state it explicitly for a scalar field.

Wick's Theorem:

\begin{eqnarray*}
T\{\Phi(x_1)\dots\Phi(x_n)\} & = & :T\{\Phi(x_1)\dots\Phi(x_n)\}: \\
& & + \left[\langle 0|T\{\Phi(x_1)\Phi(x_2)\}|0\rangle
:T\{\Phi(x_3)\dots\Phi(x_n)\}: + \mbox{~permutations}\right] \\
& & + \left[\langle 0|T\{\Phi(x_1)\Phi(x_2)\}|0\rangle \langle
0|T\{\Phi(x_3)\Phi(x_4)\}|0\rangle \times \right. \\
& & \left. \mbox{\hspace{4.0cm}} :T\{\Phi(x_5)\dots\Phi(x_n)\}: +
\mbox{~permutations}\right] \\
& &  \mbox{\hspace{1.0cm}} \vdots~\vdots \\
& & + \left\{ 	\begin{array}{l}
\left[\langle 0|T\{\Phi_1\Phi_2\}|0\rangle \dots \langle
0|T\{\Phi_{n-1}\Phi_n\}|0\rangle + \mbox{~permutations}
\right] \\
\mbox{\hspace{8.0cm} ($n$ even)} \\
\left[\langle 0|T\{\Phi_1\Phi_2\}|0\rangle \dots \langle
0|T\{\Phi_{n-2}\Phi_{n-1}\}|0\rangle \Phi_n \right. \\
\left. \mbox{\hspace{4.9cm}} + \mbox{~permutations}
\right] ~ \mbox{($n$ odd)}
\end{array}
\right.
\end{eqnarray*}

In the last line we have used the abbreviation $\Phi_i := \Phi(x_i)$ and
we will use the same when convenient.

The proof is by induction and we have already seen the validity for $n =
1, 2.$ Assume the theorem is true for $n = n$. Include an extra field
$\Phi(x_{n+1})$ and choose $t_{n+1}$ to be the earliest instant so that
it stays on the extreme right. Then,

\begin{eqnarray*}
T\{\Phi_1\dots \Phi_{n+1}\} & = & T\{\Phi_1\dots \Phi_n\}\Phi_{n+1} \\
& = & \left[:T\{\Phi_1\dots \Phi_n\}: + \sum_{perm}\langle
0|T\{\Phi_1\Phi_2\}|0\rangle:T\{\dots\}: + \dots \right]\Phi_{n+1}~ , \\
:\Phi_1\dots \Phi_n:\Phi_{n+1} & = &
\left\{\sum_{A,B}\sigma_p\prod_{i\in A}\Phi^-_i\ \prod_{j\in
B}\Phi^+_j\right\}\left(\Phi^+_{n+1} + \Phi^-_{n+1}\right) \\
& = & \sum_{A,B}\sigma_p\prod_{i\in A}\Phi^-_i\ \prod_{j\in
B}\Phi^+_j\Phi^+_{n+1} \ + \\
& & \mbox{\hspace{1.0cm}} \sum_{A,B}\sigma_p\prod_{i\in A}\Phi^-_i\
\left\{\Phi_{n+1}^- \prod_{j\in B}\Phi^+_j + \left[\prod_{j\in
B}\Phi^+_j\ , \ \Phi_{n+1}^-\right]\right\}
\end{eqnarray*}

The first two terms in the last equation have the requisite normal
ordered form for $n+1$ fields. The third term has several commutators,
$[\Phi^+_j, \Phi^-_{n+1}]$ which again are c-numbers and can be
evaluated by taking vev. That is,

\begin{eqnarray*}
[\Phi^+_j, \Phi^-_{n+1}] & = & \langle 0|(\Phi^+_j\Phi^-_{n+1} -
\Phi^-_{n+1}\Phi^+_j)|0\rangle ~ = ~ \langle
0|\Phi^+_j\Phi^-_{n+1}|0\rangle + 0 \\
& = &\langle 0|\Phi_j\Phi_{n+1}|0\rangle ~ = ~\langle
0|T\left\{\Phi_j\Phi_{n+1}\right\}|0\rangle  ~ ~ , ~ ~ \mbox{since
$t_{n+1}$ is the earliest.} \\
\therefore\ :\Phi_1\dots \Phi_n:\Phi_{n+1} & = & :\Phi_1\dots
\Phi_n\Phi_{n+1}: + \sum_j\sigma_p :\Phi_1\dots\Phi_{j-1}
\Phi_{j+1}\dots\Phi_n:\langle 0|T\left\{ \Phi_j \Phi_{n+1}
\right\}|0\rangle
\end{eqnarray*}

Thus, inserting the time ordering, the first term, $:T\{\Phi_1\dots
\Phi_n\}:\Phi_{n+1}$ gets expressed as $:T\{\Phi_1\dots
\Phi_n\Phi_{n+1}\}: + \ \mbox{\{terms which have a structure similar to
the second term\}}$ and so on. Hence the induction hypothesis allows the
expression to hold for $n+1$ and the theorem is proved by induction.

Exercise: Verify the theorem for $n = 3, 4$.

Observe that $G(x_1,\dots,x_n) =\langle 0|T\{\Phi(x_1) \dots
\Phi(x_n)\}|0\rangle$ vanishes for odd $n$ and equals
$\sum_{perm}\sigma_p \langle 0|T\{\Phi(x_1) \Phi(x_2)\}|0\rangle\dots
\langle 0|T\{\Phi(x_{n-1}) \Phi(x_n)\}|0\rangle$ for even $n$.

Thus, $2n-$point function of free fields is the sum of products of
$n$, $2-$point functions of free fields. We denote: $\boxed{\langle
0|T\{\Phi(x)\Phi(y)\}|0\rangle =: i\Delta_F(x-y) =: \mbox{(Feynman

We have the $S-$matrix elements in terms of the $n-$point functions of
interacting quantum fields, expressed in terms of vev of time ordered
products of free fields and thanks to Wick's theorem these are expressed
in terms of $2-$point functions of free fields. In short, the $S-$matrix
elements are obtained from a bunch of $2-$point functions of free
fields, together with some integrations and operations of $(\Box -
m^2)$.  It remains to have a convenient algorithm for computations.

A few remarks are in order. The normal ordering is introduced here as a
simplification step. It works because the postulated unique
Poincare invariant vacuum state is annihilated by all (free field)
annihilation operators.  This also requires the Poincare
generators of the free fields to be normal ordered[Note that
while abelian symmetries could allow a non-zero multiple of
identity for invariance (i.e.  $P^{\mu}|0\rangle =
\alpha^{\mu}|0\rangle$), non-abelian symmetry generators have to
annihilate the vacuum. Hence the Lorentz generators must
annihilate the vacuum and then the commutator of $M^{\mu\nu}$
with $P^{\lambda}$ shows that the $\alpha^{\mu} = 0$ must also hold.
Poincare group being non-abelian must have its generators being normal
ordered for the vacuum to be invariant.]. The Poincare generators of the
interacting fields also need to annihilate the vacuum. Thanks to the
assumed unitary operator linking the free and the interacting fields,
these generators too are expressed in terms of the free field
creation-annihilation operator. These expressions, in particular the
Hamiltonian, must also be normal ordered for Poincare invariance. We
will make this explicit for the interaction Hamiltonian $H_I(\Phi_{in},
\Pi_{in})$.

In the quantum optics context, the normal ordering was used since the
detectors worked by absorbing a quantum from the field and in principle,
for a detector working by emission of a quantum would require
anti-normal ordering. Such an option is not available with manifest
Poincare invariance with a unique invariant vacuum state.

An explicit example of evaluation of an $n-$point function will be
useful. Consider $G_2(x_1, x_2)$ computed to second order in $H_I$.
Consider the numerator first. All fields are `in' fields and the
`in' subscript is suppressed below.

\begin{eqnarray*}
G_2(x_1,x_2)|_{Nr} & = &\langle 0|T\{\Phi(x_1)\Phi(x_2) \left(\mathbb{1}
-i\int_{-\infty}^{\infty}dt'H_I(t') \right. \\
& & \left. \mbox{\hspace{3.0cm}} + \frac{(-i)^2}{2!}
\int_{-\infty}^{\infty} dt_1 dt_2 T\{H_I(t_1) H_I(t_2)\} \right)\}
|0\rangle \\
& = & \langle 0|T\{\phi(x_1)\Phi(x_2)\}|0\rangle - i
\int_{-\infty}^{\infty} dt_1\langle 0|T\{\Phi(x_1) \Phi(x_2) H_I(t_1)\}
|0\rangle \\
& & \mbox{\hspace{2.0cm}} + \frac{(-i)^2}{2} \int_{-\infty}^{\infty}
dt_1 dt_2 \langle 0|T\{ \Phi(x_1) \Phi(x_2) H_I(t_1) H_I(t_2) \}
\end{eqnarray*}

Now let us take a specific interaction Hamiltonian, simplest non-trivial
being $H_I(t) := g\int d^3y:\Phi(t,\vec{y})^3:$ . As noted above, for
the case of Poincare invariant vacuum, the Poincare generators and hence
in particular the Hamiltonian must be normal ordered. The time
integrations combine with these spatial integration to give a space time
integral. Thus, we get,

\begin{eqnarray*}
G_2(x_1,x_2)|_{Nr} & = & \langle 0|T\{\phi(x_1)\Phi(x_2)\}|0\rangle - ig
\int\ d^4y\langle 0|T\{\Phi(x_1) \Phi(x_2) :\Phi^3(y):\} |0\rangle \\
& & \mbox{\hspace{0.5cm}} + \frac{(-i)^2}{2}g^2 \int\ d^4y_1\int d^4y_2
\langle 0|T\{ \Phi(x_1) \Phi(x_2) :\Phi^3(y_1)::\Phi^3(y_2): \}
|0\rangle \\
& = & i\Delta_F(x_1 - x_2) - \frac{g^2}{2} \int\ d^4y_1\int d^4y_2
\langle 0|T\{ \Phi(x_1) \Phi(x_2) :\Phi^3(y_1)::\Phi^3(y_2): \}
\end{eqnarray*}

The first term is just the Feynman propagator and the $o(g)$ term
vanishes since it is a 5-point function. The non-trivial term is the
$o(g^2)$ term. This is not in the form of the Wick's theorem, in that
there are normal ordered factors. We can replace the normal ordered
product in terms of unordered product minus the product of two point
functions i.e. using the Wick's theorem in reverse form. For instance,
consider (for three distinct points)

\begin{eqnarray*}
:\Phi(y_1)\Phi(y'_1)\Phi(y''_1): & = & \Phi(y_1)\Phi(y'_1)\Phi(y''_1)
-\langle 0|\Phi(y_1)\Phi(y'_1)|0\rangle\Phi(y''_1) \\
& & \mbox{\hspace{2.0cm}} -\langle 0|\Phi(y'_1) \Phi(y''_1)|0\rangle
\Phi(y_1) -\langle 0|\Phi(y_1) \Phi(y''_1)|0\rangle \Phi(y'_1)
\end{eqnarray*}

And likewise for the second factor of $:\Phi(y_2)^3:$. If we now
substitute these in the $o(g^2)$ term and eventually take the
coincidence limit of the primed arguments, we see that the two point
functions of coincident points cancel out. The net result result is that
in writing the permutation terms, we omit the Feynman propagators of
coincident points and this holds because we use the normal ordered
interaction Hamiltonian. If we did not normal order the interaction
Hamiltonian then these terms, which are formally divergent, will remain
and will have to be handled differently. This is what will happen in the
functional integral method which will be discussed later. Keeping this
in mind, the $o(g^2)$ term under vev becomes,

\begin{eqnarray*}
\langle 0|T\{\Phi(x_1)\Phi(x_2):\Phi^3(y_1): :\Phi^3(y_2):\}|0\rangle ~
\mbox{\hspace{8.0cm}$~$} & & \\
+ \mbox{~permutations} & & \\
= (i^4)\left[\Delta_F(x_1-x_2)\Delta_F(y_1-y_2)^3 +
\Delta_F(x_1-y_1)\Delta_F(x_2-y_2)\Delta_F(y_1-y_2)^2 \right. & & \\
\left. + \Delta_F(x_1-y_2)\Delta_F(x_2-y_1)\Delta_F(y_1-y_2)^2 \right]
\end{eqnarray*}

There are no other `pairings' of space-time points as they would involve
Feynman propagators with coincident arguments.

