# Unitarity of PMNS matrix

It's a theoretical demand : $$ \begin{pmatrix} \nu_{e}\\ \nu_{\mu}\\ \nu_{\tau} \end{pmatrix} = \begin{pmatrix} U_{e1} & U_{e2} & U_{e3} \\ U_{\mu1} & U_{\mu2} & U_{\mu3} \\ U_{\tau1} & U_{\tau2} & U_{\tau3} \end{pmatrix} \begin{pmatrix} \nu_{1}\\ \nu_{2}\\ \nu_{3} \end{pmatrix} $$

You know that all states are normalized, for example : $⟨\nu_{e}|\nu_{e}⟩=1=(U_{e1}^{*}⟨\nu_{1}|+U^{*}_{e2}⟨\nu_{2}|+U^{*}_{e3}⟨\nu_{3}|)( U_{e1}|\nu_{1}⟩+U_{e2}|\nu_{2}⟩+U_{e3}|\nu_{3}⟩ )$

so

$U_{e1}^{*}U_{e1}+U_{e2}^{*}U_{e2}+U_{e3}^{*}U_{e3}=1$

You can do the same for the whole matrix and find $U^{+}U=I$

EDIT : as dmckee pointed out it's a general feature in quantum mechanics, the matrix you use to change the basis (here from mass eigenstate to flavour eigenstate) must be unitary.

It really goes deeper than just a theoretical demand on a particular domain. The time-evolution operator for *any* system must be unitary, because that preserves the total probability at one. And the PMNS matrix appears as a factor in the time evolution operation for neutrino mixing.

This is important because if I start with some state and let it evolve for a while the system must afterwards exist in *some* state which means that the sum of the probabilities taken across all final states *must* come to 1. Otherwise, things can undergo---in the words of Douglas Adams---*"a sudden and gratuitous total existence failure"*.

Nor is it acceptable to start with a single state and end up with the probability to exist in one of all possible state larger than one. What would that even mean? Sudden and gratuitous extra existence?

This was probably mentioned on the first day you started studying quantum mechanics, but it is so obvious that students often don't take much note of it.