Theoretical Spectrum of Mass of the Nucleons: New Aspects of the QM

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1. Introduction

One of the more problematic aspects in Hadronic literature [1] [2] is that one in which only a small fraction of the hadron mass seems to be associated with the bare mass of the elementary quarks. If the total bare mass of quarks (*u*,*u*,*d*) forming the proton (for literature) is around 10 MeV [3], then quarks represent only ~1% of the proton mass. The same occurs in a pion or, generally, in mesons. Nowadays, in literature [4] [5], one of the answers to this mystery of mass origin in hadrons is that the remaining mass fraction could be due to the force (gluons) binding the quarks within hadrons, *i.e.*, Quantum Chromo Dynamics (QCD). The researchers [4] [5], by particular discretization of potentials formulated in QCD, have found the values of the “bare” masses of quarks corresponding, obtained directly in lattice simulations [6]. Furthermore, these values and others of experimental masses [5] are used as input (see the pion mass) to improve the mass values and coupling parameters in the interactions and so to calculate the masses of hadrons (see later in the conclusions of this paper). Nevertheless, the procedures of calculation are very complicated, also if supported by processed programs of calculation with computers. Besides, lattice QCD is an expensive technique, because the discretization creates errors, and to remove them entails taking the lattice spacing, *a*, to zero. In a set of studies, for remedying to these complex problems of calculation and calculating with simply way the mass values of light mesons, we have used another path [7] [8] [9]. Instead of incorporate the binding energy of the quarks into appropriate potentials of the binding quanta (gluons), see the QCD, we incorporated this energy into the mass of quarks which compose the pion [7]. We have so obtained the values of “dressed” mass of the (u, d) quarks, derived, nevertheless, from a suitable “geometric structure” of the pions [7] [8] and same quarks [9]. These mass values are very greater than that of free quarks. This way, the bonded quarks inside the hadron would have enough mass to reach the hadron mass value they compose, provided that we correctly sum them up. Since the pion, the lightest meson, is made of two bonded states of quarks, its mass could be placed as base mass to determine the mass of the most massive mesons and hadrons (as it happens in literature [10] [11]. We already proved in two separate articles [8] [9] that all light mesons, following the pion, are elaborated structures of pions lattice {*π*^{+}, *π*^{−}} and {*d*,*d*}-lattice of d-quarks. This hypothesis induces us to admit an appropriate structure equation [7] [8] [9], which describes all mesons and thus also hadrons. By this equation, we can so calculate the masses of light mesons and hadrons. Moreover, the structure equation allows us to identify the possible decays of a particle and its spin value. Nevertheless, all of this (the calculation procedure) is possible only if we affirm [12] [13] [7] the quarks having a well-defined geometric structure. In this way, we cannot consider a particle (quark) anymore as an object-point, see the Quantum Relativistic Theory of fields and QCD [14] [15] ). One can achieve the idea of a geometric structure of quarks if we assume a quark made by a *non-separable* set of coupled quantum oscillators with “*Aurea*” (golden) geometric structure [12] [7]. We have referred to this as the Aureum Geometric Model (AGM) of quarks. Thanks to the quarks geometric hypothesis, it is possible to explain fundamental issues such as the origin of the mass of the hadrons, the role of gluons in constructing bonded quarks and the existence of *molecules of quarks and also pions* [8] [9]. We can achieve all of this without turning to the QCD and its calculations by colour potentials. Nevertheless, for obtaining the hadronic masses, we need to revise the mass conception in physics and to introduce a new idea of mass calculation (Ä-operation) [7]. The idea of calculation takes into account both interactions between quarks and the possible *interpenetration* [8] [9]. This last aspect is purely quantum-undulatory like the superposition of particles-waves. The latter may exist if we admit that two different structures (states) of coupled quantum oscillators (or quarks) can overlap without exchanging energy. The global mass of hadron must so take in account all the possible configurations of quark components, both the ones with interpenetration and the ones with interactions. In this way, the hypothesis of the quark geometrical structure introduces a “*new paradigm*” [8] in the phenomenology of hadronic interactions. Thanks to a few primary hypotheses, the new paradigm allows describing the hadronic phenomenology in a more structured and straight forward way of that of QCD. The first evidence is the spectrum of light mesons with mass values very closed (if not even equal) to the experimental ones that we have calculated in two papers, see ref. [8] [9]. In this paper, exactly, we will calculate, by theoretical physics aspects and suitable mathematical procedure, the masses of the nucleons, while in next study that of light baryons without strangeness (Δ°, Δ^{−}, Δ^{++}). In finally, in the conclusions section of this paper, we will show that the two paths, AGM and QCD, are physical equivalents.

2. The Geometric Structure of Particles

2.1. The Dressed Masses and Bare of Quarks (*u*, *d*)

In ref. [12], we showed that each quark (*u*, *d*) of the proton has an “aurea” geometric structure (golden structure) achieved by coupling of three quantum oscillators (vertices of a triangle), called IQuO, the acronym of Intrinsic Quantum Oscillator. This coupling consists of the IQuO quantum oscillators of the junction (junction gluons), placed along the sides. We showed that three quantum oscillators, in eigenstate *n* = 2, and connected by junction quantum oscillators (with*n* = 1, *n *= 2), can couple for origin an “aurea” triangular structure. Thanks to this conjecture, we have also given a geometric structure to mesons [9]: coupling among two golden triangles (*u*, *d*) which are bounded by gluons (binding gluons), see Figure 1.

Figure 1. Meson with gluons.

By the “aurea” hypothesis, we have calculated the mass values of quarks inside the pion (*u _{π}*,

$\{\begin{array}{l}\underset{\_}{m}\left({u}_{\pi}\right)+\underset{\_}{m}\left({d}_{\pi}\right)={m}_{\pi}^{\pm}\\ \underset{\_}{m}\left({d}_{\pi}\right)/\underset{\_}{m}\left({u}_{\pi}\right)=\varphi \end{array}$ (1)

where is *f* the golden number. This equation has solutions [*m*(*u _{π}*) = (53.31) MeV,

$\{\begin{array}{l}\left(1/2\right)\left[m\left({u}_{f}\right)+m\left({d}_{f}\right)\right]=\Delta {m}_{\pi}^{0}\\ m\left({d}_{f}\right)/m\left({u}_{f}\right)=\varphi \end{array}\Rightarrow \{\begin{array}{l}m\left({u}_{f}\right)=\left(3.51\right)\text{MeV}\\ m\left({d}_{f}\right)=\left(5.67\right)\text{MeV}\end{array}$ (2)

2.2. The Proton Structure

Also, the proton can be described by a structure equation relative to an “aurea” geometric structure: three golden triangles couple building a geometric figure which can change taking the form of a pentagon or triangle, see Figure 2. This idea [7] of aurea structure is born by the observation that between the Compton’s wave lengths (*Δ** _{pl}*) of the Planck-particle and the one of the proton (

Figure 2. Two configurations of the proton in two different times.

Where (*f*) is the “*aureus*” (golden) number.