The denominator of the $n-$point function has the same structure as the
numerator except for the $\Phi(x_1)\Phi(x_2)$ fields i.e.

\begin{eqnarray*}
G_2(x_1,x_2)|_{Dr} & = &\langle 0|0\rangle - ig \int\ d^4y\langle
0|T\{:\Phi^3(y):\} |0\rangle \\
& & \mbox{\hspace{0.5cm}} + \frac{(-i)^2}{2}g^2 \int\ d^4y_1\int d^4y_2
\langle 0|T\{ :\Phi^3(y_1)::\Phi^3(y_2): \} |0\rangle \\
& = & 1  - \frac{g^2}{2} \int\ d^4y_1\int d^4y_2 \langle 0|T\{
:\Phi^3(y_1): :\Phi^3(y_2): \} |0\rangle \\
& = & 1 + \frac{(-ig)^2}{2} (CF) \int\ d^4y_1\int d^4y_2\
\end{eqnarray*}

Here, $CF$ is the combinatorial factor - number of distinct ways
of effecting the same pairings - and equals $3!$ in the example above.

The generalization is fairly obvious. For definiteness consider the
interaction Hamiltonian density to be $H_I(\Phi(y)) :=
\Case{g}{k!} :\Phi^k(y):$.  Then, it is apparent that the numerator of a
general $n-$point function is a sum of the $m^{th}$ order contributions,
$m = 0,1,\dots$. The $m^{th}$ order contribution has the form:

$\bullet ~ \Case{(-ig)/k!)^m)}{m!}\int d^4y_1\dots d^4y_m \langle 0|
T\{\Phi(x_1)\dots\Phi(x_n) :\Phi^k(y_1): \dots: \Phi^k(y_m): \}|0\rangle
\ ;$

$\bullet ~$ The Wick's theorem gives a sum of terms, each of which is a
product of a total of $(n+km)/2$ Feynman propagators,
$i\Delta_F(x_j-y_k)$,  whose arguments give the `pairing' of space-time
points from the set: $x_1,\dots,x_n, k-\mbox{copies of} \ y_1, \dots,
k-\mbox{copies of} \ y_m$. The normal ordered form of interaction
Hamiltonian ensures that the paired points are necessarily distinct. The $x_1,\dots,x_n$ points are assumed to be distinct. A
`pairing' is also called a `Wick contraction'. The sum of terms is
generated from any particular one by permutations. If $n + km$ is odd,
the contribution vanishes;

$\bullet ~$ There would also be combinatorial factor counting distinct
ways of getting the same set of pairing (or Feynman propagators). The
denominator has similar contributions except the $x_1,\dots,x_n$ points.

There is a convenient diagrammatic representation to keep track of the
generated terms as well as associating specific factors and integrals
with them. The diagrams are the Feynman Diagrams while the
prescription to construct associated integral is given by the Feynman rules.

Procedure to generate Feynman diagrams for Green's function:

$\bullet~$ For each distinct $x_1, \dots, x_n$, draw a vertex with
a single edge.  Call these vertices as external vertices;

$\bullet~$ For normal ordered monomial of order $k$, draw a vertex $y_j$
with $k$ edges sticking out. Call these as the internal vertices;

$\bullet~$ Wick contract or pair, by joining two edges from two distinct
vertices, internal or external. If the pair has one or both vertices to
be external vertices, the line is called an external line while a
pairing with both vertices being internal, is called an internal
line. This constructs a Feynman diagram with vertices and
internal/external lines.

Procedure to associate integral with a Feynman diagram:

$\bullet~$ With each contraction, associate a Feynman propagator,

$\bullet~$ With each internal vertex, associate a factor of $(-i)\times
$coefficient of the monomial of the fields in $H_I(\Phi(y))$ and an
integral over its space-time location, $\int d^4y$.

$\bullet~$ Associate a numerical factor as a ratio. In the numerator,
count the number of distinct ways of generating the same diagram
(or the same contractions). In the denominator, put $m!$ if $m$ is the
total number of internal vertices.

For the above example of $G_2(x_1,x_2)$ to second order for $H_I =
g:\Phi(y)^3:$, here is a summary:

[above =0.4of v] (a) ;
[below left=0.4of v] (b) ;
[below right=0.4of v] (c) ;
[left =of v] (d) ;
[left =0.95of v] (d') ;
[right =of v] (e) ;
[right =0.95of v] (e') ;

*  (d') – [scalar,half
left,looseness=1.5 ] (a),
(e') – [scalar,half left,looseness=1.5] (c) ,
(v) – (a) ,
(v) – (b) ,
(v) – (c) ,
(d) – (d'),
(e) – (e'),

[c]0.5Has one loose end and vev is zero.

(v) ;
[left =1.5of v] (a) ;
[right =0.25of a] (a') ;
[left =0.1of a] (a0) \(x_1\);
[right =1.5of v] (b) ;
[left =0.25of b] (b') ;
[right =0.1of b] (b0) \(x_2\);
[above =1.0of v] (c) ;
[above =0.25of c] (c0) \(y_1\);
[below =1.0of v] (d) ;
[below =0.25of d] (d0) \(y_2\);

[below left=0.25of c] (c1) ;
[below right=0.25of c] (c2) ;
[above =0.25of c] (c3) ;

[above left=0.25of d] (d1) ;
[above right=0.25of d] (d2) ;
[below =0.25of d] (d3) ;

*  (a) – (a'),
(b') – (b),
(c) – (c3), (c) – (c1), (c) – (c2),
(d) – (d3), (d) – (d1), (d) – (d2),
(b') – [scalar,half
left,looseness=0.75,edge label=3] (d3),
(a') – [scalar,half
label=3] (c3),
(c1) – [scalar,half
label=1] (d1),
(c2) – [scalar,half
left,looseness=0.5,edge label=2] (d2),

$(-ig)^2 \left(\frac{3.3.2.1}{2!}\right)\int d^4y_1d^4y_2\
(i\Delta_F(x_1-y_1)) \times (i\Delta_F(x_2-y_2))(i\Delta_F(y_1-y_2))^2$

(v) ;
[left =1.5of v] (a) ;
[right =0.25of a] (a') ;
[right =1.5of v] (b) ;
[left =0.25of b] (b') ;
[above =1.0of v] (c) ;
[above =0.8of c] (c0) ;
[below =1.0of v] (d) ;

[below left=0.25of c] (c1) ;
[below right=0.25of c] (c2) ;
[above =0.25of c] (c3) ;

[above left=0.25of d] (d1) ;
[above right=0.25of d] (d2) ;
[below =0.25of d] (d3) ;

*  (a) – (a'),
(b') – (b),
(c) – (c3), (c) – (c1), (c) – (c2),
(d) – (d3), (d) – (d1), (d) – (d2),
(b') – [scalar,half
right,looseness=0.75] (c3),
(a') – [scalar,half
right,looseness=0.75] (d3),
(c1) – [scalar,half right,looseness=0.5] (d1),
(c2) – [scalar,half left,looseness=0.5] (d2),
(c3) – [scalar,opacity=0.1] (c0),

\left(\frac{3.3.2.1}{2!}\right)\int d^4y_1d^4y_2\ (i\Delta_F(x_1-y_2))
\times (i\Delta_F(x_2-y_1))(i\Delta_F(y_1-y_2))^2$

(a) \(x_1\);
[right = 0.25of a] (a');
[right =2.0of a] (b)\(x_2\) ;
[left =0.25of b] (b') ;

[below =1.5of a] (c) ;
[above right=0.6of c] (c1) ;
[right=0.25of c] (c2) ;
[below right=0.6of c] (c3) ;
[left =0.1of c] (c0) \(y_1\) ;
[below =0.8of c3] (c00) ;

[below =1.5of b] (d) ;
[above left =0.6of d] (d1) ;
[left =0.25of d] (d2) ;
[below left =0.6of d] (d3) ;
[right =0.1of d] (d0) \(y_2\) ;

[above =0.5of a] (a”) ;
[above =0.5of b] (b”) ;
[below =0.5of c] (c”) ;
[below =0.5of d] (d”) ;

*  (a) – (a'),
(a') – [scalar] (b'),
(b') – (b),

(c) – (c1), (c) – (c2), (c) – (c3),
(d) – (d1), (d) – (d2), (d) – (d3),

(c1) – [scalar,half left,
looseness=0.5,edge label=3] (d1),
(c2) – [scalar,edge label=2] (d2),
(c3) – [scalar,half right,
looseness=0.5,edge label=1] (d3),

(a”) – [opacity=0.1] (b”),
(c”) – [opacity=0.1] (d”),
(c3) – [opacity=0.1] (c00),

\left(\frac{1.3.2.1)}{2!}\right)\int d^4y_1d^4y_2\ (i\Delta_F(x_1-x_2))

The diagrams (a) and (b) are examples of connected diagrams while
(c) is a example of a topologically disconnected diagram. Here
(dis)connection refers to (dis)connection with external vertices.
A diagram with no external vertices is called a vacuum bubble. The
diagram in (c) has a disconnected piece which is a vacuum bubble. The
denominator diagrams are all vacuum bubbles.

It is actually possible to separate the vacuum bubble components of the
diagrams in the numerator and cancel them against those in the
denominator. The proof goes as follows.

