CP Violating Phase Originated from Right-handed Neutrino Mixing

Abstract

We propose an idea that the observed CP violation in neutrino oscillation is originated from the phase in right-handed neutrino mixing by seesaw mechanism. We add small breaking terms $M_{ij}N_{R_{i}}N_{R_{j}}$ in the model based on $A_{4}$ symmetry to generate non-diagonal right-handed neutrino mixing, which will give a small correction to TBM mixing via seesaw mechanism with nonzero reactor angle and CP violating phase. We estimate the CP violating phase by investigating the process of leptogenesis due to the decay of right-handed neutrino.

Appendices

$S_{4}$ SYMMETRY

$S_{4}$ group covers all permutations among four objects $(x_{1},x_{2},x_{3},x_{4})$, i.e. $(x_{1},x_{2},x_{3},x_{4})\rightarrow(x_{i},x_{j},x_{k},x_{l})$ and includes 24 elements as in literature [25]

$$a_{1}:(x_{1},x_{2},x_{3},x_{4}),\quad a_{2}:(x_{2},x_{1},x_{4},x_{3}),$$

$$a_{3}:(x_{3},x_{4},x_{1},x_{2}),\quad a_{4}:(x_{4},x_{3},x_{2},x_{1}),$$

$$b_{1}:(x_{1},x_{4},x_{2},x_{3}),\quad b_{2}:(x_{4},x_{1},x_{3},x_{2}),$$

$$b_{3}:(x_{2},x_{3},x_{1},x_{4}),\quad b_{4}:(x_{3},x_{2},x_{4},x_{1}),$$

$$c_{1}:(x_{1},x_{3},x_{4},x_{2}),\quad c_{2}:(x_{3},x_{1},x_{2},x_{4}),$$

$$c_{3}:(x_{4},x_{2},x_{1},x_{3}),\quad c_{4}:(x_{2},x_{4},x_{3},x_{1}),$$

$$d_{1}:(x_{1},x_{2},x_{4},x_{3}),\quad d_{2}:(x_{2},x_{1},x_{3},x_{4}),$$

$$d_{3}:(x_{4},x_{3},x_{1},x_{2}),\quad d_{4}:(x_{3},x_{4},x_{2},x_{1}),$$

$$e_{1}:(x_{1},x_{3},x_{2},x_{4}),\quad e_{2}:(x_{3},x_{1},x_{4},x_{2}),$$

$$e_{3}:(x_{2},x_{4},x_{1},x_{3}),\quad e_{4}:(x_{4},x_{2},x_{3},x_{1}),$$

$$f_{1}:(x_{1},x_{4},x_{3},x_{2}),\quad f_{2}:(x_{4},x_{1},x_{2},x_{3}),$$

$$f_{3}:(x_{3},x_{2},x_{1},x_{4}),\quad f_{4}:(x_{2},x_{3},x_{4},x_{1}),$$

(A.1)

which are written in cycle representation as

$$a_{1}:e,\quad a_{2}:(12)(34),$$

$$a_{3}:(13)(24),\quad a_{4}:(14)(32),$$

$$b_{1}:(234),\quad b_{2}:(124),\quad b_{3}:(132),\quad b_{4}:(314),$$

$$c_{1}:(243),\quad c_{2}:(231),\quad c_{3}:(134),\quad c_{4}:(142),$$

$$d_{1}:(34),\quad d_{2}:(12),\quad d_{3}:(1324),\quad d_{4}:(2314),$$

$$e_{1}:(23),\quad e_{2}:(2431),\quad e_{3}:(1342),\quad e_{4}:(14),$$

$$f_{1}:(24),\quad f_{2}:(1234),$$

$$f_{3}:(13),\quad f_{4}:(1432).$$

(A.2)