The configurations (a) and (b) belong in different times to the same particle because the u-quarks are in rotation around the sides (AD), (DB). Note the three vertices-IQuO can constitute the three scattering centers seen in the experiments of deep inelastic diffusion (e, p). In the first time, the geometric representation of quarks (AGM) can be seen differently to those of partons and QCD. In AGM, the quark is not a point-particle surrounded by clouds of gluons but a set of coupled oscillators (IQuO) by of junction oscillators (“junction gluons”). In its turn, each quark can be bound to other quarks by “*binding gluons*”. Nevertheless, studying more in-depth these new aspects, we can see the two models (QCD and AGM) having some points of connection. In fact, in QCD [14] [16] the quarks (*u, d*) would show, in the periphery, an interaction potential [*V(r) *» *kr*], with *r* the linear nucleon dimensions, expressed by an interaction strong force with modality [*F* » *k*]. This force should be similar at the interaction force in AGM. In this theory, the binding gluons in the geometric structure of two quarks (*u, d*) are equivalent to simple oscillators with elastic force [*F *»*k**x*]. Specifically, *k* is a variable elastic parameter appropriated: in the periphery the force is constant [*F* » *k*], see the Cornel potential in ref. [14]. These correlations so point out two similar paths for treating the Hadrons, even if the AGM admits the innovation of a geometric structure of quarks. This assert introduces a “*new paradigm*” [8] [9] in the representation of particles and phenomenology of their hadronic interactions. To the geometric structure of quarks (*u*, *d*), see later, we associate a structure equation expressed [7] by matrices [||*u*||, ||*d*||]: the terms of each matrix are quantum vertex-oscillators of single quark, expressed by the quantum operators (*a*, *a*^{+}), in the IQuO form. Tanks to the structure equation, we have calculated the mass of light mesons [9]. In this study, we will show that by structure equation one can determine the spectrum of the mass value of elementary baryons as the nucleons.

2.3. The Quark’s Interpenetration

The structures are not rigid because the binding gluons allow the movement of quarks: the quarks reciprocally rotate one respect to other around to a common axis. As noted [7] [9], different relative orientations between *u *quarks and* d* quarks imply a relative rotation (orbital spin), suggesting a mutual crossing of the quarks (see Figure 1). This crossing can have the meaning of a reciprocal *interpenetration *(Ä) between quarks (*q _{i}* Ä

2.4. The Correlation between Quark’s Interpenetration and Spin

The different configurations, or orientations, can induce us to think about the *orbital spin *(*s _{l}*) of each quark. To individual quark, seen like Fermion particle, one associates an

${s}_{p}=\left\{\begin{array}{c}{s}_{tot}\left(q\right)=\left[{\left(\uparrow \downarrow \right)}_{{q}_{1}},{\left(\uparrow \downarrow \right)}_{{q}_{2}},{\left(\uparrow \downarrow \right)}_{{q}_{3}}\right]\\ {s}_{tot}\left(l\right)=\left[{\left(\uparrow \downarrow \right)}_{{q}_{1}},{\left(\uparrow \downarrow \right)}_{{q}_{2}},{\left(\uparrow \downarrow \right)}_{{q}_{3}}\right]\\ {s}_{tot}\left(g\right)=\left[{\left(\uparrow \downarrow \right)}_{{q}_{1}},{\left(\uparrow \downarrow \right)}_{{q}_{2}},{\left(\uparrow \downarrow \right)}_{{q}_{3}}\right]\end{array}\right\}=\left\{{s}_{tot}\left(q\right)+{s}_{tot}\left(l\right)+{s}_{tot}\left(g\right)\right\}$ (3)

where [(*s _{tot}*(

2.5. The (Ä, Å) Operations and the (*F _{m}*

In the interpenetration of quarks and their dynamics interactions, we used, see ref. [7] [9], a new Ä*-operation* of combination (or coupling) of quarks and particles: (*a _{i}* Ä

$\begin{array}{c}{F}_{m}\left(X\right)=\left\{{\displaystyle \underset{\left(i,j\right)=1}{\overset{n}{\sum}}{F}_{m}\left[{\left({A}_{i}\right)}_{(\otimes )},{\left({B}_{j}\right)}_{(\oplus )}\right]}\right\}\\ ={\left[{\displaystyle \underset{i=1}{\overset{n}{\sum}}m{\left({A}_{i}\right)}_{(\otimes )}}\right]}_{A}+{\left[{\displaystyle \underset{j=1}{\overset{m}{\sum}}m{\left({B}_{j}\right)}_{(\oplus )}}\right]}_{B}\\ ={\left[m{\left(a\otimes b\right)}_{{A}_{1}}+\cdots +m{\left(w\otimes z\right)}_{{A}_{n}}\right]}_{A}+{\left[m{\left(a\oplus b\right)}_{{B}_{1}}+\cdots +m{\left(w\oplus z\right)}_{{B}_{m}}\right]}_{B}\end{array}$ (4)

where (*a*, *b*, …, *w*, *z*) are the base particles which build the structure (hadron). We will have the following applications:

$\left\{\begin{array}{c}{F}_{m}\left({A}_{(\otimes )}\right)={F}_{m}{\left[\left(a\right)\otimes \left(b\right)\right]}_{A}=\langle m\left(a,b\right)\rangle =\langle m\left(a\right),m\left(b\right)\rangle \\ {F}_{m}\left({B}_{(\oplus )}\right)={F}_{m}{\left[\left(a\right)\oplus \left(b\right)\right]}_{B}=m\left(\left(a\right)\oplus \left(b\right)\right)=m\left(a\right)+m\left(b\right)\end{array}\right\}$ (5)

Note that the mass of two interpenetrating particles [*a *Ä *b*] is obtained by the average value of individual masses [<*m*(*a*, *b*)>], while the mass of two interacting particles [*a *Å *b*] is obtained by the sum of the masses of each particle. To obtain the total mass of a structure, it needs to add eventual (*m _{kin}*) relativistic kinetic mass and mass defects (Δ

We will have: [*m _{tot}* =

The mass defect will be: Δ*m* = Δ*m _{g}* + Δ

Where Δ*m _{g}* is the mass defect due to binding by gluons. Nevertheless, the Δ

To obtain the mass defects (Δ*m *> 0, Δ*m *< 0) we have used [9] a Function (*F*_{Δm}) applied to structure equation for calculating the mass defects. Besides, it needs to consider that:

$\begin{array}{l}\Delta m{\left[\left(a\right)\otimes \left(b\right)\right]}_{{A}_{i}}=\left\{\begin{array}{c}0\\ \Delta m{\left(a,b\right)}_{\left(a\cap b\right)}\ne 0\end{array}\right\}\\ \Delta m{\left[\left(a\right)\oplus \left(b\right)\right]}_{{B}_{i}}=\Delta m{\left(a,b\right)}_{\text{interaction}}\end{array}$ (6)

where the (*a*, *b*) point out “base particles” of the (*A _{i}*,

Table 1. The couplings with the interaction of quarks.

Where *q _{i} *is the base particles (here we have quarks) which compose the particle

3. The Nucleon Masses

3.1. The Mass of the Proton (938.27 MeV)

In light mesons, we highlighted a mass spectrum built employing of lattices of base pions {*π*} and quarks {*d*, *d*} [9]. In the Baryons of the first octet, using a quark lattice but not pions we build the mass spectrum because the Baryons are built with three component quarks. Therefore, the fundamental element is not a couple of quarks but the individual quark and its possible coupling. The coupling of quarks with interpenetration will be expressed by Ä-operator: (*q* Ä *q* Ä *q*).The product [*u* Ä *d* Ä *u*] imply the combination of all possible configurations between quarks (*u*, *d*, *u*).