In an $n-$point function, at the $k^{th}$ order, the numerator has the

\[
Nr ~ = ~ \sum_{k=0}^{\infty}\frac{(-i)^k}{k!}\int d^4y_1\dots
d^4y_k\langle 0|T\{\Phi(x_1) \dots \Phi(x_n) :H_I(y_1): \dots:
\]

Group the contractions into two groups: (i) those that have $l$ of the
internal vertices with a contraction with at least one of the
external vertices and (ii) the remaining $(k - l)$ internal vertices
which have no segment connecting an external vertex. The above vev then
splits as,

\begin{eqnarray*}
\langle 0|T\{\Phi(x_1) \dots \Phi(x_n) :H_I(y_1): \dots:
H_I(y_k):\}|0\rangle ~ = ~ \mbox{\hspace{6.0cm}} & & \\
\langle 0|T\{\Phi(x_1) \dots \Phi(x_n) :H_I(y_1): \dots:
H_I(y_l):\}|0\rangle \times \mbox{\hspace{4.0cm}} & & \\
\langle 0|T\{:H_I(y_1): \dots: H_I(y_{k-l}):\}|0\rangle
\mbox{\hspace{2.0cm}}& &
\end{eqnarray*}

And this split can happen in $^kC_l$ ways. Hence,

\begin{eqnarray*}
Nr & = & \sum_{k=0}^{\infty} \frac{(-i)^k}{k!} \sum_{l=0}^k
\frac{k!}{l!(k-l)!}\int d^4y_1\dots d^4y_l\\
& & \mbox{\hspace{2.5cm}} \langle 0|T\{\Phi(x_1) \dots \Phi(x_n)
:H_I(y_1): \dots: H_I(y_l):\}|0\rangle \times \\
& & \mbox{\hspace{4.0cm}} \int d^4y_{l+1}\dots d^4y_{k} \langle
0|T\{:H_I(y_{l+1}): \dots: H_I(y_{k}):\}|0\rangle
\end{eqnarray*}

Putting $(-i)^k = (-i)^l(-i)^{(k-l)}$ and  interchanging the order of
summation we can write,

\begin{eqnarray*}
Nr & = & \left[\sum_{l=0}^{\infty} \frac{(-i)^l}{l!} \int d^4y_1\dots
d^4y_l \langle 0|T\{\Phi(x_1) \dots \Phi(x_n) :H_I(y_1): \dots:
H_I(y_l):\}|0\rangle \right] \\
& & \mbox{\hspace{1.0cm}} \times \left[ \sum_{k=l}^{\infty}
\frac{(-i)^{(k-l)}}{(k-l)!}
\int d^4y_{l+1}\dots d^4y_{k} \langle 0|T\{:H_I(y_{l+1}): \dots:
\end{eqnarray*}

$k \to k+l$ in the second square bracket, shows it as $\langle 0|T\{exp
-{i\int_{-\infty}^{\infty}dt H(t)}\}|0\rangle$ which is just the

The first factor is called the connected Green's function and
denoted as $G^c_n$. It consists of only the diagrams connected to
external vertices. The diagrams may be topologically disconnected. From now on we focus on the connected Green's functions and drop
the denominator.

Having gotten Feynman diagrams and Feynman integrals, we further
simplify the expression for the $S-$matrix elements.

To go from Green's functions to $S-$matrix elements (a) we operate on
the Green's function by the equation of motion differential operator for
each external vertex; (b) insert the wave function factors for each
external vertex; (c) divide by $\sqrt{Z}$ for each external vertex and
(d) integrate over the external vertices.

We have only connected diagrams. Among these are topologically
disconnected diagrams too. There is a special subclass of topologically
disconnected diagrams in which two external vertices are Wick
contracted. These represent an incoming particle which goes out without any scattering. We can separate such un-scattered processes
from the $S-$matrix (which we will do little later) and focus on
processes wherein every incoming particle necessarily scatters i.e. every external vertex is necessarily connected to an internal vertex.

Consider an external vertex $x$ connected with an internal vertex $y$.
Associated with the external vertex is a wavefunction factor
$\Case{e^{ik\cdot x}\ (or\ e^{-ik'x'})} {\sqrt{2\omega_{\vec{k}}
(2\pi)^3}}$, integration over $x$ (or $x'$), Feynman propagator
$\Delta_F(x-y)$ and $(\Box_x - m^2)$ acting on the propagator. Since the
propagator is a Green's function, $(\Box_x - m^2)\Delta_F(x-y) = -
\delta^4(x-y)$. Integration over $x$ then removes the delta function and
$e^{ik\cdot x} \to e^{ik\cdot y}$. For the external vertex of an
out-going particle, $e^{-ik'\cdot x'} \to e^{-ik'\cdot y}$. Thus the
integration over external vertices is trivially carried out and the
propagators involving an external vertex is removed - an “external leg
is said to be amputated”.

Now use the Fourier representation of the remaining propagators,
$\Delta_F(y_i - y_j) = \int \Case{d^4l}{(2\pi)^4}
\Case{e^{il\cdot(y_i-y_j)}}{l^2 + m^2 -i\epsilon}$. Each internal vertex
$y_i$ is connected to possibly several other internal vertices as well
as possibly several external vertices. For internal vertex, the Fourier
transform supplies a factor of $e^{il\cdot y_i}$ and each external
vertex provides a factor of $e^{k\cdot y_i}$. All of these combine and
the $\int d^4y_i$ then gives a $(2\pi)^4\delta^4(\sum k -\sum k' + \sum
l)$. Thus, the integration over internal vertices result in a momentum
conserving delta function.  The space-time integrations are done but we
are left with integration over the Fourier momenta which are associated
with internal lines. The delta functions trivialize many momentum
integrations leaving an overall delta function involving only the
external momenta and enforcing the momentum conservation. We also left
with some unconstrained loop momenta.

To be explicit, let $n_E, n_I$ denote the number of external and
internal lines and let $n_v$ denote the number internal vertices. We
have then $n_E + n_I$ number of momenta and $n_v$ number of conservation
equations. Thus $n_E + n_I - n_v + 1$ is the number of undetermined
momenta or the loop momenta. The $+1$ in the counting signifies the loss
of one conservation equation due to the left over delta function
enforcing conservation of external momenta.

There are no dependence on space-time points, no space-time integrations
left and the scattering matrix element is given by integration over a
bunch of loop momenta and a product of momentum space propagators
forming the integrand and of course the various factors of $i, \pi$ and
numerical combinatorial factors.

\[
\boxed{	S_{fi} \sim \delta^4(\sum k - \sum {k'}) \prod_j\int d^4l_j
\prod_{n_I}\left( \frac{1}{p^2_i(k, l) + m^2 - i\epsilon}\right) }
\]

We will elaborate the numerical factors in explicit examples. The
structure of the $S-$matrix elements (non-trivial scattering) should be
clear from the above discussion.

To separate out the trivial scattering it is customary to introduce the
the so-called $T-$matrix as,

\[
\boxed{	S := \mathbb{1} + iT ~ ~ , ~ ~\langle k'_{j'}|T|k_j\rangle :=
(2\pi)^4 \delta^4(\Sigma_{j'} k'_{j'} - \Sigma_j k_j)\ \mathcal{M}(k_j
\to k'_{j'}) }
\]

Here the $k, k'$ denote the on-shell momenta of the incoming and
outgoing particles. The delta function enforcing momentum conservation is
a consequence of translation invariance and $\mathcal(M)$ is called the
invariant matrix element. This is what is computed in practice.
The $S_{fi}$ given above is really the $iT_{fi}$.

In practice, computation of the $T-$matrix elements is only part of what
is needed to compare with experiments which measure cross-sections. To compute cross-sections, recall that the $S-$matrix
elements are the transition probability amplitudes whose non-trivial
contributions are identified as $\langle f|i\rangle = (2\pi)^4
\delta^4(k'_{total} - k_{total})(i\mathcal{M}_{i\to f})$. The
cross-section is the ratio of the number of outgoing particles per
second to the incident flux. The numerator is proportional to the
transition probability rate while the denominator is determined by the
initial state.

We have given the $S-$matrix elements in the plane wave basis which is
strictly incorrect. One should use wave packets for representing the
asymptotic states. Alternatively, a commonly used practice is to put the
system in a finite space-time box and take the limit of infinite box at
the end. This is simpler to implement in practice and suffices for most
purposes. We will use this [12] and refer to
[13] for the wave packet treatment.

§.§ Differential Cross-section for $2 \to n$ process

Imagine the scattering experiment to be enclosed in a large spatial box
of volume $V = L^3$, with periodic boundary conditions imposed on the
mode functions. Let the duration of the experiment be a large time
interval $T$. The probability of transition from an initial state
$|i\rangle$ to a final state $|f\rangle$ is given by,

\[
Prob_{i\to f} = \frac{|\langle f|i\rangle|^2}{\langle i|i\rangle \langle
f|f\rangle} = \frac{\left[(2\pi)^4\delta^4(\Sigma k'_{j'} - \Sigma
f}|^2}{\langle i|i\rangle \langle f|f\rangle}
\]

The square of the momentum conservation delta function is written with
one factor as $\delta(0)$ which is to be understood as:
$\boxed{(2\pi)^4\delta^4(0) = \int d^4x e^{i0\cdot x} = \int d^4x :=
VT}$.  The initial and final states are normalized using $\boxed{\langle
k|k\rangle := \lim_{k'\to k}\delta^3(\vec{k}' - \vec{k}) = \delta^3(0)
:= \Case{V}{(2\pi)^3}}$. Typically, the initial state consists of two particles. So let us specialize to this case of $2 \to n$
processes.  Then, $\langle i|i\rangle\langle f|f\rangle =
[\Case{V}{(2\pi)^3}]^{(n+2)}$. Therefore,

\[
\frac{Prob_{i\to f}}{T}  ~ = ~
\frac{\left[(2\pi)^4\delta^4(\Sigma k'_{j'} - \Sigma
k_j)\right]|i\mathcal{M}_{i\to f}|^2 V}{[V/(2\pi)^3]^{n+2}}
\]

Real detectors have finite aperture and hence the detected particle's
momentum is anywhere within a small window around the central value. The
corresponding probabilities are thus to be added. An estimate for such a
window follows from the box normalization we have taken. The momenta are
given by $\vec{k}_j = \Case{2\pi}{L}\vec{n}_j$. Summing over the momenta
within a window is same as summing over the integers within a window and
for large volume, $\Sigma_{\vec{n}_j} ~\approx~ \Case{V}{(2\pi)^3}\int
d^3k$, for each final state particle. This leads to the total
probability per unit time for the transition,

\begin{eqnarray*}
\frac{Prob_{2\to n}}{T}d\Gamma_n & = &
\frac{\left[(2\pi)^4\delta^4(\Sigma k'_{j'} - \Sigma
k_j)\right]|i\mathcal{M}_{2\to n}|^2 V}{[V/(2\pi)^3]^{n+2}} \prod_{j =
1}^n \left[\frac{d^3k_j}{(2\pi)^3}V\right]  \\
& = & \left\{\frac{\left[(2\pi)^4\delta^4(\Sigma k'_{j'} - k_1 -
k_2)\right]\left|i\mathcal{M}_{2\to n}\sqrt{\prod_{j=1}^{n+2}\left[
2\omega_{\vec{k}_j}(2\pi)^3\right]} \right|^2}{V(2\omega_{\vec{k}_1})
(2\omega_{\vec{k}_2})}\right\} \prod_{j = 1}^n \left[\frac{d^3k_j}
\end{eqnarray*}

In going from the first to the second line, we have divided and
multiplied by the product of $(2\omega_{\vec{k}})$ for all the $(n+2)$
particles. The last product is the $d\Gamma_n$ which is the Lorentz
invariant phase space volume.  The square root factor will get absorbed
in the invariant matrix element and will get rid of similar factors
coming from the wave functions of the incoming and outgoing particles.

