How to present and interpret the Feynman diagrams in this theory describing fermion and boson fields in a unique way, in comparison with the Feynman diagrams so far presented and interpreted?

TL;DR

The work proposes unifying the second-quantized fermion and boson fields by describing their internal spaces with Clifford-odd (fermion) and Clifford-even (boson) basis vectors in $d=(13+1)$, while keeping physical processes in $d=(3+1)$. Fermions arise from odd basis vectors organized into $2^{\frac{d}{2}-1}$ families, whereas bosons come from two orthogonal boson groups constructed from even vectors, enabling a two-tiered description of gauge and gravitational interactions. The action and resulting Feynman rules are derived from algebraic products of these basis vectors, leading to diagrams that differ from conventional ones (e.g., fermions/antifermions in the same family; two boson groups) but aim to reproduce observed massless dynamics and interactions. If successful, this approach could provide a common foundation for gravity, gauge fields, and scalar sectors, while inviting open problems on propagators, causality, and possible extensions to strings or higher dimensions.

Abstract

Although the internal spaces describing spins and charges of fermions' and bosons' second-quantised fields have such different properties, yet we can all describe them equivalently with the ``basis vectors'' which are a superposition of odd (for fermions) and even (for bosons) products of $γ^{a}$'s. In an even-dimensional internal space, as it is $d=(13+1)$, odd ``basis vectors'' appear in $2^{\frac{d}{2}-1}$ families with $2^{\frac{d}{2}-1}$ members each, and have their Hermitian conjugate partners in a separate group, while even ``basis vectors'' appear in two orthogonal groups. Algebraic multiplication of boson and fermion ``basis vectors'' determines the interactions between fermions and bosons, and among bosons themselves, and correspondingly also their action. Tensor products of the ``basis vectors'' and basis in ordinary space-time determine states for fermions and bosons, if bosons obtain in addition the space index $α$. We study properties of massless fermions and bosons with the internal spaces determined by the ``basis vectors'' while assuming that fermions and bosons are active only in $d=(3+1)$ of the ordinary space-time. We discuss the Feynman diagrams in this theory, describing internal spaces of fermion and boson fields with odd and even ``basis vectors'', respectively, in comparison with the Feynman diagrams of the theories so far presented and interpreted.

Figures (4)