$S_{4}$ is cube symmetry group and the elements can also be written as

$$a_{1}=\left(\begin{matrix}1&0&0\\ 0&1&0\\ 0&0&1\end{matrix}\right),\quad a_{2}=\left(\begin{matrix}1&0&0\\ 0&-1&0\\ 0&0&-1\end{matrix}\right),$$

$$a_{3}=\left(\begin{matrix}-1&0&0\\ 0&1&0\\ 0&0&-1\end{matrix}\right),\quad a_{4}=\left(\begin{matrix}-1&0&0\\ 0&-1&0\\ 0&0&1\end{matrix}\right),$$

$$b_{1}=\left(\begin{matrix}0&0&1\\ 1&0&0\\ 0&1&0\end{matrix}\right),\quad b_{2}=\left(\begin{matrix}0&0&1\\ -1&0&0\\ 0&-1&0\end{matrix}\right),$$

$$b_{3}=\left(\begin{matrix}0&0&-1\\ 1&0&0\\ 0&-1&0\end{matrix}\right),\quad b_{4}=\left(\begin{matrix}0&0&-1\\ -1&0&0\\ 0&1&0\end{matrix}\right),$$

$$c_{1}=\left(\begin{matrix}0&1&0\\ 0&0&1\\ 1&0&0\end{matrix}\right),\quad c_{2}=\left(\begin{matrix}0&1&0\\ 0&0&-1\\ -1&0&0\end{matrix}\right),$$

$$c_{3}=\left(\begin{matrix}0&-1&0\\ 0&0&1\\ -1&0&0\end{matrix}\right),\quad c_{4}=\left(\begin{matrix}0&-1&0\\ 0&0&-1\\ 1&0&0\end{matrix}\right),$$

$$d_{1}=\left(\begin{matrix}1&0&0\\ 0&0&1\\ 0&1&0\end{matrix}\right),\quad d_{2}=\left(\begin{matrix}1&0&0\\ 0&0&-1\\ 0&-1&0\end{matrix}\right),$$

$$d_{3}=\left(\begin{matrix}-1&0&0\\ 0&0&1\\ 0&-1&0\end{matrix}\right),\quad d_{4}=\left(\begin{matrix}-1&0&0\\ 0&0&-1\\ 0&1&0\end{matrix}\right),$$

$$e_{1}=\left(\begin{matrix}0&1&0\\ 1&0&0\\ 0&0&1\end{matrix}\right),\quad e_{2}=\left(\begin{matrix}0&1&0\\ -1&0&0\\ 0&0&-1\end{matrix}\right),$$

$$e_{3}=\left(\begin{matrix}0&-1&0\\ 1&0&0\\ 0&0&-1\end{matrix}\right),\quad e_{4}=\left(\begin{matrix}0&-1&0\\ -1&0&0\\ 0&0&1\end{matrix}\right),$$

$$f_{1}=\left(\begin{matrix}0&0&1\\ 0&1&0\\ 1&0&0\end{matrix}\right),\quad f_{2}=\left(\begin{matrix}0&0&1\\ 0&-1&0\\ -1&0&0\end{matrix}\right),$$

$$f_{3}=\left(\begin{matrix}0&0&-1\\ 0&1&0\\ -1&0&0\end{matrix}\right),\quad f_{4}=\left(\begin{matrix}0&0&-1\\ 0&-1&0\\ 1&0&0\end{matrix}\right).$$

(A.3)

Generally elements are classified by symmetry operations as follows: $T$ represents the element generated by rotating around coordinate axes which includes 3 axes $x,y,z$: rotating $\pi$ for 2-order element and $\pm\frac{1}{2}\pi$ for 4-order element; $S$ represents 2-order element generated by rotating around axes connecting the midpoints in two opposite edges which include 6 axes. $R$ represents 3-order element generated by rotating around body diagonals which include 4 axes. The elements are classified by axes as follows

$$T\ axis:\{a_{2},a_{3},a_{4},d_{3},d_{4},e_{2},e_{3},f_{2},f_{4}\},$$

$$R\ axis:\{b_{1},b_{2},b_{3},b_{4},c_{1},c_{2},c_{3},c_{4}\},$$

$$S\ axis:\{d_{1},d_{2},e_{1},e_{4},f_{1},f_{3}\},$$

(A.4)

and by the order of element as

$$h=1:\{a_{1}\},$$

$$h=2:\{a_{2},a_{3},a_{4},d_{1},d_{2},e_{1},e_{4},f_{1},f_{3}\},$$

$$h=3:\{b_{1},b_{2},b_{3},b_{4},c_{1},c_{2},c_{3},c_{4}\},$$

$$h=4:\{d_{3},d_{4},e_{2},e_{3},f_{2},f_{4}\}.$$

(A.5)

