Two-flavour neutrino oscillations

A neutrino is the only particle we routinely produce in one identity and detect in another. In the Standard Model the charged-current weak vertex couples a charged lepton to a fixed combination of neutrino mass eigenstates, so the electron neutrino born in a beta decay, or the muon neutrino born when a pion decays, is not a state of definite mass at all — it is a coherent superposition of the mass states $\nu_1$ and $\nu_2$ . For two flavours that superposition is fixed by a single number, the mixing angle $\theta$ , through a $2\times2$ rotation between the flavour basis $(\nu_e,\nu_\mu)$ and the mass basis $(\nu_1,\nu_2)$ . Set that superposition moving. Each mass eigenstate is a plane wave that carries its own quantum phase $\phi_k=E_kt-p_kL$ , and because $m_1\neq m_2$ the two phases advance at different rates. The components drift out of step as the neutrino travels, so the particular blend that spelled "electron neutrino" at the source is continuously rewritten into a different blend downstream. Project the evolved state back onto the flavour basis and the probability of finding the other flavour rises and falls sinusoidally with distance — the neutrino oscillates. Everything hard about this article lives in one phase difference, $\Delta\phi=\Delta m^2 L/2E$ , and everything useful follows from it: the oscillation probability depends only on the mass-squared difference $\Delta m^2$ and on $\sin^2 2\theta$ , never on the absolute masses. That is why a detector kilometres from a reactor, or hundreds of kilometres from an accelerator, can weigh differences of neutrino masses far too small to measure any other way.

Key takeaways

The weak (flavour) states $\nu_e,\nu_\mu$ are a rotation of the mass states $\nu_1,\nu_2$ by one mixing angle $\theta$ ; the mass states, not the flavour states, have definite mass.
Mass eigenstates propagate as plane waves and accumulate phases $\phi_k=E_kt-p_kL$ ; unequal masses give a growing phase difference $\Delta\phi=\Delta m^2 L/2E$ .
The transition probability oscillates: $P(\nu_e\to\nu_\mu)=\sin^2 2\theta\,\sin^2\!\big(\Delta m^2 L/4E\big)$ , with survival $1-P$ .
Oscillation needs both a nonzero mass-squared difference and nonzero mixing — either one alone gives $P=0$ .
Practical form: $P=\sin^2 2\theta\,\sin^2\!\big(1.27\,\Delta m^2[\mathrm{eV}^2]\,L[\mathrm{km}]/E[\mathrm{GeV}]\big)$ ; the oscillation length is $L_\mathrm{osc}=4\pi E/\Delta m^2\approx 2.48\,E[\mathrm{GeV}]/\Delta m^2[\mathrm{eV}^2]$ km.
Oscillations resolved the solar and atmospheric neutrino deficits and proved that neutrinos have mass and that lepton flavour is not separately conserved.
Key idea: a neutrino is born in a flavour state but travels as a superposition of mass states; because the heavier and lighter components keep different quantum clocks, the interference between them periodically rebuilds the wavefunction into a different flavour.

The core insight

A neutrino is born in a flavour state but travels as a superposition of mass states; because the heavier and lighter components keep different quantum-mechanical clocks, the interference between them periodically rebuilds the wavefunction into a different flavour.

Where this lives today

The two-flavour formula is the backbone of a worldwide experimental program, even though nature needs three flavours to be described fully. Long-baseline accelerator experiments — T2K in Japan and NOvA in the United States — send muon-neutrino beams hundreds of kilometres and measure both $\nu_\mu$ disappearance and $\nu_e$ appearance, using the survival dip to pin $\Delta m^2_{32}$ and $\theta_{23}$ and the appearance channel to chase the CP-violating phase and the mass ordering that lie beyond the two-flavour picture. Reactor experiments work the complementary corner: Daya Bay, RENO and Double Chooz measured the small angle $\theta_{13}$ from a percent-level $\bar\nu_e$ deficit at roughly a kilometre, and the upcoming JUNO detector will read the fine oscillatory spectral distortion at 53 km to resolve the mass ordering from vacuum oscillations alone. Atmospheric neutrinos remain a laboratory in their own right, now probed at the gigatonne scale by IceCube's DeepCore and, soon, by KM3NeT, with matter effects through the Earth in play. The flagship of the next decade, DUNE, will fire a wideband beam 1300 km through rock to a liquid-argon detector, scanning a full oscillation pattern in a single spectrum. Every one of these reads its data through the shape first written down for two flavours: an amplitude set by mixing, a frequency set by a mass-squared difference.

