Section 2 Quark Topics

#### **Chapter 2**

## Quarks Mixing in Chiral Symmetries

*Zbigniew Piotr Szadkowski*

#### **Abstract**

We discuss a subject of the quarks mixing in *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> and *SU*<sup>6</sup> ∗ *SU*<sup>6</sup> symmetries trying to calculate the quarks mixing angles and the complex phase responsible for the CP non-conservation on the basis of the Gell-Mann Oakes Renner model. Assuming symmetry breaking in a limit of exact sub-symmetries for simultaneous quarks rotations in both electric charge sub-spaces we can estimate all mention above parameters. A perfect agreement of the experimental value of the Cabibbo angles with a sum of simultaneous quarks mixing angles in doublets (u,c) and (d,s) in the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry suggests that a quarks mixing is realized in a maximal allowed range. The same assumption used for the *SU*<sup>6</sup> ∗ *SU*<sup>6</sup> and a simultaneous maximal allowed quarks mixing in both electric charge triplets (u,c,t) and (d,s,b) gives a perfect agreement with the experimental value of the Cabibbo angle and estimation on the angles Θ<sup>2</sup> and Θ<sup>3</sup> as well as a bond for the complex phase *δ*.

**Keywords:** quarks mixing, chiral symmetries, Cabibbo angle, Kobayashi-Maskawa mixing matrix, symmetry breaking

#### **1. Introduction**

#### **1.1 Quarks mixing in chiral** *SUn* **∗** *SUn* **broken symmetry in the limit of exact** *SUk* **∗** *SUk* **symmetry**

The hierarchy of chiral symmetry breaking [1–3] has been investigated since seventies of the previous century [4–8]. The symmetry breaking and mixing of quarks are connected with the rotation of quark currents and Hamiltonian densities. The determination of the rotation angle becomes an important problem. For the first time the procedure of chiral symmetry breaking, based on the Gell-Mann, Oakes, Renner (GMOR) model [9] has been used in *SU*<sup>3</sup> ∗ *SU*<sup>3</sup> symmetry in the limit of exact *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> symmetry [4] to determine the value of the Cabibbo angle [10]. The transformation of rotation is connected with the seventh generator of the *SU*<sup>3</sup> group. After the charmed particles have been discovered the *SU*<sup>3</sup> ∗ *SU*<sup>3</sup> symmetry is no longer adequate to describe the strong interactions. The *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry introduced earlier [11] to explain the behavior of charged and neutral currents becomes quite satisfactory model describing the hadron world. The problem of determining the Cabibbo angle in *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry has arisen. It is considered in [5–6] and the method of calculating the Cabibbo angle in *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry is described in [7]. It is known that the formula describing the rotation angle is not changed if the symmetry is extended. This is not unexpected because the Cabibbo angle is connected with the mixing of the d and s quarks and the rotation is performed around the seventh axis in *SU*<sup>3</sup> subspace too.

The problem of chiral *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry breaking in the limit of exact *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> symmetry is considered in [6]. Symmetry breaking is connected with the transformation of rotation around the tenth axis in *SU*<sup>4</sup> space. The rotation angle is determined in [7].

The other variant of the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry breaking in the limit of the exact *SU*<sup>3</sup> ∗ *SU*<sup>3</sup> symmetry is described in [8]. It is connected with the rotation around the fourteenth axis in *SU*<sup>4</sup> space. In this paper we introduce the general method of rotation angle description in the broken *SUn* ∗ *SUn* symmetry. The chiral *SUn* ∗ *SUn* symmetry is broken according to the GMOR model. In the first step we introduce the Hamiltonian density breaking *SUn* ∗ *SUn* symmetry but invariant under *SUk* ∗ *SUk* symmetry. In the second step we introduce quark mixing and the resulting exact symmetry is *SUk*�<sup>1</sup> ∗ *SUk*�<sup>1</sup> The particular investigation of cases like the above is not necessary.

The generalized GMOR model is used. It is assumed that by enlargement to a higher symmetry the new quantum numbers are the charges (as for example: electric charge, strangeness, charm but not isospin). Then the *SUn* ∗ *SUn* symmetry breaking Hamiltonian density can be written as a linear combination of diagonal operators *u<sup>i</sup>* .

$$H\_E = \sum\_{j=1}^n c\_{j^2 - 1} u^{j^2 - 1} \tag{1}$$

where the scalar densities *<sup>u</sup><sup>i</sup>* <sup>¼</sup> *<sup>q</sup>λ<sup>i</sup> <sup>q</sup>* and pseudo-scalar densities *vi* <sup>¼</sup> *iqλ<sup>i</sup> γ*5*q* satisfy the equal-time commutation rules

$$\left[\boldsymbol{Q}^{i},\boldsymbol{u}^{j}\right] = \dot{\boldsymbol{\mathcal{Y}}}\_{ijk}\boldsymbol{u}^{k} \qquad \left[\boldsymbol{Q}^{i},\boldsymbol{\nu}^{j}\right] = \dot{\boldsymbol{\mathcal{Y}}}\_{ijk}\boldsymbol{\nu}^{k} \tag{2}$$

$$\left[\overline{\boldsymbol{Q}}^{i},\boldsymbol{u}^{j}\right] = \dot{\boldsymbol{\mathcal{U}}}\_{ijk}\boldsymbol{\nu}^{k} \qquad \left[\overline{\boldsymbol{Q}}^{i},\boldsymbol{\nu}^{j}\right] = -\dot{\boldsymbol{\mathcal{U}}}\_{ijk}\boldsymbol{u}^{k}$$

where *fijk* are the structure constants, *dijk* - symmetric generators of the *SUn* ∗ *SUn* group. If the *SUk* ∗ *SUk* symmetry is exact then

$$
\partial^{\mu}V^{i}\_{\mu} = \partial^{\mu}A^{i}\_{\mu} = \mathbf{0} \qquad \left(i = \mathbf{1}, \mathbf{2}, \dots, k^{2} - \mathbf{1}\right) \tag{3}
$$

In the GMOR model the divergences of currents can be calculated as follows

$$
\partial^{\mu}V^{i}\_{\mu} = i \left[ H\_E, \mathbb{Q}^{i} \right] \qquad \partial^{\mu}A^{i}\_{\mu} = i \left[ H\_E, \overline{\mathbb{Q}}^{i} \right] \tag{4}
$$

We require that the *SUk* ∗ *SUk* symmetry be exact, then the following constraints are obeyed

$$c\_{j^2 - 1} = \mathbf{0} \qquad \qquad (j = 2, \dots, k) \tag{5}$$

$$\sqrt{\frac{2}{n}} c\_0 + \sum\_{j = k + 1}^n \sqrt{\frac{2}{j(j+1)}} c\_{j^2 - 1} = \mathbf{0}$$

The symmetry breaking Hamiltonian density can be written as follows

$$H\_E = c\_0 \left( u^0 - \sqrt{n - 1} u^{n^2 - 1} \right) + \sum\_{j = k + 1}^{n - 1} c\_{j^2 - 1} \left( u^{j^2 - 1} - \sqrt{\frac{n(n - 1)}{j(j - 1)}} u^{n^2 - 1} \right) \tag{6}$$

Using the standard representation of *λ* matrices one obtains

$$
\mu^0 = \sqrt{\frac{2}{n}} \sum\_{j=1}^n \overline{q}\_j q\_j \tag{7}
$$

$$\mathfrak{w}^{j^2 - 1} = \sqrt{\frac{2}{j(j - 1)}} \left( \sum\_{l=1}^{j-1} \overline{q}\_l q\_l - (j - 1)\overline{q}\_j q\_j \right) \tag{8}$$

$$u^0 - \sqrt{n-1} \ u^{n^2-1} = \sqrt{2n} \ \overline{q}\_n q\_n \tag{9}$$

$$u^{j^2-1} - \sqrt{\frac{n(n-1)}{j(j-1)}}u^{n^2-1} = \sqrt{\frac{2}{j(j-1)}} \left( (n-1) \begin{array}{c} \overline{q}\_n q\_n - j \ \overline{q}\_j q\_j - \sum\_{l=j+1}^{n-1} \overline{q}\_l q\_l \end{array} \right) \tag{10}$$

Let us note that the term *qkqk* does not exist in Eq. 11.

$$\begin{aligned} H\_E &= \left(\sqrt{2n}c\_0 + (n-1)\sum\_{j=k+1}^{n-1} c\_{j^2-1}\sqrt{\frac{2}{j(j-1)}}\overline{q}\_n q\_n\right) \\ &- \sum\_{j=k+1}^{n-1} c\_{j^2-1}\sqrt{\frac{2}{j(j-1)}}\left(j\ \overline{q}\_j q\_j + \sum\_{l=j+1}^{n-1} \overline{q}\_l q\_l\right) \end{aligned} \tag{11}$$

The chiral *SUn* ∗ *SUn* symmetry with the exact *SUk* ∗ *SUk* sub-symmetry is broken by the rotation of the *SUk* ∗ *SUk* invariant Hamiltonian density around the axis with the index *<sup>m</sup>* <sup>¼</sup> ð Þ *<sup>n</sup>* � <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>2</sup>*<sup>k</sup>* � 1.

$$H\_{\rm SB} = e^{-2iaQ'''} \ H\_E \ e^{2iaQ'''} \tag{12}$$

Only the quarks *qk* and *qn* are mixed. The *SUk* ∗ *SUk* symmetry is no longer exact. Only the term *qnqn* is rotated under transformation (12), because there is no *qkqk* term in the Hamiltonian density (11).

$$e^{-2iaQ''}H\_Ee^{2iaQ''} = \overline{q}\_n q\_n - \left(\overline{q}\_n q\_n - \overline{q}\_k q\_k\right) \quad \sin^2 a - \frac{1}{2} \left(\overline{q}\_k q\_n + \overline{q}\_n q\_k\right) \quad \sin 2a \tag{13}$$

The above consideration is limited to processes not having the change of the quantum number N connected with the *SUn* symmetry. So in the broken Hamiltonian density *HSB*ð Þ <sup>Δ</sup>*N*¼<sup>0</sup> the terms *qnqk* and *qkqn* do not appear. The broken Hamiltonian density is a linear combination of the diagonal operators *u<sup>i</sup>* only.

$$H\_{\rm SB(\Delta N=0)} = H\_E + A \sum\_{j=k}^{n-1} \left( \sqrt{\frac{j+1}{2j}} u^{(j+1)^2 - 1} - \sqrt{\frac{j-1}{2j}} u^{j^2 - 1} \right) \sin^2 a \tag{14}$$

where

$$A = \sqrt{2n} \ c\_0 + (n - 1) \sum\_{j=k+1}^{n-1} c\_{j^2 - 1} \sqrt{\frac{2}{j(j - 1)}} \tag{15}$$

In more detail Eq. (14) is given as follows:

$$H\_{\text{SB}(\Delta N=0)} = c\_0 \ u^0 - A \sqrt{\frac{k-1}{2k}} \sin^2 \alpha \ u^{k^2-1} + \sum\_{j=k+1}^{n-1} \quad \mu^{j-1} \left( c\_{j^2-1} + A \sin^2 \alpha \ \sqrt{\frac{1}{2j(j-1)}} \right) + \dotsb \tag{16}$$

$$1 - \mu^{n^2-1} \left( c\_0 \sqrt{n-1} \ + \sum\_{j=k+1}^{n-1} c\_{j^2-1} \ \sqrt{\frac{n(n-1)}{j(j-1)}} \right) - A \ \mu^{n^2-1} \ \sqrt{\frac{n}{2(n-1)}} \ \sin^2 \alpha \tag{17}$$

If the *SUk* ∗ *SUk* symmetry is exact then the pseudo-scalar mesons corresponding to the indices *<sup>j</sup>* <sup>¼</sup> 1, … , *<sup>k</sup>*<sup>2</sup> � <sup>1</sup> � � are massless [9]. After the *SUk* <sup>∗</sup> *SUk* has been broken, the *SUk*�<sup>1</sup> <sup>∗</sup> *SUk*�<sup>1</sup> symmetry is still exact, because the operator *<sup>Q</sup><sup>m</sup>* does not mix the quarks *<sup>q</sup>*1, … , *qk*�<sup>1</sup> neither with themselves nor with other quarks. The mesons corresponding to the indices *<sup>j</sup>* <sup>¼</sup> 1, … ,ð Þ *<sup>k</sup>* � <sup>1</sup> <sup>2</sup> � <sup>1</sup> � � after symmetry breaking are still massless, while the mesons corresponding to the indices *<sup>j</sup>* <sup>¼</sup> ð Þ *<sup>k</sup>* � <sup>1</sup> <sup>2</sup> , … *<sup>k</sup>*<sup>2</sup> � <sup>1</sup> � � belong to the massive multiplet ð Þ*<sup>k</sup>* <sup>1</sup> The masses of mesons are determined in the GMOR model. Before the *SUn* ∗ *SUn* symmetry is broken the masses are described by the coefficients *c*0, … ,*cn*<sup>2</sup>�<sup>1</sup> from Eq. (1). After the symmetry has been broken the new factors *c*<sup>0</sup> 0, … ,*c*<sup>0</sup> *<sup>n</sup>*2�<sup>1</sup> are obtained as the coefficients standing by the operators *ui* in the broken Hamiltonian density (16) [7].

$$c\_0' = c\_0 \tag{17}$$

$$c\_{k^2 - 1}' = A \sqrt{\frac{k - 1}{2k}} \sin^2 a$$

$$c\_{j^2 - 1}' = c\_{j^2 - 1} + A \sin^2 a \sqrt{\frac{1}{2j(j - 1)}} \quad (k < j < n)$$

$$c\_{n^2 - 1}' = -c\_0 \sqrt{n - 1} - \sum\_{j = k + 1}^{n - 1} c\_{j^2 - 1} \sqrt{\frac{n(n - 1)}{j(j - 1)}} \quad + A \sqrt{\frac{n}{2(n - 1)}} \sin^2 a$$

The masses of the mesons are determined as follows [12].

$$m\_d^2 f\_a^2 \delta\_{ab} = \sqrt{\frac{2}{n}} \left( c\_0 \, d\_{0ab} + \sum\_{j=k+1}^n c'\_{j^2-1} d\_{\{j^2-1\}ab} \right) < u^0 >\_0 \tag{18}$$

The relation between the indices a, b, j and meson states is described, for example, for the *SU*<sup>4</sup> <sup>∗</sup> *SU*<sup>4</sup> symmetry in [12, 13]. For *<sup>a</sup>* <sup>¼</sup> *<sup>b</sup>* <sup>¼</sup> *<sup>k</sup>*<sup>2</sup> � *<sup>l</sup>*, the mass of the (k) meson is given as follows

$$m\_{(k)}^2 f\_{(k)}^2 = \sqrt{\frac{2}{n}} \left( \sqrt{\frac{2}{n}} c\_0 \, \, \, + \, \sum\_{j=k+1}^n d\_{(j^2-1)ab} c'\_{j^2-1} \right) < u^0 >\_0 = \frac{1}{k} \, \sqrt{\frac{2}{n}} \, A \, \sin^2 a \, \, < u^0>\_0 \,\,\tag{19}$$

For *<sup>a</sup>* <sup>¼</sup> *<sup>b</sup>* <sup>¼</sup> *<sup>m</sup>* <sup>¼</sup> ð Þ *<sup>n</sup>* � <sup>1</sup> <sup>2</sup> <sup>þ</sup> <sup>2</sup>*<sup>k</sup>* � 1, the mass of the (n) meson is given by

$$m\_{(n)}^2 f\_{(n)}^2 = \sqrt{\frac{2}{n}} \left( \sqrt{\frac{2}{n}} c\_0 \right. \\ \left. + \sum\_{j=k+1}^n d\_{\{j^2-1\}mn} \left. c\_{j^2-1}' \right. \right) < u^0 >\_0 \tag{20}$$

Because

$$d\_{\left(j^2 - 1\right)mm} = \sqrt{\frac{1}{2j(j-1)}} \quad (j < n) \tag{21}$$

$$d\_{(n^2-1)mm} = \frac{2-n}{\sqrt{2n(n-1)}}\tag{22}$$

$$m\_{(n)}^2 f\_{(n)}^2 = \sqrt{\frac{1}{2n}} \left( \mathbf{1} - \left( \mathbf{1} - \frac{\mathbf{1}}{k} \right) \sin^2 a \right) < u^0 >\_0 \tag{23}$$

In formulas (19) and (23) to determine the masses of (k) and (n) mesons one has (n-k + 3) unknown quantities with which to deal (<*u*<sup>0</sup> >0, *c*0, *<sup>c</sup>*ð Þ *<sup>k</sup>*þ<sup>1</sup> <sup>2</sup> �1, … ,*cn*<sup>2</sup>�1, sin *<sup>α</sup>*Þ. Nevertheless the angle a is determined by the masses and decay constants of two pseudo-scalar mesons (k) and (n) only.

$$\sin^2 a = \frac{k \ m\_{(k)}^2 f\_{(k)}^2}{2 \ m\_{(n)}^2 f\_{(n)}^2 + (k - 1) \ m\_{(k)}^2 f\_{(k)}^2} \tag{24}$$

All the cases of symmetry breaking considered in [4–8] can be described by formula (24). Let us give simple examples: a) for k = 2, n = 3 a is the original Cabibbo angle Θ associated with rotation around the seventh axis in *SU*<sup>3</sup> subspace [4–7].

$$\sin^2\Theta = \frac{2}{2} \frac{m\_\pi^2 f\_\pi^2}{m\_K^2 f\_K^2 + |m\_\pi^2 f\_\pi^2|} \tag{25}$$

b) for k = 2, n = 4 and rotation around the tenth axis [6–7] one obtains

$$\sin^2 a = \frac{2 \ m\_\pi^2 f\_\pi^2}{2 \ m\_\text{D}^2 f\_\text{D}^2 + \ m\_\pi^2 f\_\text{-n}^2} \tag{26}$$

c) for k = 3, n = 4 and rotation around the fourteenth axis [7–8] one obtains

$$\sin^2 \alpha = \frac{3 \ m\_K^2 f\_K^2}{2\left(m\_D^2 f\_D^2 + m\_K^2 f\_K^2\right)}\tag{27}$$

In general the determination of the rotation angle (24) in *SUn* ∗ *SUn* symmetry is possible only if the new quantum numbers introduced by a transition to the higher symmetry are scalars of the charge type (additiv). So the Hamiltonian density (1) can be constructed as a linear combination of the diagonal operators *ui* ; only *i* ¼ *j* <sup>2</sup> � 1, *<sup>j</sup>* <sup>¼</sup> 1, … , *<sup>n</sup>* � �. The method of determining the rotation angle, discussing and interpreting the symmetry breaking is described in more detail in [7] on *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry as an example.

#### **2. Quarks mixing and the Cabibbo angle in the** *SU***<sup>4</sup> ∗** *SU***<sup>4</sup> broken symmetry**

It is known that the Cabibbo angle has been introduced into *SU*<sup>3</sup> symmetry to explain the suppression of processes in which strangeness is not conserved [4]. The Cabibbo angle is connected with the mixing of d and s quarks for weak interactions of hadrons. Its value, calculated by Oakes, does not contradict the experimental data. Before the charmed particles were discovered Glashow, Iliopoulos and Maiani [11] have suggested the generalization of a strong interaction symmetry to *SU*<sup>4</sup> [6]. The charged weak current is then given as follows

$$J\_{\mu} = \overline{q} \; \gamma\_{\mu} \; (\mathbf{1} - \gamma\_{5}) \; \text{A} \; q \tag{28}$$

where

$$A = \begin{pmatrix} 0 & 0 & \cos\Theta & \sin\Theta \\ 0 & 0 & -\sin\Theta & \cos\Theta \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \tag{29}$$

The current (28) can be expressed in another form

$$J\_{\mu} = (\overline{u}, \overline{c}) \; \gamma\_{\mu} \; (1 - \gamma\_{5}) \begin{pmatrix} \cos \Theta & \sin \Theta \\ -\sin \Theta & \cos \Theta \end{pmatrix} \begin{pmatrix} d \\ s \end{pmatrix} \tag{30}$$

so, quark mixing is described by an orthogonal matrix. On the grounds of Eq. (30) we cannot come to a conclusion about quarks in which the doublets are mixed. If the matrix A is generalized to the following form

$$A = \begin{pmatrix} 0 & 0 & \cos\Theta & \sin\Theta \\ 0 & 0 & -\sin\Theta & \cos\Theta \\ \cos\phi & \sin\phi & 0 & 0 \\ -\sin\phi & \cos\phi & 0 & 0 \end{pmatrix} \tag{31}$$

the quarks in the doublets (u, c) and (d, s) are mixed independently. The zeros in Eq. (31) are associated with the fact that the neutral currents which change the strangeness and/or charm are not observed. So, the current (28) can be given in the following form

$$J\_{\mu} = (\overline{u}, \overline{c}) \; \begin{pmatrix} \chi & (1 - \chi\_{5}) \begin{pmatrix} \cos\left(\Theta + \phi\right) & \sin\left(\Theta + \phi\right) \\ -\sin\left(\Theta + \phi\right) & \cos\left(\Theta + \phi\right) \end{pmatrix} \begin{pmatrix} d \\ s \end{pmatrix} \tag{32}$$

If the currents only are taken into consideration we cannot solve the problem if the quarks are mixed in one or both doublets. This is not unexpected because the currents are built as a bi-linear combination of quark states and the angles Θ and *ϕ*, can always be substituted the effective angle ð Þ Θ þ *ϕ* . To solve the problem the Gell-Mann, Oakes, Renner (GMOR) model [9] will be used.

