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The QCD lagrangian has a $SU(N_{f})$ vectorial symmetry, which is explicitly broken by the difference in quark current masses. If this symmetry was exact ($m_{u}=m_{d}=m_{s}$...) the charges associated with these transformations would be conserved.

In my view, these charges are the number of each quark flavour, so if the current masses were equal, processes that "create" or "destroy" flavour numbers would be forbidden?

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You may well be misunderstanding several things. The $N_f^2-1$ flavor charges do no count flavors.

It is easiest for you to see that by imagining a world of just two flavors, u and d, namely isospin, for $N_f=2$. Then the three generators/charges effect infinitesimal flavor rotations and are just the three Pauli matrices.

So you see that $\sigma_1$ just interchanges (infinitesimally; or via a $\pi/2$ finite rotation!) us with ds, and $\sigma_3$ counts us with a + sign and ds with a - sign. By the unitarity of the construction, they preserve the total number of quarks, but scramble their flavors.

In fact, the flavor projector matrices are not traceless, as the SU(N) generators/charges should be, so they cannot be built out of linear combinations of these generators. So $\bar u u$, $\bar d d$, $\bar s s$, ..., are not conserved charges (densities), even for complete degeneracy of masses.

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  • $\begingroup$ Oh, thanks. I get it! $\endgroup$ Jul 1, 2020 at 5:30

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