The physical picture

Picture the two mass eigenstates as two waves launched together from the source, perfectly in step at the moment of production. That shared starting phase is exactly what encodes the flavour: aligned in one particular way, the pair spells "muon neutrino". But the heavier state carries slightly more energy at the same momentum, so its phase turns a little faster, and the two waves slide apart as they run. When they have drifted half a beat out of step, the blend no longer spells "muon neutrino" — it spells "tau neutrino". The flavour we would measure is the running interference between two clocks ticking at almost, but not quite, the same rate. The first figure shows that drift and names the phase that drives it.

Figure

Two mass eigenstates ν₁ and ν₂ leave the source in phase, then drift apart because the heavier state turns its quantum phase faster: the accumulated difference is Δφ = (m₂² − m₁²)L / 2E = Δm² L / 2E, and where they fall out of step the flavour has changed from ν_μ toward ν_τ

Turn the drift into the quantity an experiment measures and you get the oscillation curve: the probability of finding the far-flavour, plotted against the ratio $L/E$ of baseline to energy. Its shape is fixed by just two numbers. The amplitude of the wiggle is $\sin^2 2\theta$ — how completely the flavours can interchange, set entirely by the mixing angle — and the frequency of the wiggle along the $L/E$ axis is set by $\Delta m^2$ . Experiments are built by choosing $L$ and $E$ so that the neutrino arrives near the first crest, where the conversion is largest. The second figure is that curve for the atmospheric mass splitting, drawn to scale.

Figure

Oscillation probability P(ν_μ → ν_τ) = sin²2θ · sin²(1.27 Δm² L/E) against L/E in km/GeV, for the atmospheric splitting Δm² = 2.5×10⁻³ eV² and maximal mixing sin²2θ = 1: the first maximum sits at L/E ≈ 495 km/GeV and one full oscillation length is L/E ≈ 992 km/GeV

The core equations

The whole subject is a chain of six relations, each a short step from the last. Start with the fact that flavour is a rotation of mass. For two flavours the unitary mixing matrix is a single rotation through the angle $\theta$ , tying the states the weak interaction makes and detects to the states that propagate cleanly.

\begin{pmatrix}\nu_e\\ \nu_\mu\end{pmatrix}=\begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}\nu_1\\ \nu_2\end{pmatrix} \qquad (1)

where: $\nu_e,\nu_\mu$ — flavour (weak) eigenstates, produced and detected by the charged current; $\nu_1,\nu_2$ — mass eigenstates with masses $m_1,m_2$ ; $\theta$ — mixing angle. At $\theta=0$ flavour and mass coincide and nothing oscillates. Each mass eigenstate propagates as a plane wave, picking up a phase set by its own energy and momentum. This is the only place the dynamics enters — after this it is all interference.

|\nu_k(t)\rangle=e^{-i(E_k t-p_k L)}\,|\nu_k\rangle=e^{-i\phi_k}\,|\nu_k\rangle \qquad (2)

where: $\phi_k=E_kt-p_kL$ — the phase of mass state $k$ after time $t$ over baseline $L$ ; $E_k=\sqrt{p_k^2+m_k^2}$ — its energy. Because $m_1\neq m_2$ the two phases advance unequally, and their difference is what survives. Prepare a $\nu_e$ , evolve, and project onto $\nu_\mu$ ; the modulus-squared of that amplitude is the appearance probability. This is the central result of the article.

P(\nu_e\to\nu_\mu)=\sin^2 2\theta\,\sin^2\!\left(\frac{\Delta m^2\,L}{4E}\right),\qquad \Delta m^2\equiv m_2^2-m_1^2 \qquad (3)

where: $\sin^2 2\theta$ — the oscillation amplitude, fixed by mixing; $\Delta m^2$ — mass-squared difference in $\mathrm{eV}^2$ ; $L$ — baseline; $E$ — neutrino energy. The probability depends on $\Delta m^2$ , never on $m_1$ or $m_2$ separately, and the sign of $\Delta m^2$ drops out because $\sin^2$ is even. Probability is conserved, so what does not appear as $\nu_\mu$ survives as $\nu_e$ . The survival probability is the complement, and it is what a disappearance experiment measures directly.

P(\nu_e\to\nu_e)=1-\sin^2 2\theta\,\sin^2\!\left(\frac{\Delta m^2\,L}{4E}\right) \qquad (4)

where: the two channels sum to one, $P(\nu_e\to\nu_e)+P(\nu_e\to\nu_\mu)=1$ , as unitarity demands for two flavours. A reactor experiment counts the $\bar\nu_e$ that fail to arrive; an accelerator experiment can count either channel. Restoring the factors of $\hbar$ and $c$ that natural units hid turns equation (3) into the form every experimentalist actually uses, with the baseline in kilometres and the energy in GeV.

