# Quantum electrodynamics in the null-plane causal perturbation theory II

O.A. Acevedo<EMAIL_ADDRESS>B.M. Pimentel<EMAIL_ADDRESS>Institute for Theoretical Physics (IFT), São Paulo State University (UNESP),
R. Dr. Bento Teobaldo Ferraz 271, São Paulo, SP 01140-070, Brazil

###### Abstract

We develop a complete formulation of quantum gauge invariance in light-front
dynamics for interacting theories with massless vector gauge fields in the
framework of null-plane causal perturbation theory. We apply the general
results to quantum electrodynamics, showing that the so-called “gauge terms”
present in the photon commutation distribution when quantized under the null-
plane gauge condition have no contribution in the calculation of the physical
$S$-operator matrix elements at any order. We use this result to prove the
normalizability of the theory, and to calculate the electron’s self-energy at
second order.

##  I Introduction

Light-front dynamics is one of the three forms of relativistic dynamics
discovered by Dirac in 1949 [1], complemented in 1978 with two additional
forms by Leutwyler and Stern [2]. In this dynamical form, in which the
isochronic surfaces are null planes of constant $x^{+}\propto x^{0}+x^{3}$,
the number of Poincaré generators independent of the interaction is maximum
[1], and, in the quantum theory, the vacuum state of the interacting theory is
far more simple than that of instant dynamics [3]. These peculiarities allow
the implementation of techniques which are almost impracticable in instant
dynamics, as Tamm-Dancoff’s truncation [4], which turns light-front field
theory into a very useful tool for the study of hadron physics [5].

In spite of this, the equivalence between the instant and the light-front
formulations of quantum field theory is not firmly established yet. This
problem for perturbative quantum electrodynamics (QED) was first addressed by
Ten Eyck and Rohrlich [6, 7] and by Yan [8, 9]. In their calculations,
Feynman’s amplitudes at one-loop level exhibited double-pole singularities
because of the instantaneous terms that appear in the gauge field propagator
when quantized in the null-plane gauge $A^{+}=0$; this problem was solved by
Pimentel and Suzuki [10, 11], who proposed a prescription to treat those poles
in a causal way. However, the importance of the instantaneous terms in the
fermion and gauge fields propagators is not clear yet; recent reviews on the
status quo of the gauge field propagator can be found in Refs. [12, 13]. Also,
very recently, the equivalence problem for one-loop radiative corrections was
studied in Refs. [14, 15], and the fulfilment of Ward-Takhashi’s identity at
one-loop order in Ref. [16].

Aiming to shed light to the subtleties of perturbative light-front field
theory, the authors developed the framework of null-plane causal perturbation
theory (CPT) [17, 18], an axiomatic approach to the $S$-matrix program
initiated by Heisenberg [19] in 1943, and axiomatized in the works by
Stückelberg and Rivier [20, 21] and Bogoliubov, Medvedev and Polivanov [22,
23, 24]. The detailed perturbative solution to Bogoliubov-Madvedev-Polivanov’s
axioms in instant dynamics was carried out in 1973 by Epstein and Glaser [25],
in a method in which the causality axiom plays an essential role, and its
first application to QED was done by Scharf in 1989 [26]. This approach has
the advantage of needing no regularization as the distributional character of
the quantized fields is considered. Additionally, no Feynman’s propagators
appear in loop distributions, which in light-front dynamics means that the
problems of the double poles previously referred are avoided. This program was
successfully applied to obtain the radiative corrections for Yukawa’s model
[27], directly showing the equivalence with the instant dynamics formulation
[28].

In a previous paper [29], that started the series of which the present one is
the second part, the authors started the study of QED in light-front dynamics
in the framework of null-plane CPT, in which the equivalence with instant
dynamics was accomplished for the scattering processes and vacuum
polarization, when gauge invariance is taken into account. More precisely, we
have seen that the photon quantized field operator in the null-plane gauge is
–we use latin indices $a,b,c,\cdots$ to denote the null-plane components of
vectors ($a=+,1,2,-$)–:

$\displaystyle A^{a}(x)=(2\pi)^{-3/2}$
$\displaystyle\times\sum_{\lambda=1,2}\int
d\mu(\bm{p})\varepsilon_{\lambda}(\bm{p})^{a}\left(a_{\lambda}(\bm{p})e^{-ipx}+a^{\dagger}_{\lambda}(\bm{p})e^{ipx}\right),$
(1)

in which only the physical degrees of freedom –the transversal polarization
vectors– appear. We are working with the following choice of polarization
vectors [30]:

$\varepsilon_{1}(\bm{p})^{a}=\left(0;1;0;-\frac{p_{1}}{p_{-}}\right),\
\varepsilon_{2}(\bm{p})^{a}=\left(0;0;1;-\frac{p_{2}}{p_{-}}\right),$
$\varepsilon_{+}(\bm{p})^{a}=\left(1;-\frac{p_{1}}{p_{-}};-\frac{p_{2}}{p_{-}};\frac{p_{\perp}^{2}}{2p_{-}^{2}}\right),\
\varepsilon_{-}(\bm{p})^{a}=\left(0;0;0;1\right).$ (2)

The photon emission and absorption field operators satisfy the following
commutation rule:

$\left[a_{\lambda}(\bm{p});a_{\sigma}^{\dagger}(\bm{q})\right]=2p_{-}\delta_{\lambda\sigma}\delta(\bm{p}-\bm{q}),$
(3)

from which the commutation relation for the gauge field can be derived:

$\left[A^{a}(x);A^{b}(y)\right]=:iD^{ab}(x-y),$ (4)

with the commutation distribution:

$\displaystyle D^{ab}(x)=$ $\displaystyle i(2\pi)^{-3}\int
d^{4}p\text{sgn}\left(p_{-}\right)\delta(p^{2})$
$\displaystyle\times\left(g^{ab}-\frac{p^{a}\eta^{b}+\eta^{a}p^{b}}{p_{-}}\right)e^{-ipx}.$
(5)

As we observe, the commutation distribution of the radiation field already
contains nonlocal terms (gauge terms). This is different to what happens with
the fermion field anticommutation distribution: It is covariant, and the
nonlocal term appears only in its retarded part [29]. Therefore, since in the
causal approach the causal distributions contain products of the positive- and
negative-frequency parts of the (anti)commutation distributions of the
quantized fields, the instantaneous term of the fermion field does not appear
in loop calculations, a fact that was exploited in Yukawa’s model [27] and for
the calculus of vacuum polarization in QED [29]. The situation here is
different because the positive- and negative-frequency parts of the
distribution in Eq. (I) already contain nonlocal terms, so they will appear in
the causal distributions corresponding to loop diagrams (in standard
language). As a consequence, it is of fundamental importance for the solution
of the equivalence problem to show that the “gauge terms” in Eq. (I) do not
contribute to any physical process at any order. It is clear that quantum
gauge invariance is the key for that task to be accomplished, hence the
present paper focuses on its implementation as a major part of the theory. The
technique for constructing quantum gauge theories in instant dynamics CPT was
developed by Dütsch, Hurth, Krahe and Scharf [31, 32, 33, 34], then applied by
Dütsch, Scharf and Aste [35, 36, 37] to the construction of non-Abelian gauge
theories, including the electroweak theory –see also Ref. [38]–. Other
important results regarding the uniqueness of the Yang-Mills theories can be
found in Refs. [39] and [40].

In CPT, the distributional character of quantum fields is taken into account.
As a consequence, the $S$-operator is a functional of the switching function
$g\in\mathscr{S}(\mathbb{R}^{4})$ [23] that multiply the coupling constant of
the interaction, isolating the problem of infrared divergences; it is through
the adiabatic limit $g\to 1$ that the real interaction is recovered. CPT is
constructed over the axioms of translation invariance and causality,
complemented with additional conditions as other symmetries and unitarity only
at a later stage for the normalization of the solution. The scattering
operator corresponding to an interaction regulated by
$g\in\mathscr{S}(\mathbb{R}^{4})$ is written as a formal series:

$S(g)=1+\sum\limits_{n=1}^{+\infty}\frac{1}{n!}\int dXT_{n}(X)g(X);$ (6)

with $T_{n}(X)\equiv T_{n}(x_{1};\ldots;x_{n})$, $g(X)\equiv g(x_{1})\ldots
g(x_{n})$, $dX\equiv d^{4}x_{1}\ldots d^{4}x_{n}$. This equation defines the
transition distributions $T_{n}\in\mathscr{S}^{\prime}(\mathbb{R}^{4n})$,
which are symmetrical in the coordinates $x_{1},\ldots,x_{n}$. The inverse
operator $S(g)^{-1}$ is obtained as the formal inverse of $S(g)$:

$S(g)^{-1}=1+\sum\limits_{n=1}^{+\infty}\frac{1}{n!}\int
dX\widetilde{T}_{n}(X)g(X);$ (7)
$\widetilde{T}_{n}(X)=\sum\limits_{r=1}^{n}(-1)^{r}\sum\limits_{\begin{subarray}{l}X_{1},\ldots,X_{r}\neq\emptyset\\\
X_{1}\cup\ldots\cup X_{r}=X\\\ X_{j}\cap X_{k}=\emptyset,\forall j\neq
k\end{subarray}}T_{n_{1}}(X_{1})\ldots T_{n_{r}}(X_{r}).$

As a consequence of causality, the transition distributions are
chronologically ordered –in the $x^{+}$ sense–:

$\displaystyle T_{n}(X)=T_{m}(X_{2})T_{n-m}(X_{1})\ \text{for}\ X_{1}<X_{2};$
$\displaystyle\left[T_{n}(X);T_{m}(Y)\right]=0\ \text{for}\ X\sim Y.$ (8)

Because of this, we can define the advanced distribution of order $n$ as the
following distribution:

$A_{n}(Y;x_{n})=\sum\limits_{\begin{subarray}{l}X\cup X^{\prime}=Y\\\ X\cap
X^{\prime}=\emptyset\end{subarray}}\widetilde{T}_{m}(X)T_{n-m}(X^{\prime}\cup\left\\{x_{n}\right\\}),$
(9)

and the retarded distribution of order $n$ as:

$R_{n}(Y;x_{n})=\sum\limits_{\begin{subarray}{l}X\cup X^{\prime}=Y\\\ X\cap
X^{\prime}=\emptyset\end{subarray}}T_{n-m}(X^{\prime}\cup\left\\{x_{n}\right\\})\widetilde{T}_{m}(X).$
(10)

In these distributions the $n$-point distribution appears once. Separating it
from the other terms:

$\displaystyle
A_{n}(Y;x_{n})=T_{n}(Y\cup\left\\{x_{n}\right\\})+A^{\prime}_{n}(Y;x_{n}),$
$\displaystyle
R_{n}(Y;x_{n})=T_{n}(Y\cup\left\\{x_{n}\right\\})+R^{\prime}_{n}(Y;x_{n}),$
(11)

with $A^{\prime}_{n}$ and $R^{\prime}_{n}$ the advanced subsidiary
distribution and the retarded subsidiary distribution, respectively, which do
not contain $T_{n}$. The transition distribution of order $n$ is then equal
to:

$\displaystyle T_{n}(Y\cup\left\\{x_{n}\right\\})$
$\displaystyle=R_{n}(Y;x_{n})-R^{\prime}_{n}(Y;x_{n}).$ (12)

Therefore, the $n$-point distribution can be found by obtaining the retarded
distribution of order $n$, which can be done by splitting [41, 42, 43] the
causal distribution of order $n$:

$\displaystyle D_{n}(Y;x_{n})$ $\displaystyle:=R_{n}(Y;x_{n})-A_{n}(Y;x_{n})$
$\displaystyle=R^{\prime}_{n}(Y;x_{n})-A^{\prime}_{n}(Y;x_{n}),$ (13)

which, at the light of the last equality, can be constructed with the
knowledge of the transition distributions up to order $n-1$, only. It must be
done as follows: The causal distribution has, in general, the following form:

$D_{n}(x_{1};\ldots;x_{n})=\sum_{k}d^{k}_{n}(x_{1};\ldots;x_{n})\;\bm{\mathrel{\mathop{:}}}\,\mathrel{C_{k}(u^{A})}\,\bm{\mathrel{\mathop{:}}}\;,$
(14)

with $d^{k}_{n}$ a numerical distribution and
$\;\bm{\mathrel{\mathop{:}}}\,\mathrel{C_{k}(u^{A})}\,\bm{\mathrel{\mathop{:}}}\;$
a Wick’s monomial of the quantized free field operators $u^{A}$. The support
properties of the operator-valued distribution are encoded into the numerical
distribution $d^{k}_{n}$, hence it is sufficient to split it. Using
translation invariance, define the numerical distribution
$d\in\mathscr{S}(\mathbb{R}^{4n-4})$ as:

$d(x):=d^{k}_{n}(x_{1}-x_{n};\ldots;x_{n-1}-x_{n};0),$ (15)

with $\text{supp}(d)\subseteq\Gamma^{+}_{n-1}(0)\cup\Gamma^{-}_{n-1}(0)$. It
must be split as:

$d=r-a;\ \text{supp}(r)\subseteq\Gamma^{+}_{n-1}(0),\
\text{supp}(a)\subseteq\Gamma^{-}_{n-1}(0).$ (16)

Here we are denoting:

$\displaystyle\Gamma^{+}_{n}(0):=\Big{\\{}(x_{1};\cdots;x_{n})\in\mathbb{M}^{n}\
\Big{|}\ \forall j\in\left\\{1,\cdots,n\right\\}:$ $\displaystyle
x_{j}^{+}\geq 0\wedge\ \left(\exists x_{k}\in\overline{V^{+}}(0)(k\neq
j):x_{j}\in\widetilde{V}^{+}(x_{k})\right)\Big{\\}},$

with $V^{\pm}(x)$ the interior of the future or past, respectively, light-cone
with vertex at the point $x$, $\overline{V^{\pm}}(x)$ its closure and
$\widetilde{V}^{\pm}(x)$ the union of its closure and the $x^{-}$ axis. An
analogous definition holds for $\Gamma^{-}_{n}(0)$. We are using Schwartz’s
multi-index notation [44].

To perform the splitting, it is crucial to remember that the product of a
distribution by a discontinuous function can be ill-defined if the
distribution has a singularity precisely on the discontinuity surface of the
function. In our case we then need to control the behavior of the causal
distribution near the splitting region, which is the $x^{-}$-axis. This can be
done by applying the concept of quasiasymptotics by selected variable [45]:

Definition.— Let $d\in\mathscr{S}^{\prime}(\mathbb{R}^{m})$ be a distribution,
and let $\rho$ be a continuous positive function. If the (distributional)
limit

$\lim_{s\to
0^{+}}\rho(s)s^{3m/4}d\left(sx^{+};sx^{\perp};x^{-}\right)=d_{-}(x)$ (17)

exists in $\mathscr{S}^{\prime}(\mathbb{R}^{m})$ and is non-null, then the
distribution $d_{-}$ is called the quasiasymptotics of $d$ at the $x^{-}$
axis, with regard to the function $\rho$.

