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Number of quarks operator 
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#1
Jun1114, 08:58 AM

P: 328

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 [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: $$ \hat Q_cJ/\psi\rangle\sim 0\rangle+2\bar cc\rangle+\bar c\bar ccc\rangle, $$ thus giving a number of valence quarks equal to 2. What's wrong with my calculation? Thanks a lot 


#2
Jun1114, 10:29 AM

Sci Advisor
Thanks
P: 4,160




#3
Jun1114, 10:35 AM

P: 328

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! 


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