\left(\frac{k^{\alpha}k^{\nu}}{k^{2}}\right)\right\\}$

The second term (containing $k^{\alpha}k^{\nu}$) vanishes when contracted with
the conserved electron current $j_{\nu}$. Therefore,

$\displaystyle\tilde{M}^{\mu\nu}_{{}_{full}}(k)$ $\displaystyle=$
$\displaystyle\displaystyle\frac{\tilde{M}^{\mu\nu}(k)}{1-e_{{}_{0}}^{2}\cdot\
g{{}_{p}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal P})^{2}\cdot\chi(k)}$ (97)

We define the electron charge at momentum $k$ as

$\displaystyle e^{2}(k)$ $\displaystyle:=$
$\displaystyle\displaystyle\frac{e_{{}_{0}}^{2}}{1-e_{{}_{0}}^{2}\cdot\
g{{}_{p}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal P})^{2}\cdot\chi(k)}$ (98)

The definition (99) differs from the corresponding definition in the
Heisenberg-Pauli framework [43, Equation 13.55] by the Gibbs factor in the
denominator. The physical electron charge is defined as $e(0)$, the value at
zero momentum transfer. With some abuse of notation we denote $e(0)$ as $e$.

At $O(e_{{}_{0}}^{2})$, $e_{{}_{0}}$ can be replaced by $e$ and $e^{2}(k)$ can
be written as323232As shown in (LABEL:eqnvaz), $\lim_{k\rightarrow 0}\
g{{}_{p}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal P})^{2}\,\chi(k)=0$, which is
consistent with (99).

$\displaystyle e^{2}(k)$ $\displaystyle=$
$\displaystyle\displaystyle\frac{e^{2}}{1-e^{2}\ g{{}_{p}}\,(\,{k}\,;{\cal
C}_{{}_{\cal P}},{\cal P})^{2}\chi(k)}$ (99)

Recalling that the fine structure constant $\alpha$ is related to $e$ by
$e^{2}=4\pi\alpha$, we define momentum-dependent coupling constant
$\alpha(k):=\displaystyle\frac{e^{2}(k)}{4\pi}$. With some abuse of notation
we have denoted $\alpha(0)$ as $\alpha$. From (99) we have

$\displaystyle\displaystyle\frac{\alpha(k)}{\alpha}$ $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{1-4\pi\,\alpha\,\
g{{}_{p}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal P})^{2}\,\chi(k)}\ =\
1+4\pi\,\alpha\ g{{}_{p}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal
P})^{2}\,\chi(k)+O(e_{{}_{0}}^{4})$ (100)

From Figure 2, at $O(\alpha)$ we have

$\displaystyle\Pi^{\mu\nu}(k)$ $\displaystyle=$ $\displaystyle
e^{2}\int\displaystyle\frac{d^{4}q}{(2\pi)^{4}}\
\,Tr\left\\{\gamma^{\mu}S(k+q)\gamma^{\nu}S(q)\right\\},\qquad S(q)\ =\
i\cdot\displaystyle\frac{\ g{{}_{e}}\,(\,{q}\,;{\cal C}_{{}_{\cal P}},{\cal
P})^{2}\cdot(q\\!\\!\\!/+m_{{}_{e}})}{q^{2}-m_{{}_{e}}^{2}+i\epsilon}$

Therefore,

$\displaystyle\Pi^{\mu}\,_{\mu}(k)$ $\displaystyle=$ $\displaystyle
e^{2}\int\displaystyle\frac{d^{4}q}{(2\pi)^{4}}\ \displaystyle\frac{\
g{{}_{e}}\,(\,{k+q}\,;{\cal C}_{{}_{\cal P}},{\cal P})^{2}\cdot\
g{{}_{e}}\,(\,{q}\,;{\cal C}_{{}_{\cal P}},{\cal
P})^{2}\cdot\left[\,8(k+q)\cdot
q-16m_{{}_{e}}^{2}\,\right]}{((k+q)^{2}-m_{{}_{e}}^{2}+i\epsilon)(q^{2}-m_{{}_{e}}^{2}+i\epsilon)}$
(101)

The integral is calculated numerically using the method described in Appendix
D. Since $\sqrt{|k^{2}|}$ is the only 333333 The energy scale
$\mu:=\sqrt{\rule{0.0pt}{12.0pt}|k^{2}|}$ dominates the electron mass scale
for $\mu\geq 10m_{{}_{e}}$. The other Lorentz-invariant energy scale $k\cdot
u({\cal C}_{{}_{\cal P}};{\cal C}_{{}_{\cal P}})$ depends on the momenta of
the incoming electrons, which determine ${\cal C}_{{}_{\cal P}}$. The choice
of $\mu$ as the temperature of the photon field is tantamount to asserting
that the temperature of the photon field depends only on the Lorentz-invariant
energy scales of the disturbances in the photon field, and not on the details
of how those disturbances are generated.  Lorentz-invariant energy scale that
depends only on the disturbance in the photon field, we take the ‘temperature’
$\tau$ to be $\tau=\sqrt{\rule{0.0pt}{11.0pt}|k^{2}|}$ in our calculations.
Substituting (101) into (95), we can evaluate the function $\chi(k)$, and thus
the ratio $\displaystyle\frac{\alpha(k)}{\alpha}$ shown in (100).

It is well-known [43, Equation 13.63] that for large, negative
$k^{2}/m_{{}_{e}}^{2}$, the Heisenberg-Pauli framework yields

$\displaystyle\displaystyle\frac{\alpha(k)}{\alpha}$ $\displaystyle=$
$\displaystyle
1+\displaystyle\frac{\alpha}{3\pi}\,\ln\displaystyle\frac{|k^{2}|}{m_{{}_{e}}^{2}}+O(\alpha^{2})$
(102)

Denoting $\alpha(k)$ in (102), obtained from the Heisenberg-Pauli framework,
as $\alpha_{{}_{HP}}(k)$ and the $\alpha(k)$ in (100), obtained from the new
framework, as $\alpha_{{}_{Gibbs}}(k)$, at O($\alpha$), we obtain the ratio

$\displaystyle\displaystyle\frac{\alpha_{{}_{Gibbs}}(k)}{\alpha_{{}_{HP}}(k)}$
$\displaystyle=$ $\displaystyle\displaystyle\frac{1+4\pi\,\alpha\,\
g{{}_{e}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal
P})^{2}\cdot\chi(k)}{1+\displaystyle\frac{\alpha}{3\pi}\,\ln\displaystyle\frac{|k^{2}|}{m_{{}_{e}}^{2}}},\qquad\mbox{
at }O(\alpha)$ (103)

where $\chi(k)$ is given by (95).

Figure 3: Growth of $\alpha_{{}_{Gibbs}}$ with $\sqrt{-k^{2}}/m_{{}_{e}}$ over
the range $10\leq\sqrt{-k^{2}}/m_{{}_{e}}\leq 10^{4}$, and for $k\cdot u({\cal
C}_{{}_{\cal P}};{\cal C}_{{}_{\cal P}})=200,400,600,800$ and $1000$. The
inset shows the plot for $10\leq\sqrt{-k^{2}}/m_{{}_{e}}\leq 10^{3}$.

Figure 3 shows the plot of $\frac{\alpha_{Gibbs}(k)}{\alpha}$, for
$10m_{{}_{e}}\leq\sqrt{-k^{2}}\leq 10^{4}m_{{}_{e}}$. The graphs confirm that
the $QED$ coupling constant $\alpha_{Gibbs}(k)$ increases with invariant
momentum transfer $\sqrt{\rule{0.0pt}{12.0pt}-k^{2}}$, as predicted by
previously known results [43]. Figure 4 provides a quantitative comparison
with the previously known results. The plot of
$\alpha_{Gibbs}(k)/\alpha_{HP}(k)$ shows that the discrepancy between
$\alpha_{Gibbs}(k)$, and $\alpha_{HP}(k)$ is less than 0.8% over the above
range of $\rule{0.0pt}{11.0pt}\sqrt{-k^{2}}$ and less than 0.5% at invariant
momentum transfer scale of 104 $m_{{}_{e}}$.

Figure 4: Comparison of the prediction of the new framework
($\alpha_{{}_{Gibbs}}$) with known results ($\alpha_{HP}$), for
$10\leq\sqrt{-k^{2}}/m_{{}_{e}}\leq 10^{4}$ and $k\cdot u({\cal C}_{{}_{\cal
P}};{\cal C}_{{}_{\cal P}})=200,400,600,800$ and $1000$. The inset shows the
plot for $10\leq\sqrt{-k^{2}}/m_{{}_{e}}\leq 10^{3}$.

##  7 Remarks

In this section we present a tree-level calculation in the new framework.
Subsequently, we discuss two ramifications. First, the energy density of zero-
point fluctuations of free quantum fields is finite in the new framework. As
an example, we calculate the energy density of the zero-point fluctuations of
the free Maxwell field and find that it is about three orders of magnitude
smaller343434 An analogous computation in the Heisenberg-Pauli framework
yields a divergent answer. Even with a cutoff the prediction of the
Heisenberg-Pauli framework is a few dozens of orders of magnitude larger than
the observational upper bound [93], a discrepancy that is called the
cosmological constant problem. than the current upper bound on the dark energy
density. Secondly, since the perturbative corrections are finite in the new
framework the bare parameters of a theory are not required to be infinite a
priori. Finiteness of the bare parameters restores mathematical legitimacy to
the Lagrangians of interacting quantum fields.

###  7.1 Compton Scattering

In this section we compare a tree-level prediction in the new framework with
experimental data. Specifically, we calculate the angular distribution of the
differential cross section for Compton scattering at tree level using the new
framework and compare the prediction with that of the well-known Klein-Nishina
formula [53] as well as the experimental measurements of Friedrich and
Goldhaber [24]. We perform the calculation in the laboratory frame.

In Compton scattering

$\displaystyle\gamma+e^{-}\rightarrow\gamma+e^{-}$ (104)

incoming photon and electron scatter to outgoing photon and electron as shown
in Figure 5.

Figure 5: The kinematics of Compton scattering.

We assume the scattering occurs in the $y-z$ plane in the laboratory frame. An
incoming photon of energy
$E_{\gamma}=\omega_{{}_{1}}m_{{}_{e}}c^{2}=\omega_{{}_{1}}$ ( $m_{{}_{e}}=c=1$
in natural units) collides with an electron, which is initially at rest in the
laboratory frame, and scatters to an outgoing photon of energy
$E_{\gamma}^{\prime}=\omega_{{}_{2}}m_{{}_{e}}c^{2}=\omega_{{}_{2}}$ at an
angle $\theta$ relative to the $z$-axis. The 4-momentum of the outgoing
electron in laboratory frame is denoted $q_{{}_{2}}$.

Momentum conservation and the requirement that the outgoing electron is on
mass shell determines $\omega_{{}_{2}}$ given $\omega_{{}_{1}}$ and $\theta$.
Specifically, $\omega_{{}_{2}}$ is given by

$\displaystyle\omega_{{}_{2}}$ $\displaystyle=$
$\displaystyle\displaystyle\frac{m_{{}_{e}}\omega_{{}_{1}}}{m_{{}_{e}}+\omega_{{}_{1}}(1-\cos\theta)}$
(105)

We choose $\epsilon(k_{{}_{1}},1)=(0,1,0,0)$ and
$\epsilon(k_{{}_{1}},2)=(0,0,1,0)$ to be the basis transverse polarizations
corresponding to the incoming photon and $\epsilon(k_{{}_{2}},1)=(0,1,0,0)$
and $\epsilon(k_{{}_{2}},2)=(0,0,\cos\theta,-\sin\theta)$ to be the basis
transverse polarizations corresponding to the outgoing photon. We label the
polarizations of the incoming and outgoing photons,
$\epsilon(k_{{}_{1}},r_{{}_{1}})$ and $\epsilon(k_{{}_{2}},r_{{}_{2}})$
respectively, with $r_{{}_{1}},r_{{}_{2}}=1,2$.

Figure 6: The Feynman diagrams that contribute to the tree level amplitude of
Compton scattering.

At the tree level the scattering amplitude with incoming photon (electron)
polarization353535With slight abuse of notation we use the term polarization
to denote the helicity of an electron. labeled $r_{{}_{1}}$ ($s_{{}_{1}}$) and
outgoing photon (electron) polarization $r_{{}_{2}}$ ($s_{{}_{2}}$), denoted
${\cal A}_{{}_{2}}$, receives contributions from the two Feynman diagrams
shown in Figure 6, and is given by

$\displaystyle{\cal
A}_{{}_{2}}(\left\\{k_{{}_{1}},r_{{}_{1}}\right\\},\left\\{q_{{}_{1}},s_{{}_{1}}\right\\};\left\\{k_{{}_{2}},r_{{}_{2}}\right\\},\left\\{q_{{}_{2}},s_{{}_{2}}\right\\})$
$\displaystyle=$
$\displaystyle-i(2\pi)^{4}\delta^{4}(q_{{}_{2}}+k_{{}_{2}}-q_{{}_{1}}-k_{{}_{1}})\
{\cal M}_{{}_{2}}$

where at $O(e_{{}_{0}}^{2})$ we have

$\displaystyle k_{{}_{1}}$ $\displaystyle=$
$\displaystyle(\omega_{{}_{1}},0,0,\omega_{{}_{1}}),\ k_{{}_{2}}\ =\
(\omega_{{}_{2}},0,\omega_{{}_{2}}\sin\theta,\omega_{{}_{2}}\cos\theta),$
$\displaystyle q_{{}_{1}}$ $\displaystyle=$ $\displaystyle(m_{{}_{e}},0,0,0),\
q_{{}_{2}}\ =\
(E_{{}_{2}}\,^{\prime},0,-\omega_{{}_{2}}\sin\theta,\omega_{{}_{1}}-\omega_{{}_{2}}\cos\theta),\quad
E_{{}_{2}}\,^{\prime}:=m_{{}_{e}}+\omega_{{}_{1}}-\omega_{{}_{2}}$
$\displaystyle{\cal M}_{{}_{2}}$ $\displaystyle=$ $\displaystyle-
e_{{}_{0}}^{2}\ \chi(\omega_{{}_{1}})\ \left[\,\bar{u}(q_{{}_{2}},s_{{}_{2}})\
\Gamma_{{}_{Gibbs}}\ u(q_{{}_{1}},s_{{}_{1}})\,\right]\ =\ -4\pi\alpha\
\chi(\omega_{{}_{1}})\left[\,\bar{u}(q_{{}_{2}},s_{{}_{2}})\
\Gamma_{{}_{Gibbs}}\ u(q_{{}_{1}},s_{{}_{1}})\,\right],$
$\displaystyle\chi(\omega_{{}_{1}})$ $\displaystyle:=$
$\displaystyle\displaystyle\frac{\ g{{}_{p}}\,(\,{k_{{}_{1}}}\,;{\cal F},{\cal
P})\ g{{}_{p}}\,(\,{k_{{}_{2}}}\,;{\cal F},{\cal P})\
g{{}_{e}}\,(\,{q_{{}_{1}}}\,;{\cal F},{\cal P})\
g{{}_{e}}\,(\,{q_{{}_{2}}}\,;{\cal F},{\cal
P})}{Z_{{}_{e}}\,Z_{{}_{\gamma}}\,{\cal N}}\ $ $\displaystyle=$
$\displaystyle\ g{{}_{p}}\,(\,{k_{{}_{1}}}\,;{\cal F},{\cal P})\
g{{}_{p}}\,(\,{k_{{}_{2}}}\,;{\cal F},{\cal P})\
g{{}_{e}}\,(\,{q_{{}_{1}}}\,;{\cal F},{\cal P})\
g{{}_{e}}\,(\,{q_{{}_{2}}}\,;{\cal F},{\cal P})\ =\ \
g{{}_{p}}\,(\,{k_{{}_{1}}}\,;{\cal F},{\cal P})^{2}\
g{{}_{e}}\,(\,{q_{{}_{1}}}\,;{\cal F},{\cal P})^{2}$
$\displaystyle\Gamma_{{}_{Gibbs}}$ $\displaystyle=$ $\displaystyle\
g{{}_{e}}\,(\,{q_{{}_{1}}+k_{{}_{1}}}\,;{\cal F},{\cal
P})^{2}\left[\,\frac{\epsilon\,\\!\\!\\!/(k_{{}_{2}},r_{{}_{2}})\,(q\\!\\!\\!/_{{}_{1}}+k\\!\\!\\!/_{{}_{1}}+m_{{}_{e}})\epsilon\,\\!\\!\\!/(k_{{}_{1}},r_{{}_{1}})}{(q_{{}_{1}}+k_{{}_{1}})^{2}-m_{{}_{e}}^{2}+i\epsilon}\,\right]$
(106) $\displaystyle+\ g{{}_{e}}\,(\,{q_{{}_{1}}-k_{{}_{2}}}\,;{\cal F},{\cal
P})^{2}\left[\,\frac{\epsilon\,\\!\\!\\!/(k_{{}_{2}},r_{{}_{2}})\,(q\\!\\!\\!/_{{}_{1}}-k\\!\\!\\!/_{{}_{2}}+m_{{}_{e}})\epsilon\,\\!\\!\\!/(k_{{}_{1}},r_{{}_{1}})}{(q_{{}_{1}}-k_{{}_{2}})^{2}-m_{{}_{e}}^{2}+i\epsilon}\,\right]$

where $Z_{{}_{e}}$ and $Z_{{}_{\gamma}}$ are the wavefunction renomalization
constants for electron and photon fields; ${\cal N}$ is defined in (194). At
the lowest order363636 As $e_{{}_{0}}\rightarrow 0$,
$\psi\rightarrow\psi_{{}_{\mbox{\scriptsize free}}},$ $H_{{}_{I}}\rightarrow
0$ and $U(\infty,-\infty)=1$. From (159) (and similar relation for photons)
$Z_{{}_{e}},Z_{{}_{\gamma}}\rightarrow 1$, and ${\cal N}\rightarrow 1$.  in
Compton scattering $Z_{{}_{e}}=Z_{{}_{\gamma}}={\cal N}=1$. To show the last
equality for $\chi(\omega_{{}_{1}})$, we show below that $\
g{{}_{p}}\,(\,{k_{{}_{2}}}\,;{\cal F},{\cal P})\ =\ \
g{{}_{p}}\,(\,{k_{{}_{1}}}\,;{\cal F},{\cal P})$ and $\
g{{}_{p}}\,(\,{q_{{}_{2}}}\,;{\cal F},{\cal P})=\
g{{}_{p}}\,(\,{q_{{}_{1}}}\,;{\cal F},{\cal P})$.

The Lorentz tranformation $\Lambda({\cal C}_{{}_{\cal P}},{\cal F}):{\cal
F}\rightarrow{\cal C}_{{}_{\cal P}}$ from the laboratory frame ${\cal F}$ to
the center-of-momentum frame ${\cal C}_{{}_{\cal P}}$ of the scattering
process is

$\displaystyle\Lambda({\cal C}_{{}_{\cal P}},{\cal F})$ $\displaystyle=$
$\displaystyle\left[\,\begin{array}[]{cccc}\gamma&&&-v\gamma\\\ &1&&\\\
&&1&\\\ -v\gamma&&&\gamma\end{array}\,\right],\qquad v\ =\
\displaystyle\frac{\omega_{{}_{1}}}{m_{{}_{e}}+\omega_{{}_{1}}},\quad\gamma^{-1}:=\sqrt{1-v^{2}}$
(111)

Recalling the definition in (14) and using (16) we have

$\displaystyle\xi_{\alpha}(k_{{}_{2}};{\cal F},{\cal C}_{{}_{\cal
P}})\,u^{\alpha}({\cal C}_{{}_{\cal P}};{\cal F})$ $\displaystyle=$
$\displaystyle\left[\,\uline{\Lambda({\cal C}_{{}_{\cal P}},{\cal
F})k_{{}_{2}}}\,\right]^{0}\ =\ \left[\,\Lambda({\cal C}_{{}_{\cal P}},{\cal
F})k_{{}_{2}}\,\right]^{0}$

The second equality above holds since $k_{{}_{2}}$ is on mass shell. Using the
form of $k_{{}_{2}}$, shown in (106), and the expression for $v$, shown in
(111), and (105) we have

$\displaystyle\xi_{\alpha}(k_{{}_{2}};{\cal F},{\cal C}_{{}_{\cal
P}})\,u^{\alpha}({\cal C}_{{}_{\cal P}};{\cal F})$ $\displaystyle=$
$\displaystyle\omega_{{}_{2}}\gamma(1-v\cos\theta)\ =\
\displaystyle\frac{\gamma\,m_{{}_{e}}\,\omega_{{}_{1}}}{m_{{}_{e}}+\omega_{{}_{1}}}\
=\ m_{{}_{e}}\,v\,\gamma$ (112)

Again, using (14), (16), (106) and (111) we have

$\displaystyle\xi_{\alpha}(k_{{}_{1}};{\cal F},{\cal C}_{{}_{\cal
P}})\,u^{\alpha}({\cal C}_{{}_{\cal P}};{\cal F})$ $\displaystyle=$
$\displaystyle\gamma\,\omega_{{}_{1}}(1-v)\ =\ m_{{}_{e}}v\gamma$ (113)

From (112), (113) and (LABEL:eqn081) it follows that

$\displaystyle\ g{{}_{p}}\,(\,{k_{{}_{1}}}\,;{\cal F},{\cal P})$
$\displaystyle=$ $\displaystyle\ g{{}_{p}}\,(\,{k_{{}_{2}}}\,;{\cal F},{\cal
P})$ (114)

Similarly, we have, noting that $\omega_{{}_{1}}(1-v)\ =\
\omega_{{}_{2}}(1-v\cos\theta)\ =\ m_{{}_{e}}v$, and that $q_{{}_{2}}$ is on
mass-shell

$\displaystyle\xi_{\alpha}(q_{{}_{1}};{\cal F},{\cal C}_{{}_{\cal
P}})\,u^{\alpha}({\cal C}_{{}_{\cal P}};{\cal F})$ $\displaystyle=$
$\displaystyle m_{{}_{e}}\,\gamma\ =\ \xi_{\alpha}(q_{{}_{2}};{\cal F},{\cal
C}_{{}_{\cal P}})\,u^{\alpha}({\cal C}_{{}_{\cal P}};{\cal F})$ (115)

From (17) and (115) we have

$\displaystyle\ g{{}_{e}}\,(\,{q_{{}_{1}}}\,;{\cal F},{\cal P})$
$\displaystyle=$ $\displaystyle\ g{{}_{e}}\,(\,{q_{{}_{2}}}\,;{\cal F},{\cal
P})$ (116)

From (114) and (116) it follows that $\chi$, defined in (106), is

$\displaystyle\chi$ $\displaystyle=$ $\displaystyle\
g{{}_{p}}\,(\,{k_{{}_{1}}}\,;{\cal F},{\cal P})\,\
g{{}_{p}}\,(\,{k_{{}_{2}}}\,;{\cal F},{\cal P})\,\
g{{}_{e}}\,(\,{q_{{}_{1}}}\,;{\cal F},{\cal P})\,\
g{{}_{e}}\,(\,{q_{{}_{2}}}\,;{\cal F},{\cal P})$ (117) $\displaystyle=$
$\displaystyle\left[\,\ g{{}_{p}}\,(\,{k_{{}_{1}}}\,;{\cal F},{\cal P})\,\
g{{}_{e}}\,(\,{q_{{}_{1}}}\,;{\cal F},{\cal P})\,\right]^{2}$

as claimed in (106). (117) shows that $\chi$, depends only on
$\omega_{{}_{1}}$ and is independent of $\theta$, although $k_{{}_{2}}$ and
$q_{{}_{2}}$ depend on $\theta$. Hence we write $\chi$ as
$\chi(\omega_{{}_{1}})$ in (106).

