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This question is a continuation of one of my earlier posts here and the accepted answer. Consider a two-body hypothetical decay $p^+\to e^+\gamma$ in a quantum field theory (say, a GUT). If the Lagrangian is C-violating then we have, $$\Gamma(p_L^+\to e_R^+\gamma_L)\neq\Gamma(p_L^-\to e_R^-\gamma_L)\tag{1}$$ and $$\Gamma(p_R^+\to e_L^+\gamma_R)\neq\Gamma(p_R^-\to e_L^-\gamma_R)\tag{2}.$$

  1. When we add up (1) and (2), it is quite possible that $$\Gamma(p^+\to e^+\gamma)=\Gamma(p^-\to e^-\gamma)\tag{3}$$ because the "violations" in (1) and (2) may cancel each other. In other words, (1) and (2) does not ensure that the total decay rates $\Gamma(p^+\to e^+\gamma)\neq \Gamma(p^-\to e^-\gamma)$. Is this assertion correct?

  2. If in addition, we consider CP-violation, then we have, $$\Gamma(p_L^+\to e_R^+\gamma_L)\neq\Gamma(p_R^-\to e_L^-\gamma_R)\tag{4}$$ and $$\Gamma(p_R^+\to e_L^+\gamma_R)\neq\Gamma(p_L^-\to e_R^-\gamma_L)\tag{5}.$$

Even adding (4) and (5) does not ensure $\Gamma(p^+\to e^+\gamma)\neq \Gamma(p^-\to e^-\gamma)$. Am I wrong?

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  • $\begingroup$ Did you do your Venn diagrams? No assurance of equality does not imply assurance of inequality. $\endgroup$ – Cosmas Zachos Mar 15 '17 at 19:01

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