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In the quark sector, the CKM matrix is obtained from (see p.723 of Peskin QFT) $$ V_{CKM} =U_u^\dagger U_d = \begin{bmatrix} V_{ud} & V_{us} & V_{ub} \\V_{cd} & V_{cs} & V_{cb} \\ V_{td} & V_{ts} & V_{tb} \end{bmatrix} $$ where $U_u$ is a matrix of $u,c,t$ flavor to flavor matrix. $U_d$ is a matrix of $d,s,b$ flavor to flavor matrix. The $U_u$ and $U_d$ are obtained in an attempt to diagonalizing the Higgs Yukawa term to a diagonalized form as the mass eigenstates. The $ V_{CKM}$ is the weak charge current coupling to the $W$ bosons with flavor changing process.

There are 9 degrees of freedom to parametrize $V_{CKM}$ including 3 Euler angles, and additonal 6 phases. There are 6 massive quarks so we can do chiral rotations to remove the remained 6 phases, however the overall total U(1) phase cannot have an effect. So there is a left over 1 complex phase in $V_{CKM}$ which is the source of CP violation!

Question

So how can we count the CP violation phases in neutrino flavor mixing sector for $n$ sterile neutrinos? We are given with three left-handed neutrinos: $\nu_e, \nu_\mu, \nu_\tau$. Suppose we have $$N$$ left-handed neutrinos (usually $N=3$ by any QFT textbook) and additional $$n$$ sterile neutrinos $$n=0,1,2,3,4,...$$

For example, regardless the value of $n$, we should start from thinking this matrix $$ \begin{bmatrix} V_{\nu_e e} & V_{\nu_e \mu} & V_{\nu_e \tau} \\ V_{\nu_{\mu} e} & V_{\nu_{\mu} \mu} & V_{\nu_{\mu} \tau} \\ V_{\nu_{\tau} e} & V_{\nu_{\tau} \mu} & V_{\nu_{\tau} \tau} \end{bmatrix} $$ which is not the $M_{PMNS}$ for neutrinos, to be clear.

ann marie cœur
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  • The oxymoronic matrix for leptons you wrote is meaningless, as I explained. It must be diagonal. The proper PMNS one has only one phase, like CKM, if neutrinos are Dirac, and two more if they are Majorana; for Majorana, the total is N(N-1)/2, as explained in the PDG review. – Cosmas Zachos Dec 28 '20 at 19:25
  • @Cosmas Zachos, The PDF book really helps. I just want to thank you but confirm with you some subtle point. In p252 of chap 14 of 2018 PDG on neutrino masses. See below: – ann marie cœur Jan 11 '21 at 18:25
  • I haven't thought it through, but if mixed Dirac-Majorana situations is the essence of your question, maybe you should make it explicit and dramatic in your question itself. For pure Dirac, we have (N-1)(N-2)/2 phases, as for quarks, and for Majorana N(N-1)/2; so, for N=3, one phase for Dirac and 3 phases for Majorana. – Cosmas Zachos Jan 11 '21 at 19:11
  • Thanks, Sorry I will correct my typos below. I meant (-1)(−2)/2 for the usual counting of CP violation phase for Dirac. I meant (-1)(−2)/2+(N-1)=N(-1)/2 for the counting of CP violation phase for Majorana. I am considering the mixed cases as the seesaw mechanism - I thought this should be well-known but not sure whether PDG give exactly this? – ann marie cœur Jan 11 '21 at 19:19
  • (1st) I am correct that if we have =3 left-handed neutrinos and right-handed neutrinos, and if we only allowed the Dirac fermion mass between =3 left and right-handed neutrinos, then we have $$ (−1)(−2)/2=1$$ CP violating phase? [regardless the number of right-handed neutrinos as long as we have ≥1 to give a Dirac fermion mass?] – – ann marie cœur Jan 11 '21 at 19:20
  • (2nd) suppose we also allowed the Majorana fermion mass to the right-handed neutrinos, in addition to the Dirac mass between left-handed neutrinos and right-handed neutrinos; then we can get $$−1$$ more CP violating phases? – ann marie cœur Jan 11 '21 at 19:20
  • (3rd) (i) only the right-handed neutrinos can have the Majorana fermion mass. (ii) The left-handed neutrinos cannot have the Majorana fermion mass. (iii) But we can have Dirac mass between left-handed neutrinos and right-handed neutrinos.] Also we can have the seesaw mechanism between =3 left-handed L neutrinos and right-handed R neutrinos; if the L-R have the Dirac fermion mass, and the R has the Majorana fermion mass. I hope you agree? – ann marie cœur Jan 11 '21 at 19:21
  • Excuse me - The reason I asked is due to that the PDG does not say clearly whether the counting considers both Dirac fermion mass and Majorana fermion mass together at once in the PDG formula, when we have flavors, the PDG gives: $$(−1)(−2)/2$$ for Dirac and $$(−1)(−2)/2+(−1)= (−1)/2$$ for Majorana. But for seesaw, we have both Dirac and Majorana, so how do we count then? – ann marie cœur Jan 11 '21 at 19:21
  • (I updated the question to be clear) – ann marie cœur Jan 11 '21 at 19:40

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