1 Introduction
In particle physics, the mesons are hadron subatomic particles, which consist of a quark and an anti-quark binding with strong interaction. Since the mesons are composed of quarks, their physical size is about one femtometre, about two thirds of the size of the proton or neutron. All the mesons are unstable, with the longest-lived lasting for only a few hundredths of a microsecond. The discharge mesons may collapse into the photon and both of these disintegrations indicate that the color is no longer a sideproduct.[1-5]
The potential between quarks and antiquarks in mesons can be considered in different ways, nonetheless the best practical potential is the one with all properties of quark-antiquark pair; In addition, the power of color is created by replacing the gluons between the quark and the antiquark, as each quark is influenced by the force which comes from another one; therefore, the central potential should be considered as the following configuration to be closer to the laboratory evidence.[6]
In order to study the two-quark systems known to be two-particle systems, also by Schrodinger and Dirac equation solution, various potentials in strong interactions between quarks have been proposed; however, the net harmonic oscillator potential plus the square of the distance term is applied in our calculation. The considered potential is:[7]
$$ V(r) = \frac{{k{r^2}}}{2} - \frac{\alpha }{{{r^2}}} + W\,. $$
To calculate the mass spectrum of meson, we are struggling to solve the non-relativistic model of Schrödingersequation with a QCD potential; nonetheless, some potential models are characterized by the flavor independence of the selected potential and also the presence of a confining term. The quark masses used in our model are the constituent quark masses that are the mass parameters appearing in the QCD Lagrangian.[7] They differ from current quark masses. Constituent quark masses are larger than current quark masses. Thus could result from gluonic condensate effects. Constituent quark masses are free parameters used to fit in potential models and therefore, various values assigned to them are observed in different studies in the literature.
In this form, this added term to the harmonic potential, (squared distance), provides us a precise calculation in the presence of short-range interactions, and the constant W is manually used for proportional masses beforehand in order to obtain an accurate outcome in the spectrum computation.
Following the information on this potential we can mention the constants as below:[7]
$$ k = 1.55 \times 10^{2} ( {{\rm MeV}^{3}} )\,,\\ {W_0} = - 4.9 \times 10^{3} ( {\rm MeV}^{2} )\,,\\ {\rm if} \quad ( {l = 0} ) \to 2\mu \alpha \ge 0.25\,. $$
Then the spin-spin, isospin-iso-spin and spin-isospin interactions potentials are introduced into the problem solving sentences and we obtain the contribution to the mass spectrum calculation of each meson as follows. The mass fragmentation in several layers of hydronium may be due to various factors; as a result, an ultrafine fission can be seen.
$$ H_{{s_1},{s_2}}^{( x )} = {A_s}{\Big( {\frac{1}{{\sqrt \pi {\sigma _s}}}} \Big)^3} {e^{{{ - {x^2}}}/{{\sigma _s^2}}}}( \vec{s}_1 \cdot \vec{s}_2 )\,, $$
$$ H_{{T_1},{T_2}}^{( x )} = {A_s}{\Big( {\frac{1}{{\sqrt \pi {\sigma _T}}}} \Big)^3} {e^{{{ - {x^2}}}/{{\sigma _T^2}}}}( \vec{T}_1 \cdot\vec{T}_2 )\,, $$
$$ H_{s,T}^{( x )} = {A_{s,T}}{\Big( {\frac{1}{{\sqrt \pi {\sigma _{s,T}}}}} \Big)^3} {e^{{{ - {x^2}}}/{{\sigma _{s,T}^2}}}}( \vec{T}_1\cdot \vec{T}_2 ) ( \vec{s}_1\cdot \vec{s}_2 )\,, $$