We also need the incident flux. For the initial state of two articles,
consider the laboratory frame where the particle `2' is at rest.
The particle `1' has a speed $|\vec{k}_1|/E_1$ and gives the number
density of one particle per unit volume, $1/V$. Hence, the incident
number flux is $|\vec{k}_1|/(E_1 V)$. Dividing by the flux, the exclusive, differential cross-section for $2 \to n$ process is given

\begin{eqnarray*}
d\sigma_{lab} & = & \frac{(2\pi)^4\delta^4(k_1 + k_2 -
\Sigma_jk_j)|i\tilde{\mathcal{M}}_{2\to
n}|^2}{4m_2|(\vec{k}_1)_{lab}|} d\Gamma_n  ~ \mbox{\hspace{1.0cm}
where,} \\
\tilde{\mathcal{M}}_{2\to n} & := & \mathcal{M}_{2\to n}
\sqrt{\prod_{j=1}^{n+2}\left[ 2\omega_{\vec{k}_j}(2\pi)^3\right]} ~
\mbox{\hspace{3.5cm} and,}\\
d\Gamma_n & := & \prod_{j = 1}^n \frac{d^3k_j} {2\omega_{\vec{k}}
\end{eqnarray*}

We have used $\omega_{\vec{k}_2} = m_2$ and $\omega_{\vec{k}_1} = E_1$.
Notice that all factors of the volume and the duration $T$ have
disappeared. The explicit square root factors will also cancel out.

Many basic calculations involve scattering processes with two particles
going into two particles. We have already used the initial state of two
particles to obtain the incident flux. We will now also specialize to
$n=2$ for out going particles and write the phase space integrals more

§.§.§ The Special case of $2 \to 2$ processes

Let the initial momenta be denoted by $k_1, k_2$ and the final momenta
be denoted by $k_1', k_2'$. Lorentz invariance implies that the
scattering amplitude will have Lorentz indices (tensorial or spinorial)
carried by the wavefunctions and the invariant amplitude will be a
function of Lorentz invariants. We have three independent momenta thanks
to the overall conservation due to translation invariance and we can
form 6 Lorentz scalars of the form $p_i\cdot p_j$. Of these three are
masses and the remaining three are conveniently defined as “center of
mass energy” and two types of “squared momentum transferred”. These
are known as the Mandelstam variables and are defined as ($k_1 +
k_2 = k_1' + k_2'$),

\begin{eqnarray}
s & := & - (k_1+k_2)^2 = - (k_1'+ k_2')^2 ~ ~
\mbox{\hspace{1.0cm} (Centre of Mass Energy)} \\
t & := & - (k'_1-k_1)^2 = - (k_2'- k_2)^2 ~ ~
\mbox{\hspace{1.0cm} (Squared Momentum transfer)} \\
u & := & - (k'_2-k_1)^2 = - (k_2- k'_1)^2 ~ ~
\mbox{\hspace{1.0cm} (Squared Momentum transfer)}
\end{eqnarray}

The definitions imply the Mandelstam identity: $\boxed{s + u + t =
m^2_{k_1} + m^2_{k_2} + m^2_{k'_1} + m^2_{k'_2} } $.

Laboratory and Centre of Mass Frames:

The lab frame, is defined by regarding particle 2, say, at rest while
particle 1 is incident on it.  Thus,

\begin{equation*}
\boxed{ k^{lab}_1 = (\sqrt{\vec{k}_1^2 + m_1^2}, \ \vec{k}_1^{lab})\ , \
k^{lab}_2 = (m_2, \ \vec{0})}
\end{equation*}

The lab frame is convenient for expressing the incident flux.

The center of mass frame is defined by $\vec{k}_{total} := \vec{k}_1 +
\vec{k}_2 = 0 = \vec{k}'_1 + \vec{k}'_2$. Thus,

\begin{equation*}
\boxed{k^{cm}_1 = (\sqrt{\vec{k}^2_{cm} + m_1^2}, \ \vec{k}_{cm})\ , \
k^{cm}_2 = (\sqrt{\vec{k}^2_{cm} + m_2^2}, \ -\vec{k}_{cm})}
\end{equation*}

This frame is more convenient for defining scattering angle. The common
momentum direction singles out say, the z-axis. Orienting the frame
accordingly, we take

\begin{eqnarray}
k_1 = (E_1, k_{cm}\hat{z}) & , & k_2 = (E_2, -k_{cm}\hat{z}) ~ ~
, \nonumber \\
k'_1 = (E'_1, \vec{k}') & , & k'_2 = (E'_2, -\vec{k}') ~ ~ , ~ ~
\hat{k'}\cdot\hat{z} =: cos(\Theta_{cm}).
\end{eqnarray}

The Mandelstam invariant $s$ can be used to relate the lab frame
momentum $|k_1|^{lab}$ and the center of mass momentum $|k|_{cm}$. The
invariant $s := - (k_1 + k_2)^2 = - (-m_1^2 - m^2_2 - 2E_1E_2 +
2\vec{k}_1\cdot\vec{k}_2)$ has two equivalent expressions:

\begin{eqnarray*}
s|_{lab} & = & m^2_1 + m^2_2 + 2m_2\sqrt{m^2_1 + \vec{k}_1^2} \\
s|_{cm} & = & m_1^2 + m_2^2 + 2\sqrt{m_1^2 + \vec{k}^2_{cm}}
\sqrt{m_2^2 + \vec{k}_{cm}^2} + 2\vec{k}_{cm}^2 \ .
\end{eqnarray*}

These can be solved for the momenta and give manifestly Lorentz
invariant expressions,

\begin{eqnarray}
|k_1|_{lab} & = & \frac{1}{2m_2}\sqrt{s^2 - 2s(m_1^2 + m_2^2) +
(m_1^2 - m_2^2)^2} \\
|k|_{cm} & = & \frac{1}{2\sqrt{s}}\sqrt{s^2 - 2s(m_1^2 + m_2^2) +
(m_1^2 - m_2^2)^2}
\end{eqnarray}

Phase space volume in center of mass frame: We have the
Lorentz invariant definition,

\[
d\Gamma_2 = (2\pi)^4\delta^4(k'_1+k'_2 - k_1-k_2)
\Case{d^3k'_1}{2\omega_{k'_1}(2\pi)^3}
\Case{d^3k'_2}{2\omega_{k'_2}(2\pi)^3} .
\]

In the CM frame we simplify it using: $\delta^4 \to \delta(E'_1 + E'_2 -
\sqrt{s})\delta^3(k'_1+k'_2)$ since $\vec{k}_1 + \vec{k}_2 = 0, s =
(E_1+E_2)^2$. We can remove the momentum delta function by integrating
over $\vec{k}'_2$. Denoting $d^3k'_1 := d^3k'_{cm} = dk'_{cm}(k')^2_{cm}
d\Theta_{cm}sin^2(\Theta_{cm})d\phi$, we write,

\begin{eqnarray*}
\int d^3k'_2d\Gamma_2 & = & \frac{1}{(2\pi)^2}\delta(E'_1+E'_2 -
\sqrt{s}) \frac{1}{4E'_1E'_2} dk'_{cm}(k')^2_{cm} d\Theta_{cm}
sin^2(\Theta_{cm}) d\phi
\end{eqnarray*}

Since $E'_1+E'_2 = \sqrt{(m')^2_1 + (k')^2_{cm}} + \sqrt{(m')^2_2 +
(k')^2_{cm}}$ is a function of $k'_{cm}$, we simplify the delta function
using $\delta(f(x)) = \sum_i\Case{\delta(x - x_i)}{f'(x_i)}, f(x_i) =
0$.  We have

\begin{eqnarray*}
f'(k'_{cm}) & = & \frac{k'_{cm}}{E'_1} + \frac{k'_{cm}}{E'_2} ~ = ~
k'_{cm}\frac{E'_1 + E'_2}{E'_1E'_2} = \frac{k'_{cm}\sqrt{s}}{E'_1E'_2},
\\
f(k'_{cm}) = 0 & \Rightarrow & k'_{cm} = \frac{\sqrt{s^2 - 2s( (m')^2_1
+ (m')^2_2) + ( (m')_1^2 - (m')^2_2)^2}}{2\sqrt{s}}
\end{eqnarray*}

Doing the $dk'_{cm}$ integration using the delta function gives,

\begin{equation}
\int d(k')^2_{cm}\delta(E'_1 + E'_2 - \sqrt{s}) =
\frac{(k')^2_{cm}E'_1E'_2}{k'_{cm}\sqrt{s}} ~ \mbox{which
gives}~ d\Gamma_2 =
\frac{1}{16\pi^2}d^2\Omega_{cm}\frac{k'_{cm}}{\sqrt{s}} \ .
\end{equation}

Thus, for the special case of a process with 2 particles going to 2
particles, one gets the differential cross-section as,

\begin{eqnarray}\label{DiffCrossSection}
\frac{d\sigma}{d\Omega_{cm}} & = &
\frac{1}{64\pi^2}\frac{1}{s}\frac{k'_{cm}}{k_{cm}}|\mathcal{M}|^2
\mbox{\hspace{0.5cm}where we have used\hspace{0.3cm}} 2m_2|k_1|^{lab} =
2\sqrt{s}|k|_{cm}\ , ~ ~ \mbox{and}~ ~\\
\frac{k'_{cm}}{k_{cm}} & = & \sqrt{\frac{s^2 - 2s( (m')^2_1 + (m')^2_2)
+ ((m')_1^2 - (m')^2_2)^2}{s^2 - 2s( m^2_1 + m^2_2) + (m_1^2 -
m^2_2)^2}} \ .
\end{eqnarray}

The invariant amplitude has dependence on the scattering angle
$\Theta_{cm}$ and the total cross-section is obtained by integrating
over the center of mass solid angle.

