Why is There Only One Neutral Pion?

  • Thread starter Thread starter bsaucer
  • Start date Start date
  • Tags Tags
    Neutral Pion
bsaucer
Messages
29
Reaction score
0
Why is there only one neutral pion? Shouldn't there be two of them with slightly different masses, one with u/u-bar quarks, and one with d/d-bar quarks? Do neutral pions "oscillate" in some way? After all, s/s-bar mesons, c/c-bar mesons, and b/b-bar mesons are all different.
 
Physics news on Phys.org
Because the masses of the u and d quarks are not too different, there is an approximate flavor symmetry called isospin that mixes them together. When quark assignments are given to the mesons, it's done in the limit that isospin symmetry is exact. The observed particles are very nearly in specific isospin representations.

So the neutral pion is not exactly ##(u\bar{u}-d\bar{d})/\sqrt{2}## (since the d is slightly heavier, there will be a bit less of it in the linear combination), but it's very close. There are other mesons like the ##\rho## that are also mixtures of u and d that account for the other linear combinations that are allowed by isospin.

The mass of the s is also not too much larger than that of the u and d quarks. The approximate SU(3) flavor symmetry on u, d, and s is behind the Gell-Mann - Ne'eman "eightfold way" model of the hadrons. The masses of c, b, and t are large and different enough that any higher flavor symmetry is not a good approximation.
 
The other low-lying neutral mesons with spin zero (pseudoscalar like the π0) are the eta (548 MeV) and eta prime (958 MeV). All three, π0, η and η' are linear combinations of u\overline{u}, d\overline{d} and s\overline{s}:
η = (u\overline{u} + d\overline{d} - 2s\overline{s})/√6
η' = (u\overline{u} + d\overline{d} + s\overline{s})/√3
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top