The perturbation potential, the sum of the high interaction potentials, is now defined, hence we solve the Schrodinger equation for the first potential then we consider the potential for the disturbance, and the meson spectrum will be procured. By comparing the results with the masses obtained for each meson, it can be determined that the quarks which are forming each meson are in single or triple state, and this procedure can be used to compute the states of each meson.Given the interaction and the superconducting interaction potential, we can consider the interaction term of a meson as follows:
$$ {H_{{ {\rm int}} }}( x ) ={H_0} + {H_1} \\ = U( x ) + H_{{s_1},{s_2}}^{( x )} + H_{{T_1},{T_2}}^{( x )} + H_{S,T}^{( x )}\,. $$
2 The Calculation of Mesons Binding Energy
Using the radial coordination we will predict the Schrödinger radial equation as:[17]
$$ - \frac{1}{{2\mu }}\frac{1}{{r^2}}\Big[\frac{\partial }{{\partial r}} \Big({r^2}\frac{\partial }{{\partial r}}\Big){\psi _{v,l}}(r)\Big] \quad + \Big[V(r) - {E_{v,l}} + \frac{{l(l + 1)}}{{2\mu {r^2}}}\Big]{\psi _{v,l}}(r) = 0\,. $$
Now, considering the variable alternation, which is a function of $ (r) $ as $ \psi ( r ) = ({1}/{r})\phi ( r ) $ and the substitution of that in Eq. (6) we can obtain the following term:
$$ \varphi^{\prime\prime}( r ) + 2\mu \Big[E - V(r) - \frac{{l(l + 1)}}{{2\mu {r^2}}}\Big]\varphi (r) = 0\,, $$
which the following mentioned function $\phi ( x )$ is proposed to be as below:
$$ \phi ( x ) = f( x )\exp \left[ {g( x )} \right]\,, $$
$$ {f_{v}}( x ) = \Pi_{c = 1}^{v}( {x - \alpha _{i}^{v}} ) \quad v = 1,2, \ldots\,, \\ {f_0}( x ) = 1 \quad v = 0\,. $$
Furthermore $f( x )$ is like Hermitian polynomials and the coefficient $ \alpha _i^v $ is obtained in terms of potential coefficient. Moreover, we assume and by importing it in mentioned Eq. (8) we can have:
$$ g + {(g')^2} + \frac{{f^{\prime\prime} + 2f'g}}{f} = 2\mu V(r) - 2\mu {E_{\upsilon ,l}} + \frac{{l( {l + 1} )}}{{{r^2}}}\,. $$
By maintaining Eq. (11), it is enough to set the potential interactions that we have here to be the same as mentioned pure oscillator in squared distance to achieve the relation below:
$$ g^{\prime\prime} + {(g')^2} + \frac{{f^{\prime\prime} + 2f'g}}{f} = \mu k{r^2} - \frac{{2\mu \alpha }}{{{r^2}}} \quad \, \, + \, 2\mu W - 2\mu {E_{\upsilon ,l}} + \frac{{l( {l + 1} )}}{{{r^2}}}\,. $$
Now in the ground state we choose $(f( r ) = 1 )$, then we paste the term: $ g( r ) = ({1}/{2})\beta {r^2} + \delta \ln r $ in Eq. (12) and we will reach the underneath equation:
$$ \beta - \frac{\delta }{{{r^2}}} + {\beta ^2}{r^2} + \frac{{{\delta ^2}}}{{{r^2}}} + 2\delta \beta \quad = \mu k{r^2} - \frac{{2\mu \alpha - l( {l + 1} )}}{{{r^2}}} + 2\mu W - 2\mu {E_{\nu ,l}}\,. $$
Considering the relation below based on the equation above:
$$ {\beta ^2} = \mu k \to \beta \sqrt {\mu k}\,, \\ {\delta ^2} - \delta = l( {l + 1} ) - 2\mu \alpha \to \\ \;\; \delta = \frac{1}{2} + \sqrt {{{\Big( {l + \frac{1}{2}} \Big)}^2} - 2\mu \alpha }\,, \\ 2\delta \beta + \beta = 2\mu W - 2\mu {E_{\upsilon ,l}} \to \\ \;\; {E_{\upsilon ,l}} = \frac{{ - 2\delta \beta - \beta + 2\mu W}}{{2\mu }}\,. $$
And taking $ l = 0 $ we will have the ground state energy for mesons as below:
$$ {E_{0,0}} = - \sqrt {\frac{k}{\mu }} \Big( {1 + \sqrt {\frac{1}{4} - 2\mu \alpha } } \Big) + W\,. $$
Table 1 Quark fitted mass (×103 MeV).[7] |