§ DIAGRAMMATIC RECIPE FOR S-MATRIX ELEMENTS

We now specify explicit interacting fields by giving an interaction
Lagrangian and state the Feynman rules to complete the diagrammatic
recipe for the $T-matrix$ elements. We will state the rules for the
$\Phi^4$ theory, the Yukawa theory and the Quantum electrodynamics
(QED). In the next section we will compute specific processes.  There is
a good deal of conventions of normalization etc and they have to be kept
track of carefully. The free (quadratic) part of the Lagrangian density
sets the normalization of the propagators while the interaction terms
(beyond quadratic in fields) contribute to the numerical factors. Here
are the terms in the Lagrangian densities.

\begin{eqnarray}
\mbox{Free Scalar~} & : & -\frac{1}{2} \partial^{\mu}{\Phi}
\partial_{\mu}{\Phi} -\frac{1}{2}m^2\Phi^2(x)  \ ; \\
\mbox{Free Spinor~} & : & -i\bar{\Psi}\dsl{\partial}\Psi + m
\bar{\Psi}\Psi \ ; \\
\mbox{Massless Vector~} & : & -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} ~ ~ , ~ ~
F_{\mu\nu} := \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}  \ ; \\
\mbox{Scalar self coupling~} & : & - \frac{g}{3!}\Phi^3(x) -
\frac{\lambda}{4!}\Phi(x)^4  \ ; \\
\mbox{Yukawa coupling~} & : & - g\Phi(x)\bar{\Psi}(x)\Psi(x) \ ; \\
\mbox{QED coupling~} & : & -ie A_{\mu}(x) \bar{\Psi}(x) \gamma^{\mu}
\Psi(x) \ . \\
\mbox{In the interaction~} & & \mbox{~ {\em Hamiltonian}, all coupling
terms will change signs.}
\end{eqnarray}

The propagators are the Feynman Green's functions for the free equations of
motion. Consider the massless vector field as this has a new feature.

The free Lagrangian (Maxwell) can be expressed as $\Case{1}{2}A_{\mu}(
\eta^{\mu\nu}\Box - \partial^{\mu}\partial^{\nu})A_{\nu} + $ divergence
terms. The equation of motion is $(\eta^{\mu\nu}\Box -
\partial^{\mu}\partial^{\nu})A_{\nu}(x) = 0$. Equivalently, in Fourier
space equation takes the form $-(\eta^{\mu\nu}k^2 -
k^{\mu}k^{\nu})\tilde{A}_{\nu}(k) = 0$. However, the differential
operator is not invertible and hence does not admit a Green's function!
The non-invertibility follows because every $A_{\nu}$ of the form
$\partial_{\nu}f(x)$, solves the equation. That is, $\partial_{\nu}f$ is
a non-trivial eigenvector  of the differential operator, with zero
eigenvalue. For perturbation theory though we need a propagator for
which extra terms are added to the action to break its gauge invariance
- invariance under $\delta A_{\mu}(x) = \partial_{\mu}\Lambda(x)$. This
can be done in several ways and each choice corresponds to a gauge.

A common and convenient choice is to add $-\Case{1}{2}(\partial_{\mu}
A^{\mu})^2$ term to the action. Up to a divergence term, this is just
$+\Case{1}{2}A_{\mu}\partial^{\mu}\partial^{\nu}A_{\nu}$ and precisely
cancels the term in the equation of motion operator, making it
$\eta^{\mu\nu}\Box$ which is invertible. The added term is
manifestly Lorentz invariant and the corresponding gauge is called the
Lorentz gauge. For our purposes, this gauge will suffice.  The
propagator is an inverse of the differential operator since
$(\eta^{\mu\nu}\Box_x)(D_F)_{\nu\lambda}(x-y) = \delta^{\mu}_{~\lambda}
\delta^4(x-y)$.

The QED coupling arises from the minimal substitution rule: $\vec{P} \to
\vec{P} + e\vec{A} \leftrightarrow  -i\partial_{\mu} \to -i\partial_{\mu}
+ eA_{\mu} \leftrightarrow \partial_{\mu} \to \partial_{\mu} +
ieA_{\mu}$. Substitution in the free Dirac action gives a $-ieA_{\mu}$ as
the QED coupling.

Let us gather the various factors for the scalar field in the $T-$matrix

* Each in/out particle gives a wave function factor of
$[2\omega_{\vec{k}}(2\pi)^3]^{-1/2}e^{\pm i k\cdot x}$,
an $\int d^4x$ and $(\Box_x - m^2)$ acting on the
$n-$point function. For a spinor field, the amputation
operator changes to $(-i\dsl{\partial} + m)$ and we have
additionally the $u, v, \bar{u}, \bar{v}$ spinors. For a
vector field, the amputation operator changes and we
have the polarization vectors in addition. Other factors
remain the same.

* Each order in $H_I$, gives $(-i)\int d^4y$  from the
$T-$ordered exponential and a $(m!)^{-1}$ for the
$m^{th}$ order, from the exponential;

* Each Wick contraction gives $(i\Delta_F(z-z')) := \int
\Case{d^4l}{(2\pi)^4} \Case{ie^{il\cdot(z-z')}} {l^2 +
m^2 - i\epsilon}$ ;

* Each amputation, action of $(\Box -m^2)$ gives a
$(-\delta^4(x-y))$ while each integration over $x$ or
$y$ gives $(2\pi)^4$ times a momentum conserving
$\delta^4$. The momentum integrations do not produce any

* Let $E, I, V$ denote the number of external lines (=
number of external vertices), number of internal lines
and number of internal vertices (= order of $H_I$)
respectively.  Then, the factors of $[2\omega_{\vec{k}}
(2\pi)^3] ^{-1/2}$, precisely cancel the explicit factor
we found in the amplitude $\tilde{M}_{2\to n}$. This is
due to the normalization choices which cancel out in
$\Case{|\langle f|i\rangle|^2}{\langle i|i\rangle\langle
f|f\rangle}$ and can now be dropped from both
$\tilde{M}$ and from the $T-$matrix element.

Factors of $i$: $(-i)^V(i)^I = (-1)^Vi^{V+I}$;

Factors of $2\pi$: $(2\pi)^{-4I + 4E + 4V -4}$. The last
$-4$ is because it has been taken out in the definition
of the $\mathcal{M}$ due to the overall momentum

Vertex factors, including the $i$ in the QED vertex are
to be taken case-by-case. There is the $(m!)^{-1}$ and a
combinatorial factor that will come for each diagram.
These too are taken case-by-case.

With these, we now state the Feynman rules for Feynman diagrams.

External Line:

(a) ;
[right=1.5 of a, blob] (b);
(a) – [scalar,momentum'=\(p\)] (b),

(b) ;
[left=1.5 of b, blob] (a);
(a) – [scalar,momentum'=\(p\)] (b),

(a) ;
[right=1.5 of a, blob] (b);
(a) – [fermion,momentum'=\(p\)] (b),

(b) ;
[left=1.5 of b, blob] (a);
(a) – [fermion,momentum'=\(p\)] (b),

(a) ;
[right=1.5 of a, blob] (b);
(a) – [anti fermion,momentum'=\(p\)] (b),

(b) ;
[left=1.5 of b, blob] (a);
(a) – [anti fermion,momentum'=\(p\)] (b),

(a) ;
[right=1.5 of a, blob] (b);
(a) – [photon, edge
label=\(\gamma\), momentum'=\(p\)] (b),

(b) ;
[left=1.5 of b, blob] (a);
(a) – [photon, edge
label=\(\gamma\), momentum'=\(p\)] (b),

Internal Line:

(a) ;
[right=2.0 of a] (b);
(a) – [scalar] (b),

$\frac{-i}{k^2 + m^2 - i\epsilon}$

(a) ;
[right=2.0 of a] (b);
(a) – [fermion] (b),

$\frac{(-i)(-\dsl{k} +m)}{k^2 + m^2 - i\epsilon}$

(a)\(\mu\) ;
[right=2.0 of a] (b)\(\nu\);
(a) – [photon] (b),

$\frac{-i\eta^{\mu\nu}}{k^2 - i\epsilon}$

(v) ;
[left=of v] (a);
[above right=of v] (b);
[below right=of v] (c);
(a) – [scalar] (v),
(b) – [scalar] (v),
(c) – [scalar] (v),

(v) ;
[above left=of v] (a);
[above right=of v] (b);
[below right=of v] (c);
[below left=of v] (d);
(a) – [scalar] (v),
(b) – [scalar] (v),
(c) – [scalar] (v),
(d) – [scalar] (v),

(v) ;
[left=of v] (a);
[above right=of v] (b);
[below right=of v] (c);
(v) – [scalar] (a),
(v) – [fermion] (b),
(c) – [fermion] (v),

[left=of v] (a)\(\mu\);
[above right=of v] (b);
[below right=of v] (c);
(v) – [photon] (a),
(v) – [fermion] (b),
(c) – [fermion] (v),

§ ELEMENTARY PROCESSES IN YUKAWA AND QED: NR

We begin with scattering of a fermion off another fermion, interacting
via the Yukawa coupling and compute the $T-$matrix element to the
leading order. We will take the non-relativistic limit and identify an
equivalent potential. We will consider anti-fermion scattering and
fermion-anti-fermion scattering as well. We will then compare the qed
coupling and briefly the gravitational coupling. This will lead to
appreciate the dependence the attractive/repulsive nature of the
interaction on spin of the exchanged particle.

Consider a general process depicted below together with its `expansion'.

\begin{equation}
\parbox[c]{2.85cm}{
\begin{tikzpicture}
\begin{feynman}
\vertex [blob] (v){};
\vertex [above left=of v] (a) {\(p\)};
\vertex [above right=of v] (b) {\(p'\)};
\vertex [below right=of v] (c) {\(k'\)};
\vertex [below left=of v] (d) {\(k\)};;

\diagram* { (a) -- [fermion] (v),
(v) -- [fermion] (b) ,
(v) -- [fermion] (c) ,
(d) -- [fermion] (v),
\end{feynman}
\end{tikzpicture}
%		\feynmandiagram[medium,horizontal=i1 to f2] { {i1
%	[particle=k], i2 [particle=p]} -- [fermion] c [blob] --
%	[fermion] {f1 [particle=p'], f2 [particle=k']}, };
~  = ~
\parbox[c]{2.85cm}{
\begin{tikzpicture}
\begin{feynman}
\vertex (v);
\vertex [above left=0.5of v] (a) {\(p\)};
\vertex [above right=0.5of v] (b) {\(p'\)};
\vertex [below right=0.5of v] (c) {\(k'\)};
\vertex [below left=0.5of v] (d) {\(k\)};;

\diagram* { (a) -- [fermion] (b),
(d) -- [fermion] (c),
\end{feynman}
\end{tikzpicture}
%		\feynmandiagram[small,horizontal=a to b] { a
%		[particle=k] -- [fermion] b [particle=k'] --
%	[opacity=0.0] c [particle=p'] -- [anti fermion] d [particle=p]
%-- [opacity=0.0] a, };
~  +  ~
\parbox[c]{2.85cm}{
\begin{tikzpicture}
\begin{feynman}
\vertex (v1);
\vertex [below=0.75of v1] (v2);
\vertex [above left=1.0of v1] (a) {\(p\)};
\vertex [above right=1.0of v1] (b) {\(p'\)};
\vertex [below right=1.0of v2] (c) {\(k'\)};
\vertex [below left=1.0of v2] (d) {\(k\)};;