The charged weak current in *SU*<sup>3</sup> symmetry can be written as follows

$$J\_{\mu}(\Theta) = \cos \Theta \left( J\_{\mu}^{1} + i \left. J\_{\mu}^{2} \right| + \sin \Theta \left( J\_{\mu}^{4} + i \left. J\_{\mu}^{5} \right| \right) \right. \qquad \Theta-\text{Cabibbo angle} \tag{33}$$

$$J\_{\mu} = \overline{q} \left. \gamma\_{\mu} \left( 1 - \gamma\_{5} \right) \lambda^{k} \right. \qquad q = \begin{pmatrix} u \\ d \\ s \end{pmatrix} \tag{34}$$

*c*

The current (33) can be obtained from the isospin component of the current *J* 1 *<sup>μ</sup>* <sup>þ</sup> *i J*<sup>2</sup> *μ* � � by rotation through an angle 2<sup>Θ</sup> about the seventh axis in *SU*<sup>3</sup> space according to

$$J\_{\mu}(\Theta) = e^{-2i\Theta F^{\top}} \begin{pmatrix} J\_{\mu}^{1} + i \ J\_{\mu}^{2} \end{pmatrix} e^{2i\Theta F^{\top}} \tag{35}$$

where

$$F^{k} = \int d^{3} \mathfrak{x} \,\, q^{+}(\mathfrak{x}) \,\, \frac{\lambda^{k}}{2} \,\, q(\mathfrak{x}) \,\, \tag{36}$$

The charged weak current in *SU*<sup>4</sup> symmetry (30) can be expressed in the following form

$$J\_{\mu}(\Theta) = \begin{pmatrix} \cos \Theta & \left( J\_{\mu}^{1} + i \left. J\_{\mu}^{2} \right| + \sin \Theta \ \left( J\_{\mu}^{4} + i \left. J\_{\mu}^{5} \right| - \sin \Theta \ \left( J\_{\mu}^{11} - i \left. J\_{\mu}^{12} \right) \right. \\ + \cos \Theta \ \left( J\_{\mu}^{13} - i \left. J\_{\mu}^{14} \right) \end{pmatrix} \right. \end{pmatrix} \tag{37}$$

The current (37) can be obtained by rotation of the components Δ*S* ¼ Δ*C* through an angle 2Θ about the seventh axis in *SU*<sup>4</sup> space

$$J\_{\mu}(\Theta) = e^{-2i\Theta F^{\overline{\jmath}}} \left( J\_{\mu}^{1} + i \left. J\_{\mu}^{2} + J\_{\mu}^{13} - i \left. J\_{\mu}^{14} \right| \right. \right) e^{2i\Theta F^{\overline{\jmath}}} \tag{38}$$

The transformation (38) changes the strangeness but not the charm because

$$\left[\boldsymbol{F}^{\mathsf{T}},\boldsymbol{q}\_{1}\right] = \left[\boldsymbol{F}^{\mathsf{T}},\boldsymbol{q}\_{4}\right] = \mathbf{0} \tag{39}$$

The transformation (38) is connected with the mixing of d and s quarks (as in the case of *SU*<sup>3</sup> symmetry). In the *SU*<sup>4</sup> symmetry the mixing in electric charge subspace +2/3 can be taken into consideration. This is not possible in the *SU*<sup>3</sup> symmetry where only one state with the +2/3 charge exists. The possibility of expressing the current (37) by the transformation which changes charm but not strangeness should exist. The transformation has been described by Ebrahim in [6].

$$J\_{\mu}(\phi) = e^{-2i\phi F^{0}} \left( J\_{\mu}^{1} + i \left. J\_{\mu}^{2} + J\_{\mu}^{13} - i \left. J\_{\mu}^{14} \right| \right. \right) e^{2i\phi F^{0}} \tag{40}$$

$$\left[F^{10}, q\_2\right] = \left[F^{10}, q\_3\right] = \mathbf{0} \tag{41}$$

The transformation (40) is connected with the mixing of u and c quarks. The fact that there exist two transformations giving the current (37) but connected with different generators of the *SU*<sup>4</sup> group changing strangeness or charm respectively suggests that independent mixing in both doublets is possible. It is known that the Cabibbo angle is connected with strangeness non-conservation in weak interactions. The formula describing the value of the Cabibbo angle has been obtained by Oakes [4] in the procedure of symmetry breaking. Namely the *SU*<sup>3</sup> ∗ *SU*<sup>3</sup> symmetry in the limit of the exact *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> symmetry is broken. The *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> subsymmetry is no longer exact. The symmetry is broken by the rotation of the *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> invariant Hamiltonian density through angle 2Θ about the seventh axis. Then the pion becomes massive. The symmetry breaking is connected with the mixing of d and s quarks. The rotation angle Θ, as a measure of symmetry violation, is a function of the mass and the decay constant of the pion and of the mass and the decay constant of the kaon as well (it is connected with the mixing of the strange quark and the strangeness non-conservation). If the breaking of the chiral *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> symmetry, the mass of the pion, the Cabibbo angle as well as a strangeness and charm non-conservation have a common origin then it seems that as a result of *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry breaking in the limit of the exact *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> sub-symmetry by the rotation of the *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> invariant Hamiltonian density through an angle 2*ϕ* about the tenth axis the angle *ϕ* connected with the mixing of u and c quarks as a measure of a symmetry violation should be a function of the mass and decay constant of the pion (breaking of the *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> symmetry) and a function of the mass and decay constant of a charmed meson (charm non-conservation). The cases of the separate and then simultaneously mixing of quarks in the sub-spaces of electric charge will be considered below.

If the electromagnetic mass splitting of u-d quarks is neglected the Hamiltonian density breaking the chiral *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry is given in the form

$$H = c\_0 u^0 + c\_8 u^8 + c\_{15} u^{15} \tag{42}$$

where *c*0,*c*8,*c*<sup>15</sup> are constants, *ua* (a = 0, 1, ..., 15) are the scalar components of the 4, 4 � � <sup>þ</sup> 4, <sup>4</sup> � � representation of the chiral *SU*<sup>4</sup> <sup>∗</sup> *SU*<sup>4</sup> group. On the grounds of the GMOR model the following relation for masses of the pseudo-scalar mesons can be obtained [12].

$$i < 0 \left| \left[ \overline{Q}^a, \overline{D}^b \right] \right| 0> = \delta^{ab} f\_a^2 m\_a^2 + \int \frac{dq^2}{q^2} \rho^{ab} = \tag{43}$$

$$= \left.  \left( \frac{c\_0}{2} \delta^{ab} + \frac{c\_8}{\sqrt{2}} d\_{a8b} + \frac{c\_{15}}{\sqrt{2}} d\_{a15b} \right) +$$

$$+ < u^8> \left( \frac{c\_0}{\sqrt{2}} d\_{a8b} + c\_8 d\_{a8c} d\_{b8c} + c\_{15} d\_{a8c} d\_{b15c} \right) +$$

$$+ < u\_{15}> 0 \left( \frac{c\_0}{\sqrt{2}} \right. \left. d\_{a15b} + c\_8 \left. d\_{a15c} \right. \left. d\_{b8c} + c\_{15} \left. d\_{a15c} \right. \left. \right. \right. \right)$$

where

$$\rho^{ab} = (2\pi)^3 \sum\_{n \neq a} \delta^4(p\_n - q) < 0\\|\overline{D}^a|n> < n|\overline{D}^b|0> \tag{44}$$

*f <sup>a</sup>* - decay constants, <*u<sup>i</sup>* ><sup>0</sup> - vacuum expectation value of the operator *u<sup>i</sup>* . Because the vacuum expectation values of operators *u*8, *u*<sup>15</sup> and the spectral density *δab* are proportional to the squared parameters of symmetry breaking, they are further neglected [12]. Approximately from Eq. (43) we obtain

$$m\_a^2 \; f\_a^2 \; \delta^{ab} = \frac{1}{\sqrt{2}} \left( \frac{c\_0}{\sqrt{2}} + c\_8 \; d\_{a8b} + c\_{15} \; d\_{a15b} \right) < u^0 > 0 \tag{45}$$

The masses of the mesons are given as follows

$$m\_{\pi}^{2} \int\_{-\pi}^{2} f\_{\pi}^{2} = \frac{1}{2\sqrt{3}} \left( \sqrt{3} \begin{array}{c} c\_{0} + \sqrt{2} \\ c\_{8} + c\_{15} \end{array} c\_{8} + c\_{15} \right) < u^{0} > 0$$

$$m\_{K}^{2} \int\_{K}^{2} f\_{K}^{2} = \frac{1}{2\sqrt{3}} \left( \sqrt{3} \begin{array}{c} c\_{0} - \frac{1}{\sqrt{2}} \\ c\_{8} + c\_{15} \end{array} c\_{8} + c\_{15} \right) < u^{0} > 0 \tag{46}$$

*Quarks Mixing in Chiral Symmetries DOI: http://dx.doi.org/10.5772/intechopen.95233*

$$m\_D^2 \ f\_D^2 = \frac{1}{2\sqrt{3}} \left( \sqrt{3} \ c\_0 + \frac{1}{\sqrt{2}} \ c\_8 - c\_{15} \right) < u^0 >\_{0}$$

In the limit of the exact chiral *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> sub-symmetry there is the following constraint

$$
\sqrt{3} \ c\_0 + \sqrt{2} \ c\_8 + c\_{15} = \mathbf{0} \tag{47}
$$

so the pion is massless.

Let us make some remarks. The task of the Cabibbo angle calculation in *SU*<sup>4</sup> symmetry using the procedure of symmetry breaking has been done in [6]. In Ebrahim's earlier paper [5] the parameters of the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry breaking have been found.

$$\frac{c\_8}{c\_0} = -\frac{2\sqrt{2}}{\sqrt{3}} \frac{m\_\mathrm{K}^2 f\_K^2 - m\_\mathrm{x}^2 f\_\mathrm{x}^2}{m\_\mathrm{K}^2 f\_K^2 + m\_\mathrm{D}^2 f\_\mathrm{D}^2} \qquad \frac{c\_{15}}{c\_0} = -\frac{1}{\sqrt{3}} \frac{3m\_\mathrm{D}^2 f\_\mathrm{D}^2 - m\_\mathrm{K}^2 f\_\mathrm{K}^2 - 2m\_\mathrm{x}^2 f\_\mathrm{x}^2}{m\_\mathrm{K}^2 f\_\mathrm{K}^2 + m\_\mathrm{D}^2 f\_\mathrm{D}^2} \tag{48}$$

In [6] the numerical values of parameters (48) have been used to calculate the rotation angle (interpreted as the Cabibbo angle). The *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> invariant Hamiltonian density breaking *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry has been rotated through an angle 2 Θ about the seventh axis and the coefficients of the operators *ua* (a = 0, 8, 15) have been identified with the parameters of symmetry breaking

$$\begin{aligned} H\_{\text{SB}}(\Delta \mathcal{S} = 0) &= c\_0 u^0 + \frac{\sqrt{3}}{2} c\_8 \sin^2 \Theta \cdot u^3 + c\_8 \left( 1 - \frac{3}{2} \sin^2 \Theta \right) \cdot u^8 \\ &- \left( \sqrt{3} c\_0 + \sqrt{2} c\_8 \right) \cdot u^{15} \end{aligned} \tag{49}$$

It seems to us that there are some errors in the numerical calculations of the author. The use of the numerical values of the parameters (48) has not been necessary. On the grounds of theoretical formulas only, indeed from the Eq. (7) in Ref. [5] and the Eq. (10) in A3-Ebrahim, it follows that

$$\sin^2\Theta = \frac{2m\_\pi^2 f\_\pi^2}{2m\_K^2 f\_K^2 + m\_\pi^2 f\_\pi^2} \tag{50}$$

Then the value of Θ is given by

$$\sin^2 \Theta = \left( \mathbf{0}.\mathbf{215}\right)^2 \tag{51}$$

instead of

$$
\sin^2 \Theta = -\mathbf{0}.\mathbf{0}4\tag{52}
$$

from Eqs. (10) in [6]. Formula (50) has the same form as in *SU*<sup>3</sup> symmetry. In agreement with our expectation the angle Θ is described by parameters of the pion and the strange meson.

In Ebrahim's method the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry breaking the Hamiltonian density is parametrized by the factors *c*0, *c*8, *c*15. The parameters of symmetry breaking are expressed by the masses and decay constants of the mesons and they are fixed (Eq. (7) in [5]). In the limit of the exact *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> sub-symmetry the factors *c*0, *c*8, *c*<sup>15</sup> should satisfy the constraint (47) but it is possible only if *m<sup>π</sup>* ¼ 0 namely the

parameters of symmetry breaking are not expressed by real (measured in experiment) masses of mesons. In [6] Ebrahim breaks the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry in the limit of the exact *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> sub-symmetry by the rotation of the *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> invariant Hamiltonian density through an angle 2 Θ about the seventh axis. The factors of the rotated Hamiltonian density are identified with the parameters of symmetry breaking (Eq. (7) in [5]). Solving a set of equations the author gets the factors *c*0, *c*8, *c*<sup>15</sup> dependent on the rotation angle and on the real mesons masses already. The masses of mesons standing in the formula which describes the parameters of symmetry breaking are determined by the method of symmetry breaking and they have a real value for the real realization of the symmetry breaking only. In this case the rotation angle does not matter a parameter of the symmetry violation. It seems to us that such an interpretation is not satisfactory. The expression of meson masses as a function of the rotation angle (as a measure of symmetry violation) seems to be more natural. In the present paper the other interpretation of the symmetry breaking and the method of calculating the rotation angle is proposed. We describe our method as follows.

Before the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry in the limit of the exact *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> subsymmetry is broken the masses of mesons have been expressed by the factors *c*0, *c*8, *c*<sup>15</sup> which satisfy the constraint (47). After symmetry breaking a new set of factors *c*0 0, *c*<sup>0</sup> *<sup>g</sup>* , *c*<sup>0</sup> <sup>15</sup> dependent on the old factors *c*0, *c*8, *c*<sup>15</sup> and on the rotation angle is introduced. The new factors are identified with the coefficients by the operators *ui* of the rotated Hamiltonian density (49).

$$c\_0' = c\_0 \qquad c\_g' = c\_8 \left(1 - \frac{3}{2} \sin^2 \Theta\right) \qquad c\_{15}' = -\sqrt{3}c\_0 - \sqrt{2}c\_8 \tag{53}$$

Meson masses are expressed by new factors and they are the function of the rotation angle as a measure of symmetry violation.

$$m\_x^2 f\_x^2 = \frac{1}{2\sqrt{3}} \left( \sqrt{3}c\_0' + \sqrt{2}c\_\xi' + c\_{15}' \right) < u^0 >\_0 = -\frac{\sqrt{3}}{2\sqrt{2}} c\_8 \sin^2 \Theta < u^0 >\_0 \tag{54}$$

$$m\_K^2 f\_K^2 = \frac{1}{2\sqrt{3}} \left( \sqrt{3}c\_0' - \frac{c\_8'}{\sqrt{2}} + c\_{15}' \right) < u^0 >\_0 = -\frac{\sqrt{3}}{2\sqrt{2}} c\_8 \left( 1 - \frac{1}{2} \sin^2 \Theta \right) < u^0 >\_0$$

$$m\_D^2 f\_D^2 = \frac{1}{2\sqrt{3}} \left( \sqrt{3}c\_0' + \frac{c\_8'}{\sqrt{2}} - c\_{15}' \right) < u^0 >\_0 = \left( c\_0 + \frac{\sqrt{3}}{2\sqrt{2}} \left( 1 - \frac{1}{2} \sin^2 \Theta \right) \right) < u^0 >\_0$$

It seems to be more natural that the meson masses are functions of the parameters of symmetry breaking (54) than inversely the parameters of symmetry breaking are functions of meson masses which are not consistent with the experimental data and are dependent on the method of symmetry breaking. This interpretation is consistent with the fact that the mass generation of the mesons is a consequence of symmetry breaking. From Eq. (54) we obtain the formula for the angle Θ as in Eq. (50). Let us consider the other variant of symmetry breaking described in [6]. Ebrahim, using his method, broke the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry in the limit of the exact *SU*<sup>3</sup> ∗ *SU*<sup>3</sup> symmetry by the rotation of *SU*<sup>3</sup> ∗ *SU*<sup>3</sup> invariant Hamiltonian density about the fourteenth axis in the *SU*<sup>4</sup> space. The rotation angle Θ<sup>0</sup> is identified with the Cabibbo angle. The formula describing the angle Θ<sup>0</sup> should be given as follows

$$\sin^2\Theta'=\frac{3m\_K^2f\_K^2}{2\left(m\_K^2f\_K^2+m\_D^2f\_D^2\right)}\tag{55}$$

#### *Quarks Mixing in Chiral Symmetries DOI: http://dx.doi.org/10.5772/intechopen.95233*

(in Eqs. (4a) in [8] there is the factor 3/2). The rotation of the Hamiltonian density about the fourteenth axis is considered in [14] too. The D meson is interpreted as a Goldstone boson. Putting aside the agreement of the numerical value of the angle Θ<sup>0</sup> with the experimental data it seems to us that the angle connected with the rotation about the fourteenth axis cannot be interpreted as the Cabibbo angle, because the rotation is performed inside the doublet (s, c). Then the states with the different electric charges are mixed. The interpretation that the D meson is a Goldstone boson is also unsatisfactory. If the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry is broken in such a way that the *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> sub-symmetry is still exact, so the K meson becomes massive but the pion is still massless. Such a symmetry breaking cannot be accepted, results contradict the experimental data. The next breaking of the exact *SU*<sup>3</sup> ∗ *SU*<sup>3</sup> symmetry is connected with the mixing of s and c quarks. The rotation angle cannot be interpreted as the Cabibbo angle for the reasons given above. It seems that the hierarchy of symmetry breaking is extended and the breaking of the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry taken as a whole cannot be connected with the Cabibbo angle. This is possible, however, for *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry breaking in the limit of exact *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> sub-symmetry. Then results are in agreement with our expectation.

Our method described above is used� to calculate an angle *ϕ* which is connected with the rotation about the tenth axis in *SU*<sup>4</sup> space. Then the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry is broken by the rotation of the *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> invariant Hamiltonian density through an angle 2*ϕ* about the tenth axis.