One can then show [45, 26] that for every $a>0$: $\displaystyle{\lim_{s\to
0^{+}}\rho(as)/\rho(s)=a^{\alpha}}$ for some $\alpha\in\mathbb{R}$. This
number, denoted by $\omega_{-}$, characterizes the distribution, and is called
its singular order at the $x^{-}$ axis.

In momentum space the following splitting formulas are found: For negative
singular order, $\omega_{-}<0$:

$\displaystyle\hat{r}(p)=\frac{i}{2\pi}\int\limits_{-\infty}^{+\infty}\frac{\hat{d}(p_{+}-k;\bm{p})}{k+i0^{+}}dk.$
(18)

For non-negative singular order, $\omega_{-}\geq 0$, the retarded distribution
normalized at $\left(q_{+};q_{\perp};p_{-}\right)$ is:

$\displaystyle\hat{r}_{q}(p)=\frac{i}{2\pi}\int\limits_{-\infty}^{+\infty}\frac{dk}{k+i0^{+}}\Bigg{\\{}\hat{d}(p_{+}-k;\bm{p})$
$\displaystyle-\sum\limits_{|c|=0}^{\lfloor\omega_{-}\rfloor}\frac{1}{c!}(p_{+,\alpha}-q_{+,\alpha})^{c}D^{c}_{+,\alpha}\hat{d}(q_{+}-k;q_{\perp};p_{-})\Bigg{\\}}.$
(19)

Particularly, the central solution is the one normalized at the line
$(0;0_{\perp};p_{-})$.

Finally, if $r_{1}$ and $r_{2}$ are two solutions of the splitting problem,
then they could differ by normalization terms supported at the $x^{-}$ axis.
In momentum space:

$\hat{r}_{1}(p)-\hat{r}_{2}(p)=\sum\limits_{|b|=0}^{M}\widehat{C}_{b}\left(p_{-}\right)p_{+,\perp}^{b},$
(20)

with $\widehat{C}_{b}\left(p_{-}\right)$ some distributions of the variable
$p_{-}$. The procedure of fixing them by the imposition of physical
requirements is called the normalization process.

This paper has the following structure. The construction of the gauge
invariant theory is performed in Sec. II, while its more direct consequences
for QED, Ward-Takahashi’s identities, are presented in Sec. III. With those
tools, in Sec. IV we show the calculation of the electron self-energy and its
normalization. Sec. V contains our conclusions and perspectives of future
work. In appendix, finally, we prove the normalization of the physical
$S$-matrix of null-plane QED.

##  II Quantum gauge invariance

As we have seen in Ref. [29], the gauge terms of the radiation field
commutation distribution do not contribute to Møller’s scattering, which is an
expression of gauge invariance. It is our aim in this section to prove that
they in fact do not contribute to any process, in other words, that the
scattering operator can be constructed with the covariant terms only. To do
that, we will get back to the quantization procedure of the massless vector
field –see Ap. A in Ref. [18]–. Such a procedure had the meaning of
constructing Fock’s space and the operators as well as operator-valued
distributions acting on it. This Fock’s space consists on physical states,
only, whose wavefunctions are the positive-frequency part of the classical
field solutions. As a consequence, only physical polarization states are
quantized, leading to the instantaneous term in the commutation distribution
of the field operator. If we want to use a covariant commutation distribution,
it will be necessary to quantize the non-physical polarization states as well,
and, accordingly, to extend Fock’s space in order to contain the non-physical
states created by their corresponding field operators. This extension must be
done in such a way that the physical content of the theory is not altered, as
will be shown in the following paragraphs.

###  II.1 Fock’s space extension

Vector massless fields have only two degrees of freedom, identified with the
transversal polarizations $\varepsilon_{\alpha}(x)^{a}$ ($\alpha=1,2$). Fock’s
space of these physical states will be called the physical subspace,
$\mathcal{F}_{\text{phys}}$, of the complete Fock’s space, $\mathcal{F}$,
because the physical potentials will be those which satisfy both the null-
plane and Lorenz’s gauge conditions. All the sectors of $\mathcal{F}$
different from the physical subspace are inaccessible to experimental
observation; therefore, there are no reasons to expect that the expression of
the quantized field operator has the same form as its classical version:
Fock’s space extension is an eminently mathematical process.

If the gauge conditions are not imposed to the quantized field operator, then
we must quantize all the four polarization states $\lambda=+,1,2,-$, hence we
introduce not two, but four sets of emission and absorption operators
$a_{\lambda}^{\dagger}(f)$ and $a_{\lambda}(f)$. In order to have a positive-
definite inner product in the complete Fock’s space, we impose that they must
satisfy:

$\left[a_{\lambda}(\bm{p});a_{\lambda}^{\dagger}(\bm{q})\right]=2p_{-}\delta(\bm{p}-\bm{q}).$
(21)

As a consequence, all states, including the non-transverse ones, have
positive-definite norm. The most natural extension of the quantized radiation
field operator would be:

$\displaystyle A^{a}(x)=(2\pi)^{-3/2}$ $\displaystyle\times\sum_{\lambda}\int
d\mu(\bm{p})\varepsilon_{\lambda}(\bm{p})^{a}\left(a_{\lambda}(\bm{p})e^{-ipx}+a_{\lambda}^{\dagger}(\bm{p})e^{ipx}\right),$
(22)

with the sum extended to all the polarizations, including the non-physical
ones $\lambda=+,-$. However, such a field operator has the following
commutation distribution:

$\displaystyle\left[A^{a}(x);A^{b}(y)\right]=(2\pi)^{-3}\int
d^{4}p\delta(p^{2})\Theta(p_{-})$
$\displaystyle\quad\times\left(e^{-ip(x-y)}-e^{ip(x-y)}\right)\Big{(}-g^{ab}+\frac{p^{a}\eta^{b}+\eta^{a}p^{b}}{p_{-}}$
$\displaystyle\quad+\varepsilon_{+}(\bm{p})^{a}\varepsilon_{+}(\bm{p})^{b}+\varepsilon_{-}(\bm{p})^{a}\varepsilon_{-}(\bm{p})^{b}\Big{)},$

which also exhibits instantaneous terms. In order to obtain a covariant
commutation distribution, as we need, we must only have the term $-g^{ab}$
inside the parentheses of the second line of the above equation. We will see
that this is possible with a convenient redefinition of the quantized field
operators associated to non-physical polarizations. Let us write:

$A^{a}(x)=(2\pi)^{-3/2}\sum_{\lambda}\int
d\mu(\bm{p})\varepsilon_{\lambda}(\bm{p})^{a}A_{\lambda}(\bm{p};x).$ (23)

This expression coincides with the real quantized field operator for the
physical polarization states if:

$A_{1,2}(\bm{p};x)=a_{1,2}(\bm{p})e^{-ipx}+a_{1,2}^{\dagger}(\bm{p})e^{ipx}.$
(24)

The commutator is then:

$\displaystyle\left[A^{a}(x);A^{b}(y)\right]=(2\pi)^{-3}\int
d\mu(\bm{p})d\mu(\bm{q})$
$\displaystyle\quad\times\sum_{\lambda,\lambda^{\prime}}\varepsilon_{\lambda}(\bm{p})^{a}\varepsilon_{\lambda^{\prime}}(\bm{q})^{b}\left[A_{\lambda}(\bm{p};x);A_{\lambda^{\prime}}(\bm{q};y)\right].$
(25)

Taking in mind the completeness relation of the polarization vectors, which
is:

$\sum_{\lambda,\lambda^{\prime}}g_{\lambda\lambda^{\prime}}\varepsilon_{\lambda}(\bm{p})_{a}\varepsilon_{\lambda^{\prime}}(\bm{p})_{b}=g_{ab},$
(26)

the commutation distribution will be the one we need if:

$\displaystyle\left[A_{\lambda}(\bm{p};x);A_{\lambda^{\prime}}(\bm{q};y)\right]=$
$\displaystyle\quad-2p_{-}\delta(\bm{p}-\bm{q})g_{\lambda\lambda^{\prime}}\left(e^{-i(px-
qy)}-e^{i(px-qy)}\right).$ (27)

This relation is satisfied for $\lambda=1,2$ with $A_{1,2}(\bm{p};x)$ from Eq.
(24). The other components are also obtained if we take:

$A_{\pm}(\bm{p};x)=a_{\pm}(\bm{p})e^{-ipx}-a_{\mp}^{\dagger}(\bm{p})e^{ipx}.$
(28)

Accordingly, we shall define the quantized massless vector field operator as:

$\displaystyle A^{a}(x):=(2\pi)^{-3/2}\sum_{\lambda}\int
d\mu(\bm{p})\varepsilon_{\lambda}(\bm{p})^{a}$
$\displaystyle\times\left(a_{\lambda}(\bm{p})e^{-ipx}-\sum_{\tau}g_{\lambda\tau}a_{\tau}^{\dagger}(\bm{p})e^{ipx}\right),$
(29)

which satisfies a covariant commutation relation:

$\left[A^{a}(x);A^{b}(y)\right]=ig^{ab}D_{0}(x-y),$ (30)

with $D_{0}(x)$ the massless Jordan-Pauli’s distribution. It will also be
useful to have the commutator of the negative and positive frequency parts of
the field 111The sub-index indicating the frequency sign will be put at the
right of the index of the component. For example, the positive frequency part
of the longitudinal component of the field is:
$A\indices{{}^{-}_{+}}(x)=A_{++}(x)$, and so on.:

$\displaystyle\left[A\indices{{}^{a}_{-}}(x);A\indices{{}^{b}_{+}}(y)\right]=-(2\pi)^{-3}\sum_{\lambda,\lambda^{\prime}}\int
d\mu(\bm{p})d\mu(\bm{q})$
$\displaystyle\quad\times\varepsilon_{\lambda}(\bm{p})^{a}\varepsilon_{\lambda^{\prime}}(\bm{q})^{b}\sum_{\tau^{\prime}}g_{\lambda^{\prime}\tau^{\prime}}\left[a_{\lambda}(\bm{p});a_{\tau^{\prime}}^{\dagger}(\bm{q})\right]e^{-i(px-
qy)}$ $\displaystyle=-(2\pi)^{-3}\int
d\mu(\bm{p})\sum_{\lambda\lambda^{\prime}}g_{\lambda\lambda^{\prime}}\varepsilon_{\lambda}(\bm{p})^{a}\varepsilon_{\lambda^{\prime}}(\bm{p})^{b}e^{-ip(x-y)}$
$\displaystyle=ig^{ab}D_{+}(x-y).$ (31)

And, analogously:

$\left[A\indices{{}^{a}_{+}}(x);A\indices{{}^{b}_{-}}(y)\right]=ig^{ab}D_{-}(x-y).$
(32)

This is the desired result. Now Fock’s space contains photons with the four
polarization degrees, among which only the transversal ones can constitute
asymptotically free states. In this extended space the positive-definite
Hamiltonian operator is:

$\bm{\mathsf{P}}_{+}=\int
d\mu(\bm{p})p_{+}\sum_{\lambda}a_{\lambda}^{\dagger}(\bm{p})a_{\lambda}(\bm{p}),$
(33)

because with it the field operator of Eq. (II.1) satisfies Heisenberg’s
equation of motion:

$\displaystyle i\partial_{+}A^{a}(x)$
$\displaystyle=(2\pi)^{-3}\sum_{\lambda}\int
d\mu(\bm{p})p_{+}\varepsilon_{\lambda}(\bm{p})^{a}$
$\displaystyle\quad\times\left(a_{\lambda}(\bm{p})e^{-ipx}+\sum_{\tau}g_{\lambda\tau}a_{\tau}^{\dagger}(\bm{p})e^{ipx}\right)$
$\displaystyle=\left[A^{a}(x);\bm{\mathsf{P}}_{+}\right].$ (34)

What remains in order to prove that the gauge terms in the radiation field
commutation distribution in Eq. (I) do not contribute is to show that the
physics does not change when the extended field is used. The proof of this
statement is the topic of the following paragraphs of this section.

###  II.2 Poincaré’s invariance of the physical subspace

Let us consider the null-plane and Lorenz’s gauge conditions when applied to
the quantized field operator. Since the “$+$” component of all polarization
vectors, except the one of $\varepsilon_{+}(\bm{p})^{a}$, is null [see Eq.
(2)], we have that:

$\displaystyle A^{+}(x)=(2\pi)^{-3/2}$ $\displaystyle\times\int
d\mu(\bm{p})\varepsilon_{+}(\bm{p})^{+}\left(a_{+}(\bm{p})e^{-ipx}-a_{-}^{\dagger}(\bm{p})e^{ipx}\right).$
(35)

Also, the divergence of the quantized field operator is:

$\displaystyle\partial_{a}A^{a}(x)=-i(2\pi)^{-3/2}\sum_{\lambda}\int
d\mu(\bm{p})p_{a}\varepsilon_{\lambda}(\bm{p})^{a}$
$\displaystyle\times\left(a_{\lambda}(\bm{p})e^{-ipx}+\sum_{\tau}g_{\lambda\tau}a_{\tau}^{\dagger}(\bm{p})e^{ipx}\right).$
(36)

But, as it can be seen in Eq. (2), the polarization vectors are chosen in such
a way that:

$\varepsilon_{1,2}(\bm{p})_{-}=0\ ,\
\varepsilon_{1,2}(\bm{p})_{+}=-\frac{p_{\alpha}\varepsilon_{1,2}(\bm{p})^{\alpha}}{p_{-}},$
(37)

which imply that they are orthogonal to the momentum $p^{a}$:
$p_{a}\varepsilon_{1,2}(\bm{p})^{a}=0$. Therefore, in the sum in Eq. (II.2)
only the polarization states $\lambda=+,-$ contribute:

$\displaystyle\partial_{a}A^{a}(x)=-i(2\pi)^{-3/2}\sum_{\lambda=+,-}\int
d\mu(\bm{p})p_{a}\varepsilon_{\lambda}(\bm{p})^{a}$
$\displaystyle\times\left(a_{\lambda}(\bm{p})e^{-ipx}+\sum_{\tau}g_{\lambda\tau}a_{\tau}^{\dagger}(\bm{p})e^{ipx}\right),$
(38)

and only the emission and absorption field operators
$a_{\pm}^{\dagger}(\bm{p})$ and $a_{\pm}(\bm{p})$ appear. Since every state in
$\mathcal{F}_{\text{phys}}$ has polarizations $\lambda=1,2$, Eqs. (II.2) and
(II.2) mean that this subspace can be characterized by the accomplishment of
the gauge conditions as matrix elements:

$\forall\Phi,\Psi\in\mathcal{F}_{\text{phys}}:\
\left(\Phi;A^{+}(x)\Psi\right)=0\ \wedge\
\left(\Phi;\partial_{a}A^{a}(x)\Psi\right)=0.$ (39)

Let $f=(f_{a})$ be a wave-function in the extended one-particle Hilbert’s
space; it is transformed under a Poincaré’s transformation by the classical
field law:

$\left(U(a;\Lambda)f\right)(x)=\Lambda^{-1}f\left(\Lambda^{-1}(x-a)\right).$
(40)

Over the space of these functions we define the operator $A(f)$ according to:

$A(f):=\int d^{4}xf_{a}(x)A^{a}(x);$ (41)

this is the smearing of the operator-valued distribution $A^{a}(x)$ over the
test-function $f_{a}$. The operator of Poincaré’s transformation which acts on
the extended Fock’s space is the operator $\bm{\mathsf{U}}(a;\Lambda)$ such
that:

$\bm{\mathsf{U}}(a;\Lambda)A(f)\bm{\mathsf{U}}(a;\Lambda)^{-1}=A\left(U(a;\Lambda)f\right).$
(42)

Hence, using Eqs. (40) and (41) we obtain the transformation law of the
quantized field operator:

$\bm{\mathsf{U}}(a;\Lambda)A^{a}(x)\bm{\mathsf{U}}(a;\Lambda)^{-1}=\left(\Lambda^{-1}\right)\indices{{}^{a}_{b}}A^{b}(\Lambda
x+a).$ (43)

The application of the “$\dagger$” adjoint to this equation allows us to see
that the operator $\bm{\mathsf{U}}(a;\Lambda)$ is not unitary because
$A^{a}(x)$ is not Hermitian. This does not violate Wigner’s theorem:
Poincaré’s transformations are symmetries in the real world, so that they must
be symmetries in the physical subspace only. In order to show that this is the
case, denote by $\mathcal{F}_{L}$ the subspace of $\mathcal{F}$ which
satisfies Lorenz’s gauge condition. In Ref. [47] it is shown that this gauge
condition is compatible with the null-plane one in the free case –which is
always the case in CPT–, and, equally important, that these two conditions
determine the gauge completely –i.e., there is no remnant gauge symmetry after
the imposition of them–; therefore, we can affirm:

$\displaystyle\forall\Phi^{\prime}\in$ $\displaystyle\mathcal{F}_{L}:\
\exists!\Phi\in\mathcal{F}_{\text{phys}}:$
$\displaystyle\Phi^{\prime}_{a}(x)=\Phi_{a}(x)+\partial_{a}\Lambda(x),\
\square\Lambda(x)=0.$ (44)

We denote the projection onto the physical subspace by: $\Phi=P\Phi^{\prime}$.
Now, the inner product in the one-particle space is:

$\displaystyle(f;g)_{1}$ $\displaystyle=\sum_{a}i\int
f_{a}(x)^{*}\overleftrightarrow{\partial}_{-}g_{a}(x)d^{3}\bm{x}$
$\displaystyle=\sum_{a}\int
d\mu(\bm{p})\hat{f}_{a}(\bm{p})^{*}\hat{g}_{a}(\bm{p}).$ (45)

Let $\Phi,\Psi\in\mathcal{F}_{\text{phys}}$ be one-particle states, with wave-
functions $\Phi_{a}$ and $\Psi_{a}$, respectively. They satisfy the null-plane
gauge: $\Phi^{+}=0=\Psi^{+}$, hence their inner product can be put in
Poincaré’s invariant form:

$\displaystyle(\Phi;\Psi)$
$\displaystyle=\sum_{\alpha}i\int\Phi_{\alpha}(x)^{*}\overleftrightarrow{\partial}_{-}\Psi_{\alpha}(x)d^{3}\bm{x}$
$\displaystyle=-\sum_{\alpha}i\int\Phi_{\alpha}(x)^{*}\overleftrightarrow{\partial}_{-}\Psi^{\alpha}(x)d^{3}\bm{x}$
$\displaystyle=-(\Phi_{a};\Psi^{a}).$ (46)

Let us apply now a reference frame transformation. In the general case the
value of $\Phi^{+}=0$ changes, and the transformed states $\Phi^{\prime}$ and
$\Psi^{\prime}$ are no more in $\mathcal{F}_{\text{phys}}$, but they are in
$\mathcal{F}_{L}$ yet. Applying the operator $P$ to obtain again states in
$\mathcal{F}_{\text{phys}}$:

$\widetilde{\Phi}=P\bm{\mathsf{U}}(a;\Lambda)\Phi=:\widetilde{\bm{\mathsf{U}}}(a;\Lambda)\Phi,\
\widetilde{\Psi}=\widetilde{\bm{\mathsf{U}}}(a;\Lambda)\Psi.$ (47)

Therefore, the operator which transforms the states in
$\mathcal{F}_{\text{phys}}$ is not $\bm{\mathsf{U}}(a;\Lambda)$, but
$\widetilde{\bm{\mathsf{U}}}(a;\Lambda)$, and it is this one that must be
unitary on $\mathcal{F}_{\text{phys}}$. Effectively it is: Since
$\widetilde{\Phi}$ and $\Phi^{\prime}$ are related by a gauge transformation,
as well as $\widetilde{\Psi}$ and $\Psi^{\prime}$, we write:

$\Phi^{\prime}_{a}=\widetilde{\Phi}_{a}+\partial_{a}\chi,\
\Psi^{\prime}_{a}=\widetilde{\Psi}_{a}+\partial_{a}\Lambda.$ (48)

Then the inner product in Eq. (II.2), which is Poincaré’s invariant, is equal
to:

$\displaystyle(\Phi;\Psi)=-(\Phi^{\prime}_{a};\Psi^{\prime a})$
$\displaystyle=\left(\widetilde{\Phi};\widetilde{\Psi}\right)-\left(\partial_{a}\chi;\widetilde{\Psi}^{a}\right)-\left(\widetilde{\Phi}_{a};\partial^{a}\Lambda\right)-\left(\partial_{a}\chi;\partial^{a}\Lambda\right).$
(49)

It is easy to see that the second and third terms in the last equality are
null, because $\widetilde{\Phi}$ and $\widetilde{\Psi}$ satisfy the gauge
conditions and because they are assumed to vanish at infinity (asymptotic
conditions). Using again the asymptotic conditions, the last term takes the
form:

$\displaystyle(\partial_{a}\chi;\partial^{a}\Lambda)=(\partial_{+}\chi;\partial_{-}\Lambda)+(\partial_{-}\chi;\partial_{+}\Lambda)-(\partial_{\alpha}\chi;\partial_{\alpha}\Lambda)$
$\displaystyle=-(\partial_{+}\partial_{-}\chi;\Lambda)-(\chi;\partial_{+}\partial_{-}\Lambda)+\frac{1}{2}(\partial_{\perp}^{2}\chi;\Lambda)+\frac{1}{2}(\chi;\partial_{\perp}^{2}\Lambda)$
$\displaystyle=-\frac{1}{2}(\square\chi;\Lambda)-\frac{1}{2}(\chi;\square\Lambda),$

which are null as established in Eq. (II.2). In conclusion:

$(\Phi;\Psi)=\left(\widetilde{\Phi};\widetilde{\Psi}\right)=\left(\widetilde{\bm{\mathsf{U}}}(a;\Lambda)\Phi;\widetilde{\bm{\mathsf{U}}}(a;\Lambda)\Psi\right),$
(50)

and Poincaré’s transformations operator in $\mathcal{F}_{\text{phys}}$,
$\widetilde{\bm{\mathsf{U}}}(a;\Lambda)$, is a unitary operator. This shows
that the physical subspace is the same in all reference frames.

###  II.3 Quantum gauge transformations

As we have said, the extension of Fock’s space is not a physical procedure,
but a mathematical one. Accordingly, we now have to impose that the physical
content of the theory is independent of this extension.

Definition.— A quantum gauge transformation is a transformation of the
quantized field, or equivalently, of the quantized field operator, which
depends continuously on a constant parameter $\lambda$ and has the form

$A^{\prime}(f)=e^{-i\lambda Q}A(f)e^{i\lambda Q}\ \Leftrightarrow\ A^{\prime
a}(x)=e^{-i\lambda Q}A^{a}(x)e^{i\lambda Q},$ (51)

form which warrants (1) that the commutation distribution of the transformed
field operator does not change, and (2) that the quantized field operator
still satisfies the equation of motion $\square A^{\prime a}(x)=0$. Also, we
impose the condition that $e^{i\lambda Q}$ is an operator which leaves
invariant the states of $\mathcal{F}_{\text{phys}}$. The operator $Q$,
generator of the quantum gauge transformations, is called the gauge charge
operator.

The most immediate consequence of this definition is that the gauge charge
operator annihilates physical states, hence:

$\mathcal{F}_{\text{phys}}\subseteq\text{Ker}(Q).$ (52)

But we do not know yet under what conditions the equality holds in Eq. (52).
Let us start by establishing:

Lemma: The gauge charge operator $Q$, as well as its adjoint $Q^{\dagger}$, is
constructed with non-physical emission and absorption operators, only.

Proof: In first place, let us see that the non-physical states are orthogonal
to the physical ones; in effect, let $a_{\text{phys}}^{\dagger}$ and
$a_{\text{nph}}^{\dagger}$ be emission operators of physical and non-physical
particles, respectively. Since they commute:

$\displaystyle\left(a_{\text{phys}}^{\dagger}\Omega;a_{\text{nph}}^{\dagger}\Omega\right)=\left(\Omega;a_{\text{phys}}a_{\text{nph}}^{\dagger}\Omega\right)$
$\displaystyle\quad=\left(\Omega;a_{\text{nph}}^{\dagger}a_{\text{phys}}\Omega\right)=0.$
(53)

Consider now two states $\Phi,\Psi\in\mathcal{F}_{\text{phys}}$:

$0=(\Phi;Q\Psi)=\left(Q^{\dagger}\Phi;\Psi\right)\ \Rightarrow\
Q^{\dagger}\Phi\in\mathcal{F}_{\text{phys}}^{\perp}.$ (54)

Now, every operator acting on Fock’s space can be written as a function of
emission and absorption operators. Particularly, in order to $Q$ to annihilate
every physical state it is mandatory that all the terms in it have an
absorption operator of a non-physical particle at the right. Let us suppose
that one of the terms has a physical emission operator at the left:

$Q\sim a_{\text{phys}}^{\dagger}a_{\text{nph}}+\ldots$ (55)

Then, applying the adjoint to a physical state $\Phi$:

$Q^{\dagger}\Phi\sim a_{\text{nph}}^{\dagger}a_{\text{phys}}\Phi+\ldots,$ (56)

which in general still contains physical particles, in contradiction with Eq.
(54). Hence, $Q$ could not contain physical emission operators, and the same
is clearly true for physical absorption operators. From this, it follows that
$Q^{\dagger}$ also contains non-physical emission and absorption operators,
hence $\forall\Phi\in\mathcal{F}_{\text{phys}}$: $Q^{\dagger}\Phi=0$.
$\blacksquare$

As a consequence of this lemma, Eq. (52) must be substituted by:

$\mathcal{F}_{\text{phys}}\subseteq\text{Ker}(Q)\cap\text{Ker}\left(Q^{\dagger}\right)=\text{Ker}\left(\left\\{Q;Q^{\dagger}\right\\}\right).$
(57)

Expanding the exponentials in Eq. (51) as a series in the parameter $\lambda$,
we will find that:

$\displaystyle A^{\prime a}(x)=$ $\displaystyle
A^{a}(x)-i\lambda\left[Q;A^{a}(x)\right]$
$\displaystyle-\frac{\lambda^{2}}{2}\left[Q;[Q;A^{a}(x)]\right]+\mathscr{O}(\lambda^{3}).$
(58)

By virtue of the lemma, all the commutators in Eq. (II.3) are non-null only
for the non-physical part of the quantized field operator $A^{a}(x)$; in other
words, the quantum gauge transformation does not modify the dynamical
(physical) part of the quantized field, as required, and all the quantities
constructed with $A^{a}(x)$ will have the same matrix elements on the physical
subspace: This is quantum gauge invariance.

Now, every operator $Q$, constructed with non-physical emission and absorption
operators in such a way that an emission operator is at the left and an
absorption operator is at the right of every term, can be used as a generator
of a quantum gauge transformation. However, such a general gauge charge could
originate a transformed gauge field which is a composite operator. We must
impose that $Q$ is a quadratic operator in order to maintain $A^{a}(x)$ as a
simple field operator.

