How is the mixing of neutrinos related to eigenstates? Can someone please explain to me why the following statement is true:
"Neutrino mixing phenomena arise from the noncoincidence of energy-propagation eigenstate and the weak (interaction) eigenstate bases"
This is a statement from an article about sterile neutrinos, but this particular statement is regarding neutrinos in general (active and sterile).
How was it deduced that because these eigenstates do not coincide, neutrinos must mix?
 A: Let's take the case where there are only 2 kinds of neutrino's, since this is the easiest. When neutrino's propagate, they do so as the propagating eigenstates of the Hamiltonian. However, when detecting a neutrino we detect their mass eigenstate, not the propagating one.
Let the propagating eigenstates be $|\nu_1>$ and $|\nu_2>$ with mass eigenstates
$|\nu_e>$ and $|\nu_\mu>$. One can write one set of states as a linear combination of the other one (mixing):
\begin{align} 
|\nu_e> &= \cos(\theta) |\nu_1> + \sin(\theta) |\nu_2> \\
|\nu_\mu> &= -\sin(\theta) |\nu_1> + \cos(\theta) |\nu_2>  
\end{align}
This is a simple two-state system.
Let's say we prepare a state as $|\psi(0)> = |\nu_e>$, then this state will propagate according to
\begin{equation}
|\psi(\vec{x}, t)> = \cos(\theta) |\nu_1> e^{-ip_1 \cdot x} + \sin(\theta) e^{-ip_2 \cdot x} |\nu_2>
\end{equation}
where the evolution in the propagation eigenbasis is given by plane waves.
When writing $p_i \cdot x = E_i t - p_i L$ where $L$ is the traversed distance. It's actually already clear that the pobability of having a $|\nu_e>$ again, which is given by $|<\nu_e|\psi(\vec{x},t>|^2$ will depend on both $t$ and $L$. This is where the oscillations come in. Since the total probability stays 1, there will be values of $L, t$ for which one of the probabilities becomes 0 and the other 1.
A: I think the best analogy is birefringence (for 2 flavors). You have 2 orthogonal states of polarization (subbing for flavor), $H$ and $V$, and say electrons only interact with $H$ and muons with $V$, and never the twain shall meet.
Until they start propagating, in a birefringent medium. It has two eigenstates (the fast axis and the orthogonal slow axis, with two different indices of refraction). The index of refraction is sort of a like and effective mass, as it affects the speed of propagation. Moreover, $\Delta n$ plays the role of $\Delta m^2$, the difference in the masses (squared) of the mass eigenstates.
A "non-coincidence of eigenstates" means the crystal is not aligned:
$$ |H\rangle = \cos\theta|F\rangle+\sin\theta|S\rangle$$
$$ |V\rangle = -\sin\theta|F\rangle+\cos\theta|S\rangle$$
It would be a good exercise to fully develop the analogy, but in once $|H\rangle$ starts propagating, there is a phase shift between the $|F\rangle$ and $|S\rangle$ that cause the plane of polarization to oscillate between $H$ and $V$.
Throw in an third flavor, and of course, it gets a bit more complicated, but the idea is the same.