Recall [Ä = Ä U Å], see ref. [7] [8] [9], where the Ä-operation expresses the interpenetration between quarks while the Å-operation expresses them eventual interactions through gluons and photons. Nevertheless, we incorporate the elastic tensions of binding gluons into opportune mass values, see the quarks’ masses [*m*(*u _{π}*),

${\left[\left({u}_{1}d\right),\left({u}_{1}{u}_{2}\right),\left(d{u}_{2}\right)\right]}_{A1},{\left[{u}_{1}\left(d{u}_{2}\right),d\left({u}_{1}{u}_{2}\right),{u}_{2}\left({u}_{1}d\right)\right]}_{A2}$ _{ }

The property of operation (Ä) which describes the combinations of three quarks is:

$\left[A\text{\hspace{0.17em}}\underset{\_}{\otimes}\text{\hspace{0.17em}}B\text{\hspace{0.17em}}\underset{\_}{\otimes}\text{\hspace{0.17em}}C\right]=\left\{\left[\left(A\right)\otimes \left(B\oplus C\right)\right]\oplus \left[\left(B\right)\otimes \left(C\oplus A\right)\right]\oplus \left[\left(C\right)\otimes \left(A\oplus B\right)\right]\right\}$ (7)

The geometric representation of proton admits two configurations, Figure 3.

Figure 3. Proton configurations.

Note the configuration 1(b) shows an electric charge equal to one, along the propagation side BC.

Also if we give a structure of coupling oscillators to the proton, the same one needs to associate a wave function to the proton and, thus, a field of quantum oscillators, expressed by annihilation operator (*a*) and creation (*a ^{+}*). The representative quanta of the proton propagate along the axis passing through B and C (see Figure 3); this axis consists of a chain of quantum oscillators belonging to the diagonal of junction of a lattice of pairs [(

Let us go to determining the structure equation of the proton. Through geometric structure, see Figure 1 and Figure 3, one can deduce that a quark (*q _{i}*) interpenetrates the others two quark (

$\begin{array}{l}\left[{u}_{1}\text{\hspace{0.17em}}\underset{\_}{\otimes}\text{\hspace{0.17em}}d\text{\hspace{0.17em}}\underset{\_}{\otimes}\text{\hspace{0.17em}}{u}_{2}\right]\\ ={\left\{{\left[\left({u}_{1}\right)\otimes \left(d\oplus {u}_{2}\right)\right]}_{{A}_{1}}\oplus {\left[\left(d\right)\otimes \left({u}_{2}\oplus {u}_{1}\right)\right]}_{{A}_{2}}\oplus {\left[\left({u}_{2}\right)\otimes \left({u}_{1}\oplus d\right)\right]}_{{A}_{3}}\right\}}_{{A}_{3}}\end{array}$ (8)

This relation tells us that the quarks reciprocally interpenetrate, but at the same time, they interact between them. Therefore, the structure equation of the proton is:

$\left(p\right)=\left(\left\{3{\kappa}_{p}\left[{u}_{1}\underset{\_}{\otimes}\text{\hspace{0.17em}}d\text{\hspace{0.17em}}\underset{\_}{\otimes}\text{\hspace{0.17em}}{u}_{2}\right]\right\}\right)$ (9)

where the number 3 point out the 3-dimensionality of a baryon (three free degrees) while to the mesons we assign a 2-dimensionality or two free degrees. The *k** _{p}* coefficient is in connection to the global elasticity of the system of quantum oscillators (IQuO) of the geometric structure of the quarks into nucleons: we highlight that the increase of the number of vertices in a geometric structure makes increase the elastic tension between the quantum oscillators of structure.

The Å-operation in eq. 8 not cancels the no-locality of superposition of three states (Y* _{A}*) associated with

$\begin{array}{c}p=3{\kappa}_{p}\left\{{\left[{u}_{1}\otimes \left(d\oplus {u}_{2}\right)\right]}_{{A}_{1}}\oplus {\left[d\otimes \left({u}_{2}\oplus {u}_{1}\right)\right]}_{{A}_{2}}\oplus {\left[{u}_{2}\otimes \left({u}_{1}\oplus d\right)\right]}_{{A}_{3}}\right\}\\ =\left\{3{\kappa}_{p}\left[{p}_{a}\oplus {p}_{b}\oplus {p}_{c}\right]\right\}\end{array}$ (10)

where (*a*, *b*, *c*) are the indices of the following relations:

$\begin{array}{l}{p}_{a}={\left[{u}_{1}\otimes \left(d\oplus {u}_{2}\right)\right]}_{a}={\left[\left({u}_{1}\otimes d\right)\oplus \left({u}_{1}\otimes {u}_{2}\right)\right]}_{a}\\ {p}_{b}={\left[d\otimes \left({u}_{2}\oplus {u}_{1}\right)\right]}_{b}={\left[\left(d\otimes {u}_{2}\right)\oplus \left(d\otimes {u}_{1}\right)\right]}_{b}\\ {p}_{c}={\left[{u}_{2}\otimes \left({u}_{1}\oplus d\right)\right]}_{c}={\left[\left({u}_{2}\otimes {u}_{1}\right)\oplus \left({u}_{2}\otimes d\right)\right]}_{c}\end{array}$ (11)

Now, we calculate partial mass of the proton, using the properties of operations (Ä, Å), see the tables in ref. [9] and Function *F _{m}* in Equations (4) and (5). In first, it is:

$\begin{array}{l}{F}_{m}\left({p}_{a}\right)=m\left({p}_{a}\right)=m{\left[\left({u}_{1}\otimes d\right)\oplus \left({u}_{1}\otimes {u}_{2}\right)\right]}_{a}=m\left(\langle {u}_{1},d\rangle \right)+m\left(\langle {u}_{1},{u}_{2}\rangle \right)\\ {F}_{m}\left({p}_{b}\right)=m\left({p}_{b}\right)=m{\left[\left(d\otimes {u}_{2}\right)\oplus \left(d\otimes {u}_{1}\right)\right]}_{b}=m\left(\langle d,{u}_{2}\rangle \right)+m\left(\langle d,{u}_{1}\rangle \right)\\ {F}_{m}\left({p}_{c}\right)=m\left({p}_{c}\right)=m{\left[\left({u}_{2}\otimes {u}_{1}\right)\oplus \left({u}_{2}\otimes d\right)\right]}_{c}=m\left(\langle {u}_{2},{u}_{1}\rangle \right)+m\left(\langle {u}_{2},d\rangle \right)\end{array}$ (12)

With sum:

$\begin{array}{c}m\left({p}_{a,b,c}\right)=m\left({p}_{a}\right)+m\left({p}_{b}\right)+m\left({p}_{c}\right)\\ =2m\left(\langle {u}_{1},d\rangle \right)+2m\left(\langle {u}_{2},d\rangle \right)+2m\left(\langle {u}_{1},{u}_{2}\rangle \right)\end{array}$ (13)

This because is
$\left[A\otimes B\right]=\left[B\otimes A\right]$. Nevertheless, note the quarks masses (*u*,d) of proton could be different to one of pion:
$\left[m\left({u}_{p}\right),m\left({d}_{p}\right)\right]\ne \left[m\left({u}_{\pi}\right),m\left({d}_{\pi}\right)\right]$