\diagram* {
(a) -- [fermion] (v1),
(v1) -- [fermion] (b),
(v1) -- [scalar,momentum=\(p'-p\)] (v2),
(v2) -- [fermion] (c),
(d) -- [fermion] (v2),
\end{feynman}
\end{tikzpicture}
%		\feynmandiagram[small,horizontal=i1 to f2] { i1
%			[particle=p] -- [fermion] c1 -- [fermion]
%			f2[particle=p'] , c1 -- [scalar,
%			momentum=\(p'-p\)] c2, i2 [particle=k'] -- [anti
%			fermion] c2 -- [anti fermion] f3[particle=k] ,
%		};
~ ~ ~ ~ +
\parbox[c]{2.85cm}{
\begin{tikzpicture}
\begin{feynman}
\vertex (v1);
\vertex [below=1.0of v1] (v2);
\vertex [above left=0.8of v1] (a) {\(p\)};
\vertex [above right=0.8of v1] (b) {\(p'\)};
\vertex [below right=0.8of v2] (c) {\(k'\)};
\vertex [below left=0.8of v2] (d) {\(k\)};;

\diagram* {
(a) -- [fermion] (v1),
(v1) -- [fermion] (c),
(v1) -- [scalar,momentum'=\(k'-p\)] (v2),
(v2) -- [fermion] (b),
(d) -- [fermion] (v2),
\end{feynman}
\end{tikzpicture}
%		\feynmandiagram[small,horizontal=i1 to f3] { i1
%			[particle=p] -- [fermion] c1 -- [fermion]
%			f3[particle=k'] , c1 -- [scalar,
%			momentum=\(k'-p\)] c2, i2 [particle=p'] -- [anti
%			fermion] c2 -- [anti fermion] f2[particle=k] ,
%		};
}   ~ ~ +  ~ \dots
\end{equation}

The first term ($o(g^0)$) denotes `no scattering'. There are two
contributions at $o(g^2)$ corresponding to exchange of two out-going
fermions. The overall momentum conservation delta function is the same,
enforcing $p + k = p' + k'$. The Feynman rules give the expression as,

\begin{eqnarray}
i\mathcal{M} & = & (ig)^2\left\{ \bar{u}(p')u(p) \frac{-i}{(p'-p)^2 +
m_{\varphi}^2 - i\epsilon} \bar{u}(k')u(k) \right. \nonumber \\
& & \mbox{\hspace{2.0cm}} \left. -  \bar{u}(k')u(p) \frac{-i}{(k'-p)^2 +
m_{\varphi}^2 - i\epsilon} \bar{u}(p')u(k) \right\}
\end{eqnarray}

The relative minus sign between the two terms is due to the
T-ordering definition for the fermions. The overall sign of the amplitude
is determined by the convention adopted for ordering the initial/final
state labels for the fermions. See [13, 12].
Notice how the fermion arrows are followed.

Note: If we had anti-fermion scattering, then all the
fermion arrows will be reversed and their momenta will be denoted as
minus the previous momenta[Thus, we may adopt a convention that
the diagram displays the fermion arrow and the momentum is also
in the same direction. For a fermion the momentum is $p$ and for
anti-fermion it is $-p$.]. Apart from the reversal of fermion arrows,
the $u(p), \bar{u}(p')$ spinors go to $v(p'), \bar{v}(p)$ spinors.

If it is a fermion-anti-fermion scattering, then the second exchange
diagram will be absent.

A specific scattering arrangement may permit the initial and the final
state fermions to be distinguishable. Then only one of the two
scattering diagrams will contribute.

Note: The reduction formula for fermions gave a factor of
$2m$ for the wavefunctions since we had normalized the spinors as
$\bar{u}(p,\sigma)u(p,\sigma') = \delta_{\sigma,\sigma'}$. With the
Feynman rules we have adopted in the table the wavefunctions have only
the spinors. This is equivalent to using the normalization:
$\bar{u}(p,\sigma)u(p,\sigma') = 2m\delta_{\sigma,\sigma'} = -
\bar{v}(p,\sigma)v(p,\sigma')$, $m$ is of course the fermion mass.

With this noted, the amplitude becomes,

\begin{eqnarray}
i\mathcal{M} & = & (ig)^2 (-i) (2m)^2 \left\{
\frac{\delta_{\sigma_{p},\sigma_{p'}}
\delta_{\sigma_{k},\sigma_{k'}}} {(p'-p)^2 + m_{\varphi}^2 -
-~\frac{\delta_{\sigma_{p},\sigma_{k'}} \delta_{\sigma_{p'},\sigma_{k}}}
{(k'-p)^2 + m_{\varphi}^2 - i\epsilon} \right\}
\end{eqnarray}

Noting that the scalar field momentum is just the momentum transfer in
both the diagrams, we denote it by $q$. And $q^2 = -(q^0)^2 +
|\vec{q}|^2 = |\vec{q}|^2$ in both cases. We thus write the amplitude
more conveniently as,

\begin{eqnarray*}
i\mathcal{M} & = & \frac{+ig^2\ 4m^2}{|\vec{q}|^2 + m_{\varphi}^2 -
i\epsilon} \left\{ \delta_{\sigma_{p},\sigma_{p'}}
\delta_{\sigma_{k},\sigma_{k'}} - \delta_{\sigma_{p},\sigma_{k'}}
\delta_{\sigma_{p'},\sigma_{k}} \right\}.
\end{eqnarray*}

As noted above, only one of the two terms will contribute if the
fermions are distinguishable.

In the non-relativistic scattering theory, we have $\langle
\vec{k}'|S|\vec{k}\rangle := \langle \vec{k}'|\vec{k}\rangle - 2\pi i
\delta(E_{k'} - E_k)\langle \vec{k}'|T|\vec{k}\rangle$ with,

\[
\langle \vec{k}'|T|\vec{k}\rangle =\langle
\vec{k}'|H'|\vec{k}\rangle =\langle
\vec{k}'|V(\vec{q})|\vec{k}\rangle ~ = ~ V(|\vec{q} = \vec{k}' -
\vec{k}|)\ .
\]

We can read-off the potential as,

\begin{equation}
V(|q|) =  - \frac{4m^2 g^2}{|\vec{q}|^2 +
m^2_{\varphi}}\left(\delta \delta - \delta \delta\right)  := -
\frac{g^2_{eff}}{\vec{q}^2 + m_{\varphi}^2}.
\end{equation}

The inverse Fourier transform gives,

\begin{eqnarray}
V(\vec{x}) & = & -g^2_{eff}\int \frac{d^3q}{(2\pi)^3}
\frac{e^{i\vec{q}\cdot \vec{x}}}{\vec{q}^2 + m^2_{\varphi}} ~ =
~ - \frac{g^2_{eff}}{(2\pi)^2}\int_0^{\infty} dq \frac{q^2}{q^2 +
\nonumber \\
& = & - \frac{g^2_{eff}}{4\pi^2(ir)}\int_{-\infty}^{\infty}dq
\frac{qe^{iqr}}{q^2 + m^2_{\varphi}} ~ \mbox{\hspace{2.0cm}
contour integrate to pick up $q = im_{\varphi}$,} \nonumber \\
\therefore V(r) & = & - \frac{g^2_{eff}}{4\pi}\
\frac{e^{-m_{\varphi}r}}{r} ~ \mbox{\hspace{4.0cm} (Yukawa
\end{eqnarray}

This is a simple illustration how an effective potential can be inferred
from the underlying relativistically specified interaction. Several
remarks are in order.

Remark: The most important point is the sign of the
potential which renders it attractive. But is the sign

We have already noted that the overall sign of the amplitude is
determined by the convention adopted for ordering of the fermion labels
in the in/out states while the relative sign is due to the Pauli
principle. When the fermions are distinguishable, either of the two
diagrams may be chosen, there is no preference. But the diagrams
contribute opposite signs!

The convention arises as follows [13]. We have taken
$|p,k\rangle \sim b^{\dagger}_pb^{\dagger}_k|0\rangle$ (sequence of
creation operators follows label order in the in-state) and
$\langle p',k'| \sim \langle 0|b_{k'} b_{p'}$ which is opposite to that
of the in-state, but consistent with the Hermitian conjugation. Hence,

\begin{eqnarray*}
\langle p',k'|P,k\rangle & \sim & \langle 0 |b_{k'} b_{p'} b^{\dagger}_p
b^{\dagger}_k |0\rangle ~ = ~ \langle 0|b_{k'}b^{\dagger}_{k}|0\rangle
\delta^3(p'-p) - \langle 0|b_{k'} b^{\dagger}_{p}b_{p}
b^{\dagger}_{k}|0\rangle \\
& \approx & \delta^3(k'-k)\delta^3(p'-p) - \delta^3(p'-k)\delta^3(k'-p)
\end{eqnarray*}

Actually there are $\bar{\Psi}\Psi$ fields in between, but being a
bilinear it does not change the relative sign. This explains the overall

In a non-relativistic comparison, $|q| \ll m_{\varphi}$, the small
momentum transfer means large spatial separation and hence fermions are
distinguishable. If so, there is no Pauli principle or anti-commutation
and no relative sign either! So either of the two diagrams will give the
same answer.

Ideally, we should consider one of the fermions to be very massive
to mimic a source of potential (means $\vec{q} \to \vec{0} $ limit) and
match all the factors for a precise comparison.

The effective coupling contains the Kronecker deltas enforcing
conservation of the spin projections. Hence, the Yukawa potential
preserves spin projection. For matching with non-relativistic
normalization we have to take $2m_{fermion} \delta_{\sigma_{p},
\sigma_{p'}} \to 1$. This is needed for inferring the strength of
the Yukawa potential.

Remark: Suppose we consider fermion-anti-fermion scattering.
Then we will have one of the Kronecker deltas to have a minus sign.
However, in this case, $b^{\dagger} \to d^{\dagger}$ and doing the
contractions with $\Psi's$ from the Yukawa coupling, we pick up another
minus sign. Hence the overall sign does not change. Hence, the Yukawa
potential between fermion-anti-fermion is also attractive. The
same holds for anti-fermion scattering.

Thus, the inferred Yukawa potential is always attractive. Its underlying
QFT description is an exchange of a (virtual) massive scalar particle of
mass $m_{\varphi}$. Yukawa proposed this potential as a model for binding
of nucleons and working with the properties of nucleons, estimated the
scalar mass to be about 200 MeV which is close to the mass of the pion.
Of course pions come in three varieties, $\pi^{\pm}, \pi^0$ while we
have taken only a single scalar.