The differential cross section in the laboratory frame ${\cal F}$ for
prespecified polarizations and helicities of the incoming and outgoing photons
and electrons is given by [86, Equations 11.9, 11.12, 11.22, 11.23]

$\displaystyle(d\sigma_{r_{{}_{1}},r_{{}_{2}},s_{{}_{1}},s_{{}_{2}}})_{{}_{Gibbs}}$
$\displaystyle=$
$\displaystyle\left[\,\int_{0}^{\infty}d|\vec{k}_{{}_{2}}|\,|\vec{k}_{{}_{2}}|^{2}\int
d\vec{q}_{{}_{2}}\left\\{\displaystyle\frac{\delta^{4}(k_{{}_{1}}+q_{{}_{1}}-k_{{}_{2}}-q_{{}_{2}})\cdot|{\cal
M}_{{}_{2}}|^{2}}{(16\,\omega_{{}_{1}}\,\omega_{{}_{2}}\,m_{{}_{e}}\,E_{{}_{2}}\,^{\prime})(2\pi)^{2}}\right\\}\,\right]d\Omega$

where $d\Omega:=\sin\theta\,d\theta\,d\varphi$. Performing the integration
with respect to $\vec{q}_{{}_{2}}$, and setting $\kappa:=|\vec{k}_{{}_{2}}|$,
we have

$\displaystyle(d\sigma_{r_{{}_{1}},r_{{}_{2}},s_{{}_{1}},s_{{}_{2}}})_{{}_{Gibbs}}$
$\displaystyle=$
$\displaystyle\left[\,\int_{0}^{\infty}d\kappa\,\kappa^{2}\,\left\\{\frac{\delta\left(\omega_{{}_{1}}+m_{{}_{e}}-\kappa-\sqrt{m_{{}_{e}}^{2}+\omega_{{}_{1}}^{2}+\kappa^{2}-2\omega_{{}_{1}}\kappa\cos\theta}\right)|{\cal
M}_{{}_{2}}|^{2}}{(16\,\omega_{{}_{1}}\,\omega_{{}_{2}}\,m_{{}_{e}}\,E_{{}_{2}}\,^{\prime})(2\pi)^{2}}\right\\}\,\right]d\Omega$

Using (105) and a property373737 $\displaystyle\delta(f(x))$ $\displaystyle=$
$\displaystyle\sum_{x_{{}_{i}}}\displaystyle\frac{\delta(x-x_{{}_{i}})}{|f^{\prime}(x_{{}_{i}})|},\qquad\mbox{
where }f(x_{{}_{i}})=0$ of $\delta$-function we note that

$\displaystyle\delta\left(\omega_{{}_{1}}+m_{{}_{e}}-\kappa-\sqrt{m_{{}_{e}}^{2}+\omega_{{}_{1}}^{2}+\kappa^{2}-2\omega_{{}_{1}}\kappa\cos\theta}\right)$
$\displaystyle=$
$\displaystyle\delta(\kappa-\omega_{{}_{2}})\cdot\left(\frac{E_{{}_{2}}\,^{\prime}\omega_{{}_{2}}}{m_{{}_{e}}\omega_{{}_{1}}}\right),$
(118)

Performing the integration with respect to $\kappa$ using (118), averaging
over the possible polarizations of the incoming photon, and summing over the
possible spins of the incoming and outgoing electrons and the possible
polarizations of the outgoing photon we have,

$\displaystyle\left(\frac{d\sigma}{d\Omega}\right)_{{}_{Gibbs}}$
$\displaystyle=$
$\displaystyle\displaystyle\frac{1}{\left(16\,m_{{}_{e}}^{2}\right)(2\pi)^{2}}\left(\frac{\omega_{{}_{2}}}{\omega_{{}_{1}}}\right)^{2}\left[\,\left(\frac{1}{2}\right)\sum_{r_{{}_{1}},r_{{}_{2}}=1}^{2}\sum_{s_{{}_{1}},s_{{}_{2}}=1}^{2}|{\cal
M}_{{}_{2}}|^{2}\,\right]$

At the tree level, using (106), we have

$\displaystyle\left(\frac{d\sigma}{d\Omega}\right)_{{}_{Gibbs}}$
$\displaystyle=$
$\displaystyle\displaystyle\frac{\alpha^{2}}{8m_{{}_{e}}^{2}}\left(\frac{\omega_{{}_{2}}}{\omega_{{}_{1}}}\right)^{2}\,{\cal
T}_{{}_{Gibbs}},\quad\mbox{ where }$ $\displaystyle\quad{\cal T}_{{}_{Gibbs}}$
$\displaystyle:=$
$\displaystyle(\chi(\omega_{{}_{1}}))^{2}\sum_{r_{{}_{1}},r_{{}_{2}}=1}^{2}\sum_{s_{{}_{1}},s_{{}_{2}}=1}^{2}|\bar{u}(q_{{}_{2}},s_{{}_{2}})\
\Gamma_{{}_{Gibbs}}\ u(q_{{}_{1}},s_{{}_{1}})|^{2}$ (119)

Using the trace identities we have

$\displaystyle{\cal T}_{{}_{Gibbs}}$ $\displaystyle=$
$\displaystyle\sum_{r_{{}_{1}},r_{{}_{2}}=1}^{2}\left[\left\\{\
g{{}_{e}}\,(\,{k_{{}_{1}}+q_{{}_{1}}}\,;{\cal F},{\cal
P})^{4}\right\\}\left[\,\frac{T_{11}}{4(q_{{}_{1}}\cdot
k_{{}_{1}})^{2}}\,\right]+\left\\{\
g{{}_{e}}\,(\,{q_{{}_{1}}-k_{{}_{2}}}\,;{\cal F},{\cal
P})^{4}\right\\}\left[\,\frac{T_{22}}{4(q_{{}_{1}}\cdot
k_{{}_{2}})^{2}}\,\right]\right.$ (120) $\displaystyle\left.+\left\\{2\
g{{}_{e}}\,(\,{k_{{}_{1}}+q_{{}_{1}}}\,;{\cal F},{\cal P})^{2}\
g{{}_{e}}\,(\,{q_{{}_{1}}-k_{{}_{2}}}\,;{\cal F},{\cal
P})^{2}\right\\}\left[\,\displaystyle\frac{T_{12}}{4(q_{{}_{1}}\cdot
k_{{}_{1}})(q_{{}_{1}}\cdot
k_{{}_{2}})}\,\right]\right](\chi(\omega_{{}_{1}}))^{2}$

where, using the abbreviations
$\epsilon_{{}_{2}}:=\epsilon(k_{{}_{2}},r_{{}_{2}}),$ and
$\epsilon_{{}_{1}}:=\epsilon(k_{{}_{1}},r_{{}_{1}})$,

$\displaystyle T_{11}$ $\displaystyle=$ $\displaystyle 16(k_{{}_{1}}\cdot
q_{{}_{1}})(\epsilon_{{}_{2}}\cdot k_{{}_{1}})^{2}+8(q_{{}_{1}}\cdot
k_{{}_{1}})(q_{{}_{1}}\cdot k_{{}_{2}})$ $\displaystyle T_{12}$
$\displaystyle=$ $\displaystyle
16(\epsilon_{{}_{1}}\cdot\epsilon_{{}_{2}})^{2}(q_{{}_{1}}\cdot
k_{{}_{1}})(q_{{}_{1}}\cdot k_{{}_{2}})+8(k_{{}_{1}}\cdot
q_{{}_{1}})(\epsilon_{{}_{1}}\cdot k_{{}_{2}})^{2}-8(q_{{}_{1}}\cdot
k_{{}_{2}})(\epsilon_{{}_{2}}\cdot k_{{}_{1}})^{2}$
$\displaystyle-4(q_{{}_{1}}\cdot k_{{}_{2}})^{2}+4(k_{{}_{1}}\cdot
k_{{}_{2}})^{2}-4(k_{{}_{1}}\cdot q_{{}_{1}})^{2}$ $\displaystyle T_{22}$
$\displaystyle=$ $\displaystyle-16(\epsilon_{{}_{1}}\cdot
k_{{}_{2}})^{2}(q_{{}_{1}}\cdot k_{{}_{2}})+8(q_{{}_{1}}\cdot
k_{{}_{1}})(q_{{}_{1}}\cdot k_{{}_{2}})$

The Gibbs factors in (120) are calculated using (17) and (LABEL:eqn081).

We obtain the corresponding definitions and relations in the Heisenberg-Pauli
framework by setting the Gibbs factors to $\
g{{}_{e}}\,(\,{q_{{}_{1}}}\,;{\cal F},{\cal P})=\
g{{}_{p}}\,(\,{k_{{}_{1}}}\,;{\cal F},{\cal P})=\
g{{}_{e}}\,(\,{q_{{}_{2}}}\,;{\cal F},{\cal P})=\
g{{}_{e}}\,(\,{k_{{}_{2}}}\,;{\cal F},{\cal P})=\
g{{}_{e}}\,(\,{q_{{}_{1}}+k_{{}_{1}}}\,;{\cal F},{\cal P})=\
g{{}_{e}}\,(\,{q_{{}_{1}}-k_{{}_{2}}}\,;{\cal F},{\cal P})=1$ in (106) and in
(120). Thus,

$\displaystyle\Gamma_{{}_{HP}}$ $\displaystyle:=$
$\displaystyle\left[\,\frac{\epsilon\,\\!\\!\\!/(k_{{}_{2}},r_{{}_{2}})\,(q\\!\\!\\!/_{{}_{1}}+k\\!\\!\\!/_{{}_{1}}+m_{{}_{e}})\epsilon\,\\!\\!\\!/(k_{{}_{1}},r_{{}_{1}})}{(q_{{}_{1}}+k_{{}_{1}})^{2}-m_{{}_{e}}^{2}+i\epsilon}\,\right]+\left[\,\frac{\epsilon\,\\!\\!\\!/(k_{{}_{2}},r_{{}_{2}})\,(q\\!\\!\\!/_{{}_{1}}-k\\!\\!\\!/_{{}_{2}}+m_{{}_{e}})\epsilon\,\\!\\!\\!/(k_{{}_{1}},r_{{}_{1}})}{(q_{{}_{1}}-k_{{}_{2}})^{2}-m_{{}_{e}}^{2}+i\epsilon}\,\right]$
$\displaystyle{\cal T}_{{}_{HP}}$ $\displaystyle=$
$\displaystyle\sum_{r_{{}_{1}},r_{{}_{2}}=1}^{2}\left\\{\left[\,\frac{T_{11}}{4(q_{{}_{1}}\cdot
k_{{}_{1}})^{2}}\,\right]+\left[\,\frac{T_{22}}{4(q_{{}_{1}}\cdot
k_{{}_{2}})^{2}}\,\right]+2\left[\,\displaystyle\frac{T_{12}}{4(q_{{}_{1}}\cdot
k_{{}_{1}})(q_{{}_{1}}\cdot k_{{}_{2}})}\,\right]\right\\}$ (121)

The differential cross section

$\displaystyle\left(\frac{d\sigma}{d\Omega}\right)_{{}_{HP}}$ $\displaystyle=$
$\displaystyle\displaystyle\frac{\alpha^{2}}{8m_{{}_{e}}^{2}}\left(\frac{\omega_{{}_{2}}}{\omega_{{}_{1}}}\right)^{2}{\cal
T}_{{}_{HP}}\ =\
\displaystyle\frac{\alpha^{2}}{8m_{{}_{e}}^{2}}\left(\frac{\omega_{{}_{2}}}{\omega_{{}_{1}}}\right)^{2}\left[\,\displaystyle\frac{2(k_{{}_{1}}\cdot
k_{{}_{2}})^{2}}{(q_{{}_{1}}\cdot k_{{}_{1}})(q_{{}_{1}}\cdot
k_{{}_{2}})}+8(\epsilon_{{}_{1}}\cdot\epsilon_{{}_{2}})^{2}\,\right]$ (122)

is the well-known Klein-Nishina formula383838 In the laboratory frame, using
(105), we have $k_{{}_{1}}\cdot k_{{}_{2}}\ =\
\omega_{{}_{1}}\omega_{{}_{2}}(1-\cos\theta)\ =\
m\,\omega_{{}_{1}}\,\omega_{{}_{2}}\left(\displaystyle\frac{1}{\omega_{{}_{2}}}-\displaystyle\frac{1}{\omega_{{}_{1}}}\right)$,
$q_{{}_{1}}\cdot k_{{}_{1}}\ =\ m_{{}_{e}}\omega_{{}_{1}}$, and
$q_{{}_{1}}\cdot k_{{}_{2}}\ =\ m_{{}_{e}}\omega_{{}_{2}}$. Therefore, (122)
can be rewritten as
$\displaystyle\left(\frac{d\sigma}{d\Omega}\right)_{{}_{HP}}$ $\displaystyle=$
$\displaystyle\displaystyle\frac{\alpha^{2}}{4m_{{}_{e}}^{2}}\left(\frac{\omega_{{}_{2}}}{\omega_{{}_{1}}}\right)^{2}\left[\,\displaystyle\frac{\omega_{{}_{1}}}{\omega_{{}_{2}}}+\displaystyle\frac{\omega_{{}_{2}}}{\omega_{{}_{1}}}-2+4(\epsilon_{{}_{1}}\cdot\epsilon_{{}_{2}})^{2}\,\right]$
which is the familiar form of the Klein-Nishina formula [92, Equation 8.7.39].
[43, 53].

Figure 7: Angular distribution, expressed as a fraction of the value at
$\theta=90^{o}$. The theoretical predictions of the new framework (Gibbs) and
the Klein-Nishina formula are tree-level approximations. The incident photon
energy is $\omega_{{}_{1}}=0.173\ m_{{}_{e}}c^{2}$. The experimental data is
from Friedrich and Goldhaber [24].

For large $\omega_{{}_{1}}$ (incident photon energy) pair production is known
to mask the Compton scattering [68]. Experimental studies of the Compton
scattering have focused largely on the angular distribution of differential
scattering cross section for unpolarized incident photons. The widely
referenced experimental data is due to Friedrich and Goldhaber [24, Page 703,
Table 1] for incident photon energy of
$\omega_{{}_{1}}=0.173\,m_{{}_{e}}c^{2}$. They present intensity of scattered
radiation at angle $\theta$ as a fraction of the intensity at $\theta=90^{o}$.
In order to compare the prediction of the new framework with that of Friedrich
and Goldhaber’s data, we have plotted the differential cross section at angle
$\theta$ as a fraction of the value at $\theta=90^{o}$ for both the new
framework—Equation (119)—and the Heisenberg-Pauli framework (Klein-Nishina
formula)—Equation (122)—in Figure 7. Figure 7 also shows Goldhaber’s
experimental measurements, taken at $\theta=10^{o},20^{o},\ldots,160^{o}$.

###  7.2 Zero-Point Fluctuations and the Energy Density of Vacuum

In the following calculation we estimate the contribution that the zero-point
fluctuations of a free Maxwell field makes to the energy density of vacuum.
The corresponding calculation within the Heisenberg-Pauli framework leads to
the prediction that the energy density of zero-point fluctuations of the
electromagnetic field is infinite (or very large if one uses a $UV$ cutoff),
giving rise to the infamous discrepancy between the predicted and observed
values of the cosmological constant [93]. The following argument can be
adapted to other free quantum fields to verify that the contribution of the
zero-point fluctuations of a typical quantum field to the energy density of
vacuum is small. We work in the Lorenz gauge and use the natural units
$c=\epsilon_{{}_{0}}=\hbar=1$.

Since $\partial_{\mu}A^{\mu}=0$ in the Lorenz gauge, the Maxwell Lagrangian
can be written as

$\displaystyle{\cal L}_{{}_{EM}}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{2}\left(\partial_{\mu}A^{\mu}\right)^{2}$

${\cal L}_{{}_{EM}}$ can be rewritten as

$\displaystyle{\cal L}_{{}_{EM}}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\left(\partial_{\mu}A_{\nu}\,\partial^{\mu}A^{\nu}\right)+\partial_{\mu}\left\\{A_{\nu}\partial^{\nu}A^{\mu}-A^{\mu}\partial_{\nu}A^{\nu}\right\\}\equiv-\frac{1}{2}\left(\partial_{\mu}A_{\nu}\,\partial^{\mu}A^{\nu}\right):=\tilde{}{\cal
L}_{{}_{EM}}$

Since ${\cal L}_{{}_{EM}}$ and $\tilde{}{\cal L}_{{}_{EM}}$ differ by a total
derivative, assuming that the surface integral of
$W^{\mu}:=A^{\mu}\partial_{\nu}A^{\nu}-A_{\nu}\partial^{\nu}A^{\mu}$ vanishes
at the boundary of spacetime, the Maxwell action can be equivalently written
using $\tilde{}{\cal L}_{{}_{EM}}$.

The Hamiltonian of the free Maxwell field in the Lorenz gauge is

$\displaystyle H$ $\displaystyle=$ $\displaystyle\int
d\,\vec{x}\left\\{\pi^{\mu}\dot{A}_{\mu}-\tilde{}{\cal
L}_{{}_{EM}}\right\\},\qquad\pi^{\mu}=-\dot{A}^{\mu}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int
d\,\vec{x}\left\\{-\dot{A}^{\mu}\dot{A}_{\mu}+\partial_{{}_{i}}A_{\mu}\,\partial^{i}A^{\mu}\right\\}$

Using (45), (46), (47) and (48) we obtain

$\displaystyle H$ $\displaystyle=$ $\displaystyle
H_{physical}+H_{unphysical}+H_{zero-point}$ (123) $\displaystyle H_{physical}$
$\displaystyle:=$
$\displaystyle\int\displaystyle\frac{d\vec{k}}{(2\pi)^{3}(2\omega(\vec{k}))}\
\left\\{\ g{{}_{p}}(\,\uline{\vec{k}}\,;{\cal C}_{{}_{\cal P}},{\cal
P})\right\\}^{2}\cdot\omega(\vec{k}\,)\displaystyle\sum_{\lambda=1}^{2}{\bf\hat{a}}^{\dagger}(\vec{k},\lambda){\bf\hat{a}}(\vec{k},\lambda)$
$\displaystyle H_{unphysical}$ $\displaystyle:=$
$\displaystyle\int\displaystyle\frac{d\vec{k}}{(2\pi)^{3}(2\omega(\vec{k}))}\
\left\\{\ g{{}_{p}}(\,\uline{\vec{k}}\,;{\cal C}_{{}_{\cal P}},{\cal
P})\right\\}^{2}\cdot\omega(\vec{k}\,)\left\\{{\bf\hat{a}}^{\dagger}(\vec{k},3){\bf\hat{a}}(\vec{k},3)-{\bf\hat{a}}^{\dagger}(\vec{k},0){\bf\hat{a}}(\vec{k},0)\right\\}$
$\displaystyle H_{zero-point}$ $\displaystyle:=$ $\displaystyle 2\int
d\,\vec{k}\left\\{\ g{{}_{p}}(\,\uline{\vec{k}}\,;{\cal C}_{{}_{\cal P}},{\cal
P})\right\\}^{2}\cdot\omega(\vec{k}\,)\cdot\delta^{3}(0)$ (124)

The positive definiteness of the Hamiltonian $H$ in (123) would be assured for
physical states if we stipulate that only the states that contain equal
numbers of scalar and longitudinal photons are physical. That is, an arbitrary
physical state $\left.\left|\,\psi_{physical}\,\right\rangle\right.$ must
satisfy the following condition:

$\displaystyle\left\langle\psi_{physical}\,\left|\,{\bf\hat{a}}^{\dagger}(\vec{k},0){\bf\hat{a}}(\vec{k},0)\
\right|\psi_{physical}\,\right\rangle$ $\displaystyle=$
$\displaystyle\left\langle\psi_{physical}\,\left|\,{\bf\hat{a}}^{\dagger}(\vec{k},3){\bf\hat{a}}(\vec{k},3)\
\right|\psi_{physical}\,\right\rangle$ (125)

for all $\vec{k}$. The condition (125) as well as the Lorenz gauge condition
$\partial_{\mu}A^{\mu}=0$ in a weak form

$\displaystyle\left\langle\psi_{physical}\,\left|\,\partial_{\mu}A^{\mu}\
\right|\psi_{physical}\,\right\rangle$ $\displaystyle=$ $\displaystyle 0$
(126)

are satisfied by $\left.\left|\,\psi_{physical}\,\right\rangle\right.$ if it
satisfies the Gupta-Bleuler condition [6, 31, 46]

$\displaystyle\partial^{\mu}A_{\mu}^{+}\left.\left|\,\psi_{{}_{physical}}\,\right\rangle\right.$
$\displaystyle=$ $\displaystyle 0$ (127)

where $A_{\mu}^{+}(x)$ is the part of $A^{\mu}(x)$ that contains the positive-
energy modes. That is,

$\displaystyle A_{\mu}^{+}(x)$ $\displaystyle=$
$\displaystyle\int\displaystyle\frac{d\vec{k}}{(2\pi)^{3}(2\omega(\vec{k}))}\
\ \ g{{}_{p}}(\,\uline{\vec{k}}\,;{\cal C}_{{}_{\cal P}},{\cal P})\
\displaystyle\sum_{\lambda=0}^{3}\epsilon^{\mu}(\vec{k},\lambda){\bf\hat{a}}(\vec{k},\lambda)e^{\displaystyle-i\uline{k}x}$

Clearly vacuum $\left.\left|\,0\,\right\rangle\right.$ satisfies (125) and is
a physical state. Further,
$\left\langle\,0\left|\,H_{{}_{physical}}\,\right|0\,\right\rangle\ =\
\left\langle\,0\left|\,H_{{}_{unphysical}}\,\right|0\,\right\rangle\ =\ 0$.
Therefore, only $H_{{}_{zero-point}}$ contributes to the vacuum energy
density. Noting that the volume of space393939 The statement is to be
interpreted as a limit. One starts the quantization in a finite box, with a
well-defined volume, and considers the limit of the energy density as the
volume of the box goes to infinity.  is $(2\pi)^{3}\cdot\delta^{3}(0)$, the
energy density of the zero-point fluctuations of the electromagnetic field, in
frame ${\cal C}_{{}_{\cal P}}$ that is not moving relative to the cosmic
microwave background (CMB), is

$\displaystyle{\cal E}$ $\displaystyle=$ $\displaystyle
2\int\displaystyle\frac{d\,\vec{k}}{(2\pi)^{3}}\cdot|\vec{k}|\cdot\
g{{}_{p}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal
P})^{2}=\displaystyle\frac{1}{\pi^{2}}\int_{0}^{\infty}d\kappa\cdot\kappa^{3}\cdot\left[\,\frac{2}{e^{\displaystyle\kappa/\tau+\tau/\kappa}-1}\,\right]^{1/2},\qquad\kappa:=|\vec{k}|$
$\displaystyle=$ $\displaystyle 12.7217\cdot\tau^{4}$

where $\tau$ is the temperature (energy scale) at which the photon field is at
equilibrium with the other quantum fields. The energy density can be written
in SI units by introducing the factors of $\hbar$ and $c$ to get

$\displaystyle{\cal E}$ $\displaystyle=$
$\displaystyle\left(\frac{12.7217}{\hbar^{3}\,c^{3}}\right)\tau^{4}$ (128)

A natural estimate of the temperature characterizing the equilibrium404040
Over the time scales of our observation we may assume that the temperature of
the Cosmic Microwave Background (CMB) is constant and characterizes the
thermal equilibrium between photons and the other quantum fields.  between the
photon field and other quantum fields over cosmic distance scales is the
cosmic microwave background temperature (energy scale) of
$\tau_{{}_{CMB}}=0.23482\times 10^{-3}eV$. Using $\tau=\tau_{{}_{CMB}}$ in
(91) leads to an estimate that the zero-point fluctuations of the free
electromagnetic field contribute about 5.02 MeV/$m^{3}$, or equivalent mass
density of about 10 electrons/$m^{3}$, to the energy density of vacuum. The
above rough estimate does not contradict WMAP’s latest estimate [70] that dark
energy contributes at most 3.96 GeV/$m^{3}$ (equivalent to about 7760
electrons/m3) to the vacuum energy density of the cosmos.