The couplings [[*u*_{1}(*du*_{2}), *d*(*u*_{1}*u*_{2}), *u*_{2}(*u*_{1}*d*)]_{A}_{2}] in proton induces us to think that the pair (*u _{p}*,

$\left({u}_{p},{d}_{p}\right)={k}_{p}\left({u}_{\pi},{d}_{\pi}\right),\left({k}_{p}>0\right)$

We have already said that the elastic tension (*k*) between the IQuO of sides and the ones of vertices is function of them number; then, it is intuitive considering the parameter*k* given by the rapport between *n _{v}* number vertices

[(*A*(*d*), *B*(*d*), *C*(*d*), *D*(*u*_{1}), *E*(*u*_{2})] of geometric structure (in proton is *n _{v}* = 5) and number of total (

· In b-config.: [(*B*(*d*,*u*_{1}), *C*(*d*,*u*_{1}); (*B*(*d*,*u*_{2}), *C*(*d*,*u*_{2}); (*B*(*u*_{1},*u*_{2}), *C*(*u*_{1},*u*_{2})] ó (*n _{elv}*) = 6

· In a-config.: [(*B*(*d*,*u*_{1}), *C*(*d*,*u*_{2}); (*B*(*d*,*u*_{2}); *B*(*u*_{1},*u*_{2},*d*)] ó (*n _{elv}*) = 6

Where *B*(*u*_{1}, *u*_{2}, *d*) is a triple vertex. It follows: [*k** _{p}* =

$\begin{array}{c}m\left(p\right)=\left\{3{\kappa}_{p}\left[m\left({p}_{a}\right)+m\left({p}_{b}\right)+m\left({p}_{c}\right)\right]\right\}\\ =3{\kappa}_{p}\left[2m\left(\langle {u}_{1},d\rangle \right)+2m\left(\langle {u}_{2},d\rangle \right)+2m\left(\langle {u}_{1},{u}_{2}\rangle \right)\right]\\ =3{\kappa}_{p}\left[4m\left(\langle u,d\rangle \right)+2m\left(\langle {u}_{1},{u}_{2}\rangle \right)\right]\\ =3{\kappa}_{p}\left[4\left(69.79\right)+2\left(53.31\right)\right]\text{MeV}\end{array}$ (14)

Having that [*k** _{p}* = (5/6)], it follows:

$\begin{array}{c}m\left(p\right)=3\left(\frac{5}{6}\right)\left[4m\left(\langle u,d\rangle \right)+2m\left(\langle u,u\rangle \right)\right]\\ =\left[10\left(69.79\right)+5\left(53.31\right)\right]\text{MeV}=\left(\text{964}\text{.45}\right)\text{MeV}\end{array}$ (15)

In the structure equation, one can identify the various defects of mass. The electromagnetic mass defects will be Δ*m** _{g}*(

We can admit that: $\Delta {m}_{\gamma}{\left({q}_{i},{q}_{j}\right)}_{p}=3{\kappa}_{p}\Delta {m}_{\gamma}{\left({q}_{i},{q}_{j}\right)}_{\pi}$. Recall that [7]:

$\Delta {m}_{\gamma}\left({u}_{\pi},{d}_{\pi}\right)=m\left({\pi}^{\pm}\right)-m\left({\pi}^{0}\right)=\Delta m\left({\epsilon}_{{\pi}^{\pm}}\left(\gamma \right)\right)=\left(4.59\right)\text{MeV}/{c}^{2}$ (16)

With

$\Delta m\left({\epsilon}_{{\pi}^{\pm}}\left(\gamma \right)\right)=\frac{1}{2}\left[{m}_{\gamma}\left(u\right)+{m}_{\gamma}\left(d\right)\right]={m}_{\gamma}\left(\langle u,d\rangle \right)$ (17)

where *m** _{g}*(

$\left\{\begin{array}{c}{m}_{\gamma}\left(u\right)=\left(3.51\right)\text{MeV}/{c}^{2}=\Delta {m}_{\gamma}\left(u,u\right)\\ {m}_{\gamma}\left(d\right)=\left(\text{5}\text{.68}\right)\text{MeV}/{c}^{2}=\Delta {m}_{\gamma}\left(d,d\right)\end{array}\right\}$ (18)

where we have so calculated Δ*m** _{g}*(

We could calculate the global mass defect Δ*m** _{g}*(

1) Using the *F*_{Δm}-function

2) Using the matrix *A _{ij}* of the mass defects Δ

Here, we choose the matrix *A _{ij}* of mass defects. Then, we build the matrix

If the interpenetrating quark pair interacts to other quark,
$\left[\left({q}_{i}\otimes {q}_{j}\right)\oplus {q}_{j}\right]$, then it is:
$\left[{\left({q}_{i}\otimes {q}_{j}\right)}_{\oplus}\right]\ne \left[{\left({q}_{j}\otimes {q}_{i}\right)}_{\oplus}\right]$. This aspect gives the no-commutation of field operators (*q _{i}*,

Table 2. The couplings with the interaction of quarks in the proton.

For calculating the mass defect, we sum all mass defects relative to elements of the matrix. We are turning to Equations (16)-(18):

$\begin{array}{c}\Delta {m}^{*}\left(p\right)={\left[4\Delta m\left({\left(u\otimes d\right)}_{\oplus}\right)\right]}_{G}+{\left[2\Delta m\left({\left({u}_{1}\otimes {u}_{2}\right)}_{\oplus}\right)\right]}_{B}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left[\Delta m\left({\left({u}_{1}\otimes {u}_{1}\right)}_{\oplus}\right)+\Delta m\left({\left({u}_{2}\otimes {u}_{2}\right)}_{\oplus}\right)+\Delta m\left({\left(d\otimes d\right)}_{\oplus}\right)\right]}_{R}\\ =\left\{\left[{\left(18.36\right)}_{G}-{\left(7,02\right)}_{B}\right]\text{MeV}+{\left[{\left(R\left({u}_{1}\right)+R\left({u}_{2}\right)+R\left(d\right)\right)}_{\oplus}\right]}_{R}\right\}\\ =\left(11.34\right)\text{MeV}+R\left({q}_{i}\right)\end{array}$ (19)

Recall the $\Delta {m}_{\gamma}{\left({q}_{i},{q}_{j}\right)}_{p}>0$, while $\Delta {m}_{\gamma}{\left({q}_{i},{q}_{i}\right)}_{p}<0$.

For finding the values of mass defects *R*(*u _{i}*) and

$\begin{array}{c}{R}_{\oplus}=\left[\left(R\left({u}_{1}\right)+R\left({u}_{2}\right)+R\left(d\right)\right)\right]\text{}\\ ={\left[{\left({r}_{1}\Delta m{\left({u}_{(\uparrow )}\otimes {u}_{(\uparrow )}\right)}_{{u}_{1}}+{r}_{2}\Delta m{\left({u}_{(\downarrow )}\otimes {u}_{(\downarrow )}\right)}_{{u}_{2}}+{r}_{3}\Delta m{\left({d}_{(\uparrow )}\otimes {d}_{(\uparrow )}\right)}_{d}\right)}_{\oplus}\right]}_{R}\end{array}$ (20)

The arrows of the subscripts in terms of (*u*) and (*d*) point out the spin, see Equation (3).