Consider fermion scattering in QED, by replacing the Yukawa coupling by
the QED coupling. The relevant diagrams are:

\begin{eqnarray*}
\parbox[c]{3.0cm}{
\begin{tikzpicture}
\begin{feynman}
\vertex (v1); %{\(\mu\)};
\vertex [below=1.0of v1] (v2); %{\(\nu\)};
\vertex [above left=0.8of v1] (a) {\(p\)};
\vertex [above right=0.8of v1] (b) {\(p'\)};
\vertex [below right=0.8of v2] (c) {\(k'\)};
\vertex [below left=0.8of v2] (d) {\(k\)};;

\diagram* {
(a) -- [fermion] (v1),
(v1) -- [fermion] (b),
(v1) -- [photon,momentum=\(p'-p\)] (v2),
(v2) -- [fermion] (c),
(d) -- [fermion] (v2),
\end{feynman}
\end{tikzpicture}
%		\feynmandiagram[small,horizontal=i1 to f2] { i1
%			[particle=p] -- [fermion] c1 -- [fermion]
%			f2[particle=\(p'\)] , c1 [label=180:\(\mu\)] --
%			[photon, momentum=\(p'-p\)] c2
%			[label=180:\(\nu\)] , i2 [particle=\(k'\)] --
%			[anti fermion] c2 -- [anti fermion]
%		f3[particle=k], };
& \mbox{\hspace{1.0cm} + \hspace{1.0cm}} &
\parbox[c]{3.0cm}{
\begin{tikzpicture}
\begin{feynman}
\vertex (v1); %{\(\mu\)};
\vertex [below=1.0of v1] (v2); %{\(\nu\)};
\vertex [above left=0.8of v1] (a) {\(p\)};
\vertex [above right=0.8of v1] (b) {\(p'\)};
\vertex [below right=0.8of v2] (c) {\(k'\)};
\vertex [below left=0.8of v2] (d) {\(k\)};;

\diagram* {
(a) -- [fermion] (v1),
(v1) -- [fermion] (c),
(v1) -- [photon,momentum'=\(k'-p\)] (v2),
(v2) -- [fermion] (b),
(d) -- [fermion] (v2),
\end{feynman}
\end{tikzpicture}
%		\feynmandiagram[small,horizontal=i1 to f3] { i1
%			[particle=p] -- [fermion] c1 -- [fermion]
%			f3[particle=\(k'\)] , c1  [label=180:\(\mu\)]--
%			[photon, momentum=\(k'-p\)] c2
%			[label=180:\(\nu\)], i2 [particle=\(p'\)] --
%			[anti fermion] c2 -- [anti fermion]
%		f2[particle=k] , };
\end{eqnarray*}

The corresponding expression is,

\begin{eqnarray}
i\mathcal{M} & = & (ie)^2\left\{ \bar{u}(p')\gamma^{\mu}u(p)
\frac{-i\eta_{\mu\nu}}{(p'-p)^2 - i\epsilon} \bar{u}(k')\gamma^{\nu}u(k)
\right. \nonumber \\
& & \mbox{\hspace{2.0cm}} \left. -  \bar{u}(k')\gamma^{\mu}u(p)
\frac{-i\eta_{\mu\nu}}{(k'-p)^2 - i\epsilon} \bar{u}(p')\gamma^{\nu}u(k)
\right\}  \ ,
\end{eqnarray}

with the same relative minus sign.

As before, consider the case of non-relativistic limit with
distinguishable fermions and consider only the first diagram.  The
denominator simplifies the same way: $(p'-p)^2 - i\epsilon \to
|\vec{q}|^2 - i\epsilon$. In the numerator we have $[
\bar{u}(p',\sigma') \gamma^{\mu} u(p,\sigma)][ \bar{u}(k',\rho')
\gamma_{\mu} u(k,\rho)]$.

Claim: In the NR limit, $\bar{u}(p',\sigma') \gamma^{\mu}
u(p,\sigma) \to \bar{u}\gamma^0u$.

Proof: The spinor $u$ satisfies, $(\dsl{p} + m)u = 0$.
Consider say the Dirac representation of the gamma matrices. Then in
terms of the two component notation we have,

\begin{eqnarray*}
\left( \begin{array}{cc} -\omega_{\vec{p}} + m &
\vec{p}\cdot\vec{\sigma} \\ -\vec{p}\cdot\vec{\sigma} & \omega_{\vec{p}}
+ m \end{array}\right) \left( \begin{array}{c} u_1 \\ u_2
\end{array}\right) & = & 0  \Rightarrow
- \vec{p}\cdot\vec{\sigma}u_1 = (\omega_{\vec{p}} + m)u_2 ,
\vec{p}\cdot\vec{\sigma}u_2 = (-\omega_{\vec{p}} + m)u_1
\end{eqnarray*}

The NR limit $\omega_{\vec{p}} \approx m + \frac{\vec{p}^2}{2m}$ then
implies $u_2 \sim o(|p|/m) u_1$.  Also, in the two component notation,
we have

\begin{eqnarray*}
\bar{u}'\gamma^i u & = & ( (u_1')^{\dagger}, (u_2')^{\dagger} )
\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)
\left(\begin{array}{cc} 0 & \sigma^i \\ -\sigma^i & 0
\end{array}\right)
\left( \begin{array}{c} u_1 \\ u_2 \end{array}\right)  =
(u_1')^{\dagger}\sigma^iu_2 + (u_2')^{\dagger}\sigma^i u_1 ~
\sim ~ o(p/m)	\ ; \\
\bar{u}'\gamma^0 u & = & ( (u_1')^{\dagger}, (u_2')^{\dagger} )
\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)
\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)
\left( \begin{array}{c} u_1 \\ u_2 \end{array}\right)  =
(u_1')^{\dagger}u_1 + (u_2')^{\dagger}u_2 \sim o(1) + o(|p|/m) .
\end{eqnarray*}

This proves the claim.

Hence the numerator approximates to $[\bar{u}(\vec{p}',\sigma') \gamma^0
u(\vec{p},\sigma)] [\bar{u}(\vec{k}',\rho') \gamma^0 u(\vec{k},\rho)]$
and we get,

\begin{equation}
i\mathcal{M} \approx
\frac{ie^2\eta_{00}[u^{\dagger}(\vec{p}'\sigma')
u(\vec{p},\sigma)] [u^{\dagger}(\vec{k}',\rho')
u(\vec{k},\rho)]}{\vec{q}^2 - i\epsilon} \ .
\end{equation}

Since the square brackets are positive ($p' \approx p, k' \approx k$ for
small momentum transfer) and the $\eta_{00} < 0$, relative to the Yukawa
matrix element we have an opposite sign and hence, the effective
potential from the QED coupling will be repulsive for fermion
scattering. The form of the potential itself can be obtained from the
Yukawa one by taking $m_{\varphi} \to 0$ and as expected, we get the
Coulomb potential, $V(r) = \Case{(e')^2}{4\pi} =:
\Case{\alpha}{4\pi}$. We have absorbed the normalization factors in $e'$.
By redefining the original $e$, we can take $e' \to e$, the measured
electric charge (in the natural units). The constant $\alpha :=
\Case{e^2}{4\pi} \approx \Case{1}{137}$ is called the fine
structure constant.

If we consider fermion-anti-fermion scattering via the QED coupling, the
$u \to v$ does not introduce a negative sign since it is
$u^{\dagger}u$ and not $\bar{u}u$. The $b^{\dagger} \to d^{\dagger}$
introduces a minus sign as before and hence the overall sign changes.
Hence, for fermion-anti-fermion, the potential is attractive. For
anti-fermion scattering of course the potential is repulsive.

Note: It seems the sign depends on the $\eta_{00}$ which is
convention dependent. However, the signs in the propagator will also
change accordingly and the sign of the potential will be metric
signature independent.

Note: For a tensorial interaction like gravity, helicity 2,
we will have two factors of $\eta_{00}$ and the overall sign will not change. The Newtonian potential will be attractive and
will be so for fermion-anti-fermion scattering as well. This argument is
to be taken as heuristic since we need to be explicit about the
gravitational coupling to infer the numerator factors of the spinors
which are sensitive to $u \to v$ changes.

These examples indicate that exchange of quanta can be interpreted as
giving an equivalent potential (and hence force) in the non-relativistic
limit (where the concept of potential is meaningful). The qualitative
properties of attractive/repulsive are already encoded in the Feynman
rules. The comparison with effective potential also allows the coupling
parameters to be determined experimentally.

§ BASIC QED PROCESSES

The quantum electrodynamics has basic scattering of two charged
particles, the scattering of light by a charged particle,
particle-anti-particle annihilation and pair creation. These are
analyzed as: (i) electron-muon scattering, (ii) electron-positron
scattering as well as production of muon-anti-muon, (iii)
electron-photon scattering (Compton scattering) and (iv)
electron-positron annihilation into two photons. We will evaluate these
in the leading $e^2$ approximation. We will also compute the
cross-sections for comparison with experiments. These cross-sections
will be for the un-polarized particles in the initial state, for
which we will average the cross section over the initial
spins/polarizations. We will also not detect the final particles
spin/polarizations and hence sum the cross-section over the final

Here is the list of the processes with their diagrams and the
corresponding invariant amplitudes.

$e^-(p) \mu^-(k) \to e^-(p') \mu^-(k')$     (for $\mu$ replaced by $e$ it
is Moller Scattering)

[below=1.0of v1] (v2);
[above left=0.8of v1] (a) \(e^-(p)\);
[above right=0.8of v1] (b) \(e^-(p')\);
[below right=0.8of v2] (c) \(\mu^-(k')\);
[below left=0.8of v2] (d) \(\mu^-(k)\);;

(a) – [fermion] (v1),
(v1) – [fermion] (b),
(v1) – [photon,momentum'=\(q\)] (v2),
(v2) – [fermion] (c),
(d) – [fermion] (v2),

[c]10.0cm$ \hfill i\mathcal{M} =
[\bar{u}(p')(ie\gamma^{\mu})u(p)] \left[\frac{-i
\eta_{\mu\nu}}{q^2 - i\epsilon}\right]
[\bar{u}(k')(ie\gamma^{\mu})u(k)]\hfill $

$e^+(p') e^-(p) \to \mu^+(k') \mu^-(k)$     (for $\mu$ replaced by $e$ it
is Bhabha Scattering)

[right=0.6of v1] (v2);
[above left=0.8of v1] (a) \(e^+(p')\);
[below left=0.8of v1] (d) \(e^-(p)\);
[above right=0.8of v2] (b) \(\mu^+(k')\);
[below right=0.8of v2] (c) \(\mu^-(k)\);;

(v1) – [fermion] (a),
(d) – [fermion] (v1),
(v1) – [photon,momentum=\(q\)] (v2),
(v2) – [fermion] (b),
(c) – [fermion] (v2),

[c]10.0cm$ \hfill i\mathcal{M} =
[\bar{v}(p')(ie\gamma^{\mu})u(p)] \left[\frac{-i
\eta_{\mu\nu}}{q^2 - i\epsilon}\right]
[\bar{u}(k')(ie\gamma^{\mu})v(k)] \hfill $

In the next two processes, there are two diagrams contributing, obtained
by exchanging the initial and final state photons in the Compton
scattering and the two final state photons in the annihilation process.
Although the diagrams may not look `crossed', they are!

