Derivation Notes on Soliton Quark Meson Coupling Model

Derivation Notes on Soliton Quark Meson Coupling Model

The model
The Lagrange of a spin $$1/2$$ field with Lorentz, parity, isospin, chiral invariance and renormalizability reads (with $$m_q =0$$ )


 * $$\begin{align}

&\mathcal{L}_{\sigma}=\bar\Psi\left(i\gamma_{\mu}(\partial^{\mu}-g_vV^{\mu})-m_q-g_s(\sigma+i\gamma_5\vec\tau\cdot\vec\pi)\right)\Psi +\frac{1}{2}(\partial_{\mu}\sigma\partial^{\mu}\sigma+\partial_{\mu}\pi\partial^{\mu}\pi) \\ &-\frac{\lambda^2}{2}(\sigma^2+\pi^2-\kappa^2)^2-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}m_v^2V_{\mu}V^{\mu} \end{align}$$

A simple Lagrangian of spin $$1/2$$ field coupling to a scalar field with soliton solution reads


 * $$\begin{align}

\mathcal{L}_{FL}=\bar\Psi\left(i\gamma_{\mu}(\partial^{\mu}-m_q-g_{\phi}\phi)\right)\Psi+\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-U(\phi) \end{align}$$

with


 * $$\begin{align}

U(\phi)=\frac{1}{2}a\phi^2+\frac{1}{3!}b\phi^3+\frac{1}{4!}c\phi^4 \end{align}$$

where we identify the spin $$1/2$$ field to be of quark, and explicitly introduce a smaller quark mass. We note that it is the form of the potential of $$U(\phi)$$  guarantees the soliton solution. It is required that $$U(\phi)$$  possesses two minimums, one absolute minimum at  $$\phi=\phi_{vacuum}$$  and another at  $$\phi=0$$. $$g\phi_{vacuum}$$ should be bigger comparing to hadron mass  $$M$$ (total energy of soliton solution) so that the soliton solution is stable against decaying into plane wave solutions (with  $$\phi=\phi_{vacuum}$$  everywhere and plane wave solution for quark fields).

Our model is aiming at combining the two ingredients, namely, the Chiral symmetry provided by $$\sigma$$  and  $$\pi$$  as the solition solution provided by  $$\phi$$. A cheap shot is to directly combine the two Lagrange by simply inserting $$U(\phi)$$  into  $$\mathcal{L}_{\sigma}$$  and replace  $$\sigma$$  with  $$\phi $$, as below


 * $$\begin{align}

&\mathcal{L}_{\sigma\to\phi}=\bar\Psi\left(i\gamma_{\mu}(\partial^{\mu}-g_vV^{\mu})-m_q-g_{\phi}(\phi+i\gamma_5\vec\tau\cdot\vec\pi)\right)\Psi +\frac{1}{2}(\partial_{\mu}\phi\partial^{\mu}\phi+\partial_{\mu}\pi\partial^{\mu}\pi) \\ &-\frac{\lambda^2}{2}(\phi^2+\pi^2-\kappa^2)^2-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}m_v^2V_{\mu}V^{\mu}-U(\phi) \end{align}$$

However, this will explicitly break the chiral symmetry of the model, therefore undermine the first ingredient. For this reason, in this work, we will introduce gluonball potential into the Lagrange as a workaround, instead of explicitly incorporating $$U(\phi)$$  into our model.

Inspired by the gluonball potential of the form


 * $$\begin{align}

V'_G=-m_0^2\left(\frac{G}{G_0}\right)^2(\sigma^2+\pi^2)+\lambda_2(\sigma^2+\pi^2)^2+\frac{m_G^2}{16\Lambda^2}(4G^4\log(\frac{G}{\Lambda})-G^4) \end{align}$$

we rewrite our Lagrange as following


 * $$\begin{align}

&\mathcal{L}_{\sigma,G}=\bar\Psi\left(i\gamma_{\mu}\partial^{\mu}-m_q-g_s(\sigma+i\gamma_5\vec\tau\cdot\vec\pi)-g_G(\frac{G}{G_0})\right)\Psi +\frac{1}{2}(\partial_{\mu}\sigma\partial^{\mu}\sigma+\partial_{\mu}\pi\partial^{\mu}\pi+\partial_{\mu}G\partial^{\mu}G)-V_G \end{align}$$

