Null-plane causal perturbation theory

Abstract

We reconstruct causal perturbation theory in a way compatible with light-front dynamics on the basis of Rohrlich’s invariant null-plane coordinatization. We show that, when the causality axiom is referred to the $x^+$ time coordinate, the retarded and advanced distributions are possibly non-null on the entire $x^-$-axis, despite which Bogoliubov–Medvedev–Polivanov’s axioms still have a perturbative solution whose practical implementation is exposed in detail. Our results are then applied to the construction of the free fields Feynman’s propagators, for which instantaneous terms arise in the fermion and massless vector cases, and then to Yukawa’s model for the interaction of a neutral pseudo-scalar field with nucleons.

Quantization of fields on the null-plane

In quantum mechanics a particle is described by a wave-function. If the description is relativistic, such wave-functions are the positive frequency part of the so-called relativistic classical fields, this is to say, solutions of some relativistic equation of motion. The restriction to consider only the positive frequency part of the solution is done because in the real world the energy of the particle must be positive; the frequency and the energy are related by Schrödinger’s equation, which is an axiom in the quantum theory. In instant dynamics this could be confusing: We know that every component of every field satisfies Klein–Gordon–Fock’s equation, but it is impossible to put that equation in Schrödinger’s form. In light-front dynamics this issue is clear: Klein–Gordon–Fock’s equation can be put in Schrödinger’s form,

$$\begin{aligned} i\partial _+\varphi _1(x)=P_+\varphi _1(x), \end{aligned}$$

(244)

with the Hamiltonian:

$$\begin{aligned} P_+=\frac{i}{2}\partial _-^{-1}(\partial _\perp ^2-m^2)=\frac{i}{4}\int dv\text {sgn}(x^--v)(\partial _\perp ^2-m^2). \end{aligned}$$

(245)

All the solutions $\varphi _1$ of this equation constitute the one-particle Hilbert space ${\mathcal {H}}_1$. This is sufficient if we want to treat free particles only, but when the particles are subjected to interaction we need to consider the possibility that they change their identity and their number. Of course, such changes cannot be described in detail: the very concept of particle is then lost; but we can describe the asymptotically free incoming and outgoing particles of a scattering process. And that could be done only if one works on a space containing states describing different number of particles.

Definition

Fock’s space ${\mathcal {F}}^{\pm }$ of bosons or fermions, respectively, is the direct sum of Hilbert’s spaces of symmetrized or anti-symmetrized wave-functions of n particles, ${\mathcal {H}}_n^{\pm }$, extended to all non-negative values of n:

$$\begin{aligned} {\mathcal {F}}^{\pm }:=\bigoplus \limits _{n=0}^{+\infty }{\mathcal {H}}_n^{\pm }. \end{aligned}$$

(246)

Their elements are:

$$\begin{aligned} \Phi \equiv \left( \varphi _0;\varphi _1;\cdots ;\varphi _n;\cdots \right) ;\quad \varphi _n\in {\mathcal {H}}_n^{\pm }. \end{aligned}$$

(247)

Particularly, the vacuum state of zero particles is:

$$\begin{aligned} \Omega :=\left( 1;0;\cdots \right) . \end{aligned}$$

(248)

Fock’s space ${\mathcal {F}}^{\pm }$ is equipped with the interior product $\left( \bullet ;\bullet \right)$ defined as:

$$\begin{aligned} (\Phi ;\Psi ):=\sum \limits _{n=0}^{+\infty }(\varphi _n;\psi _n)_n, \end{aligned}$$

(249)

and norm induced by it:

$$\begin{aligned} \Vert \Phi \Vert ^2:=\sum \limits _{n=0}^{+\infty }\Vert \varphi _n\Vert _n^2=\sum \limits _{n=0}^{+\infty }(\varphi _n;\varphi _n)_n, \end{aligned}$$

(250)

with the restriction—over the states—that its value to be finite.

The passage from states of a given number of particles to another is done by the emission and absorption operators.

