 Solution

Assume coherent production and detection, relativistic neutrinos, two active neutrino flavours, propagation in vacuum, negligible packet separation and no absorption. Invert the given real lepton mixing matrix to obtain $|{\nu }_{\mu }⟩=c|{\nu }_2⟩-s|{\nu }_3⟩$ and $|{\nu }_{\tau }⟩=s|{\nu }_2⟩+c|{\nu }_3⟩$, where $c=\cos \theta$, $s=\sin \theta$. With phases ${\varphi }_a=E_{a}t-{\mathbf{p}}_a\cdot \mathbf{x}$ in natural units,Using the paper's opposite sign convention conjugates this probability amplitude and leaves the probability unchanged. For equal momentum, $E_a\simeq p+m_a^2/(2p)$ and $t\simeq L$, so $\Delta \varphi =\Delta m^{2}L/(2E)$. The equal-energy convention gives the same leading result from the momentum difference; evaluating each component at a different arrival time would incorrectly double the phase.Since $\hslash c=1.97327\times 10^{-7}\mathrm{eV}\mathrm{m}$,This gives the stated factor $1.27$. Neutrino oscillation in this approximation measures $|\Delta m^2|$, not the absolute neutrino mass or its ordering.