For a general gauge charge, Eq. (57) is satisfied with the symbol “$\subset$”;
the equality holds if $\left\\{Q;Q^{\dagger}\right\\}$ is an operator in which
all the non-physical absorption operators appear. In order to obtain such a
gauge charge, note that $[Q;A^{a}(x)]$ is always a linear combination of non-
physical emission and absorption operators. Therefore, regarding Eq. (II.3),
define the field operator $u(x)$ by:

$i\partial_{a}u(x):=\left[Q;A_{a}(x)\right].$ (59)

Eq. (II.3) adopts the form:

$A^{\prime
a}(x)=A^{a}(x)+\lambda\partial^{a}u(x)-\frac{i\lambda^{2}}{2}\left[Q;\partial^{a}u(x)\right]+\mathscr{O}(\lambda^{3}).$
(60)

But, by definition, $A^{\prime a}(x)$ must satisfy Klein-Gordon-Fock’s
equation, which is possible if $u(x)$ satisfies:

$\square u(x)=0.$ (61)

The solution for this equation is:

$u(x)=\int\limits_{y^{+}=y_{0}^{+}}u(y)\overleftrightarrow{\partial}_{-}^{y}D_{0}(y-x).$
(62)

Deriving Eq. (62) and recognizing the commutation distribution of the extended
gauge field operator of Eq. (30):

$i\partial_{a}u(x)=\left[\int\limits_{y^{+}=y_{0}^{+}}\partial_{b}A^{b}(y)\overleftrightarrow{\partial}_{-}u(y)d^{3}\bm{y};A_{a}(x)\right],$
(63)

from which we identify, by comparison with Eq. (59), that the gauge charge
operator is:

$Q=\int\limits_{x^{+}=x_{0}^{+}}\partial_{a}A^{a}(x)\overleftrightarrow{\partial}_{-}u(x)d^{3}\bm{x}.$
(64)

On the other hand, Eq. (61) implies that the field operator $u(x)$ has the
form:

$u(x)=(2\pi)^{-3/2}\int
d\mu(\bm{p})\left(c_{2}(\bm{p})e^{-ipx}+c_{1}^{\dagger}(\bm{p})e^{ipx}\right).$
(65)

The field operators $c_{1}(\bm{p})$ and $c_{2}(\bm{p})$ are, at the moment,
unknown, but they belong to non-physical “particles”222This is, of course, an
abuse of language, because the concept of particle can only be defined in the
asymptotically free region, whereas a “non-physical particle” cannot reach
that region.. Joining Eqs. (II.2) and (65) into Eq. (64) and by an elemental
integration, we find:

$\displaystyle Q=$ $\displaystyle\sum\limits_{\lambda=+,-}\int
d\mu(\bm{p})p_{a}\varepsilon_{\lambda}(\bm{p})^{a}$
$\displaystyle\times\left(a_{\lambda}(\bm{p})c_{1}^{\dagger}(\bm{p})-\sum_{\tau}g_{\lambda\tau}a_{\tau}^{\dagger}(\bm{p})c_{2}(\bm{p})\right).$
(66)

By construction, this is the most general form that a quadratic gauge charge
operator can have. This formula reveals: Since $Q$ must have absorption
operators to the right, and this does not occur in the first term of Eq.
(II.3), we must commute $a_{\lambda}(\bm{p})$ with $c_{1}^{\dagger}(\bm{p})$;
but, in doing that, if $c_{1}(\bm{p})$ were one of $a_{+}(\bm{p})$ or
$a_{-}(\bm{p})$, then $Q$ would gain a constant term and does not annihilate
the physical states. This is impossible. In consequence, it is necessary to
extend Fock’s space even more in order to contain field operators
$c_{1}(\bm{p})$ and $c_{2}(\bm{p})$ and their adjoints, associated to non-
physical particles and which are different from $a_{+}(\bm{p})$ and
$a_{-}(\bm{p})$. The quantized field operator $u(x)$, as a new non-physical
field, is called ghost field. As we see, the ghost field is indispensable for
quantum gauge invariance. Since $c_{1}^{\dagger}(\bm{p})$ corresponds to a new
particle, it can be simply commutated with $a_{\lambda}(\bm{p})$. From this in
Eq. (II.3) it follows that:

$\displaystyle Q=$ $\displaystyle\sum\limits_{\lambda=+,-}\int
d\mu(\bm{p})p_{a}\varepsilon_{\lambda}(\bm{p})^{a}$
$\displaystyle\times\left(c_{1}^{\dagger}(\bm{p})a_{\lambda}(\bm{p})-\sum_{\tau}g_{\lambda\tau}a_{\tau}^{\dagger}(\bm{p})c_{2}(\bm{p})\right).$
(67)

The explicit form of the polarization vectors of the massless vector field are
given in Eq. (2), from which it follows that:

$p_{a}\varepsilon_{+}(\bm{p})^{a}=p_{+}-\frac{p_{\perp}^{2}}{2p_{-}}=\frac{p^{2}}{2p_{-}},\
p_{a}\varepsilon_{-}(\bm{p})^{a}=p_{-}.$ (68)

Introducing this into Eq. (II.3) and taking in mind that for the massless
field it is $p^{2}=0$, we arrive at the final expression:

$Q=\int
d\mu(\bm{p})p_{-}\left(c_{1}^{\dagger}(\bm{p})a_{-}(\bm{p})-a_{+}^{\dagger}(\bm{p})c_{2}(\bm{p})\right).$
(69)

The adjoint operator is:

$Q^{\dagger}=\int
d\mu(\bm{p})p_{-}\left(a_{-}^{\dagger}(\bm{p})c_{1}(\bm{p})-c_{2}^{\dagger}(\bm{p})a_{+}(\bm{p})\right).$
(70)

In order to characterize the physical subspace we are interested in the
anticommutator of $Q$ and $Q^{\dagger}$ [see Eq. (57)]. Using Eqs. (69) and
(70):

$\displaystyle\left\\{Q;Q^{\dagger}\right\\}=\int
d\mu(\bm{p})d\mu(\bm{q})p_{-}q_{-}\Big{(}2p_{-}\delta(\bm{p}-\bm{q})c_{1}^{\dagger}(\bm{p})c_{1}(\bm{p})$
$\displaystyle+2p_{-}\delta(\bm{p}-\bm{q})c_{2}^{\dagger}(\bm{p})c_{2}(\bm{p})+\left\\{c_{1}(\bm{q});c_{1}^{\dagger}(\bm{p})\right\\}a_{-}^{\dagger}(\bm{q})a_{-}(\bm{p})$
$\displaystyle+\left\\{c_{2}(\bm{p});c_{2}^{\dagger}(\bm{q})\right\\}a_{+}^{\dagger}(\bm{p})a_{+}(\bm{q})\Big{)}.$
(71)

The first two terms in this equation have the desired form. For the other two,
if $c_{1}(\bm{p})$ and $c_{2}(\bm{p})$ were boson operators, then the
anticommutator $\left\\{Q;Q^{\dagger}\right\\}$ would have terms of the form
$a_{-}^{\dagger}(\bm{q})c_{1}^{\dagger}(\bm{p})c_{1}(\bm{q})a_{-}(\bm{p})$,
and there will be non-physical states annihilated by
$\left\\{Q;Q^{\dagger}\right\\}$, for example, a one particle state with
longitudinal polarization. As this is not required and contradicts our
hypothesis on the physical degrees of freedom of the gauge field, the
quantized field operator $u$ must be a fermion field, with emission and
absorption field operators subjected to the anticommutation rules:

$\displaystyle\left\\{c_{1}(\bm{p});c_{1}^{\dagger}(\bm{q})\right\\}=2p_{-}\delta(\bm{p}-\bm{q}),$
$\displaystyle\left\\{c_{2}(\bm{p});c_{2}^{\dagger}(\bm{q})\right\\}=2p_{-}\delta(\bm{p}-\bm{q}).$
(72)

And, hence:

$\left\\{u(x);\widetilde{u}(y)\right\\}=-iD_{0}(x-y).$ (73)

The field $\widetilde{u}(x)$ is called antighost field, and is given by:

$\widetilde{u}(x)=(2\pi)^{-3/2}\int
d\mu(\bm{p})\left(-c_{1}(\bm{p})e^{-ipx}+c_{2}^{\dagger}(\bm{p})e^{ipx}\right).$
(74)

Being that way, Eq. (71) adopts the final form:

$\displaystyle\left\\{Q;Q^{\dagger}\right\\}$ $\displaystyle=\int
d\mu(\bm{p})p_{-}^{2}\Big{(}a_{-}^{\dagger}(\bm{p})a_{-}(\bm{p})+a_{+}^{\dagger}(\bm{p})a_{+}(\bm{p})$
$\displaystyle+c_{1}^{\dagger}(\bm{p})c_{1}(\bm{p})+c_{2}^{\dagger}(\bm{p})c_{2}(\bm{p})\Big{)}.$
(75)

And now we can affirm:

$\mathcal{F}_{\text{phys}}=\text{Ker}\left(\left\\{Q;Q^{\dagger}\right\\}\right).$
(76)

Also, as a consequence of Eq. (69), and by direct calculation, it is immediate
to see that the gauge charge operator is nilpotent:

$Q^{2}=\frac{1}{2}\left\\{Q;Q\right\\}=0.$ (77)

The fermion statistics of the ghost particles implies, in Eq. (60), that the
parameter $\lambda$ is a number following a Grassmannian multiplication rule,
so the series ends in the term linear in $\lambda$: The quantum gauge
transformation of the operator $F(x)$, containing gauge, ghosts and anti-
ghosts field operators, is exactly:

$F^{\prime}(x)=e^{-i\lambda Q}F(x)e^{i\lambda Q}=F(x)-i\lambda
QF(x)+iF(x)\lambda Q.$ (78)

Since $\lambda$ is a Grassmann’s number, we cannot simply commute it with
$F(x)$ in the last term of Eq. (78); in putting it at the left of $F(x)$ we
must write a factor $(-1)^{n_{F}}$, with $n_{F}$ the so-called ghost number of
the operator $F(x)$, equal to the difference between the number of ghosts and
anti-ghost fields contained in it. Therefore, we define the gauge variation of
the operator $F(x)$ as:

$d_{Q}F(x):=\left[Q;F(x)\right]_{(-1)^{1+n_{F}}}.$ (79)

From Eq. (78) it follows that the exact gauge transformation is given by:

$F^{\prime}(x)=F(x)-i\lambda d_{Q}F(x).$ (80)

And from this equation we see that, necessarily, $d_{Q}$ changes the
statistics of the operator $F(x)$, in such a way that $\lambda d_{Q}F(x)$
maintains the one of $F(x)$.

Let us enumerate some properties of the gauge variation $d_{Q}$. Firstly, it
satisfies the following rule for the product of operators:

$d_{Q}(F(x)G(y))=(d_{Q}F(x))G(y)+(-1)^{n_{F}}F(x)(d_{Q}G(y)).$ (81)

Secondly, it is nilpotent:

$d_{Q}^{2}=0.$ (82)

Thirdly, from Eq. (59) they follow the gauge variations of the field
operators:

$d_{Q}A\indices{{}^{a}_{\pm}}=i\partial^{a}u_{\pm},$ (83) $d_{Q}u=0,$ (84)
$d_{Q}\widetilde{u}_{\pm}=-i\partial_{a}A\indices{{}^{a}_{\pm}}.$ (85)

And fourthly, it satisfies the following result, proved in Ref. [38]:

Lemma: Let $G$ be a Wick’s monomial containing gauge fields $A$, ghosts $u$
and anti-ghosts $\widetilde{u}$. The gauge variation of $G$ commutes with its
normal ordering:

$d_{Q}\;\bm{\mathrel{\mathop{:}}}\,\mathrel{G}\,\bm{\mathrel{\mathop{:}}}\;=\;\bm{\mathrel{\mathop{:}}}\,\mathrel{d_{Q}G}\,\bm{\mathrel{\mathop{:}}}\;.$
(86)

###  II.4 Quantum gauge invariance of the scattering operator

The concept of quantum gauge invariance was already introduced in the
commentaries after Eq. (II.3): It is the independence of the physical
quantities from the mathematical operation of extending Fock’s space. This
kind of symmetries frequently occurs in physics; examples of it are the
principle of general covariance, the one of classical gauge symmetry, et
cetera. In the $S$-matrix program the fundamental physical quantity is the
transition amplitude between two physical states. Then [38]:

Definition.— Let $P$ be the projection operator onto the physical subspace
$\mathcal{F}_{\text{phys}}$. Two scattering operators $S(g)$ and
$S^{\prime}(g)$ are called physically equivalent if for every states
$\Phi,\Psi\in\mathcal{F}$:

$\lim_{g\to 1}\left(\Phi;PS(g)P\Psi\right)=\lim_{g\to
1}\left(\Phi;PS^{\prime}(g)P\Psi\right),$ (87)

if the adiabatic limit exists.

The condition of existence of the adiabatic limit is essential, because only
in that case the bilinear form $\left(\Phi;PS(g)P\Psi\right)$ has physical
meaning; such existence will be assumed in the present discussion. Inserting
in this definition the formal series of Eq. (6) we obtain a perturbative
version of it:

Definition.— Two scattering operators $S(g)$ and $S^{\prime}(g)$ are
(perturbatively) physically equivalent at the $n$-th order if their
corresponding $n$-point distributions satisfy –the limit is weak in the sense
of Eq. (87)–:

$\text{w-}\lim_{g\to 1}\int
dX\left(PT_{n}(X)P-PT^{\prime}_{n}(X)P\right)g(X)=0,$ (88)

if the adiabatic limit exists for the order $n$ of the perturbation series.

Eq. (88) has the general solution
$PT_{n}(X)P=PT^{\prime}_{n}(X)P+\partial(\ldots)$, with $\partial(\ldots)$
meaning the divergence of the quantity $(\ldots)$. In effect, the integration
by parts implies that in such case the left-hand-side of Eq. (88) is equal to:

$\text{w-}\lim_{g\to 1}\int dX(\ldots)\partial g(X),$ (89)

which is null in the adiabatic limit in which $g$ turns to a constant. On the
other hand, in the expression $PT_{n}(X)P$, $T_{n}(X)$ is defined modulo a
gauge variation, because Eq. (76) implies that: $PQ=0=QP$, and, as a
consequence: $Pd_{Q}XP=PQX-XQP=0$. As a result, we have that the distributions
$T_{n}$ and $T^{\prime}_{n}$ are physically equivalent if they differ by terms
which are divergences or gauge variations. This we denote by the symbol
“$\sim$”:

$T^{\prime}_{n}\sim T_{n}\ :\Leftrightarrow\
T^{\prime}_{n}=T_{n}+d_{Q}(\ldots)+\partial(\ldots).$ (90)

Being that way, the matrix elements of the $S(g)$ operator between states of
the physical subspace, at any order, are trivially quantum gauge invariant,
because $d_{Q}T_{n}$ is physically equivalent to the null distribution: $0\sim
d_{Q}T_{n}$. Quantum gauge invariance of $T_{1}(x)$, therefore, is assured
whatever it was. But this is not sufficient: $T_{1}(x)$ must be constructed in
such a way that the second order distribution $T_{2}(x_{1};x_{2})$ is also
physically equivalent to $T^{\prime}_{2}(x_{1};x_{2})$ constructed with
$T^{\prime}_{1}(x)=T_{1}(x)-i\lambda d_{Q}T_{1}(x)$. This is a consistency
condition. And, clearly, the same must be true for the higher order $T_{n}$
distributions. In a word, we must impose that quantum gauge invariance is not
destroyed in the inductive procedure of CPT.

Definition.— An interacting theory generated by $T_{1}(x)$ is a quantum gauge
theory if it is physically equivalent to the theory generated by
$T_{1}(x)-i\lambda d_{Q}T_{1}(x)$ at all orders of the perturbation series.