Definition

The emission operator of a particle with wave-function f, with $f\in {\mathcal {H}}_1$, is the operator $a^*(f):\text {Dom}(a^*(f))\subseteq {\mathcal {F}}^{\pm }\rightarrow \mathcal F^{\pm }$, defined as:

$$\begin{aligned} (a^*(f)\Phi )_0:=0;\quad (a^*(f)\Phi )_n:=\sqrt{n}S_n^{\pm }(f\otimes \varphi _{n-1})\quad (n\in {\mathbb {N}}). \end{aligned}$$

(251)

The absorption operator of a particle with wave-function f, with $f\in {\mathcal {H}}_1$, is the operator $a(f):{\mathcal {F}}^{\pm }\rightarrow {\mathcal {F}}^{\pm }$, defined such that:

$$\begin{aligned} a(f)\Omega :=0;\quad (a(f)\Phi )_n(\varvec{x}_1;\cdots ;\varvec{x}_n):=\sqrt{n+1}\left( f(\varvec{x});\varphi _{n+1}(\varvec{x};\varvec{x}_1;\cdots ;\varvec{x}_n)\right) _{1,\varvec{x}}. \end{aligned}$$

(252)

The Hermiticity of the (anti-)symmetrization operators $S^{\pm }_n$ implies the following relation between the emission and absorption operators:

$$\begin{aligned} a^*(f)=a(f)^\dagger \equiv a^\dagger (f), \end{aligned}$$

(253)

The (anti-)commutation relations of the emission and absorption operators are found by using their definitions given in Eqs. (251) and (252). Let $f,g\in {\mathcal {H}}_1$, $\Phi \in \text {Dom}\left( [a(f);a^\dagger (g)]_{{\mp }}\right) \subset {\mathcal {F}}^{\pm }$. By direct calculus:

$$\begin{aligned}&\displaystyle \left[ a(f);a^\dagger (g)\right] _{\mp }\Phi =(f;g)_1\Phi , \end{aligned}$$

(254)

$$\begin{aligned}&\displaystyle \left[ a(f);a(g)\right] _{\mp }\Phi =0;\quad \left[ a^\dagger (f);a^\dagger (g)\right] _{\mp }\Phi =0\, . \end{aligned}$$

(255)

As we see, these relations consider the commutator in ${\mathcal {F}}^+$ (bosons space) and the anti-commutator in ${\mathcal {F}}^-$ (fermions space). Under the assumption that ${\mathcal {H}}_1$ is separable, it is possible to introduce a countable and orthonormal basis of functions $f_j$:

$$\begin{aligned} (f_j;f_k)_1=\delta _{jk}, \end{aligned}$$

(256)

in function of which every $f\in {\mathcal {H}}_1$ is expanded as:

$$\begin{aligned} f(\varvec{x})=\sum _j(f_j;f)_1f_j(\varvec{x}). \end{aligned}$$

(257)

Therefore, defining the emission and absorption operators on those basis functions:

$$\begin{aligned} a^\dagger _j:=a^\dagger (f_j),\quad a_j:=a(f_j), \end{aligned}$$

(258)

we obtain that $a_j$ and $a^\dagger _k$ satisfy the following relations [see Eqs. (254), (255) and (256)]:

$$\begin{aligned} {[}a_j;a^\dagger _k]_{\mp }=\delta _{jk},\quad [a_j;a_k]_{{\mp }}=0=\left[ a^\dagger _j;a^\dagger _k\right] _{\mp }. \end{aligned}$$

(259)

Now we must specify the interior product in Eq. (254) in such a way that it is relativistic invariant.

Definition

Let ${\mathcal {H}}_1$ be the one-particle Hilbert’s space. Consider $f,g\in {\mathcal {H}}_1$, and let ${{\hat{f}}}$ and ${{\hat{g}}}$ be their Fourier’s transforms. The interior product of them is defined by:

$$\begin{aligned} (f;g)_1:=\int {\mathrm{d}}\mu _m(p){{\hat{f}}}(p)^*{{\hat{g}}}(p), \end{aligned}$$

(260)

with ${\mathrm{d}}\mu _m(p)$ the relativistic invariant measure:

$$\begin{aligned} {\mathrm{d}}\mu _m(p):=\delta (p^2-m^2)\Theta (p_-)d^4p=\left. \Theta (p_-)\frac{d^3\varvec{p}}{2p_-}\right| _{p_+=E};\quad E:=\frac{\omega _p^2}{|2p_-|}=\frac{p_\perp ^2+m^2}{|2p_-|}. \end{aligned}$$