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At the beginning, we set the $W$-value to fit each meson with proportional masses, so that the empirical experiment has an appropriate proximity, then we will consider an average $W$ for all the mentioned meson structures. We can witness an example for Charmonium below:
$$ M_{c \bar{c }} = 3.182 + 3.812 - \sqrt {\frac{k}{\mu }} + w_{c \bar{c }}\,. $$
We choose the $ {w_{c \bar{c} }} = - 4356 ( {\rm MeV} ) $ so the mass can be the exact same quantity as the following number $ {M_{c \bar{c} }} = 2983 ( {\rm MeV} ) $.[21] We shall do the similar stages for other structures and we can acquire various values on $W$ which have been transitioned in Table 2.[21]
Table 2 The W constant calculated for every meson manually. |
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The last one is named $ \phi = s \bar{s } $, which is not the intentions of our computation as a reason of the light mass it has in comparison with other structures; however, it is probably a conclusive proof on our theory, which does not work on light mesons class that we will find out later on this article. Finally the interval can be easily summed up with 6 more constants, and their division into 6 equals to $ {W_{\rm refrence}} = {W_{\rm ref}} = - 4.306 \times 10^{3}( {\rm MeV} ), $ hence new masses are exposed within Table 3.
Table 3 Meson masses (×103 MeV) by the constant Wref . |
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By repeating the calculations for all the particles with this reference constant, we can observe that the obtained constant has a suitable value. Moving on this computation, we can manage to have the amounts of ground state energy for mesons either by the reference constant $(W)$ which is going to be achieved by the same direction we applied for masses although this time with Eq. (12) for ground state energy. The percentage of their very small errors both in masses and ground state energy in the next section is displayed as a table.
3 Calculation of Meson's Masses in the Excited State
In order to calculate the meson mass in the first excited state we should consider:
$$ f( r ) = r - \beta\,, \quad g( r ) = \frac{1}{2}\beta {r^2} + \delta \ln r\,. $$
In this case:
$$ g^{\prime\prime} + g' + \frac{{f' + 2f'g'}}{f} = \mu k{r^2} - \frac{{2\mu \alpha }}{{{r^2}}} \quad + 2\mu w - 2\mu {E_{\upsilon ,l}} + \frac{{l(l + 1)}}{{{r^2}}}\,, $$
which we can interpret the equation above as below:
$$ g' = \beta r + \frac{\delta }{r} \Rightarrow g^{\prime2} = {\beta ^2}{r^2} + \frac{{{\delta ^2}}}{{{r^2}}} + 2\beta \delta\,, \\ g^{\prime\prime} = \beta - \frac{\delta }{{{r^2}}}\,, \quad f'( r ) = 1 \,, \quad f^{\prime\prime} = 0\,. $$
As a result it is led to:
$${\beta ^2} = \mu k \Rightarrow \beta = \sqrt {\mu k}\,, $$
$$3\beta + 2\beta \delta = 2\mu w - 2\mu {E_{\upsilon ,l}}\,, $$
$$- {\beta ^2} - 2{\beta ^2}\delta = - 2\mu w\beta + 2\mu \beta {E_{\upsilon ,l}}\,, $$
$${\delta ^2} + \delta = - 2\mu \alpha + l( {l + 1} )\,, $$
$$- {\delta ^2} + \delta = 2\mu \alpha - l( {l + 1} )\,. $$
Henceforth, the energy relation will be written as below:
$$ 4\mu {E_{1,l}} = 4\mu w - 4\beta - 4\beta \delta\,, \quad {E_{1,l}} = w - \sqrt {\frac{k}{\mu }} - \sqrt {\frac{k}{\mu }} \delta\,. $$
Now, by multiplying a negative in Eq. (20e), and making it a perfect square, $ \delta $ would be written as below:
$$ \delta = \frac{1}{2} + \sqrt {{{\Big( {l + \frac{1}{2}} \Big)}^2} - 2\mu \alpha }\,. $$
Given the above equations, if $ l = 0 $ then $ \delta = {1}/{2} $ is derived.