In ref. [8] [9] to elements in red of *A _{ij}* matrix in mesons, we gave the values of (1, 15) MeV. In the proton, we assume the same value (see the couplings (

$\begin{array}{c}{s}_{p}=\left\{\begin{array}{c}{s}_{t}\left(q\right)=\left[{(\uparrow )}_{{u}_{1}},{(\downarrow )}_{{u}_{2}},{\left(\uparrow ,\downarrow \right)}_{{d}_{1}}\right]\equiv {s}_{q}\left({u}_{1}\right)+{s}_{q}\left({u}_{2}\right)+{s}_{q}\left({d}_{1}\right)={s}_{q}\left({d}_{1}\right)=\pm \left(1/2\right)\\ {s}_{t}\left(l\right)=\left[{(\uparrow )}_{{u}_{1}},{(\downarrow )}_{{u}_{2}},{\left(\uparrow ,\downarrow \right)}_{{d}_{1}}\right]\equiv {s}_{l}\left({u}_{1}\right)+{s}_{l}\left({u}_{2}\right)+{s}_{l}\left({d}_{1}\right)={s}_{l}\left({d}_{1}\right)=\pm \left(1/2\right)\\ {s}_{t}\left(g\right)=\left[{(\downarrow )}_{{u}_{1}},{(\uparrow )}_{{u}_{1}},{\left(\downarrow ,\uparrow \right)}_{{d}_{1}}\right]\equiv {s}_{g}\left({u}_{1}\right)+{s}_{g}\left({u}_{2}\right)+{s}_{g}\left({d}_{1}\right)={s}_{g}\left({d}_{1}\right)=\mp \left(1/2\right)\end{array}\right\}\\ ={s}_{q}\left({d}_{1}\right)=\pm \left(1/2\right)\end{array}$ (21)

Note that [*s*(*l*) + *s*(*g*) = 0] and [(*s _{tot}*(

$\begin{array}{l}{\left(R\left({u}_{1}\right)+R\left({u}_{2}\right)+R\left(d\right)\right)}^{\prime}\\ ={\left[\left({{r}^{\prime}}_{1}\Delta {m}^{\prime}{\left({u}_{(\uparrow )}\otimes {u}_{(\uparrow )}\right)}_{{u}_{1}}+{{r}^{\prime}}_{2}\Delta {m}^{\prime}{\left({u}_{(\downarrow )}\otimes {u}_{(\downarrow )}\right)}_{{u}_{2}}+{{r}^{\prime}}_{3}\Delta {m}^{\prime}{\left({d}_{(\uparrow )}\otimes {d}_{(\uparrow )}\right)}_{d}\right)\right]}^{\text{'}}\\ ={\left[\left({\left(-1\right)}_{{u}_{1}}+{\left(+1\right)}_{{u}_{2}}+{\left(-1\right)}_{d}\right)\right]}_{R}\left(1.15\right)\text{MeV}={\left[\left({0}_{u,u}-{1}_{d}\right)\right]}_{R}\left(1.15\right)\text{MeV}\end{array}$ (22)

where we have calculated the eigenvalues of $\left[R\left(q\right)=r\Delta m{\left({q}_{i}\otimes {q}_{i}\right)}_{\oplus}\right]$ :

${R}^{\prime}\left(q\right)={r}^{\prime}\left[\Delta {m}^{\prime}\left({q}_{i}\otimes {q}_{i}\right)\right]$ with eigenvalues *r’* = ±1 and
$\left[\Delta {m}^{\prime}\left({q}_{i}\otimes {q}_{i}\right)\right]=\left(1.15\right)\text{MeV}$

Exactly, we have

1) ${r}^{\prime}\Delta {m}^{\prime}\left({u}_{1}\otimes {u}_{1}\right)=-{r}^{\prime}\Delta {m}^{\prime}{\left({u}_{2}\otimes {u}_{2}\right)}^{\prime}$ because the spins are opposite

2) ${r}^{\prime}\Delta {m}^{\prime}{\left({u}_{1}\otimes {u}_{1}\right)}^{\prime}=-{r}^{\prime}\Delta {m}^{\prime}{\left(d\otimes d\right)}^{\prime}$ because quarks are reciprocally in relative movement

Then we have:

$\Delta {m}^{*}\left(p\right)=\left\{{\left[\left(11.34\right)\right]}_{GB}-\left[{\left(1.15\right)}_{R}\right]\right\}\text{MeV}=\left(10.19\right)\text{MeV}$ (23)

Nevertheless, the value (1, 15) MeV is only approximate, see ref [9]. With this approximation we obtain the total mass defect, being explicit the elasticity parameter *k*_{p}*,* see Equation (15):

$\Delta m\left(p\right)=3{\kappa}_{p}\left[\Delta {m}^{*}\left(p\right)\right]=\left(\frac{5}{2}\right)\left[\left(10.19\right)\right]\text{MeV}=\left(25.48\right)\text{MeV}$ (24)

Then, the total mass of the proton is:

${m}_{tot}\left(p\right)=m\left(p\right)-\Delta {m}_{\gamma}\left(p\right)=\left[\left(964.45\right)-\left(\text{25}\text{.48}\right)\right]\text{MeV}=\left(\text{938}\text{.97}\right)\text{MeV}$ (25)

Value very next to the experimental.

The state with [(*s _{tot}*)

${s}_{{N}^{+}}=\left\{\begin{array}{c}{s}_{t}\left(q\right)=\left[{\left(\uparrow ,\downarrow \right)}_{{u}_{1}},{\left(\uparrow ,\downarrow \right)}_{{u}_{2}},{\left(\uparrow ,\downarrow \right)}_{{d}_{1}}\right]\equiv {s}_{q}\left({u}_{1}{d}_{1}{u}_{2}\right)=\pm \left(3/2\right)\\ {s}_{t}\left(l\right)=\left[{\left(\uparrow ,\downarrow \right)}_{{u}_{1}},{\left(\uparrow ,\downarrow \right)}_{{u}_{2}},{\left(\uparrow ,\downarrow \right)}_{{d}_{1}}\right]\equiv {s}_{l}\left({d}_{1}\right)=\pm 3\\ {s}_{t}\left(g\right)=\left[{\left(\downarrow ,\uparrow \right)}_{{u}_{1}},{\left(\downarrow ,\uparrow \right)}_{{u}_{1}},{\left(\downarrow ,\uparrow \right)}_{{d}_{1}}\right]\equiv {s}_{g}\left({d}_{1}\right)=\mp 3\end{array}\right\}={s}_{q}=\pm \left(3/2\right)$ (26)

In the next paper, we will show the structure equation of resonance *N*^{+}(3/2) and its calculation of mass.

3.2. The Neutron

We might think to follow geometric structures of the neutral state with three quarks (*d*, *u*, *d*) (Figure 4):

Figure 4. The two base configurations of the neutron.