(261)

This measure ${\mathrm{d}}\mu _m(p)$ is defined on the “upper mass-shell”

$$\begin{aligned} {\mathcal {M}}^+:=\left\{ p\in {\mathbb {R}}^4\ \big |\ p^2=m^2\ \wedge \ p_->0\right\} , \end{aligned}$$

(262)

so we identify the space ${\mathcal {H}}_1$ with the space of square-integrable functions on ${\mathcal {M}}^+$ with respect to the measure ${\mathrm{d}}\mu _m$:

$$\begin{aligned} {\mathcal {H}}_1:=L^2({\mathcal {M}}^+;{\mathrm{d}}\mu _m). \end{aligned}$$

(263)

We present the following results without proof—the proof in instant dynamics is given in Ref. [64]; in light-front dynamics the proof follows with simple modifications.

Theorem

Every irreducible representation of the (anti-)commutation relations of Eqs. (254) and (255), with vacuum state $\Omega$, are equivalent to Fock’s representation.

Corollary

Every bounded operator acting on Fock’s space $\mathcal F^{\pm }$ can be expressed in function of the emission and absorption operators $a^\dagger (f)$ and a(f).

Also, Eq. (261) implies that, when one restricts the wave-functions to the null-plane, the measure of the interior product becomes independent of the mass. This leads to the following theorem [16]:

Theorem

Let ${\mathcal {F}}_\nu ^{\pm }$ ($\nu =1,2$) be two Fock’s spaces corresponding to particles with different masses $m_\nu$, with $m_1\ne m_2$. The Fock’s spaces ${\mathcal {F}}_1^{\pm }$ and $\mathcal F_2^{\pm }$ are unitarily equivalents.

The next step in the construction of the quantum theory is the adequate election of the basis functions in ${\mathcal {H}}_1$. If the basis is chosen in a such a way to satisfy the equations and constraints of the classical field theory, then we will have no problem in recognizing that their evolution according to the Hamiltonian $P_+$ is relativistic. Our strategy will be the following: We will write a basis for the physical solutions of Klein–Gordon–Fock’s equation; with them we will construct the complete multi-component fields by using the constraint equations and the adequate statistics. In order to do that, we remember that the initial value problem in the null-plane has the solution for the positive frequency part [5, 6]:

$$\begin{aligned} f(x)=\left\langle iD_+(x-y)^*;f(y)\right\rangle _y, \end{aligned}$$

(264)

with $\left\langle \bullet ;\bullet \right\rangle$ meaning that the distribution is applied to the function according to the formal rule given by the interior product $(\bullet ;\bullet )_1$, differentiating the notation in order to clarify that $\left\langle \bullet ;\bullet \right\rangle$ is not an interior product, but the application of a distribution to a function. Also, $D_+$ is the positive frequency part of Jordan-Pauli’s distribution. The expansion of f in the basis $\left\{ f_j\right\}$ implies the following completeness relation:

$$\begin{aligned} \sum _j f_j(x)f_j(y)^*=-iD_+(x-y). \end{aligned}$$

(265)

The mapping from the function f to the operators a(f) and $a^\dagger (f)$ is making by:

Definition

Let $f\in {\mathcal {H}}_1$ be a wave-function. The emission and absorption field operators in real space, $a^\dagger (\varvec{x})$ and $a(\varvec{x})$, respectively, are the operator-valued distributions which lead f to the operators $a^\dagger (f)$ and a(f), respectively, by means of the rules:

$$\begin{aligned} a^\dagger (f):=\left\langle f^*;a^\dagger (\varvec{x})\right\rangle ;\quad a(f):=\left\langle f;a(\varvec{x})\right\rangle , \end{aligned}$$

(266)

Additionally, if $f\in {\mathscr {S}}({\mathbb {R}}^3)$, we define the emission and absorption field operators in momentum space, $a^\dagger (\varvec{p})$ and $a(\varvec{p})$, respectively, as the distributional Fourier’s transforms of the corresponding field operators in real space.