One can derive similar estimates for free massive boson and fermion fields.
The estimates show that the zero-point energy densities of free fields do not
diverge in our framework. We will present a calculation of the total
contribution of the zero-point energies of the known interacting fields to
vacuum energy density elsewhere.

###  7.3 Bare and Physical Parameters

In the Heisenberg-Pauli framework, the perturbative corrections that must be
applied to a bare parameter—such as mass—to obtain its physical value turn out
to be infinite. If one starts with a finite value for a bare parameter, then
one reaches an absurd conclusion that its experimentally measured physical
value—such as the mass of an electron—is infinite.

The workaround that has been adopted to circumvent the above unphysical
prediction414141In more colorful words, the workaround is used to“cure” the
“sick” theory [100]. is to assume that the bare parameters that occur in a
Lagrangian—such as the $m_{{}_{0}}$ and $\lambda_{{}_{0}}$ in (129) below—are
themselves infinite. A Lagrangian constructed using parameters that are
assumed to be infinite a priori is—borrowing Dirac’s words—not mathematically
sensible.

In the new framework the perturbative corrections to the bare parameters are
finite at every order of perturbation theory. Consequently, the bare
parameters can be taken to be finite a priori, restoring the Lagrangian of a
theory to the status of a mathematically sensible entity. We demonstrate the
finiteness of the corrections by calculating, as examples, the leading
correction to the mass parameter of the $\varphi^{4}$ scalar field theory and
the leading correction to the coupling constant in $QED$.

####  7.3.1 Bare mass parameter in $\varphi^{4}$ theory

Consider the interacting scalar field described by the Lagrangian

$\displaystyle{\cal L}$ $\displaystyle:=$
$\displaystyle\displaystyle\frac{1}{2}\
\partial_{\mu}\varphi\,\partial^{\mu}\varphi-\displaystyle\frac{1}{2}\,m_{{}_{0}}^{2}\,\varphi^{2}-\displaystyle\frac{\lambda_{{}_{0}}}{4!}\,\varphi^{4}$
(129)

where $m_{{}_{0}}$ and $\lambda_{{}_{0}}$ are the bare mass and coupling
constant. We work in the center-of-momentum frame ${\cal C}_{{}_{\cal P}}$ of
a process of interest. The physical mass $m$, that is the pole of the full
propagator, is given by the relation

$\displaystyle G_{{}_{2}}(x-y)$ $\displaystyle:=$
$\displaystyle\left\langle\,0\left|\,T\left\\{\varphi_{{}_{I}}(x)\varphi_{{}_{I}}(y)\exp\left\\{-i\int{\cal
H}_{{}_{\mbox{\scriptsize
int}}}(z)d^{4}z\right\\}\right\\}\,\right|0\,\right\rangle$ (130)
$\displaystyle:=$ $\displaystyle
i\,\int\displaystyle\frac{d^{4}k}{(2\pi)^{4}}\ \displaystyle\frac{\left[\,\
g{{}_{b}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal
P})\,\right]^{2}\,e^{\displaystyle-ik(x-y)}}{k^{2}-m^{2}+i\epsilon}$

where

$\displaystyle{\cal H}_{{}_{I}}$ $\displaystyle:=$
$\displaystyle\displaystyle\frac{\lambda_{{}_{0}}}{4!}\,\varphi_{{}_{I}}^{4}+\delta
m^{2}\,\varphi_{{}_{I}}^{2},\qquad\delta m^{2}:=m_{{}_{0}}^{2}-m^{2}$

The subscript $I$, which indicates that we are working in the interaction
picture, will be dropped hereafter. We note that at
$O((\lambda_{{}_{0}})^{0})$, $\delta m^{2}=0$. Therefore, the expansion of
$\delta m^{2}$ in powers of $\lambda_{{}_{0}}$ is

$\displaystyle\delta m^{2}$ $\displaystyle=$ $\displaystyle(\delta
m^{2})_{{}_{1}}\,(\lambda_{{}_{0}})^{1}+(\delta
m^{2})_{{}_{2}}\,(\lambda_{{}_{0}})^{2}+\ldots$

Adding the terms at $O((\lambda_{{}_{0}})^{0})$ and
$O((\lambda_{{}_{0}})^{1})$, we have

$\displaystyle G_{{}_{2}}^{(1)}(x-y)$ $\displaystyle=$ $\displaystyle
S(x-y)-i\lambda_{{}_{0}}\int
d^{4}z\left\\{S(x-z)\left[\,\displaystyle\frac{1}{2}\,S(0)+(\delta
m^{2})_{{}_{1}}\,\right]S(z-y)\right\\}$

where $S(x)$ is the free field propagator corresponding to the physical mass
$m$,

$\displaystyle S(x):=i\,\int\displaystyle\frac{d^{4}k}{(2\pi)^{4}}\
\cdot\displaystyle\frac{e^{\displaystyle-ikx}\left\\{\
g{{}_{b}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal
P})\right\\}^{2}}{k^{2}-m^{2}+i\epsilon}$

Performing the $z$-integration we get

$\displaystyle G_{{}_{2}}^{(1)}(x-y)$ $\displaystyle=$
$\displaystyle\int\displaystyle\frac{d^{4}k}{(2\pi)^{4}}\ e^{\displaystyle-
ik(x-y)}\left\\{\tilde{S}(k)-\tilde{S}(k)\left\\{i\lambda_{{}_{0}}\left[\,\displaystyle\frac{1}{2}\,S(0)+(\delta
m^{2})_{{}_{1}}\,\right]\right\\}\tilde{S}(k)\right\\},$ (131)

where

$\displaystyle\tilde{S}(k):=\displaystyle\frac{(i)\left\\{\
g{{}_{b}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal
P})\right\\}^{2}}{k^{2}-m^{2}+i\epsilon}$

The error in rewriting (131) as (LABEL:eqnbpp3) is of
$O((\lambda_{{}_{0}})^{2})$ and therefore at $O((\lambda_{{}_{0}})^{1})$, we
can rewrite $G_{{}_{2}}^{(1)}(x-y)$ as in (LABEL:eqnbpp3).

$\displaystyle G_{{}_{2}}^{(1)}(x-y)$ $\displaystyle=$
$\displaystyle\int\displaystyle\frac{d^{4}k}{(2\pi)^{4}}\ e^{\displaystyle-
ik(x-y)}\tilde{S}(k)\cdot\displaystyle\frac{1}{1+\left\\{i\lambda_{{}_{0}}\left[\,\displaystyle\frac{1}{2}\,S(0)+(\delta
m^{2})_{{}_{1}}\,\right]\right\\}\tilde{S}(k)},$ $\displaystyle=$
$\displaystyle\int\displaystyle\frac{d^{4}k}{(2\pi)^{4}}\
\cdot\displaystyle\frac{(i)\left\\{\ g{{}_{b}}\,(\,{k}\,;{\cal C}_{{}_{\cal
P}},{\cal P})\right\\}^{2}\cdot e^{\displaystyle-
ik(x-y)}}{k^{2}-m^{2}+i\epsilon+\left\\{i\lambda_{{}_{0}}\left[\,\displaystyle\frac{1}{2}\,S(0)+(\delta
m^{2})_{{}_{1}}\,\right]\right\\}\,(i)\,\left\\{\ g{{}_{b}}\,(\,{k}\,;{\cal
C}_{{}_{\cal P}},{\cal P})\right\\}^{2}}$

$G_{{}_{2}}^{(1)}(x-y)$, the propagator at $O((\lambda_{{}_{0}})^{1})$, agrees
with the the full propagator $G_{{}_{2}}(x-y)$ in (130), if we insist that

$\displaystyle(\delta m^{2})_{{}_{1}}+\displaystyle\frac{1}{2}S(0)$
$\displaystyle=$ $\displaystyle 0\quad\implies\quad(\delta m^{2})_{{}_{1}}\ =\
-\displaystyle\frac{1}{2}\,S(0)$

Therefore the correction $\delta m^{2}$ at $O((\lambda_{{}_{0}})^{1})$, that
is $\lambda_{{}_{0}}(\delta m^{2})_{{}_{1}}$, is given by

$\displaystyle m_{{}_{0}}^{2}-m^{2}$ $\displaystyle=$
$\displaystyle-\displaystyle\frac{\lambda_{{}_{0}}}{2}\,\left\\{i\int\displaystyle\frac{d^{4}k}{(2\pi)^{4}}\
\displaystyle\frac{\left\\{\ g{{}_{b}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal
P})\right\\}^{2}}{k^{2}-m^{2}+i\epsilon}\right\\}$ (133)

Performing the $k^{0}$ integration and using spherical coordinates we get, at
$O((\lambda_{{}_{0}})^{1})$, setting $\kappa=|\,\vec{k}\,|$,

$\displaystyle m_{{}_{0}}^{2}-m^{2}$ $\displaystyle=$
$\displaystyle-\lambda_{{}_{0}}\pi\int\displaystyle\frac{d\,\vec{k}}{(2\pi)^{4}}\displaystyle\frac{\left\\{\
g{{}_{b}}(\,\uline{\vec{k}}\,;{\cal C}_{{}_{\cal P}},{\cal
P})\right\\}^{2}}{2\omega(\vec{k}\,)}\ =\
-\displaystyle\frac{\lambda_{{}_{0}}}{8\pi^{2}}\int
d\kappa\displaystyle\frac{\kappa^{2}}{\sqrt{\rule{0.0pt}{10.0pt}\kappa^{2}+m^{2}}\left[\,e^{\sqrt{\rule{0.0pt}{10.0pt}\kappa^{2}+m^{2}}/\tau}-1\,\right]^{1/2}}$

We have taken $d_{{}_{b}}=1$ (see (18)) since the mode-expansion of a free
scalar field $\varphi$ has only one creation operator for fixed 3-momentum
$\vec{k}$. Setting $y=\sqrt{\kappa^{2}+m^{2}}/\tau$, and assuming
$\lambda_{{}_{0}}>0$, we have

$\displaystyle 0\ <\ m^{2}-m_{{}_{0}}^{2}$ $\displaystyle=$
$\displaystyle\displaystyle\frac{\lambda_{{}_{0}}\tau^{2}}{8\pi^{2}}\int_{m/\tau}^{\infty}\displaystyle\frac{dy\
\sqrt{y^{2}-(m/\tau)^{2}}}{\left[\,e^{y}-1\,\right]^{1/2}}\ <\
\displaystyle\frac{\lambda_{{}_{0}}\tau^{2}}{8\pi^{2}}\int_{0}^{\infty}\displaystyle\frac{y\
dy}{\left[\,e^{y}-1\,\right]^{1/2}}\ =\
\displaystyle\frac{\lambda_{{}_{0}}\tau^{2}}{8\pi^{2}}(4.3552)$

where $\tau$, the temperature at which the field $\varphi$ is at equilibrium
with its reservoir, is taken to be finite. Rewriting the above inequalities we
get

$\displaystyle-\displaystyle\frac{4.3552\,\lambda_{{}_{0}}\,\tau^{2}}{8\pi^{2}}\
<\ m_{{}_{0}}^{2}-m^{2}<0\quad\implies\quad
m^{2}-\displaystyle\frac{4.3552\,\lambda_{{}_{0}}\,\tau^{2}}{8\pi^{2}}\ <\
m_{{}_{0}}^{2}\ <\ m^{2}$

at $O((\lambda_{{}_{0}})^{1})$. The correction to the bare $m_{{}_{0}}$ at
$O((\lambda_{{}_{0}})^{1})$ is finite, if the bare parameter
$\lambda_{{}_{0}}$ is chosen to be finite. Since the physical mass, $m$, is a
finite quantity the bare mass occurring in the Lagrangian can also be taken to
be a finite quantity, at $O((\lambda_{{}_{0}})^{1})$, as claimed.

####  7.3.2 Bare coupling constant in QED

Inverting (98) and recalling that the fine structure constant $\alpha\ =\
\displaystyle\frac{e^{2}}{4\pi}$, we have

$\displaystyle e_{{}_{0}}^{2}$ $\displaystyle=$
$\displaystyle\displaystyle\frac{4\pi\,\alpha}{1+4\pi\,\alpha\,g_{{}_{p}}(0)^{2}\,\chi(0)}$
(134)

From (95) and (101), and recalling that $e_{{}_{0}}^{2}$ was replaced by
$e^{2}$ in (101) at 1-loop approximation, we have

$\displaystyle\chi(k)$ $\displaystyle=$
$\displaystyle\displaystyle\frac{I(k)}{k^{2}},\quad
I(k):=-\displaystyle\frac{i}{3}\int\displaystyle\frac{d^{4}q}{(2\pi)^{4}}\
\displaystyle\frac{\ g{{}_{e}}\,(\,{k+q}\,;{\cal C}_{{}_{\cal P}},{\cal
P})^{2}\cdot\ g{{}_{e}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal
P})^{2}\left[\,8(k+q)\cdot
q-16m_{{}_{e}}^{2}\,\right]}{((k+q)^{2}-m_{{}_{e}}^{2}+i\epsilon)(q^{2}-m_{{}_{e}}^{2}+i\epsilon)}$

Numerical evaluation shows that $I(0)\approx-0.1635$.

As in Section 6, we take $k^{2}<0$. In order to take the limit $k\rightarrow
0$, we take $k=(k^{0},\hat{\eta}\ \sqrt{(k^{0})^{2}+\mu^{2}})$, where
$\hat{\eta}$ is an arbitrary unit 3-vector, and $\mu$ is a real number. Then,

$\displaystyle g_{{}_{p}}(0)^{2}\,\chi(0)$ $\displaystyle=$
$\displaystyle(-0.1635)\cdot\lim_{k^{0}\rightarrow
0}\left\\{\lim_{\mu\rightarrow
0}\left[\,\frac{2}{e^{\sqrt{\mu^{2}+(k^{0})^{2}}/\mu+\mu/\sqrt{\mu^{2}+(k^{0})^{2}}}-1}\,\right]^{1/2}\cdot\displaystyle\frac{1}{\mu^{2}}\right\\}\
=\ 0$

As in Section 6, we have taken the ‘temperature’ to be the
$\mu=\sqrt{-k^{2}}$. Therefore, we conclude that in our framework, at 1-loop,
the bare coupling constant in QED is $e_{{}_{0}}=\sqrt{4\pi\alpha}=e$, which
is a finite number.

##  8 Conclusion

Figure 8: The rate of change of $\alpha_{{}_{Gibbs}}$ with
$\sqrt{-k^{2}}/m_{{}_{e}}$ is shown in the main figure. The behavior of the
derivatives at large $\sqrt{-k^{2}}/m_{{}_{e}}$ is resolved by plotting
$-\log_{10}\left[\,\displaystyle\frac{\partial\alpha_{{}_{Gibbs}}}{\partial\sqrt{-k^{2}}/m_{{}_{e}}}\,\right]$
against $\sqrt{-k^{2}}/m_{{}_{e}}$ in the inset.

As shown in Figure 8 the derivative of $\alpha_{{}_{Gibbs}}$ with respect to
the invariant momentum transfer scale $\mu=\sqrt{|k^{2}|}$ appears to decay at
large $\mu$, with the following caveats. First, the plot in Figure 8 pertains
to 1-loop approximation. Secondly, the plot is over a finite range, and for a
finite set of values of $k\cdot u_{{}_{{\cal C}}}=200,400,600,800$ and $1000$.

The plots in Figures 3 and 8 raise the following question: does
$\alpha_{{}_{Gibbs}}(k)$ asymptote to a finite value as
$\sqrt{-k^{2}}/m_{{}_{e}}\rightarrow\infty$? An affirmative answer to the
above question suggests that QED is $UV$-complete in the new framework.

Figure 9: Plot of $\xi(k)$.

A related question pertains to the asymptotic behavior of the function

$\displaystyle\xi(k):=\ g{{}_{p}}\,(\,{k}\,;{\cal C}_{{}_{\cal P}},{\cal
P})^{2}\,\chi(k)$ (137)

that appears in (98). Although Figure 3 implicitly shows the variation of
$\xi(k)$, for convenience we have plotted $\xi(k)$ over the range
$10\leq\sqrt{|k^{2}|}/m_{{}_{e}}\leq 10^{4}$ in Figure 9. The plot in Figure 9
raises the question: is $\xi(k)$ bounded above by $(4\pi\alpha)^{-1}$? An
affirmative answer to the above question implies that $QED$ does not have the
infamous Landau pole in the new framework. Further, it will also establish
that in the new framework $QED$ is not trivial [43, 59].

The implications of the new framework to the renormalization group and
nonrenormalizable theories will be presented elsewhere.

##  9 Acknowledgment

I thank Alan Guth, Jeffrey Goldstone and Samir Mathur for helpful
conversations.

##  Appendix A Appendix: Grand Canonical Distribution

In this appendix we summarize the salient details of the Grand Canonical
Distribution, derived by Josiah Willard Gibbs [27, 52]. Consider a system
${\cal S}$ that is in thermal and diffusive contact with a reservoir ${\cal
R}$. We assume that ${\cal S}$ and ${\cal R}$ have fixed volumes $V_{{}_{{\cal
S}}}$ and $V_{{}_{{\cal R}}}$ and that the system-reservoir complex ${\cal C}$
forms a closed system. Further, assume that the complex has a very large but
fixed number $N$ of particles of a single species, fixed energy $E$ and fixed
volume $V$. The set of microstates that the complex can inhabit—called the set
of accessible microstates—will be denoted $\Omega(E,N,V)$. $W=|\Omega(E,N,V)|$
denotes the cardinality of $\Omega(E,N,V)$. The accessible microstates of
${\cal C}$ will be denoted ${\cal M}^{({\cal C})}_{{}_{1}},\ldots,{\cal
M}^{({\cal C})}_{{}_{W}}$.

The probability of finding the complex ${\cal C}$ in microstate ${\cal
M}_{k}^{({\cal C})}$ at time $t$ will be denoted $P_{k}(t)$, for
$k=1,\ldots,W$. We assume that over a time interval $T_{{}_{{\cal C}}}$—called
the relaxation time—the complex reaches equilibrium, owing to the interactions
among its particles (modes). Thereafter, $P_{k}(t)$ becomes time-independent
with $P_{k}(t)=1/W$ for $k=1,2\ldots W$. That is, after the system reaches
equilibrium we may assume that it obeys the fundamental postulate of
statistical mechanics, also called the postulate of equal a priori
probabilities 424242 Let $H:=\sum_{k=1}^{W}P_{k}(t)\ln P_{k}(t)$. Then
Boltzmann’s $H$-theorem asserts that under fairly general assumptions
$\displaystyle\frac{dH}{dt}\leq 0$ for a closed system, with equality
occurring when $P_{k}(t)=1/W$. In other words, the $H$ function for a closed
system decreases until the system reaches a configuration in which the
postulate of equal a priori probabilities holds. We assume that the
assumptions used in the $H$-theorem hold. .

Consider the set of microstates ${\cal A}_{{}_{k}}$ of the complex in which
the system is in a definite (single) microstate ${\cal M}^{({\cal
S})}_{{}_{k}}(E_{{}_{S}},N_{{}_{S}},V_{{}_{S}})$ with energy $E_{{}_{S}}$ and
particle number $N_{{}_{S}}$. The subscript $k$ specifies the single
microstate of interest among possibly multiple microstates of the system in
all of which the system has energy $E_{{}_{S}}$, particle number $N_{{}_{S}}$
and volume $V_{{}_{S}}$. We will assume that $|{\cal A}_{{}_{k}}|$ is
independent of $k$–that is the reservoir has the same number of accessible
microstates corresponding to each of the degenerate microstates of the system
in which the system has energy $E_{{}_{S}}$, particle number $N_{{}_{S}}$ and
volume $V_{{}_{S}}$. We observe that $|{\cal
A}_{{}_{k}}|=W_{{}_{R}}(E-E_{{}_{S}},N-N_{{}_{S}},V_{{}_{R}})$, where
$W_{{}_{R}}(E-E_{{}_{S}},N-N_{{}_{S}},V_{{}_{R}})$ is the number of
microstates of the reservoir in which the reservoir has energy $E-E_{{}_{S}}$,
particle number $N-N_{{}_{S}}$ and volume $V_{{}_{R}}$, with the system in a
fixed microstate ${\cal M}^{({\cal
S})}_{{}_{k}}(E_{{}_{S}},N_{{}_{S}},V_{{}_{S}})$.

