How can we calculate the PMNS matrix in general?

Question

If we only introduce Dirac mass terms for neutrinos, we have analogous to the quark mass matrices, a normal $3 \times 3$ neutrino mass matrix $M_\nu$, written in the basis: $(\nu_L^1,\nu_L^2,\nu_L^3)M_\nu (\nu_R^1,\nu_R^2,\nu_R^3)^T$.

If we introduce Majorana masses for three heavy neutrinos, we can use the usual seesaw formula:

$$M_\nu = - m M^{-1} m^T$$ and then get the normal PMNS matrix by diagonalizing the $3 \times 3$ matrix $M_\nu$. However, what about more general scenarios?

In general if both Diarc and Majorana mass terms exist, we can't simply define a $3 \times 3$ mass matrix for the neutrinos. Instead, it will be a $6\times 6$ matrix:

$$ (\nu_L^1,\nu_L^2,\nu_L^3,\nu_R^1,\nu_R^2,\nu_R^3)M_\nu (\nu_L^1,\nu_L^2,\nu_L^3,\nu_R^1,\nu_R^2,\nu_R^3)^T .$$

This is necessary since only this way we are able to include all mass terms that appear in the Lagrangian.

The matrix $U$ that diagonalizes this $6 \times 6$ matrix is also a $6 \times 6$ matrix.

For definiteness, let's consider a scenario where only one neutrino gets a heavy Majorana mass $M$. The mass matrix then reads

$$ M_\nu = \left( \begin{array}{cccccc} 0 & 0 & 0 & a & d & g \\ 0 & 0 & 0 & b & e & h \\ 0 & 0 & 0 & c & f & i \\ a & b & c & 0 & 0 & 0 \\ d & e & f & 0 & 0 & 0 \\ g & h & i & 0 & 0 & M \\ \end{array} \right) .$$

How do we know which $3\times 3$ submatrix of the big $6 \times 6$ matrix that diagonalizes $M_\nu$ corresponds to our low-energy PMNS matrix? (For simplicity, let's assume we work in a basis where the charged lepton mass matrix is already diagonal.)

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Since $M_\nu$ is symmetric we can diagonalize it with an orthogonal matrix $ U M_\nu U^T = diag$, where

$$ U=\left( \begin{array}{cc} A & B \\ C & D \\ \end{array} \right) ,$$ and

$$ diag = \left( \begin{array}{cccccc} M & 0 & 0 & 0 & 0 & 0 \\ 0 & \text{m1} & 0 & 0 & 0 & 0 \\ 0 & 0 & \text{m1} & 0 & 0 & 0 \\ 0 & 0 & 0 & \text{m2} & 0 & 0 \\ 0 & 0 & 0 & 0 & \text{m2} & 0 \\ 0 & 0 & 0 & 0 & 0 & m \\ \end{array} \right) , $$

where $M\gg m1, m2 \gg m$.

Answer 1

You start with the $W^{\pm}$ interaction term, \begin{equation} W^{-}_\mu\bar{e}_{kL}\gamma^\mu\nu_{kL}+\mathrm{h.c.} \end{equation} Whenever you diagonalize the neutrino mass term with whatever mass matrix you like you represent $\nu_{kL}$ and $\bar{\nu}_{kR}$ as superpositions of the observed light neutrinos $\nu^{(l)}_{j}$ and heavy ones $\nu^{(h)}_{j}$, \begin{equation} \nu_{kL}=U_{kL,j}^{(l)}\nu^{(l)}_{j}+U_{kL,j}^{(h)}\nu^{(h)}_{j} \end{equation} Assuming that the charged lepton mass matrix is already diagonal, this turns the interaction term above into, \begin{equation} W_\mu^{-}\bar{e}_{kL}\gamma^\mu U^{(l)}_{kL,j}\nu_j^{(l)}+W_\mu^{-}\bar{e}_{kL}\gamma^\mu U^{(h)}_{kL,j}\nu_j^{(h)}+\mathrm{h.c.} \end{equation} Then as you can see it is $U_{kL,j}^{(l)}$ that plays the role of the PMNS matrix. Similarly $U_{kL,j}^{(h)}$ plays the role of its analog for the interaction with heavy neutrinos. It's important to note that generally speaking $U_{kL,j}^{(l)}$ is not unitary by itself, only the full $6\times 6$ matrix is.