# How are neutrino energy eigenstates different to the momentum eigenstates?

Neutrino flavour eigenstates can be expressed (approximately) in terms of their mass eigenstates, leading to neutrino oscillations.

$|\nu_e\rangle = \cos \theta |\nu_1\rangle - \sin \theta |\nu_2\rangle$

$|\nu_\mu\rangle = \sin \theta |\nu_1\rangle + \cos \theta |\nu_2\rangle$

Where, I understand that the 1, 2 states are mass eigenstates and the e, $\mu$ states are momentum eigenstates. If the two states are different, does that not imply that $[p, H] \neq 0$, and therefore the expectation value of momentum is not constant?

I could have oversimplified/not understood something, so thanks for your help in advance!

• Couple of typesetting details to pay attention to. @CosmasZachos fixed the function names for you, replacing sin and cos with \sin and\cos (compare $sin \theta$ to $\sin \theta$). And I fixed the angle brackets; plain > is type set as a relational operator, while \ranlge is a bracket (compare $|a>$ to $|a\rangle$). – dmckee --- ex-moderator kitten Apr 14 '18 at 15:46

No, flavor eigenstates are not momentum eigenstates. How did you reach this absurd impression? In some, old fashioned, notations, they are both taken to have the same momentum, so it drops out of the problem and is then reinserted as energy. But then it is universal, and of course conserved.

To avoid this misunderstanding, it is easiest to stick to Lipkin's fixed energy picture, where the two physical, propagating, components, 1,2, are produced and detected with the same energy E and thus have slightly different momenta, $p_1^2 - p_2^2= m_2^2 - m_1^2$. That is, $$p_1-p_2= \frac{p_1^2- p_2^2}{p_1+p_2}=\frac{m_2^2-m_1^2}{p_1+p_2}\approx \frac{\Delta m_{21}^2}{2E} ~.$$

The mass eigenstates are then the momentum and energy eigenstates, $$|\nu_i (t,L)\rangle = e^{i p_iL -iE t} |\nu_i (0,0)\rangle,$$ so that $$|\nu_\mu (t,L)\rangle =\sin\theta ~e^{i(p_1L-Et)} |\nu_1\rangle + \cos\theta ~e^{i(p_2L-Et)} |\nu_2\rangle .$$ (You might check directly this is not a $\hat p$ eigenstate!)

It then follows that the transition probability amounts to $$|\langle \nu_\mu (t,L) |\nu_e (0,0)\rangle|^2= \sin^2 \theta \cos^2\theta ~~|1-e^{i(p_1-p_2)L}|^2 =\sin^2(2\theta)~\sin^2 \left(\frac{\Delta m_{21}^2 L}{4E}\right) .$$

It might help to think of the flavor eigenstates as "convenience detection packages", an artifact of our observation techniques, rather than real particles zipping about, like 1,2, which do everything you'd expect of a particle in your course. And yes, there is a bit of a slop in the different momenta you are tracking, corresponding to the same energy--it's just that the corresponding states were produced in sloppy/unspecified locations in the decay horn, or wherever... Not worth worrying about.

Edit in response to comment: The OP points out Sakurai's (or Sakurai-Napolitano) book (a superb one!) talks about a flavor eigenstate that's an eigenstate of momentum. Ahem! Yes, it is unfortunate. It sticks to the old convention that the flavor eigenstate has fixed momentum, not energy, as above.

Now, 1 and 2 have a common momentum, and so νμ has the very same momentum as well. In fact, in that setup, everything has the same, conserved, fixed momentum; but the energies must differ for the mass eigenstates, and the flavor eigenstates are not energy eigenstates anymore, as the phases in the plane waves are now $i(pL-tE_i$. Now, you repeat the above for t instead of L, and then convert t to L with the speed of light at the detector, at the very end. It is equivalent, but much more awkward, since they managed to confuse you. Here, then, you monitor in t the oscillating mean energy of the combination of the two energy eigenstates 1,2. But t differences amount to L differences, so...

• Thanks for such a quick and detailed response. I imagine that there is a mistake in the book I took this from (Sakurai, Modern Quantum Mechanics pp. 78) "Suppose we prepare at time at time t = 0, a momentum eigenstate of one flavour of neutrino, say $|\nu_e\rangle$...", should be flavour eigenstate, I suppose. – Tom E. Apr 14 '18 at 17:17