Eqs. (257) and (266) imply that we can write the emission and absorption field operators as:

$$\begin{aligned} a^\dagger (\varvec{x})=\sum _j a^\dagger _j f_j(\varvec{x})^*,\quad a(\varvec{x})=\sum _j a_jf_j(\varvec{x}), \end{aligned}$$

(267)

and in momentum space:

$$\begin{aligned} a^\dagger (\varvec{p})=\sum _j {{\hat{a}}}^\dagger _j {{\hat{f}}}_j(\varvec{p})^*,\quad a(\varvec{p})=\sum _j {{\hat{a}}}_j{{\hat{f}}}_j(\varvec{p}). \end{aligned}$$

(268)

From these relations, jointly with the expansions in Eq. (259) and the completeness relation given in Eq. (265):

$$\begin{aligned} \left[ a(\varvec{x});a^\dagger (\varvec{y})\right] _{\mp }=-iD_+(0;\varvec{x}-\varvec{y}),\quad [a(\varvec{x});a(\varvec{y})]_{\mp }=0=\left[ a^\dagger (\varvec{x});a^\dagger (\varvec{y})\right] _{\mp }. \end{aligned}$$

(269)

Definition

The quantized field operator is the operator-valued distribution u(x) defined as:

$$\begin{aligned} u(x):=a(x)+a^\dagger (x). \end{aligned}$$

(270)

Its part corresponding to the absorption field operator is called “negative frequency part”, denoted $u_-$; the corresponding to the emission field operator, “positive frequency part”, denoted $u_+$.

To that definition many forms can be given. Using the expansion showed in Eq. (267):

$$\begin{aligned} u(x)=\sum _j\left( a_jf_j(x)+a_j^\dagger f_j(x)^*\right) . \end{aligned}$$

(271)

Here we see that the dependence with the coordinates is contained in the functions $f_j(x)$. Since $\left\{ f_k\right\}$ is basis of ${\mathcal {H}}_1$, Eq. (271) implies that the quantized field operator satisfies the equation of motion of the classical field. Also, writing the function $f_j(x)$ in Eq. (271) in function of their Fourier’s transform,

$$\begin{aligned} f_j(x)=(2\pi )^{-3/2}\int {\mathrm{d}}\mu (\varvec{p}){{\hat{f}}}_j(\varvec{p})e^{-ipx}, \end{aligned}$$

(272)

we obtain the following wave-packet expansion for the quantized field operator:

$$\begin{aligned} u(x)=(2\pi )^{-3/2}\sum _j\int {\mathrm{d}}\mu (\varvec{p})\left( a_j{{\hat{f}}}_j(\varvec{p})e^{-ipx}+a^\dagger _j{{\hat{f}}}(\varvec{p})^*e^{ipx}\right) ;\quad p_+=\frac{\omega _p}{|2p_-|}. \end{aligned}$$

(273)

Finally, by applying Fourier’s transformation directly to the emission and absorption field operators a(x) and $a^\dagger (x)$ in Eq. (270):

$$\begin{aligned} u(x)=(2\pi )^{-3/2}\int {\mathrm{d}}\mu (\varvec{p})\left( a(\varvec{p})e^{-ipx}+a^\dagger (\varvec{p})e^{ipx}\right) . \end{aligned}$$

(274)

Once again the field operators $a^\dagger (\varvec{p})$ and $a(\varvec{p})$ can be expanded as in Eq. (268), with the same result as in Eq. (273); this shows that, in general: ${{\hat{a}}}_j=a_j$ and ${{\hat{a}}}_j^\dagger =a_j^\dagger$; this is not surprising, because $a_j$ and $a_j^\dagger$ are independent of the coordinates and momenta.

