Physics Asked by user176263 on February 22, 2021
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
$$left|psi(t)right>=aleft|uparrowright>+bleft|downarrowright>,$$
where $left|uparrowright>$ is up-spin state and $left|downarrowright>$ 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
$$left|Psi(t)right>=aleft|nu_e(0)right>+bleft|nu_{mu}(0)right>.$$
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?
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).
Answered by anna v on February 22, 2021
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 $|uparrowrangle$ state a slightly higher energy eigenvalue than that of the $|downarrowrangle$ state. The relevant eigenvalues for the phase differences of the two states are $epsilon_pm$, respectively. Thus, for $$ |psi(0)rangle=costheta |uparrowrangle+sintheta |downarrowrangle ~~~~leadsto |psi(t)rangle=e^{-iepsilon_+ t/hbar } ~costheta |uparrowrangle+e^{-iepsilon_- t/hbar } sintheta |downarrowrangle . $$ 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, $$ |nu_e rangle = costheta |nu_1 rangle+ sintheta |nu_2 rangle |nu_{mu} rangle = -sintheta |nu_1 rangle+ costheta |nu_2 rangle . $$
For ultra relativistic propagation of "comoving" mass eigenstates, the simple kinematics yields $$ E_iapprox E+ {m_i^2over 2E}, $$ so that, nondimensionalizing $hbar$ and c as is customary in HEP, $$ |nu_e (t) rangle = e^{-i (E+ m_1^2/2E)t} costheta |nu_1 rangle+ e^{-i (E+ m_2^2/2E)t}sintheta |nu_2 rangle |nu_{mu} (t)rangle = -e^{-i (E+ m_1^2/2E)t}sintheta |nu_1 rangle+ e^{-i (E+ m_2^2/2E)t}costheta |nu_2 rangle , $$ 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.
Answered by Cosmas Zachos on February 22, 2021
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