Let us start by studying the second order distribution, $T_{2}(x_{1};x_{2})$.
We introduce the following notation: $T_{1}(x)=T_{1}^{0}(x)-i\lambda
d_{Q}T_{1}^{0}(x)$; $T_{2}^{0}(x_{1};x_{2})$ is the two-points distribution
coming from $T_{1}^{0}(x)$; $T_{2}(x_{1};x_{2})$ is the one corresponding to
$T_{1}(x)$, and so on. As we have already known, the construction of the
transition distribution starts by determining the subsidiary distributions:

$\displaystyle A^{\prime}_{2}(x_{1};x_{2})=-T_{1}(x_{1})T_{1}(x_{2}),$
$\displaystyle R^{\prime}_{2}(x_{1};x_{2})=-T_{1}(x_{2})T_{1}(x_{1}).$ (91)

Substituting in them the expression of $T_{1}(x)$ we have that the second
order causal distribution, $D_{2}=R^{\prime}_{2}-A^{\prime}_{2}$, is:

$\displaystyle D_{2}(x_{1};x_{2})=D_{2}^{0}(x_{1};x_{2})-i\lambda
d_{Q}D_{2}^{0}(x_{1};x_{2})$ $\displaystyle+2i\lambda
T_{1}^{0}(x_{1})d_{Q}T_{1}^{0}(x_{2})-2i\lambda
T_{1}^{0}(x_{2})d_{Q}T_{1}(x_{1}).$ (92)

From it we extract the retarded part, $R_{2}(x_{1};x_{2})$, by means of the
splitting procedure:

$\displaystyle R_{2}(x_{1};x_{2})=R_{2}^{0}(x_{1};x_{2})-i\lambda
d_{Q}R_{2}^{0}(x_{1};x_{2})$
$\displaystyle+2i\lambda\text{ret}\left\\{T_{1}^{0}(x_{1})d_{Q}T_{1}^{0}(x_{2})-T_{1}^{0}(x_{2})d_{Q}T_{1}^{0}(x_{1})\right\\},$
(93)

with $\text{ret}\left\\{\ldots\right\\}$ meaning the retarded part of
$\left\\{\ldots\right\\}$. The two-points distribution
$T_{2}=R_{2}-R^{\prime}_{2}$ is then:

$\displaystyle T_{2}(x_{1};x_{2})=T_{2}^{0}(x_{1};x_{2})-i\lambda
d_{Q}T_{2}^{0}(x_{1};x_{2})$
$\displaystyle+2i\lambda\Big{(}\text{ret}\left\\{T_{1}^{0}(x_{1})d_{Q}T_{1}^{0}(x_{2})-T_{1}^{0}(x_{2})d_{Q}T_{1}^{0}(x_{1})\right\\}$
$\displaystyle-T_{1}^{0}(x_{2})d_{Q}T_{1}^{0}(x_{1})\Big{)}.$ (94)

From here it follows that $T_{2}$ will be physically equivalent to $T_{2}^{0}$
if and only if the second line in Eq. (94) is a divergence or a gauge
variation. The first possibility is realized if $T_{1}^{0}(x)$ is a gauge
variation: $T_{1}^{0}(x)=d_{Q}(\ldots)$; this case has no physical interest.
The second possibility holds when $T_{1}^{0}(x)$ is a divergence:

$d_{Q}T_{1}^{0}(x)=i\partial_{a}T^{a}_{1/1}(x).$ (95)

And, in that way, the theory maintains quantum gauge invariance at second
order:

$T_{2}\sim T_{2}^{0}.$ (96)

The $T_{1/1}^{a}(x)$ distribution is called Q-vertex. Hence, Eq. (95) is a
necessary condition in every quantum gauge theory. It implies, also, that
$T_{1}(x)$ is not only gauge invariant as a bilinear form in the physical
subspace, but in the whole Fock’s space.

In order to see if Eq. (95) is also a sufficient condition in order to the
theory to be quantum gauge invariant we must study the following orders in the
perturbation series, which will be easier after establishing the consequences
of Eq. (95). As we know, the $n$-point distributions are chronological
products of $T_{1}$ distributions –ordered according to the null-plane time
$x^{+}$–:

$T_{n}(x_{1};\ldots;x_{n})=\mathfrak{T}_{+}\left\\{T_{1}(x_{1})\ldots
T_{1}(x_{n})\right\\}.$ (97)

Since every $T_{1}$ has boson character, when we apply the gauge variation
operator to $T_{n}$ we find:

$d_{Q}T_{n}=\sum\limits_{l=1}^{n}\mathfrak{T}_{+}\left\\{T_{1}(x_{1})\ldots
d_{Q}T_{1}(x_{l})\ldots T_{1}(x_{n})\right\\}.$ (98)

And with Eq. (95):

$d_{Q}T_{n}=i\sum\limits_{l=1}^{n}\partial_{a}^{x_{l}}T^{a}_{n/l}(x_{1};\ldots;x_{n});$
(99) $T^{a}_{n/l}:=\mathfrak{T}_{+}\left\\{T_{1}(x_{1})\ldots
T_{1/1}^{a}(x_{l})\ldots T_{1}(x_{n})\right\\}.$

Now, $T_{1}(x)$ is in general a Wick’s polynomial:

$T_{1}(x)=T_{1}^{0}(x)+T_{1}^{1}(x)+\ldots+T_{1}^{p}(x),$ (100)

with every $T_{1}^{j}(x)$ a Wick’s monomial. Considering Eq. (97), the
$T_{n}(x_{1};\ldots;x_{n})$ distribution has the general form:

$T_{n}(x_{1};\ldots;x_{n})=\sum\limits_{i_{1},\ldots,i_{n}\in\left\\{0;\ldots;p\right\\}}T_{n}^{i_{1}\ldots
i_{n}}(x_{1};\ldots;x_{n});$ (101)

with the superindex $i_{j}$ indicating the one-point distribution
$T_{1}^{i_{j}}$ corresponding to the point $x_{j}$, this is to say:

$\displaystyle T_{n}^{i_{1}\ldots i_{n}}(x_{1};\ldots;x_{n}):=$
$\displaystyle\quad\mathfrak{T}_{+}\left\\{T_{1}^{i_{1}}(x_{1})\ldots
T_{1}^{i_{j}}(x_{j})\ldots T_{1}^{i_{n}}(x_{n})\right\\}.$ (102)

With this we can already establish the following results, which are general
and contain, as a particular case, the sufficiency of Eq. (95) as a condition
for quantum gauge invariance. We only enunciate them without proof, because
they are a generalization of our results at second order and follow with
simple modifications from the proof given by Dütsch in instant dynamics in
Ref. [39] –see also Ref. [38]–.

Theorem: Let $T_{1}(x)$ be a distribution of the form:

$T_{1}(x)=T_{1}^{0}(x)+T_{1}^{2}(x);\ T_{1}^{2}(x)=\partial_{a}T_{1}^{4a}(x).$
(103)

If the gauge variation of $T_{1}^{0}(x)$ is a divergence,

$d_{Q}T_{1}^{0}(x)=i\partial_{a}T_{1}^{1a}(x),$ (104)

then, in Eq. (II.4) notation:

$T_{n}(x_{1};\ldots;x_{n})=T_{n}^{0\ldots
0}(x_{1};\ldots;x_{n})+\partial(\ldots).$ (105)

According to this theorem, any term in $T_{1}(x)$ which is a divergence can be
disregarded without altering the physics at any order in the perturbation
series. In particular, this is valid for the divergence of the Q-vertex, which
means that Eq. (95) is a sufficient condition for quantum gauge invariance at
all orders.

Theorem: Let $T_{1}(x)$ be a distribution of the form:

$T_{1}(x)=T_{1}^{0}(x)+T_{1}^{3}(x)\quad;\quad
T_{1}^{3}(x)=d_{Q}T_{1}^{5}(x).$ (106)

If for all order $n$ the gauge variation $T_{n}^{0\ldots 0}$ is a divergence,

$d_{Q}T_{n}^{0\ldots 0}=\partial(\ldots),$ (107)

then $T_{n}$ and $T_{n}^{0\ldots 0}$ are physically equivalents:

$T_{n}(x_{1};\ldots;x_{n})=T_{n}^{0\ldots
0}(x_{1};\ldots;x_{n})+\partial(\ldots)+d_{Q}(\ldots).$ (108)

Note that Eq. (107) is always respected once $T_{1}(x)$ is appropriately
chosen [see Eq. (99)], when all the coordinates correspond to different times.
On the other hand, Eq. (99) could be invalid only because of instantaneous
terms –this is to say, when there are some points corresponding to the same
time–, but they can always be cancelled by a convenient choice of the
normalization terms. Therefore, all the terms in $T_{1}(x)$ which are gauge
variations can be neglected without any physical implication.

###  II.5 Unitarity of the scattering operator in $\mathcal{F}_{\text{phys}}$

There is, still, one point that must be clarified, which is if the extension
of Fock’s space and quantum gauge invariance is compatible with the axiom of
unitarity of the scattering operator. Remember that such unitarity, being a
physical property, must be imposed on $\mathcal{F}_{\text{phys}}$ only. As
this issue has already appeared in our study of Poincaré’s transformations,
let us start by considering Eq. (43), from which it follows that there is a
possibility for $\bm{\mathsf{U}}(a;\Lambda)$ to be pseudo-unitary on the whole
Fock’s space according to some adjoint “$K$”, in the sense:

$\bm{\mathsf{U}}(a;\Lambda)^{K}=\bm{\mathsf{U}}(a;\Lambda)^{-1},$ (109)

if the extended quantized field operator $A^{a}(x)$ is Hermitian according to
it: $A^{a}(x)^{K}=A^{a}(x)$. This is:

$\displaystyle A^{a}(x)=(2\pi)^{-3/2}$ $\displaystyle\times\sum_{\lambda}\int
d\mu(\bm{p})\varepsilon_{\lambda}(\bm{p})^{a}\left(a_{\lambda}(\bm{p})e^{-ipx}+a_{\lambda}^{K}(\bm{p})e^{ipx}\right).$
(110)

Comparison with Eq. (II.1) leads to the equalities:

$a_{\lambda}^{K}(\bm{p})=-\sum_{\tau}g_{\lambda\tau}a_{\tau}^{\dagger}(\bm{p}),$
(111)

this is to say: $\quad a_{1,2}^{K}(\bm{p})=a_{1,2}^{\dagger}(\bm{p})$ and
$a_{\pm}^{K}(\bm{p})=-a_{\mp}^{\dagger}(\bm{p})$. In instant dynamics with the
physical space containing photon wavefunctions in the radiation gauge, one can
find explicitly that a similar adjoint exists by using the definition
$B^{K}=(-1)^{\mathsf{N}_{0}}B^{\dagger}(-1)^{\mathsf{N}_{0}}$, with
$\mathsf{N}_{0}$ the number of scalar particles operator [26, 38]. In light-
front dynamics, instead, while not finding its explicit expression, we will
prove that such an adjoint “$K$” does exist, by considering the following
theorem [49]:

Theorem: Let $(\bullet;\bullet)_{K}$ be a bilinear form in Hilbert’s space
$\mathcal{H}$. If it is bounded, i.e.:

$\exists C\in\mathbb{R}_{+}:\ \forall f,g\in\mathcal{H}:\
\left|(f;g)_{K}\right|\leq C\|f\|\|g\|\ ,$ (112)

then there is a bounded linear operator $\eta:\mathcal{H}\to\mathcal{H}$, with
$\|\eta\|\leq C$ and $\text{Dom}(\eta)=\mathcal{H}$, such that:

$(f;g)_{K}=(f;\eta g).$ (113)

Moreover, $(\bullet;\bullet)_{K}$ is Hermitian333This is to say, for every two
vectors $f,g\in\mathcal{H}$: $(f;g)_{K}=(g;f)^{*}_{K}$. if and only if the
operator $\eta$ is self-adjoint, $\eta^{\dagger}=\eta$.

We start by defining the following operators, for test functions
$f\in\mathscr{S}(\mathbb{R}^{3})$:

$\displaystyle a^{a}_{\lambda}(f):=\int
d\mu(\bm{p})\hat{f}(\bm{p})^{*}\varepsilon_{\lambda}(\bm{p})^{a}a_{\lambda}(\bm{p}),$
$\displaystyle a^{a}_{\lambda}(f)^{\dagger}:=\int
d\mu(\bm{p})\hat{f}(\bm{p})\varepsilon_{\lambda}(\bm{p})^{a}a_{\lambda}^{\dagger}(\bm{p}).$
(114)

They generate the states:

$\Phi^{a_{1}\cdots
a_{n}}_{\lambda_{1}\cdots\lambda_{n}}(f_{1};\cdots;f_{n}):=a^{a_{1}}_{\lambda_{1}}(f_{1})^{\dagger}\cdots
a^{a_{n}}_{\lambda_{n}}(f_{n})^{\dagger}\Omega.$ (115)

Similarly, having Eq. (111) as inspiration, we define the operators:

$a^{a}_{\lambda}(f)^{K}:=-\int
d\mu(\bm{p})\hat{f}(\bm{p})\varepsilon_{\lambda}(\bm{p})^{a}\sum_{\tau}g_{\tau\lambda}a_{\tau}^{\dagger}(\bm{p}),$
(116)

and the states generated by them are denoted:

$\widetilde{\Phi}^{a_{1}\cdots
a_{n}}_{\lambda_{1}\cdots\lambda_{n}}(f_{1};\cdots;f_{n}):=a^{a_{1}}_{\lambda_{1}}(f_{1})^{K}\cdots
a^{a_{n}}_{\lambda_{n}}(f_{n})^{K}\Omega.$ (117)

We define now the bilinear form:

$\displaystyle\left(\Phi_{\lambda_{1}\cdots\lambda_{n}}(f_{1};\cdots;f_{n});\Phi_{\sigma_{1}\cdots\sigma_{m}}(g_{1};\cdots;g_{m})\right)_{K}:=$
$\displaystyle\quad\left(\Phi_{\lambda_{1}\cdots\lambda_{n}}(f_{1};\cdots;f_{n});\widetilde{\Phi}_{\sigma_{1}\cdots\sigma_{m}}(g_{1};\cdots;g_{m})\right).$
(118)

This definition, given only for the states of Eq. (115), is suficient because
they generate Fock’s space and because it is defined with regard to the inner
product; its evaluation for more general states is defined by linear
continuation.

