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.

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user agemO