Consider a field theory given by the following Lagrangian $$\mathcal{L}_{int}=y\overline{\psi_1}\psi_2\phi+y^*\overline{\psi}_2\psi_1\phi^\dagger$$ where $\phi$ is a complex scalar field, and $\psi_1,\psi_2$ are two different fermion fields. If the bosonic quanta of $\phi$ are represented by $B,\bar{B}$ and that of $\psi_1$ and $\psi_2$ are respectively $f_1,\bar{f}_1$ and $f_2,\bar{f}_2$ (an overbar represents antiparticle). If the coupling is complex i.e., $y\neq y^*$, can the process $B\rightarrow f_1\bar{f}_2$ and its CP-conjugate $\bar{B}\rightarrow \bar{f}_1f_2$ have the same decay rate?

I have computed the tree-level decay rates for both the processes. But both the expressions contain $|y|^2$, and if we assume mass of the particle and antiparticle are the same, then the decay rates come out to be equal i.e., $\Gamma(B\rightarrow f_1\bar{f}_2)=\Gamma(\bar{B}\rightarrow \bar{f}_1f_2)$. This implies there is no CP-violation.

My questions are: (i) Can you say apriori whether this theory will be CP-violating or CP-conserving?

(ii) Is there a chance of CP-violation when higher order Feynman diagram contributions are taken into account?

(iii) Does the existence of a complex coupling $y$ guarantee that the theory will be CP-violating?

(iv) If this theory is CP-conserving, can someone provide an idea of minimally modifying this theory so as to incorporate CP-violation in decays(or scatterings) starting from a Lagrangian.


1 Answer 1


The phase of a single constant $y$ is unphysical and may be completely eliminated by the redefinition $$ y \to t\cdot \exp(i\alpha), \quad \phi\to \phi\cdot \exp(-i\alpha)$$ which doesn't change the Lagrangian and for a given $\alpha$, $y$ will be real positive. One could also get rid of the phase by transforming the fermions.

So there can't be a CP-violation in this simple theory; there's no CP-violation at any order. In fact, one needs at least three generations of quarks (or leptons) to produce the CP-violation in this way. Then the mass matrix (or Yukawa coupling matrix) may still be transformed by several phase transformations of this kind but they're no longer enough to map a generic $U(3)$ matrix to the real $O(3)$ form, and therefore a physical CP-violating phase remains in the CKM matrix.

  • $\begingroup$ Dear @Lubos, I think, there is a very minor typo in your first equation. You meant $y\to t e^{i\alpha}$ instead of $y\to y e^{i\alpha}$. I think. $\endgroup$
    – SRS
    Commented Oct 29, 2018 at 14:55

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