# Ultrarelativistic limit for neutrinos: Why is this approximation working?

In this section of the Wikipedia article on neutrino oscillations, a neutrino mass eigenstate $$\left|\nu_i\right>$$ is written as

$$\left|\nu_i(t)\right> = e^{-i(E_it-\vec p_i\cdot\vec{x})} \left|\nu_i(0)\right>$$

where $$E_i$$ is the energy of the mass eigenstate $$i$$ (rest energy + kinetic energy, as I understand). Then the first approximation is made, using $$|\vec p_i| = p_i \gg m_i$$ to obtain

$$E_i = \sqrt{p_i^2 + m_i^2} \approx p_i + \frac{m_i^2}{2p_i}.$$

I'm fine with that step since that is just a Taylor expansion. But the next approximation confuses me:

$$p_i + \frac{m_i^2}{2p_i} \approx E + \frac{m_i^2}{2E}$$

where $$E$$ is "the total energy of the particle". How is this energy $$E$$ different than the energy $$E_i$$ from above? Even if $$E$$ is the "total energy", this sounds like we would expect that $$E_i$$ is smaller than the "total" energy $$E$$. But our approximation gives $$E_i \approx E + \frac{m_i^2}{2E}$$, so $$E_i$$ should be larger than $$E$$.

• As explained in an answer, use p, the common reference momentum, as a synonym for E. Oct 15 at 13:43

The conceit is that your wave packet consisting of, say, two mass eigenstates, with $$m_1 are produced in entanglement at the meson decay "source" at a sloppy point, so with common momentum p. Since their masses differ, slightly, so do their energies and hence their velocities, $$E_i=\sqrt{p^2+m_i^2}\approx p+ \frac {m_i^2}{2p}.$$
So the wavepacket simulacrum ~ $$e^{-itE_1} +ae^{-itE_2}$$ evolves like $$e^{-it(p+m_1^2/2p)} \left (1+ ae^{-it(m_2^2-m_1^2)/2p}\right ),$$ so with "total" energy $$E\approx p$$ unobservable in the pure collective phase of the wavepacket, and differing from the momentum by $$O(m^2)$$, so we might as well put it in the denominator of the interesting oscillation phase following a in lieu of the momentum, since this only contributes slop of $$O(m^4)$$!
Again, $$t\approx L$$ for the wave packet, so the interesting oscillating probability at the detection point amounts to $$\left |1+ae^{-iL(m_2^2-m_1^2)/2E}\right |^2=1+a^2 + 2a \sin \left ( {L(m_1^2-m_2^2)\over 2E}\right ) ~.$$