Attending at first at the one-particle states, the direct calculus shows that:

$\displaystyle\left(\Phi_{\lambda}(f);\Phi_{\sigma}(g)\right)_{K}=$
$\displaystyle-g_{\lambda\sigma}\int
d\mu(\bm{p})\hat{f}(\bm{p})^{*}\hat{g}(\bm{p})\left(\sum_{a}\varepsilon_{\lambda}(\bm{p})^{a}\varepsilon_{\sigma}(\bm{p})^{a}\right).$
(119)

By comparison, the inner product between the same states is:

$\displaystyle\left(\Phi_{\lambda}(f);\Phi_{\sigma}(g)\right)=$
$\displaystyle\delta_{\lambda\sigma}\int
d\mu(\bm{p})\hat{f}(\bm{p})^{*}\hat{g}(\bm{p})\left(\sum_{a}\varepsilon_{\lambda}(\bm{p})^{a}\varepsilon_{\sigma}(\bm{p})^{a}\right).$
(120)

So we see that for the physical polarizations $\lambda=1,2$, the bilinear form
$(\bullet;\bullet)_{K}$ reduces to the inner product, which satisfies Eq.
(112) with $C=1$ because of Cauchy-Schwarz’s inequality. The other case in
which the bilinear product is non-null (modulo symmetry operations) is:

$\displaystyle\left(\Phi_{+}(f);\Phi_{-}(g)\right)_{K}=-\int
d\mu(\bm{p})\hat{f}(\bm{p})^{*}\hat{g}(\bm{p})$
$\displaystyle\qquad\quad\times\left(\sum_{a}\varepsilon_{+}(\bm{p})^{a}\varepsilon_{-}(\bm{p})^{a}\right)$
$\displaystyle\qquad=-\int
d\mu(\bm{p})\frac{p_{\perp}^{2}}{2p_{-}^{2}}\hat{f}(\bm{p})^{*}\hat{g}(\bm{p}),$
(121)

in which we have used the explicit form of the polarization vectors. On the
other hand, from Eq. (121) we obtain:

$\displaystyle\left\|\Phi_{+}(f)\right\|^{2}=\int
d\mu(\bm{p})\left(1+\frac{p_{\perp}^{2}}{2p_{-}^{2}}\right)^{2}\left|\hat{f}(\bm{p})\right|^{2},$
$\displaystyle\left\|\Phi_{-}(g)\right\|^{2}=\int
d\mu(\bm{p})\left|\hat{g}(\bm{p})\right|^{2}.$ (122)

Eqs. (121) and (II.5), together with Cauchy-Schwarz’s inequality in
$L^{2}(\mathcal{M},\mu)$, imply that:

$\displaystyle\left|\left(\Phi_{+}(f);\Phi_{-}(g)\right)_{K}\right|$
$\displaystyle=\left|\int
d\mu(\bm{p})\left(\frac{p_{\perp}^{2}}{2p_{-}^{2}}\hat{f}(\bm{p})\right)^{*}\hat{g}(\bm{p})\right|$
$\displaystyle=\left|\left(\frac{p_{\perp}^{2}}{2p_{-}^{2}}\hat{f}(\bm{p});\hat{g}(\bm{p})\right)_{L^{2}}\right|$
$\displaystyle\leq\left\|\frac{p_{\perp}^{2}}{2p_{-}^{2}}\hat{f}\right\|\left\|\hat{g}\right\|$
$\displaystyle\leq\left\|\Phi_{+}(f)\right\|\left\|\Phi_{-}(g)\right\|.$ (123)

Therefore, also in this case Eq. (112) is satisfied with $C=1$. Now we can
generalize to the case in which the states are linear combinations of
different polarizations, as for the same polarization:
$\Phi^{a}_{\lambda}(f_{1})+\Phi^{a}_{\lambda}(f_{2})=\Phi^{a}_{\lambda}(f_{1}+f_{2})$.
Hence, since each polarization has only one corresponding polarization such
that the bilinear form is non-null, as indicated by Eq. (II.5), we will have,
for example:

$\displaystyle\left|\left(\Phi_{\lambda_{1}}+\Phi_{\lambda_{2}};\Phi_{\sigma_{1}}+\Phi_{\sigma_{2}}\right)_{K}\right|$
$\displaystyle\quad=\left|\left(\Phi_{\lambda_{1}};\Phi_{\sigma_{1}}\right)_{K}+\left(\Phi_{\lambda_{2}};\Phi_{\sigma_{2}}\right)_{K}\right|$
$\displaystyle\quad\leq\left|\left(\Phi_{\lambda_{1}};\Phi_{\sigma_{1}}\right)_{K}\right|+\left|\left(\Phi_{\lambda_{2}};\Phi_{\sigma_{2}}\right)_{K}\right|$
$\displaystyle\quad\leq\left\|\Phi_{\lambda_{1}}\right\|\left\|\Phi_{\sigma_{1}}\right\|+\left\|\Phi_{\lambda_{2}}\right\|\left\|\Phi_{\sigma_{2}}\right\|$
$\displaystyle\quad\leq\left(\left\|\Phi_{\lambda_{1}}\right\|^{2}+\left\|\Phi_{\lambda_{2}}\right\|^{2}\right)^{1/2}\left(\left\|\Phi_{\sigma_{1}}\right\|^{2}+\left\|\Phi_{\sigma_{2}}\right\|^{2}\right)^{1/2}.$

The last line contains the norm of the states
$\Phi^{a}_{\lambda_{1}}+\Phi^{a}_{\lambda_{2}}$ and
$\Phi^{b}_{\sigma_{1}}+\Phi^{b}_{\sigma_{2}}$, because the polarization
vectors –and then the corresponding states– are mutually orthogonal –see Eq.
(II.5)–. The same applies when other polarizations are present, so even in the
most general linear combination:

$\left|\left(\Phi;\Psi\right)_{K}\right|\leq\|\Phi\|\|\Psi\|.$ (124)

As the constant is $C=1$, its value will not be modified when tensor products
are considered, so Eq. (124) is valid for any $\Phi,\Psi\in\mathcal{F}$. The
above theorem then implies that there is an operator $\eta$, defined over the
entire Fock’s space, such that:

$(\Phi;\Psi)_{K}=(\Phi;\eta\Psi),\ \|\eta\|\leq 1.$ (125)

Even more, from Eq. (II.5) it is clear that the bilinear form
$(\bullet;\bullet)_{K}$ is Hermitian,

$\left(\Phi_{\lambda}(f);\Phi_{\sigma}(g)\right)^{*}_{K}=\left(\Phi_{\sigma}(g);\Phi_{\lambda}(f)\right)_{K},$
(126)

from which it follows that the operator $\eta$ is self-adjoint:

$\eta^{\dagger}=\eta.$ (127)

Utilizing Eqs. (115) and (117) it is possible to show that:

$\left(\widetilde{\Phi}_{\lambda}(f);\widetilde{\Phi}_{\sigma}(g)\right)=\left(\Phi_{\lambda}(f);\Phi_{\sigma}(g)\right),$
(128)

while Eqs. (125)-(127) imply:

$\displaystyle\left(\widetilde{\Phi}_{\lambda}(f);\widetilde{\Phi}_{\sigma}(g)\right)=\left(\widetilde{\Phi}_{\lambda}(f);\Phi_{\sigma}(g)\right)_{K}=\left(\widetilde{\Phi}_{\lambda}(f);\eta\Phi_{\sigma}(g)\right)$
$\displaystyle\quad=\left(\eta\Phi_{\sigma}(g);\widetilde{\Phi}_{\lambda}(f)\right)^{*}=\left(\eta\Phi_{\sigma}(g);\Phi_{\lambda}(f)\right)^{*}_{K}$
$\displaystyle\quad=\left(\eta\Phi_{\sigma}(g);\eta\Phi_{\lambda}(f)\right)^{*}=\left(\Phi_{\lambda}(f);\eta^{2}\Phi_{\sigma}(g)\right).$
(129)

The comparison of Eqs. (128) and (129) leads to establish that:

$\eta^{2}=1.$ (130)

Other important properties are the following. Since the bilinear form
$(\bullet;\bullet)_{K}$ reduces to the inner product between physical states
–transverse polarizations–, we will have that:

$\forall\Phi\in\mathcal{F}_{\text{phys}}:\ \eta\Phi=\Phi.$ (131)

This is valid, particularly, for the vacuum state: $\eta\Omega=\Omega$. Also,
from Eqs. (II.5) and (125):

$\left(\Phi;\eta a^{\dagger}_{\lambda_{1}}(f_{1})\cdots
a^{\dagger}_{\lambda_{n}}(f_{n})\Omega\right)=\left(\Phi;a^{K}_{\lambda_{1}}(f_{1})\cdots
a^{K}_{\lambda_{n}}(f_{n})\Omega\right).$ (132)

Introducing the identity $1=\eta^{2}$ between any two emission operators in
the left-hand-side of this equation, we arrive at the rigorous definition of
the adjoint “$K$”: Let $B$ be an operator:

$B^{K}:=\eta B^{\dagger}\eta.$ (133)

This definition and the properties of the operator $\eta$ found above allow
one to show that the adjoint “$K$” is an involution:

$(A+B)^{K}=A^{K}+B^{K},\ (AB)^{K}=B^{K}A^{K},$ $(B^{K})^{K}=B,\
(cB)^{K}=c^{*}B^{K}\ (c\in\mathbb{C}).$ (134)

In this way, we have shown that there is an adjoint “$K$” such that the
extended radiation field is pseudo-Hermitian, and accordingly, such that the
operator $\bm{\mathsf{U}}(a;\Lambda)$ is pseudounitary in the entire Fock’s
space.

Finally, in order that the pseudounitarity according to “$K$” to be maintained
even after a quantum gauge transformation as in Eq. (60), one needs to impose
also the pseudo-Hermiticity of $u(x)$:

$c_{1}^{K}(\bm{p})=c_{2}^{\dagger}(\bm{p})\ ,\
c_{2}^{K}(\bm{p})=c_{1}^{\dagger}(\bm{p}).$ (135)

With this it follows that the antighost field is anti-pseudo-Hermitian under
“$K$”: $\widetilde{u}(x)^{K}=-\widetilde{u}(x)$. Also, the ghost fields can be
written now as:

$u(x)=(2\pi)^{-3/2}\int
d\mu(\bm{p})\left(c_{2}(\bm{p})e^{-ipx}+c_{2}^{K}(\bm{p})e^{ipx}\right);$
(136) $\widetilde{u}(x)=(2\pi)^{-3/2}\int
d\mu(\bm{p})\left(-c_{1}(\bm{p})e^{-ipx}+c_{1}^{K}(\bm{p})e^{ipx}\right).$
(137)

The following result is immediate –its proof is identical to the proof of
unitarity in the physical Fock’s space (see Ref. [18])–.

Theorem: Let $T_{1}(x)$ be the one-point distribution of a quantum gauge
theory. If $T_{1}$ satisfies the perturbative pseudo-unitarity condition,

$\widetilde{T}_{1}(x)=T_{1}(x)^{K},$ (138)

then the $n$-point distributions can be constructed, by the inductive
procedure of CPT, so as to satisfy the same pseudo-unitarity condition.

Since the adjoints “$K$” and “$\dagger$” coincide in
$\mathcal{F}_{\text{phys}}$, the above theorem implies that the unitarity in
the physical subspace is not destroyed in the extension of Fock’s space
process.

To finish this section, we want to say a word about the bilinear form
$(\bullet;\bullet)_{K}$. It corresponds to the “undefinite metric” used in
Gupta-Bleuler’s quantization procedure. If one adopts it as the “inner
product” (and therefore uses pseudo-Hilbert or Krein spaces), the non-
transverse states have zero norm, as can be seen from Eq. (II.5). This is
similar to what happens in instant dynamics, but in that case the metrics is
$\text{diag}(+1;-1;-1-1)$, hence one non-transverse mode has negative norm
while the other has a positive one. We think that it is very important to show
that pseudo-Hilbert spaces are unnecessary in the sense that there is no
reason to say that the bilinear form $(\bullet;\bullet)_{K}$ is the inner
product: The extended Fock’s space is a perfectly defined Hilbert space and
the bilinear form $(\bullet;\bullet)_{K}$, which is not an inner product,
simply defines a different involution, which coincides with the “$\dagger$”
adjoint in the physical subspace. In this sense, both physics and mathematics
are put on a solid and safe ground.

##  III Ward-Takahashi’s identities for QED

Now we turn to the application of the general results just founded to null-
plane QED. The procedure to obtain Ward-Takahashi’s identities is as in the
instant dynamics formulation [38]. Recall [29] that the one-point distribution
for QED only contains the interaction between the gauge field and the matter
fields:

$T_{1}(x)=iej^{a}(x)A_{a}(x),$ (139)

with $j^{a}(x)$ the matter fields current. In that case, the gauge variation
of $T_{1}$ is equal to

$\displaystyle d_{Q}T_{1}(x)$ $\displaystyle=-ej^{a}(x)\partial_{a}u(x)$
$\displaystyle=\partial_{a}\left(-ej^{a}(x)u(x)\right)+e\left(\partial_{a}j^{a}(x)\right)u(x),$
(140)

which reduces to a divergence if the current $j^{a}$ is conserved; in such
case, the second term in the above equation is null. Hence: The field-current
coupling in CPT holds with the conserved current of the free matter fields.
For a fermion field, it is:

$j^{a}(x)=\;\bm{\mathrel{\mathop{:}}}\,\mathrel{\overline{\psi}(x)\gamma^{a}\psi(x)}\,\bm{\mathrel{\mathop{:}}}\;.$
(141)

The Q-vertex for QED is thus:

$T_{1/1}^{a}(x)=iej^{a}(x)u(x)=ie\;\bm{\mathrel{\mathop{:}}}\,\mathrel{\overline{\psi}(x)\gamma^{a}\psi(x)}\,\bm{\mathrel{\mathop{:}}}\;u(x).$
(142)

Now we want to explore the consequences of gauge invariance in the higher
order terms $T_{n}$. Since there is no selfinteraction term of the radiation
field in $T_{1}$, at each point only a gauge field can appear, and the $T_{n}$
distribution will have the general form:

$T_{n}(x_{1};\cdots;x_{n})=\;\bm{\mathrel{\mathop{:}}}\,\mathrel{T_{l}^{a}(x_{1};\cdots;x_{n})A_{a}(x_{l})}\,\bm{\mathrel{\mathop{:}}}\;+\cdots,$
(143)

in which neither $T_{l}^{a}$ nor the ellipsis include another gauge operator
at the point $x_{l}$, $A_{a}(x_{l})$. Being that way, the gauge variation of
$T_{n}$ is:

$d_{Q}T_{n}(x_{1};\cdots;x_{n})=i\;\bm{\mathrel{\mathop{:}}}\,\mathrel{T_{l}^{a}(x_{1};\cdots;x_{n})\partial_{a}^{x_{l}}u(x_{l})}\,\bm{\mathrel{\mathop{:}}}\;+\cdots,$
(144)

and this time the ellipsis does not include any ghost field at the point
$x_{l}$, $u(x_{l})$. According to Eq. (99), which is the general condition for
gauge invariance at order $n$, the gauge variation in Eq. (144) must reduce to
a divergence. Leibniz’s rule implies that this is possible –note that this
procedure is identical to that in Eq. (III)–, and

$d_{Q}T_{n}(x_{1};\cdots;x_{n})=i\partial_{a}^{x_{l}}\left(\;\bm{\mathrel{\mathop{:}}}\,\mathrel{T^{a}_{l}(x_{1};\cdots;x_{n})u(x_{l})}\,\bm{\mathrel{\mathop{:}}}\;\right)+\cdots,$
(145)

whenever the Ward-Takahashi’s identities hold:

$\partial_{a}^{x_{l}}\;\bm{\mathrel{\mathop{:}}}\,\mathrel{T^{a}_{l}(x_{1};\cdots;x_{n})}\,\bm{\mathrel{\mathop{:}}}\;=0.$
(146)

These equations are valid at every point $x_{l}$ at which there is a radiation
field. If there is no radiation field at a given point, the gauge variation
will be automatically null, and the corresponding term, gauge invariant.