*HSB*ð Þ¼ <sup>Δ</sup>*<sup>C</sup>* <sup>¼</sup> <sup>0</sup> *<sup>c</sup>*0*u*<sup>0</sup> <sup>þ</sup> ffiffi 2 <sup>p</sup> *<sup>c</sup>*<sup>0</sup> <sup>þ</sup> ffiffiffi 3 p <sup>2</sup> *<sup>c</sup>*<sup>8</sup> � � sin <sup>2</sup> *<sup>ϕ</sup> <sup>u</sup>*<sup>3</sup> <sup>þ</sup> *<sup>c</sup>*<sup>8</sup> 1 2 ffiffi 2 p 4 ffiffi <sup>3</sup> <sup>p</sup> *<sup>c</sup>*<sup>0</sup> <sup>þ</sup> ffiffi 2 <sup>p</sup> *<sup>c</sup>*<sup>8</sup> � � sin <sup>2</sup> *ϕ* � � *<sup>u</sup>*<sup>8</sup><sup>þ</sup> � þ � ffiffiffi 3 <sup>p</sup> *<sup>c</sup>*<sup>0</sup> <sup>þ</sup> ffiffi 2 <sup>p</sup> *<sup>c</sup>*<sup>8</sup> <sup>þ</sup> 4 ffiffiffi <sup>3</sup> <sup>p</sup> *<sup>c</sup>*<sup>0</sup> <sup>þ</sup> ffiffi 2 <sup>p</sup> *<sup>c</sup>*<sup>8</sup> � � sin <sup>2</sup> *ϕ* � � *<sup>u</sup>*<sup>15</sup> (56)

Using the factors from the Hamiltonian density (56) the masses of mesons are given as follows

$$m\_x^2 f\_\pi^2 = \left(c\_0 + \frac{\sqrt{6}}{4}\right) \sin^2 \phi \quad < u^0>\_0 \tag{57}$$

$$m\_K^2 f\_K^2 = -\left(\frac{\sqrt{6}}{4} - \frac{1}{2}\left(c\_0 + \frac{\sqrt{6}}{4}c\_8\right) \sin^2 \phi\right) < u^0>\_0$$

$$m\_D^2 f\_D^2 = \left(1 - \frac{1}{2}\sin^2 \phi\right)\left(c\_0' + \frac{\sqrt{6}}{4}c\_8\right) < u^0>\_0$$

so

$$\sin^2 \phi = \frac{2m\_\pi^2 f\_\pi^2}{2m\_\text{D}^2 f\_D^2 + m\_\pi^2 f\_\pi^2} \tag{58}$$

In agreement with our expectation the angle *ϕ* is a function of the mass of the pion (as a measure of the *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> violation) and is connected with the parameters of the charmed meson (mixing in (u, c) doublet). For the mass *mD* = 1862 MeV and *f <sup>D</sup>= f <sup>π</sup>* = 0.974 [13] one gets

$$
\sin^2 \phi = 0.076\tag{59}
$$

#### *Quantum Chromodynamic*

The small value of the angle *ϕ* is the effect of the large mass of the charmed quark. From (59) results that only mixing in the (u, c) system is excluded, the value of the angle *ϕ* contradicts the experimental data. The simultaneous mixing in both doublets are, however, still possible. Fritzsch [15] considers also the mixing in (u, c) system. The mixing angle is calculated on the grounds of quark masses and does not contradict the results obtained above. Although the value of the angle *ϕ* is relatively small, it is significant: the sum of the angles Θ þ *ϕ* is larger than the value of the angle measured experimentally, called Cabibbo angle. This fact cannot be explained by the limits of experimental errors. Let us note that the angles (50) and (58) are calculated for the case where quarks are mixed separately. The angles from formula (32) cannot be identified with those from Eqs. (50) and (58). In the case of simultaneous mixing in both doublets the relation between the angles is more complicated. To find the relation, the *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> invariant Hamiltonian density is rotated through an angle 2*ϕ* about the tenth axis and afterwards by an angle 2Θ about the seventh axis. The sequence of the rotations is insignificant, because

$$\left[F^{\vec{l}}, F^{10}\right] = \mathbf{0} \tag{60}$$

The rotated Hamiltonian density is given by

$$\begin{aligned} H\_{\text{SB}}(\Delta \mathcal{S} = \Delta \mathcal{C} = 0) &= c\_0 u^0 + \left( \frac{\sqrt{3}}{2} \begin{array}{c} c\_8 \ \sin^2 \Theta + \left( \sqrt{2} \ c\_0 + \frac{\sqrt{3}}{2} c\_8 \right) \sin^2 \phi \end{array} \right) u^3 + \\\\ &+ \left( \frac{1}{\sqrt{3}} \left( \sqrt{2} c\_0 + \frac{\sqrt{3}}{2} c\_8 \right) \sin^2 \phi + c\_8 \left( 1 - \frac{3}{2} \sin^2 \Theta \right) \right) \ u^8 + \\\\ &+ \left( -\sqrt{3} \ c\_0 - \sqrt{2} \ c\_8 + \frac{2\sqrt{2}}{\sqrt{3}} \left( \sqrt{2} c\_0 + \frac{\sqrt{3}}{2} c\_8 \right) \sin^2 \phi \right) \ u^{15} \end{aligned} \tag{61}$$

The meson masses are given as follows

$$m\_{\pi}^{2}f\_{\pi}^{2} = \frac{1}{2\sqrt{3}}\left(\sqrt{6}\left(\sqrt{2}\ c\_{0} + \frac{\sqrt{3}}{2}\ c\_{8}\right)\sin^{2}\phi - \frac{3}{\sqrt{2}}\ c\_{8}\sin^{2}\Theta\right) \\ < u^{0}>\_{0} \quad \text{(62)}$$

$$m\_{K}^{2}f\_{K}^{2} = \frac{1}{2\sqrt{3}}\left(\frac{3}{\sqrt{6}}\left(\sqrt{2}\ c\_{0} + \frac{\sqrt{3}}{2}\ c\_{8}\right)\sin^{2}\phi - \frac{3}{\sqrt{2}}\ c\_{8}\left(1 - \frac{1}{2}\sin^{2}\Theta\right)\right) \\ < u^{0}>\_{0}$$

$$m\_{D}^{2}f\_{D}^{2} = \frac{1}{2\sqrt{3}}\left(2\sqrt{3}\ c\_{0} - \frac{3}{\sqrt{6}}\left(\sqrt{2}c\_{0} + \frac{\sqrt{3}}{2}c\_{8}\right)\sin^{2}\phi + \frac{3}{\sqrt{2}}\ c\_{8}\left(1 - \frac{1}{2}\sin^{2}\Theta\right)\right) \\ < u^{0}>\_{0}$$

Now the angles Θ and *ϕ* cannot be described independently. The following relation is obeyed.

$$\begin{aligned} \left(2\ \ m\_{\pi}^{2}f\_{\pi}^{2} + 2\ \left(m\_{K}^{2}f\_{K}^{2} + m\_{D}^{2}f\_{D}^{2}\right)\ \sin^{2}\Theta \ \sin^{2}\phi &=& \ \end{aligned} \tag{63}$$

$$= \left(2\ \ m\_{K}^{2}f\_{K}^{2} + m\_{\pi}^{2}f\_{\pi}^{2}\right)\ \sin^{2}\Theta + \left(2\ \ m\_{D}^{2}f\_{D}^{2} + m\_{\pi}^{2}f\_{\pi}^{2}\right)\ \sin^{2}\phi$$

or equivalently

$$\mathbf{1} + \left(\frac{\mathbf{1}}{\sin^2 \Theta} + \frac{\mathbf{1}}{\sin^2 \phi} - \mathbf{1}\right) \sin^2 \Theta \quad \sin^2 \phi = \frac{\sin^2 \Theta}{\sin^2 \Theta\_0} + \frac{\sin^2 \phi}{\sin^2 \phi\_0} \tag{64}$$

*Quarks Mixing in Chiral Symmetries DOI: http://dx.doi.org/10.5772/intechopen.95233*

where

$$\sin^2\Theta\_0 = \frac{2m\_\pi^2 f\_\pi^2}{2m\_K^2 f\_K^2 + m\_\pi^2 f\_\pi^2} \qquad \sin^2\phi\_0 = \frac{2m\_\pi^2 f\_\pi^2}{2m\_D^2 f\_D^2 + m\_\pi^2 f\_\pi^2} \tag{65}$$

The angles Θ and *ϕ* from Eq. (64) concern a simultaneous mixing in doublets (d, s) and (u, c) respectively and they can be identified with those from Eq. (32). The condition (64) limits the values of the angles Θ and *ϕ*. The maximal values of the angles Θ<sup>0</sup> and *ϕ*<sup>0</sup> are given by Eq. (65). The value of the function

$$f(\Theta, \phi) = \sin\left(\Theta + \phi\right) \tag{66}$$

is also limited. A numerical calculation shows that there is an extremum (a maximum) of function (66) on the condition (64) for

$$
\Theta\_m = 0.20452 \qquad \phi\_m = 0.02575 \qquad \sin \left( \Theta\_m + \phi\_m \right) = 0.2282 \tag{67}
$$

It is worth noticing that the extremum of function (66) on condition (64) can be identified with the measured Cabibbo angle. It is not excluded that symmetry breaking is realized in the maximal allowed case, so the effective angle of mixing would correspond to the maximum of the function (66).

#### **3. Bonds for the Kobayashi-Maskawa mixing parameters in a model with hierarchical symmetry breaking**

A simultaneous mixing in (d, s) and (u, c) sectors has also been taken into account [15, 16], but due to the large mass of the c quark, the influence of the mixing in the (u, c) sector can be treated as a perturbation. At the six-quark level the quark mixing is described by three Cabibbo-like flavor mixing angles and the phase parameter responsible for CP-non-conservation [17]. The charged weak current in the *SU*<sup>6</sup> ∗ *SU*<sup>6</sup> chiral symmetry

$$J\_{\mu} = (\overline{u}, \overline{c}, \overline{t}) \; \begin{array}{c} \chi\_{\mu} \ (1 - \chi\_{5}) \ U \end{array} \begin{pmatrix} d \\ s \\ b \end{pmatrix} \tag{68}$$

is described by a unitary matrix U, which can be put in 21 different forms [18], however only the standard Kobayashi-Maskawa matrix [19] will be used further.

$$U = \begin{pmatrix} c\_1 & s\_1 c\_3 & s\_1 s\_3 \\ -s\_1 c\_2 & c\_1 c\_2 c\_3 - s\_2 s\_3 e^{i\delta} & c\_1 c\_2 s\_3 + s\_2 c\_3 e^{i\delta} \\ s\_1 s\_2 & -c\_1 s\_2 c\_3 - c\_2 s\_3 e^{i\delta} & -c\_1 s\_2 s\_3 + c\_2 s\_3 e^{i\delta} \end{pmatrix} \tag{69}$$

where *si* ¼ sin Θ*i*, *ci* ¼ cos Θ*i*. The matrix (69) can be expressed as follows

$$U = \begin{pmatrix} 1 & 0 & 0 \\ 0 & c\_2 & s\_2 \\ 0 & -s\_2 & c\_2 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \epsilon^{i\delta} \end{pmatrix} \begin{pmatrix} c\_1 & s\_1 & 0 \\ -s\_1 & c\_1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & c\_3 & s\_3 \\ 0 & -s\_3 & c\_3 \end{pmatrix} \tag{70}$$
 
$$U = U\_2$$
  $U\_\delta$   $U\_1$   $U\_3$ 

and it can mix quarks either in the negative or in the positive electric charge subspace.

A simultaneous mixing in both spaces was also considered [10]. From the form of the matrix (70) the following variants of the quark mixing are allowed:

$$A \quad : \quad U = U\_2(\mathfrak{s} - b) \text{ } U\_\delta \text{ } U\_1(d - \mathfrak{s}) \text{ } U\_3(\mathfrak{s} - b) \tag{72}$$

$$B \quad : \quad U = U\_2(\mathfrak{c} - \mathfrak{t}) \ U\_\delta \ U\_1(d - \mathfrak{s}) \ U\_3(\mathfrak{s} - \mathfrak{b}) \tag{73}$$

$$\mathbf{C} \quad : \quad U = U\mathbf{\hat{z}}(\mathbf{c} - \mathbf{t}) \text{ U} \boldsymbol{\delta} \cdot \mathbf{U} \mathbf{\hat{z}}(\mathbf{u} - \mathbf{c}) \text{ U} \mathbf{\hat{z}}(\mathbf{s} - \mathbf{b}) \tag{74}$$

$$D \quad : \quad U = U\_2(\mathfrak{c} - \mathfrak{t}) \text{ U}\_{\delta} \text{ U}\_1(\mathfrak{u} - \mathfrak{c}) \text{ U}\_3(\mathfrak{c} - \mathfrak{t}) \tag{75}$$

where *Uk*ð Þ *x* � *y* denotes the mixing of x and y quarks by the matrix *Uk*. It is known that the Cabibbo angle cannot be explained by the mixing in the (u-c) sector only [15, 16], so the variants C and D must be rejected. Let us examine the variant B.

The charged weak current (68) with the matrix (69) for the variant B can be expressed as follows

$$J\_{\mu} = R \lrcorner J\_{\mu}(\mathbf{0}) \parallel R^{-1} \tag{76}$$

where

$$J\_{\mu}(\mathbf{0}) = (\overline{u}, \overline{c}, \overline{t}) \; \begin{array}{c} \chi\_{\mu} \ (1 - \chi\_{5}) \ I \end{array} \begin{pmatrix} d \\ s \\ b \end{pmatrix} \tag{77}$$

$$R = e^{-2i\Theta\_3 \int\_0^{2i} e^{-2i\Theta\_1 \int\_0^{\gamma}} e^{-i\delta X} e^{2i\Theta\_2 \int\_0^{\beta 2}}}\tag{78}$$

$$X = \frac{4}{\sqrt{10}} \,\mathrm{Q}^{24} - \frac{1}{\sqrt{15}} \,\mathrm{Q}^{35} \tag{79}$$

where *Q<sup>k</sup>* is the 6 ∗ 6 matrix representation of the k-th generator of *SU*<sup>6</sup> group. To get the values of the angles Θ*<sup>i</sup>* the Gell-Mann-Oakes-Renner model will be used [9]. If the electromagnetic mass splitting of u-d quarks is neglected the Hamiltonian density breaking the chiral *SU*<sup>6</sup> ∗ *SU*<sup>6</sup> symmetry is given as follows

$$H\_0 = c\_0 u^0 + c\_8 u^8 + c\_{15} u^{15} + c\_{24} u^{24} + c\_{35} u^{35} \tag{80}$$

where *c*0, … ,*c*<sup>35</sup> are the symmetry breaking parameters, *u<sup>i</sup>* (i = 0,1, ..., 35) are the scalar components Of the 6, 6 � � <sup>þ</sup> 6, <sup>6</sup> � � representation of the chiral *SU*<sup>6</sup> <sup>∗</sup> *SU*<sup>6</sup> group. From the GMOR model, neglecting the vacuum expectation values of operators *uk* (k = 8, 15, 24, 35) and the spectral density *ρab* as proportional to the squared parameters of the symmetry breaking [12, 16], we get the approximate relation for masses of the pseudo-scalar mesons

$$m\_a^2 \, \int\_a^2 \, \delta^{ab} = \frac{1}{\sqrt{3}} \left( \frac{c\_0}{\sqrt{3}} + c\_8 \, \, d\_{a8b} + c\_{15} \, \, d\_{a15b} + c\_{24} \, \, d\_{a24b} + c\_{35} \, \, d\_{a35b} \right) < u^0 > 0 \,\tag{81}$$

where *f <sup>a</sup>* are the decay constants, *daib* - symmetric constants of the *SU*<sup>6</sup> group, < *u*<sup>0</sup> ><sup>0</sup> - the vacuum expectation value of the operator *u*0. From (81) we obtain

$$\pi = m\_{\pi}^{2} \left. f\_{\pi}^{2} \right. \\ = \frac{1}{\sqrt{3}} \left( \frac{c\_{0}}{\sqrt{3}} + \frac{c\_{8}}{\sqrt{3}} + \frac{c\_{15}}{\sqrt{6}} + \frac{c\_{24}}{\sqrt{10}} + \frac{c\_{35}}{\sqrt{15}} \right) \\ < u^{0} > 0 \tag{82}$$

$$K = m\_K^2 \left. f\_K^2 \right. \\ = \frac{1}{\sqrt{3}} \left( \frac{c\_0}{\sqrt{3}} - \frac{c\_8}{2\sqrt{3}} + \frac{c\_{15}}{\sqrt{6}} + \frac{c\_{24}}{\sqrt{10}} + \frac{c\_{35}}{\sqrt{15}} \right) \\ < u^0 > 0 \tag{83}$$

$$D = m\_D^2 \ f\_D^2 \ = \frac{1}{\sqrt{3}} \left( \frac{c\_0}{\sqrt{3}} + \frac{c\_8}{2\sqrt{3}} - \frac{c\_{15}}{\sqrt{6}} + \frac{c\_{24}}{\sqrt{10}} + \frac{c\_{35}}{\sqrt{15}} \right) \ < u^0 > 0 \tag{84}$$

$$B = m\_B^2 \left. f\_B^2 \right. \\ = \frac{1}{\sqrt{3}} \left( \frac{c\_0}{\sqrt{3}} + \frac{c\_8}{2\sqrt{3}} + \frac{c\_{15}}{2\sqrt{6}} - \frac{3c\_{24}}{2\sqrt{10}} + \frac{c\_{35}}{\sqrt{15}} \right) \\ < u^0 > 0 \tag{85}$$

$$T = m\_T^2 \left. f\_T^2 \right. \\ = \frac{1}{\sqrt{3}} \left( \frac{c\_0}{\sqrt{3}} + \frac{c\_8}{2\sqrt{3}} + \frac{c\_{15}}{2\sqrt{6}} + \frac{c\_{24}}{2\sqrt{10}} - \frac{2c\_{35}}{\sqrt{15}} \right) \\ < u^0 > 0 \tag{86}$$

By the symmetry breaking, the massless quark *x* can become massive if it is mixed with the other massive *y*. The rotation angle is then described by the masses of pseudo-scalar mesons. If the *SUn* ∗ *SUn* symmetry with the exact *SUk* ∗ *SUk* subsymmetry is broken to the exact *SUk*�<sup>1</sup> ∗ *SUk*�<sup>1</sup> symmetry, the rotation angle is a function of masses of a pseudo-scalar meson belonging to n-multiplet of the *SUn* ∗ *SUn* group and the meson which has become massive [19]. We demand the quarks to become massive due to the hierarchical symmetry breaking, so the highest exact symmetry of the Hamiltonian density, which Can be assumed, is *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> (at least one quark in the each sector must be massive). Oakes and the others [4, 20, 21] in order to get the Cabibbo angle value in the *SU*<sup>3</sup> ∗ *SU*<sup>3</sup> or *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry, have rotated the Hamiltonian density breaking the chiral symmetry in the same way as the weak charged current. In a model with hierarchical symmetry breaking such a procedure cannot be used. Let us notice that from the form (5) of the rotation operator R it follows that the quarks are mixed in the following sequence: (c-t), a phase rotation, (d-s), (s-b), so for the exact *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry the massless quarks d and s would be mixed as the first (in the negative electric charge subspace) and then the generation of their masses would not be possible. The quark s would become massive in the next stage of the symmetry breaking after the mixing with the massive quark b. So, in order to get the massive both d and s quarks, they should be mixed in the inverse sequence. In the first stage of the symmetry breaking the exact *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry is broken to the exact *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> symmetry, in the 2*nd* stage even the *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> symmetry is no longer exact. The next mixing stages are connected either with the mass generation of the c quark (variant B) or with the repeated mixing of massive s and b quarks (variant A). In our procedure the Hamiltonian density breaking the chiral *SU*<sup>6</sup> ∗ *SU*<sup>6</sup> symmetry will be rotated in the inverse sequence in comparison with the rotation of the weak charged current.

$$H\_{\rm SB} = R\_1 \ H\_0 \ R\_1^{-1} \tag{87}$$

where

$$R\_1 = e^{2i\begin{array}{cccc}\Theta\_2 & Q^{2i} \\ \end{array}} e^{-iX} \begin{array}{cccc}\delta & -2i & \Theta\_1 & Q^{7} \\ \end{array} e^{-2i\begin{array}{cccc}\Theta\_3 & Q^{21} \\ \end{array}} \tag{88}$$

The exact *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry implies the following relations

$$c\_8 = c\_{15} = \mathbf{0} \qquad \sqrt{\mathbf{5}}$$
  $c\_0 + c\_{35} = \mathbf{0} \tag{89}$ 

So, the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> invariant Hamiltonian density is given as

$$H\_E = c\_0 \left( u^0 - \sqrt{5} \ u^{35} \right) + c\_{24} \left( u^{24} - \sqrt{\frac{3}{2}} u^{35} \right) \tag{90}$$

or equivalently

$$H\_E = P \begin{array}{c} \overline{q}\_6 \ q\_6 - V \ \overline{q}\_5 \ q\_5 \end{array} \tag{91}$$

where

$$P = \sqrt{c\_0} + V \qquad V = \frac{5}{\sqrt{10}} \quad c\_{24} \tag{92}$$

The symmetry-breaking Hamiltonian density

$$H\_{SB} = R\_1 \ H\_E \ R\_1^{-1} \tag{93}$$

retaining the flavor-conservation part only is given as follows

$$H\_{\rm SB} = \overline{q}\_6 q\_6 P c\_2^2 - \overline{q}\_5 q\_5 V c\_3^2 + \overline{q}\_4 q\_4 P s\_2^2 - \overline{q}\_3 q\_3 V c\_1^2 s\_3^2 - \overline{q}\_2 q\_2 V s\_1^2 s\_3^2 \tag{94}$$

Let us notice that the phase transformation does not produce terms *qi qi* since the operator (79) commutes with the scalar components *uk*. The flavor-conservation on each stage of the symmetry breaking has been assumed. The Hamiltonian density (94) can be written as a function of the operators if, so the coefficients of *uk*<sup>0</sup> s are given as

$$
\mathcal{L}'\_0 = \mathcal{c}\_0 \tag{95}
$$

$$
\sigma\_{34}' = \frac{V}{2} \left| \begin{array}{c} s\_1^2 \ s\_3^2 \end{array} \right. \tag{96}
$$

$$c\_8' = \frac{V}{2\sqrt{3}} \quad \left(2\begin{array}{c} c\_1^2 \ s\_3^2 \ -s\_1^2 \ s\_3^2 \end{array}\right) \tag{97}$$

$$c'\_{15} = -\frac{1}{2\sqrt{6}} \left( \mathfrak{J} \ P \ s\_2^2 + V \ s\_3^2 \right) \tag{98}$$

$$c'\_{24} = \frac{1}{2\sqrt{10}} \quad \left(4\ \text{V} - 5\ \text{V}\ s\_3^2 + P\ s\_2^2\right) \tag{99}$$

$$\sigma\_{35}' = -\frac{1}{2\sqrt{15}} \begin{pmatrix} \mathfrak{5} \ P + V - \mathfrak{6} \ P \ \mathfrak{s}\_2^2 \end{pmatrix} \tag{100}$$

Now, after the symmetry breaking, the pseudo-scalar masses (82–86) will be described as functions of the coefficients *ci* <sup>0</sup> (i = 0, 3, 8, 15, 24, 35) [7, 16].

$$
\pi = Z \text{ V } \mathfrak{s}\_1^2 \text{ s}\_3^2 \tag{101}
$$

$$K = Z \quad V \ s\_3^2 \quad \left(1 - \frac{1}{2} \; s\_1^2\right) \tag{102}$$

$$D = -Z \left( P \ s\_2^2 - \frac{1}{2} \right) V \ s\_1^2 \ s\_3^2 \Big) \tag{103}$$

$$B = Z \quad V \; \left( \mathbf{1} - s\_3^2 \; \left( \mathbf{1} - \frac{\mathbf{1}}{2} \; s\_1^2 \right) \right) \tag{104}$$

*Quarks Mixing in Chiral Symmetries DOI: http://dx.doi.org/10.5772/intechopen.95233*

$$T = -Z \left( P \ c\_2^2 - \frac{1}{2} \ N \ s\_1^2 \ s\_3^2 \right) \tag{105}$$

where

$$Z = -\frac{1}{2\sqrt{3}} \quad \text{<} \mathbf{u}^0 > \mathbf{0} \tag{106}$$

The Cabibbo angle Θ<sup>1</sup> is expressed in the same form as at four-quark level in the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry [7, 16],

$$s\_1^2 = \frac{\pi}{2 \ K + \pi} \tag{107}$$

Because

$$s\_1^2 \ s\_3^2 = \frac{\pi}{K + B} \quad \Rightarrow \quad s\_3^2 = \frac{K + \frac{\pi}{2}}{K + B} \tag{108}$$