with


 * $$\begin{align}

V_G=\frac{\lambda^2}{2}(\sigma^2+\pi^2-\kappa^2)^2-\epsilon\sigma+\frac{m_G^2}{16\Lambda^2}(4G^4\log(\frac{G}{\Lambda})-G^4) \end{align}$$

where we have (1) omitted the Lorentz-vector field $$V^{\mu}$$  which is not important in our approach, and (2) we did not introduce the effective scalar field  $$\phi$$  in Friedberg-Lee model, the (3) introduction of gluonball potential takes into account the potential term  $$-\frac{\lambda^2}{2}(\sigma^2+\pi^2-\kappa^2)^2$$, also the gluonball possess a kinetic energy term $$ \partial_{\mu}G\partial^{\mu}G$$  and (5) it is coupled to the quark field by the linear form  $$-g_G\bar\Psi\Psi(\frac{G}{G_0})$$  which is essential to restore the soliton solution (6) explicit Chiral violation term  -\epsilon\sigma  is essential to break the degeneracy therefore the soliton is nontopological. Following discussion of the gluonball potential using specific parameterization shows that it is possible to have solition solution due the gluonball degree of freedom.

About, the first four plots show what is expected to happen outside the solition (nucleon) bag where the quark field contribution has been omitted, the next four plots show what is expected to happen inside the solition bag, where the interaction term of quark field and gluonball gives rise to the shift of absolute vacuum.

Following this line, we would expect the following behavior of the fields. (1) Gluonball potential guarantees the solition solution, therefore $$G$$  is big outside the bag, its minimal is shifted to a smaller value inside the big due to the presence of quark field. (2) The absolute minimal of gluon field $$G_0 \sim \Lambda \sim 200 \text{MeV}$$. (3) Since the vacuum expectation value of an peseudo-scalr is zero, we assume pion field will be zero inside as well as ourside the bag. (4) The value of \sigma  field depends on the linear sigma model, so that outside the bag  \sigma=f_{\pi} = 93 \text{MeV}, inside the bag it value is determined by the modified potential.

Equation of motion and numerical results
Giving the Lagrange


 * $$\begin{align}

&\mathcal{L}_{\sigma,G}=\bar\Psi\left(i\gamma_{\mu}\partial^{\mu}-m_q-g_s(\sigma+i\gamma_5\vec\tau\cdot\vec\pi)-g_G(\frac{G}{G_0})\right)\Psi +\frac{1}{2}(\partial_{\mu}\sigma\partial^{\mu}\sigma+\partial_{\mu}\pi\partial^{\mu}\pi+\partial_{\mu}G\partial^{\mu}G) \\ &-\frac{\lambda^2}{2}(\sigma^2+\pi^2-\kappa^2)^2+\epsilon\sigma-\frac{m_G^2}{16\Lambda^2}(4G^4\log(\frac{G}{\Lambda})-G^4) \end{align}$$

we adopt semi-classical approximation which treats quark fields as wave functions, but meson fields classically, i.e., only as functions of spatial coordinates. In this case the Euler-Lagrange equation of our model reads


 * $$\begin{align}

&(-i\vec\alpha\cdot \vec\nabla+\beta(m_q+g_s(\sigma+i\gamma_5\vec\tau\cdot\vec\pi)+g_G(\frac{G}{G_0}))\Psi=\varepsilon\Psi \\ &-\nabla^2\sigma+2\lambda^2(\sigma^2+\pi^2-\kappa^2)\sigma-\epsilon=-g_s\Psi^+\beta\Psi \\ &-\nabla^2\vec\pi+2\lambda^2(\sigma^2+\pi^2-\kappa^2)\vec\pi=-ig_s\Psi^+\beta\gamma_5\vec\tau\Psi \\ &-\nabla^2G+\frac{m_G^2}{\Lambda^2}G^3\log(\frac{G}{\Lambda})=-g_G\Psi^+\beta(\frac{1}{G_0})\Psi \end{align}$$