Note, in the b-configuration, along the diagonal (A’B’), the sum of the electric charge of quarks is zero. In previous works [9] for calculating the mass values of mesons, we needed to admit the presence of a background lattice of pairs {*d*, *d*}, which comes included in the structure equation of mesons. The same it also occurs in the light neutral nucleon, where a couple (*d*, *d*), belonging to background lattice, is being in “loan” for being incorporated in them structure: then, it needs add, into calculations of them mass, a quarks’ pair (*d*, *d*). Therefore, the structure equation could be:

$\left(n\right)=3{\kappa}_{n}\left(\left\{{\left(d\otimes \underset{\_}{d}\right)}_{A}\text{\hspace{0.17em}}\underset{\_}{\otimes}\text{\hspace{0.17em}}{\left[{d}_{1}\underset{\_}{\otimes}\text{\hspace{0.17em}}u\text{\hspace{0.17em}}\underset{\_}{\otimes}\text{\hspace{0.17em}}{d}_{2}\right]}_{B}\right\}\right)$ (27)

With configuration, see Figure 5:

Figure 5. The neutron configuration.

The *k** _{n}* is the elasticity coefficient modulating the masses of quarks (

The vertices number [*A*, *B*, *C*, *D*, *F*, *G*, *E*] is *n _{v}* = 7, while the number of total (

[(*A*(*d*_{1}, *u*), *C*(*d*_{1}, *u*); *A*(*d*_{1}, *d*_{2}), *C*(*d*_{1}, *d*_{2}); *A*(*d*_{2}, *u*), *C*(*d*_{2}, *u*); *A*(*d*_{3}, *d*_{3}), *C*(*d*_{3}, *d*_{3})] or number 8

We find: [*k** _{n}* =

Recall, the increasing of number of vertices in a geometric structure makes increase the elastic tension between the quantum oscillators of structure. Processing the structure Equation (27), we will have:

$\begin{array}{c}n=3{\kappa}_{n}\{{\left[\left(d\otimes \underset{\_}{d}\right)\right]}_{A}\text{\hspace{0.17em}}\underset{\_}{\otimes}\text{\hspace{0.17em}}\{{\left[{d}_{1}\otimes \left(u\oplus {d}_{2}\right)\right]}_{{B}_{1}}\oplus {\left[u\otimes \left({d}_{2}\oplus {d}_{1}\right)\right]}_{{B}_{2}}\oplus {\left[{d}_{2}\otimes \left({d}_{1}\oplus u\right)\right]}_{{B}_{3}}\}\}\\ =3{\kappa}_{n}\left\{{\left[\left(d\otimes \underset{\_}{d}\right)\right]}_{A}\text{\hspace{0.17em}}{\underset{\_}{\otimes}}_{}{\left[{N}_{a}\oplus {N}_{b}\oplus {N}_{c}\right]}_{B}\right\}\end{array}$ (28)

The mass calculation uses *F _{m}*-function. In this case, with the presence of the pair (

$\left[{\underset{\_}{\otimes}}_{Fm}={\left(\otimes \text{U}\oplus \right)}_{Fm}={\oplus}_{\otimes}\right]$

That is the operation (Å_{Ä}) point out interaction with interpenetration of components. Then, it is:

$\begin{array}{c}m\left(n\right)=3{\kappa}_{n}{F}_{m}\left({\left[\left(d\otimes \underset{\_}{d}\right)\right]}_{A}\underset{\_}{\otimes}{\left[{n}_{a}\oplus {n}_{b}\oplus {n}_{c}\right]}_{B}\right)\\ =3{\kappa}_{n}m({\left[\left(d\otimes \underset{\_}{d}\right)\right]}_{A}{\oplus}_{\otimes}\{{\left[{d}_{1}\otimes \left(u\oplus {d}_{2}\right)\right]}_{{B}_{1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.05em}}\oplus {\left[u\otimes \left({d}_{2}\oplus {d}_{1}\right)\right]}_{{B}_{2}}\oplus {{\left[{d}_{2}\otimes \left({d}_{1}\oplus u\right)\right]}_{{B}_{3}}\})}_{\otimes}\end{array}$

$\begin{array}{c}=3{\kappa}_{n}\{m{\left(d\otimes \underset{\_}{d}\right)}_{A}+m(\{{\left[{d}_{1}\otimes \left(u\oplus {d}_{2}\right)\right]}_{{B}_{1}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.05em}}\oplus {\left[u\otimes \left({d}_{2}\oplus {d}_{1}\right)\right]}_{{B}_{2}}\oplus {{\left[{d}_{2}\otimes \left({d}_{1}\oplus u\right)\right]}_{{B}_{3}}\})\}}_{\otimes}\\ ={\left[3{\kappa}_{n}m\left(d\right)\right]}_{A}+{\left\{3{\kappa}_{n}\left[m\left({n}_{a}\right)+m\left({n}_{b}\right)+m\left({n}_{c}\right)\right]\right\}}_{B}\\ =3{\kappa}_{n}m{\left(n\right)}_{A}+3{\kappa}_{n}m{\left(n\right)}_{B}\end{array}$ (29)

With

$\begin{array}{l}m\left({n}_{a}\right)=m{\left[\left({d}_{1}\otimes u\right)\oplus \left({d}_{1}\otimes {d}_{2}\right)\right]}_{a}=m\left(\langle {d}_{1},u\rangle \right)+m\left(\langle {d}_{1},{d}_{2}\rangle \right)\\ m\left({n}_{b}\right)=m{\left[\left(u\otimes {d}_{2}\right)\oplus \left(u\otimes {d}_{1}\right)\right]}_{b}=m\left(\langle u,{d}_{2}\rangle \right)+m\left(\langle u,{d}_{1}\rangle \right)\\ m\left({n}_{c}\right)=m{\left[\left({d}_{2}\otimes {d}_{1}\right)\oplus \left({d}_{2}\otimes u\right)\right]}_{\text{c}}\text{}=m\left(\langle {d}_{2},{d}_{1}\rangle \right)+m\left(\langle {d}_{2},u\rangle \right)\end{array}$ (30)

The state with spin *s* = 1/2 is the neutron, see Figure 5.

If the neutron spin is (1/2) and *u*-quark rotates around axis AC, then the two *d*-quarks have opposite spin to it. Recall the (*d*, *d*)* _{g}* pair is at spin

$\begin{array}{l}{s}_{n}=\left\{\begin{array}{c}{s}_{t}\left(q\right)=\left[{\left(\downarrow \downarrow \right)}_{\left({d}_{1},{d}_{2}\right)},{\left(\uparrow ,\downarrow \right)}_{{u}_{1}}\right],\left[{\left(\uparrow \uparrow \right)}_{{\left(d,\underset{\_}{d}\right)}_{\gamma}}\right]\equiv \left[{s}_{\gamma}\left(d,\underset{\_}{d}\right)+{s}_{q}\left({d}_{1},{d}_{2}\right)\right]+{s}_{q}\left({u}_{1}\right)=\pm \left(1/2\right)\\ {s}_{t}\left(l\right)=\left[{\left(\downarrow \downarrow \right)}_{\left({d}_{1},{d}_{2}\right)},{\left(\uparrow ,\downarrow \right)}_{{u}_{1}}\right],\left[{\left(\uparrow \uparrow \right)}_{{\left(d,\underset{\_}{d}\right)}_{\gamma}}\right]\equiv {s}_{l}\left({u}_{1}\right)=\pm \left(1/2\right)\\ {s}_{t}\left(g\right)=\left[{\left(\uparrow \uparrow \right)}_{\left({d}_{1},{d}_{2}\right)},{\left(\downarrow ,\uparrow \right)}_{{u}_{1}}\right],\left[{\left(\downarrow \downarrow \right)}_{{\left(d,\underset{\_}{d}\right)}_{\gamma}}\right]\equiv {s}_{g}\left({u}_{1}\right)=\mp \left(1/2\right)\end{array}\right\}\\ \text{}={s}_{q}\left({u}_{1}\right)=\pm \left(1/2\right)\end{array}$ (31)

