Spectrum of two color QCD

31/10/2008

In a preceding post (see here) I have shown how SU(2) QCD can be reduced to a theory easily amenable to perturbation treatment. This case is quite easy as, due to the gauge group, algebra is not too much involved. At the leading order one gets the following set of equations

i\gamma^0\partial_\tau q_0+\frac{1}{2\sqrt{2}}\phi_0\Sigma q_0=0

\partial_\tau^2\phi_0+\phi_0^3=0

that are very easy to solve and so, we can get an understanding of the mass spectrum both for \phi and the quark field. At a first glance we easily recognize that, at this order, is clear that QCD has a chiral symmetry and this symmetry arises naturally from the strength of the coupling g.

At this stage we want just to write down the mass spectra. For the gluon field we have

m_n=(2n+1)\frac{\pi}{2K(i)}\sqrt{\sigma}

exactly as for SU(3). The only change here should be the value of the string tension \sigma.Indeed, the folowing relation should hold \sigma_{SU(2)}=\sqrt{2/3}\sigma_{SU(3)} and so, if \sigma_{SU(3)}=(440MeV)^2 one will have \sigma_{SU(2)}\approx (398MeV)^2. This is in perfect agreement with lattice evidence (see here). This lattice evidence is quite old and should be pursued further. For a quark we have the spectrum

M_n=n\frac{\pi}{2K(i)}\sqrt{\frac{\sigma}{4\pi\alpha_s}}

and as for the glueball spectrum we have n=0,1,2,\ldots and K(i) an elliptic integral. From this we say that the lowest state in the spectrum will have zero mass, the pion, and this is just a manifestation of the above approximate chiral symmetry.

The next step will be to go to higher orders and correct these results. But we see that, already at leading order, we conclude that the glueball spectrum must manifest itself at an experimental level exactly as happens to hadronic spectrum. Any other correction to it is just higher order.

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