Hi everyone. In QFT one usually defines the "number of valence quarks" of a certain particle via the operator:(adsbygoogle = window.adsbygoogle || []).push({});

$$

\hat N_{val}=\sum_f |\hat Q_f|,$$

where:

$$

\hat Q_f=\int d^3x \bar \psi_f\gamma_0\psi_f.$$

According to this definition I expected, for example, for the [itex]J/\psi[/itex] to have [itex]N_{val}=0[/itex], i.e. the same quantum numbers as the vacuum. However, I can't understand what I am doing wrong. Very roughly speaking, in terms of creation/annhilation operators we have:

$$

\hat Q_c\sim (a_{\bar c}+a_c^\dagger)(a_c+a_{\bar c}^\dagger)=a_{\bar c}a_c+a_{\bar c}a^\dagger_{\bar c}+a_ca_c^\dagger+a^\dagger_c a^\dagger_{\bar c}.

$$

Hence, when applied to the particle [itex]|J/\psi\rangle=|\bar c c\rangle[/itex] is should give me:

$$

\hat Q_c|J/\psi\rangle\sim |0\rangle+2|\bar cc\rangle+|\bar c\bar ccc\rangle,

$$ thus giving a number of valence quarks equal to 2. What's wrong with my calculation?

Thanks a lot

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Number of quarks operator

Loading...

Similar Threads for Number quarks operator |
---|

I Number of fission neutrons in ENSDF |

I Excited hadrons v. fundamental particles |

I Low quantum numbers, high energy, and distance scales. |

I Baryon number nonconservation in the early universe |

**Physics Forums | Science Articles, Homework Help, Discussion**