Neutrino Oscillations in Vacuum

Atmospheric Neutrinos

Neutrinos get produced via cosmic rays (accelerated protons, He) interacting with the atmosphere.

$R_{\frac{\mu }{e}}=\frac{N_{{\nu }_{\mu }}+N_{{\bar{\nu }}_{\mu }}}{N_{{\nu }_e}+N_{{\bar{\nu }}_e}}\sim 2$

Two Neutrino Oscillation

A neutrino has 3 flavours (${\nu }_e,{\nu }_{\mu },{\nu }_{\tau }$) and 3 mass states (${\nu }_1,{\nu }_2,{\nu }_3$).

But, the flavour of a neutrino changes depending on where the neutrino was detected (distance/energy dependence). So, the system is described by a linear superposition of the 3 mass states.

Consider a charged current interaction.

The Lagrangian for this interaction is

${\mathcal{L}}_{CC}\supset \frac{-ig}{\sqrt{2}}{\bar{l}}_{\alpha }{\gamma }^{\mu }{P}_L{\nu }_{\alpha }=\frac{-ig}{\sqrt{2}}{\bar{l}}_{\alpha }{\gamma }^{\mu }{P}_L{U}_{\alpha i}{\nu }_i$ where α (= e, μ, τ) is the flavour index and i (= 1, 2, 3) denotes the mass index.

U is the PMNS mixing matrix:

$\left(\begin{array}{c}{\nu }_e\\ {\nu }_{\mu }\\ {\nu }_{\tau }\end{array}\right)=\underset{\text{PMNS matrix}}{\underbrace{\left(\begin{array}{ccc}{u}_{e_1}& u_{e_2}& u_{e_3}\\ u_{{\mu }_1}& u_{{\mu }_2}& u_{{\mu }_3}\\ u_{{\tau }_1}& u_{{\tau }_2}& u_{{\tau }_3}\end{array}\right)}}\left(\begin{array}{c}{\nu }_1\\ {\nu }_2\\ {\nu }_3\end{array}\right)$

For example, $|{\nu }_e⟩=u_{e_1}|{\nu }_1⟩+u_{e_2}|{\nu }_2⟩+u_{e_3}|{\nu }_3⟩$

Hence, the neutrino travels as a linear superposition until it is measured. Then, the wavefunction collapses and neutrinos are measured a flavour state.

$\hat{H}\psi =i\frac{d\psi }{dt}=E\psi$

$\psi (x,t)=\varphi (x)e^{iEt}$

The wavefunction evolves in time due to the phase shift and this allows for neutrino oscillations, i.e., the phenomenon of neutrino flavour transformation.

Two Neutrino Mixing

$|{\nu }_1(t)⟩=|{\nu }_1⟩e^{i(p_1.x-E_{1}t)}$

$|{\nu }_2(t)⟩=|{\nu }_2⟩e^{i(p_2.x-E_{2}t)}$

$\left(\begin{array}{c}{\nu }_e\\ {\nu }_{\mu }\end{array}\right)=\left(\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right)\left(\begin{array}{c}{\nu }_1\\ {\nu }_2\end{array}\right)$

$U^{†}U=1_{2\times 2}$

Hence, $U=\left(\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right)$

In the two-neutrino case, consider that a ${\nu }_e$ is created at t=0. Then,

$|\psi (0)⟩=|{\nu }_e⟩=cos\theta |{\nu }_1⟩+sin\theta |{\nu }_2⟩$

At time t,

$|\psi (L,t)⟩=cos\theta |{\nu }_1⟩e^{-i{\varphi }_1}+sin\theta |{\nu }_2⟩e^{-i{\varphi }_2}$

with ${\varphi }_i=-\overrightarrow{p_i}L+E_{i}t$