Let $\sigma_{{}_{R}}=\log W_{{}_{R}}$. Expanding $\sigma_{{}_{R}}$ in Taylor
series we have

$\displaystyle\sigma_{{}_{R}}(E-E_{{}_{S}},N-N_{{}_{S}},V_{{}_{R}})$
$\displaystyle=$
$\displaystyle\sigma_{{}_{R}}(E,N,V_{{}_{R}})-\displaystyle\frac{\partial\sigma_{{}_{R}}(E,N,V_{{}_{R}})}{\partial
E}\cdot
E_{{}_{S}}-\displaystyle\frac{\partial\sigma_{{}_{R}}(E,N,V_{{}_{R}})}{\partial
N}\cdot N_{{}_{S}}+\ldots$ $\displaystyle:=$
$\displaystyle\sigma_{{}_{R}}(E,N,V_{{}_{R}})-\displaystyle\frac{E_{{}_{S}}}{\tau}+\displaystyle\frac{\mu
N_{{}_{S}}}{\tau}+\ldots$

where the fundamental temperature434343The fundamental temperature, which we
use in the discussion hereafter, has the units of energy. It is related to the
temperature in Kelvin scale by the Boltzman constant as $\tau=k_{{}_{B}}T$,
where $k_{{}_{B}}$ is the Boltzman constant and $T$ the temperature in Kelvin
scale. $\tau$ and chemical potential $\mu$ of the reservoir are defined by the
relations $1/\tau:=\partial\sigma_{{}_{R}}(E,N,V_{{}_{R}})/\partial E$ and
$\mu/\tau:=-\partial\sigma_{{}_{R}}(E,N,V_{{}_{R}})/\partial N$. Then, the
probability of finding the system in a definite state ${\cal M}^{({\cal
S})}_{{}_{k}}(E_{{}_{S}},N_{{}_{S}},V_{{}_{S}})$ at time $t>T_{{}_{{\cal
C}}}$, denoted $Prob({\cal M}^{({\cal S})}(t)={\cal M}^{({\cal
S})}_{{}_{k}}\left(E_{{}_{S}},N_{{}_{S}},V_{{}_{S}}\right)),$ is

$\displaystyle Prob\left[\,{\cal M}^{({\cal S})}(t)\ =\ {\cal M}^{({\cal
S})}_{{}_{k}}\left(E_{{}_{S}},N_{{}_{S}},V_{{}_{S}}\right)\,\right]\hskip
216.81pt$ $\displaystyle=$
$\displaystyle\displaystyle\frac{e^{{}^{\sigma_{{}_{R}}(E-E_{{}_{S}},N-N_{{}_{S}},V_{{}_{R}})}}}{W}\approx\displaystyle\frac{W_{{}_{R}}(E,N,V_{{}_{R}})}{W}\cdot
e^{{}^{-\left(E_{{}_{S}}-\mu N_{{}_{S}}\right)/\tau}}$ (138) $\displaystyle:=$
$\displaystyle{\cal Z}e^{{}^{-\left(E_{{}_{S}}-\mu
N_{{}_{S}}\right)/\tau}},\qquad{\cal Z}:=W_{{}_{R}}(E,N,V_{{}_{R}})/W$

The approximation is a good estimate if $E_{{}_{S}}\ll E$ and $N_{{}_{S}}\ll
N$. The term $e^{{}^{-\left(E_{{}_{S}}-\mu N_{{}_{S}}\right)/\tau}}$ is called
the Gibbs factor in the literature [52]. The probability distribution shown in
(138) is called the Grand Canonical Distribution.

The Gibbs factor can be understood as follows. The energy and particle number
in the system constantly fluctuate owing to the exchange of energy and
particles between the system and the reservoir. The probability of a
fluctuation that puts the system in a definite state with energy $E_{{}_{S}}$
and particle number $N_{{}_{S}}$ is proportional to the Gibbs factor.

The argument leading to (138) does not depend on the nature of the microscopic
interactions among the particles (modes) within the complex. Hence, (138) is
broadly applicable to any system that is in thermal and diffusive contact with
its reservoir. In particular, the arguments leading to the Grand Canonical
Distribution must be broadly applicable to a quantum field (system) that is in
thermal and diffusive contact with its reservoir, and, the probability of a
fluctuation of a quantum field, as viewed in a fixed Lorentz frame, must be
exponentially suppressed by the energy of the fluctuation, as shown in (138).

A noteworthy aspect is that the Grand Canonical Distribution is based on the
fundamental postulate of statistical mechanics, which becomes applicable only
when the constituent particles (modes) of the system-reservoir complex
interact to drive the complex towards equilibrium. Thus the Grand Canonical
Distribution can be regarded as an emergent law that arises due to the
collective interactions among the constituent modes of a complex. While the
current framework of quantum field theory is based on the details of the
microscopic interactions among modes, it is not constrained by the emergent
laws—such as the Grand Canonical Distribution— that arise as a consequence of
the interactions among modes.

##  Appendix B Appendix: Asymptotic Operators of Interacting Fields

In this appendix we describe the construction of single-particle creation and
annihilation operators for an interacting spinor field $\psi(x)$. Such
construction is well-known in the absence of Gibbs factors; for example, see
[86]. The following discussion describes the construction in the presence of
Gibbs factors, while providing more rigorous proofs of the arguments outlined
in [86]. The construction is then used to establish the modified LSZ reduction
formula for QED that includes the Gibbs factors. While the following
discussion is focused on QED the argument can be adapted to apply to all
quantum fields.

###  B.1 Single-particle Creation and Annihilation Operators

From (LABEL:eqn601) the creation operator for free electrons in frame ${\cal
F}$ for a scattering process ${\cal P}$ is

$\displaystyle{\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}(\vec{k},s)\ =\ \displaystyle\frac{1}{\ g{{}_{e}}\,(\,{k}\,;{\cal
F},{\cal P})}\int d\,\vec{x}\ e^{\displaystyle-i\uline{k}x}\
\bar{\psi}_{{}_{\mbox{\scriptsize free}}}(x)\ \gamma^{0}\ u(\uline{k},s)$
(139)

${\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize free}}}(\vec{k},s)$ is time-
independent. $\uline{k}$ denotes an on-shell 4-momentum with
$\uline{k}^{2}=m_{{}_{e}}^{2}$, where $m_{{}_{e}}$ is the physical mass of the
electron. Acting on the vacuum of the free-field of mass $m_{{}_{e}}$, denoted
$\left.\left|\,0_{{}_{F}}\,\right\rangle\right.$, the creation operators
${\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize free}}}(\vec{k},s)$ and
${\bf\hat{b}}^{\dagger}_{{}_{\mbox{\scriptsize free}}}(\vec{k},s)$ create
positive-energy and negative-energy eigenstates of on-shell 4-momentum
$\uline{k}$ and $z$-spin444444The discrete label $s$ represents the
$z$-component of spin, called $z$-spin hereafter. $s$, denoted
$\left.\left|\,\uline{k},s;+\,\right\rangle\right.$ and
$\left.\left|\,\uline{k},s;-\,\right\rangle\right.$ respectively. Thus,

$\displaystyle\left.\left|\,\uline{k},s;+\,\right\rangle\right.$
$\displaystyle:=$ $\displaystyle{\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}(\vec{k},s)\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.;\qquad\left.\left|\,\uline{k},s;-\,\right\rangle\right.\
:=\ {\bf\hat{b}}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}(\vec{k},s)\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.$ (140)

Our objective, in the next two sections, is to use the interacting field
$\psi(x)$ to construct operators that produce single-particle states
$\left.\left|\,\uline{k},s;+\,\right\rangle\right.$ and
$\left.\left|\,\uline{k},s;-\,\right\rangle\right.$, by acting on not the
vacuum $\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.$ of the free-field
but on the vacuum $\left.\left|\,0\,\right\rangle\right.$ of the interacting
field.

###  B.2 Asymptotic Operators

We define an operator that has the same form as in (139) except for the
substitution of the free-field $\psi_{{}_{\mbox{\scriptsize free}}}(x)$ with
the interacting field $\psi(x)$ as

$\displaystyle{\hat{\bf a}^{\dagger}(\vec{k},s,t)}$ $\displaystyle:=$
$\displaystyle\displaystyle\frac{1}{\ g{{}_{e}}\,(\,{\uline{\vec{k}}}\,;{\cal
F},{\cal P})}\int d\,\vec{x}\ e^{\displaystyle-i\uline{\vec{k}}x}\
\bar{\psi}(x)\ \gamma^{0}\ u(\uline{\vec{k}},s)$ (141)

The operator ${\bf\hat{a}}^{\dagger}(\vec{k},s,t)$ is time-dependent.
$\uline{\vec{k}}$ is on-shell with $(\uline{\vec{k}})^{2}=m_{e}^{2}$, where
$m_{e}$ is the physical mass of the electron.

We show below that as $t\rightarrow\pm\infty$ the operator defined in (141)
acting on the interacting vacuum $\left.\left|\,0\,\right\rangle\right.$
produces a state that is proportional to the 1-particle state
$\left.\left|\,\uline{\vec{k}},\,s,\,+\,\right\rangle\right.$ as shown in
(165).

For $n\geq 2$, if $\left.\left|\,P;n\,\right\rangle\right.$ is an $n$-particle
eigenstate of the momentum operator $\hat{P}$ of the interacting theory with
$\hat{P}\left.\left|\,P;n\,\right\rangle\right.\ =\
P\left.\left|\,P;n\,\right\rangle\right.$, then $P^{2}\geq 4m_{{}_{e}}^{2}$.
The following Lemma is a more rigorous version of the argument outlined in
[86].

###### Lemma 1

For $n\geq 2$, let $\left.\left|\,P;n\,\right\rangle\right.$ denote an
$n$-particle eigenstate of $\hat{P}$, the 4-momentum operator 454545 In
analogy with the normal ordering for free fields, the 4-momentum operator of
the interacting theory $\hat{P}$ is defined as
$\hat{P}^{\mu}:=\hat{P}_{cl}^{\mu}-\left\langle\,0\left|\,\hat{P}_{c}^{\mu}\,\right|0\,\right\rangle$,
where $\hat{P}_{cl}^{\mu}:=\int d^{3}x\ T^{\mu 0}$, where $T$ is the energy-
momentum tensor constructed from the field. We assume that the vacuum of the
interacting theory is a simultaneous eigenstate of $\hat{P}^{\mu}_{cl}$ and
$(\hat{P}_{cl})^{\mu}(\hat{P}_{cl})_{\mu}$ operators with the lowest
eigenvalue of the $(\hat{P}_{cl})^{\mu}(\hat{P}_{cl})_{\mu}$ operator. Since
$\left.\left|\,0\,\right\rangle\right.$ is assumed to be an eigenstate of
$\hat{P}^{\mu}_{cl}$, say
$\hat{P}^{\mu}_{cl}\left.\left|\,0\,\right\rangle\right.\ =\
p^{\mu}\left.\left|\,0\,\right\rangle\right.$, we have
$\hat{P}^{\mu}\left.\left|\,0\,\right\rangle\right.\ =\ 0.$ of the interacting
field, with
$\hat{P}\left.\left|\,P;n\,\right\rangle\right.=P\left.\left|\,P;n\,\right\rangle\right.$.
Then

$\displaystyle\lim_{t\rightarrow\pm\infty}\left.\left\langle\,P;n\,\right.\right|{\hat{\bf
a}^{\dagger}(\vec{k},s,t)}\left.\left|\,0\,\right\rangle\right.$
$\displaystyle=$ $\displaystyle 0$

where ${\hat{\bf a}^{\dagger}(\vec{k},s,t)}$ is the operator in the Heisenberg
picture, defined in (141), and $\left.\left|\,0\,\right\rangle\right.$ is the
vacuum state of the interacting field defined as
$\hat{P}\left.\left|\,0\,\right\rangle\right.=0$.

Proof: We define a selector function $S:\mathbb{R}^{3}\rightarrow\mathbb{R}$
as follows.

$\displaystyle S(\vec{x}\ ;\ \vec{x}_{{}_{0}},\ \epsilon)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle\frac{1}{\left({4\pi\epsilon^{3}}/{3}\right)},&|\vec{x}-\vec{x}_{{}_{0}}\
|\leq\epsilon\\\ 0,&|\vec{x}-\vec{x}_{{}_{0}}\ |>\epsilon>0\end{array}\right.$

Let $P=(p^{0},\vec{p}\,)$ and let $P^{2}=M^{2}\geq 4m_{{}_{e}}^{2}$. For
$\epsilon>0$, consider the following superpositions of eigenstates of
$\hat{P}$, denoted $\left.\left|\,\psi\,;\
\vec{p}\,,\,\epsilon\,\right\rangle\right.$.

$\displaystyle\left.\left|\,\psi\,;\
\vec{p}\,,\,\epsilon\,\right\rangle\right.$ $\displaystyle:=$
$\displaystyle\int d\,\vec{q}\ \ S(\vec{q}\,;\,\vec{p}\,,\,\epsilon)\
\left.\left|\,Q;n\,\right\rangle\right.,\qquad
Q=\left(\left[\,M^{2}+|\,\vec{q}\,|\,^{2}\,\right]^{1/2},\ \vec{q}\,\right),$
$\displaystyle\hat{A}(\vec{k},\,s,\,t;\epsilon)$ $\displaystyle:=$
$\displaystyle\int d\,\vec{l}\ \ S(\vec{l}\,;\,\vec{k}\,,\,\epsilon)\ \
{\bf\hat{a}}^{\dagger}(\vec{l},s,t)$ (143)

Using (141), noting that $\psi(x)=e^{\displaystyle i\hat{P}x}\ \psi(0)\
e^{\displaystyle-i\hat{P}x}$ and setting
$\omega(\vec{q};M):=\sqrt{|\,\vec{q}\,|\,^{2}+M^{2}}$ we have

$\displaystyle\left.\left\langle\,Q;n\,\right.\right|{\bf\hat{a}}^{\dagger}(\vec{l},s,t)\left.\left|\,0\,\right\rangle\right.\hskip
433.62pt$

$\displaystyle=$ $\displaystyle\displaystyle\frac{1}{\
g{{}_{e}}\,(\,{\uline{\vec{l}}}\,;{\cal F},{\cal P})}\int d\,\vec{x}\
e^{\displaystyle-i\underline{\uline{\vec{l}}}x}\left.\left\langle\,Q;n\,\right.\right|e^{\displaystyle
i\hat{P}x}\bar{\psi}(0)\left.\left|\,0\,\right\rangle\right.\,\gamma^{0}\,u(\uline{\vec{l}},s)$
(144) $\displaystyle=$ $\displaystyle\delta^{3}(\vec{l}-\vec{q})\
e^{\displaystyle i(\omega(\vec{q}\,;\,M)-\omega(\vec{q}\,;\,m_{e}))t}\
\left[\,\frac{(2\pi)^{3}\,\left.\left\langle\,Q;n\,\right.\right|\bar{\psi}(0)\left.\left|\,0\,\right\rangle\right.\,\gamma^{0}\,u(\uline{\vec{q}},s)}{\
g{{}_{e}}\,(\,{\uline{\vec{q}}}\,;{\cal F},{\cal P})}\,\right]$
$\displaystyle:=$ $\displaystyle\delta^{3}(\vec{l}-\vec{q})\ e^{\displaystyle
i(\omega(\vec{q}\,;\,M)-\omega(\vec{q}\,;\,m_{e}))t}\ W(\uline{\vec{q}},s)$

In the second step above, $\omega(\vec{l};m_{{}_{e}})$ has been replaced by
$\omega(\vec{q};m_{{}_{e}})$ noting the presence of the
$\delta^{3}(\vec{l}-\vec{q})$. Also we note that
$\uline{\vec{q}}=(\omega(\vec{q},m_{e}),\vec{q})$ and $W(\uline{\vec{q}},s)$
is the abbreviation of the expression within brackets in the preceding
expression. We assume the calculation pertains to a scattering process ${\cal
P}$ in frame ${\cal F}$.

Using (143) and (144) and performing the integration with respect to $\vec{l}$
we have

$\displaystyle\left.\left\langle\,\psi\,;\,\vec{p},\,\epsilon\,\right.\right|\hat{A}(\vec{k},\,s,\,t;\epsilon)\left.\left|\,0\,\right\rangle\right.=\int
d\,\vec{q}\cdot S(\vec{q}\,;\,\vec{p}\,,\,\epsilon)\cdot
S(\vec{q}\,;\,\vec{k}\,,\,\epsilon)\cdot e^{\displaystyle
i(\omega(\vec{q}\,;\,M)-\omega(\vec{q}\,;\,m_{e}))t}\ W(\uline{q},s)$

(145)

The presence of the selector functions inside the integral ensures that the
integrand is nonzero only in the intersection of the two balls of radius
$\epsilon$ centered at $\vec{p}$ and $\vec{k}$. We note that
$W(\uline{\vec{q}},s)$ is finite in the region where the product of the
selector functions is nonzero; the Gibbs factor, is nonzero since it has a
zero only at $|\vec{q}\ |\rightarrow\infty$. The interacting field
$\bar{\psi}(0)$ acting on the interacting vacuum
$\left.\left|\,0\,\right\rangle\right.$, is expected to produce single-
particle states of the interacting theory as well as multi-particle states as
well. We assume that
$\left.\left\langle\,Q;n\,\right.\right|\bar{\psi}(0)\left.\left|\,0\,\right\rangle\right.$,
the coefficient of the multi-particle state
$\left.\left|\,Q;n\,\right\rangle\right.$ when
$\bar{\psi}(0)\left.\left|\,0\,\right\rangle\right.$ is expressed as a linear
combination of the eigenstates of $\hat{P}$, is finite.

If $\vec{p}\neq\vec{k}$, then we can make the product of selector functions
vanish by choosing $\epsilon<\displaystyle\frac{|\vec{k}-\vec{p}\ |}{2}$, and
hence the whole integral vanishes. Since $\lim_{\epsilon\rightarrow
0}S(\vec{q};\vec{p},\epsilon)\ =\ \delta^{3}(\vec{q}-\vec{p})$, we have
$\lim_{\epsilon\rightarrow 0}\left.\left|\,\psi\,;\
\vec{p}\,,\,\epsilon\,\right\rangle\right.\ =\
\left.\left|\,P;n\,\right\rangle\right.$, $\lim_{\epsilon\rightarrow
0}\hat{A}(\vec{k},\,s,\,t;\epsilon)\ =\ {\bf\hat{a}}^{\dagger}(\vec{q},s,t)$.
Taking the limit $\epsilon\rightarrow 0$, the claim in the Lemma follows.

Therefore, we will assume that $\vec{p}\ =\ \vec{k}$, and perform the integral
(145) in spherical coordinates $(\rho,\theta,\varphi)$ ($\rho=|\,\vec{q}\,\
|$). We set

$\displaystyle h(\rho;\vec{k},\epsilon)$ $\displaystyle:=$
$\displaystyle\rho^{2}\int\ d\theta\cdot\sin\theta\cdot d\varphi\cdot
S((\rho,\theta,\phi)\,;\,\vec{k}\,,\,\epsilon)^{2}\cdot
W((\rho,\theta,\phi),s)$

Owing to the presence of the selector function,
$h(\rho;\vec{k},\epsilon)\equiv 0$ for $\rho>|\,\vec{k}\,\
|+\epsilon:=\rho_{{}_{{max}}}$. Then the integral (145) can be rewritten as

$\displaystyle\left.\left\langle\,\psi\,;\,\vec{p},\,\epsilon\,\right.\right|\hat{A}(\vec{k},\,s,\,t;\epsilon)\left.\left|\,0\,\right\rangle\right.$
$\displaystyle=$ $\displaystyle\int_{0}^{\rho_{{}_{{max}}}}d\rho\cdot
h(\rho;\,\vec{k},\,\epsilon)\cdot e^{\displaystyle i\,b(\rho;M,m_{e})\,t}$
(146)

where
$b(\rho,M,m_{e}):=\left[\,\rho^{2}+M^{2}\,\right]^{1/2}-\left[\,\rho^{2}+m_{e}^{2}\,\right]^{1/2}$.
$b(\rho,M,m_{e})$ is a monotonically decreasing function of $\rho$ for $M\geq
2\,m_{e}$. Therefore, if we set $z=b(\rho,M,m_{e})$, then $\rho$ can be
written as function of $z$ as $\rho=c(z)$. Further, at fixed $\epsilon>0$,
$\displaystyle\frac{h(\rho;\vec{k},\epsilon)}{b^{\prime}(\rho,M,m_{e})}$ is
bounded over the interval [$0,\rho_{{}_{{max}}}$]. That is, we can find a
$0<\chi(\epsilon)<\infty$ such that
$\displaystyle\frac{|h(\rho;\vec{k})|}{|\,b^{\prime}(\rho,M,m_{e})|}<\chi(\epsilon)$
for $0\leq\rho\leq\rho_{{}_{{max}}}$. Therefore,

$\displaystyle\left|\left.\left\langle\,\psi\,;\,\vec{p},\,\epsilon\,\right.\right|\hat{A}(\vec{k},\,s,\,t;\epsilon)\left.\left|\,0\,\right\rangle\right.\right|$
$\displaystyle\leq$
$\displaystyle\chi(\epsilon)\cdot\left|\,\int^{b(\rho_{{}_{{max}}},M,m_{e})}_{b(0,M,m_{e})}\
e^{\displaystyle izt}\cdot dz\right|$ (147) $\displaystyle=$
$\displaystyle\chi(\epsilon)\cdot\displaystyle\frac{\left|\,e^{\displaystyle
ib(\rho_{{}_{{max}}},M,m_{e})t}-e^{\displaystyle
ib(0,M,m_{e})t}\right|}{|\,t\,|}\hskip 25.29494pt$

From (147) we see that

$\displaystyle\lim_{t\rightarrow\pm\infty}\left|\left.\left\langle\,\psi\,;\,\vec{p},\,\epsilon\,\right.\right|\hat{A}(\vec{k},\,s,\,t;\epsilon)\left.\left|\,0\,\right\rangle\right.\right|$
$\displaystyle=$ $\displaystyle 0$

for every $\epsilon>0$. Taking the limit $\epsilon\rightarrow 0$ we then
obtain

$\displaystyle\lim_{\epsilon\rightarrow
0}\left\\{\lim_{t\rightarrow\pm\infty}\left|\left.\left\langle\,\psi\,;\,\vec{p},\,\epsilon\,\right.\right|\hat{A}(\vec{k},\,s,\,t;\epsilon)\left.\left|\,0\,\right\rangle\right.\right|\right\\}$
$\displaystyle=$
$\displaystyle\lim_{t\rightarrow\pm\infty}\left\\{\lim_{\epsilon\rightarrow
0}\left|\left.\left\langle\,\psi\,;\,\vec{p},\,\epsilon\,\right.\right|\hat{A}(\vec{k},\,s,\,t;\epsilon)\left.\left|\,0\,\right\rangle\right.\right|\right\\}$
$\displaystyle=$
$\displaystyle\lim_{t\rightarrow\pm\infty}\left.\left\langle\,P;n\,\right.\right|{\hat{\bf
a}^{\dagger}(\vec{k},s,t)}\left.\left|\,0\,\right\rangle\right.\ =\ 0$

as claimed.

Thus, the state
${\bf\hat{a}}^{\dagger}(\vec{k},s,\pm\infty)\left.\left|\,0\,\right\rangle\right.$
has a vanishing overlap with every multi-particle state
$\left.\left|\,P;n\,\right\rangle\right.$, $n\geq 2$; we have used the
abbreviation ${\bf\hat{a}}^{\dagger}(\vec{k},s,\pm\infty)\ :=\
\lim_{t\rightarrow\pm\infty}{\bf\hat{a}}^{\dagger}(\vec{k},s,t)$.

Similarly, one can prove the following Lemma 2.

###### Lemma 2

For $n\geq 2$, let $\left.\left|\,P;n\,\right\rangle\right.$ denote an
$n$-particle eigenstate of $\hat{P}$, the 4-momentum operator of the
interacting field, with
$\hat{P}\left.\left|\,P;n\,\right\rangle\right.=P\left.\left|\,P;n\,\right\rangle\right.$.
Then

$\displaystyle\lim_{t\rightarrow\pm\infty}\left.\left\langle\,P;n\,\right.\right|{\hat{\bf
a}(\vec{k},s,t)}\left.\left|\,0\,\right\rangle\right.$ $\displaystyle=$
$\displaystyle 0$

where ${\hat{\bf a}(\vec{k},s,t)}$ is the operator in the Heisenberg picture,
is defined as

$\displaystyle{\hat{\bf a}(\vec{k},s,t)}\ =\ \displaystyle\frac{1}{\
g{{}_{f}}\,(\,{\uline{\vec{k}}}\,;{\cal F},{\cal P})}\int d\,\vec{x}\
e^{\displaystyle i\uline{\vec{k}}x}\ \bar{u}(\vec{k},s)\ \gamma^{0}\ \psi(x)$

and $\left.\left|\,0\,\right\rangle\right.$ is the vacuum state of the
interacting field defined as $\hat{P}\left.\left|\,0\,\right\rangle\right.=0$.