The most important property of the quantized field operators is the (anti-)commutation relation between their negative and positive frequency parts. Using the expression in Eq. (271) and Eqs. (259) and (265):

$$\begin{aligned} \left[ u_-(x);u_+(y)\right] _{\mp }=-iD_+(x-y). \end{aligned}$$

(275)

This commutator appears in a central rôle in CPT and was called “Wick’s contraction”: It comes from the solution of the classical field theory Goursat’s problem. If the same were calculated from Eq. (274), then we would obtain:

$$\begin{aligned} \left[ u_-(x);u_+(y)\right] _{\mp }=(2\pi )^{-3}\int {\mathrm{d}}\mu (\varvec{p}){\mathrm{d}}\mu (\varvec{q})\left[ a(\varvec{p});a^\dagger (\varvec{q})\right] _{\mp } \left. e^{-i(px-qy)}\right| _{p_+=E_p,q_+=E_q}. \end{aligned}$$

(276)

The comparison between Eqs. (275) and (276) allows the obtention of $\left[ a(\varvec{p});a^\dagger (\varvec{q})\right] _{\mp }$.

Now we turn to the study of the scalar, fermion and massless vector fields.

1.1 Neutral scalar field

In this case we need to introduce one set of emission and absorption operators, only, $a^\dagger (f)$ and a(f). The quantized field operator is:

$$\begin{aligned} \varphi (x)=(2\pi )^{-3/2}\int {\mathrm{d}}\mu (\varvec{p})\left( a(\varvec{p})e^{-ipx}+a^\dagger (\varvec{p})e^{ipx}\right) . \end{aligned}$$

(277)

Their commutation distribution will be Jordan-Pauli’s distribution, because it solves Goursat’s problem in the classical theory. Then:

$$\begin{aligned} \left[ \varphi _-(x);\varphi _+(y)\right] =-iD_+(x-y), \end{aligned}$$

(278)

and:

$$\begin{aligned} {[}\varphi (x);\varphi (y)]=-iD(x-y). \end{aligned}$$

(279)

Jordan-Pauli’s distribution is:

$$\begin{aligned} D(x)=i(2\pi )^{-3}\int d^4p\delta (p^2-m^2)\text {sgn}(p_-)e^{-ipx}; \end{aligned}$$

(280)

their positive and negative frequency parts are, respectively, given by:

$$\begin{aligned} D_+(x)=i(2\pi )^{-3}\int {\mathrm{d}}\mu _m(p)e^{-ipx},\quad D_-(x)=-D_+(-x). \end{aligned}$$

(281)

From Eqs. (278), (281) and (276) we obtain:

$$\begin{aligned} \int {\mathrm{d}}\mu (\varvec{p})\left. e^{-ip(x-y)}\right| _{p_+=E_p}=\int {\mathrm{d}}\mu (\varvec{p}){\mathrm{d}}\mu (\varvec{q})\left[ a(\varvec{p});a^\dagger (\varvec{q})\right] \left. e^{-i(px-qy)}\right| _{p_+=E_p,q_+=E_q}, \end{aligned}$$

(282)

which implies:

$$\begin{aligned} \left[ a(\varvec{p});a^\dagger (\varvec{q})\right] =2p_-\delta (\varvec{p}-\varvec{q}). \end{aligned}$$

(283)

Particularly, from Eq. (279) we can obtain the commutation relation at equal times. From Eq. (280), by integrating the variables $p^+$ and $p^\perp$, we find:

$$\begin{aligned} D(0;\varvec{x})=\frac{1}{2\pi }\delta (x^\perp )\int \limits _0^{+\infty }dp_-\frac{\sin (p_-x^-)}{p_-}=\frac{1}{4}\text {sgn}(x^-)\delta (x^\perp ), \end{aligned}$$

(284)

hence:

$$\begin{aligned} \left[ \varphi (x^+;\varvec{x});\varphi (x^+;\varvec{y})\right] =-\frac{i}{4}\text {sgn}(x^--y^-)\delta (x^\perp -y^\perp ). \end{aligned}$$

(285)

Here we observe the non-commutativity of the field in the $x^-$-axis, which does not violate causality. Additionally, taking the derivative with respect to $y^-$ in Eq. (285):

$$\begin{aligned} \left[ \varphi (x^+;\varvec{x});\partial _-\varphi (x^+;\varvec{y})\right] =-\frac{i}{2}\delta (\varvec{x}-\varvec{y}). \end{aligned}$$

(286)