In an agreement with our prediction the angle Θ<sup>3</sup> connected with the mixing of s and b quarks is expressed by the parameters of the strange and beautiful mesons. The angle Θ<sup>1</sup> however, connected with mixing d and s quarks and breaking of the *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> symmetry is expressed by the masses of the pion and the kaon. The angle Θ<sup>2</sup> connected with the mixing in the (c-t) sector is given as

$$s\_2^2 = \frac{D - \frac{\pi}{2}}{D + T - \pi} \tag{109}$$

Let us notice that if we do not demand the flavor-conservation on each stage of the symmetry breaking, after the rotation around the 21*st* axis the terms *q*5*q*3, *q*3*q*<sup>5</sup> in the broken Hamiltonian density arise. In the second stage (the rotation around the 7*th* axis) there will be in *HSB* the following terms: *q*5*q*3, *q*3*q*5, *q*2*q*3, *q*3*q*2, *q*5*q*2, *q*2*q*5. Because

$$\begin{bmatrix} X, \overline{q}\_3 \ q\_5 \end{bmatrix} = i \begin{array}{c} \delta \ \overline{q}\_3 \ q\_5 \end{array} \qquad \begin{bmatrix} X, \overline{q}\_5 \ q\_3 \end{bmatrix} = -i \begin{array}{c} \delta \ \overline{q}\_5 \ q\_3 \end{array} \tag{110}$$

after the phase rotation there will arise in the *HSB* the following terms: *q*<sup>3</sup> *q*<sup>5</sup> *e<sup>i</sup><sup>δ</sup>*, *q*<sup>5</sup> *q*<sup>3</sup> *e*�*i<sup>δ</sup>*, .... In the variant B the matrix *U*<sup>2</sup> has mixed c and t quarks so that in the flavor-conservation part of the broken Hamiltonian density the phase factor *e<sup>i</sup><sup>δ</sup>* cannot appear. But if the matrix *U*<sup>2</sup> mixes s and b quarks again, due to the following relations

$$e^{-2i\Theta\_2Q^{2\natural}}\overline{q}\_3q\_5e^{2i\Theta\_2Q^{2\natural}} = \overline{q}\_3q\_5c\_2^2 - \overline{q}\_5q\_3s\_2^2 + \frac{1}{2}\left(\overline{q}\_3q\_5 - \overline{q}\_3q\_3\right) \quad \sin 2\Theta\_2 \tag{111}$$

$$e^{-2i\Theta\_2} \cdot \stackrel{\text{Q}^2}{\overline{q}} \overline{q}\_5 q\_3 \mathcal{e}^{2i\Theta\_2 \, \text{Q}^{21}} = \overline{q}\_5 q\_3 \mathcal{e}\_2^2 - \overline{q}\_3 q\_5 \mathcal{s}\_2^2 + \frac{1}{2} \left( \overline{q}\_5 q\_5 - \overline{q}\_3 q\_3 \right) \sin 2\Theta\_2 \tag{112}$$

the following terms: *q*<sup>5</sup> *q*<sup>5</sup> *e<sup>i</sup><sup>δ</sup>*, *q*<sup>5</sup> *q*<sup>5</sup> *e*�*i<sup>δ</sup>*, *q*<sup>3</sup> *q*<sup>3</sup> *ei<sup>δ</sup>*, *q*<sup>3</sup> *q*<sup>3</sup> *e*�*i<sup>δ</sup>* appear in the broken Hamiltonian density.

We assume that the symmetry is broken by the quarks mixing in the following sequence: (s-b), (d-s), a phase rotation, (s-b) and the flavor will not be conserved in the intermediate stages of the symmetry breaking, but it will be conserved in the

#### *Quantum Chromodynamic*

broken symmetry taken as a whole. The assumptions given above are consistent with the variant A. Let us take it into account.

We assume the exact *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry. The Hamiltonian density is given by Eq. (91). After symmetry breaking, the flavor conserving part of the broken Hamiltonian density *HSB* is given as

$$H\_{\left(\Delta \overline{F} = 0\right)} = \overline{q}\_6 q\_6 P - \overline{q}\_5 q\_5 V \left(a - A\right) - \overline{q}\_3 q\_3 V \left(\beta + A\right) - \overline{q}\_2 q\_2 V \gamma \tag{113}$$

where

$$a = c\_1^2 s\_2^2 s\_3^2 + c\_2^2 c\_3^2 \tag{114}$$

$$
\beta = c\_1^2 c\_2^2 s\_3^2 + s\_2^2 c\_3^2 \tag{115}
$$

$$
\gamma = s\_1^2 s\_3^2 \tag{116}
$$

$$A = \frac{1}{2} \cos \Theta\_1 \sin 2\Theta\_2 \sin 2\Theta\_3 \cos \delta \tag{117}$$

Since

$$a + \beta + \gamma = \mathbf{1} \tag{118}$$

*α* can be eliminated from (113).

The coefficients by the operators *uk* are as follows

$$
\mathcal{c}'\_0 = \mathcal{c}\_0 \tag{119}
$$

$$
\sigma\_3' = \frac{V}{2}\gamma \tag{120}
$$

$$c\_8' = \frac{V}{2\sqrt{3}} \left(2\beta + 2A - \gamma\right) \tag{121}$$

$$
\sigma\_{\rm 15}' = -\frac{V}{2\sqrt{\mathsf{G}}} \left( \beta + A + \gamma \right) \tag{122}
$$

$$c'\_{24} = \frac{V}{2\sqrt{10}} \left( 4 - 5\left(\beta + A + \gamma\right) \right) \tag{123}$$

$$
\omega'\_{35} = -\sqrt{5}\varepsilon\_0 - \sqrt{\frac{3}{2}}\varepsilon\_{24} \tag{124}
$$

Let us notice that the functions *β* and A occur in Eqs. (121–123) as a sum *β* þ *A* only. So, only two functions can be expressed independently. Because there is no mixing in the positive electric charge subspace we shall not use relations describing mesons D and T. The following relations are obeyed

$$
\pi = ZV\gamma \tag{125}
$$

$$K = ZV\left(\beta + A + \frac{\gamma}{2}\right) \tag{126}$$

$$B = ZV\left(1 - \beta - A - \frac{\gamma}{2}\right) \tag{127}$$

*Quarks Mixing in Chiral Symmetries DOI: http://dx.doi.org/10.5772/intechopen.95233*

so we immediately obtain

$$s\_1^2 s\_3^2 = \frac{\pi}{K + B} \tag{128}$$

as in the variant B, but at the moment the angles Θ<sup>1</sup> and Θ<sup>3</sup> cannot be calculated separately. Putting the experimental value

$$
\cos\Theta\_1 = 0.9737 \qquad (\sin\Theta\_1 = 0.2278) \tag{129}
$$

as an input [22], we get

$$
\sin \Theta\_3 = 0.136 \qquad (\Theta\_3 = 7.8^\circ) \tag{130}
$$

for

$$m\_{\pi} = 0.139 \text{ GeV} \tag{131}$$

$$m\_K = 0.495 \text{ GeV} \qquad f\_K = 1.28 \quad [21] \tag{132}$$

$$m\_B = 5.2 \text{ GeV} \qquad \qquad f\_B = 0.86 \quad [\text{?} \text{2}] \tag{133}$$

The angle Θ<sup>3</sup> was calculated by Fritzsch [15] also for the following quark masses ratios:

$$m\_u \quad : \ m\_d \quad : \ m\_s \quad : \ m\_c = 1 \quad : 1.78 \quad : \ 35.7 \quad : 285$$

and the limit for the angle Θ<sup>2</sup>

$$
\Theta\_2 < \sqrt{\frac{m\_c}{m\_t}} = 0.33\tag{135}
$$

For the assumptions given above Fritzsch obtained the following boundary

$$
\sin \Theta\_3 < 0.09 \qquad (\Theta\_3 < 5^\circ) \tag{136}
$$

However there is no agreement between descriptions of the quark masses ratios. The other authors [23] give smaller difference between quark masses

$$m\_u \quad :\ m\_d \quad :\ m\_s \quad :\ m\_c = 1 \quad :\ 1.1 \quad :\ 6.4 \quad :\ 23.6 \tag{137}$$

Thus for the ratio (137) we get the following limit for the angle Θ<sup>3</sup>

$$
\sin \Theta\_3 < 0.163 \tag{138}
$$

The value of the angle Θ<sup>3</sup> (130) is consistent with the boundary (138). The value (130) is close to the value given by Białas [24] and consistent with results in [25, 26] as well as the experimental boundary:

$$
\sin \Theta\_3 < 0.42 \tag{24}
$$

$$|\sin\Theta\_3| = 0.28\begin{cases} +0.21\\ -0.28 \end{cases} \quad \text{[19]}\tag{140}$$

Let us consider the relation between the angle Θ<sup>2</sup> and the phase parameter *δ*. From (125)–(127) we obtain

*Quantum Chromodynamic*

$$
\beta + A = \frac{K - \frac{\pi}{2}}{K + B} \tag{141}
$$

or equivalently

$$\frac{K + \frac{\pi}{2}}{K + B} - \frac{\chi}{s\_1^2} = s\_2^2 \left( 1 + \chi \: \left( 1 - \frac{2}{s\_1^2} \right) \right) + A \tag{142}$$

denoting

$$\xi = \frac{\left(K + \frac{\pi}{2}\right) - \frac{\pi}{s\_1^2}}{K + B} \tag{143}$$

$$\eta = \mathbf{1} + \frac{\boldsymbol{\pi}}{K + B} \tag{144}$$

$$
\rho = \frac{1}{2} \quad \cos \Theta\_1 \quad \sin 2\Theta\_3 \tag{145}
$$

we get

$$\cos \delta = \frac{\xi - s\_2^2}{\rho} \frac{\eta}{\sin 2\Theta\_2} \tag{146}$$

It is worth noting that if we take the constraint on the Cabibbo angle Θ<sup>1</sup> from the four- quark level [7, 16], which is the same as given by Eq. (107), the parameter *ξ* (143) will be exactly equal to zero, hence we get

$$\cos \delta = -\frac{\eta}{2\rho} \tan \Theta\_2 \tag{147}$$

Because ∣*cos δ*∣ ≤1, so from (147)

$$|\Theta\_2| < \left| \arctan \frac{2\rho}{\eta} \right| \tag{148}$$

and we get also a boundary on the angle Θ<sup>2</sup>

$$\sin \Theta\_2 < 0.265 \qquad (\Theta\_2 < 15.4^\circ) \tag{149}$$

The value (149) is in a good agreement with the results given by Fritzsch [15], Białas [24], Shrock, Treiman, Wang [22], Barger, Long, Pakvasa [25] and experimental limits [27], respectively:

$$9^\* < \Theta\_2 < 19^\* \tag{150}$$

$$
\sin \Theta\_2 = 0.23 \tag{151}
$$

$$|\sin\Theta\_2| < 0.25 \qquad (m\_l = 15 \text{ GeV}) \tag{152}$$

$$\sin \Theta\_2 < 0.5 \qquad \left(m\_t = \text{30 GeV}\right) \tag{153}$$

The Eq. (146) can be written as follows

$$\left(\varkappa^2 \left(\eta^2 + 4\right)\rho^2 \cos^2\delta\right) - 2\varkappa\left(\xi\eta \; \; \; + 2\rho^2 \cos^2\delta\right) + \xi^2 = 0\tag{154}$$

*Quarks Mixing in Chiral Symmetries DOI: http://dx.doi.org/10.5772/intechopen.95233*

where

$$\propto = \sin^2 \Theta\_2 \tag{155}$$

To get a real value of the angle Θ<sup>2</sup> the determinant of the square Eq. (154) cannot be negative, so

$$\mathbf{1}\mathbf{1}\boldsymbol{\delta}\rho^2\cos^2\delta\left(\xi\eta+\rho^2\cos^2\delta-\xi^2\right)\geq\mathbf{0}\tag{156}$$

hence

$$1 \ge \cos^2 \delta \ge \frac{\xi(\xi - \eta)}{\rho^2} \tag{157}$$

For (129, 130) we get

$$
\eta = 0.9647 \qquad \rho = 0.1309 \tag{158}
$$

If the parameter *ξ*. which can be identified with a change of the Cabibbo angle description by a transition to the higher symmetries, is slightly less than zero, the phase parameter *δ* will be bounded (∣*δ*∣ should be nearly zero, as the Cabibbo angle description should not change strongly by a transition to higher symmetries, on the other hand the Eq. (157) gives a boundary on the parameter

$$
\xi > -0.0175\tag{159}
$$

From (147)

$$\operatorname{sign}(\cos \delta) = -\operatorname{sign}(\tan \Theta\_2) \tag{160}$$

so, for the angle Θ<sup>2</sup> lying in the first quadrant, it follows *<sup>π</sup>* <sup>2</sup> < *δ*<*π* and from (159) there is a lower limit for the phase *δ*. For an input given by the Eqs. (129, 130, 132) we get

$$
\xi = 0.002 \tag{161}
$$

so there is no boundary on *δ*, since sign *ξ* = + 1. Let us notice, that a small change of the *f <sup>K</sup>* can change the sign of the parameter *ξ*. Following Fuchs [28], in a chiral perturbation theory at the *SU*<sup>3</sup> ∗ *SU*<sup>3</sup> level

$$\frac{f\_K}{f\_\pi} = 1 + \frac{3\left(m\_K^2 - m\_\pi^2\right)}{64\,\pi^2 f\_\pi^2} \ln\frac{\Lambda}{4\,\mu^2} + O(\varepsilon) \tag{162}$$

where *μ*<sup>2</sup> is the average meson squared mass and Λ is a cut-off parameter, which is estimated to be near 4*m*<sup>2</sup> *<sup>N</sup>*, it implies

$$\frac{f\_K}{f\_x} = 1.15\tag{163}$$

$$
\xi = -0.00179 \tag{164}
$$

$$\cos^2 \delta > 0.1\tag{165}$$

Taking into account (160) we obtain

$$109^\circ < \delta < 180^\circ \tag{166}$$

Since *f <sup>K</sup>* is treated as a variable and can depend on the energy scale via Λ parameter and the symmetry breaking parameters *ε*, the boundary of the phase due to the Eqs. (143, 157) can be expected. For sign (cos *δ*) = �1 there is a lower limit of the angle Θ<sup>2</sup> also. A variant cos *δ*>0 is allowed but the angle Θ<sup>2</sup> corresponding to this variant is too severely limited and it is not consistent with the experimental data [27].

We have shown that the weak mixing angles at the six-quark level can be estimated in terms of the masses of pseudo-scalar mesons. The calculation of mixing angles is possible by using the hierarchical symmetry breaking leading to a quark masses generation. A number of independent mixing angles that can be calculated on the ground of the given above model is equal to a number of degrees of freedom connected with the symmetry breaking and the quarks mixing in the fixed electric charge subspace (let us notice that in the variant B after the rotation around the 21*st* axis and next around the 7*th* one, even the exact *SU*<sup>2</sup> ∗ *SU*<sup>2</sup> symmetry did not remain; however the angle Θ<sup>2</sup> connected with the mixing in the positive electric charge subspace could be calculated). An assumption that in the hierarchical symmetry breaking the flavor does not have to be conserved on each stage of the symmetry breaking, while it is conserved in the broken symmetry taken as a whole, has allowed the author to introduce to the broken Hamiltonian density a phase angle responsible for CP-non-conservation. The experimental value of the Cabibbo angle treated as an input has allowed the author to calculate the angle Θ<sup>3</sup> and to find the relation connecting the angle Θ<sup>2</sup> and the phase parameter *δ*. Limits of trigonometric functions values imply boundaries on the angle Θ<sup>2</sup> and the phase *δ*. The kaon decay constant is a sensitive parameter, which can introduce CP-non-conservation to the chiral perturbation theory. Boundaries for the angle Θ<sup>2</sup> and the phase *δ* vs. *f <sup>K</sup>* can be also found.

#### **4. The standard six-quark model with a hierarchical symmetry breaking**

The simultaneous mixing of quarks in both negative and positive electric charge sub-spaces is considered. Quark mixing in each space is described by the Kobayashi-Maskawa matrix. In order to get a right number of independent mixing parameters only one angle *θ*<sup>7</sup> common for both sub-spaces has been adjusted. Since the electromagnetic mass splitting of u and d quarks has been taken into account the real K-M mixing angles can be calculated explicitly. As an input only meson masses and *f <sup>x</sup>* factors (treated as factors in matrix elements between one meson state and vacuum according to PCAC) are needed. Physical quark mixing is realized for maximal allowed symmetry breaking and it corresponds to vanishing of *θ*7, which implies that only quark mixings with mass generation are permitted. Bounds of the phase *δ* have been also found.

The Kobayashi-Maskawa mixing matrix (69) is usually considered to mix quarks in the negative electric charge subspace. It can be written also as (70) and it can mix quarks either in the negative or in the positive electric charge subspace. A simultaneous mixing in both spaces [13] was also taken onto account.

Let us assume that quarks are mixed in both sub-spaces simultaneously, then the charged weak current is expressed as follows

$$J\_{\mu} = (\overline{u}, \overline{c}, \overline{t}) \; \begin{array}{c} \chi\_{\mu} \ (1 - \chi\_{5}) \ U\_{+} \ U\_{-} \end{array} \begin{pmatrix} d \\ s \\ s \\ b \end{pmatrix} \tag{167}$$

where

$$U\_{+} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & c\_2 & s\_2 \\ 0 & -s\_2 & c\_2 \end{pmatrix} \begin{pmatrix} c\_5 & s\_5 & 0 \\ -s\_5 & c\_5 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & e^{i\delta\_2} \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & c\_6 & s\_6 \\ 0 & -s\_6 & c\_6 \end{pmatrix} \tag{168}$$
 
$$U\_{-} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & c\_7 & s\_7 \\ 0 & -s\_7 & c\_7 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & e^{i\delta\_1} \end{pmatrix} \begin{pmatrix} c\_1 & s\_1 & 0 \\ -s\_1 & c\_1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & c\_3 & s\_3 \\ 0 & -s\_3 & c\_3 \end{pmatrix} \tag{169}$$

The matrices *U*<sup>þ</sup> and *U*� mix quarks in spaces with charges +2/3 and � 1/3 respectively. The Kobayashi-Maskawa mixing matrix is parametrized by four independent parameters only, while in the product of the matrices (168) and (169) in the current (167) there are eight mixing angles. In order to get the effective mixing matrix in the K-M form with the right number of independent mixing parameters, we must adjust the angles in such a way as to get the effective matrix with only four independent angles. We shall demand the following elements *U*11, *U*12, *U*13, *U*21, *U*<sup>31</sup> of the effective matrix to be real and the complex phase to exist in the elements *U*22, *U*23, *U*32, *U*<sup>33</sup> only, as in the original K-M matrix. The only solution is

$$
\theta\_{\mathsf{6}} = -\theta\_{\mathsf{7}} \tag{170}
$$

Hence in the matrix *U*<sup>þ</sup> *U*� there will be effectively only four parameters: *θ*2, *θ<sup>C</sup>* ¼ *θ*<sup>1</sup> þ *θ*5, *θ*<sup>3</sup> and *δ* ¼ *δ*<sup>1</sup> þ *δ*2. The current (167) can be expressed as follows

$$J\_{\mu} = R\_2 \,\, R\_1 \,\, J\_{\mu}(\mathbf{0}) \,\, R\_1^{-1} \,\, R\_2^{-1} \tag{171}$$

where

$$J\_{\mu}(\mathbf{0}) = (\overline{\boldsymbol{u}}, \overline{\boldsymbol{c}}, \overline{\boldsymbol{t}}) \ \boldsymbol{\chi}\_{\mu} \ (\mathbf{1} - \boldsymbol{\chi}\_{5}) \ \boldsymbol{I} \ \begin{pmatrix} \boldsymbol{d} \\ \boldsymbol{s} \\ \boldsymbol{b} \end{pmatrix} \tag{172}$$

$$R\_1 = e^{-2i\,\theta\_3 \, Q^{21}} e^{-2i\,\theta\_1 \, Q^{\overline{\gamma}}} \ e^{-iX\delta\_1} \, e^{-2i\,\theta\_\overline{\gamma} \, Q^{21}} \tag{173}$$

$$R\_2 = e^{-2i\theta\_2 Q^{32}} e^{-iY\delta\_2} \ e^{-2i\theta\_5 Q^{10}} e^{-2i\theta\_7 Q^{32}} \tag{174}$$

where X = (79) and *<sup>Y</sup>* <sup>¼</sup> ffiffiffi <sup>15</sup> <sup>p</sup> <sup>3</sup> *<sup>Q</sup>*35. *<sup>Q</sup><sup>k</sup>* is the 6 <sup>∗</sup> 6 matrix representation of the k-th generator of *SU*<sup>6</sup> group. In variants A (72) and B (73), because of quark mixing in (d, s, b) sector only, the electromagnetic mass splitting of *u* and *d* quarks was neglected. For the simultaneous mixing in both (d, s, b) and (u, c, t) sectors the calculation of the angles *θ<sup>i</sup>* explicitly is not possible (see below formulas (187) and (188). The Hamiltonian density breaking the chiral *SU*<sup>6</sup> ∗ *SU*<sup>6</sup> symmetry is given as follows

$$H\_0 = \sum\_{j=1}^6 \left. c\_{j^2 - 1} \right. \left. u^{j^2 - 1} \right. \tag{175}$$

where *ci* are the symmetry breaking parameters, *u<sup>i</sup>* - the scalar components of the 6, 6 � � <sup>þ</sup> 6, <sup>6</sup> � � of the chiral *SU*<sup>6</sup> <sup>∗</sup> *SU*<sup>6</sup> group. From the GMOR model we obtain the following relations for masses of pseudo-scalar mesons for *SU*<sup>6</sup> ∗ *SU*<sup>6</sup> symmetry:

*<sup>π</sup>* <sup>¼</sup> *<sup>m</sup>*<sup>2</sup> *<sup>π</sup> f* 2 *<sup>π</sup>* <sup>¼</sup> *<sup>Z</sup> <sup>c</sup>*<sup>0</sup> ffiffiffi <sup>3</sup> <sup>p</sup> <sup>þ</sup> *c*8 ffiffiffi <sup>3</sup> <sup>p</sup> <sup>þ</sup> *c*15 ffiffiffi 6 p þ *c*<sup>24</sup> ffiffiffiffiffi <sup>10</sup> <sup>p</sup> <sup>þ</sup> *c*35 ffiffiffiffiffi <sup>15</sup> <sup>p</sup> � � (176) *<sup>K</sup>*<sup>þ</sup> <sup>¼</sup> *<sup>m</sup>*<sup>2</sup> *<sup>K</sup>*<sup>þ</sup> *f* 2 *<sup>K</sup>*<sup>þ</sup> <sup>¼</sup> *<sup>Z</sup> <sup>c</sup>*<sup>0</sup> ffiffiffi <sup>3</sup> <sup>p</sup> <sup>þ</sup> *c*3 <sup>2</sup> � *<sup>c</sup>*<sup>8</sup> 2 ffiffiffi <sup>3</sup> <sup>p</sup> <sup>þ</sup> *c*15 ffiffiffi 6 p þ *c*<sup>24</sup> ffiffiffiffiffi <sup>10</sup> <sup>p</sup> <sup>þ</sup> *c*35 ffiffiffiffiffi <sup>15</sup> <sup>p</sup> � � *<sup>K</sup>*<sup>0</sup> <sup>¼</sup> *<sup>m</sup>*<sup>2</sup> *<sup>K</sup>*<sup>0</sup> *f* 2 *<sup>K</sup>*<sup>0</sup> <sup>¼</sup> *<sup>Z</sup> <sup>c</sup>*<sup>0</sup> ffiffiffi <sup>3</sup> <sup>p</sup> � *<sup>c</sup>*<sup>3</sup> <sup>2</sup> � *<sup>c</sup>*<sup>8</sup> 2 ffiffiffi <sup>3</sup> <sup>p</sup> <sup>þ</sup> *c*15 ffiffiffi 6 p þ *c*<sup>24</sup> ffiffiffiffiffi <sup>10</sup> <sup>p</sup> <sup>þ</sup> *c*35 ffiffiffiffiffi <sup>15</sup> <sup>p</sup> � � *<sup>D</sup>*<sup>þ</sup> <sup>¼</sup> *<sup>m</sup>*<sup>2</sup> *<sup>D</sup>*<sup>þ</sup> *f* 2 *<sup>D</sup>*<sup>þ</sup> <sup>¼</sup> *<sup>Z</sup> <sup>c</sup>*<sup>0</sup> ffiffiffi <sup>3</sup> <sup>p</sup> � *<sup>c</sup>*<sup>3</sup> 2 þ *c*8 2 ffiffiffi 3 <sup>p</sup> � *<sup>c</sup>*<sup>15</sup> ffiffiffi 6 p þ *c*<sup>24</sup> ffiffiffiffiffi <sup>10</sup> <sup>p</sup> <sup>þ</sup> *c*35 ffiffiffiffiffi <sup>15</sup> <sup>p</sup> � � *<sup>D</sup>*<sup>0</sup> <sup>¼</sup> *<sup>m</sup>*<sup>2</sup> *<sup>D</sup>*<sup>0</sup> *f* 2 *<sup>D</sup>*<sup>0</sup> <sup>¼</sup> *<sup>Z</sup> <sup>c</sup>*<sup>0</sup> ffiffiffi <sup>3</sup> <sup>p</sup> � *<sup>c</sup>*<sup>3</sup> 2 þ *c*8 2 ffiffiffi <sup>3</sup> <sup>p</sup> � *<sup>c</sup>*<sup>15</sup> ffiffiffi 6 p þ *c*<sup>24</sup> ffiffiffiffiffi <sup>10</sup> <sup>p</sup> <sup>þ</sup> *c*35 ffiffiffiffiffi <sup>15</sup> <sup>p</sup> � � *<sup>B</sup>*<sup>þ</sup> <sup>¼</sup> *<sup>m</sup>*<sup>2</sup> *<sup>B</sup>*<sup>þ</sup> *f* 2 *<sup>B</sup>*<sup>þ</sup> <sup>¼</sup> *<sup>Z</sup> <sup>c</sup>*<sup>0</sup> ffiffiffi <sup>3</sup> <sup>p</sup> <sup>þ</sup> *c*3 2 þ *c*8 2 ffiffiffi <sup>3</sup> <sup>p</sup> <sup>þ</sup> *c*15 2 ffiffiffi 6 <sup>p</sup> � <sup>3</sup>*c*<sup>24</sup> 2 ffiffiffiffiffi <sup>10</sup> <sup>p</sup> <sup>þ</sup> *c*35 ffiffiffiffiffi <sup>15</sup> <sup>p</sup> � � *<sup>T</sup>*<sup>þ</sup> <sup>¼</sup> *<sup>m</sup>*<sup>2</sup> *<sup>T</sup>*<sup>þ</sup> *f* 2 *<sup>T</sup>*<sup>þ</sup> <sup>¼</sup> *<sup>Z</sup> <sup>c</sup>*<sup>0</sup> ffiffiffi 3 <sup>p</sup> � *<sup>c</sup>*<sup>3</sup> 2 þ *c*8 2 ffiffiffi <sup>3</sup> <sup>p</sup> <sup>þ</sup> *c*15 2 ffiffiffi 6 p þ *c*<sup>24</sup> 2 ffiffiffiffiffi <sup>10</sup> <sup>p</sup> � <sup>2</sup>*c*<sup>35</sup> ffiffiffiffiffi <sup>15</sup> <sup>p</sup> � �

In a model with hierarchical symmetry breaking the highest exact symmetry, which can be assumed, is the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> one. At least one quark in each sector must be massive. Following the procedure described in [29] the Hamiltonian density breaking the chiral *SU*<sup>6</sup> ∗ *SU*<sup>6</sup> symmetry will be rotated in the opposite direction by comparison with the rotation of the weak charged current.

$$H\_{\rm SB} = R\_{21} \ R\_{11} \ H\_E \ R\_{11}^{-1} \ R\_{21}^{-1} \tag{177}$$

where

$$R\_{11} = e^{-2i\,\theta\gamma\,Q^{21}} e^{-iX\delta\_1} e^{-2i\,\theta\_1\,Q^{7}} e^{-2i\,\theta\_3\,Q^{21}}\tag{178}$$
 
$$R\_{21} = e^{-2i\,\theta\_1\,Q^{32}} e^{-iY\delta\_2} \begin{array}{c} e^{2i\,\theta\_5\,Q^{10}} \ e^{2i\,\theta\_2\,Q^{32}}\end{array}$$

The exact *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry implies that

$$
\omega\_{\mathfrak{J}} = \mathfrak{c}\mathfrak{s} = \mathfrak{c}\mathfrak{s} = \sqrt{\mathfrak{F}}\mathfrak{c}\mathfrak{o} + \mathfrak{c}\mathfrak{s} = \mathfrak{0} \tag{179}
$$

The *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> invariant Hamiltonian density is given as

$$H\_E = P\overline{q}\_6 q\_6 - V\overline{q}\_5 q\_5 \tag{180}$$

where

$$P = \sqrt{12}c\_0 + V \qquad \qquad V = \frac{5}{\sqrt{10}}c\_{24} \tag{181}$$

We shall assume that in the model with hierarchical symmetry breaking the flavor will not be conserved in the intermediate stages of the symmetry breaking, but it will be conserved in the broken symmetry taken as a whole. The symmetry breaking Hamiltonian density retaining onlyÂ� the flavor-conserving part is given as follows

$$H\_{\left(\Delta F=0\right)} = \overline{q}\_{\xi}q\_{\xi}P(\lambda - M) - \overline{q}\_{\xi}q\_{\xi}V(a - A) + \overline{q}\_{4}q\_{4}P(\rho + M) - \overline{q}\_{3}q\_{3}V(\beta + A) - \overline{q}\_{1}q\_{1}V\gamma$$

$$+ \overline{q}\_{1}q\_{1}P\tau \tag{182}$$

where

$$\begin{aligned} \alpha &= c\_1^2 s\_3^2 s\_7^2 + c\_3^2 c\_7^2 \\ \beta &= c\_1^2 s\_3^2 c\_7^2 + c\_3^2 s\_7^2 \\ \gamma &= s\_1^2 s\_3^2 \\ A &= \frac{1}{2} \sin 2\theta\_7 \sin 2\theta\_3 c\_1 \cos \delta\_1 \\ \lambda &= s\_2^2 c\_5^2 s\_7^2 + c\_2^2 c\_7^2 \\ \rho &= s\_2^2 c\_5^2 c\_7^2 + c\_2^2 s\_7^2 \\ \tau &= s\_2^2 s\_5^2 \\ \mathcal{M} &= -\frac{1}{2} \sin 2\theta\_7 \sin 2\theta\_2 c\_5 \cos \delta\_2 \end{aligned} \tag{184}$$

The broken Hamiltonian density (182) can be expressed as a function of operators *u<sup>k</sup>* (k = 0, 3, 8, 15, 24, 35). The coefficients of the operators *uk* are as follows

$$\begin{aligned} c\_0' &= c\_0 \\ c\_3' &= \frac{1}{2} \left( P\tau + V\gamma \right) \\ c\_8' &= \frac{1}{2\sqrt{3}} \left( P\tau + V\left( 2\beta' - \gamma \right) \right) \\ c\_{15}' &= \frac{1}{2\sqrt{6}} \left( P\left( \tau - 3\rho' \right) - V\left( \beta' + \gamma \right) \right) \\ c\_{24}' &= \frac{1}{2\sqrt{10}} \left( P\left( \tau + \rho' \right) - V\left( 5\beta' + 5\gamma - 4 \right) \right) \\ c\_{35}' &= \frac{1}{2\sqrt{15}} \left( 6P\left( \tau + \rho' \right) - 5P - V \right) \end{aligned} \tag{181}$$

where

$$
\beta' = \beta + A \qquad \qquad \rho' = \rho + M \tag{186}
$$

After symmetry breaking the pseudo-scalar masses (176) will be described as functions of the coefficients *ci*' [16].

$$\begin{aligned} \pi &= \frac{Z}{2} \left( P\tau - V\gamma \right) \\ K^+ &= \frac{Z}{2} \left( P\tau - V\beta' \right) \\ D^0 &= \frac{Z}{2} P \left( \rho' + \tau \right) \\ B^+ &= \frac{Z}{2} \left( P\tau + V\left( \theta' + \gamma - 1 \right) \right) \end{aligned} \tag{187}$$
 
$$\begin{aligned} B^+ &= \frac{Z}{2} \left( P\tau + V\left( \theta' + \gamma - 1 \right) \right) \end{aligned} \qquad \qquad \begin{aligned} T^+ &= \frac{Z}{2} \left( P\left( 1 - \tau - \rho' \right) - V\gamma \right) \end{aligned}$$

From (183), (184), (186) and (187) we get

$$\begin{aligned} \rho' &= \frac{K^0 + K^+ - \pi}{2B^+ + 3K^0 - K^+ - \pi} & \qquad \gamma &= \frac{K^0 - K^+ + \pi}{2B^+ + 3K^0 - K^+ - \pi} \\\\ \rho' &= \frac{D^0 + D^+ - \pi}{2T^+ + 3D^0 - D^+ - \pi} & \qquad \tau &= \frac{D^0 - D^+ + \pi}{2T^+ + 3D^0 - D^+ - \pi} \end{aligned} \tag{188}$$

(contrary to the case of mixing in (d, s, b) sector only (variant A in [29] the electromagnetic mass splitting of u and d quarks cannot be neglected; if we put arbitrarily *c*<sup>3</sup> = 0 in Eq. (175) as in variants A and B in [29], the parameters *γ*, *β*<sup>0</sup> *τ*, *ρ*<sup>0</sup> could not be calculated separately. We would obtain only three nonlinear relations connecting these parameters with meson masses). Since

$$a + \beta + \gamma = \lambda + \rho + \tau = \mathbf{1} \tag{189}$$

putting (183, 184, 186) to (188) and eliminating *θ*<sup>2</sup> and *θ*<sup>3</sup> from the obtained set of four equations we get

$$f\_1(\theta\_1, \delta\_1) = \tan \theta\_{\overline{\!\!\!}} = -f\_{\overline{\!\!\!\!}}(\theta\_{\overline{\!\!\!\!}}, \delta\_{\overline{\!\!\!\!\!}}) \tag{190}$$

where

$$f\_1(\theta\_1, \delta\_1) = \frac{B\_1 \mp \sqrt{B\_1^2 - A\_1 C\_1}}{A\_1} \tag{191}$$

$$A\_1 = s\_1^2(1 - \beta') - \gamma \qquad B\_1 = \sqrt{\gamma} \sqrt{s\_1^2 - \gamma} c\_1 \cos \delta\_1 \qquad C\_1 = \gamma - s\_1^2(\beta' + \gamma) \tag{192}$$

$$f\_{\mathfrak{F}}(\theta\_{\mathfrak{F}}, \delta\_{\mathfrak{Z}}) = \frac{B\_{\mathfrak{F}} \mp \sqrt{B\_{\mathfrak{F}}^2 - A\_{\mathfrak{F}} C\_{\mathfrak{F}}}}{A\_{\mathfrak{F}}} \tag{193}$$

$$A\_5 = s\_5^2(1 - \rho') - \tau \qquad B\_5 = \sqrt{\tau} \sqrt{s\_5^2 - \tau} c\_5 \cos \delta\_2 \qquad C\_5 = \tau - s\_5^2(\rho' + \tau) \tag{194}$$

We considered in [16] the simultaneous mixing in (d, s) and (u, c) sectors in the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry. The mixing angles Θ and *ϕ* could not be calculated separately, however the nonlinear formula connecting both angles and pseudo-scalar masses was found

$$2\pi + 2(K+D)\sin^2\Theta\sin^2\phi = (2K+\pi)\sin^2\Theta + (2D+\pi)\sin^2\phi\tag{195}$$

A numerical calculation showed that there is an extremum (a maximum) of the function (66) with condition (195) for the angles Θ*<sup>m</sup>* þ *ϕ<sup>m</sup>* very close to the experimentally measured Cabibbo angle. This fact suggests that the symmetry breaking is realized in the maximal allowed case, so the effective angle of mixing would correspond to the maximum of function (66). As in [16] we shall look for the extremum of the function

$$f(\theta\_1, \theta\_5) = \sin\left(\theta\_1 + \theta\_5\right) \tag{196}$$

with condition (190). The following set of equations must be obeyed

$$f\_1(\theta\_1, \delta\_1) + f\_\S(\theta\_\delta, \delta\_2) = \mathbf{0} \qquad \frac{\partial f\_1(\theta\_1, \delta\_1)}{\partial \delta\_1} = \mathbf{0} \tag{197}$$

*Quarks Mixing in Chiral Symmetries DOI: http://dx.doi.org/10.5772/intechopen.95233*

$$\frac{\partial f\_1(\theta\_1, \delta\_1)}{\partial \theta\_1} - \frac{\partial f\_\S(\theta\_\5, \delta\_2)}{\partial \theta\_\5} = 0 \qquad \qquad \frac{\partial f\_\S(\theta\_\5, \delta\_2)}{\partial \delta\_2} = 0$$

From (197) we get

$$\mathbf{C\_1 = 0} \qquad \mathbf{C\_5 = 0} \tag{198}$$

respectively, which implies that the separation constant

$$
\tan \theta\_{\overline{\!\!\!=}} = \mathbf{0} \tag{199}
$$

This means that the maximal allowed symmetry breaking occurs only for independent mixing of quarks in both sectors.

Let us consider the action of the operators *R*<sup>11</sup> and *R*<sup>21</sup> on quarks. The operator *R*<sup>11</sup> mixes quarks in the negative electric charge subspace in the following sequence: (s-b)(*θ*3), (d-s)(*θ*1), a phase rotation (*δ*1), (s-b)(*θ*7), however the operator *R*<sup>21</sup> mixes quarks as follows: (c-t) (*θ*2), (u-c)(*θ*5). a phase rotation (*δ*2), (c-t)(*θ*7). By the exact *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> symmetry only b and t quarks are massive. After the symmetry breaking a massless quark can become massive if it mixes with the other massive one. By the mixing in the sector with the charge - 1/3 the quark s has become massive in the first stage of the hierarchical symmetry breaking, after mixing with the quark b (the rotation on the angle *θ*<sup>3</sup> generated by the operator *Q*21), the quark d has become massive in the second stage after mixing with the already massive quarks (the rotation on the angle *θ*1). The next rotation by the angle *θ*7, and mixing of s and b quarks are not connected with the symmetry breaking, because the mixing quarks have been already massive. There is analogical situation in the sector with the charge +2/3. The c and u quarks have become massive due to the hierarchical symmetry breaking (rotations on angles *θ*<sup>2</sup> and *θ*5, respectively), however the rotation by the angle *θ*<sup>7</sup> and mixing of c and t quarks are also not connected with the symmetry breaking. Thus, from (199) it results that the physical quark mixing is realized only in the symmetry breaking with the quark masses generation. Putting (199) to (183) and (184) and comparing with (188) we get

$$\sin^2\theta\_1 = \frac{K^0 - K^+ + \pi}{2K^0} \qquad\qquad\qquad\sin^2\theta\_5 = \frac{D^0 - D^+ + \pi}{2D^0} \tag{200}$$

$$\sin^2\theta\_3 = \frac{2K^0}{2B^+ + 3K^0 - K^+ - \pi} \qquad\qquad\sin^2\theta\_2 = \frac{2D^0}{2T^+ + 3D^0 - D^+ - \pi}$$

so *θ<sup>C</sup>* ¼ *θ*<sup>1</sup> þ *θ*<sup>5</sup> depends on the parameters of mesons belonging only to the *SU*<sup>4</sup> multiplet. Let us notice that in comparison to the variant A in [29], taking into account the quark mixing in the (u, c, t) sector allowed the author to calculate the Cabibbo angle from the model and the angles *θ*<sup>2</sup> and *θ*3. Let us compare the value of the calculated angle *θ<sup>C</sup>* ¼ *θ*<sup>1</sup> þ *θ*<sup>5</sup> realized for the maximal symmetry breaking with the experimentally measured Cabibbo angle value [22].

$$
\cos \theta = 0.9737 \pm 0.0025 \tag{201}
$$

The well known values of meson masses were taken from [30]. *f <sup>π</sup>*, *f <sup>K</sup>*<sup>þ</sup> , ... were assumed as the factors in the matrix elements between one meson state and the vacuum according to PCAC, so for meson multiplets with the isospin 1/2 the factors for charged and neutral mesons are the same [31]. There exist many conjectures concerning the values of *f <sup>x</sup>*� They widely differ in magnitude, depending on the particular approach to the estimation of the matrix element <0∣*vx*∣*x*> and so far

they have no reliable experimental support. Only in the case of *f <sup>K</sup>* there is a fair consensus that the value is around 1.28 [12, 13, 31, 32]. For a calculation we took as *f <sup>D</sup>* for comparison's sake values significantly different

$$f\_D = 0.974 \quad \text{[10]} \qquad f\_D = 0.65 \quad \text{[29]} \tag{202}$$

which gives

$$\cos\left(\theta\_1 + \theta\_5\right) = 0.9799 \qquad \cos\left(\theta\_1 + \theta\_5\right) = 0.9709 \tag{203}$$

respectively, very close to the experimental value (201), as in the case of the *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> broken symmetry [16]. It seems to us that such a well agreement in both *SU*<sup>4</sup> ∗ *SU*<sup>4</sup> and *SU*<sup>6</sup> ∗ *SU*<sup>6</sup> symmetries is not accidental and the symmetry breaking is indeed realized for the maximal allowed case.

Putting (198) to (197) we find the relation connecting both phase parameters

$$\cos \delta\_2 = \xi \cos \delta\_1 \tag{204}$$

where

$$\xi = \sqrt{\frac{\chi \rho' (1 - \chi - \beta) \left(\rho' + \tau\right)}{\tau \beta' \left(1 - \tau - \rho'\right) \left(\beta' + \gamma\right)}}\tag{205}$$

The effective phase parameter *δ* ¼ *δ*<sup>1</sup> þ *δ*<sup>2</sup> is bounded for *ξ* 6¼ 1. Indeed, the Eq. (204) has solution only for

$$|\delta| \begin{cases} > \arccos \frac{1}{\xi} & \text{if } (\xi > 1) \\ > \arccos \left(\xi\right) & \text{if } (\xi < 1) \end{cases} \tag{206}$$

It is worth noticing that even for *ξ* ! ∞ or *ξ* ! 0 the second and third quadrant for *δ* is still allowed.