If one only is interested in the ground state of the quark field with $$\kappa =1$$, by introducing


 * $$\begin{align}

\Psi\equiv \left( \begin{array}{c} u \\ i\left(\frac{\vec \sigma\cdot\vec r}{r}\right)v \end{array}\right)\chi_m  =\left( \begin{array}{c} u  \\ i\left(\vec \sigma\cdot\hat r\right)v \end{array}\right)\chi_m \end{align}$$

where $$\vec\sigma$$  are the Pauli matrices and $$\chi_m=\left( \begin{array}{c} 1 \\ 0 \end{array}\right)$$ or $$\left( \begin{array}{c} 0 \\ 1 \end{array}\right)$$, one may further write the set of equations in terms of quark field components. As an example, we derive explicitly all the components for the equation of motion of u quark reads, where we take $$\vec \pi \to (0,0,\pi_3)$$  and denoting  $$\pi_3$$  by  $$\tilde \pi$$, and noting that isospin degree  $$\vec \tau$$  makes $$u$$ quark different from $$d$$ quark. Therefore one has


 * $$\begin{align}

&-i\left( \begin{array}{cc} 0 & \sigma\cdot \nabla  \\ \sigma\cdot \nabla & 0 \end{array}\right) \left( \begin{array}{c} u \\ i\left(\vec \sigma\cdot\hat r\right)v \end{array}\right)  = \left( \begin{array}{c} (\sigma\cdot\nabla)(\sigma\cdot\hat r)v \\ -i(\sigma\cdot\nabla)u \end{array}\right)\\ &\left( \begin{array}{cc} 1 &   \\ & -1 \end{array}\right)g_s \sigma \left( \begin{array}{c} u \\ i\left(\vec \sigma\cdot\hat r\right)v \end{array}\right)  = \left( \begin{array}{c} g_s\sigma u \\ -i(\sigma\cdot\hat r)g_s\sigma v \end{array}\right) \\ &\left( \begin{array}{cc} 1 &   \\ & -1 \end{array}\right)g_s i \left( \begin{array}{cc} &  1  \\ 1 & \end{array}\right) \tilde \pi\left( \begin{array}{c} u \\ i\left(\vec \sigma\cdot\hat r\right)v \end{array}\right)  = \left( \begin{array}{c} -(\sigma\cdot\hat r)g_s \tilde \pi v \\ -ig_s \tilde \pi u \end{array}\right) \end{align}$$

We have been working explicitly in Pauli-Direc representation


 * $$\begin{align}

&\alpha^i=\left( \begin{array}{cc} 0 & \sigma^i \\ \sigma^i & 0  \end{array}\right) \\ &\beta=\left( \begin{array}{cc} I & 0 \\ 0 & -I  \end{array}\right) \\ &\gamma^{\mu}=(\beta,\beta\alpha) \end{align}$$

or


 * $$\begin{align}

&\gamma^0=\beta, \gamma^i=\beta\alpha^i=\left( \begin{array}{cc} 0 & \sigma^i \\ -\sigma^i & 0  \end{array}\right) \gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3=\left( \begin{array}{cc} 0 & I \\ I & 0  \end{array}\right) \\ &\sigma^1=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0  \end{array}\right) , \sigma^2=\left( \begin{array}{cc} 0 & -i \\ i & 0  \end{array}\right) , \sigma^3=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1  \end{array}\right) \end{align}$$

and using the relations


 * $$\begin{align}

(\sigma\cdot A)(\sigma\cdot B)=A\cdot B+\sigma (A\times B) \end{align}$$

one has


 * $$\begin{align}

& (\sigma\cdot \nabla)(\sigma\cdot \hat r)v=\nabla \cdot (\hat r v) = \hat r \nabla v +\frac{2}{r} v \\ & (\sigma\cdot \hat r)(\sigma\cdot \nabla)u=\hat r \cdot \nabla u \\ & (\sigma\cdot \hat r)^2=(\sigma\cdot \hat r)(\sigma\cdot \hat r)=1 \end{align}$$

where


 * $$\begin{align}

\nabla \hat r = \nabla \left( \frac{\vec r}{r} \right) = \frac{2}{r} \end{align}$$

