Quarkonia and Dirac spectra

28/08/2008

In these days we are discussing at length the question of heavy quarkonia, that is bound states of heavy quark-antiquark and we have got a perfect agreement for their ground states assuming a potential in the form $V(r)=-\frac{\alpha_s}{r}+0.8762499705\alpha_s\sqrt{\sigma}$

being $\sigma=(0.44GeV)^2$ the string tension for Yang-Mills theory. This potential was derived here and here. We derived it in the limit of small distances and this means that excited states and states with higher angular momentum can fail to be recovered and the full potential without any approximation should be used instead. Anyhow, our derivation of ground states was in the non-relativistic approximation. We want to check here the solution of Dirac equation to get a complete confirmation of our results and, as an added bonus, we will derive also the mass of $B_c$ that is a bottom-charm meson. As said we cannot do better as to go higher excited states we need to solve Dirac equation with the full potential, an impossible task unless we recur to numerical computations.

So, let us write down the Dirac spectrum for a heavy quark-antiquark state: $M(n,j)=\frac{3}{2}m_q+\frac{m_q}{2}\frac{1}{\sqrt{1+\frac{\alpha_s^2}{(n-\delta_j)^2}}}+0.8762499705\alpha_s\sqrt{\sigma}$

being $\delta_j=j+\frac{1}{2}-\sqrt{(j+\frac{1}{2})^2-\alpha_s^2}$.

We apply this formula to charmonium, bottomonium and toponium obtaining $m_{\eta_c}=2977$ MeV

against the measured one $m_{\eta_c}=2979.8\pm 1.2$ MeV and $m_{\eta_b}=9387.5$ MeV

against the measured one $9388.9 ^{+3.1}_ {-2.3} (stat) +/- 2.7(syst)$ MeV and, finally $m_{\eta_t}=344.4$ GeV

that confirms our preceding computation. The agreement is absolutely striking. But we can do better. We consider a bottom-charm meson $B_c$ and the Dirac formula $M(n,j)=m_c+m_b-\frac{m_cm_b}{m_c+m_b}+\frac{m_cm_b}{m_c+m_b}\frac{1}{\sqrt{1+\frac{\alpha_s^2}{(n-\delta_j)^2}}}$ $+0.8762499705\alpha_s\sqrt{\sigma}$

obtaining $m_{B_c}=6.18$ GeV

against the PDG average value $6.286\pm 0.005$ GeV the error being about 2%!

Our conclusion is that, at least for the lowest states, our approximation is extremely good and confirms the constant originating from our form of gluon propagator that was the main aim of all these computations. The implications are that quarkonia could be managed with our full potential and Dirac equation on a computer, a task surely easier than solving full QCD on a lattice.

Fermions and massless scalar field

30/07/2008

As I always do to take trace of some computations I am carrying on here and there, I put them on the blog. This time I devised to solve Dirac equation using the massive solution of the massless scalar field given here and here: $\phi(x)=\mu\left(2 \over \lambda\right)^{1 \over 4}{\rm sn}(p\cdot x,i).$ $\mu$ is an arbitrary parameter, $\lambda$ the coupling and ${\rm sn}$ the snoidal Jacobi elliptical function. An arbitrary phase $\varphi$ can be added but we take it to be zero in order to keep formulas simpler. Now, we couple this field to a massless fermion field and one has to solve Dirac equation $(i\gamma\partial+\beta \phi)\psi=0$

where we have used the following Yukawa model for the coupling $L_{int}=\Gamma\bar\psi\psi\phi$

so that $\beta=\Gamma\mu\left(2 \over \lambda\right)^{1 \over 4}$. Dirac equation with such a field can be solved exactly to give $\psi(x)=e^{-iq\cdot x}e^{-i\frac{p\cdot q}{m_0^2}p\cdot x-\beta\frac{\gamma\cdot p}{m_0^2}[\ln({\rm dn}(p\cdot x,i)-i{\rm cn}(p\cdot x,i))-\ln(1-i)]}u_q$

being $m_0=\mu (\lambda /2)^{1\over 4}$ the mass acquired by the scalar field and ${\rm dn}$ and ${\rm cn}$ two other Jacobi elliptical functions. This formula says us an interesting thing, that is there is a fermion excitation with zero mass unless a mass is initially given to the fermion. Such a conclusion is reminiscent of the pion status in QCD. So, the computation may seem involved but the conclusion is quite rewarding!