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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 dckee 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'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'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 $U^{+}U=I$

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 dckee 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'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 now 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 $U^{+}U=I$

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 $U^{+}U=I$