• Figure 1: An electron, with the internal space described by $\hat{b}^{1 \dagger}_{1}$ and with the momentum $\vec{p}_1$ in ordinary space, Table \ref{['Table Clifffourplet.']}, radiates the photon with the "basis vector" ${}^{II}\hat{\cal A}^{1 \dagger}_{3} (\equiv \stackrel{03}{[-i]} \stackrel{12}{[+]}\stackrel{56}{[+]}\equiv (\hat{b}^{1 \dagger }_{1})^{\dagger}\, *_{A}\, \hat{b}^{1 \dagger }_{1}$), with the momentum $\vec{p}_3$, while the electron, $\hat{b}^{1 \dagger }_{1}$, continues its way with a smaller momentum $\vec{p}_2$, Fig. \ref{['bIIAb']}. This diagram is representing also the electron in the usual theories, except that photons are not presented in our way. For the electron with the "basis vector", $e^{-\dagger}_{L} (\equiv \stackrel{03}{[- i]} \stackrel{12}{[+]} \stackrel{56}{(-)} \stackrel{78}{(+)} \stackrel{9\, 10}{(+)} \stackrel{11\,12}{(+)} \stackrel{13\,14}{(+)})$, from Table 2 in Ref. n2025Bled, the photon with the "basis vector" ${}^{II}\hat{\cal A}^{\dagger}(\equiv (e^{-\dagger}_{L})^{\dagger}$$\, *_A\, e^{-\dagger}_{L} \equiv \stackrel{03}{[- i]} \stackrel{12}{[+]}$$\stackrel{56}{[+]} \stackrel{78}{[-]} \stackrel{9\, 10}{[-]} \stackrel{11\,12}{[-]} \stackrel{13\,14}{[-]})$ takes away the momentum.
• Figure 2: A positron, with the internal space described by $\hat{b}^{3 \dagger}_{1}$ and with the momentum $\vec{p}_1$ in ordinary space, Table \ref{['Table Clifffourplet.']}, radiates the photon with the "basis vector" ${}^{II}\hat{\cal A}^{1 \dagger}_{3} (\equiv \stackrel{03}{[-i]} \stackrel{12}{[+]}\stackrel{56}{[+]}\equiv (\hat{b}^{3 \dagger }_{1})^{\dagger}\, *_{A}\, \hat{b}^{3 \dagger }_{1}=(\hat{b}^{1 \dagger }_{1})^{\dagger}\, *_{A}\, \hat{b}^{1 \dagger }_{1}$ ), with the momentum $\vec{p}_3$, while the positron, $\hat{b}^{3 \dagger }_{1}$, continues its way with smaller momentum $\vec{p}_2$, Fig. \ref{['bIIAbpos']}. For the positron with the "basis vector", $e^{+\dagger}_{R} (\equiv \stackrel{03}{(+i)} \stackrel{12}{[+]} \stackrel{56}{[+]} \stackrel{78}{[-]} \stackrel{9\, 10}{[-]} \stackrel{11\,12}{[-]} \stackrel{13\,14}{[-]})$, from Table 2 in Ref. n2025Bled, the photon with the "basis vector" ${}^{II}\hat{\cal A}^{\dagger} (\equiv (e^{+\dagger}_{R})^{\dagger}$$\, *_A\, e^{+\dagger}_{R} \equiv \stackrel{03}{[- i]} \stackrel{12}{[+]}$$\stackrel{56}{[+]} \stackrel{78}{[-]} \stackrel{9\, 10}{[-]} \stackrel{11\,12}{[-]} \stackrel{13\,14}{[-]})$ takes away the momentum. The corresponding Feynman diagram in the usual theories, representing the positron should have the arrows for the positron turned back, $\uparrow$ should be turned into $\downarrow$.
• Figure 3: The left-hand side represents the path of the electron, $e^{-\dagger}_{L}$, which radiates a photon $(e^{-\dagger}_{L})^{\dagger}\,*_{A}\, e^{-\dagger}_{L}$, and continues its way straight to the right, up to a positron, $e^{+\dagger}_{R}$ coming up. They both radiate a photon $(e^{-\dagger}_{L})^{\dagger}\,*_{A}\, e^{-\dagger}_{L}$ and $(e^{+\dagger}_{R})^{\dagger}\,*_{A}\, e^{+\dagger}_{R}$ (both are of the same kind) and remain without momenta in the quantum vacuum. It can also happen the opposite: The positron, $e^{+\dagger}_{R}$, radiates a photon $(e^{+\dagger}_{R})^{\dagger}\,*_{A}\,e^{+\dagger}_{R}$, and continues its way straight to the left, up to an electron, $e^{-\dagger}_{L}$ coming up from the left hand side. Both radiate a photon $(e^{-\dagger}_{L})^{\dagger}\,*_{A}\,e^{-\dagger}_{L}$ of the same kind. Both remain without momentum in the quantum vacuum.
• Figure 4: The electron $e^{- \dagger}_L$ radiates a photon ${}^{I}\hat{{\cal A}}^{\dagger}_{phee^{\dagger}} (\equiv e^{-\dagger}_{L}\,*_{A}\,(e^{-\dagger}_{L})^{\dagger})$, and goes to the right to the vacuum. The positron $e^{+ \dagger}_L$, radiates a photon ${}^{I}\hat{{\cal A}}^{\dagger}_{phpp^{\dagger}} (\equiv e^{+\dagger}_{R}\,*_{A}\,(e^{+\dagger}_{R})^{\dagger}$, and turning to the left remains with electron in the vacuum.