Proof: The argument used in the proof of Lemma 1 can be applied, essentially
unchanged, except for the following small modification. In (144) we need to
replace the definition of $W(\uline{\vec{q}},s)$ with the following
definition,

$\displaystyle W(\uline{\vec{q}},s)$ $\displaystyle:=$
$\displaystyle\left[\,\frac{(2\pi)^{3}\,\bar{u}(\uline{\vec{q}},s)\gamma^{0}\left.\left\langle\,Q;n\,\right.\right|\psi(0)\left.\left|\,0\,\right\rangle\right.\,}{\
g{{}_{e}}\,(\,{\uline{\vec{q}}}\,;{\cal F},{\cal P})}\,\right]$

Next, we show that the the state
${\bf\hat{a}}^{\dagger}(\vec{k},s,\pm\infty)\left.\left|\,0\,\right\rangle\right.$
has a vanishing overlap with the vacuum
$\left.\left|\,0\,\right\rangle\right.$ of the interacting theory.

###### Lemma 3

The operator ${\bf\hat{a}}^{\dagger}(\vec{k},\,s,\,t)$, defined in (141), has
a vanishing vacuum expectation value with respect to the vacuum of the
interacting theory. Specifically,

$\displaystyle\lim_{t\rightarrow\pm\infty}\left\langle\,0\left|\,{\bf\hat{a}}^{\dagger}(\vec{k},\,s,\,t)\,\right|0\,\right\rangle$
$\displaystyle=$ $\displaystyle 0$

Proof: Using (141), we have

$\displaystyle\left\langle\,0\left|\,{\bf\hat{a}}^{\dagger}(\vec{k},\,s,\,t)\,\right|0\,\right\rangle$
$\displaystyle=$ $\displaystyle\displaystyle\frac{1}{\
g{{}_{e}}\,(\,{\uline{\vec{k}}}\,;{\cal F},{\cal P})}\int d\,\vec{x}\
e^{\displaystyle-i\uline{\vec{k}}x}\
\left\langle\,0\left|\,\bar{\psi}(x)\,\right|0\,\right\rangle\ \gamma^{0}\
u(\uline{\vec{k}},s)$ $\displaystyle=$ $\displaystyle\displaystyle\frac{1}{\
g{{}_{e}}\,(\,{\uline{\vec{k}}}\,;{\cal F},{\cal P})}\int d\,\vec{x}\
e^{\displaystyle-i\uline{\vec{k}}x}\
\left\langle\,0\left|\,\bar{\psi}(0)\,\right|0\,\right\rangle\ \gamma^{0}\
u(\uline{\vec{k}},s)$ $\displaystyle=$ $\displaystyle 0$

The last step follows from
$\left\langle\,0\left|\,\psi(0)\,\right|0\,\right\rangle=0$, which is a
consequence of the Lorentz invariance of the vacuum
$\left.\left|\,0\,\right\rangle\right.$ of the interacting theory (see [86,
Equation 41.22]).

Lemmas 1 and 3 show that the state
${\bf\hat{a}}^{\dagger}(\vec{k},s,\pm\infty)\left.\left|\,0\,\right\rangle\right.$
has a vanishing overlap with $n-$particle states where $n\neq 1,n\geq 0$. On
the other hand, as we show in the following lemma, the state
${\bf\hat{a}}^{\dagger}(\vec{k},s,\pm\infty)\left.\left|\,0\,\right\rangle\right.$
has a nonvanishing overlap with exactly one single-particle state.

###### Lemma 4

Let $\left.\left|\,\uline{p}\,,r;+\,\right\rangle\right.$ denote the time-
independent positive-energy single-particle eigenstate of the 4-momentum
operator $\hat{P}$ of the interacting field of an electron, in the Heisenberg
picture, with
$\hat{P}\left.\left|\,\uline{p}\,,r;+\,\right\rangle\right.=\uline{p}\left.\left|\,\uline{p}\,,r;+\,\right\rangle\right.$.
$r$ labels the $z$-spin. Then, noting that
$\uline{p}=\left(\sqrt{m_{{}_{e}}^{2}+|\,\vec{p}\,|^{2}}\,,\vec{p}\,\right)$,
in Lorentz frame ${\cal F}$, for scattering process ${\cal P}$464646 The
relation
${\bf\hat{a}}^{\dagger}(\vec{k},s,+\infty)-{\bf\hat{a}}^{\dagger}(\vec{k},s,-\infty)\
=\ -i\int
d^{4}xe^{\displaystyle-i\uline{k}x}\bar{\psi}(i\overleftarrow{\cancel{\partial}}+m)u(\uline{k},s)$
is not in conflict with (148) since  $\displaystyle\int
d^{4}xe^{\displaystyle-i\uline{k}x}\left.\left\langle\,\uline{p}\,,r;+\,\right.\right|\bar{\psi}(x)\left.\left|\,0\,\right\rangle\right.(i\overleftarrow{\cancel{\partial}}+m_{{}_{e}})u(\uline{k},s)\hskip
289.07999pt$ $\displaystyle=$ $\displaystyle\int d^{4}x\,e^{\displaystyle
i(\uline{p}-\uline{k})x}\left.\left\langle\,\uline{p}\,,r;+\,\right.\right|\,\bar{\psi}(0)\left.\left|\,0\,\right\rangle\right.(i\overleftarrow{\cancel{\partial}}+m_{{}_{e}})\,u(\vec{k},s)$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}dx^{0}\,W(\uline{p},r;+)\,(-\uline{p}\\!\\!\\!/+m_{{}_{e}})\,u(\uline{p},s)\,(2\pi)^{3}\,\delta^{3}(\vec{k}-\vec{p})$
$\displaystyle=$ $\displaystyle 0$ We have used the abbreviation
$W(\uline{p},r;+):=\left.\left\langle\,\uline{p}\,,r;+\,\right.\right|\bar{\psi}(x)\left.\left|\,0\,\right\rangle\right.$.
The last equality follows from  $(p\\!\\!\\!/-m_{{}_{e}})u(\uline{p},s)\ =\
0$. The $\delta^{3}(\vec{k}-\vec{p})$ factor enables us to replace
$u(\uline{k},s)$ with $u(\uline{p},s)$. ,

$\displaystyle\left.\left\langle\,\uline{p}\,,r;+\,\right.\right|\
{\bf\hat{a}}^{\dagger}(\vec{k},s,\pm\infty)\left.\left|\,0\,\right\rangle\right.$
$\displaystyle=$ $\displaystyle\left[\,\frac{\alpha(\uline{p},s;{\cal F})}{\
g{{}_{e}}\,(\,{\uline{p}}\,;{\cal F},{\cal P})}\,\right](2\pi)^{3}\cdot
2\omega(\vec{k}\,)\cdot\delta^{3}(\vec{k}-\vec{p})\cdot\delta_{r,s},\hskip
21.68121pt$ (148)

where $\quad\alpha(\uline{p},s;{\cal F})\in\mathbb{C}$.

Proof: From (141) we have

$\displaystyle A_{+}(\uline{p},r;\uline{k},s;t)$ $\displaystyle:=$
$\displaystyle\left.\left\langle\,\uline{p}\,,r;+\,\right.\right|{\bf\hat{a}}^{\dagger}(\vec{k},s,t)\left.\left|\,0\,\right\rangle\right.$
$\displaystyle=$ $\displaystyle\displaystyle\frac{1}{\
g{{}_{e}}\,(\,{\uline{k}}\,;{\cal F},{\cal P})}\int
d\,\vec{x}\,\left.\left\langle\,\uline{p}\,,r;+\,\right.\right|\psi^{\dagger}(\vec{x},t)\left.\left|\,0\,\right\rangle\right.e^{\displaystyle-i\uline{k}x}\,u(\uline{k},s)$
$\displaystyle=$ $\displaystyle\displaystyle\frac{1}{\
g{{}_{e}}\,(\,{\uline{k}}\,;{\cal F},{\cal P})}\int
d\,\vec{x}\,\left.\left\langle\,\uline{p}\,,r;+\,\right.\right|\psi^{\dagger}(0)\left.\left|\,0\,\right\rangle\right.e^{\displaystyle
i(\underline{p}-\uline{k})x}\,u(\uline{k},s)$ $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{\ g{{}_{e}}\,(\,{\uline{p}}\,;{\cal
F},{\cal
P})}\left\\{(2\pi)^{3}\,\delta^{3}(\vec{k}-\vec{p})\,\left.\left\langle\,\uline{p}\,,r;+\,\right.\right|\psi^{\dagger}(0)\left.\left|\,0\,\right\rangle\right.\,u(\uline{k},s)\right\\}$
(149)

From (149) we conclude that the state
${\bf\hat{a}}^{\dagger}(\uline{k},s,\pm\infty)\left.\left|\,0\,\right\rangle\right.$
has a non-vanishing overlap only with a single-particle state that has
$3$-momentum $\vec{k}$.

Define

$\displaystyle W(\uline{p},r;+)$ $\displaystyle=$ $\displaystyle\left\langle
0\,\left|\,\psi(0)\ \right|\uline{p},r;+\,\right\rangle$ (150)

Under a Lorentz transformation $\Lambda:{\cal F}\rightarrow{\cal
F}\,^{\prime}$ we have $\psi^{\prime}(0)\ =\ S[\Lambda]\,\psi(0)$ and from
Lemma 7 (Appendix C),

$\displaystyle\left.\left|\,\uline{p},r;+\,\right\rangle\right._{{}_{{\cal
F}}}\ =\
\sum_{s=1}^{2}D^{\tiny(u)}_{s,r}\,\left.\left|\,\Lambda\uline{p},s;+\,\right\rangle\right._{{}_{{\cal
F}\,^{\prime}}}$ (151)

$\psi$ and $\psi^{\prime}$ are the interacting spinor fields seen by observers
at rest in ${\cal F}$ and ${\cal F}\,^{\prime}$ respectively. Inserting (151)
into (150) we get

$\displaystyle W(\uline{p},r;+)$ $\displaystyle=$
$\displaystyle\sum_{s=1}^{2}S^{-1}[\Lambda]\left.\left\langle\,0\,\right.\right|\psi^{\prime}(0)\left.\left|\,\Lambda\uline{p},s;+\,\right\rangle\right._{{}_{{\cal
O}\,^{\prime}}}D^{\tiny(u)}_{s,r}\ =\ \sum_{s=1}^{2}S^{-1}[\Lambda]\cdot
W(\Lambda\uline{p},s;+)\cdot D^{\tiny(u)}_{s,r}$

From (210) we know that the spinors transform under Lorentz transformation
$\Lambda$ as

$\displaystyle u(\uline{p},r)$ $\displaystyle=$
$\displaystyle\sum_{s=1}^{2}\,S^{-1}[\Lambda]\cdot u(\Lambda\uline{p},s)\cdot
D^{\tiny(u)}_{s,r};\qquad v(\uline{p},r)\ =\
\sum_{s=1}^{2}\,S^{-1}[\Lambda]\cdot v(\Lambda\uline{p},s)\cdot
D^{(v)}_{s,r}\qquad$

Since $u(\uline{p},r),\,v(\uline{p},r),\ r=1,2$ form a basis of the 4-spinor
space we can expand

$\displaystyle W(\uline{p},r;+)$ $\displaystyle=$
$\displaystyle\sum_{s=1}^{2}\,\left\\{\lambda_{r,s}u(\uline{p},s)+\beta_{r,s}v(\uline{p},s)\right\\}$
(154)

From the constraint that $W(\uline{p},r;+)$ must transform as in
(LABEL:eqn7002) for all proper orthochronous Lorentz transformations, and
knowing that the basis spinors transform as in (LABEL:eqn7005) under proper
orthochronous Lorentz transformations, we conclude that $\beta_{{}_{r,s}}=0$
and $\lambda_{r,s}=\delta_{r,s}\,\xi(\uline{p},r;{\cal F})$, where
$\xi(\uline{p},r;{\cal F})\in\mathbb{C}$. That is,

$\displaystyle W(\uline{p},r,+)$ $\displaystyle=$
$\displaystyle\xi(\uline{p},r;{\cal F})\cdot u(\uline{p},r)$ (155)

Inserting (155) into (149) and defining $\alpha(\uline{p},r;{\cal
F}):=\xi^{*}(\uline{p},r;{\cal F})$ establishes the claim in the Lemma.

The argument used to establish Lemma 4 can be used to establish the following
related assertions.

###### Corollary 1

Let $\left.\left|\,\uline{p}\,,r;-\,\right\rangle\right.$ denote the time-
independent negative-energy single-particle eigenstate of the 4-momentum
operator $\hat{P}$ of the interacting field of an electron, in the Heisenberg
picture, with
$\hat{P}\left.\left|\,\uline{p}\,,r;-\,\right\rangle\right.=\uline{p}\left.\left|\,\uline{p}\,,r;-\,\right\rangle\right.$.
$r$ labels the $z$-spin. Then, noting that
$\uline{p}=\left(\sqrt{m_{{}_{e}}^{2}+|\,\vec{p}\,|^{2}}\,,\vec{p}\,\right)$,
we have in Lorentz frame ${\cal F}$, for scattering process ${\cal P}$,

$\displaystyle\left.\left\langle\,\uline{p}\,,r;-\,\right.\right|\
{\bf\hat{b}}^{\dagger}(\uline{k},s,\pm\infty)\left.\left|\,0\,\right\rangle\right.$
$\displaystyle=$
$\displaystyle\displaystyle\frac{\tilde{}\alpha(\uline{p},s;{\cal
F})\cdot(2\pi)^{3}\cdot
2\omega(\vec{k}\,)\cdot\delta^{3}(\vec{k}-\vec{p})\cdot\delta_{r,s}}{\
g{{}_{e}}\,(\,{p}\,;{\cal F},{\cal
P})},\quad\tilde{}\alpha(\uline{p},s)\in\mathbb{C}$

###### Corollary 2

Let $\left.\left|\,\uline{p},r;\pm\,\right\rangle\right.$ denote the time-
independent positive-energy/negative-energy single-particle eigenstate of the
4-momentum operator $\hat{P}$ of the interacting field of an electron, in the
Heisenberg picture, with
$\hat{P}\left.\left|\,\uline{p}\,,r;\pm\,\right\rangle\right.=\uline{p}\left.\left|\,\uline{p}\,,r;\pm\,\right\rangle\right.$.
$r$ labels the $z$-spin. Then, noting that
$\uline{p}=\left(\sqrt{m_{{}_{e}}^{2}+|\,\vec{p}\,|^{2}}\,,\vec{p}\,\right)$,
in Lorentz frame ${\cal F}$, for scattering process ${\cal P}$,

$\displaystyle\left\langle\uline{p},r;-\,\left|\,{\bf\hat{a}}^{\dagger}(\uline{k},s,\pm\infty\
\right|0\,\right\rangle$ $\displaystyle=$
$\displaystyle\left\langle\uline{p},r;+\,\left|\,{\bf\hat{b}}^{\dagger}(\uline{k},s,\pm\infty\
\right|0\,\right\rangle\ =\ 0;$

###  B.3 Wavefunction Renormalization

Let $\psi$ be the interacting field of an electron in QED and
$\hat{P}_{{}_{\psi}}$ its 4-momentum operator. As is well known the spectrum
of $\hat{P}^{2}_{{}_{\psi}}$ has three sectors: an eigenvalue of 0
corresponding to the vacuum state, an eigenvalue of $m_{{}_{e}}^{2}>0$
corresponding to the single-particle states, and a continuum of eigenvalues in
the range [ $4m_{{}_{e}}^{2},\infty)$ corresponding to the multi-particle
states.

Consider

$\displaystyle\Delta(\psi)$ $\displaystyle:=$
$\displaystyle\left\langle\,0\left|\,\bar{\psi}(x)\,\psi(x)\,\right|0\,\right\rangle$

where $\left.\left|\,0\,\right\rangle\right.$ represents the vacuum of the
interacting field. Since $\bar{\psi}\psi$ is a Lorentz scalar and
$\left.\left|\,0\,\right\rangle\right.$ is invariant under Lorentz
transformation, $\Delta(\psi)$ is a Lorentz invariant. Using $\psi(x)\ =\
e^{\displaystyle
i\hat{P}_{{}_{\psi}}x}\,\psi(0)\,e^{\displaystyle-i\hat{P}_{{}_{\psi}}x}$

$\displaystyle\Delta(\psi)$ $\displaystyle=$
$\displaystyle\textstyle\sum_{{\mbox{\tiny(all)}}}\left.\left\langle\,0\,\right.\right|\bar{\psi}(x)\left.\left|\,n\,\right\rangle\right.\left.\left\langle\,n\,\right.\right|\psi(x)\left.\left|\,0\,\right\rangle\right.\
=\
\textstyle\sum_{{\mbox{\tiny(all)}}}\left.\left\langle\,0\,\right.\right|\bar{\psi}(0)\left.\left|\,n\,\right\rangle\right.\left.\left\langle\,n\,\right.\right|\psi(0)\left.\left|\,0\,\right\rangle\right.\qquad$
(156)

where $\textstyle\sum_{{\mbox{\tiny(all)}}}$ denotes sum over all the
eigenstates of $\hat{P}_{{}_{\psi}}$. $\left.\left|\,n\,\right\rangle\right.$
represents a generic eigenstate of $\hat{P}_{{}_{\psi}}$. (156) shows that
$\Delta(\psi)$ is independent of $x$, and depends only on $\psi$ justifying
the notation $\Delta(\psi)$.

If we restrict the sum to span only single-particle states we obtain

$\displaystyle\Delta_{sp}(\psi)$ $\displaystyle=$
$\displaystyle\textstyle\sum_{{\mbox{\tiny(sp)}}}\left\langle
0\,\left|\,\bar{\psi}(x)\ \right|n\,\right\rangle\left\langle
n\,\left|\,\psi(x)\ \right|0\,\right\rangle\ =\
\Delta_{sp}^{+}(\psi)+\Delta_{sp}^{-}(\psi),$
$\displaystyle\Delta_{sp}^{+}(\psi)$ $\displaystyle:=$
$\displaystyle\sum_{r=1}^{2}\int
d^{4}p\left.\left\langle\,0\,\right.\right|\bar{\psi}(x)\left.\left|\,p,r;+\,\right\rangle\right.\left.\left\langle\,p,r;+\,\right.\right|\psi(x)\left.\left|\,0\,\right\rangle\right.\delta(p^{2}-m_{{}_{e}}^{2})\,\theta(p^{0})$
$\displaystyle\Delta_{sp}^{-}(\psi)$ $\displaystyle:=$
$\displaystyle\sum_{r=1}^{2}\int
d^{4}p\left.\left\langle\,0\,\right.\right|\bar{\psi}(x)\left.\left|\,p,r;-\,\right\rangle\right.\left.\left\langle\,p,r;-\,\right.\right|\psi(x)\left.\left|\,0\,\right\rangle\right.\delta(p^{2}-m_{{}_{e}}^{2})\,\theta(p^{0})\qquad$
(157)

where $\textstyle\sum_{{\mbox{\tiny(sp)}}}$ represents sum over single-
particle states. We will abbreviate

$\displaystyle\Delta_{{}_{sp}}(\psi)$ $\displaystyle=$
$\displaystyle\sum_{r=1}^{2}\int
d^{4}p\left.\left\langle\,0\,\right.\right|\bar{\psi}(x)\left.\left|\,p,r;\pm\,\right\rangle\right.\left.\left\langle\,p,r;\pm\,\right.\right|\psi(x)\left.\left|\,0\,\right\rangle\right.\delta(p^{2}-m_{{}_{e}}^{2})\,\theta(p^{0})$

Consider a proper orthochronous Lorentz transformation $\Lambda$. From Lemma 7
(Appendix C) we have

$\displaystyle\sum_{r=1}^{2}\left.\left|\,p,r;+\,\right\rangle\right.\left.\left\langle\,p,r;+\,\right.\right|$
$\displaystyle=$
$\displaystyle\displaystyle\sum_{r=1}^{2}\left[\,\displaystyle\sum_{r^{\prime}=1}^{2}D^{\tiny(u)}_{r^{\prime},r}\left.\left|\,\uline{\Lambda
p},r^{\prime};+\,\right\rangle\right.\,\right]\left[\,\left[\,\displaystyle\sum_{s^{\prime}=1}^{2}\left(D^{\tiny(u)}\right)^{-1}_{r,s^{\prime}}\left.\left\langle\,\uline{\Lambda
p},s^{\prime};+\,\right.\right|\,\right]\,\right]$ (158) $\displaystyle=$
$\displaystyle\displaystyle\sum_{s^{\prime}=1}^{2}\left.\left|\,\uline{\Lambda
p},s^{\prime};+\,\right\rangle\right.\left.\left\langle\,\uline{\Lambda
p},s^{\prime};+\,\right.\right|$

Using (158) and noting that $\bar{\psi}(x)\psi(x)$ is a Lorentz scalar we
conclude that $\Delta_{sp}^{+}(\psi)$ is a Lorentz invariant. A similar
argument shows that $\Delta_{sp}^{-}(\psi)$ is a Lorentz invariant as well,
establishing the Lorentz invariance of $\Delta_{{}_{sp}}(\psi)$.