In all two cases, the two quarks (*d*_{1}, *d*_{2}) cannot interpenetrate each other, because

They have parallel spins: then, it is (*d*_{1} Ä *d*_{2}) º 0 and, thus, [*m*(*d*_{1} Ä *d*_{2}) = 0], see Table 2 in ref [9]. Being [*m*(*d*_{1} Ä *d*_{2}) = 0] it follows, see Equation (30):

$\begin{array}{l}m\left({n}_{a}\right)=m{\left[\left({d}_{1}\otimes u\right)\right]}_{a}=m\left(\langle {d}_{1},u\rangle \right)\\ m\left({n}_{b}\right)=m{\left[\left(u\otimes {d}_{2}\right)\oplus \left(u\otimes {d}_{1}\right)\right]}_{b}=m\left(\langle u,{d}_{2}\rangle \right)+m\left(\langle u,{d}_{1}\rangle \right)\\ m\left({n}_{c}\right)=m{\left[\left({d}_{2}\otimes u\right)\right]}_{c}=m\left(\langle {d}_{2},u\rangle \right)\end{array}$ (32)

Thus:

$m{\left({n}_{a,b,c}\right)}_{B}={\left[m\left({n}_{a}\right)+m\left({n}_{b}\right)+m\left({n}_{c}\right)\right]}_{B}={\left[2m\left(\langle {d}_{1},u\rangle \right)+2m\left(\langle u,{d}_{2}\rangle \right)\right]}_{B}$ (33)

and

$\begin{array}{c}m{\left(n\right)}_{B}=\left\{\left[m\left({n}_{a}\right)+m\left({n}_{b}\right)+m\left({n}_{c}\right)\right]\right\}=\left[2m\left(\langle {d}_{1},u\rangle \right)+2m\left(\langle u,{d}_{1}\rangle \right)\right]\\ =\left[4\left(69.79\right)\right]\text{MeV}=\left(279.16\right)\text{MeV}\end{array}$ (34)

It follows

$\begin{array}{c}m\left(n\right)=3{\kappa}_{n}\left[m{\left(n\right)}_{A}+m{\left(n\right)}_{B}\right]=3{\kappa}_{n}\left[m\left(d\right)+m{\left(n\right)}_{B}\right]\\ =3{\kappa}_{n}\left[\left(86.26\right)+\left(279.16\right)\right]\text{MeV}\\ =3\left(\frac{7}{8}\right)\left(365.42\right)\text{MeV}=\left(959.23\right)\text{MeV}\end{array}$ (35)

As in the proton, also here we insert the interpenetration of a quark with itself. This last aspect comes highlighted by the matrix *A _{ij}*.We build the matrix

The matrix also admits the elements *A _{ii}*¹ 0. Then, the matrix is, see Table 3.

Table 3. The couplings with the interaction of quarks in a neutron.

Here we have omitted the subscript (Å), see Table 1 in the proton. The mass defect is:

$\begin{array}{c}\Delta {m}^{*}\left(n\right)=\left[\Delta m{\left(6\left(u\otimes d\right)\right)}_{G1}+\Delta m{\left(6\left(\underset{\_}{d}\otimes d\right)\right)}_{G2}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left[\Delta m{\left(6\left({d}_{i}\otimes {d}_{j}\right)\right)}_{B1}+2\Delta m{\left(\left(u\otimes \underset{\_}{d}\right)\right)}_{B2}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+[2\Delta m\left({\left({d}_{i}\otimes {d}_{i}\right)}_{1,2}\right)+\Delta m\left({\left({d}_{i}\otimes {d}_{i}\right)}_{\gamma}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\Delta m\left({\left({\underset{\_}{d}}_{i}\otimes {\underset{\_}{d}}_{i}\right)}_{\gamma}\right)+{\Delta m\left(\left({u}_{i}\otimes {u}_{i}\right)\right)]}_{R}\end{array}$ (36)

For the spin (*s* = 1/2), see the table of the neutron spin, we can have the term *R’*:

$\begin{array}{c}{R}^{\prime}\left(n\right)=[2\Delta m\left({\left({d}_{i}\otimes {d}_{i}\right)}_{1,2}\right)+\Delta m\left({\left({d}_{i}\otimes {d}_{i}\right)}_{\gamma}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\Delta m\left({\left({\underset{\_}{d}}_{i}\otimes {\underset{\_}{d}}_{i}\right)}_{\gamma}\right)+{\Delta m\left(\left({u}_{i}\otimes {u}_{i}\right)\right)]}_{{R}^{\prime}}\\ =[{{r}^{\prime}}_{1}\Delta {m}^{\prime}{\left({d}_{(\downarrow )}\otimes {d}_{(\downarrow )}\right)}_{{d}_{1}}+{{r}^{\prime}}_{2}\Delta {m}^{\prime}{\left({d}_{(\downarrow )}\otimes {d}_{(\downarrow )}\right)}_{{d}_{2}}+{{r}^{\prime}}_{3}\Delta {m}^{\prime}{\left({d}_{(\uparrow )}\otimes {d}_{(\uparrow )}\right)}_{\gamma}\end{array}$

$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{{r}^{\prime}}_{4}\Delta {m}^{\prime}{\left({\underset{\_}{d}}_{(\uparrow )}\otimes {\underset{\_}{d}}_{(\uparrow )}\right)}_{\gamma}+{{r}^{\prime}}_{5}\Delta {m}^{\prime}\left({u}_{(\downarrow )}\otimes {u}_{(\downarrow )}\right)]\\ =\left[\left(-{2}_{\left({d}_{1},{d}_{2}\right)}+{1}_{\gamma}-{1}_{\gamma}-{1}_{u}\right){\left(1.15\right)}_{R}\right]\text{MeV}=\left[\left(-3\right){\left(1.15\right)}_{R}\right]\text{MeV}\end{array}$ (37)

Then, it is

$\begin{array}{c}\Delta {m}^{*}\left(n\right)=\left[\Delta m{\left(6\left(u\otimes d\right)\right)}_{G1}+\Delta m{\left(6\left(\underset{\_}{d}\otimes d\right)\right)}_{G2}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left[\Delta m{\left(6\left({d}_{i}\otimes {d}_{j}\right)\right)}_{B1}+2\Delta m{\left(\left(u\otimes \underset{\_}{d}\right)\right)}_{B2}\right]+R\left(n\right)\\ =\left[\text{6}{\left(\text{4}\text{.59}\right)}_{G1}+\text{6}{\left(\text{5}\text{.68}\right)}_{G2}\right]\text{MeV}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left[6{\left(5.68\right)}_{B1}+2{\left(\text{4}\text{.59}\right)}_{B2}\right]\text{MeV}+\left[-{\left(3.45\right)}_{R}\right]\text{MeV}\\ =\left(14.91\right)\text{MeV}\end{array}$ (38)

The pair (*d*, *d*) in the structure of neutron shields the electromagnetic interactions between quarks, therefore it weakens the mass defect. The presence of the pair *d*-quark along diagonal, see Figure 5, splits all mass defects, see ref. [9], then it follows:

$\Delta {m}^{**}\left(n\right)={\left[\Delta {m}^{*}\left(n\right)\right]}_{\left(d,\underset{\_}{d}\right)}=\left(\frac{1}{2}\right)\left[\Delta {m}^{*}\left(n\right)\right]=\left(7.46\right)\text{MeV}$ (39)

Like the proton, in the neutron being explicit the elasticity parameter *k** _{n}*, it is, see Equation (24):

$\Delta m\left(n\right)=3{\kappa}_{n}{\left[\Delta {m}^{**}\left(n\right)\right]}_{\left(d,\underset{\_}{d}\right)}=3\left(\frac{7}{8}\right)\left(7.46\right)\text{MeV}=\left(19.58\right)\text{MeV}$ (40)

Thus, it is:

${m}_{tot}\left(n\right)=m\left(n\right)-\Delta {m}_{\gamma}\left(n\right)=\left\{\left(959.23\right)-\left(19.58\right)\right\}\text{MeV}=\left[\left(939.65\right)\right]\text{MeV}/{c}^{2}$ (41)

Next to that experimental m(n) = (939.57) MeV.

