In the system of up-spin and down-spin states of an electron, we can write a general state of electron at time $t$ as


where $\left|\uparrow\right>$ is up-spin state and $\left|\downarrow\right>$ is down-spin state. We also say that $e^{-}$ can oscillate between these two spin states (correct me if I am wrong). Now coming to neutrino oscillation phenomenon, we can write a general state of the neutrino as


Now, my question is, why do we need mass eigenstates ($\nu_1$ and $\nu_2$) to describe neutrino oscillations, while we don't need such type of thing to explain the oscillation (again I would say, correct me if I am wrong) between $e^{-}$'s up-spin state and down-spin state?


2 Answers 2


1) the spin states for the electron are two possible eigen values +1/2 or -1/2 .

2) the neutrino states involve two masses for the neutrino

Spin is a conserved angular momentum variable.

Mass is not a conserved quantity.

A free particle carries a number of conserved quantum numbers and momentum and energy and angular momentum are also conserved. When leaving an interaction vertex , from angular momentum conservation it will be in one of the two spin states, and cannot change it until it interacts again. The state you have written describes an electron only if angular momentum is not an eigenstate of the interacting system under consideration.

Because mass is not a conserved quantity but only the total energy momentum vector leaving the interaction vertex, and the masses of neutrinos are not associated with charges ( charge is a conserved quantum number) oscillations can happen.( It is related mathematically to how virtual particles have to conserve quantum numbers but the exchanged particle is off mass shell).

  • $\begingroup$ can you please explain one of your statements viz, "The state you have written describes an electron only if angular momentum is not an eigenstate of the interacting system under consideration." $\endgroup$
    – user176263
    Feb 25, 2018 at 5:39
  • $\begingroup$ I mean that we do not know the interaction from which it came., whence the spin would be unique. It is a free non interacting electron, which when measured will give one of the two spin states, because spin is a quantum number. It is not the spin that is arbitrary. Mass is not a quantum number. There is no format by which an up spin can become a down spin without and interaction. Have read en.wikipedia.org/wiki/Neutral_particle_oscillation $\endgroup$
    – anna v
    Feb 25, 2018 at 6:19

You do very much need "such type of [a] thing" in both cases, as you are really doing the same type of calculation, really.

For the static electron flipflop, you are normally assuming a hamiltonian with an asymmetric term like $S_z$ giving the $|\uparrow\rangle$ state a slightly higher energy eigenvalue than that of the $|\downarrow\rangle$ state. The relevant eigenvalues for the phase differences of the two states are $\epsilon_\pm$, respectively. Thus, for $$ |\psi(0)\rangle=\cos\theta |\uparrow\rangle+\sin\theta |\downarrow\rangle ~~~~\leadsto \\ |\psi(t)\rangle=e^{-i\epsilon_+ t/\hbar } ~\cos\theta |\uparrow\rangle+e^{-i\epsilon_- t/\hbar } \sin\theta |\downarrow\rangle . $$ Proceed to compute the oscillating transition amps for $\langle \psi(t)|\psi(0)\rangle $, etc...

For neutrino oscillations, sticking to your two-flavor model,
$$|\Psi(0)\rangle= |\nu_e \rangle, ~~~\hbox {or} ~~ |\nu_{\mu} \rangle ,$$ as you start from the production moment associated with a muon or an electron respectively. But these are not the mass, and hence hamiltonian, eigenstates.

Instead, they are linear combinations of these 1,2 eigenstates, $$ |\nu_e \rangle = \cos\theta |\nu_1 \rangle+ \sin\theta |\nu_2 \rangle \\ |\nu_{\mu} \rangle = -\sin\theta |\nu_1 \rangle+ \cos\theta |\nu_2 \rangle . $$

For ultra relativistic propagation of "comoving" mass eigenstates, the simple kinematics yields $$ E_i\approx E+ {m_i^2\over 2E}, $$ so that, nondimensionalizing $\hbar$ and c as is customary in HEP, $$ |\nu_e (t) \rangle = e^{-i (E+ m_1^2/2E)t} \cos\theta |\nu_1 \rangle+ e^{-i (E+ m_2^2/2E)t}\sin\theta |\nu_2 \rangle \\ |\nu_{\mu} (t)\rangle = -e^{-i (E+ m_1^2/2E)t}\sin\theta |\nu_1 \rangle+ e^{-i (E+ m_2^2/2E)t}\cos\theta |\nu_2 \rangle , $$ from each of which you compute the transition amps $\langle \nu_e (t) |\nu_e (0)\rangle$, etc, as before.

The takeaway is that states which are not eigenstates of the hamiltonian flip flop as they propagate, the common feature of oscillations, here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.