#### **Appendix**

From the Gell-Mann Oakes Renner model for *SU*<sup>6</sup> ∗ *SU*<sup>6</sup> symmetry we obtain the following relation for masses of pseudo-scalar mesons

$$m\_a^2 f\_a^2 \delta^{ab} + \left[ \frac{dq^2}{q^2} \rho^{ab} = i < 0 \right] \overline{Q}^a, \overline{D}^b |0> \\ = \sum\_{i=1}^6 \left( \sum\_{j=1}^6 c\_{j^2-1} d\_{i^2-1, a\varepsilon} d\_{j^4-1, b\varepsilon} \right) < u^{i^2-1} >\_0 \tag{207}$$

where

$$\rho^{ab} = (2\pi)^3 \sum\_{n \neq a} \delta^4(p\_n - q) < 0\\|\overline{D}^4|n> < n|\overline{D}^b|0> \tag{208}$$

*dabc* symmetric constants of the *SU*<sup>6</sup> group, < *ui* ><sup>0</sup> - vacuum expectation value of the operator *u<sup>i</sup>* , *<sup>Q</sup><sup>i</sup>* � *<sup>Q</sup><sup>i</sup>* ¼ Ð *d*3*xV<sup>α</sup>* <sup>0</sup>ð Þ� *<sup>x</sup>* <sup>Ð</sup> *d*3 *xA<sup>α</sup>* <sup>0</sup>ð Þ *x* - the generators of the *SU*<sup>6</sup> ∗ *SU*<sup>6</sup> group

*Quarks Mixing in Chiral Symmetries DOI: http://dx.doi.org/10.5772/intechopen.95233*

$$D^{\mathfrak{a}} = \partial^{\mathfrak{a}} V\_{\mu}^{\mathfrak{a}}(\mathfrak{x}) \qquad \overline{D}^{\mathfrak{a}} = \partial^{\mathfrak{a}} A\_{\mu}^{\mathfrak{a}}(\mathfrak{x}) \qquad \qquad [\mathfrak{1}2] \tag{209}$$

Because the vacuum expectation values of operators *u<sup>i</sup>* : i = 3, 8, 15, 24, 35 and the spectral density *ρab* are proportion to the squared parameters Of symmetry breaking, they were neglected. Approximately we obtain

$$m\_a^2 f\_a^2 = \frac{1}{\sqrt{3}} \left( \sum\_{j=1}^6 c\_{j^2 - 1} d\_{j^2 - 1, a, a} \right) < u\_0 >\_0 \tag{210}$$

Because the symmetric constants of *SU*<sup>6</sup> group: *d*<sup>113</sup> ¼ *d*<sup>223</sup> ¼ *d*<sup>333</sup> ¼ 0 the masses of neutral and charged pions are not differentiated, however there is the electromagnetic mass splitting of the other meson multiplets (see Eq. (176)).

The experimental data [30] gives

$$
\Delta m\_K = m\_{K^0} - m\_{K^+} = 4.00 \text{\textdegree MeV} \qquad \Delta m\_D = m\_{D^0} - m\_{D^+} = -5.3 \text{ MeV} \tag{211}
$$

so

$$
\text{sign } \Delta m\_K = -\text{sign } \Delta m\_D \tag{212}
$$

Let us notice that from (17613) we get

$$\text{sign}\ \left(\boldsymbol{K}^{0} - \boldsymbol{K}^{+}\right) = -\text{sign}\ \left(\boldsymbol{D}^{0} - \boldsymbol{D}^{+}\right)\tag{213}$$

so the direction of the electromagnetic mass splitting by the factor *c*<sup>3</sup> responsible for this effect is consistent with the experimental data. On the other hand

$$
\Delta \text{sign } \Delta m\_{\pi} = m\_{\pi^{\circ}} - m\_{\pi^{+}} = -4.603 \,\text{MeV} \tag{214}
$$

so the electromagnetic mass splitting of pions is of the same order as kaons or D mesons. It suggests that the neglected terms in approximate formula (207) are of the order of the factor *c*3. This means that such an approximation does not generate error greater than the electromagnetic ma s splitting of pion in meson masses description.

#### **Author details**

Zbigniew Piotr Szadkowski Faculty of Physics and Applied Informatics, University of Łódź, 90-236 Pomorska 149, Łódź, Poland

\*Address all correspondence to: zbigniew.szadkowski@fis.uni.lodz.pl

© 2021 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/ by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### **References**

[1] E.S. Abers, B.W. Lee, Physics Reports 9, Issue 1 (1973)

[2] S. Weinberg, Phys. Rev. Lett. 19, 1264 (1967)

[3] A. Salam, Rev. of Modern Phys. 52, 539 (1980)

[4] R. J. Oakes, Phys. Lett. 298, 683 (1969).

[5] A. Ebrahim, Lett. Nuovo Cimento 19, 225 (1977).

[6] A. Ebrahim, Lett. Nuovo Cimento 19, 437 (1977).

[7] Z. Szadkowski, Acta Phys. Pol. B13, 247 (1982).

[8] A. Ebrahim, Phys. Lett. 698, 229 (1977).

[9] M. Gell-Mann, R. J. Oakes, B. Renner, Phys. Rev. 175, 2195 (1968).

[10] N. Cabibbo, Phys. Rev. Lett. 10, 531 (1963).

[11] S. L. Glashow, J. Iliopoulos, L. Maiani, Phys. Rev. D2, 1285 (1970).

[12] V. de Alfaro, S. Fubini, G. Furlan, C. Rossetti, Currents in Hadron Physics, North Holland Publishing Company 1973, p. 509–514.

[13] K. P. Das, N. G. Deshpande, Phys. Rev. D19, 3387 (1979).

[14] V. P. Gautam, R. Bagchi, Acta Phys. Pol. B9, 1017 (1978).

[15] H. Fritzsch, Nucl. Phys. BISS, 189 (1979).

[16] Z. Szadkowski, Acta Phys. Pol. Bl4, 23 (1983).

[17] Y. Fujimoto, Lett. Nuovo Cimento 29, 283 (1980).

[18] R. Mignani, Lett. Nuovo Cimento 28, 529 (1980).

[19] M. Kobayashi, K. Maskawa, Prog. Theor. Phys. 49, 652 (1973).

[20] N. Cabibbo, L. Maiani, Phys. Lett. 28B, 131 (1968).

[21] R. Gatto, G. Sartori, M. Tonin, Phys. Lett. 28B, 128 (1968).

[22] R. E. Shrock, S. B. Treiman, L. L. Wang, Phys. Rev. Lett. 12, 1589 (1979).

[23] B. Bagchi, V. P. Gautam, A. Nandy, Phys. Rev. 019, 3380 (1979).

[24] A. Białas, Acta Phys. Pol. B12, 897 (1981).

[25] V. Barger, W. F. Long, S. Pakvasa, Phys. Rev. Lett. 42, 1585 (1979).

[26] E. Ma, Phys. Lett. 96B, 115 (1980).

[27] K. Kleinknecht, B. Renk, Z. Phys. C-Particles and Fields 16, 7 (1983).

[28] N. H. Fuchs, Lett. Nuovo Cimento 27, 21 (1980).

[29] Z. Szadkowski, Acta Phys. Pol. B16, 1067 (1985).

[30] Particle Data Group, Phys. Lett. 1118, (1982).

[31] S. L. Adler, R. F. Dashen, Current Algebras and Applications to Particle Physics, W. A. Benjamin Inc. N. Y.- Amsterdam 1968, p. 125–142.

[32] S. Fajfer, Z. Stipcevic, Lett. Nuovo Cimento 29, 207 (1980).

#### **Chapter 3**

## The Inter-Nucleon Up-to-Down Quark Bond and Its Implications for Nuclear Binding

*Nancy Lynn Bowen*

#### **Abstract**

This paper describes an interesting and potentially significant phenomenon regarding the properties of up and down quarks within the nucleus, specifically how the possible internucleon bonding of these quarks may affect the bonding energy of the nuclear force. A very simple calculation is used, which involves a bond between two internucleon up and down quarks. This simple calculation does not specify the shape or structure for the nucleus, rather this calculation only examines the energy of all possible internucleon up-to-down bonds that may be formed within a quantum nucleus. A comparison of this calculated binding energy is made to the experimental binding energy with remarkably good results. The potential significance and implications of this noteworthy finding are discussed.

**Keywords:** nuclear binding energy, nuclear force, nuclear bond, quarks, up quark, down quark, internucleon bond, quantum

#### **1. Introduction**

The nuclear force is defined as the force which binds the protons and neutrons together within a nucleus. One of the currently accepted models of the nuclear force is the liquid drop model [1]. This model of the nuclear force uses the Weizsäcker formula to predict the binding energies of nuclides. The Weizsäcker formula is a curve-fitting formula that uses five parameters, plus one conditional logic statement, in order to achieve its results [2]. These parameters are selected to empirically curve-fit an equation to match the experimental data. The liquid drop model is considered to be a "semi-classical" model of the nuclear force, rather than a quantum model [3].

Another currently accepted model of the nuclear force is the shell model, which uses magic numbers to explain certain nuclear behavior. The nuclear shell model is similar to the electronic shell model, which describes the electrons orbiting around an atom. However, the nuclear shell model does not predict the nuclear binding energy, rather the shell model defers back to the Weizsäcker formula for binding energy calculations.

A third currently accepted model of the nuclear force is the residual chromodynamic force (RCDF) model, also known as the residual strong force model.

Before describing this residual chromodynamic force, it is useful to mention a few specifics about quantum chromodynamics (QCD). Quantum chromodynamics postulates that the three valance quarks of protons and neutrons possess an attribute called "color charge." Historically, a contradiction of the quantum mechanical basis of nucleon properties with the Pauli Exclusion Principle led to the concept of the color charge for quarks [4]. The color charges of the quarks are considered to be either red, green, or blue. The words red, green, and blue are simply the names of the color charges and do not imply any type of physically visual hue for the quarks. Also, the term "charge", when referring specifically to the color charge, is not related to electric charge, which unfortunately can often be a point of confusion. Quantum chromodynamics states that a very strong bond is formed among the three color charges of the quarks inside the nucleon [5]. Both protons and neutrons have all three colors inside the nucleon.

The residual chromodynamic force model assumes that the chromodynamic force also has a weaker *residual* force outside of the nucleon. The RCDF model states that this residual force forms an internucleon bond, binding the nucleons together. The internucleon bond is formed by the residual chromodynamic force of the quarks outside of the nucleons. This is shown, in an illustrative representation, in **Figure 1**.

In **Figure 1**, the bold black line represents the chromodynamic force inside the nucleon, and the dotted gray line represents the residual chromodynamic force between two nucleons. (Note that quarks are considered to be point-like particles. Thus this drawing is not meant to be a scaled representation of the quarks, rather it is meant for illustrative purposes only.) The residual chromodynamic bond can be between any two quarks of different colors, such as between a red and blue, a green and red, or a blue and green. The residual chromodynamic bond can be between a neutron and a neutron, a proton and a neutron, or a proton and another proton.

While the RCDF model is considered to be the mechanism for nuclear bonding, the model is unable to duplicate the experimental binding energy curve. This inability of the RCDF model to reproduce this nuclear behavior is currently attributed to the extreme difficulty of modeling the multi-body interactions of the three color charges [6, 7]. This difficulty with the derivation of the nuclear binding forces from the residual chromodynamic force model is two-fold. First, each nucleon consists of three quarks, which means that a system of two nucleons is already a sixbody problem. Second, because the chromodynamic force between quarks inside the nucleons has the feature of being very strong compared to the residual chromodynamic force outside the nucleons, this disproportionate ratio of strength makes a converging solution for the complicated mathematical calculations difficult to find. For nuclides with a small number of nucleons, the problem can be solved with

*The Inter-Nucleon Up-to-Down Quark Bond and Its Implications for Nuclear Binding DOI: http://dx.doi.org/10.5772/intechopen.94377*

brute-force computing power by putting each of the quarks into a four-dimensional lattice of discrete points: three dimensions of space and one of time. This method is known as lattice quantum chromodynamics, or lattice QCD. This brute-force method for the computer calculations in lattice QCD iterates the position of each quark by assigning an x, y, z, and t position to it, calculating the resulting forces on each quark, allowing their position to change as a result of these forces, and then iterating this procedure until a resulting converging solution is found. If a converging solution is found, then these calculations are able to determine the binding energy of the nuclide in question. These computer calculations are done through extremely complex mathematical models, often using Monte-Carlo simulations [8]. Because of these computational difficulties, modeling the binding energies of only the smallest nuclides has been achieved.

However, such calculations are computationally expensive, requiring very large computers. Because of these complications, this modeling method is not normally used as a standard nuclear physics tool [7] .Thus, the RCDF model remains largely unverified when testing its binding energy predictions against experimental data.

#### **2. Properties of up and down quarks**

Besides having the attribute of color charge, there is also another attribute of quarks called flavor. From QCD theory, we know there are six different flavors of quarks: up, down, strange, charm, top, and bottom. Of these six different flavors, only two flavors are found in the stable matter of neutrons and protons: the up and down quarks [9]. (The terms of up and down do not imply any specific orientation with regard to spatial direction, and are simply the names of these types of quarks).

An up quark has an electric charge that is +2/3 the charge of a proton, and it also contains a positive magnetic moment. The up quark has a spin of ½ and a mass of about 0.3% of the proton. The color of an up quark can be either red, green, or blue. A down quark has an electric charge that is 1/3 the charge of a proton, and it contains a negative magnetic moment, which is anti-parallel to of the spin of the nuclide. The down quark has a spin of ½, and a mass of about 0.6% of the proton. The color of a down quark can be either red, green, or blue.

The magnetic moments of an up quark is estimated to be +1.85 and the magnetic moments of a down quark is estimated to be 0.97, both in units of nuclear magnetons. The electric charges of the proton and neutron are completely contained within the quarks. The proton is comprised of two up quarks and one down quark, giving it a net charge of one (2/3 + 2/3 -1/3 = 1). The neutron is comprised of one up quark and two down quarks, giving it a net charge of zero (2/3 -1/3 -1/3 = 0). **Figure 2** illustrates these properties.

**Figure 2.** *An illustrative representation of the up and down quarks in a proton and neutron.*

**Figure 3.** *Possible bonds in the RCDF model. A bond can be formed regardless of the flavor (up or down) of the quark.*

The quarks inside of a proton and neutron have both attributes of flavor (up or down) and color (red, green, or blue). Thus, each quark inside of a proton or neutron is one of six types: up and red, up and green, up and blue, down and red, down and green, or down and blue [5]. Since both the neutron and the proton contain all three different colors, there is no difference between the proton and the neutron with regard to the attribute of color charges. The only difference in the quark characteristics between a proton or a neutron resides in the number of up and down quarks. Therefore, any bond between the different colors is also inherently a bond between some combination of the up and down quarks. Hence, the quantum assumptions that are made in the RCDF model about the possibility of an internucleon bond between the residual colors of quarks are also inherently applicable to the formation of an internucleon bond between up and down quarks.

**Figure 3** shows three possible bonds, all of which are allowed in the RCDF model: a bond between two up quarks, between two down quarks, and between an up and a down quark. In the RCDF model, as long as the bond is between different colors of quarks, the up or down flavor of the quarks, is considered relatively unimportant.

Although it is considered relatively unimportant in the RCDF model, the up or down flavor of the quarks does indeed cause an energy difference among the three types of bonds that are illustrated in **Figure 3**. If there is an internucleon bond between two up quarks or between two down quarks, then the intrinsic electromagnetic force between these quarks is repulsive. Conversely, if there is an internucleon bond between an up quark and a down quark, then the intrinsic electromagnetic force between these quarks is attractive. Among the three types of bonds shown in **Figure 3**, this inherent difference in the electromagnetic energy may cause the two repulsive bonds to be less probable or less stable, producing a situation in which the up-to-down quark bond would be more prevalent in stable matter.

#### **3. An additional constraint for internucleon quark-to-quark binding**

As mentioned previously, the color charges of the quarks contained within a nucleon do not inherently distinguish between a neutron or proton; it is only the up and down attribute of the quarks that distinguish between the two types of stable nucleons. Thus, an examination of an internucleon bond being formed only between an up and a down quark is an appropriate possibility to explore. Specifically, this additional constraint is that not only must the internucleon quark-toquark bond be between different colors, but also it must be between only an up and a down quark; specifically, it cannot be between two up quarks or two down quarks. If this quite reasonable constraint is made to the RCDF model, a quick calculation of *The Inter-Nucleon Up-to-Down Quark Bond and Its Implications for Nuclear Binding DOI: http://dx.doi.org/10.5772/intechopen.94377*

the allowed bonds can be easily made. By using the currently accepted RCDF concept of the internucleon quark-to-quark bond, and applying this additional constraint, in which bonds are only formed between up and down quarks, an interesting and potentially significant set of data emerges.

For any given nuclide, the number of internucleon up-to-down quark pairs can be determined, based on how many up and down quarks each nuclide has. This calculation, as shown in Eq. (1), is made for each nuclide.

$$Number\_{up\ quarks} = (Z \times \mathbf{2}) + (N \times \mathbf{1})$$

$$Number\_{down\ quarks} = (Z \times \mathbf{1}) + (N \times \mathbf{2})\tag{1}$$

$$Number\_{possible\ pairs} = \text{the smaller of } \text{ these two numbers}$$

For simplicity of this very quick and easy calculation, it is assumed that every bonded pair of up-to-down quarks has the same bonding energy. Thus, just for this simple calculation, the equation for the calculated binding energy (CBE) of a nonquantum nuclide is the number of internucleon up-to-down quark pairs times the binding energy per pair, as shown in Eq. (2).

$$\text{CBE} = (number\ of\ pairs) \times (biniding\ energy\ per\ bonded\ pair) \tag{2}$$

For a representative sample of stable nuclides, this information is also shown in **Table 1**. For values of mass number with two stable nuclides, such as A = 40, both stable nuclides are shown. The following information is listed:


**Figure 4** is a plot for this same a representative sample of nuclides, showing both the experimental binding energy per nucleon (EBE/A) and the non-quantum calculated binding energy per nucleon (CBE/A) for an object with a fixed energy (6.000 MeV) per bond. For this quick calculation, neither the type of bond nor the structure of these bonds comes into consideration. Simply stated, this is a



*The Inter-Nucleon Up-to-Down Quark Bond and Its Implications for Nuclear Binding DOI: http://dx.doi.org/10.5772/intechopen.94377*


#### **Table 1.**

*A, Z, N, EBE, EBE/a, # up quarks, # down quarks, # possible bonds, and non-quantum calculated binding energy (classical CBE) for a representative sample of nuclides.*

#### **Figure 4.**

*In blue, a plot of the experimental binding energy (EBE) per nucleon. In orange, a plot of the calculated binding energy (CBE) per nucleon, based on the number of possible non-quantum up-to-down quark pairs and a fixed binding energy per bonded pair.*

theory-independent calculation of the number of possible bonded pairs times a fixed binding energy per bonded pair.

#### **4. Quantum considerations**

A nucleus is a quantum object, and being so, certain quantum rules must apply. A known phenomenological feature of the nuclear force is the QCD hard-core repulsion. The hard-core repulsion states that nucleons, such as a proton or neutron, cannot overlap in their spatial location [11, 12]. If too many bonds are formed for either <sup>2</sup> H or <sup>3</sup> H or <sup>3</sup> He, overlap will occur. This overlap is illustrated in **Figure 5**.

To prevent this overlap, hydrogen <sup>2</sup> H can have only one bond instead of two or three. Similarly, helium <sup>3</sup> He (as well as hydrogen <sup>3</sup> H) can have only three bonds instead of four or five. Three other nuclides are subject to this constraint, those with odd-odd configurations: <sup>6</sup> Li, 10B, and 14N. Specifically, the odd neutron and the odd proton cannot bond twice to either each other or to another nucleon. Other stable nuclides are not affected by the application of this rule, since there are enough nucleons to prevent an overlap from occurring for the larger nuclides.

*The Inter-Nucleon Up-to-Down Quark Bond and Its Implications for Nuclear Binding DOI: http://dx.doi.org/10.5772/intechopen.94377*

#### **Figure 5.**

*An illustration of the overlapping nucleons if too many bonds are attempted. For the nuclide <sup>2</sup> H, if two or three bonds are attempted, indicated by the red bonds, an overlap occurs. Similarly, <sup>3</sup> He can form only three bonds. If four or five are attempted, an overlap occurs in three dimensions.*

Quantum mechanics also states there can be no net electric dipole moment for the nuclide [13, 14]. For this second quantum rule, three more bonds must be subtracted from the number of bonds available, in order to remove the electric dipole moment. Without stating any specific configuration for the nuclide, this reduction of bonds can be best understood from the fact that the electric charge distribution of the nuclide must not have a net asymmetry in electrical charge for any of the three spatial dimensions, x, y, or z. To prevent an electric dipole moment, a bond is broken in each of these three dimensions, so that the net charge is symmetric about the x, y, and z axes. This quantum requirement removes three of the classically-allowed bonds. This rule applies to all stable nuclides, except for the three very smallest stable nuclides, <sup>2</sup> H, <sup>3</sup> He, and <sup>4</sup> He.

The inclusion of these two quantum rules is shown in **Table 2**. The first 8 columns of **Table 2** are similar to the first 8 columns of **Table 1**. Also shown in **Table 2** is the number of possible quantum bonds for each nuclide, taking into consideration the two above mentioned quantum rules. The last three columns of **Table 2** show the quantum calculated binding energy, the CBE/A, and the percent error of that calculated energy, as compared with the experimental binding energy.