4. Conclusions

In this paper, some aspects introduce a new paradigm in the phenomenology of the interactions. Describing the particles as a set of coupling quantum oscillators allows us of understanding more in-depth the Standard Model (SM) and its descriptions about the interactions, decay, spin and dynamics of particles. Besides, the AGM adds the possibilities of calculating [8] [9] the masses of hadronic particles without turning to the QCD. Not only but AGM extends the same QM, because it adds original quantum aspects as:

· The interpenetration between quarks;

· The connection between interpenetration and spin;

· The interpenetration of a quark with itself;

· The not-locality of a quark in time or the time as a quantum operator.

AGM allows explaining some aspects of the phenomenology of hadronic interactions which in contemporary literature do not have a conclusive understanding yet. For instance:

1) The spin question of the proton (mesons) has a clarifying explanation in Equations (3) and (21).

2) The moment questions of protons (parton model) [21], where the fraction of moment carried by the quarks is: (*P*_{quark}/*P _{p}*) » (1/2). This result is a consequence of the presence of gluons in the proton. The cross-sections confirm the fact that the gluons bring a missing moment in hadron collisions. AGM explains this result, considering that along the side BC in Figure 3, the quantum oscillators of the junction of vertices B and C are just gluons. The side BC is along the propagation axis of proton-structure; thus, the moment along this axis (

3) Some researchers, see ref. [5], have shown, with a calculation based on QCD, that quarks and gluons determine four separate contributions to the proton’s mass: (a) the quark condensate (∼9%), (b) the quark energy (~32%), (c) the gluonic field strength energy (∼37%), and (d) the anomalous gluonic contribution (~23%). In AGM, the orbital rotations around the diagonal axis of quarks with their gluons of binding and junction are in accordance to this partition of mass energy (with (b), (c) and (d)). The strange quark condensate (9%) could be, instead, according to the interpenetration of quarks with themselves (or auto-interpenetration); that recalls the self-force of classic electrodynamics.

4) The problematic phenomenology of the hadron jets [22] that reveal the close connection between quarks and gluons. In collisions at high energies the existence of gluons is in the presence of 3 coplanar hadronic jets in the system of the center of mass of the annihilation* *

$\left({e}^{+}{e}^{-}\right):{e}^{+}+{e}^{-}\to q+q+g\to {X}_{hadrons}$ (*)

Recall both Quark and gluons, are never visible in the detector because they can only exist confined within the hadrons. In the theory of AGM is possible to interpret the jets and their hadronization as consequence of presence of lattice {*d*,*d*} U {*u*,*u*}. In fact, we write the reaction (*) as:

$\left({e}^{+}+{e}^{-}\right)\to \left[\left(\gamma +\gamma \right)\underset{\_}{\otimes}\left\{d,\underset{\_}{d}\right\}\otimes \left\{u,\underset{\_}{u}\right\}\right]\to q+q+g\to {X}_{hadrons}$

while, in annihilation process with proton-antiproton:

$\left({p}^{+}\underset{\_}{\otimes}{p}^{-}\right)\underset{\_}{\otimes}\left[\left\{d,\underset{\_}{d}\right\}\otimes \left\{u,\underset{\_}{u}\right\}\right]\to {\left[{N}^{+}+{N}^{-}\right]}_{\left\{d,\underset{\_}{d}\right\}}+{\left[{X}_{hadrons}\right]}_{\left\{d,\underset{\_}{d}\right\}}$ _{ }

Which we show in Figure 6.

By structure equations and geometric aspect of hadron, it is possible to find the decay probabilities.

The interpenetration of a quark with itself Δ*m*(*q _{i}* Ä

Figure 6. Production of three jets.

not-local time and, thus, time operator or the time as an observable.

Further, AGM allows explaining some problematic aspects in the hadron production accompanied by the production of the *W*, *Z* bosons:

$pp\to q{q}^{-}+X+Z\to {l}^{+}{l}^{-}+X$

The appearance of vector bosons in hadronic jets or the transformations in beta decay induce us to introduce a hypothesis of structure in the boson particles. For example, one of the issues not yet fully clarified is the transformation of leptonic particles into quarks as happens in the production of hadronic jets:

$\left({e}^{-}+{e}^{+}\right)\to \gamma +\gamma \to q+q\to hadrons$

How is it possible for photons to create quarks? Recall that photons are the intermediaries of the EM force. Then, we ask us because they can produce particles like quarks, provided with an “additional” agent charge, which is the color charge.

A comprehensive answer could then be to assign a structure also to a “real” elementary particle (that is not composed of sub-particles) and that it is possible to transform a structure into another only if the particles are geometric structures. Thus, there would be a mechanism of topological transformations on geometrical structures that would thus be transformed one into each other. In this way, we introduce a new paradigm in the phenomenology of strong interactions that open new perspectives for resolving the various problems of these interactions.

Note speaking of structures of quantum oscillators into particles implies (see Figure 5) a structure of quantum oscillator. On this basis, we can discuss “substructure” of quarks [19] [20] with the physical awareness that this sub-structure is realizable only through “particular” quantum oscillators. The vertex-oscillators for connecting to the other vertices through junction oscillators have to be able to have in turn a structure of “hooks”: this induces us to talk about a “sub-structure” into quantum oscillator which is highlighted only into quantum oscillator coupled to other oscillators. Indication of a “composite structure” in quantum oscillator derives from its wave function. Therefore, we can describe the quantum oscillator with two sub-units of oscillation or “sub-oscillators”. Not only but more components in an oscillator encourages us to believe that the energy of the “quanta” should be distributed between these oscillating components. The presence of more components in an oscillator causes the splitting of its quanta of energy into two, and more, sub-oscillators: this introduces the idea of half-quanta (“semi-quanta”) or individually half-quantum (“semi-quantum”). A quantum oscillator with a sub-structure constituted by sub-unit of oscillation, or “sub-oscillators”, and “semi-quanta” is an oscillator of type “IQuO” [19] [20] [7]. To treat the particles as IQuO structures can explain the origin of some fundamental physics greatness as the electric charge, spin, isospin and color charge [23]. So, the IQuO idea constitutes a new paradigm in physics which allow us to describe with depth the physical phenomenon of particles and to open new descriptive scenarios of interactions between particles.

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