# What is neutrino oscillation

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1. Jul 24, 2014

### Greg Bernhardt

Definition/Summary

There are three masses of neutrino (they are unknown, but the differences of their squares is known approximately).

There are three flavours of neutrino: electron- muon- and tau-, and a neutrino when it is created must be one of those three flavours.

Each flavour is a superposition (a combination) of the three different masses. For a given energy, the wave-functions of the three different masses have different periods, and so those three wave-functions get out of synch with each other, and so the magnitude of the sum of those three wave-functions will oscillate.

So the probability (= that magnitude squared) of being detected in the same flavour as when it was created will oscillate, returning periodically to 1 (and the probability of being detected in either of the other two flavours will also oscillate, returning periodically to 0).

This "neutrino oscillation" does not involve any "decay" or "virtual exchange" or interaction: it is an intrinsic property of the wave-function of a single neutrino.

A neutrino cannot oscillate into an anti-neutrino.

A neutrino cannot be observed directly. A neutrino can only be detected (or created) if at the same time a charged lepton (electron muon or tauon), or anti-lepton, is created or destroyed: it is then named after that charged lepton.

Equations

Extended explanation

Observation:

Measurement of neutrino oscillation must be done over a known distance ("base-line") after the creation of a known flavour of neutrino (or anti-neutrino).

For example, over the distance from the Sun to the Earth, after electron neutrinos are created in the Sun; over the height of the atmosphere, after muon neutrinos are created in collisions of "cosmic-ray" particles with the atmosphere; over hundreds of kilometres after muon neutrino beams are created in particle accelerators; and over tens of kilometres after electron anti-neutrinos are created in nuclear reactors.

Distance:

The wavelength of the oscillation (the distance over which the probability of detecting the neutrino in the same flavour in which it was created returns to 100%) is proportional to the energy: it is approximately 33,000 km per GeV.

Two pendulums (coupled harmonic oscillator):

Place two identical pendulums on the same wall, and start the left one swinging. Even without any connection between them (other than the wall itself), energy will gradually transfer from the left to the right pendulum, until for an instant only the right pendulum is moving, and then the energy will transfer back again, and so on.

So the pure motion of one pendulum (in that situation) is not an eigenstate.

The two eigenstates (also called modes) are obtained by releasing both pendulums at the same time and from the same height, either on the same side or on opposite sides.

Release them on the same side ("in-synch"), and they will always be parallel (as if they were joined at the bottom).

Release them on the opposite side ("opposite-synch"), and they will always be mirror images of each other.

Release them in any other way, and they will behave as if they were in a combination of the two modes, continually changing so as to move towards being purely in-synch, then purely opposite-synch, and so on.

Similarly, two non-identical pendulums will have two (more complicated) modes, and (unless they start in one of the two modes) will oscillate between them.

In particular, release only one pendulum, and the system will continually change so as to move towards being purely in-synch, then purely the other pendulum moving, then purely opposite-synch, then purely the original pendulum moving, and so on.

Ultimately, this is because the potential energy of the system can (for small oscillations) be written in the form:
$V = (m_1,m_2)\left( \begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \end{array} \right) \left( \begin{array}{c} m_1 \\ m_2 \end{array} \right)$
where $m_1$and $m_2$ are the "amounts" of the two modes (so, in that basis, the pure modes are $(1,0)$ and $(0,1)$).

If $f_1$and $f_2$ are the vectors for pure motion of one pendulum or the other, then we can write:
$\left( \begin{array}{c} m_1 \\ m_2 \end{array} \right) = \left( \begin{array}{cc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{array} \right) \left( \begin{array}{c} f_1 \\ f_2 \end{array} \right)$
for some "mixing" angle $\theta$ (which, for identical pendulums, is 45°).

(This angle has nothing to do with the angle of either pendulum, it is simply a factor which describes how to mix the two flavours.)

f1 and f2 correspond to flavours, and m1 and m2 correspond to masses.

Three pendulums (coupled harmonic oscillator):

Similarly, for three pendulums, there will be three modes, and we can write the potential energy in the form:

$V = (m_1,m_2,m_3)\left( \begin{array}{ccc} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3\end{array} \right) \left( \begin{array}{c} m_1 \\ m_2 \\ m_3 \end{array} \right)$
where
$\left( \begin{array}{c} m_1 \\ m_2 \\ m_3 \end{array} \right) = \left( U \right) \left( \begin{array}{c} f_1 \\ f_2 \\ f_3 \end{array} \right)$
where U is a fairly complicated 3x3 matrix (see here) involving three "mixing" angles $\theta_1\ \theta_2$ and $\theta_3$ (and possibly also a phase factor $\delta$).

Unlike the two-pendulum case, releasing only one pendulum will generally never result in that pendulum being stationary: some (but not all) its energy will gradually be transferred to both the other pendulums, and gradually back again, and so on.

Three flavours of neutrino:

The electron- muon- and tau- neutrino flavours correspond to three pendulums, of which two (muon- and tau-) are fairly similar.

The three neutrino masses (the heavier two of which are fairly similar) correspond to three modes of the pendulums (of which two have fairly similar energies).

The combined energy of the pendulums corresponds to the total energy of a neutrino.

The angles of the pendulums correspond to the phases of the waves of the three masses.

We can imagine that eg the muon pendulum will start swinging only if a muon hits the wall (and is destroyed). Or if something much bigger hits the wall and creates an anti-muon. Similarly, the three pendulums will stop swinging (simultaneously) only if a charged anti-lepton hits the wall, or a charged lepton is created.

Playing dice with neutrinos:

A neutrino can be thought of as a six-sided loaded die that looks honest on the outside, but the outside is just a light hollow frame, and a smaller solid box-shaped die (but not a cube) is fixed inside the frame so that its faces are not parallel to the outside faces.

The six outside faces of the die are marked $\epsilon\ \mu\ \tau\ \epsilon\ \mu\ \tau$ … (an anti-die would be marked $\ \bar{\epsilon}\ \bar{\mu}\ \bar{\tau}\ \bar{\epsilon}\ \bar{\mu}\ \bar{\tau}$).

We'll shake it inside a shaker, so that it is spinning, then throw it, northwards, at the inside of a frustrum of a cone whose narrowest part is just large enough to let the die through square-on … so it can only emerge in one of the three flavours.

But when it emerges, it will keep spinning, and because it is loaded it will precess, and the original outside face will not face north again except at the end of an exact number of precession periods.

If we place a second frustrum in the way, again facing northwards, the die will not necessarily emerge with the original outside face (except at the end of an exact number of precession periods).

(This is different from a pair of polarisers … if light emerges polarised from the first of two identical and parallel polarisers, it will automatically emerge polarised the same way from the second polariser … polarised light stays polarised the same way.)

* This entry is from our old Library feature, and was originally created by tiny-tim.

Last edited by a moderator: Jul 27, 2014