As before for this simple calculation, the calculated binding energy is the number of bonds times a fixed energy per bond. The energy per bond is the only selected parameter; for this simple calculation, it is 6.000 MeV per bond. These plots take into consideration the quantum rules of hard-core repulsion and zero electric dipole moment. These data are plotted in **Figures 6** and **7**. In **Figure 6**, a representative sample of all of the stable nuclides is shown, out to lead 204Pb. In **Figure 7**, only the first 60 nuclides are plotted, to show the detail. As before, when there is more than one stable nuclide for a given mass number, these additional points are plotted as well.

To reiterate, this is a very quick and easy calculation, only involving a simple numerical count of quantum-allowed internucleon up-to-down quark pairs. This calculation does not specify the arrangement of the nucleons or the mechanism of the bond. It is simply a count of the quantum-allowed up-to-down quark bonds.


*Quantum Chromodynamic*


#### *The Inter-Nucleon Up-to-Down Quark Bond and Its Implications for Nuclear Binding DOI: http://dx.doi.org/10.5772/intechopen.94377*


*Quantum Chromodynamic*


*The Inter-Nucleon Up-to-Down Quark Bond and Its Implications for Nuclear Binding DOI: http://dx.doi.org/10.5772/intechopen.94377*

> **Table 2.**

*A representative sample of nuclides, showing quantum-allowed bonded up-to-down quark pairs for each nuclide.*

#### **Figure 6.**

*A plot of the experimental nuclear binding energy per nucleon (blue) and the simple quantum calculated binding energy (orange).*

#### **Figure 7.**

*A plot of the experimental nuclear binding energy per nucleon (blue) and the simple quantum calculated binding energy (orange) for only the first 60 nuclides.*

Other than being quantum, this calculation is theory independent, and as such, it is not subject to theoretical criticisms or theoretical differences of opinion.

#### **5. Discussion**

The excellent reproduction of the experimental data for these calculated results is impressive, especially considering that there is only one empirically-selected variable for this calculation, the value of 6.000 MeV for the bond energy, instead of the five empirically-selected variables for the Weizsäcker formula. This reproduction of the experimental data is especially impressive considering that other currently accepted nuclear theories cannot easily duplicate this curve.

In terms of the possible mechanism for the bond, the residual chromodynamic force between the color charges for internucleon quark-to-quark bonding is one possibility. Another possibility for this bond becomes apparent when it is recalled

#### *The Inter-Nucleon Up-to-Down Quark Bond and Its Implications for Nuclear Binding DOI: http://dx.doi.org/10.5772/intechopen.94377*

that the up quark has an electric charge +2/3 the charge of a proton, the down quark has an electric charge of 1/3 the charge of a proton, and both quarks carry a magnetic moment. These electromagnetic properties of the up and down quarks create a strong attractive electromagnetic force between the up and the down quarks; the strength of this electromagnetic force is dependent only on the minimum proximity between the up and down quarks engaged in the bond. Historically, it was believed that the strength of the electromagnetic force had an upper limit, based on the misconceptions that protons were homogeneously charged and that quarks did not exist. However, these misconceived notions are invalid when quarks, which contain all of the electric charge for the nucleons, are taken into consideration.

The internuclear quark-to-quark bond is most likely some combination of both the electromagnetic charge and the color charge of the quarks, but the relative percentages of these two contributions is not postulated here. Regardless of the relative percentages, the electromagnetic component of this bond should not be ignored–as is usually the case in current theories. When any internucleon quark-toquark bond is considered, the electromagnetic component must be taken into full account, rather than being considered relatively unimportant. A more detailed analysis of the electromagnetic contribution of this internucleon up-to-down quark bonding can easily be made by using the standard electromagnetic Eqs. A detailed analysis would include the addition of the energy due to all electric charges interacting with each other. In other words, this would be a double summation of the interaction for each electric charge of each quark with every other electric charge on all other quarks [15]. This double summation calculation would inherently include the Coulomb energy of the net repulsive electric energies among the protons.

Similarly, a more detailed electromagnetic analysis would also include the variation of the electromagnetic bond due to the vector orientation of the magnetic moments of the quarks. The energy of the magnetic moments interacting with each other should be included, which again would be a double summation for the magnetic interaction for all of the magnetic moment vectors [16]. Finally, the kinetic energy of the quantum spin of the nuclide should also be included in this more detailed binding energy calculation [17, 18]. However, for this more detailed and accurate calculation to be done, the lowest energy configuration of the nuclide must be determined and specified before the electromagnetic interaction energies can be accurately calculated.

#### **6. Conclusion**

An extremely simple calculation of the internucleon up-to-down quark bonding has been made, giving excellent results in duplicating the nuclear binding energy curve, using only one parameter rather than five. The resulting errors for nuclides going up to lead 204Pb are only few percent. The average error, going from A = 10 to A = 60, is only 2.32% with a standard deviation for that error of only 1.91%. Also, due to the inherent similarities of this concept to the currently accepted residual chromodynamic force model, with its quark-to-quark internucleon bonding, the existence of an internucleon up-to-down quark bond cannot be relegated as implausible.

An obvious implication of these results is that a significant part of the nuclear force is electromagnetic. To some, this may be an unexpected implication, but not unfeasible, especially when the electromagnetic attraction of the up-to-down quarks is considered. If one only considers, as is the case historically, that

#### *Quantum Chromodynamic*

homogenously charged protons cannot bond to other homogeneously charged protons, then the concept that the nuclear force could be partially electromagnetic is deemed implausible. However, with the understanding that the electrical charges of the up and down quarks are able to attract each other and bond to each other, and given that the RCDF allows a quark-to-quark internucleon bond to occur, such restrictions about the nuclear force being partly electromagnetic are no longer relevant.

The excellent reproduction of experimental binding energy data with only one empirically-selected variable strongly suggests that the internucleon up-to-down quark bonding is a concept that should be seriously considered and more thoroughly examined by nuclear physicists.

### **Author details**

Nancy Lynn Bowen Colorado Mountain College, Glenwood Springs, CO, USA

\*Address all correspondence to: nbowen@coloradomtn.edu

© 2020 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/ by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

*The Inter-Nucleon Up-to-Down Quark Bond and Its Implications for Nuclear Binding DOI: http://dx.doi.org/10.5772/intechopen.94377*

#### **References**

[1] Gamow G. Gamow's description of the liquid drop model. Royal Society. 1929;123:373-390.

[2] VonWeizsäcker C. Zur theorie der kernmassen. *Zeitschrift für Physik*. 1935; 96:431-458.

[3] Basdevant J, Rich J, Spiro M. Fundamentals in Nuclear Physics. New York: Springer Science + Business Media, LLC; 2005. 67 p.

[4] Greenberg O. Color Charge Degree of Freedom in Particle Physics. In: Greenberger D, Hentschel K, Weinert F, editors. Compendium of Quantum Physics. Berlin, Heidelberg: Springer; 2009. p. 109-111. DOI: https://doi.org/ 10.1007/978-3-540-70626-7\_32

[5] Eisberg R, Resnick R. Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles. New York: Wiley; 1985. 683 p.

[6] Meisner U. Modern theory of nuclear forces AD 2006. European Physics Journal. 2007;A31:397-402. DOI: 10.1140/epja/i2006-10170-1.

[7] Machleidt R. Nuclear Forces. Scholarpedia*,* 9(1):30710, 2014. DOI: 10: 4249/scholarpedia.30710.

[8] Duane S, Kennedy A, Pendleton B, Roweth D. Physics Letters B. 1987;195: 216.

[9] Griffiths J. Introduction to Elementary Particle. New York: John Wiley & Sons; 1987. 47 p.

[10] National Nuclear Data Center. NuDat 2 database [Internet]. 2020. Available from: http://www.nndc.bnl. gov/nudat2/ [Accessed 2019]

[11] Wilczek F. Particle physics: hardcore revelations. Nature. 2007;445: 156-157. DOI: 10.1038/445156a

[12] Ishii N, Aoki S, Hatsuda T. Nuclear force from lattice QCD. Physics Review Letters. 2007;99:022001. DOI: 10.1103/ PhysRevLett.99.022001

[13] McGervey J. Introduction to Modern Physics. New York: Academic Press; 1971. 479 p.

[14] Bertulani C. Nuclear Physics in a Nutshell. Princeton, New Jersy: Princeton University Press; 2007. 105 p.

[15] Owen G. Electromagnetic Theory. Boston: Allen and Bacon; 1963. 23 p.

[16] Yosida K. Theory of Magnetism. New York: Springer; 1996. 13 p.

[17] Krane K. Introductory Nuclear Physics. New York: Wiley; 1988. 144 p.

[18] McGervey J. Introduction to Modern Physics. New York: Academic Press; 1971. 498 p.

## **Chapter 4** Spectrosopic Study of Baryons

*Zalak Shah and Ajay Kumar Rai*

#### **Abstract**

Baryons are the combination of three quarks(antiquarks) configured by qqq(*qqq*). They are fermions and obey the Pauli's principal so that the total wave function must be anti-symmetric. The SU(5) flavor group includes all types of baryons containing zero, one, two or three heavy quarks. The Particle Data Froup (PDG) listed the ground states of most of these baryons and many excited states in their summary Table. The radial and orbital excited states of the baryons are important to calculate, from that the Regge trajectories will be constructed. The quantum numbers will be determined from these slopes and intersects. Thus, we can help experiments to determine the masses of unknown states. The other hadronic properties like decays, magnetic moments can also play a very important role to emphasize the baryons. It is also interesting to determine the properties of exotic baryons nowadays.

**Keywords:** baryons, potential model, mass spectra, Regge trajectories, decays

#### **1. Introduction**

The particle physics has a remarkable track record of success by discovering the basic building blocks of matter and their interactions. Everything in the observed universe is found to be made from a few basic building blocks called *fundamental particles*, governed by four fundamental forces. All of these are encapsulated in the **Standard Model (SM)**. The Standard Model has been established in early 1970s and so far it is the most precise theory ever made by mankind. In the Standard Model elementary particles are considered to be the constituents of all observed matter. These elementary particles are quarks and leptons and the force carrying particles, such as gluons and W, Z bosons. All stable matter in the universe is made from particles that belong to the first generation; any heavier particles quickly decay to the next most stable level. The concept of the quark was first proposed by Murray Gell-Mann and George Zweig in 1937. The quarks have very strong interaction with each other, that is a reason, they always stuck inside composite system. But quarks interact with leptons weakly, the best example of this is protons, neutrons and electrons of atomic nuclei. The flavored quarks combine together in various aggregates called hadrons. The Standard Model includes 12 elementary particles of half-spin known as fermions. They follow the Pauli exclusion principle and each of them have a corresponding anti-particle. There are also twelve integer-spin particles which mediate interactions between these particles known as Bosons. They obey the Bose-Einstein statistics. The classification of bosons and fermions along with hadrons are drawn in **Figure 1**.

The Quantum Chromodynamics (QCD) as the theory of strong interactions was successfully used to explore spectroscopic parameters and decay channels of hadrons during last five decades. The interaction is governed by massless spin 1 objects

**Figure 1.** *The classifications of particles.*

called gluons. Quarks inside the hadrons exchange gluons and create a very strong color force field. To conserve color charge, quarks constantly change their color by exchanging gluons with other quarks. As the quarks within a hadron get closer together, the force of containment gets weaker so that it asymptotically approaches zero for close confinement. The quarks in close confinement are completely free to move about. This condition is referred to as "asymptotic freedom". An essential requirements for the progress in hadronic physics is the full usage of present facilities and development of new ones, with a clear focous on experiments that provide geniune insight into the inner workings of QCD.

Hadron spectroscopy is a tool to reveal the dynamics of the quark interactions within the composite systems. The short-lived hadrons and missing excited states could be identified through the possible decays of the resonance state. The experimentally discovered states are listed in summary tables of Particle Data Group [1]. The worldwide experiments such as LHCb, BELLE, BARBAR, CDF, CLEO are main source of identification of heavy baryons so far and especially LHCb and Belle experiments have provided the new excited states in heavy baryon sector recently [2]. The various phenomenological approaches for spectroscopy is all about to use the potential and establish the excited resonances. These approches are, relativistic quark model, HQET, QCD Sum rules, Lattice-QCD, Regge Phenomology, and many Phenomenological models [3–8]. An overview to the current status of research in the field of baryon physics from an experimental and theoretical aspects with a view to provide motivation and scope for the present chapter. The present study covers the baryons with one heavy and two light quarks [9–13]; two heavy and one light quark [14, 15] as well as three heavy quarks [16–18]. We also like to discuss the spectroscopy of nucleons [19, 20]. The decay properties, magnetic moments and Regge Trajectories are also discussed.

#### **2. Light, heavy flavored baryons and exotics**

In the case of baryons, when three same quark combines, definitely their electric charge, spin, orbital momentum would be same. This might violates the Pauli exclusion principal stated *"no two identical fermions can be found in the same quantum state at the same time"*. However, the color quantum number of each quark is different so that the exclusion principle would not be violated. One of the most significant aspects of the baryon spectrum is the existence of almost degenerate levels of different charges which have all the characteristics of isospin multiplets, quartets, triplets, doublets and singlets. A more general charge formula that encompasses all these nearly degenerate multiplets is

*Spectrosopic Study of Baryons DOI: http://dx.doi.org/10.5772/intechopen.97639*

$$Q = I\_3 + \frac{1}{2}Y, \qquad \qquad Y = B + S + C/B' \tag{1}$$

where Q, *I*3, *Y*, B, C, S, B<sup>0</sup> are referred as charge, isospin, hypercharge, baryon number, charm, strangeness and bottomness,respectively. The strangeness of baryon is always negative.

The baryons are strongly interacting fermions made up of three quarks and have 1 <sup>2</sup> integer spin. They obey the Pauli exclusion principle, thus the total wave function must be anti-symmetric under the interchange of any two quarks. Since all observed hadrons are color singlets, the color component of the wave function must be completely anti-symmetric.

$$\begin{array}{ll} \text{For Octet,} & \begin{cases} \overrightarrow{s}\_{u} \cdot \overrightarrow{s}\_{s} + \overrightarrow{s}\_{d} \cdot \overrightarrow{s}\_{s} = -\textbf{1} & \text{symmetric} \\ \overrightarrow{s}\_{u} \cdot \overrightarrow{s}\_{s} + \overrightarrow{s}\_{d} \cdot \overrightarrow{s}\_{s} = \textbf{0} & \text{antisymmetric} \end{cases} \\\\ \text{For Decplet,} & \begin{cases} \overrightarrow{s}\_{u} \cdot \overrightarrow{s}\_{s} + \overrightarrow{s}\_{d} \cdot \overrightarrow{s}\_{s} = \frac{\textbf{1}}{2} & \text{symmetric} \\ \overrightarrow{s}\_{u} \cdot \overrightarrow{s}\_{s} + \overrightarrow{s}\_{d} \cdot \overrightarrow{s}\_{s} = \frac{\textbf{3}}{2} & \text{antisymmetric} \end{cases} \end{array}$$

Here, *s* ! *<sup>u</sup>*, *s* ! *<sup>d</sup>* and *s* ! *<sup>s</sup>* are spin of *u*, *d*, *s* quarks.

It was considered that u, d and s are the sole elementary quarks. The symmetry group to consider three flavors of quark is done by SU(3) symmetry group. SU(3) flavor symmetry of light quarks. Each of these symmetries refers to an underlying threefold symmetry in strong interaction physics. SU(3) is the group symmetry transformations of the 3-vector wavefunction that maintain the physical constraint that the total probability for finding the particle in one of the three possible states equals 1. A Young diagram is the best way to represnts the symmetriesconsists of an array of boxes arranged in one or more left-justified rows, with each row being at least as long as the row beneath. The correspondence between a diagram and a multiplet label is: The top row juts out *α* boxes to the right past the end of the second row, the second row juts out *β* boxes to the right past the end of the third row, etc. The representation is shown in **Figures 2** and **3**.

The flavor wave functions of baryon states can then be constructed to be members of SU(3) multiplets as [5].

$$\mathfrak{Z}\otimes\mathfrak{Z}\otimes\mathfrak{Z} = \mathfrak{1}\mathfrak{O}\_{\mathbb{S}}\oplus\mathfrak{8}\_{M}\oplus\mathfrak{8}\_{M}\oplus\mathfrak{1}\_{A}$$

Murray Gell-Mann introduced the Eightfold geometrical pattern for mesons and baryons in 1962 [21]. The eight lighest baryons fit into the hexagonal array with two particle in center are called baryon octet. A triangular array with 10 particles are called the baryon decuplet. Moreover, the antibaryon octet and decuplet also exist

**Figure 2.** *The representation of the multiplets (1,0), (0,1), (0,0), (1,1), and (3,0) in SU(3) diagrams.*

**Figure 3.** *A standard young tableaux for* N *= 3.*

with opposite charge and strangeness [22]. Baryons having only *u* and *d* quarks are called nucleons *<sup>N</sup>* and <sup>Δ</sup> resonances. The *proton* and *neutron* have spin (*I*<sup>3</sup> <sup>¼</sup> <sup>1</sup> 2 ) and <sup>Δ</sup> particles have spin (*I*<sup>3</sup> <sup>¼</sup> <sup>3</sup> 2 ). The four possible four combinations of the symmetric wave function gives four <sup>Δ</sup> particles; <sup>Δ</sup>þþ (uuu, *<sup>I</sup>*<sup>3</sup> <sup>¼</sup> <sup>3</sup>*=*2), <sup>Δ</sup><sup>þ</sup> (uud, *<sup>I</sup>*<sup>3</sup> =1/2), <sup>Δ</sup><sup>0</sup> (udd, *I*<sup>3</sup> =1/2) and Δ� (ddd, *I*<sup>3</sup> =3/2). Particles with combination of *u*, *d* and *s* quarks are called hyperons; Λ, Σ, Ξ and Ω. While discussing heavy sector baryons, we need baryons having heavy quark(s) combination. Any *s* quark(s) of hyperon baryons can be replaced by heavy quark (c, b) in heavy baryon particles. The added heavy quark(s) will be added to the suffix of the particular baryon. The particles are also depend on isospin quantum number, such that Σ and Ξ baryons have isospin triplets and doublets, respectively. Λ*<sup>c</sup>* and Σ*<sup>c</sup>* (Λ*<sup>b</sup>* and Σ*b*) are formed by replacing one *s* quark. For Ξ baryon, replacement of one s quark gives Ξ*<sup>c</sup>* (Ξ*b*) and the particles Ξ*cc*, Ξ*bb* and Ξ*bc* found while replacing two *s* quarks. The biggest family is found for Ω particle. Replacement of one *s* quark gives Ω*<sup>c</sup>* (Ω*b*); two *s* quarks replace to provide Ω*cc*, Ω*bb* and Ω*bc*; all three quark replacement with *s* quark give Ω*ccc*, Ω*bbb*, Ω*bbc* and Ω*ccb* particles.

SU(4) group includes all of the baryons containing zero, one, two or three heavy *Q* (charm or beauty) quarks with light u, d and s quarks. The number of particles in a multiplet, N=N(*α*, *β*, *γ*) is

$$N = \frac{(a+1)}{1} \cdot \frac{(\beta+1)}{1} \cdot \frac{(\gamma+1)}{1} \cdot \frac{(a+\beta+2)}{2} \cdot \frac{(\beta+\gamma+2)}{2} \cdot \frac{(a+\beta+\gamma+3)}{3} \tag{2}$$

It is clear from Eq. (2) that multiplets that are conjugate to one another have the same number of particles, but so can other multiplets. The multiplets (3,0,0) and (1,1,0) each have 20 particles. This multiplet structure is expected to be repeated for every combination of spin and parity which provides a very rich spectrum of baryonic states. The multiplet numerology of the tensor product of three fundamental representation is given as:

$$4 \otimes 4 \otimes 4 = 20 \oplus 20'\_1 \oplus 20'\_2 \oplus \overline{4}.\tag{3}$$

Representation shows totally symmetric 20-plet, the mixed symmetric 20<sup>0</sup> -plet and the total anti-symmetric 4 multiplet. The charm baryon multiplets are presented in **Figure 4**. The ground levels of SU(4) group multiplets are SU(3) decuplet, octet, and singlet, respectively. These baryon states can be further decomposed according to the heavy quark content inside. According to the symmetry, the heavy baryons belong to two different SU(3) flavor representations: the symmetric sextet 6*<sup>s</sup>* and anti-symmetric anti-triplet 3*A*. It can also be represented by young tableaux (refer **Figure 2**). The observed resonances of all light and heavy baryons are listed in PDG (2020) baryon summary **Table 1**.

*Spectrosopic Study of Baryons DOI: http://dx.doi.org/10.5772/intechopen.97639*

#### **Figure 4.**

*The baryon multiplets in SU(4) group. (a) the spin 3/2 20-plet, (b) the spin 1/2 20*<sup>0</sup> *-plet and (c)* 4 *singlet [1].*


#### **Table 1.**

*Our predicted baryonic states are compared with experimental unknown (JP) excited states.*

#### **2.1 Exotics: non-conventional hadrons**

Apart from the simplest pairings of quarks-anti quarks in formation of mesons and baryons, there are many observed states that do not fit into this picture. Numerous states have recently been found and some of those have exotic quark structures. Some of these exotics states are experimentally explained as *tetraquarks*

#### *Quantum Chromodynamic*

(contains two-quarks and two anti-quarks) and *pentaquarks* (contains four-quarks plus an anti-quark) states with active gluonic degrees of freedom (hybrids), and even states of pure glue (glueballs) so far. Many experiments Belle, Barabar, CLEO, BESIII, LHCb, ATLAS, CMS and DO collaborations are working on the investigation of these exotic states. The theoretical approaches such as effective field theories of QCD, various quark moels, Sum rules, Lattice QCD, etc. also preicted many states of the exotic states.