Next consider a free Dirac field, also of mass $m_{{}_{e}}$. Let
$\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.$ denote the vacuum of the
free field corresponding to mass $m_{{}_{e}}$. With a slight abuse of notation
we continue to use the symbol $\Delta$ for the free field as well. Thus,

$\displaystyle\Delta(\psi_{{}_{{\mbox{\tiny free}}}})$ $\displaystyle:=$
$\displaystyle\left\langle\,0_{{}_{F}}\,\left|\,\bar{\psi}_{{}_{{\mbox{\tiny
free}}}}(x)\,\psi_{{}_{{\mbox{\tiny
free}}}}(x)\,\right|\,0_{{}_{F}}\,\right\rangle$ $\displaystyle=$
$\displaystyle\textstyle\sum_{{\mbox{\tiny(all)}}}\left\langle
0_{{}_{F}}\,\,\left|\,\bar{\psi}_{{}_{{\mbox{\tiny free}}}}(x)\
\right|n_{{}_{\mbox{\scriptsize free}}}\,\right\rangle\left\langle
n_{{}_{\mbox{\scriptsize free}}}\,\left|\,\psi_{{}_{{\mbox{\tiny free}}}}(x)\
\right|0_{{}_{F}}\,\,\right\rangle$ $\displaystyle=$
$\displaystyle\textstyle\sum_{{\mbox{\tiny(sp)}}}\left\langle
0_{{}_{F}}\,\,\left|\,\bar{\psi}_{{}_{{\mbox{\tiny free}}}}(x)\
\right|n_{{}_{\mbox{\scriptsize free}}}\,\right\rangle\left\langle
n_{{}_{\mbox{\scriptsize free}}}\,\left|\,\psi_{{}_{{\mbox{\tiny free}}}}(x)\
\right|0_{{}_{F}}\,\,\right\rangle$ $\displaystyle=$
$\displaystyle\Delta_{{}_{sp}}(\psi_{{}_{{\mbox{\tiny free}}}})$
$\displaystyle=$ $\displaystyle\sum_{r=1}^{2}\int
d^{4}p\left.\left\langle\,0_{{}_{F}}\,\,\right.\right|\bar{\psi}_{{}_{{\mbox{\tiny
free}}}}(x)\left.\left|\,p,r;\pm\,\right\rangle\right.\left.\left\langle\,p,r;\pm\,\right.\right|\psi_{{}_{{\mbox{\tiny
free}}}}(x)\,|\,0_{{}_{F}}\,;\,m_{{}_{e}}\,\rangle\delta(p^{2}-m_{{}_{e}}^{2})\,\theta(p^{0})$

$\left.\left|\,n_{{}_{\mbox{\scriptsize free}}}\,\right\rangle\right.$ is a
generic eigenstate of $\hat{P}_{{}_{\psi_{{}_{{\mbox{\tiny free}}}}}}$. As
before, $\textstyle\sum_{{\mbox{\tiny(all)}}}$ represents the sum over all
eigenstates of $\hat{P}_{{}_{\psi_{{}_{{\mbox{\tiny free}}}}}}$ while
$\textstyle\sum_{{\mbox{\tiny(sp)}}}$ is the sum over its single-particle
eigenstates. $\Delta(\psi_{{}_{{\mbox{\tiny
free}}}})=\Delta_{{}_{sp}}(\psi_{{}_{{\mbox{\tiny free}}}})$ since
$\left.\left\langle\,0_{{}_{F}}\,\,\right.\right|\bar{\psi}_{{}_{{\mbox{\tiny
free}}}}(0)\left.\left|\,n_{{}_{\mbox{\scriptsize
free}}}\,\right\rangle\right.$ vanishes when
$\left.\left|\,n_{{}_{\mbox{\scriptsize free}}}\,\right\rangle\right.$ is
either $\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.$ or a multi-particle
state. We note that $\Delta(\psi_{{}_{{\mbox{\tiny free}}}})$ and
$\Delta_{{}_{sp}}(\psi_{{}_{{\mbox{\tiny free}}}})$ are also independent of
the spacetime point $x$ at which they are evaluated. Using (LABEL:eqn11) and
(27) we have

$\displaystyle\Delta_{sp}(\psi_{{}_{{\mbox{\tiny free}}}})$ $\displaystyle=$
$\displaystyle\Delta(\psi_{{}_{{\mbox{\tiny free}}}})\ =\
4m_{{}_{e}}\int\displaystyle\frac{d\vec{k}}{(2\pi)^{3}(2\omega(\vec{k}))}\
\left[\,\ g{{}_{e}}\,(\,{k}\,;{\cal F},{\cal P})\,\right]^{2}\ >0.$

We note that $\Delta_{{}_{sp}}(\psi_{{}_{{\mbox{\tiny free}}}})$ is a Lorentz
invariant, and hence so is

$\displaystyle
Z:=\displaystyle\frac{\Delta_{{}_{sp}}(\psi)}{\Delta_{{}_{sp}}(\psi_{{}_{{\mbox{\tiny
free}}}})}.$ (159)

Thus we have

$\displaystyle\Delta_{sp}(\psi)$ $\displaystyle=$ $\displaystyle Z\
\Delta_{sp}(\psi_{{}_{{\mbox{\tiny free}}}})\ =\
\Delta_{sp}(\sqrt{Z}\cdot\psi_{{}_{{\mbox{\tiny free}}}})$ (160)

that is,

$\displaystyle\sum_{r=1}^{2}\int
d^{4}p\left.\left\langle\,0\,\right.\right|\bar{\psi}(x)\left.\left|\,p,r;\pm\,\right\rangle\right.\left.\left\langle\,p,r;\pm\,\right.\right|\psi(x)\left.\left|\,0\,\right\rangle\right.\delta(p^{2}-m_{{}_{e}}^{2})\,\theta(p^{0})\hskip
72.26999pt$ (161)

$\displaystyle=$ $\displaystyle\sum_{r=1}^{2}\int
d^{4}p\left.\left\langle\,0_{{}_{F}}\,\,\right.\right|\sqrt{Z}\,\bar{\psi}_{{}_{{\mbox{\tiny
free}}}}(x)\left.\left|\,p,r;\pm\,\right\rangle\right.$ $\displaystyle\hskip
144.54pt\left.\left\langle\,p,r;\pm\,\right.\right|\sqrt{Z}\,\psi_{{}_{{\mbox{\tiny
free}}}}(x)\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.\delta(p^{2}-m_{{}_{e}}^{2})\,\theta(p^{0})$

Let ${\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize free}}}(\vec{k},s)$ denote
the normalized creation operator of the positive-energy single-particle state
corresponding to the free field of mass $m_{{}_{e}}$.
${\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize free}}}(\vec{k},s)$ acts on the
vacuum of the free field $\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.$ to
create the single-particle state. Noting that
$\left.\left|\,p,r;+\,\right\rangle\right.\ =\
{\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}(\vec{p},r)\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.$, and using
the the anticommutation relations (27), we can rewrite (148) as

$\displaystyle\left\langle\uline{p},r;+\,\left|\,{\bf\hat{a}}^{\dagger}(\vec{k},s,\pm\infty)\
\right|0\,\right\rangle$ $\displaystyle=$
$\displaystyle\left[\,\frac{\alpha(\uline{p},s)}{\ g{{}_{e}}\,(\,{p}\,;{\cal
F},{\cal
P})}\,\right]\left\langle\uline{p},r;+\,\left|\,{\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}(\vec{k},s)\ \right|0_{{}_{F}}\,\right\rangle\hskip 36.135pt$ (162)

Using (139) and (141), we can rewrite (162) as

$\displaystyle\lim_{t\rightarrow\pm\infty}\displaystyle\frac{1}{\
g{{}_{e}}\,(\,{k}\,;{\cal F},{\cal P})}\int d\,\vec{x}\
e^{\displaystyle-i\uline{k}x}\
\left.\left\langle\,\uline{p},r;+\,\right.\right|\bar{\psi}(x)\left.\left|\,0\,\right\rangle\right.\
\gamma^{0}\ u(\uline{k},s)\hskip 216.81pt$ $\displaystyle=$
$\displaystyle\lim_{t\rightarrow\pm\infty}\displaystyle\frac{1}{\
g{{}_{e}}\,(\,{k}\,;{\cal F},{\cal P})}\int d\,\vec{x}\
e^{\displaystyle-i\uline{k}x}\
\left.\left\langle\,\uline{p},r;+\,\right.\right|\left[\,\frac{\alpha(\uline{p},s)}{\
g{{}_{e}}\,(\,{p}\,;{\cal F},{\cal
P})}\,\right]\bar{\psi}_{{}_{\mbox{\scriptsize
free}}}(x)\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.\ \gamma^{0}\
u(\uline{k},s)$

Since ${\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize free}}}(\vec{k},s)$ is
time-independent, one can take the limit $t\rightarrow\pm\infty$ on the right
hand side as well.

The relations (161) and (LABEL:eqneebhanta) suggest the identification
$\displaystyle\frac{\alpha(\uline{p},s)}{\ g{{}_{e}}\,(\,{p}\,;{\cal F},{\cal
P})}\equiv\sqrt{Z}$, using which we can rewrite (162) as

$\displaystyle\left\langle\uline{p},r;+\,\left|\,{\bf\hat{a}}^{\dagger}(\vec{k},s,\pm\infty)\
\right|0\,\right\rangle$ $\displaystyle=$
$\displaystyle\sqrt{Z}\left\langle\uline{p},r;+\,\left|\,{\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}(\vec{k},s)\ \right|0_{{}_{F}}\,\,\right\rangle$ (164)

We label the operators in the asymptotic past and future the IN and OUT
operators. That is, we use the abbreviations

$\displaystyle{\bf\hat{a}}^{\dagger}(\vec{k},s;\mbox{\scriptsize IN})$
$\displaystyle:=$
$\displaystyle\left(\frac{1}{\sqrt{Z}}\right)\lim_{t\rightarrow-\infty}\
{\bf\hat{a}}^{\dagger}(\vec{k},s,t)$
$\displaystyle{\bf\hat{a}}^{\dagger}(\vec{k},s;\mbox{\scriptsize OUT})$
$\displaystyle:=$
$\displaystyle\left(\frac{1}{\sqrt{Z}}\right)\lim_{t\rightarrow+\infty}\
{\bf\hat{a}}^{\dagger}(\vec{k},s,t)$ (165)

Using (164) and (165) we have

$\displaystyle\left\langle\uline{p},r;+\,\left|\,{\bf\hat{a}}^{\dagger}(\vec{k},s;\mbox{\scriptsize
OUT})\ \right|0\,\right\rangle\ =\
\left\langle\uline{p},r;+\,\left|\,{\bf\hat{a}}^{\dagger}(\vec{k},s;\mbox{\scriptsize
IN})\ \right|0\,\right\rangle$ $\displaystyle=$
$\displaystyle\left\langle\uline{p},r;+\,\left|\,{\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}(\vec{k},s)\ \right|0_{{}_{F}}\,\,\right\rangle$

with a similar statement for
${\bf\hat{a}}^{\dagger}(\vec{k},s;\mbox{\scriptsize OUT})$. Thus, the
operators ${\bf\hat{a}}^{\dagger}(\vec{k},s;\mbox{\scriptsize IN})$ and
${\bf\hat{a}}^{\dagger}(\vec{k},s;\mbox{\scriptsize OUT})$ create single-
particle states when they act on the vacuum
$\left.\left|\,0\,\right\rangle\right.$ of the interacting field. The
completeness of the set of $n$-particle eigenstates of the 4-momentum
operator, together with Lemmas 1, 2 and 4 imply that

$\displaystyle{{\bf\hat{a}}^{\dagger}(\vec{k},s;\mbox{\scriptsize
OUT})}\left.\left|\,0\,\right\rangle\right.\ =\
{{\bf\hat{a}}^{\dagger}(\vec{k},s;\mbox{\scriptsize
IN})}\left.\left|\,0\,\right\rangle\right.$ $\displaystyle=$
$\displaystyle\xi\cdot{{\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}(\vec{k},s)}\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.$

where $\xi$ is a scale factor. From (LABEL:eqngonaon28) we conclude $\xi\ =\
1$ and thus we have

$\displaystyle{{\bf\hat{a}}^{\dagger}(\vec{k},s;\mbox{\scriptsize
OUT})}\left.\left|\,0\,\right\rangle\right.\ =\
{{\bf\hat{a}}^{\dagger}(\vec{k},s;\mbox{\scriptsize
IN})}\left.\left|\,0\,\right\rangle\right.$ $\displaystyle=$
$\displaystyle{{\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}(\vec{k},s)}\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.:=\left.\left|\,\uline{k},\,s,\,+\,\right\rangle\right.$
(167)

Similarly we conclude that

$\displaystyle{{\bf\hat{b}}^{\dagger}(\vec{k},s;\mbox{\scriptsize
OUT})}\left.\left|\,0\,\right\rangle\right.\ =\
{{\bf\hat{b}}^{\dagger}(\vec{k},s;\mbox{\scriptsize
IN})}\left.\left|\,0\,\right\rangle\right.$ $\displaystyle=$
$\displaystyle{{\bf\hat{b}}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}(\vec{k},s)}\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.:=\left.\left|\,\uline{k},\,s,\,-\,\right\rangle\right.$
(168)

where

$\displaystyle{\bf\hat{b}}^{\dagger}(\vec{k},s;\mbox{\scriptsize IN})$
$\displaystyle:=$
$\displaystyle\left(\frac{1}{\sqrt{Z}}\right)\lim_{t\rightarrow-\infty}\
{\bf\hat{b}}^{\dagger}(\vec{k},s,t)$
$\displaystyle{\bf\hat{b}}^{\dagger}(\vec{k},s;\mbox{\scriptsize OUT})$
$\displaystyle:=$
$\displaystyle\left(\frac{1}{\sqrt{Z}}\right)\lim_{t\rightarrow+\infty}\
{\bf\hat{b}}^{\dagger}(\vec{k},s,t)$ (169)

Thus, ${\bf\hat{a}}^{\dagger}(\vec{k},\,s,\,\mbox{\scriptsize IN}),$
${\bf\hat{a}}^{\dagger}(\vec{k},\,s,\,\mbox{\scriptsize OUT}),$
${\bf\hat{b}}^{\dagger}(\vec{k},\,s,\,\mbox{\scriptsize IN})$ and
${\bf\hat{b}}^{\dagger}(\vec{k},\,s,\,\mbox{\scriptsize OUT})$ create single-
particle states when they act on the vacuum of the interacting field.

While the asymptotic operators $\hat{a}^{\dagger}$ and $\hat{b}^{\dagger}$
create single-particle states, the asymptotic operators $\hat{a}$ and
$\hat{b}$ annihilate the vacuum of the interacting theory, as shown in the
following Lemma.

###### Lemma 5

Let

$\displaystyle{\hat{\bf a}(\vec{k},s,t)}\ =\ \displaystyle\frac{1}{\
g{{}_{f}}\,(\,{\uline{\vec{k}}}\,;{\cal F},{\cal P})}\int d\,\vec{x}\
e^{\displaystyle i\uline{\vec{k}}x}\ \bar{u}(\vec{k},s)\ \gamma^{0}\ \psi(x)$
(170)

and let $\left.\left|\,0\,\right\rangle\right.$ be the vacuum of the
interacting theory. Then,

$\displaystyle\lim_{t\rightarrow\pm\infty}{\hat{\bf
a}(\vec{k},s,t)}\left.\left|\,0\,\right\rangle\right.$ $\displaystyle=$
$\displaystyle 0.$

Proof: We assume that the set of $n$-particle eigenstates of the 4-momentum
operator, with $n=0,1,2\ldots$, forms a complete set. Thus, it is sufficient
to show that

$\displaystyle\lim_{t\rightarrow\pm\infty}\left.\left\langle\,n\,\right.\right|{\hat{\bf
a}(\vec{k},s,t)}\left.\left|\,0\,\right\rangle\right.$ $\displaystyle=$
$\displaystyle 0$ (171)

where $\left.\left|\,n\,\right\rangle\right.$ is an $n$-particle eigenstate of
the 4-momentum operator with $n=0,1,2,\ldots$. Lemma 2 establishes (171) for
$n\geq 2$. Since $\left\langle\,0\left|\,\psi(x)\,\right|0\,\right\rangle\ =\
0$, due to Lorentz invariance (see [86, Equation 41.22]), using (170) the
claim in (171) follows for $n=0$. Therefore, we focus on single particle
states $\left.\left|\,p,r,\pm\,\right\rangle\right.$. As shown in [86,
Equation 41.24],

$\displaystyle\left.\left\langle\,\uline{p},\,r,\,-\,\right.\right|\psi(x)\left.\left|\,0\,\right\rangle\right.$
$\displaystyle=$ $\displaystyle\kappa\,v(\vec{p},r)\,e^{\displaystyle
i\uline{p}x}$ (172)

due to Lorentz invariance474747The Minkowski signature used in [86] is
(-,+,+,+), while we use (+,-,-,-).; also see (155) above. $\kappa$ is a
constant. $\uline{p}\ =\ (\sqrt{|\vec{p}\,|^{2}+m_{{}_{e}}^{2}},\vec{p})$.
Using (170) and (172) we have

$\displaystyle\left.\left\langle\,\uline{p},\,r,\,-\,\right.\right|{\hat{\bf
a}(\vec{k},s,t)}\left.\left|\,0\,\right\rangle\right.$ $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{\ g{{}_{f}}\,(\,{\uline{\vec{k}}}\,;{\cal
F},{\cal P})}\int d\,\vec{x}\ e^{\displaystyle i\vec{k}x}\ \bar{u}(\vec{k},s)\
\gamma^{0}\ \left\langle\uline{p},\,r,\,-\,\left|\,\psi(x)\
\right|0\,\right\rangle$ $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{\ g{{}_{f}}\,(\,{\uline{\vec{k}}}\,;{\cal
F},{\cal P})}\int d\,\vec{x}\ e^{\displaystyle i\uline{\vec{k}}x}\
\bar{u}(\vec{k},s)\ \gamma^{0}\ \left[\,\kappa\,v(\vec{p},r)\,e^{\displaystyle
i\uline{p}x}\,\right]$ $\displaystyle=$ $\displaystyle\displaystyle\frac{1}{\
g{{}_{f}}\,(\,{\uline{\vec{k}}}\,;{\cal F},{\cal
P})}\left[\,(2\pi)^{3}\delta^{3}(\vec{k}+\vec{p})\,\right]e^{\displaystyle
2i\omega(\vec{k},m_{{}_{e}})t}$ $\displaystyle\bar{u}(\vec{k},s)\ \gamma^{0}\
\left[\,\kappa\,v(-\vec{k},r)\,\right]$ $\displaystyle=$ $\displaystyle 0$

where $\omega(\vec{k},m_{{}_{e}}):=\sqrt{|\vec{k}\,|^{2}+m_{{}_{e}}^{2}}$. In
the last step we have used $u^{\dagger}(\vec{k},s)v(-\vec{k},r)\ =\ 0$.

Finally, as stated in [86, Equation 41.23],

$\displaystyle\left\langle\uline{p},\,r,\,+\,\left|\,\psi(x)\
\right|0\,\right\rangle$ $\displaystyle=$ $\displaystyle 0$ (173)

due to charge conservation. We summarize the proof of (173) presented in [86]
below. The global gauge transformation $\psi(x)\rightarrow e^{\displaystyle
i\alpha}\,\psi(x),\ \alpha\in\mathbb{R}$ is a symmetry in $QED$. Therefore,
from Wigner’s Theorem [98] we know that associated with every gauge
transformation $\psi(x)\rightarrow e^{\displaystyle i\alpha}\psi(x)$ is a
unitary transformation $U(\alpha):=e^{\displaystyle-i\alpha Q}$ on the space
of physical states, where $Q$ is a Hermitian operator. Gauge invariance of
transition amplitudes induced by the field $\psi(x)$ then implies that

$\displaystyle e^{\displaystyle-i\alpha Q}\,\psi(x)\,e^{\displaystyle i\alpha
Q}\ =\ e^{\displaystyle i\alpha}\,\psi(x)$ (174)

from which we get

$\displaystyle[\psi(x),Q]$ $\displaystyle=$ $\displaystyle\psi(x)$ (175)

Taking the conjugate of (175) and noting that $Q$ is Hermitian we get

$\displaystyle[\psi^{\dagger}(x),Q]$ $\displaystyle=$
$\displaystyle-\psi^{\dagger}(x)$ (176)

Gauge invariance of vacuum requires that $e^{\displaystyle-i\alpha
Q}\left.\left|\,0\,\right\rangle\right.=\left.\left|\,0\,\right\rangle\right.$
for all $\alpha$, from which we conclude that

$\displaystyle Q\left.\left|\,0\,\right\rangle\right.$ $\displaystyle=$
$\displaystyle 0.$ (177)

Using (167) and (141) we have

$\displaystyle Q\left.\left|\,\vec{k},\,s,\,+\,\right\rangle\right.$
$\displaystyle=$ $\displaystyle
Q\,{\bf\hat{a}}^{\dagger}(\vec{k},\,s,\,+)\left.\left|\,0\,\right\rangle\right.\
=\ \displaystyle\frac{1}{\ g{{}_{e}}\,(\,{\uline{\vec{k}}}\,;{\cal F},{\cal
P})}\int
d\,\vec{x}e^{\displaystyle-i\uline{\vec{k}}x}\left[\,Q\,\psi^{\dagger}(x)\left.\left|\,0\,\right\rangle\right.\,\right]u(\vec{k},s)$
(178) $\displaystyle=$ $\displaystyle\displaystyle\frac{1}{\
g{{}_{e}}\,(\,{\uline{\vec{k}}}\,;{\cal F},{\cal P})}\int
d\,\vec{x}e^{\displaystyle-i\uline{\vec{k}}x}\left[\,\psi^{\dagger}(x)\left.\left|\,0\,\right\rangle\right.\,\right]u(\vec{k},s)$
$\displaystyle=$
$\displaystyle\left.\left|\,\vec{k},\,s,\,+\,\right\rangle\right.$

In the penultimate step above we have used (176) and (177). A similar argument
shows that

$\displaystyle Q\left.\left|\,\vec{k},\,s,\,-\,\right\rangle\right.$
$\displaystyle=$
$\displaystyle-\left.\left|\,\vec{k},\,s,\,-\,\right\rangle\right.$ (179)

We can prove (173) as follows. Using (178), the hermiticity of $Q$ and (174)
we have for all $\alpha\in\mathbb{R}$,

$\displaystyle\left\langle\uline{p},\,r,\,+\,\left|\,\psi(x)\
\right|0\,\right\rangle$ $\displaystyle=$ $\displaystyle e^{\displaystyle
i\alpha}\left\langle\uline{p},\,r,\,+\,\left|\,e^{\displaystyle i\alpha
Q}\,\psi(x)\,e^{\displaystyle-i\alpha Q}\ \right|0\,\right\rangle\ =\
e^{\displaystyle 2i\alpha}\left\langle\uline{p},\,r,\,+\,\left|\,\psi(x)\
\right|0\,\right\rangle$

from which (173) follows.