Elements are classified into conjugate classes as

$$C_{1}(h=1):\{a_{1}\},$$

$$C_{3}(h=2):\{a_{2},a_{3},a_{4}\},$$

$$C_{6}(h=2):\{d_{1},d_{2},e_{1},e_{4},f_{1},f_{3}\},$$

$$C_{8}(h=3):\{b_{1},b_{2},b_{3},b_{4},c_{1},c_{2},c_{3},c_{4}\},$$

$$C_{6}^{\prime}(h=4):\{d_{3},d_{4},e_{2},e_{3},f_{2},f_{4}\}.$$

(A.6)

Permutation group or symmetric group $S_{n}$ includes two generators which could be chosen one cycle of $n$ length, e.g. $(123\cdots n)$ and one cycle of its neighboring objects e.g. $(12)$. It could be easily verified by proving $(j\quad j+1)=(123\cdots n)(j-1\quad j)(123\cdots n)^{-1}$. For example, from Eq. (32), we could choose $d_{1}$ and corresponding $n$-length cycle $f_{2}$ as generators of $S_{4}$.

Appendix B

$A_{4}$ SYMMETRY

$A_{4}$ group covers all even permutations of four objects, which is subgroup of $S_{4}$ and includes 12 elements as follows

$$a_{1}=\left(\begin{matrix}1&0&0\\ 0&1&0\\ 0&0&1\end{matrix}\right),\quad a_{2}=\left(\begin{matrix}1&0&0\\ 0&-1&0\\ 0&0&-1\end{matrix}\right),$$

$$a_{3}=\left(\begin{matrix}-1&0&0\\ 0&1&0\\ 0&0&-1\end{matrix}\right),\quad a_{4}=\left(\begin{matrix}-1&0&0\\ 0&-1&0\\ 0&0&1\end{matrix}\right),$$

$$b_{1}=\left(\begin{matrix}0&0&1\\ 1&0&0\\ 0&1&0\end{matrix}\right),\quad b_{2}=\left(\begin{matrix}0&0&1\\ -1&0&0\\ 0&-1&0\end{matrix}\right),$$

$$b_{3}=\left(\begin{matrix}0&0&-1\\ 1&0&0\\ 0&-1&0\end{matrix}\right),\quad b_{4}=\left(\begin{matrix}0&0&-1\\ -1&0&0\\ 0&1&0\end{matrix}\right),$$

$$c_{1}=\left(\begin{matrix}0&1&0\\ 0&0&1\\ 1&0&0\end{matrix}\right),\quad c_{2}=\left(\begin{matrix}0&1&0\\ 0&0&-1\\ -1&0&0\end{matrix}\right),$$

$$c_{3}=\left(\begin{matrix}0&-1&0\\ 0&0&1\\ -1&0&0\end{matrix}\right),\quad c_{4}=\left(\begin{matrix}0&-1&0\\ 0&0&-1\\ 1&0&0\end{matrix}\right).$$

(B.1)

$A_{4}$ group is the symmetry group of regular tetrahedron which generators generally are called $S,T$. $T$ represents 3-order element generated by rotating around axes passing through vertices which include 4 axes, which can be classified as: element by rotating $\frac{2}{3}\pi$ clockwise and element by rotating $\frac{2}{3}\pi$ anticlockwise. $S$ represents 2-order element generated by rotating around axes connecting the midpoints in two opposite edges which include 3 axes.

All 12 elements can be classified by axes as follows

$$S\ \text{axis: }\{a_{2},a_{3},a_{4}\},$$

$$T\ \text{axis: }\{b_{1},b_{2},b_{3},b_{4},c_{1},c_{2},c_{3},c_{4}\}.$$

(B.2)

The generators could be chosen $S=a_{2},T=b_{1}$ which satisfy $S^{2}=T^{3}=1$.

More can refer to literature [25].