The two pentaquarks *P*<sup>þ</sup> *<sup>c</sup>* ð Þ 4380 and *P*<sup>þ</sup> *<sup>c</sup>* ð Þ 4450 , discovered in 2015 by the LHCb collaboration, in the *J=ψpK* invarnt mass distribution [23]. The newly observed states, *P*<sup>þ</sup> *<sup>c</sup>* ð Þ 4440 , *P*<sup>þ</sup> *<sup>c</sup>* ð Þ 4457 , *P*<sup>þ</sup> *<sup>c</sup>* ð Þ 4312 were investigated via different methods in 2019 by LHCb. These states are considered in various recent studies and the majority suggested as negative spin parity quantum number [24]. The investigations of pentaquark states resulted in support of different possibilities for their subsubstructures leaving their structures still ambiguous. Therefore to discriminate their sub-structure we need further theoretical and experimental investigations.

R. Jaffe obtained six-quark states built of only light u, d, and s quarks called as *dibaryon* or *hexaquark* that belong to flavor group *SU <sup>f</sup>*ð Þ3 . Using for analysis the MIT quark-bag model, Jaffe predicted existence of a H-dibaryon, i.e., a flavorsinglet and neutral six-quark uuddss bound state with isospin–spin-parity *I J<sup>P</sup>* <sup>¼</sup> 0 0<sup>þ</sup> ð Þ [25]. In the past fifteen years, new states have been observed called the XYZ states, different from the ordinary hadrons. Some of them, like the charged states, are undoubtedly exotic. Theoretical study include the phenomenological quark model to exotics, non-relativistic effective field theories and lattice QCD calculations and enormous experimental studies we can see on XYZ states.

As a hadronic molecule [26, 27], deuteron has been well-established loosely bound state of a proton and a neutron. Ideally, the large masses of the heavy baryons reduce the kinetic energy of the systems, which makes it easier to form bound states. Such a system is approximately non- relativistic. Therefore, it is very interesting to study the binding of two heavy baryons *dibaryon* and a combination of heavy baryon and an antibaryon *baryonium*.

#### **3. Specroscopic properties**

Hadron spectroscopy is a key to strong interactions in the region of quark confinement and very useful for understanding the hadron as a bound state of quarks and gluons. Any system within a standard model becomes difficult to deal considering all the interaction of quark-quark, quark-gluon and gluon-gluon. This is the reason for using constituent quark mass incorporating all the other effects in the form of some parameters. The bound state heavy baryons can be studied in the QCD motivated potential models treating to the non relativistic Quantum mechanics. A Constituent Quark Model is a modelization of a baryon as a system of three quarks or anti-quarks bound by some kind of confining interaction. The present study deals with the Hypercentral Constituent Quark Model(hCQM), an effective way to study three body systems is through consideration of Jacobi coordinates.

The hypercentral approach has been applied to solve bound states and scattering problems in many different fields of physics and chemistry. The basic idea of the hypercentral approach to three-body systems is very simple. The two relative coordinates are rewritten into a single six dimensional vector and the non-relativistic *Schrödinger* equation in the six dimensional space is solved. The potential expressed in terms of the hypercentral radial co-ordinate, takes care of the three body interactions effectively [28]. The ground states and some of the excited states of light

baryons has also been affirmed theoretically by this scheme. It is interesting to identify the mass spectrum of heavy baryons(singly, doubly and triply) in charm as well as bottom sector and then to the light sector. We consider a nonrelativistic Hamiltonian given by

$$H = \frac{P\_x^2}{2m} + V(\mathbf{x})\tag{4}$$

where, *<sup>m</sup>* <sup>¼</sup> <sup>2</sup>*mρm<sup>λ</sup> mρ*þ*m<sup>λ</sup>* , is the reduced mass and *m<sup>ρ</sup>* and *m<sup>λ</sup>* are reduced masses of

Jacobi co-ordinates *ρ* ! and *λ* ! . *x* is the six dimensional radial hyper central coordinate of the three body system. Non-relativistically, this interaction potential, *V x*ð Þ consists of a central term *Vc*ð Þ*r* and spin dependent part *VSD*ð Þ*r* . The central part *Vc*ð Þ*r* is given in terms of vector (Couloumb) plus scalar (confining) terms as

$$V\_c(r) = V\_V + V\_S = -\frac{2}{3}\frac{a\_t}{r} + \beta r^\nu \tag{5}$$

The short-distance part of the static three-quark system, arising from one-gluon exchange within baryon, is of Coulombic shape. Here, we can observe that the strong running coupling constant (*αs*) becomes smaller as we decrease the distance, the effective potential approaches the lowest order one-gluon exchange potential given in Eq. (4) as r ! 0. So, for short distances, one can use the one gluon exchange potential, taking into account the running coupling constant *αs*. We employ the Coulomb plus power potential (CPP*v*) with varying power index *ν*, as there is no definite indication for the choices of *ν* that works at different hadronic sector. The values of potential index *ν* is varying from 0.5 to 2.0; in other words, S.R (1/2), linear (1.0), 3/2 power- law (1.5) and quadratic (2.0) potentials are taken into account in case of singly heavy baryon. The hypercentral potential V(x) as the color coulomb plus power potential with first order correction and spin-dependent interactionare written as,

$$V(\mathbf{x}) = V^0(\mathbf{x}) + \left(\frac{\mathbf{1}}{m\_\rho} + \frac{\mathbf{1}}{m\_\lambda}\right) V^{(1)}(\mathbf{x}) + V\_{SD}(\mathbf{x}) \tag{6}$$

where *<sup>V</sup>*<sup>0</sup>ð Þ *<sup>x</sup>* is given by the sum of hyper Coulomb(hC) interaction and a confinement term

$$V^{(0)}(\mathbf{x}) = \frac{\boldsymbol{\pi}}{\boldsymbol{\pi}} + \beta \mathbf{x}^{\boldsymbol{\nu}} \tag{7}$$

and first order correction as similar to the one given by [29],

$$V^{(1)}(\mathbf{x}) = -C\_F C\_A \frac{a\_s^2}{4\varkappa^2} \tag{8}$$

Here, *CF* and *CA* are the Casimir charges of the fundamental and adjoint representation. For computing the mass difference between different degenerate baryonic states, we consider the spin dependent part of the usual one gluon exchange potential (OGEP). The spin-dependent part, *VSD*ð Þ *x* contains three types of the interaction terms, such as the spin–spin term *VSS*ð Þ *x* , the spin-orbit term *V<sup>γ</sup>S*ð Þ *x* and tensor term *VT*ð Þ *x* . Considering all isospin splittings, the ground and excited state masses are determined for all heavy baryon system. The radial excited states from 2S–5S and orbital excited states from 1P-5P, 1D-4D and 1F–2F are calculated using the hCQM formalism. These mass spectra can be found in our Refs. [10–18].

The correction used in the potential have its dominant effect on potential energy. In our calculation, the effect of the correction to the potential energy part is decreasing as mass of the system increasing. For example, if noticed the maximum effect of heavy Ξ baryon family then, for the radial excited states of Ξ baryons are Ξ*c*(3.0%) > Ξ*cc*(3.0%) >Ξ*b*(0.41%) >Ξ*bc*(0.3%) >Ξ*bb*(0.16%). The orbital excited states are (1.4–3.5%) for Ξ*c*, (0.1–0.4%) for Ξ*<sup>b</sup>* and in case of doubly heavy region the error is rising in order of baryons Ξ*bb* >Ξ*bc* > Ξ*cc*. Singly heavy baryons show the effect from (0.1–1.7%), Doubly heavy baryons show the effect from (0.1–1.0%) and Triply heavy baryons show the effect from (0.2–0.9%) in orbital excited states. For better idea, we shown the effect of masses with and without first order corrections in case of Σ*<sup>c</sup>* baryon (See, **Figure 5**). In the Figure, we plotted the graph of potenial index vs. mass. The radial excited states 2S–5S and the orbital excited states 1P-5P are shown for Σþþ *<sup>c</sup>* , Σ<sup>0</sup> *<sup>c</sup>* and Σ<sup>þ</sup> *<sup>c</sup>* baryons.

**Figure 5.** *Spectra of* Σ *triplets in S and P state with and without first order correction [13].*

#### *Spectrosopic Study of Baryons DOI: http://dx.doi.org/10.5772/intechopen.97639*

Recently, we calculate the masses of *N* <sup>∗</sup> and Δ resonances using the hCQM model by adding the first-order correction to the potential. The complete mass spectra with individual states graphically compared with the experimental states in **Figure 6**. We can observe that, many states are in accordance with the experimental resonances. We also predicted the *J <sup>P</sup>* values of unknown states.

#### **3.1 Regge trajectoy**

We can say that, the mass spectra of hadrons can be conveniently described through Regge trajectories. These trajectories will aid in identifying the quantum number of particular resonance states. The important three properties of Regge Trajectories are: Linearity, Divergence and Parallelism.The higher excited radial and orbital states mass calculation enable to construct the Regge trajectories in the (n, *M*<sup>2</sup> ) and (J, *M*<sup>2</sup> ) planes. These graphical representation is useful in assigning *J P* value to the experimental unkown states. We are getting almost linear, parallel and equidistance lines in each case of the baryons.

Some of the obtained masses are plotted in accordance with quantum number as well as with natural and unnatural parities. For the singly heavy baryons, the trajectory is shown for Ξ*<sup>c</sup>* doublets baryons (See, **Figure 7**). For the doubly heavy baryons, the spectra of Ξ*bc* and Ω*bc* are less determined till the date. Thus, we plotted their trajetories in **Figure 8**. The natural and unnatural parities are shown in (J, *M*<sup>2</sup> ) for all triply heavy baryons; Ω*ccc*, Ω*bbb*, Ω*bbc* and Ω*ccb* (See, **Figures 9** and **10**). The rest trajetories of all heavy baryon families in both planes can be found in our articles.

#### **3.2 Decays: strong, radiative and weak**

The particles which are known to us decay by a similar sort of dissipation. Those who decay rapidly are unstable and take a long time are metastable. Some particles, like electron, three lightest neutrinos(and their antiparticles), the photon are stable

#### **Figure 6.**

*The resonances are predicted from the first radial excited state (2S) to the orbital excited state (2F) for N* <sup>∗</sup> *baryons [19].*

**Figure 7.**

*Variation of mass for* Ξ<sup>0</sup> *<sup>c</sup> and* Ξ<sup>þ</sup> *<sup>c</sup> with different states. The (M*<sup>2</sup> ! *n) Regge trajectories for J<sup>P</sup> values* <sup>1</sup><sup>þ</sup> <sup>2</sup> *,* <sup>1</sup>� <sup>2</sup> *and* 5þ <sup>2</sup> *are shown from bottom to top. Available experimental data are also given with particle names [13].*

partices (*never decays*). The observations of decays of baryon and meson resonances afford a valuable guidance in assigning to the correct places in various symmetry schemes. The correct isotopic spin assignment is likely to be implied by the experimental branching ratio into different charge states of particles produced by the decay,while experimental decay widths provide a means of extracting phenomenological coupling constants. For the success of a particular model, it is required to produce not only the mass spectra but also the decay properties of these baryons.

**Figure 8.** *The doubly charm-beauty baryons in (M*<sup>2</sup> ! *n) plane [14, 15].*

For the success of a particular model, it is required to produce not only the mass spectra but also the decay properties of these baryons. The masses obtained from the hypercentral Constitute Quark Model (hCQM) are used to calculate the radiative and the strong decay widths. Such calculated widths are reasonably close to

other model predictions and experimental observations (where available) [30]. The effective coupling constant of the heavy baryons is small, which leads to their strong interactions perturbatively and makes it easier to understand the systems containing only light quarks. The Heavy Hadron Chiral Perturbation Theory (HHCPT) describes the strong interactions in the low-energy regime by an exchange of light Goldstone boson. Some of the strong P -wave couplings among the s-wave baryons and S-wave couplings between the s-wave and p-wave baryons are shown in **Table 2** with PDG values.

*Spectrosopic Study of Baryons DOI: http://dx.doi.org/10.5772/intechopen.97639*

#### **Figure 10.**

*Parent and daughter (J, M*<sup>2</sup>*) Regge trajectories for triply heavy charm-beauty baryons with natural (first) and unnatural (second) parities [17, 18].*

The electromagnetic properties are one of the essential key tools in understanding the internal structure and geometric shapes of hadrons. In the present study, the magnetic moments of heavy flavor and light flavor baryons are computed based on


#### **Table 2.**

*Several strong one-pion decay rates (in MeV) [30].*

the non-relativistic hypercentral constituent quark model using the spin-flavor wave functions of the constituent quark and their effective masses [11, 13]. Generally, the meaning of the constituent quark mass corresponds to the energy that the quarks have inside the color singlet hadrons, we call it as the effective mass. The magnetic moment of baryons are obtained in terms of the spin, charge and effective mass of the bound quarks. The study has been performed for all singly, doubly and triply heavy baryon systems for positive parity *J <sup>P</sup>* <sup>¼</sup> <sup>1</sup> 2 <sup>þ</sup>, <sup>3</sup><sup>þ</sup> 2 .

$$
\mu\_B = \sum\_i \left< \phi\_{\mathcal{f}} \left| \mu\_{ix} \right| \phi\_{\mathcal{f}} \right> \tag{9}
$$

where

$$\mu\_i = \frac{e\_i \sigma\_i}{2m\_i^{\ell\overline{\ell}}} \tag{10}$$

*ei* is a charge and *σ<sup>i</sup>* is the spin of the respective constituent quark corresponds to the spin flavor wave function of the baryonic state. The effective mass for each of the constituting quark *meff <sup>i</sup>* can be defined as [31].

$$m\_i^{eff} = m\_i \left( \mathbf{1} + \frac{\langle H \rangle}{\sum\_i m\_i} \right) \tag{11}$$

where, h i *<sup>H</sup>* =E+ *Vspin* � �. Using these equations, we calculate magnetic moments of singly, doubly and triply heavy baryons. The spin flavor wave function [32] *ϕsf* of all computed heavy flavor baryons are given in **Table 3**.

The electromagnetic radiative decay width is mainly the function of radiative transition magnetic moment *μ<sup>B</sup>*<sup>0</sup> *<sup>C</sup>*!*Bc* (in *μN*) and photon energy (*k*) [12] as

$$
\Gamma\_{\gamma} = \frac{k^3}{4\pi} \frac{2}{2J+1} \frac{e}{m\_p^2} \mu\_{B\_c \to B\_c'}^2 \tag{12}
$$

where *mp* is the mass of proton, *J* is the total angular momentum of the initial baryon ð Þ *Bc* . Some radiative decays are mentioned below:

*Spectrosopic Study of Baryons DOI: http://dx.doi.org/10.5772/intechopen.97639*


Weak decays of heavy hadrons play a crucial role to understand the heavy quark physics. In these decays the heavy quark acts as a spectator and the light quark inside heavy hadron decays in weak interaction [33]. The transition can be *s* ! *u* or *d* ! *u* depending on the available phase space. Since the heavy quark is spectator in such case, one can investigate the behavior of light quark system. These kind of small phase space transition could be possible in semi-electronic, semi-muonic and


#### **Table 3.**

*Magnetic moments (in nuclear magnetons) with spin-flavor wavefunctions of JP=*<sup>1</sup><sup>þ</sup> <sup>2</sup> , <sup>3</sup><sup>þ</sup> <sup>2</sup> *are listed for nucleons (light) and singly, doubly, triply(heavy) baryons.*


**Table 4.**

*Semi-electronic decays in s* ! *u transition for charm baryons are listed [13].*

non leptonic decays of the heavy baryons and mesons. We calculate here, the semielectronic decays for strange-charm heavy flavor baryons Ω*c*, Ξ*c*, Ξ*<sup>b</sup>* and Ξ*<sup>b</sup>* using our spectral parameters.

The differential decay rates for exclusive semi-electronic decays are given by [33],

$$\frac{d\Gamma}{dw} = \frac{G\_F^2 \mathcal{M}^5 |V\_{CKM}|^2}{\mathbf{1} \mathbf{9} \mathbf{2} \pi^3} \sqrt{w^2 - \mathbf{1}} P(w) \tag{13}$$

where *P w*ð Þ contains the hadronic and leptonic tensor. Assuming that the form factors are slowly varying functions of the kinematic variables, we may replace all form factors by their values at variable *w*=1. The calculated semi-electronic decays for Ξ*c*, Ω*c*, Ξ*<sup>b</sup>* and Ω*<sup>b</sup>* baryons are listed in **Table 4**. We can observe that our results are in accorance with ref. [33] for singly heavy baryons.

#### **4. Current cenario in the field of baryons**

Baryons with heavy quarks provide a beautiful laboratory to test our ideas of QCD. As the heavy quarks mass increases its motion decreases and the baryons properties are increasingly governed by the dynamics of the light quark and approach a universal limit. As we discussed, many theoretically approaches are calculating and obtaning the masses and decay widths of heavy baryons. We study the mass spectroscopy of ligt and heavy baryons and their properties. A few number of excited states for the singly heavy baryons have also been reported along with their ground states. The singly charmed baryons are, Λ*c*ð Þ 2286 <sup>þ</sup>, Λ*c*ð Þ 2595 <sup>þ</sup>, <sup>Λ</sup>*c*ð Þ <sup>2625</sup> <sup>þ</sup>, <sup>Λ</sup>*c*ð Þ <sup>2880</sup> <sup>þ</sup>, <sup>Λ</sup>*c*ð Þ <sup>2940</sup> <sup>þ</sup>, <sup>Λ</sup>*c*ð Þ <sup>2765</sup> <sup>þ</sup>,Λ*c*ð Þ <sup>2860</sup> <sup>þ</sup>, <sup>Σ</sup>*c*ð Þ <sup>2455</sup> þþ,þ,0, <sup>Σ</sup>*c*ð Þ <sup>2520</sup> þþ,þ,0, <sup>Σ</sup>*c*ð Þ <sup>2800</sup> þþ,þ,0,Ξ*c*ð Þ <sup>2468</sup> <sup>þ</sup>,0, <sup>Ξ</sup><sup>0</sup> *<sup>c</sup>*ð Þ <sup>2580</sup> <sup>þ</sup>,0, <sup>Ξ</sup><sup>0</sup> *<sup>c</sup>*ð Þ <sup>2645</sup> <sup>þ</sup>,0, <sup>Ξ</sup>*c*ð Þ <sup>2790</sup> <sup>þ</sup>,0, <sup>Ξ</sup>*c*ð Þ <sup>2815</sup> <sup>þ</sup>,0, <sup>Ξ</sup><sup>0</sup> *<sup>c</sup>*ð Þ <sup>2930</sup> <sup>0</sup> , <sup>Ξ</sup>*c*ð Þ <sup>2980</sup> <sup>þ</sup>,0, <sup>Ξ</sup>*c*ð Þ <sup>3055</sup> <sup>þ</sup>, <sup>Ξ</sup>*c*ð Þ <sup>3080</sup> <sup>þ</sup>,0, <sup>Ξ</sup><sup>0</sup> *<sup>c</sup>*ð Þ <sup>3123</sup> <sup>þ</sup>, <sup>Ω</sup>*c*ð Þ <sup>2695</sup> <sup>0</sup> , <sup>Ω</sup>*c*ð Þ <sup>2770</sup> <sup>0</sup> . The singly beauty baryons are <sup>Λ</sup>*b*ð Þ <sup>5619</sup> <sup>0</sup> , <sup>Λ</sup>*b*ð Þ <sup>5912</sup> <sup>0</sup> , <sup>Λ</sup>*b*ð Þ <sup>5920</sup> <sup>0</sup> , <sup>Σ</sup>*b*ð Þ <sup>5811</sup> <sup>þ</sup>, <sup>Σ</sup>*b*ð Þ <sup>5816</sup> �, <sup>Σ</sup>*b*ð Þ <sup>5832</sup> <sup>þ</sup>, <sup>Σ</sup>*b*ð Þ <sup>5835</sup> �, <sup>Ξ</sup>*b*ð Þ <sup>5790</sup> �,0, <sup>Ξ</sup>*b*ð Þ <sup>5945</sup> <sup>0</sup> , Ξ*b*ð Þ 5955 �, Ξ*b*ð Þ 5935 0�. And now, LHCb experiment has identified new resonances(excited states) for singly heavy baryons (See, **Table 5**). The experimental state and masses are in first two columns. The third column shows our predicted masees and in the forth column we assign the states with *J <sup>P</sup>* values. We can observe that, apart from first radial and orbital excited states, we also have D state resonances in heavy



#### **Table 5.**

*The newly observed baryonic states are listed with the observed mass in column 1 & 2. Our predicted baryonic states are compared with J<sup>P</sup> value and masses.*

sector. According to the SU(3) symmetry we also have doubly and triply baryons in charm as well as bottom sector. Among which, the evidence of 1S state for doubly heavy baryons Ξ<sup>þ</sup> *cc* and Ξþþ *cc* are observed by the SELEX and the LHCb experiments respectively. The Belle and CDF collaboration had also observed some of the particles. The fulture experiments like Panda, Belle-II and BES-II are expected to give more results soon.

*Quantum Chromodynamic*

### **Author details**

Zalak Shah\* and Ajay Kumar Rai Department of Physics, Sardar Vallabhbhai National Institute of Technology, Surat, Gujarat, India

\*Address all correspondence to: zalak.physics@gmail.com

© 2021 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/ by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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