Finally, using (173), we have

$\displaystyle\left\langle\uline{p},\,r,\,+\,\left|\,{\bf\hat{a}}(\vec{k},\,s,\,t)\
\right|0\,\right\rangle$ $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{\ g{{}_{f}}\,(\,{\uline{\vec{k}}}\,;{\cal
F},{\cal P})}\int d\,\vec{x}\ e^{\displaystyle i\uline{\vec{k}}x}\
\bar{u}(\vec{k},s)\ \gamma^{0}\
\left\langle\uline{p},\,r,\,+\,\left|\,\psi(x)\ \right|0\,\right\rangle\ =\ 0$

Using a similar argument we can prove

###### Lemma 6

Let

$\displaystyle{\hat{\bf b}(\vec{k},s,t)}\ =\ \displaystyle\frac{1}{\
g{{}_{f}}\,(\,{\uline{\vec{k}}}\,;{\cal F},{\cal P})}\int d\,\vec{x}\
e^{\displaystyle i\uline{\vec{k}}x}\ \bar{\psi}(x)\,\gamma^{0}\,u(\vec{k},s)$
(180)

and let $\left.\left|\,0\,\right\rangle\right.$ be the vacuum of the
interacting theory. Then,

$\displaystyle\lim_{t\rightarrow\pm\infty}{\hat{\bf
b}(\vec{k},s,t)}\left.\left|\,0\,\right\rangle\right.$ $\displaystyle=$
$\displaystyle 0.$

Defining

$\displaystyle{\bf\hat{a}}(\vec{k},\,s,\,\mbox{\scriptsize IN})$
$\displaystyle:=$
$\displaystyle\left(\frac{1}{\sqrt{Z}}\right)\lim_{t\rightarrow-\infty}{\bf\hat{a}}(\vec{k},\,s,\,t);\qquad{\bf\hat{a}}(\vec{k},\,s,\,\mbox{\scriptsize
OUT}):=\left(\frac{1}{\sqrt{Z}}\right)\lim_{t\rightarrow\infty}{\bf\hat{a}}(\vec{k},\,s,\,t)$
$\displaystyle{\bf\hat{b}}(\vec{k},\,s,\,\mbox{\scriptsize IN})$
$\displaystyle:=$
$\displaystyle\left(\frac{1}{\sqrt{Z}}\right)\lim_{t\rightarrow-\infty}{\bf\hat{b}}(\vec{k},\,s,\,t);\qquad{\bf\hat{b}}(\vec{k},\,s,\,\mbox{\scriptsize
OUT}):=\left(\frac{1}{\sqrt{Z}}\right)\lim_{t\rightarrow\infty}{\bf\hat{b}}(\vec{k},\,s,\,t)$

we can summarize Lemmas 5 and 6 as

$\displaystyle{\bf\hat{a}}(\vec{k},\,s,\,\mbox{\scriptsize
IN})\left.\left|\,0\,\right\rangle\right.$ $\displaystyle=$
$\displaystyle{\bf\hat{a}}(\vec{k},\,s,\,\mbox{\scriptsize
OUT})\left.\left|\,0\,\right\rangle\right.\ =\
{\bf\hat{b}}(\vec{k},\,s,\,\mbox{\scriptsize
IN})\left.\left|\,0\,\right\rangle\right.\ =\
{\bf\hat{a}}(\vec{k},\,s,\,\mbox{\scriptsize
OUT})\left.\left|\,0\,\right\rangle\right.\ =\ 0$

###  B.4 IN and OUT fields

Using (167), (168), (LABEL:eqnwhew1) and the anticommutation relations (27)
for creation and annihilation operators of free fields we have

$\displaystyle\left\langle\,0\left|\,\left\\{{\bf\hat{a}}(\vec{q},r,\mbox{\scriptsize
IN}),{\bf\hat{a}}^{\dagger}(\vec{k},s,\mbox{\scriptsize
IN})\right\\}\,\right|0\,\right\rangle$ $\displaystyle=$
$\displaystyle\left\langle\,0\left|\,{\bf\hat{a}}(\vec{q},r,\mbox{\scriptsize
IN}){\bf\hat{a}}^{\dagger}(\vec{k},s,\mbox{\scriptsize
IN})\,\right|0\,\right\rangle$ $\displaystyle=$
$\displaystyle\left\langle\,0_{{}_{F}}\,\left|\,{\bf\hat{a}}_{{}_{\mbox{\scriptsize
free}}}(\vec{q},r){\bf\hat{a}}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}(\vec{k},s)\,\right|\,0_{{}_{F}}\,\right\rangle$ $\displaystyle=$
$\displaystyle\left\langle\,0_{{}_{F}}\,\left|\,\left\\{{\bf\hat{a}_{{}_{\mbox{\scriptsize
free}}}}(\vec{q},r),{\bf\hat{a}^{\dagger}_{{}_{\mbox{\scriptsize
free}}}}(\vec{k},s)\right\\}\,\right|\,0_{{}_{F}}\,\right\rangle$
$\displaystyle=$
$\displaystyle(2\pi)^{3}\,2\omega(\vec{k}\,)\,\delta^{3}(\vec{q}-\vec{k})\,\delta_{{}_{r,s}}$

Assuming that the set of eigenstates of the 4-momentum operator forms a
complete set, we can expand the state
$\left\\{{\bf\hat{a}}(\vec{q},r,\mbox{\scriptsize
IN}),{\bf\hat{a}}^{\dagger}(\vec{k},s,\mbox{\scriptsize
IN})\right\\}\left.\left|\,0\,\right\rangle\right.$ as a linear combination of
the eigenstates of the 4-momentum operator. Noting that the vacuum state
$\left.\left|\,0\,\right\rangle\right.$ does not overlap with either the
single-particle or multi-particle eigenstates of the 4-momentum operator, we
conclude that

$\displaystyle\left\\{{\bf\hat{a}}(\vec{q},r,\mbox{\scriptsize
IN}),{\bf\hat{a}}^{\dagger}(\vec{k},s,\mbox{\scriptsize IN})\right\\}$
$\displaystyle=$
$\displaystyle(2\pi)^{3}\,2\omega(\vec{k}\,)\,\delta^{3}(\vec{q}-\vec{k})\,\delta_{{}_{r,s}}$
(183)

Similarly, one can show that

$\displaystyle\left\\{{\bf\hat{b}}(\vec{q},r,\mbox{\scriptsize
IN}),{\bf\hat{b}}^{\dagger}(\vec{k},s,\mbox{\scriptsize IN})\right\\}$
$\displaystyle=$
$\displaystyle(2\pi)^{3}\,2\omega(\vec{k}\,)\,\delta^{3}(\vec{q}-\vec{k})\,\delta_{{}_{r,s}}$
$\displaystyle\left\\{{\bf\hat{a}}(\vec{q},r,\mbox{\scriptsize
IN}),{\bf\hat{a}}(\vec{k},s,\mbox{\scriptsize IN})\right\\}$ $\displaystyle=$
$\displaystyle\left\\{{\bf\hat{b}}(\vec{q},r,\mbox{\scriptsize
IN}),{\bf\hat{b}}^{\dagger}(\vec{k},s,\mbox{\scriptsize IN})\right\\}\ =\ 0$
$\displaystyle\left\\{{\bf\hat{a}}(\vec{q},r,\mbox{\scriptsize
OUT}),{\bf\hat{a}}^{\dagger}(\vec{k},s,\mbox{\scriptsize OUT})\right\\}$
$\displaystyle=$
$\displaystyle\left\\{{\bf\hat{b}}(\vec{q},r,\mbox{\scriptsize
OUT}),{\bf\hat{b}}^{\dagger}(\vec{k},s,\mbox{\scriptsize OUT})\right\\}\ =\
(2\pi)^{3}\,2\omega(\vec{k}\,)\,\delta^{3}(\vec{q}-\vec{k})\,\delta_{{}_{r,s}}$
$\displaystyle\left\\{{\bf\hat{a}}(\vec{q},r,\mbox{\scriptsize
IN}),{\bf\hat{a}}(\vec{k},s,\mbox{\scriptsize IN})\right\\}$ $\displaystyle=$
$\displaystyle\left\\{{\bf\hat{b}}(\vec{q},r,\mbox{\scriptsize
IN}),{\bf\hat{b}}^{\dagger}(\vec{k},s,\mbox{\scriptsize IN})\right\\}\ =\ 0$
(184)

Not surprisingly, the anticommutation relations (183) and (184) have the same
form as the anticommutation relations (27) for the creation and annihilation
operators of the free Dirac field.

Using the IN and OUT creation and annihilation operators one can construct the
IN and OUT fields as follows:

$\displaystyle\psi(x;{\cal F},{\cal P},\mbox{\scriptsize
IN}):=\int\displaystyle\frac{d\vec{k}\,\ g{{}_{f}}(\,\uline{\vec{k}}\,;{\cal
F},{\cal P})}{(2\pi)^{3}\
(2\omega(\vec{k}\,))}\left\\{\sum_{s=1}^{2}\left[\,u(\vec{k},s)\
{\bf\hat{a}}(\vec{k},s,\mbox{\scriptsize IN})\
e^{\displaystyle-i\uline{\vec{k}}x}+v(\vec{k},s)\
{\bf\hat{b}}^{\dagger}(\vec{k},s,\mbox{\scriptsize IN})\ e^{\displaystyle
i\uline{\vec{k}}x}\,\right]\right\\}$ $\displaystyle\bar{\psi}(x;{\cal
F},{\cal P},\mbox{\scriptsize IN}):=\int\displaystyle\frac{d\vec{k}\,\
g{{}_{f}}(\,\uline{\vec{k}}\,;{\cal F},{\cal P})}{(2\pi)^{3}\
(2\omega(\vec{k}\,))}\left\\{\sum_{s=1}^{2}\left[\,\bar{u}(\vec{k},s)\
{\bf\hat{a}}^{\dagger}(\vec{k},s,\mbox{\scriptsize IN})\ e^{\displaystyle
i\uline{\vec{k}}x}+\bar{v}(\vec{k},s)\
{\bf\hat{b}}(\vec{k},s,\mbox{\scriptsize IN})\
e^{\displaystyle-i\uline{\vec{k}}x}\,\right]\right\\}$
$\displaystyle\psi(x;{\cal F},{\cal P},\mbox{\scriptsize
OUT}):=\int\displaystyle\frac{d\vec{k}\,\ g{{}_{f}}(\,\uline{\vec{k}}\,;{\cal
F},{\cal P})}{(2\pi)^{3}\
(2\omega(\vec{k}\,))}\left\\{\sum_{s=1}^{2}\left[\,u(\vec{k},s)\
{\bf\hat{a}}(\vec{k},s,\mbox{\scriptsize OUT})\
e^{\displaystyle-i\uline{\vec{k}}x}+v(\vec{k},s)\
{\bf\hat{b}}^{\dagger}(\vec{k},s,\mbox{\scriptsize OUT})\ e^{\displaystyle
i\uline{\vec{k}}x}\,\right]\right\\}$ $\displaystyle\bar{\psi}(x;{\cal
F},{\cal P},\mbox{\scriptsize OUT}):=\int\displaystyle\frac{d\vec{k}\,\
g{{}_{f}}(\,\uline{\vec{k}}\,;{\cal F},{\cal P})}{(2\pi)^{3}\
(2\omega(\vec{k}\,))}\left\\{\sum_{s=1}^{2}\left[\,\bar{u}(\vec{k},s)\
{\bf\hat{a}}^{\dagger}(\vec{k},s,\mbox{\scriptsize OUT})\ e^{\displaystyle
i\uline{\vec{k}}x}+\bar{v}(\vec{k},s)\
{\bf\hat{b}}(\vec{k},s,\mbox{\scriptsize OUT})\
e^{\displaystyle-i\uline{\vec{k}}x}\,\right]\right\\}$

where $\uline{\vec{k}}$ is on mass shell with $(\uline{\vec{k}})^{2}\ =\
m_{{}_{e}}^{2}$; $m_{{}_{e}}$ is the physical mass of the electron. Clearly
$\psi(x;{\cal F},{\cal P},\mbox{\scriptsize IN})$ and $\psi(x;{\cal F},{\cal
P},\mbox{\scriptsize OUT})$ satisfy the Dirac equation for free fields. So
$\psi(x;{\cal F},{\cal P},\mbox{\scriptsize IN})$ and $\psi(x;{\cal F},{\cal
P},\mbox{\scriptsize OUT})$ can be regarded as free fields. Under a proper
orthochronous Lorentz transformation $\Lambda:{\cal F}\rightarrow{\cal
F}^{\prime}$ the fields transform as

$\displaystyle\psi(x;{\cal F},{\cal P},\mbox{\scriptsize{I}N})$
$\displaystyle\rightarrow$ $\displaystyle\psi^{\prime}(\Lambda x;{\cal
F}^{\prime},{\cal P},\mbox{\scriptsize{I}N})\ =\ S[\Lambda]\,\psi(x;{\cal
F},{\cal P},\mbox{\scriptsize{I}N})$ $\displaystyle\psi(x;{\cal F},{\cal
P},\mbox{\scriptsize{O}UT})$ $\displaystyle\rightarrow$
$\displaystyle\psi^{\prime}(\Lambda x;{\cal F}^{\prime},{\cal
P},\mbox{\scriptsize{O}UT})\ =\ S[\Lambda]\,\psi(x;{\cal F},{\cal
P},\mbox{\scriptsize{O}UT})$ (186)

where $S$ is the spinor representation of the Lorentz group.

###  B.5 LSZ Reduction

The scattering amplitude of a process such as

$\displaystyle{\cal P}:\qquad
e^{-}(p_{{}_{1}},r_{{}_{1}})+e^{+}(p_{{}_{2}},r_{{}_{2}})\longrightarrow
e^{-}(q_{{}_{1}},s_{{}_{1}})+e^{+}(q_{{}_{2}},s_{{}_{2}})$ (187)

in frame ${\cal F}$ would then be given by

$\displaystyle{\cal A}({\cal P},{\cal F})$ $\displaystyle=$
$\displaystyle\left\langle
0\,\left|\,{\bf\hat{b}}(q_{{}_{2}},s_{{}_{2}};\mbox{\scriptsize
OUT})\cdot{\bf\hat{a}}(q_{{}_{1}},s_{{}_{1}};\mbox{\scriptsize
OUT})\cdot{\bf\hat{b}}^{\dagger}(p_{{}_{2}},r_{{}_{2}};\mbox{\scriptsize
IN})\cdot{\bf\hat{a}}^{\dagger}(p_{{}_{1}},r_{{}_{1}};\mbox{\scriptsize IN})\
\right|0\,\right\rangle$ (188)

Using (165), (169) and (LABEL:eqninoutdfn2), we can rewrite (188) as

$\displaystyle{\cal A}({\cal P},{\cal F})$ $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{(\sqrt{Z})^{4}}\left\langle
0\,\left|\,{\bf\hat{b}}(q_{{}_{2}},s_{{}_{2}},+\infty)\cdot{\bf\hat{a}}(q_{{}_{1}},s_{{}_{1}},+\infty)\cdot{\bf\hat{b}}^{\dagger}(p_{{}_{2}},r_{{}_{2}},-\infty)\cdot{\bf\hat{a}}^{\dagger}(p_{{}_{1}},r_{{}_{1}},-\infty)\
\right|0\,\right\rangle$ $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{(\sqrt{Z})^{4}}\left\langle
0\,\left|\,T\left\\{{\bf\hat{b}}(q_{{}_{2}},s_{{}_{2}},+\infty)\cdot{\bf\hat{a}}(q_{{}_{1}},s_{{}_{1}},+\infty)\cdot{\bf\hat{b}}^{\dagger}(p_{{}_{2}},r_{{}_{2}},-\infty)\cdot{\bf\hat{a}}^{\dagger}(p_{{}_{1}},r_{{}_{1}},-\infty)\right\\}\
\right|0\,\right\rangle$

Using (141) and well-known calculations [86] we obtain, in an arbitrary frame
${\cal F}$,

$\displaystyle{\bf\hat{a}}^{\dagger}(\uline{k},s,+\infty)-{\bf\hat{a}}^{\dagger}(\uline{k},s,-\infty)$
$\displaystyle=$ $\displaystyle-\displaystyle\frac{i}{\
g{{}_{e}}\,(\,{k}\,;{\cal F},{\cal P})}\int d^{4}x\cdot
e^{\displaystyle-i\uline{k}x}\cdot\bar{\psi}(x)\cdot(i\overleftarrow{\partial\\!\\!\\!\\!/}+m)\cdot
u(\uline{k},s)$

with similar relations for ${\bf\hat{a}},{\bf\hat{b}}$ and
${\bf\hat{b}}^{\dagger}$. Inserting (LABEL:eqn744), and similar relations for
${\bf\hat{a}},{\bf\hat{b}}$ and ${\bf\hat{b}}^{\dagger}$, into (LABEL:eqn743)
and observing that ${\bf\hat{a}}(\uline{k},s,\pm\infty)$ and
${\bf\hat{b}}(\uline{k},s,\pm\infty)$ annihilate the vacuum
$\left.\left|\,0\,\right\rangle\right.$ we obtain the LSZ reduction formula in
the presence of the Gibbs factors

$\displaystyle{\cal A}({\cal P},{\cal F})\hskip 397.48499pt$ $\displaystyle=$
$\displaystyle\displaystyle\frac{(-i)^{2}(i)^{2}}{(\sqrt{Z})^{4}\cdot\
g{{}_{e}}\,(\,{p_{{}_{1}}}\,;{\cal F},{\cal P})\,\
g{{}_{e}}\,(\,{p_{{}_{2}}}\,;{\cal F},{\cal P})\,\
g{{}_{e}}\,(\,{q_{{}_{1}}}\,;{\cal F},{\cal P})\,\
g{{}_{e}}\,(\,{q_{{}_{2}}}\,;{\cal F},{\cal P})}\int
d^{4}x_{{}_{1}}\,d^{4}x_{{}_{2}}\,d^{4}y_{{}_{1}}\,d^{4}y_{{}_{2}}\left\\{\rule{0.0pt}{14.0pt}\right.$
$\displaystyle
e^{\displaystyle-i\left(\underline{p_{{}_{1}}}x_{{}_{1}}+\underline{p_{{}_{2}}}x_{{}_{2}}-\underline{q_{{}_{1}}}y_{{}_{1}}-\underline{q_{{}_{2}}}y_{{}_{2}}\right)}\left[\,\rule{0.0pt}{11.0pt}\bar{u}(\underline{q_{{}_{1}}},s_{{}_{1}})\,(i\partial\\!\\!\\!\\!/_{y_{{}_{1}}}-m)\,\right]\left[\,\rule{0.0pt}{11.0pt}\bar{v}(\underline{p_{{}_{2}}},r_{{}_{2}})\,(i\partial\\!\\!\\!\\!/_{x_{{}_{2}}}-m)\,\right]$
$\displaystyle\left\langle\,0\left|\,T\left\\{\bar{\psi}(y_{{}_{2}})\psi(y_{{}_{1}})\psi(x_{{}_{2}})\bar{\psi}(x_{{}_{1}})\right\\}\,\right|0\,\right\rangle\left[\,(i\overleftarrow{\partial\\!\\!\\!\\!/}_{x_{{}_{1}}}+m)\,\rule{0.0pt}{11.0pt}u(\underline{p_{{}_{1}}},r_{{}_{1}})\,\right]\left[\,(i\overleftarrow{\partial\\!\\!\\!\\!/}_{y_{{}_{2}}}+m)\,\rule{0.0pt}{11.0pt}v(\underline{q_{{}_{2}}},s_{{}_{2}})\,\right]\left.\rule{0.0pt}{14.0pt}\right\\}$

A generalized version of (LABEL:eqn745) is presented in (36).

###  B.6 Vacuum Correlation Functions

The fields and the vacuum state shown in (LABEL:eqn745) are in the Heisenberg
picture. For completeness we mention the perturbative expansion of the vacuum
correlation functions in the interaction picture.

As is well known, the 4-point function

$\displaystyle G$ $\displaystyle:=$ $\displaystyle\left\langle
0\,\left|\,T\left\\{\bar{\psi}(y_{{}_{2}})\psi(y_{{}_{1}})\psi(x_{{}_{2}})\bar{\psi}(x_{{}_{1}})\right\\}\
\right|0\,\right\rangle$ (192)

in (LABEL:eqn745) is evaluated in a perturbative expansion in the interaction
picture. As shown in [38, 43, 45, 74, 86], $G$ can be rewritten as

$\displaystyle G$ $\displaystyle=$ $\displaystyle\displaystyle\frac{1}{{\cal
N}}\ \langle\,0_{{}_{F}}\,|\
T\left\\{\bar{\psi}_{{}_{I}}(y_{{}_{2}})\psi_{{}_{I}}(y_{{}_{1}})\psi_{{}_{I}}(x_{{}_{2}})\bar{\psi}_{{}_{I}}(x_{{}_{1}})e^{\displaystyle-i\int{\cal
H}^{\prime}_{{}_{I}}(x)\,d^{4}x}\right\\}\ |\,0_{{}_{F}}\rangle$

where $\psi_{{}_{I}}$ represents a field in the interaction picture, and
${\cal H}_{{}_{I}}^{\prime}(x)$ represents the interaction Hamiltonian density
in the interaction picture. In the interaction picture the time evolution of
the fields is governed by $H_{{}_{0}}$, the unperturbed Hamiltonian
corresponding to a free electron with mass $m_{{}_{e}}$—the physical mass of
the electron—and the free photon field.
$\left.\left|\,0_{{}_{F}}\,\right\rangle\right.=\left.\left|\,0_{{}_{F}};electron\,\right\rangle\right.\otimes\left.\left|\,0_{{}_{F}};photon\,\right\rangle\right.$
is the vacuum state of $H_{{}_{0}}$. The normalizing constant is

$\displaystyle{\cal N}$ $\displaystyle=$ $\displaystyle\langle\,0_{{}_{F}}\,|\
Te^{\displaystyle-i\int{\cal H}^{\prime}_{{}_{I}}(x)\,d^{4}x}\
|\,0_{{}_{F}}\,\rangle$ (194)

Proof of (LABEL:eqnvcf3): From the Adiabatic Theorem [43, Page 156] we have

$\displaystyle\left[\,\frac{U(0,-\infty)\,|\,0_{{}_{F}}\,\rangle}{\langle
0_{{}_{F}}\,\,|\,U(0,-\infty)\,|\,0_{{}_{F}}\,\rangle}\,\right]$
$\displaystyle=$ $\displaystyle
c_{{}_{1}}\left.\left|\,0\,\right\rangle\right.$

where $U$ is the time-evolution operator in the interaction picture.
$\left.\left|\,0\,\right\rangle\right.$ is the unit-norm vacuum of the full
Hamiltonian $H$ and $\left.\left|\,0_{{}_{F}}\,\,\right\rangle\right.$ the
unit-norm vacuum of the free field Hamiltonian $H_{{}_{0}}$ with mass
$m_{{}_{e}}$. Adapting the proof of the Adiabatic Theorem we show that

$\displaystyle\left[\,\frac{\langle 0_{{}_{F}}\,\,|\,U(0,\infty)}{\langle
0_{{}_{F}}\,\,|\,U(0,\infty)\,|\,0_{{}_{F}}\,\rangle}\,\right]$
$\displaystyle=$ $\displaystyle
c_{{}_{2}}\left.\left\langle\,0\,\right.\right|$ (195)

Define

$\displaystyle H_{{}_{\epsilon}}$ $\displaystyle:=$ $\displaystyle
H_{{}_{0}}+\lambda
e^{\displaystyle-\epsilon|t|}H^{\prime},\qquad\epsilon>0,\qquad\lim_{\epsilon\rightarrow
0^{+}}H_{{}_{\epsilon}}\ =\ H$ $\displaystyle U^{+}_{{}_{\epsilon}}$
$\displaystyle:=$ $\displaystyle
T\left\\{e^{\displaystyle-i\int_{0}^{\infty}\,\lambda\,e^{\displaystyle-\epsilon\tau}H^{\prime}(\tau)\,d\tau}\right\\}$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-i\lambda)^{n}}{n!}\int_{0}^{\infty}dt_{{}_{1}}\ldots\int_{0}^{\infty}dt_{{}_{n}}e^{\displaystyle-\epsilon(t_{{}_{1}}+\ldots+t_{{}_{n}})}T\left\\{H^{\prime}(t_{{}_{1}})\ldots
H^{\prime}(t_{{}_{n}})\right\\}$

We use the abbreviations

$\displaystyle Z_{{}_{n}}$ $\displaystyle:=$ $\displaystyle
T\left\\{H^{\prime}(t_{{}_{1}})\ldots
H^{\prime}(t_{{}_{n}})\right\\},\qquad\int_{n}:=\int_{0}^{\infty}dt_{{}_{1}}\ldots\int_{0}^{\infty}dt_{{}_{n}}e^{\displaystyle-\epsilon(t_{{}_{1}}+\ldots+t_{{}_{n}})}$

