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  Charge conjugation matrix in baryon current

+ 3 like - 0 dislike

In his paper *Calculation of baryon masses in quantum chromodynamics* ([ScienceDirect](http://www.sciencedirect.com/science/article/pii/0550321381902595)), B.L. Ioffe considers currents describing baryons. In equation (13) he gives an interpolating current for the isobar $\Delta^{++}$,
$$\eta_\mu (x) = \left( u^a(x) C \gamma_\mu u^b(x) \right) u^c(x) \varepsilon^{abc},$$
where $u^a(x)$ is an up quark field of colour $a$ and $C$ is the "charge conjugation matrix".
This current has the proper isospin of $3/2$ (three up quarks) and has the correct spin (carries one Lorentz index and implicitly one spinor index at $u^c$).

How does $C$ come into this equation though? If I understand the expression correctly, $u^a(x)$ is implicitly transposed so that the expression in the big parentheses is just a number in spinor space. So I do not think that $u^a(x) C$ is some other way of writing $\bar u^a(x)$ – in which case the current would consist of two quarks and one antiquark. What is $C$ doing then?

Furthermore, how is this $C$ related to the charge conjugation matrix (say $\tilde C$ to distinguish) introduced e.g. in *Peskin/Schroeder*? $\tilde C$ exchanges particles with antiparticles by $\tilde C a_{\mathbf p}^s \tilde C = b_{\mathbf p}^s$ and $\tilde C b_{\mathbf p}^s \tilde C = a_{\mathbf p}^s$ resulting in e.g.
$$\tilde C \psi \tilde C = - i \left(\bar\psi \gamma^0\gamma^2\right)^T$$
and seems to always appear in pairs.

asked Nov 5, 2014 in Theoretical Physics by PassfishSwordword (15 points) [ no revision ]

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