$${E_{1,0}} = w - \frac{3}{2}\sqrt {\frac{k}{\mu }}\,, $$
$$ M_{q \bar{q} }^* = {m_q} + {m_{ \bar{q} }} - \frac{3}{2}\sqrt {\frac{k}{\mu }} + w\,, $$
in which quark quantities and masses can be used from the previous sections as follows:[7]
$$ k =1.55 \times 10^{2} ( {\rm MeV}^3 )\,, \\ {m_s}=2.725 \times 10^{3} ( {\rm MeV} )\,, \\ {m_c}=3.812\times 10^{3} ( {\rm MeV} )\,, \\ {m_b}=7.093\times 10^{3} ( {\rm MeV} )\,. $$
Considering a new reference constant $(W)$ is an obligation, which has been applied in previous section, toward having precise quantities on both mass and energy. Now, as before, we obtain a manual value for $W$ in order to receive the closest answer to the experimental measure. Thus for strange $B$ meson we have:[22]
$$ M_{B_{S}^{0}}^{*} = 9818 - \frac{3}{2}( {280.59} ) + {w_{s \bar{b }}}\,. $$
$$ w_{c \bar{s} }^* = - 3750.82 ( {\rm MeV} )\,, $$
$$ w_{c \bar{c} }^* = - 3557.14 ( {\rm MeV} )\,, $$
$$ w_{b \bar{b} }^* = - 3873.42 ( {\rm MeV} )\,. $$
In this part, as it has been investigated in Sec. 2, by averaging on the four obtained equations, we introduce a reference constant though in an excited state:
$$ W_{\rm ref}^* = - 3790.775 ( {\rm MeV} )\,. $$
The overriding point in this calculation is the quark masses, which should be constructed of the similar fitted cited masses as we mentioned earlier in the first table. In addition, the meson mass of the applied structures in the excited states must exist. Therefore, we are only calculating the masses and energies for mentioned structures given in Eq. (25). Based on Eq. (23) for excited state energy we shall have the values represented in Table 4.[22-25]
Table 4 The ground state meson masses according to Refs. [21{25]. |
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Regarding the values obtained from calculations in Secs. 2 and 3, the percentage of errors of listed meson's masses in the ground state and the excited state are available to be shown in Tables 5--7. (the following tables.)
4 Investigating the Spin-Spin Interaction Potential Impact
In this regard, in order to refine the solutions, we will consider the interaction sentence as a perturbation in our calculations:
$$ {H_{{S_1}\cdot{S_2}}} = {A_s}{\Big(\frac{1}{{\sqrt {\pi {\sigma _s}} }}\Big)^3}{e^{{{ - {x^2}}}/{{{\sigma ^2}_s}}}}( \vec{s}_1 \cdot \vec{s}_2)\,, $$
where $ S_{1} $ and $ S_{2} $ are the quark and antiquark spins, respectively. This interaction potential can be written as:
$$ \delta({S_1}\cdot{S_2}) = \frac{{\langle \psi |H|\psi \rangle }}{{\langle \psi |\psi \rangle }}\,, $$
which $ \langle {H_{\rm in}}\rangle $ can be derived as below:
$$ \langle {H_{\rm in}}\rangle = \int {{\psi _\gamma }{H_{\rm in}}} {\psi _\gamma }{d^3}x\,. $$
And so far we define $ \Psi ( r ) $ and $ \phi ( r ) $ as below:
$$ \Psi ( r ) = \frac{1}{r}\phi ( r )\,, \quad \phi ( r ) = f( r ){e^{g( r )}} = {r^\delta }{e^{({\beta }/{2}){r^2}}}\,. $$
Knowing $ \delta = {1}/{2} + \sqrt {{l^2} + l + {1}/{4} - 2\mu \alpha } $ and considering the $2\mu \alpha = {1}/{4}, l = 0 $ we should have $ \delta = {1}/{2} $ then we can write:
$$ \phi ( r ) \!=\! {r^{{1}/{2}}}{e^{\sqrt {({{\mu k}}/{2})} {r^2}}}\,, \quad \Psi ( r ) \!=\! {N_0}{r^{{{ - 1}}/{2}}}{e^{\sqrt {({{\mu k}}/{2})} {r^2}}}\,. $$
By replacing the achieved functions in the following mentioned interaction
$$ \delta _S^{( 1 )} = \frac{{\int _0^\infty {{\psi _\gamma }{H_s}\psi _\gamma ^*d r} }}{{\int _0^\infty {{\psi _\gamma }\psi _\gamma ^*d r} }} $$
we can reach to the main formula:
$$ \delta _S^{( 1 )} = \frac{{{A_s}( \vec{s}_1\cdot \vec{s}_2 )}}{{{{( {\sqrt {\pi {\delta _s}} } )}^3}}}\frac{{\int _0^\infty {{r^{ - 1}}} {e^{ - {{{r^2}}}/{{{\sigma _s}^2}} + \sqrt {\mu k} {r^2}}}d r}}{{\int _0^\infty {{r^{ - 1}}{e^{\sqrt {\mu k} {r^2}}}d r} }}\,. $$
In this part we need to solve the integrals which by default limitations, the answer of the numerator integral and the total fraction will be divergent; hence to unravel this problem, the potential must be set to zero with the reference constant for the meson, in order to obtain an approximate numerical value for meson radius. Because of the strong interaction among these particles, they do not interact with distance far from their radius; therefore, we consider this distance as physical infinity, (physically unlimited).
Considering $ \alpha = {1}/{{8\mu }} $ and computing this amount for charmonium. Given the numerical values for $K$, $ W_{\rm ref} $ used in Sec. 2 to obtain an approximate radius for an infinite physical value, we have set the potential for this calculated value to zero, which is followed as: $ r = 0.012. $ Thus we can write the first perturbation as below:
$$ \delta _{S}^{( 1 )} = \frac{{{A_s}( \vec{s}_1\cdot\vec{s}_2 )}}{{{{( {\sqrt {\pi {\delta _s}} } )}^3}}}\frac{{\int _0^{0.012} {{r^{ - 1}}} {e^{ - {{{r^2}}}/{{{\sigma _s}^2}} + \sqrt {\mu k} {r^2}}}d r}}{{\int _0^{0.012} {{r^{ - 1}}{e^{\sqrt {\mu k} {r^2}}}d r} }}\,. $$
The integral relation of the numerator in the fraction is solved in the variable change method, which ultimately leads us to the following answer as a perturbation given below:
$$ \frac{1}{2}\Big[ {\ln | u | + \sum _{i = 1}^\infty {\frac{{{u^i}}}{{i\cdot i!}}} } \Big]\,. $$
By taking the first sentence given in these series to be calculated, the result is as follows:
$$ \int _0^{0.012} {{r^{ - 1}}} {e^{\sqrt {\mu k} {r^2}}}d r = \frac{1}{2}[ {\ln ( {\sqrt {\mu k} {r^2}} ) + ( {\sqrt {\mu k} {r^2}} )} ]\,. $$
Considering the amounts of given constants below and substituting them in Eq. (37) we can calculate the desired term:
$${A_s} = 67.4\;{( {\rm fm} )^2} = 2696000\;{( {\rm MeV} )^2}\,, \\ {\sigma _s} = 2.87\;( {\rm fm} ) = 574\;( {\rm MeV} )\,, \\ k = 1.55 \times 10^{2} ({\rm MeV}^{3} )\,, \quad m = 200\;( {\rm MeV} )\,. $$
5 Investigating the Isospin-Isospin Interaction Potential Impact
The same applied method in earlier section can be used here both by taking the amounts $ {A_{{sI}}} $ and $ {\sigma _{I}} $ as below:
$$ {A_{{sI}}} = 51.7\;{( {\rm fm} )^2} = 2068000 {( {\rm MeV} )^2}\,, \\ {\sigma _{I}} = 3.45\;( {\rm fm} ) = 690 ( {\rm MeV} )\,. $$
The Potential perturbation can be written as the following:
$$ \delta_I^{( 1 )} = \frac{{{A_I}( \vec{I}_1 \cdot \vec{I}_2 )}}{{{{( {\sqrt {\pi {\delta _I}} } )}^3}}}\frac{{\int _0^{0.012} {{r^{ - 1}}} {e^{ - {{{r^2}}}/{{{\sigma _I}^2}} + \sqrt {\mu k} {r^2}}}d r}}{{\int _0^{0.012} {{r^{ - 1}}{e^{\sqrt {\mu k} {r^2}}}d r} }}\,. $$
6 Investigating the Spin-Isospin Interaction Potential Impact
Potential perturbation of the interaction is to be calculated as follows:
$$ \delta_I^{( 1 )} = \frac{{{A_{sI}}( \vec{I}_1 \cdot \vec{I}_2 )( \vec{S}_1 \cdot \vec{S}_2 )}}{{{{( {\sqrt {\pi {\delta _{SI}}} } )}^3}}} \\ \times \frac{{\int _0^{0.012} {{r^{ - 1}}} {e^{ - {{{r^2}}}/{{{\sigma _{SI}}^2}} + \sqrt {\mu k} {r^2}}}d r}}{{\int _0^{0.012} {{r^{ - 1}}{e^{\sqrt {\mu k} {r^2}}}d r} }}\,. $$
In which the next given fraction is our perturbation potential:
$$ H_{_{S\cdot I}}^\alpha = \frac{{{A_{\rm SI}}( \vec{I}_1\cdot\vec{I}_2 ) (\vec{S}_1\cdot\vec{S}_2 )}}{{{{( {\sqrt {\pi {\delta _{SI}}} } )}^3}}}\,. $$
Considering the amounts of $ A_{\rm SI} $ and $ \sigma_{SI} $ as below:
$$ {A_{{\rm SI}}} = - 106.2\;{( {\rm fm} )^2} = - 4248000\;{( {\rm MeV} )^2}\,, \\ {\sigma_{SI}} = 2.31\;( {\rm fm} ) = 462\;( {\rm MeV} )\,. $$
We shall have the results of entire applied interaction potentials as it is shown in Table 8.
Table 8 Investigating the values of energy in the excited states.[22_25] |
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$$ {M_{q \bar{q} }} = {m_q} + {m_{ \bar{q} }} + {E_{0,0}} + \delta _S^{( 1 )} + \delta _I^{( 1 )} + \delta _{SI}^{( 1 )}. $$
The results for the calculated mass in each structure and their error percentage with their experimental values are obtained in Table 9.
Table 9 Calculated masses for mesons, including the spin and isospin interaction potentials and their percentage error with references. |
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7 Results
The application of the mentioned potentials has led us to remark, that by considering the proportional fitted masses; in addition, by fetching the manual constants, we can approximate the outcomes to the empirical results obtained; furthermore, the values given as constants of this potential could be applied in physical computation, to calculate the masses of mesons with the main mentioned potential assistance. Howbeit, the significant matter in numerical calculations of interaction potentials such as spin-spin, isospin-isospin, and isospin-spin is the applied perturbation category which is not practical for those masons that they have been considered as light weighted class. The same approach is principally confirmed, as the main potential utilization for the single light meson was not efficient and its error rate was far higher in analogy with the empirical references, nevertheless, in the direction of this article purpose, on calculating the mass spectrum of heavy mesons, the deployed approach was an adequate route toward reducing our numerical computational error, which by the usage of the theoretical applied trend, the mentioned error is now close to zero for even some heavy mesons.