In a Hamiltonian analysis, the conjugate momenta to the field $\varphi$ is $\pi =\partial _-\varphi$, so Eq. (286) is the canonical commutation relation at equal times. Equation (286), however, is unusual in the sense that it contains a factor of 1/2, which is due to the fact that $\varphi$ and $\pi$ are not dynamically independent in light-front dynamics [6]. It is usual that systems which are not constrained in instant dynamics turn to be constrained in the light-front one; we remember to the reader, in this sense, that the stability group of the null-plane is bigger than that of the spatial plane and the number of dynamical generators is less [2, 3]. Equation (286) is the same relation found by the Dirac–Bergmann algorithm [127]; we have found it without using this algorithm because we have constructed Fock’s space with an appropriate basis of one-particle wave-functions, which already satisfy all the constraints of the theory. Finally, we see that the commutation relations at equal times require only the knowledge of the field on the null-plane.

1.2 Fermion field

Consider now Dirac’s fermion field, so we will use anti-commutation relations. Since it describes two particles, each one with two polarization states, as it has spin 1/2, we will use four sets of emission and absorption operators: each set for each polarization of each particle. We start by remembering the reader that the classical Dirac’s field is decomposed into two parts by the projection operators $\Lambda _{({\pm })}$:¹⁴

$$\begin{aligned} \psi (x)=\psi _{(+)}(x)+\psi _{(-)}(x);\quad \psi _{({\pm })}:=\Lambda _{({\pm })}\psi , \quad \Lambda _{({\pm })}:=\frac{1}{\sqrt{2}}\gamma ^0\gamma ^{\pm }. \end{aligned}$$

(287)

From them, only $\psi _{(+)}$ is dynamical and follows Klein–Gordon–Fock’s equation. The projection $\psi _{(-)}$, on the other hand, is obtained from $\psi _{(+)}$ by means of the constraint equation:

$$\begin{aligned} \psi _{(-)}=\frac{1}{i\sqrt{2}\partial _-}\gamma ^0(m-i\gamma ^\perp \partial _\perp )\psi _{(+)}. \end{aligned}$$

(288)

If we decompose $\psi _{(+)}(x)$ in the basis of the space projected by $\Lambda _{(+)}$:

$$\begin{aligned} \psi _{(+)}(x)=\left( \begin{matrix} 0\\ 1\\ 0\\ 0 \end{matrix}\right) \alpha (x)+\left( \begin{matrix} 0\\ 0\\ 1\\ 0 \end{matrix}\right) \beta (x), \end{aligned}$$

(289)

then we can consider $\alpha (x)$ and $\beta (x)$ as fermion charged scalar fields, so their corresponding quantized field operators are:

$$\begin{aligned}&\alpha (x)=(2\pi )^{-3/2}\int {\mathrm{d}}\mu (\varvec{p})\left( b_1(\varvec{p})e^{-ipx}+d_1^\dagger (\varvec{p})e^{ipx}\right) , \end{aligned}$$

(290)

$$\begin{aligned}&\beta (x)=(2\pi )^{-3/2}\int {\mathrm{d}}\mu (\varvec{p})\left( b_{-1}(\varvec{p})e^{-ipx}+d_{-1}^\dagger (\varvec{p})e^{ipx}\right) . \end{aligned}$$

(291)

They are subjected to the following anti-commutation relations:

$$\begin{aligned} \left\{ \alpha (x);{{\widetilde{\alpha }}}(y)\right\} =-iD(x-y),\quad \left\{ \beta (x);{{\widetilde{\beta }}}(y)\right\} =-iD(x-y), \end{aligned}$$

(292)

and for the emission and absorption field operators ($s={\pm }1$):

$$\begin{aligned} \left\{ b_s(\varvec{p});b_s^\dagger (\varvec{q})\right\} =2p_-\delta (\varvec{p}-\varvec{q}),\quad \left\{ d_s(\varvec{p});d_s^\dagger (\varvec{q})\right\} =2p_-\delta (\varvec{p}-\varvec{q}). \end{aligned}$$

(293)