Particularly, application of Ward-Takahashi’s identity to vacuum polarization
leads to the condition of transversality to the momentum, which in real space
is expressed as

$\partial_{a}^{x_{1}}\Pi^{ab}(x_{1};x_{2})=0=\partial_{b}^{x_{2}}\Pi^{ab}(x_{1};x_{2}),$

and which has been implicitly used in our previous paper, Ref. [29], in order
to normalize the vacuum polarization scalar only, leaving unchanged the tensor
structure.

##  IV Electron’s self-energy

In this section we will use the obtained results to calculate electron’s self-
energy, using only the covariant part of the radiation field commutation
distribution. As was shown, such restriction has no influence in the matrix
elements of the scattering operator between physical states. Electron’s self-
energy comes from the causal distribution (see Ref. [29]):

$\displaystyle D_{2}^{(SE)}(x_{1};x_{2})=$
$\displaystyle\;\bm{\mathrel{\mathop{:}}}\,\mathrel{\overline{\psi}(x_{1})d(y)\psi(x_{2})}\,\bm{\mathrel{\mathop{:}}}\;$
$\displaystyle-\;\bm{\mathrel{\mathop{:}}}\,\mathrel{\overline{\psi}(x_{2})d(-y)\psi(x_{1})}\,\bm{\mathrel{\mathop{:}}}\;,$
(147)

with:

$\displaystyle
d(y):=-e^{2}\gamma^{a}\left[d^{+}(y)+d^{-}(y)\right]\gamma_{a};$
$\displaystyle d^{\pm}(y):=S_{\pm}(y)D_{0+}(\pm y).$ (148)

We will start by calculating Fourier’s transform of $d^{-}$, which is:

$\displaystyle\hat{d}^{-}(p)$ $\displaystyle=(2\pi)^{-2}\int
d^{4}q\widehat{S}_{-}(p-q)\widehat{D}_{0+}(-q)$
$\displaystyle=(2\pi)^{-4}\left[(\not{p}+m)I_{1}-\gamma^{a}I_{2a}\right],$
(149)

with the following definitions of the various integrals:

$I_{1}=\int
d^{4}q\Theta(-q_{-})\Theta(q_{-}-p_{-})\delta(q^{2})\delta\left((p-q)^{2}-m^{2}\right),$
(150) $I_{2a}=\int
d^{4}q\Theta(-q_{-})\Theta(q_{-}-p_{-})\delta(q^{2})\delta\left((p-q)^{2}-m^{2}\right)q_{a}.$
(151)

The calculus of these integrals is simplified in the reference frame in which
$p=(p_{+};0_{\perp};p_{-})$, that exists because $q\in\overline{V^{-}}$ and
$p-q\in V^{-}$, so that $p=q+(p-q)\in V^{-}$; this implies, in particular,
that $p_{+},p_{-}<0$. In that reference frame:

$\left.(p-q)^{2}-m^{2}\right|_{q^{2}=0}=2p_{+}p_{-}-2p_{+}q_{-}-2p_{-}q_{+}-m^{2}.$
(152)

Then, using the properties of Dirac’s delta distributions:

$\displaystyle\Theta(-q_{-})\delta(q^{2})\delta\left((p-q)^{2}-m^{2}\right)=\frac{1}{|2p_{-}|}\Theta(-q_{-})$
$\displaystyle\times\Theta(q_{-}-A)\Theta(2p_{+}p_{-}-m^{2})\delta\left(q_{+}-\frac{q_{\perp}^{2}}{2q_{-}}\right)$
$\displaystyle\times\delta\left[q_{\perp}^{2}-\frac{2p_{+}}{p_{-}}(A-q_{-})q_{-}\right],$
(153)

with:

$A=\frac{2p_{+}p_{-}-m^{2}}{2p_{+}}.$ (154)

Since $p_{+}<0$, it is trivial that $A>p_{-}$, so that Heaviside’s function
$\Theta(q_{-}-p_{-})$ appearing in the integrals of Eqs. (150) and (151) is
redundant. Also, in the chosen reference frame it is clear, by symmetry
arguments, that:

$I_{2\alpha}=0.$ (155)

Then, the only integrals that we need to calculate are: $I_{1}$, $I_{2+}$ and
$I_{2-}$. The integration in the variables $q_{+}$ and $q_{\perp}^{2}$ is
immediate by using the supports of Dirac’s delta distributions in Eq. (IV); we
obtain:

$I_{1}=\frac{\pi}{|2p_{-}|}\Theta(-p_{-})\Theta(2p_{+}p_{-}-m^{2})\int\limits_{A}^{0}dq_{-},$
(156)
$I_{2+}=\frac{\pi}{|2p_{-}|}\Theta(-p_{-})\Theta(2p_{+}p_{-}-m^{2})\frac{p_{+}}{p_{-}}\int\limits_{A}^{0}dq_{-}(A-q_{-}),$
(157)
$I_{2-}=\frac{\pi}{|2p_{-}|}\Theta(-p_{-})\Theta(2p_{+}p_{-}-m^{2})\int\limits_{A}^{0}dq_{-}q_{-},$
(158)

The integration in the $q_{-}$ variable is now elementary:

$I_{1}=\frac{\pi}{2}\Theta(-p_{-})\Theta(2p_{+}p_{-}-m^{2})\left(1-\frac{m^{2}}{2p_{+}p_{-}}\right),$
(159)
$I_{2\pm}=\frac{p_{\pm}}{2}\left(1-\frac{m^{2}}{2p_{+}p_{-}}\right)I_{1}.$
(160)

Substituting these results into Eq. (149) and multiplying by $\gamma^{a}$ by
the left and by $\gamma_{a}$ by the right we obtain:

$\displaystyle\gamma^{a}$
$\displaystyle\hat{d}^{-}(p)\gamma_{a}=(2\pi)^{-3}\Theta(-p_{-})\Theta(2p_{+}p_{-}-m^{2})$
$\displaystyle\times\left(1-\frac{m^{2}}{2p_{+}p_{-}}\right)\left\\{m-\frac{\not{p}}{4}\left(1+\frac{m^{2}}{2p_{+}p_{-}}\right)\right\\}.$
(161)

In an analogous manner, it can be found:

$\displaystyle\gamma^{a}$
$\displaystyle\hat{d}^{+}(p)\gamma_{a}=-(2\pi)^{-3}\Theta(p_{-})\Theta(2p_{+}p_{-}-m^{2})$
$\displaystyle\times\left(1-\frac{m^{2}}{2p_{+}p_{-}}\right)\left\\{m-\frac{\not{p}}{4}\left(1+\frac{m^{2}}{2p_{+}p_{-}}\right)\right\\}.$
(162)

Replacing Eqs. (161) and (162) into Eq. (IV) we finally arrive at the causal
distribution:

$\displaystyle\hat{d}(p)=$ $\displaystyle
e^{2}(2\pi)^{-3}\text{sgn}(p_{-})\Theta(2p_{+}p_{-}-m^{2})\left(1-\frac{m^{2}}{2p_{+}p_{-}}\right)$
$\displaystyle\times\left\\{m-\frac{\not{p}}{4}\left(1+\frac{m^{2}}{2p_{+}p_{-}}\right)\right\\}.$
(163)

Because of the polynomial factorization theorem [29], in order to obtain its
retarded part we write it in the following form:

$\displaystyle\hat{d}(p)=$ $\displaystyle e^{2}(2\pi)^{-3}(2p_{+}p_{-}-m^{2})$
$\displaystyle\times\left\\{2p_{+}p_{-}m-\frac{\not{p}}{4}(2p_{+}p_{-}+m^{2})\right\\}\hat{d}_{1}(p),$
(164)

with:

$\hat{d}_{1}(p)=\text{sgn}(p_{-})\Theta(2p_{+}p_{-}-m^{2})\frac{1}{(2p_{+}p_{-})^{2}}.$
(165)

Now we only need to split this distribution, whose singular order at the
$x^{-}$ axis is negative 444The reason for this value of the singular order is
the integral
$\int\rho(s)\Theta\left(\frac{p}{s}-a\right)\frac{s^{2}}{p^{2}}\hat{f}(p)dp=s\rho(s)\int\limits_{a}^{+\infty}\hat{f}(sq)\frac{dq}{q^{2}},$
which goes to $\hat{f}(0)/a$ as $s\to 0$ provided we choose the function
$\rho(s)=s^{-1}$ –this defines the value $\omega_{-}=-1$ of the distribution
$\hat{d}_{1}$–. In that case, in the limit of $s\to 0$:
$\rho(s)\Theta\left(\frac{p}{s}-a\right)\frac{s^{2}}{p^{2}}\to\frac{\delta(p)}{a}.$
This is a departure from power-counting, which assigns a value $-2$ for the
degree of $d_{1}$. This departure is irrelevant in this case because the
formula for obtaining the retarded distribution is the same for all negative
values of the singular order. Nonetheless, to determine the correct value of
the singular order is extremely important as it can lead to different physical
results if the singular order is non-negative [58].:

$\omega_{-}\left[d_{1}\right]=-1.$ (166)

Accordingly, its splitting is done via Eq. (18). Performing the change
$s=-2kp_{-}+2p_{+}p_{-}$:

$\displaystyle\hat{r}_{1}(p)$
$\displaystyle=\frac{i}{2\pi}\int\frac{dk}{k+i0^{+}}\text{sgn}(p_{-})\Theta(-2kp_{-}+2p_{+}p_{-}-m^{2})$
$\displaystyle\quad\times\frac{1}{(-2kp_{-}+2p_{+}p_{-})^{2}}$
$\displaystyle=\frac{i}{2\pi}\Bigg{\\{}\text{V.p.}\int\limits_{m^{2}}^{+\infty}\frac{ds}{s^{2}(2p_{+}p_{-}-s)}$
$\displaystyle\quad-i\pi\text{sgn}(p_{-})\Theta(2p_{+}p_{-}-m^{2})\frac{1}{(2p_{+}p_{-})^{2}}\Bigg{\\}}$
$\displaystyle=\frac{i}{2\pi}\frac{1}{(2p_{+}p_{-})^{2}}\Bigg{\\{}\log\left(\left|\frac{2p_{+}p_{-}-m^{2}}{m^{2}}\right|\right)+\frac{2p_{+}p_{-}}{m^{2}}$
$\displaystyle\quad-i\pi\text{sgn}(p_{-})\Theta(2p_{+}p_{-}-m^{2})\Bigg{\\}}.$
(167)

Therefore we obtain the retarded distribution:

$\displaystyle\hat{r}(p)=\frac{ie^{2}}{(2\pi)^{4}}\Bigg{\\{}\left(1-\frac{m^{2}}{2p_{+}p_{-}}\right)\left[m-\frac{\not{p}}{4}\left(1+\frac{m^{2}}{2p_{+}p_{-}}\right)\right]$
$\displaystyle\times\left[\log\left(\left|\frac{2p_{+}p_{-}-m^{2}}{m^{2}}\right|\right)-i\pi\text{sgn}(p_{-})\Theta(2p_{+}p_{-}-m^{2})\right]$
$\displaystyle+\frac{m^{2}\not{p}}{4(2p_{+}p_{-})}-m-\frac{2p_{+}p_{-}}{m}(\not{p}-m)\Bigg{\\}}.$
(168)

In this expression, the last two terms in the second line have the form of
normalization terms, so they can be replaced by arbitrary values, having in
mind that the singular order at the $x^{-}$ axis of the complete causal
distribution in Eq. (163) is $\omega_{-}\left[d\right]=+1$. Also, the
subsidiary retarded distribution is
$\hat{r}^{\prime}(p)=-e^{2}\gamma^{a}\hat{d}^{-}(p)\gamma_{a}$; its value is
given in Eq. (161). Therefore, if we define the fermion self-energy $\Sigma$
such that the transition distribution is:

$\displaystyle T_{2}^{(SE)}$
$\displaystyle(x_{1};x_{2})=i\;\bm{\mathrel{\mathop{:}}}\,\mathrel{\overline{\psi}(x_{1})\Sigma(x_{1}-x_{2})\psi(x_{2})}\,\bm{\mathrel{\mathop{:}}}\;$
$\displaystyle+i\;\bm{\mathrel{\mathop{:}}}\,\mathrel{\overline{\psi}(x_{2})\Sigma(x_{2}-x_{1})\psi(x_{1})}\,\bm{\mathrel{\mathop{:}}}\;,$
(169)

then:

$\widehat{\Sigma}(p)=-i\hat{t}_{2}^{(SE)}(p)=-i\left(\hat{r}(p)-\hat{r}^{\prime}(p)\right),$
(170)

so that ($C_{0}$ and $C_{1}$ are normalization constants), in Lorentz’s
covariant form we finally get:

$\displaystyle\widehat{\Sigma}(p)=$
$\displaystyle\frac{e^{2}}{(2\pi)^{4}}\Bigg{\\{}\left(1-\frac{m^{2}}{p^{2}}\right)\left[m-\frac{\not{p}}{4}\left(1+\frac{m^{2}}{p^{2}}\right)\right]$
$\displaystyle\left[\log\left(\left|\frac{p^{2}-m^{2}}{m^{2}}\right|\right)-i\pi\Theta(p^{2}-m^{2})\right]$
$\displaystyle+\frac{m^{2}\not{p}}{4p^{2}}+C_{0}+C_{1}\not{p}\Bigg{\\}}.$
(171)

This distribution is the same of that in instant dynamics [26].