$e^-(p) \gamma(k) \to e^-(p') \gamma(k') $     (Compton Scattering)

[right=0.6of v1] (v2);
[above left=0.8of v1] (a) \(e^-(p)\);
[below left=0.8of v1] (d) \(\gamma(k)\);
[above right=0.8of v2] (b) \(e^-(p')\);
[below right=0.8of v2] (c) \(\gamma(k')\);;

(a) – [fermion] (v1),
(v1) – [fermion,momentum=\(p+k\)] (v2),
(v2) – [fermion] (b),
(d) – [photon] (v1),
(c) – [photon] (v2),

\begin{eqnarray*} 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]
\\
& & \mbox{\hspace{2.0cm}} \times
(ie\gamma^{\mu})u(p) \varepsilon^*_{\mu}
\end{eqnarray*}

[right=1.0of v1] (v2);
[above left=0.8of v1] (a) \(e^-(p)\);
[below left=0.8of v1] (d) \(\gamma(k)\);
[above right=0.8of v2] (b) \(e^-(p')\);
[below right=0.8of v2] (c) \(\gamma(k')\);;

(a) – [fermion] (v1),
(v1) – [fermion,momentum=\(p'-k\)] (v2),
(v2) – [fermion] (b),
(c) – [photon] (v1),
(d) – [photon] (v2),

\begin{eqnarray*} 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]
\\
& & \mbox{\hspace{2.0cm}} \times
(ie\gamma^{\mu})u(p) \varepsilon_{\mu}
\end{eqnarray*}

$e^+(p') e^-(p) \to \gamma(k) \gamma(k') $     (Annihilation

[below=0.6of v1] (v2);
[above left=0.8of v1] (a) \(e^-(p)\);
[above right=0.8of v1] (b) \(\gamma(k)\);
[below left=0.8of v2] (d) \(e^+(p')\);
[below right=0.8of v2] (c) \(\gamma(k')\);;

(a) – [fermion] (v1),
(v1) – [fermion,momentum=\(p-k\)] (v2),
(v2) – [fermion] (d),
(b) – [photon] (v1),
(c) – [photon] (v2),

\begin{eqnarray*} i\mathcal{M} & = &
\varepsilon_{\nu}(\vec{k}',\lambda')
\bar{v}(p')(ie\gamma^{\nu})
\left[\frac{-i (-(\dsl{p} - \dsl{k}) +
m)}{(p-k)^2 + m^2_e - i\epsilon}\right]
\\
& & \mbox{\hspace{2.0cm}} \times
(ie\gamma^{\mu})u(p) \varepsilon_{\mu}
\end{eqnarray*}

[below=1.0of v1] (v2);
[above left=0.8of v1] (a) \(e^-(p)\);
[above right=0.8of v1] (b) \(\gamma(k)\);
[below left=0.8of v2] (d) \(e^+(p')\);
[below right=0.8of v2] (c) \(\gamma(k')\);;

(a) – [fermion] (v1),
(v1) – [fermion,momentum'=\(p-k\)] (v2),
(v2) – [fermion] (d),
(c) – [photon] (v1),
(b) – [photon] (v2),

\begin{eqnarray*} i\mathcal{M} & = &
\varepsilon_{\nu}(\vec{k},\lambda)
\bar{v}(p')(ie\gamma^{\nu})
\left[\frac{-i (-(\dsl{k} - \dsl{p}') +
m)}{(p-k')^2 + m^2_e - i\epsilon}\right]
\\
& & \mbox{\hspace{2.0cm}} \times
(ie\gamma^{\mu})u(p) \varepsilon_{\mu}
\end{eqnarray*}

Since we will be computing un-polarized (average over initial
spin/polarizations), inclusive (sum over final spin/polarizations)
cross-sections, we have to evaluate $\Case{1}{2} \Case{1}{2}
\sum_{\sigma,\sigma'} |\mathcal{M}|^2$ for the fermion scattering and
likewise for the other cases. For the Compton scattering and pair
annihilation where two diagrams contribute, we need to add them before taking the mod-square.

§.§ Electron-muon processes:

Consider the electron-muon scattering. We have,

\begin{eqnarray*}
(i\mathcal{M})(-i\mathcal{M}^*) & = & \frac{e^4}{(q^2 - i\epsilon)(q^2 +
i\epsilon)} \{(\bar{u}(p')\gamma^{\mu}u(p)) (u^{\dagger}(p)
(\gamma^{\nu})^{\dagger} (\gamma^0)^{\dagger} u(p'))\} \\
& & \mbox{\hspace{3.0cm}} \times \{(\bar{u}(k')\gamma_{\mu}u(k))
(u^{\dagger}(k) (\gamma_{\nu})^{\dagger} (\gamma^0)^{\dagger} u(k'))\}
\end{eqnarray*}

The spins, $\sigma,\sigma', \rho, \rho'$ are implicit in the above
expression.  Summing over all of these replaces $u\bar{u}$ using the
completeness relation. Each of the braces can be expressed as traces
over the gamma matrices. With the new normalizations used, the
completeness relations take the form,

\begin{equation} \label{NewCompletenessReln}
\boxed{\sum_{\sigma}u(p,\sigma)\bar{u}(p,\sigma) = -\dsl{p} + m ~ , ~
\sum_{\sigma}v(p,\sigma)\bar{v}(p,\sigma) = -\dsl{p} - m}.
\end{equation}

Thus the electron and muon spin sums become,

\[
\{\dots\}_{p,p'} = Tr[\gamma^{\mu}(-\dsl{p} +
m_e)\gamma^{\nu}(-\dsl{p}' + m_e] ~ ~ , ~ ~
\{\dots\}_{k,k'} = Tr[\gamma_{\mu}(-\dsl{k} +
m_{\mu})\gamma_{\nu}(-\dsl{k}' + m_{\mu}] \ .
\]
This leads to,

\begin{eqnarray}
\frac{1}{4}\sum_{\sigma,\sigma',\rho,\rho'}
|\mathcal{M}|^2_{e\mu\to e\mu}  & = & \frac{1}{4}\frac{e^4}{(q^2
- i\epsilon)(q^2 + i\epsilon)} Tr[\gamma^{\mu}(-\dsl{p} +
m_e)\gamma^{\nu}(-\dsl{p}' + m_e] \nonumber \\
& & \mbox{\hspace{3.0cm}}\times Tr[\gamma_{\mu}(-\dsl{k} +
m_{\mu})\gamma_{\nu}(-\dsl{k}' + m_{\mu}]
\end{eqnarray}

For the $e^+ e^- \to \mu^+ \mu^-$ we will have,

\begin{eqnarray*}
(i\mathcal{M})(-i\mathcal{M}^*) & = & \frac{e^4}{(q^2 - i\epsilon)(q^2 +
i\epsilon)} \{(\bar{v}(p')\gamma^{\mu}u(p)) (u^{\dagger}(p)
(\gamma^{\nu})^{\dagger} (\gamma^0)^{\dagger} v(p'))\} \\
& & \mbox{\hspace{3.0cm}} \times \{(\bar{u}(k')\gamma_{\mu}v(k))
(v^{\dagger}(k) (\gamma_{\nu})^{\dagger} (\gamma^0)^{\dagger} u(k'))\}
\end{eqnarray*}

This is the same as before except the $p', k'$ spinors are $v$ spinors.
Using the corresponding completeness relation we get,

\begin{eqnarray}
\frac{1}{4}\sum_{\sigma,\sigma',\rho,\rho'}
|\mathcal{M}|^2_{e^+e^-\to \mu^+\mu^-}  & = &
\frac{1}{4}\frac{e^4}{(q^2 - i\epsilon)(q^2 + i\epsilon)}
Tr[\gamma^{\mu}(-\dsl{p} + m_e)\gamma^{\nu}(-\dsl{p}' - m_e]
\nonumber \\
& & \mbox{\hspace{3.0cm}}\times Tr[\gamma_{\mu}(-\dsl{k} +
m_{\mu})\gamma_{\nu}(-\dsl{k}' - m_{\mu}]
\end{eqnarray}

The Compton scattering and annihilation processes have two diagrams each
and each of these has a single Dirac trace. But each diagram also
has photon polarization completeness relations. Incidentally, for Moller
and Bhabha processes, there are `crossed diagrams' too. We will do these

To proceed with the two processes above, we need to traces of 2 and 4
gamma matrices. Here are the relevant formulae.

\begin{eqnarray}\label{DiracTraces}
Tr\mathbb{1} & = & 4 ; \\
Tr \gamma^{\mu}\gamma^{\nu} & = & -4 \eta^{\mu\nu} ; \\
Tr \gamma^{\mu}\gamma^{\nu}\gamma^{\alpha}\gamma^{\beta} & = &
4\left(\eta^{\mu\nu}\eta^{\alpha\beta} - \eta^{\mu\alpha}\eta^{\nu\beta}
+ \eta^{\mu\beta}\eta^{\nu\alpha}\right) \\ \nonumber
\end{eqnarray}

Using these, we get

\begin{eqnarray*}
Tr[\gamma^{\mu}(-\dsl{p} + m_e)\gamma^{\nu}(-\dsl{p}' + m_e] & =
& p_{\alpha}p'_{\beta} Tr[\gamma^{\mu} \gamma^{\alpha}
\gamma^{\nu} \gamma^{\beta}] +
m_e^2Tr[\gamma^{\mu}\gamma^{\nu}] \\
& = & 4(p^{\mu}p^{'\nu} - \eta^{\mu\nu}p\cdot p' +
p^{\nu}p^{'\mu}) - 4m_e^2\eta^{\mu\nu} ~ ~ ~ ~
\mbox{and,} \\
Tr[\gamma_{\mu}(-\dsl{k} +
m_{\mu})\gamma_{\nu}(-\dsl{k}' + m_{\mu}] & = &
k_{\alpha}k'_{\beta} Tr[\gamma_{\mu}
\gamma^{\alpha} \gamma_{\nu} \gamma^{\beta}] +
m_{\mu}^2Tr[\gamma_{\mu}\gamma_{\nu}] \\
& = & 4(k_{\mu}k'_{\nu} - \eta_{\mu\nu}k\cdot k' +
k_{\nu}k'_{\mu}) - 4m_{\mu}^2\eta_{\mu\nu}.
\end{eqnarray*}

Dotting the traces gives,

\begin{eqnarray}
Tr[\dots]Tr[\dots] & = & 32[(p\cdot k)(p'\cdot k') + (p\cdot k')(p'\cdot
k) + m_{\mu}^2p\cdot p'+ m^2_ek\cdot k'+ 2 m_e^2m_{\mu}^2]
\mbox{\hspace{1cm}} \nonumber \\
\therefore \frac{1}{4}\sum_{spins} |\mathcal{M}|^2_{e\mu\to e\mu}  & = &
\frac{8e^4}{q^4} [(p\cdot k)(p'\cdot k') + (p\cdot k')(p'\cdot k) +
m^2_{\mu}p\cdot p' + m_{e}^2k\cdot k' + 2 m_e^2m_{\mu}^2]
\end{eqnarray}

where $q  =  p' -p$ .

For the $e^+ e^- \to \mu^+ \mu^-$ process, the only difference is
$m^2_e, m^2_{\mu}$ changing signs. Thus we have,

\begin{equation}
\frac{1}{4}\sum_{spins} |\mathcal{M}|^2_{e^+e^-\to\mu^+\mu^-}  =
\frac{8e^4}{q^4} [(p\cdot k)(p'\cdot k') + (p\cdot k')(p'\cdot k) -