Although the quantization is complete, it is convenient to use Eq. (288) to reconstruct the entire $\psi (x)$ field and write the anti-commutation relation in function of it. Substituting Eqs. (290) and (291) into Eq. (289), we obtain $\psi _{(+)}$. Then, using Eq. (288) we find $\psi _{(-)}$; their sum—as in Eq. (287)—is the quantized field operator $\psi (x)$. In this procedure, no degree of freedom is increased, and the complete field is implicitly written in function of the dynamical degrees of freedom of the theory, $\alpha (x)$ and $\beta (x)$. We obtain:

$$\begin{aligned} \psi (x)=(2\pi )^{-3/2}\sum _s\int {\mathrm{d}}\mu (\varvec{p})\sqrt{2p_-}\left( u_s(\varvec{p})b_s(\varvec{p})e^{-ipx}+v_s(\varvec{p})d_s^\dagger (\varvec{p})e^{ipx}\right) . \end{aligned}$$

(294)

And for the adjoint field:

$$\begin{aligned} {{\overline{\psi }}}(x)=(2\pi )^{-3/2}\sum _s\int {\mathrm{d}}\mu (\varvec{p})\sqrt{2p_-}\left( {{\overline{v}}}_s(\varvec{p})d_s(\varvec{p})e^{-ipx}+\overline{u}_s(\varvec{p})b_s^\dagger (\varvec{p})e^{ipx}\right) , \end{aligned}$$

(295)

with the four-component functions $u, {{\overline{u}}}$ and $v,\overline{v}$ normalized so as to satisfy the sum rules:

$$\begin{aligned} \sum _s u_s(\varvec{p}){\overline{u}}_s(\varvec{p})= & {} \frac{E\gamma ^++|p_-|\gamma ^-+p_\perp \gamma ^\perp +m}{|2p_-|}, \end{aligned}$$

(296)

$$\begin{aligned} \sum _s v_s(\varvec{p}){{\overline{v}}}_s(\varvec{p})= & {} \frac{E\gamma ^++|p_-|\gamma ^-+p_\perp \gamma ^\perp -m}{|2p_-|}. \end{aligned}$$

(297)

With the expressions given in Eqs. (294) and (295) and using the anti-commutation rules of Eq. (293) we find:

(298)

and:

(299)

Therefore, since $\Theta (p_-)-\Theta (-p_-)=\text {sgn}(p_-)$, the anti-commutator of Dirac’s field with its adjoint is:

(300)

We identify here the distribution S(x) which in the classical case solves Goursat’s problem for Dirac’s equation:

$$\begin{aligned} \left\{ \psi (x);{{\overline{\psi }}}(y)\right\} =-iS(x-y), \end{aligned}$$

(301)

with:

(302)

1.3 Neutral massless vector field

As it is established at the classical level, the neutral massless vector field $A^a$ has only two degrees of freedom, corresponding to the transversal components $A^\alpha$ ($\alpha =1,2$), obeying Klein–Gordon–Fock’s equation:

$$\begin{aligned} \square A^\alpha =0\, . \end{aligned}$$

(303)

The component $A^+$ is zero—this is the null-plane gauge condition, imposed in order to have a well-defined initial value problem [6], and the component $A^-$ can be obtained from Lorenz’s gauge condition, implied by the null-plane one in the free case:

$$\begin{aligned} A^+=0,\quad A^-=-\partial _-^{-1}\partial _\alpha A^\alpha \, . \end{aligned}$$

(304)

By construction, only physical states are in Fock’s space, so that only the potentials $A^a$ satisfying Eqs. (303) and (304) can be one-particle wave-functions. The covariance of Fock’s space is assured by the fact that the null-plane gauge condition completely determines the gauge.

Accordingly with our quantization program, the field operators associated to the components $A_1$ and $A_2$ are:

$$\begin{aligned} A_1(x)= & {} (2\pi )^{-3/2}\int {\mathrm{d}}\mu (\varvec{p})\left( a_1(\varvec{p})e^{-ipx}+a_1^\dagger (\varvec{p})e^{ipx}\right) , \end{aligned}$$

(305)

$$\begin{aligned} A_2(x)= & {} (2\pi )^{-3/2}\int {\mathrm{d}}\mu (\varvec{p})\left( a_2(\varvec{p})e^{-ipx}+a_2^\dagger (\varvec{p})e^{ipx}\right) . \end{aligned}$$

(306)