In order to fix the normalization constants $C_{0}$ and $C_{1}$ we study
electron’s self-energy insertions into Compton’s scattering, following the
path shown for Yukawa’s model in Ref. [27]; we find that the complete fermion
propagator is the one which solves the equation:

$\widehat{S}_{\text{tot}}=\hat{t}_{2}^{(C)}\left(1+(2\pi)^{4}\widetilde{\Sigma}\widehat{S}_{\text{tot}}\right),$
(172)

with $\hat{t}_{2}^{(C)}$ the normalized fermion Feynman’s propagator (without
instantaneous term), as it is the one that appear in the transition
distribution for Compton’s scattering at second order [29]:

$\hat{t}_{2}^{(C)}(p)=(2\pi)^{-2}\frac{1}{\not{p}-m+i0^{+}}.$ (173)

The solution is:

$\widehat{S}_{\text{tot}}(p)=(2\pi)^{-2}\frac{1}{\not{p}-\left(m+(2\pi)^{2}\widetilde{\Sigma}(p)\right)+i0^{+}}.$
(174)

Then one imposes the normalization conditions: (1) The parameter $m$ is the
fermion’s physical mass; (2) the parameter $e$ is the physical value of the
electric charge. These conditions, respectively, are translated into:

$\lim_{\not{p}\to m}\widetilde{\Sigma}(p)=0,\ \lim_{\not{p}\to
m}\frac{d\widetilde{\Sigma}(p)}{d\not{p}}=0.$ (175)

The first of them can be directly imposed, since $\widetilde{\Sigma}$ in Eq.
(171) does not diverge on the mass shell; it leads to the relation:

$C_{0}=-m\left(C_{1}+\frac{1}{4}\right).$ (176)

The second normalization condition, however, cannot be directly imposed, as
the derivative of $\widetilde{\Sigma}$ is singular on the mass shell. Such
infrared problem is present in every formalism: non-analyticity is a general
property when a massless distribution is part of the convolution (when an
internal line corresponds to a massless particle) [23]. There are some partial
solutions to this problem in the literature, e.g., to introduce a non-null
mass for the photon [23, 54], or to use Pauli-Villars’ regularization [55];
the dependence on the unphysical parameters, however, cannot be posteriorly
eliminated. Fortunately, one can avoid such complications by using Ward-
Takahashi’s identities to show that the normalization of self-energy and that
of the vertex function are not independent: they are related in such a way
that the normalization of one of them is compensated by the normalization of
the other so that the electric charge is not changed. Because of this, the
normalization constant $C_{1}$ does not contain any physics and can be chosen
to satisfy a different normalization condition in which infrared divergences
are not present. For example, normalizing at $p=0$:

$\lim_{\not{p}\to 0}\frac{d\widetilde{\Sigma}(p)}{d\not{p}}=0\ \Rightarrow\
C_{1}=\frac{1}{8}.$ (177)

##  V Conclusions

We have shown an explicit construction of a quantum gauge invariant theory in
light-front dynamics, which proves the independence of the physical scattering
operator of the gauge terms in the massless gauge field commutation
distribution. For such a task it was necessary to extend Fock’s space to
contain also non-physical states, not only coming from the non-physical
polarization states of the gauge field, but also from massless fermion ghost
fields. The scattering operator matrix elements between physical states are
then independent of the gauge terms of the commutation distribution whenever
the normalization terms in the splitting of the causal distribution are chosen
so as to satisfy Ward-Takahashi’s identities. In particular, this imposes a
restriction over the one-point distribution $T_{1}$, similarly as classical
gauge invariance dictates the form of the Lagrangian interaction density by
Utiyama’s minimal coupling prescription.

With the aid of the result just commented, we have shown (see the appendix)
the normalizability of the physical scattering matrix of null-plane QED, as
well as that no selfinteraction terms for the fields will appear at any order.
Additionally, we have calculated electron’s self-energy by utilizing the
covariant part of the radiation field commutation distribution, showing by
direct comparison the equivalence with instant dynamics. Its normalization has
been done by studying its insertions into Compton’s scattering. In that way, a
relation between the two undetermined constants which appear in its expression
follows from the imposition of the physical value of the mass of the lepton.
On the other hand, the physical value of the coupling constant cannot be
imposed on the mass shell because of the infrared divergences, hence one
imposes a condition at a different normalization point. To show that such a
change in the normalization point does not affect the physical value of the
electric charge needs a study of the vertex function of null-plane QED, to
which the self-energy is tied by Ward-Takahashi’s identities. Such a study of
the vertex function, including the derivation of the electron’s gyromagnetic
ratio, will be done in the third part of this series.

###### Acknowledgements.

O.A.A. thanks CAPES-Brazil for total financial support; B.M.P. thanks CNPq-
Brazil for partial financial support.

##  Appendix A Normalizability of the null-plane QED4

In this appendix we will address the problem of determining the singular order
of a general transition distribution. Since it is that order which determines
the number of unknown coefficients, it defines the normalization problem in
CPT 555The problem of renormalizability of null-plane QED in the standard
approach was addressed in Refs. [56] and [57].. Consider, then, a causal
distribution of the order $n=r+s$, which comes from the product of the
transition distributions $T_{r}^{1}(x_{1};\ldots;x_{r})$ and
$T_{s}^{2}(y_{1};\ldots;y_{s})$ by means of $l_{p}$ photon contractions and
$l_{f}$ fermion ones; its numerical part will be:

$\displaystyle d$ $\displaystyle(x_{1};\ldots;x_{r};y_{1};\ldots;y_{s})\propto
t_{1}(x_{1};\ldots;x_{r})$
$\displaystyle\times\prod\limits_{j=1}^{l_{p}}D_{ab+}(x_{r_{j}}-y_{s_{j}})\prod\limits_{m=1}^{l_{f}}S_{+}(x_{r_{m}}-y_{s_{m}})$
$\displaystyle\times t_{2}(y_{1};\ldots;y_{s}).$ (178)

In this expression, $t_{1}$ and $t_{2}$ are the numerical distributions of
$T_{r}^{1}$ and $T_{s}^{2}$, respectively; $\left\\{x_{r_{j}}\right\\}$ and
$\left\\{y_{s_{j}}\right\\}$ are the points at which the photon contractions
take place, while $\left\\{x_{r_{m}}\right\\}$ and
$\left\\{y_{s_{m}}\right\\}$ are the ones at which the fermion contractions
occur. The symbol “$\propto$” stands instead of the equality because the
ordering of the fermion’s matrix commutation distributions are not taken into
account. Because of translation invariance we can use relative coordinates:

$\displaystyle\xi_{j}:=x_{j}-x_{r}\quad(j=1,\ldots,r-1),$
$\displaystyle\lambda_{j}:=y_{j}-y_{s}\quad(j=1,\ldots,s-1),$
$\displaystyle\lambda:=x_{r}-y_{s},$ (179)

so that, defining the vectors $\bm{\xi}=(\xi_{1};\ldots;\xi_{r-1})$ and
$\bm{\lambda}=(\lambda_{1};\ldots;\lambda_{s-1})$:

$\displaystyle d$ $\displaystyle(\bm{\xi};\bm{\lambda};\lambda)\propto
t_{1}(\bm{\xi})\prod\limits_{j=1}^{l_{p}}D_{ab+}(\xi_{r_{j}}-\lambda_{s_{j}}+\lambda)$
$\displaystyle\times\prod\limits_{m=1}^{l_{f}}S_{+}(\xi_{r_{m}}-\lambda_{r_{m}}+\lambda)t_{2}(\bm{\lambda}).$
(180)

As a next step we go to momentum space via Fourier’s transformation:

$\displaystyle\hat{d}(\bm{p};\bm{q};q)\propto$ $\displaystyle\int
d^{4(r-1)}\bm{\xi}d^{4(s-1)}\bm{\lambda}d^{4}\lambda
d(\bm{\xi};\bm{\lambda};\lambda)$ $\displaystyle\times
e^{i\bm{p}\bm{\xi}+i\bm{q}\bm{\lambda}+iq\lambda},$ (181)

by means of which we obtain:

$\displaystyle\hat{d}$
$\displaystyle(\bm{p};\bm{q};q)\propto\int\prod_{j,m}d^{4}k_{m}d^{4}h_{j}\widehat{D}_{ab+}(h_{j})\widehat{S}_{+}(k_{m})$
$\displaystyle\quad\times\left[\int
d^{4(r-1)}\bm{\xi}t_{1}(\bm{\xi})e^{i\bm{p}\bm{\xi}-i\left(\sum_{m}k_{m}\xi_{r_{m}}+\sum_{j}h_{j}\xi_{r_{j}}\right)}\right]$
$\displaystyle\quad\times\left[\int
d^{4(s-1)}\bm{\lambda}t_{2}(\bm{\lambda})e^{i\bm{q}\bm{\lambda}-i\left(\sum_{m}k_{m}\lambda_{s_{m}}+\sum_{j}h_{j}\lambda_{s_{j}}\right)}\right]$
$\displaystyle\quad\times\left[\int d^{4}\lambda
e^{i\left(q-\sum_{m}k_{m}-\sum_{j}h_{j}\right)}\right]$
$\displaystyle\propto\int\prod_{j,m}d^{4}k_{m}d^{4}h_{j}\widehat{D}_{ab+}(h_{j})\widehat{S}_{+}(k_{m})\hat{t}_{1}(\bm{p}-\bm{k}_{r}-\bm{h}_{r})$
$\displaystyle\quad\times\hat{t}_{2}(\bm{q}+\bm{k}_{s}+\bm{h}_{s})\delta\left[q-\left(\sum_{j}h_{j}+\sum_{m}k_{m}\right)\right],$
(182)

with:

$(\bm{p}-\bm{k}_{r}-\bm{h}_{r})_{j}=\left\\{\begin{array}[]{c
l}p_{j}-k_{r_{j}}-h_{r_{j}}&\text{if $\xi_{r_{j}}$ is contracted}\\\
p_{j}&\text{if $\xi_{r_{j}}$ is not contracted}\end{array}\right.$

and similarly for $\bm{q}+\bm{k}_{s}+\bm{h}_{s}$. Now, as it was shown for
Yukawa’s model in Ref. [18], the convolution of quasiasymptotics at the
$x^{-}$ axis generally does not exist, so that we redefine the integration
variables by scaling all their four components: $\widetilde{k}_{m}=sk_{m}$ and
$\widetilde{h}_{j}=sh_{j}$. Eq. (182) then reads:

$\displaystyle\hat{d}$ $\displaystyle(\bm{p};\bm{q};q)\propto
s^{-4(l_{p}+l_{f})}\int\prod\limits_{j,m}d^{4}\widetilde{k}_{m}d^{4}\widetilde{h}_{j}\widehat{D}_{ab+}\left(\frac{\widetilde{h}_{j}}{s}\right)$
$\displaystyle\times\widehat{S}_{+}\left(\frac{\widetilde{k}_{m}}{s}\right)\hat{t}_{1}\left(\bm{p}-\frac{\widetilde{\bm{k}_{r}}+\widetilde{\bm{h}_{r}}}{s}\right)\hat{t}_{2}\left(\bm{q}+\frac{\widetilde{\bm{k}_{s}}+\widetilde{\bm{h}_{s}}}{s}\right)$
$\displaystyle\times\delta\left[1-\frac{1}{s}\left(\sum_{j}\widetilde{h}_{j}+\sum_{m}\widetilde{k}_{m}\right)\right].$
(183)

According to Sec. II, for all physical purposes the commutation distribution
of the radiation field in Eq. (183) can be taken as:

$\widehat{D}_{ab}(p)=g_{ab}\widehat{D}_{0}(p),$ (184)

with $\widehat{D}_{0}(p)$ Jordan-Pauli’s distribution of zero mass.
Accordingly, Eq. (183) is:

$\displaystyle\hat{d}_{0}$ $\displaystyle(\bm{p};\bm{q};q)\propto
s^{-4(l_{p}+l_{f})}\int\prod\limits_{j,m}d^{4}\widetilde{k}_{m}d^{4}\widetilde{h}_{j}\widehat{D}_{0+}\left(\frac{\widetilde{h}_{j}}{s}\right)$
$\displaystyle\times\widehat{S}_{+}\left(\frac{\widetilde{k}_{m}}{s}\right)\hat{t}_{1}\left(\bm{p}-\frac{\widetilde{\bm{k}_{r}}+\widetilde{\bm{h}_{r}}}{s}\right)\hat{t}_{2}\left(\bm{q}+\frac{\widetilde{\bm{k}_{s}}+\widetilde{\bm{h}_{s}}}{s}\right)$
$\displaystyle\times\delta\left[1-\frac{1}{s}\left(\sum_{j}\widetilde{h}_{j}+\sum_{m}\widetilde{k}_{m}\right)\right].$
(185)

In order to obtain the singular order of this causal distribution we need to
evaluate the limit:

$\lim_{s\to
0}s^{\omega_{-}}\hat{d}\left(\frac{\bm{p}}{s_{*}};\frac{\bm{q}}{s_{*}};\frac{q}{s_{*}}\right);\
\frac{p}{s_{*}}\equiv\left(\frac{p_{+}}{s};\frac{p_{\perp}}{s};p_{-}\right).$
(186)

Then:

$\displaystyle\hat{d}_{0}$
$\displaystyle\left(\frac{\bm{p}}{s_{*}};\frac{\bm{q}}{s_{*}};\frac{q}{s_{*}}\right)\propto
s^{-4(l_{p}+l_{f})}\int\prod\limits_{j,m}d^{4}\widetilde{k}_{m}d^{4}\widetilde{h}_{j}\widehat{D}_{0+}\left(\frac{\widetilde{h}_{j}}{s}\right)$
$\displaystyle\times\widehat{S}_{+}\left(\frac{\widetilde{k}_{m}}{s}\right)\hat{t}_{1}\left(\frac{\bm{p}}{s_{*}}-\frac{\widetilde{\bm{k}_{r}}+\widetilde{\bm{h}_{r}}}{s}\right)\hat{t}_{2}\left(\frac{\bm{q}}{s_{*}}+\frac{\widetilde{\bm{k}_{s}}+\widetilde{\bm{h}_{s}}}{s}\right)$
$\displaystyle\times\delta\left[\frac{q}{s_{*}}-\frac{1}{s}\left(\sum_{j}\widetilde{h}_{j}+\sum_{m}\widetilde{k}_{m}\right)\right].$
(187)

In this form, by exactly the same steps given for Yukawa’s model in Ref. [18]
it can be proven that:

Lemma: The parts of the causal distribution which do not contain the gauge
terms of the commutation distribution of the radiation field have the
following singular order at the $x^{-}$ axis:

$\omega_{-}=4-N-\frac{3}{2}M,$ (188)

with $N$ the number of external radiation field operators, and $M$ the number
of external fermion field ones.

We conclude that the physical $S$-operator of null-plane QED is normalizable.

Before ending this section we want to mention that, according to Eq. (188),
the only possibilities for $\omega_{-}\geq 0$ are those shown in Tab. 1.

$N$ | $M$ | $\omega_{-}$ | Process
---|---|---|---
0 | 0 | 4 | Vacuum-to-vacuum distribution
0 | 2 | 1 | Electron’s self-energy
1 | 0 | 3 |  $=0$ by Furry’s theorem
1 | 2 | 0 | Vertex function
2 | 0 | 2 | Vacuum polarization
3 | 0 | 1 |  $=0$ by Furry’s theorem
4 | 0 | 0 | Light-light scattering
Table 1: Non-negative singular order distributions.

As we see, it is impossible to have electron selfinteracting terms, while it
is, in principle, permitted to have a photon selfinteracting term of the type
$\;\bm{\mathrel{\mathop{:}}}\,\mathrel{A^{4}}\,\bm{\mathrel{\mathop{:}}}\;$,
because the singular order at the $x^{-}$ axis of the distributions
corresponding to light-light scattering is $\omega_{-}=0$, so a normalization
term is allowable for them. It is possible to show, however, that Ward-
Takahashi’s identities [Eq. (146)] forbid the presence of such selfinteraction
term. This makes rigorous the arguments given in Ch. 13 of Ref. [52],
according to which, in conventional QFT, the potential ultraviolet divergence
for light-light scattering is eliminated by means of gauge invariance.

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