I am aware that flavor $\neq$ mass eigenstate, which is how mixing happens, but whenever someone talks about neutrino oscillations they tend to state without motivation that when neutrinos are actually propagating, they are doing so in a mass eigenstate. Presumably this is glossed over because it is a deep and basic artifact of quantum mechanics that I'm missing, but I'm having trouble coming up with it.
I had some help here where they say
The mass eigenstates are the free-particle solution to the wave equation[…]
but I suppose "why is that" could be a more basic reformulation of my question! Why can't mass be time-dependent in the wave equation, yielding (I would think) eigenstates that don't have well-defined mass?
Best Answer
It is not true in any sense that neutrinos "have to propagate as mass eigenstates". Quite on the contrary, one of the universal postulates of quantum mechanics – which are valid everywhere in the Universe – is the superposition principle. Whenever the states $|\psi_1\rangle$ and $|\psi_2\rangle$ are allowed, the general linear superposition $$ a_1|\psi_1\rangle + a_2|\psi_2\rangle $$ is equally allowed as well. The only special feature of the energy (Hamiltonian's) eigenstates is the fact that their time dependence is "simple", $$|\psi(t)\rangle = \exp(Et/i\hbar) |\psi(0)\rangle $$ This time-dependence boils down to the fact that the Hamiltonian is the operator that generates time translations. Equivalently, it's the operator that appears in Schrödinger's equation or Heisenberg's equations of motion. The reason why energy is connected with the evolution in time is ultimately linked to the deepest definition of the energy we have: via Noether's theorem, energy is the quantity that has to exist and be conserved as a direct implication of the time-independence of the laws of physics.
Note that the energy eigenstates only evolve by changing their overall phase. The phase isn't directly observable which implies that all observable properties of a Hamiltonian eigenstate are actually static in time; they are time-independent. This property makes them very useful as a basis of the Hilbert space: we may immediately solve Schrödinger's equation for this basis i.e. determine how a general state (when written as a combination of energy eigenstates) evolves in time.
These statements are completely general. They're true for neutrinos, too. Neutrino states may also be expanded into a superposition of energy eigenstates which makes the time dependence "easy". In fact, as long as we neglect the interactions, the neutrinos' wave functions evolve in time a a combination of plane waves $$ \exp(i\vec k \cdot \vec x - i Et) $$ where $E^2-|\vec k|^2=m^2$ is the squared mass. These plane waves are only a useful basis of all the allowed spacetime-dependent wave functions of the neutrinos if $m^2$ is effectively an ordinary number. In general, $m^2$ is another operator that acts on the Hilbert space of states of a neutrino. However, on the subspace of this Hilbert space of eigenstates of $\hat m^2$ with a given eigenvalue $m^2$, the operator $\hat m^2$ may be replaced by the eigenvalue $m^2$. The corresponding plane waves that fully dictate how the neutrino's wave function depends on space and time have a simple form in this case and all allowed wave functions may be written as linear superpositions of these simple plane waves.