Putting all pieces together, one gets


 * $$\begin{align}

\left( \begin{array}{c} (\sigma\cdot\nabla)(\sigma\cdot\hat r)v \\ -i(\sigma\cdot\nabla)u \end{array}\right)  + \left( \begin{array}{c} g_s\sigma u \\ -i(\sigma\cdot\hat r)g_s\sigma v \end{array}\right)  + \left( \begin{array}{c} -(\sigma\cdot\hat r)g_s \tilde \pi v \\ -ig_s \tilde \pi u \end{array}\right)  + \left( \begin{array}{c} g_G(G/G_0) u \\ -i(\sigma\cdot\hat r)g_G(G/G_0) v \end{array}\right)  =\left( \begin{array}{c} \varepsilon u  \\ i\left(\vec \sigma\cdot\hat r\right) \varepsilon v \end{array}\right) \end{align}$$

This can be simplified if one introduces


 * $$\begin{align}

\tilde \pi \equiv (\sigma\cdot \hat r) \pi \end{align}$$

so that


 * $$\begin{align}

\left( \begin{array}{c} (\sigma\cdot\nabla)(\sigma\cdot\hat r)v \\ -i(\sigma\cdot\nabla)u \end{array}\right)  + \left( \begin{array}{c} g_s\sigma u \\ -i(\sigma\cdot\hat r)g_s\sigma v \end{array}\right)  + \left( \begin{array}{c} -g_s \pi v \\ -i(\sigma\cdot\hat r)g_s \pi u \end{array}\right)  + \left( \begin{array}{c} g_G(G/G_0) u \\ -i(\sigma\cdot\hat r)g_G(G/G_0) v \end{array}\right)  =\left( \begin{array}{c} \varepsilon_1 u  \\ i\left(\vec \sigma\cdot\hat r\right) \varepsilon_1 v \end{array}\right) \end{align}$$

which is equivalent to the following set of equations for quarks, we use shorthand subscript 1 to denote up and 2 to denote down


 * $$\begin{align}

& \frac{du_1}{dr}=(-\varepsilon_1-g_s\sigma-\frac{g_GG}{G_0})v_1-g_s\pi u_1 \\ & \frac{dv_1}{dr}+\frac{2}{r}v_1=(\varepsilon_1-g_s\sigma-\frac{g_GG}{G_0})u_1+g_s\pi v_1 \\ & \frac{du_2}{dr}=(-\varepsilon_2-g_s\sigma-\frac{g_GG}{G_0})v_1+g_s\pi u_1 \\ & \frac{dv_2}{dr}+\frac{2}{r}v_2=(\varepsilon_2-g_s\sigma-\frac{g_GG}{G_0})u_2-g_s\pi v_2 \end{align}$$

to be solved consistently with the equations of three mesons, namely


 * $$\begin{align}

&\frac{d^2\sigma}{dr^2}+\frac{2}{r}\frac{d\sigma}{dr}-2\lambda^2\sigma(\sigma^2+\pi^2-\kappa^2)+\epsilon=g_s N_1(u_1^2-v_1^2)+g_sN_2(u_1^2-v_1^2) \\ &\frac{d^2\pi}{dr^2}+\frac{2}{r}\frac{d\pi}{dr}-2\lambda^2\pi(\sigma^2+\pi^2-\kappa^2)=-2g_s N_1u_1v_1+2g_sN_2u_2v_2 \\ & \frac{d^2G}{dr^2}+\frac{2}{r}\frac{dG}{dr}-\frac{m_G^2}{\Lambda^2}G^3\log(\frac{G}{\Lambda})=\frac{g_G}{G_0} N_1(u_1^2-v_1^2)+\frac{g_G}{G_0}N_2(u_1^2-v_1^2) \end{align}$$

The present result shows no stability, the result is for $$\sigma$$  field only