They satisfy the commutation relations:

$$\begin{aligned} \left[ A_{1-}(x);A_{1+}(y)\right] =-iD_+(x-y),\quad \left[ A_{2-}(x);A_{2+}(y)\right] =-iD_+(x-y). \end{aligned}$$

(307)

From Eq. (307) we obtain:

$$\begin{aligned} \left[ A_1(x);A_1(y)\right] =-iD(x-y),\quad \left[ A_2(x);A_2(y)\right] =-iD(x-y), \end{aligned}$$

(308)

and:

$$\begin{aligned} \left[ a_1(\varvec{p});a_1^\dagger (\varvec{q})\right] =2p_-\delta (\varvec{p}-\varvec{q}),\quad \left[ a_2(\varvec{p});a_2^\dagger (\varvec{q})\right] =2p_-\delta (\varvec{p}-\varvec{q}). \end{aligned}$$

(309)

As in the fermion field case, it will be useful to write an expression for the complete vector field, for which we use the second of Eqs. (304), from that the component $A^-(x)$ can be found:

$$\begin{aligned} A^-(x)=-(2\pi )^{-3/2}\int \frac{{\mathrm{d}}\mu (\varvec{p})}{p_-}\left[ \left( p_1a_1(\varvec{p})+p_2a_2(\varvec{p})\right) e^{-ipx}+\left( p_1a_1^\dagger (\varvec{p})+p_2a_2^\dagger (\varvec{p})\right) e^{ipx}\right] , \end{aligned}$$

(310)

where we have identified the “−” component of the polarization vectors $\varepsilon _{1,2}(\varvec{p})^-$. For completeness we give the expression of the polarization vectors, which can be found in a classical analysis:

$$\begin{aligned} \varepsilon _1(\varvec{p})^a=\left( 0;1;0;-\frac{p_1}{p_-}\right) ,\quad \varepsilon _2(\varvec{p})^a=\left( 0;0;1;-\frac{p_2}{p_-}\right) , \end{aligned}$$

$$\begin{aligned} \varepsilon _+(\varvec{p})^a=\left( 1;-\frac{p_1}{p_-};-\frac{p_2}{p_-};\frac{p_\perp ^2}{2p_-^2}\right) ,\quad \varepsilon _-(\varvec{p})^a=\left( 0;0;0;1\right) . \end{aligned}$$

(311)

Being that way, we can write the quantized field operator:

$$\begin{aligned} A^a(x)=(2\pi )^{-3/2}\sum _{\lambda =1,2}\int {\mathrm{d}}\mu (\varvec{p})\varepsilon _\lambda (\varvec{p})^a\left( a_\lambda (\varvec{p})e^{-ipx}+a^\dagger _\lambda (\varvec{p})e^{ipx}\right) . \end{aligned}$$

(312)

With the aid of the commutation relations of Eq. (309) we can already calculate the commutation distribution of the quantized field operators in Eq. (312):

$$\begin{aligned} \left[ A^a(x);A^b(y)\right]&=(2\pi )^{-3}\int {\mathrm{d}}\mu (\varvec{p})\sum _{\lambda =1,2}\varepsilon _\lambda (\varvec{p})^a\varepsilon _\lambda (\varvec{p})^b\left. \left( e^{-ip(x-y)}-e^{ip(x-y)}\right) \right| _{p_+=E}\nonumber \\&=(2\pi )^{-3}\int d^4p\delta (p^2)\Theta \left( p_-\right) \left( -g^{ab}+\frac{p^a\eta ^b+\eta ^a p^b}{p_-}\right) \left( e^{-ip(x-y)}-e^{ip(x-y)}\right) \nonumber \\&=-(2\pi )^{-3}\int d^4p\text {sgn}\left( p_-\right) \delta (p^2)\left( g^{ab}-\frac{p^a\eta ^b+\eta ^ap^b}{p_-}\right) e^{-ip(x-y)}. \end{aligned}$$

(313)

In these expressions, the light-like vector $\eta ^a$ is defined as:

$$\begin{aligned} (\eta ^a):=(0;0^\perp ;1),\quad (\eta _a)=(1;0_\perp ;0). \end{aligned}$$

(314)

Therefore, defining the distribution $D^{ab}(x)$ by:

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

(315)

we found that it is:

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

(316)