Noting that in the interaction picture $[H_{{}_{0}},H^{\prime}(t)]\ =\
-i\,\displaystyle\frac{\partial H^{\prime}(t)}{\partial t}$

$\displaystyle[H_{{}_{0}},U^{+}_{{}_{\epsilon}}]$ $\displaystyle=$
$\displaystyle-i\sum_{n=1}^{\infty}\displaystyle\frac{(-i\lambda)^{n}}{n!}\int_{n}\left(\displaystyle\frac{\partial}{\partial
t_{{}_{1}}}+\ldots+\displaystyle\frac{\partial}{\partial
t_{{}_{n}}}\right)Z_{n}\ =\
-\lambda\sum_{n=1}^{\infty}\displaystyle\frac{(-i\lambda)^{n-1}}{(n-1)!}\int_{n}\left(\displaystyle\frac{\partial
Z_{n}}{\partial t_{{}_{1}}}\right)$

Integrating by parts we have

$\displaystyle[H_{{}_{0}},U^{+}_{{}_{\epsilon}}]$ $\displaystyle=$
$\displaystyle\lambda
U_{{}_{\epsilon}}^{+}H^{\prime}(0)-i\,\epsilon\,\lambda\displaystyle\frac{\partial
U_{{}_{\epsilon}}^{+}}{\partial\lambda}$ (196)

We note that

$\displaystyle\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}(H_{{}_{0}}-E_{{}_{0}})$
$\displaystyle=$ $\displaystyle\langle
0_{{}_{F}}\,\,|\,[U_{{}_{\epsilon}}^{+},H_{{}_{0}}]$ $\displaystyle=$
$\displaystyle\langle 0_{{}_{F}}\,\,|\,\left\\{-\lambda
U_{{}_{\epsilon}}^{+}H^{\prime}(0)+i\,\epsilon\,\lambda\displaystyle\frac{\partial
U_{{}_{\epsilon}}^{+}}{\partial\lambda}\right\\}$ $\displaystyle\langle
0_{{}_{F}}\,\,|\,\left[\,U_{{}_{\epsilon}}^{+}(H(0)-E_{{}_{0}})-i\,\epsilon\,\lambda\,\displaystyle\frac{\partial
U_{{}_{\epsilon}}^{+}}{\partial\lambda}\,\right]$ $\displaystyle=$
$\displaystyle 0$ (197)

Setting
$\left.\left\langle\,X_{{}_{\epsilon}}\,\right.\right|:=\displaystyle\frac{\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}}{\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}\,|\,0_{{}_{F}}\,\rangle}$ we have

$\displaystyle\displaystyle\frac{\partial\left.\left\langle\,X_{{}_{\epsilon}}\,\right.\right|}{\partial\lambda}$
$\displaystyle=$ $\displaystyle\displaystyle\frac{1}{\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}\,|\,0_{{}_{F}}\,\,\rangle}\langle
0_{{}_{F}}\,\,|\,\displaystyle\frac{\partial
U_{{}_{\epsilon}}^{+}}{\partial\lambda}$ $\displaystyle-\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}\cdot\displaystyle\frac{1}{(\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}\,|\,0_{{}_{F}}\,\,\rangle)^{2}}\displaystyle\frac{\partial\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}\,|\,0_{{}_{F}}\,\,\rangle}{\partial\lambda}$
$\displaystyle=$ $\displaystyle\displaystyle\frac{1}{\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}\,|\,0_{{}_{F}}\,\,\rangle}\langle
0_{{}_{F}}\,\,|\,\displaystyle\frac{\partial
U_{{}_{\epsilon}}^{+}}{\partial\lambda}-\left.\left\langle\,X_{{}_{\epsilon}}\,\right.\right|\
\displaystyle\frac{\partial\ln\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}\,|\,0_{{}_{F}}\,\,\rangle}{\partial\lambda}$

Using (197) we have

$\displaystyle\left.\left\langle\,X_{{}_{\epsilon}}\,\right.\right|(H(0)-E_{{}_{0}})-i\epsilon\lambda\displaystyle\frac{\partial\left.\left\langle\,X_{{}_{\epsilon}}\,\right.\right|}{\partial\lambda}$
$\displaystyle=$ $\displaystyle
i\,\epsilon\,\lambda\left.\left\langle\,X_{{}_{\epsilon}}\,\right.\right|\
\displaystyle\frac{\partial\ln\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}\,|\,0_{{}_{F}}\,\,\rangle}{\partial\lambda}$

Taking the limit $\epsilon\rightarrow 0$ and assuming that
$\displaystyle\frac{\partial\left.\left\langle\,X_{{}_{\epsilon}}\,\right.\right|}{\partial\lambda}$
is finite we get

$\displaystyle\left.\left\langle\,X_{{}_{0}}\,\right.\right|H(0)$
$\displaystyle=$
$\displaystyle\left.\left\langle\,X_{{}_{0}}\,\right.\right|\left(E_{{}_{0}}+\Delta
E_{{}_{0}}\right),\qquad\Delta
E_{{}_{0}}:=\lambda\left[\,\lim_{\epsilon\rightarrow 0}(i\,\epsilon\,)\
\displaystyle\frac{\partial\ln\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}\,|\,0_{{}_{F}}\,\,\rangle}{\partial\lambda}\,\right]$

showing that $\left.\left|\,X_{{}_{0}}\,\right\rangle\right.$ is an
eigenvector of $H(0)$, with eigenvalue $E_{{}_{0}}+\Delta E_{{}_{0}}$.

In [43, Equation 8.92] it is shown that the smallest eigenvalue of the full
Hamiltonian $H$, namely $E$, corresponding to the vacuum state
$\left.\left|\,0\,\right\rangle\right.$, is given by

$\displaystyle E$ $\displaystyle=$ $\displaystyle
E_{{}_{0}}+\lambda\langle\,\,0_{{}_{F}}\,|H^{\prime}\left.\left|\,0\,\right\rangle\right.$
(199)

From (197) we have

$\displaystyle\Delta E_{{}_{0}}$ $\displaystyle=$
$\displaystyle\displaystyle\frac{1}{\left\langle
0_{{}_{F}}\,\left|\,U^{+}_{{}_{0}}\
\right|0_{{}_{F}}\,\right\rangle}\left[\,\lim_{\epsilon\rightarrow
0}\left\\{(i\,\epsilon\,\lambda)\,\displaystyle\frac{\partial\left\langle
0_{{}_{F}}\,\left|\,U^{+}_{{}_{\epsilon}}\
\right|0_{{}_{F}}\,\right\rangle}{\partial\lambda}\right\\}\,\right]$ (200)
$\displaystyle=$ $\displaystyle\displaystyle\frac{1}{\left\langle
0_{{}_{F}}\,\left|\,U^{+}_{{}_{0}}\
\right|0_{{}_{F}}\,\right\rangle}\left[\,\lim_{\epsilon\rightarrow
0}\left\\{\,\left\langle
0_{{}_{F}}\,\left|\,U^{+}_{{}_{\epsilon}}(H(0)-E_{{}_{0}})\
\right|0_{{}_{F}}\,\right\rangle\right\\}\,\right]$ $\displaystyle=$
$\displaystyle\lambda\left\langle X_{{}_{0}}\,\left|\,H^{\prime}\
\right|0_{{}_{F}}\,\right\rangle\ =\ \lambda\left\langle
0_{{}_{F}}\,\left|\,H^{\prime}\ \right|X_{{}_{0}}\,\right\rangle$

In the last step we have used the hermiticity of $H^{\prime}$. From
(LABEL:eqnwev), (199) and (200) we conclude that
$\left.\left|\,X_{{}_{0}}\,\right\rangle\right.$ is proportional to the vacuum
state $\left.\left|\,0\,\right\rangle\right.$ of the full Hamiltonian, and
$E_{{}_{0}}+\Delta E_{{}_{0}}$ represents the corresponding eigenvalue of the
full Hamiltonian $H$.

Therefore, we conclude that

$\displaystyle\lim_{\epsilon\rightarrow 0}\displaystyle\frac{\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}}{\langle
0_{{}_{F}}\,\,|\,U_{{}_{\epsilon}}^{+}\,|\,0_{{}_{F}}\,\,\rangle}$
$\displaystyle=$ $\displaystyle\displaystyle\frac{\langle
0_{{}_{F}}\,\,|\,U(\infty,0)}{\langle
0_{{}_{F}}\,\,|\,U(\infty,0)\,|\,0_{{}_{F}}\,\,\rangle}\ =\
c_{{}_{2}}\left.\left\langle\,0\,\right.\right|$ (201)

We recall that for an operator ${\cal O}$,

$\displaystyle{\cal O}_{{}_{H}}(t)$ $\displaystyle=$ $\displaystyle
U(0,t){\cal O}_{{}_{I}}(t)U(t,0)$

Therefore, setting
$t_{{}_{2}}^{\prime}:=(y_{{}_{2}})^{0},t_{{}_{1}}^{\prime}:=(y_{{}_{1}})^{0},\
t_{{}_{2}}:=(x_{{}_{2}})^{0},\ t_{{}_{1}}:=(x_{{}_{1}})^{0}$, and assuming
without loss of generality that
$t_{{}_{2}}^{\prime}>t_{{}_{1}}^{\prime}>t_{{}_{2}}>t_{{}_{1}}$, $G$ in (192)
can be rewritten as

$\displaystyle G$ $\displaystyle=$
$\displaystyle\left\langle\,0\left|\,U(0,t_{{}_{2}}^{\prime})\bar{\psi}_{{}_{I}}(y_{{}_{2}})U(t_{{}_{2}}^{\prime},t_{{}_{1}}^{\prime})\psi_{{}_{I}}(y_{{}_{1}})U(t_{{}_{1}}^{\prime},t_{{}_{2}})\psi_{{}_{I}}(x_{{}_{2}})U(t_{{}_{2}},t_{{}_{1}})\bar{\psi}_{{}_{I}}(x_{{}_{1}})U(t_{{}_{1}},0)\,\right|0\,\right\rangle$
$\displaystyle=$ $\displaystyle\langle\,0\,|\,U(0,\infty)$ $\displaystyle
U(\infty,t_{{}_{2}}^{\prime})\bar{\psi}_{{}_{I}}(y_{{}_{2}})U(t_{{}_{2}}^{\prime},t_{{}_{1}}^{\prime})\psi_{{}_{I}}(y_{{}_{1}})U(t_{{}_{1}}^{\prime},t_{{}_{2}})\psi_{{}_{I}}(x_{{}_{2}})U(t_{{}_{2}},t_{{}_{1}})\bar{\psi}_{{}_{I}}(x_{{}_{1}})U(t_{{}_{1}},-\infty)$
$\displaystyle\hskip 325.215ptU(-\infty,0)\,0\,\rangle$

From (195) and (201) we have

$\displaystyle G$ $\displaystyle=$
$\displaystyle\displaystyle\frac{\left\langle\,0_{{}_{F}}\,\right.\left.\left|T\left\\{\bar{\psi}_{{}_{I}}(y_{{}_{2}})\psi_{{}_{I}}(y_{{}_{1}})\psi_{{}_{I}}(x_{{}_{2}})\bar{\psi}_{{}_{I}}(x_{{}_{1}})U(\infty,-\infty)\right\\}\right.\right|\left.0_{{}_{F}}\,\right\rangle}{c_{{}_{2}}\,\langle\,0_{{}_{F}}\,\,|\,U(\infty,0)\,|\,0_{{}_{F}}\,\,\rangle\,c_{{}_{1}}\,\langle
0_{{}_{F}}\,\,|\,U(0,-\infty)\,|\,0_{{}_{F}}\,\,\rangle}$

Dividing $G$ by

$\displaystyle 1$ $\displaystyle=$ $\displaystyle\langle 0|0\rangle\ =\
\displaystyle\frac{\langle\,0_{{}_{F}}\,\,|\,U(\infty,0)U(0,-\infty)\,|\,0_{{}_{F}}\,\,\rangle}{c_{{}_{2}}\,\langle\,0_{{}_{F}}\,\,|\,U(\infty,0)\,|\,0_{{}_{F}}\,\,\rangle\,c_{{}_{1}}\,\langle
0_{{}_{F}}\,\,|\,U(0,-\infty)\,|\,0_{{}_{F}}\,\,\rangle}$

we obtain

$\displaystyle G$ $\displaystyle=$
$\displaystyle\displaystyle\frac{\left\langle\,0_{{}_{F}}\,\right.\left.\left|T\left\\{\bar{\psi}_{{}_{I}}(y_{{}_{2}})\psi_{{}_{I}}(y_{{}_{1}})\psi_{{}_{I}}(x_{{}_{2}})\bar{\psi}_{{}_{I}}(x_{{}_{1}})U(\infty,-\infty)\right\\}\right.\right|\left.0_{{}_{F}}\,\right\rangle}{\langle\,0_{{}_{F}}\,\,|\,U(\infty,-\infty)\,|\,0_{{}_{F}}\,\,\rangle}$

which completes the proof of (LABEL:eqnvcf3).

##  Appendix C Appendix: Lorentz Invariance of Scattering Amplitudes

Consider the electron-positron scattering process ${\cal P}$ in frame ${\cal
F}$, shown in (187). The amplitude ${\cal A}({\cal P},{\cal F})$ of ${\cal P}$
is shown in (LABEL:eqn745). In this Appendix we establish the Lorentz
invariance of (187). A similar argument can be used to establish the Lorentz
invariance of the amplitude of the more general process (36).

We start by showing how the creation and annihilation operators of a free
field transform under a Lorentz transformation. Consider two frames ${\cal F}$
and ${\cal F}\,^{\prime}$ that are related by a proper orthochronous Lorentz
transformation $\Lambda$. Specifically, if the coordinates of an event
measured in ${\cal F}$ and ${\cal F}\,^{\prime}$ yield 4-vectors $x$ and
$x^{\prime}$ respectively, then $x^{\prime}\ =\ \Lambda x$, where $\Lambda\ =\
\exp\left\\{(i/2)\ \omega_{\mu\nu}M^{\mu\nu}\right\\}$, and $M^{\mu\nu}$ are
the generators of the quantum Lorentz algebra. Then the Dirac field $\psi(x)$
and $\psi^{\prime}(x^{\prime})$ as seen in ${\cal F}$ and ${\cal
F}\,^{\prime}$ are related as

$\displaystyle\psi^{\prime}(x^{\prime})\ =\ S[\Lambda]\,\psi(x),$ (202)

where $S[\Lambda]\ =\ \exp\left\\{(i/2)\
\omega_{\mu\nu}\Sigma^{\mu\nu}\right\\}$, and $\Sigma^{\mu\nu}\ :=\
(i/4)[\gamma^{\mu},\gamma^{\nu}]$.

If $\psi(x)$ is a free field then

$\displaystyle\psi^{\prime}(x^{\prime})\ =\ S[\Lambda]\ \psi(x)$
$\displaystyle=$ $\displaystyle\int\displaystyle\frac{d^{4}k}{(2\pi)^{3}}\
\theta(k^{0})\ \delta(k^{2}-m_{{}_{e}}^{2})\ g{{}_{f}}\,(\,{k}\,;{\cal
F},{\cal P})$ (203)
$\displaystyle\left\\{S[\Lambda]\,u(\uline{k})\,{\bf\hat{a}}(\uline{k})\,e^{\displaystyle-i\uline{k}x}+S[\Lambda]\,v(\uline{k})\,\left[\,{\bf\hat{b}}^{\dagger}(\uline{k})\,\right]^{{}^{T}}\,e^{\displaystyle
i\uline{k}x}\right\\}$ $\displaystyle=$
$\displaystyle\int\displaystyle\frac{d\,\vec{k}}{(2\pi)^{3}}\ \
g{{}_{f}}\,(\,{\uline{\vec{k}}}\,;{\cal F},{\cal P})$
$\displaystyle\left\\{S[\Lambda]\,u(\uline{\vec{k}})\,{\bf\hat{a}}(\uline{\vec{k}})\,e^{\displaystyle-i\uline{\vec{k}}x}+S[\Lambda]\,v(\uline{\vec{k}})\,\left[\,{\bf\hat{b}}^{\dagger}(\uline{\vec{k}})\,\right]^{{}^{T}}\,e^{\displaystyle
i\uline{\vec{k}}x}\right\\}\qquad$

where

$\displaystyle
u(\uline{\vec{k}}):=\left[\,\begin{array}[]{cc}u(\uline{\vec{k}},1)&u(\uline{\vec{k}},2)\end{array}\,\right],\
{\bf\hat{a}}(\uline{\vec{k}}):=\left[\,\begin{array}[]{c}{\bf\hat{a}}(\uline{\vec{k}},1)\\\
{\bf\hat{a}}(\uline{\vec{k}},2)\end{array}\,\right],$ $\displaystyle
v(\uline{\vec{k}}):=\left[\,\begin{array}[]{cc}v(\uline{\vec{k}},1)&v(\uline{\vec{k}},2)\end{array}\,\right],\
\left[\,{\bf\hat{b}}^{\dagger}(\uline{\vec{k}})\,\right]^{{}^{T}}:=\left[\,\begin{array}[]{c}{\bf\hat{b}}^{\dagger}(\uline{\vec{k}},1)\\\
{\bf\hat{b}}^{\dagger}(\uline{\vec{k}},2)\end{array}\,\right]$

$u(\uline{\vec{k}})$ contains the first two columns of
$(\uline{\vec{k}}\\!\\!\\!/+m_{{}_{e}})$, which are linearly independent484848
The third and fourth columns of $(\uline{\vec{k}}\\!\\!\\!/+m_{{}_{e}})$ are
linearly dependent on the first two columns.. Similarly, if
$q:=\Lambda\uline{\vec{k}}$, then $u(q)$ comprises the first two columns of
$(q\\!\\!\\!/\ +m_{{}_{e}})$494949 which are linearly independent. Again the
third and fourth columns of $(q\\!\\!\\!/\,+m_{{}_{e}})$ are linearly
dependent on its first two columns.. Recalling that

$\displaystyle S^{-1}[\Lambda]\,\gamma^{\mu}S[\Lambda]\ =\ \Lambda^{\mu}\
_{\nu}\gamma^{\nu}$

we observe that $(q\\!\\!\\!/\,+m_{{}_{e}})S[\Lambda]\ =\
S[\Lambda](\uline{\vec{k}}\\!\\!\\!/+m_{{}_{e}})$. Therefore,
$(q\\!\\!\\!/\,+m_{{}_{e}})$ and
$S[\Lambda]\,(\uline{\vec{k}}\\!\\!\\!/+m_{{}_{e}})$ have the same column
space if one allows for complex coefficients in the linear combinations.
Therefore, $S[\Lambda]u(\uline{\vec{k}},1)$ and
$S[\Lambda]u(\uline{\vec{k}},2)$ can be written as linear combinations of
$u(q,1^{\prime})$ and $u(q,2^{\prime})$ as follows

$\displaystyle S[\Lambda]\
\left[\,\begin{array}[]{cc}u(\uline{\vec{k}},1)&u(\uline{\vec{k}},2)\end{array}\,\right]$
$\displaystyle=$
$\displaystyle\left[\,\begin{array}[]{cc}u(q,1^{\prime})&u(q,2^{\prime})\end{array}\,\right]\left[\,\begin{array}[]{cc}D^{\tiny(u)}_{1^{\prime},1}&D^{\tiny(u)}_{1^{\prime},2}\\\\[21.68121pt]
D^{\tiny(u)}_{2^{\prime},1}&D^{\tiny(u)}_{2^{\prime},2}\end{array}\,\right],$
(209) $\displaystyle D^{\tiny(u)}_{s^{\prime},s}$ $\displaystyle:=$
$\displaystyle\displaystyle\frac{1}{2m_{{}_{e}}}\,\bar{u}(\Lambda\uline{\vec{k}},s^{\prime})\,S[\Lambda]\,u(\uline{\vec{k}},s)$
(210)

The spin indices in ${\cal F}^{\prime}$ are written as $1^{\prime}$ and
$2^{\prime}$.

In abbreviated form we have

$\displaystyle S[\Lambda]\,u(\uline{\vec{k}})\ =\
u(\Lambda\uline{\vec{k}})\,D^{\tiny(u)}$ (211)

where the matrix $D^{\tiny(u)}$ is defined in (210). The superscript $(u)$ is
used to distinguish the transformation of $u$-spinors from that of the
$v$-spinors.

A similar argument show that

$\displaystyle S[\Lambda]v(\uline{\vec{k}})$ $\displaystyle=$ $\displaystyle
v(\Lambda\uline{\vec{k}})D^{(v)},\qquad
D^{(v)}_{s^{\prime},s}:=-\displaystyle\frac{1}{2m_{{}_{e}}}\cdot\bar{v}(\Lambda\uline{\vec{k}},s^{\prime})\,S[\Lambda]\,v(\uline{\vec{k}},s)$
(212)

Expanding the field $\psi^{\prime}(x^{\prime})$ in frame ${\cal F}^{\prime}$
we obtain505050 Since the matrices $q\\!\\!\\!/\pm m_{{}_{e}}$ are nonsingular
if and only if $q$ is not on mass shell, we take $u(q)\ =\
\left[\,\begin{array}[]{cc}u(q,1)&u(q,2)\end{array}\,\right]$, $v(q)\ =\
\left[\,\begin{array}[]{cc}v(q,1)&v(q,2)\end{array}\,\right]$ when $q$ is on
mass-shell and $u(q)\ =\ v(q)\ =\
\left[\,\begin{array}[]{cc}0&0\end{array}\,\right]$ when $q$ not on mass
shell. The presence of the $\delta(k^{2}-m_{{}_{e}}^{2})$ inside the integral
in (213) makes the case of $q$ not being on mass shell irrelevant.

$\displaystyle\psi^{\prime}(x^{\prime})$ $\displaystyle=$
$\displaystyle\int\displaystyle\frac{d^{4}k^{\prime}}{(2\pi)^{3}}\
\theta((k^{\prime})^{0})\ \delta((k^{\prime})^{2}-m^{2})\
g{{}_{f}}\,(\,{k^{\prime}}\,;{\cal F}^{\prime},{\cal P})$ (213)
$\displaystyle\left\\{u(k^{\prime})\,{\bf\hat{a}}(k^{\prime})\,e^{\displaystyle-
ik^{\prime}x^{\prime}}+v(k^{\prime})\,\left[\,{\bf\hat{b}}^{\dagger}(k^{\prime})\,\right]^{{}^{T}}\,e^{\displaystyle
ik^{\prime}x^{\prime}}\right\\}\hskip 36.135pt$

Under an orthochronous Lorentz transformation $\Lambda$, $k^{\prime}\ =\
\Lambda k$ and $x^{\prime}\ =\ \Lambda x$, $(k^{\prime})^{2}=k^{2}$,
$k^{\prime}x^{\prime}=kx$ and $\theta((k^{\prime})^{0})\ =\ \theta(k^{0})$.
Further the Jacobian of the Lorentz transformation is 1. From (23), $\
g{{}_{f}}\,(\,{k^{\prime}}\,;{\cal F}^{\prime},{\cal P})=\
g{{}_{f}}\,(\,{k}\,;{\cal F},{\cal P})$.

Therefore, changing the variable of integration from $k^{\prime}\rightarrow
k=\Lambda^{-1}k^{\prime}$, (213) can be rewritten as