# Number of quarks operator

by Einj
Tags: number, operator, quarks
 P: 321 Hi everyone. In QFT one usually defines the "number of valence quarks" of a certain particle via the operator: $$\hat N_{val}=\sum_f |\hat Q_f|,$$ where: $$\hat Q_f=\int d^3x \bar \psi_f\gamma_0\psi_f.$$ According to this definition I expected, for example, for the $J/\psi$ to have $N_{val}=0$, 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 $|J/\psi\rangle=|\bar c c\rangle$ 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
 Quote by Einj $$\hat Q_c\sim (a_{\bar c}+a_c^\dagger)(a_c+a_{\bar c}^\dagger)=a_{\bar c}a_c+a_{\bar c}a^\dagger_{\bar c}+a_ca_c^\dagger+a^\dagger_c a^\dagger_{\bar c}.$$
Isn't it $N_c = a_c^\dagger a_c$ for the quarks and $N_{\bar c} = a_{\bar c}^\dagger a_{\bar c}$ for the antiquarks? And then $N = N_c - N_{\bar c}$ is zero for the J/ψ state.
 P: 321 Oh I think I got it. You are right. The point is that in the canonical quantization you need to write the operators using the "good order" prescription. I also wrote it incorrectly, we should have: $$\hat Q_c\sim(a_{\bar c}+a_c^\dagger)(a_c+a^\dagger_{\bar c})=a_{\bar c}a_c+a_{\bar c}a_{\bar c}^\dagger+a_c^\dagger a_c+a_c^\dagger a_{\bar c}^\dagger.$$ In order to have the right order (annihilation on the left) you need to anticommute the second term, thus obtaining the extra minus sign. Thanks!