# 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?

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
$$$$|\psi(\vec{x}, t)> = \cos(\theta) |\nu_1> e^{-ip_1 \cdot x} + \sin(\theta) e^{-ip_2 \cdot x} |\nu_2>$$$$ 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.

• So that means that because of the possibility of values between 0 and 1 there is always the chance that it oscillates between flavours? Commented Sep 3, 2020 at 16:55
• Yeah, kinda, it oscillates between flavours because the probabilities change. One can have a 50% of finding it in either state or eg. 100% probability of finding an electron-neutrino at certain $(L,t)$. The fundamental reason is still that the mass eigenstates are not equal to the propagating ones. Commented Sep 3, 2020 at 17:03
• Thank you so much for your help. I just have one last question: So the coincidence of the eigenstates would happen if the dot product would result on the RHS going back to its original form? Commented Sep 3, 2020 at 17:08
• No, even if the phase factors would vanish, there is still the $\sin(\theta)$ and $\cos(\theta)$. But the interference of both states do make it happen that at certain $(L,t)$ only one component remains. Normally one defines the relative phase between the 2 components so that only one components gets a phase factor. This phase factor is dependent on $E_1, E_2$ of course. A more detailed explanation and computation of this relative phase can be found in almost every particle physics book, I recommend Mark Thomson's Modern Particle Physics. Commented Sep 3, 2020 at 17:26
• Glad I could help! Commented Sep 3, 2020 at 21:56

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.