P=\sin^2 2\theta\,\sin^2\!\left(1.267\,\frac{\Delta m^2[\mathrm{eV}^2]\,L[\mathrm{km}]}{E[\mathrm{GeV}]}\right) \qquad (5)

where: the constant $1.267$ packages $\hbar c$ and the unit conversions (derived below); the same number holds for $L$ in metres and $E$ in MeV. This is the working formula behind every oscillation plot. Finally, the distance over which the pattern repeats defines the oscillation length, the natural yardstick for choosing a baseline.

L_\mathrm{osc}=\frac{4\pi E}{\Delta m^2}\approx 2.48\,\frac{E[\mathrm{GeV}]}{\Delta m^2[\mathrm{eV}^2]}\ \mathrm{km} \qquad (6)

where: $L_\mathrm{osc}$ — one full period of the oscillation; the first appearance maximum sits at $L_\mathrm{osc}/2$ . Matching $L/E$ to $L_\mathrm{osc}$ for a given $\Delta m^2$ is the whole art of designing an oscillation experiment.

Derivation AFrom flavour mixing to the oscillation probability

The goal is to turn the physical story into equation (3) with no gaps: start from a neutrino produced as a pure flavour state, write it in the mass basis where propagation is simple, let each mass eigenstate accumulate its plane-wave phase, project the evolved state back onto the other flavour, and square the amplitude. Every step is elementary quantum mechanics of a two-state system; the only physics input is that the mass eigenstates, not the flavour states, are what propagate with definite phase. We keep two flavours throughout so the algebra stays a single rotation.

Step 1. Produce a $\nu_e$ at $L=0$ , $t=0$ . Using equation (1), write it in the mass basis: $|\nu(0)\rangle=|\nu_e\rangle=\cos\theta\,|\nu_1\rangle+\sin\theta\,|\nu_2\rangle$ . Why: the weak interaction makes a flavour state, but only the mass states have a definite energy–momentum relation, so we must change basis before we can propagate anything.

Step 2. Evolve each mass component with its own phase from equation (2): $|\nu(t)\rangle=\cos\theta\,e^{-i\phi_1}|\nu_1\rangle+\sin\theta\,e^{-i\phi_2}|\nu_2\rangle$ . Why: superposition means each eigenstate evolves independently; the two phases $\phi_1,\phi_2$ are the only things that change, and their difference is the entire physics.

Step 3. Ask for the amplitude to detect a $\nu_\mu$ . With $\langle\nu_\mu|=-\sin\theta\,\langle\nu_1|+\cos\theta\,\langle\nu_2|$ and orthonormal mass states, $\langle\nu_\mu|\nu(t)\rangle=\sin\theta\cos\theta\,(e^{-i\phi_2}-e^{-i\phi_1})$ . Why: the detector, like the source, couples to a flavour state, so the observable amplitude is the projection of the evolved state onto $\nu_\mu$ — and it vanishes unless the two phases differ.

Step 4. Square the amplitude. Using $|e^{-i\phi_2}-e^{-i\phi_1}|^2=2-2\cos(\phi_1-\phi_2)=4\sin^2\!\big(\tfrac{\phi_1-\phi_2}{2}\big)$ and $\sin^2\theta\cos^2\theta=\tfrac14\sin^2 2\theta$ , the factors of four cancel: $P(\nu_e\to\nu_\mu)=\sin^2 2\theta\,\sin^2\!\big(\tfrac{\Delta\phi}{2}\big)$ , with $\Delta\phi=\phi_1-\phi_2$ . Why: this is the interference term made real — the mixing sets how much can interfere, and the phase difference sets where in the cycle the interference lands.

Step 5. Insert the phase difference $\Delta\phi=\Delta m^2 L/2E$ from the next derivation to reach $P(\nu_e\to\nu_\mu)=\sin^2 2\theta\,\sin^2\!\big(\Delta m^2 L/4E\big)$ . Why: with the phase evaluated, the abstract two-state result becomes the concrete oscillation formula, depending only on $\Delta m^2$ , $\theta$ , and $L/E$ .

Go deeper · Derivation B · 1 minThe propagation phase and why equal energy, equal momentum, and equal velocity all agree

Derivation BThe propagation phase and why equal energy, equal momentum, and equal velocity all agree

The aim is to justify the single most important line in the subject, $\Delta\phi=\Delta m^2 L/2E$ , and to defuse a genuine confusion: the raw phase $\phi_k=E_kt-p_kL$ mixes an energy term and a momentum term, and the two mass eigenstates strictly have different energies, momenta, times of flight, and velocities. The reassuring fact is that in the ultra-relativistic limit that always applies to real neutrinos, three different simplifying assumptions — equal momenta, equal energies, or equal velocities — deliver the identical answer, and a full wave-packet treatment confirms it. We show the two cleanest routes converge.

Step 1. Write the difference of phases exactly: $\Delta\phi=\phi_1-\phi_2=(E_1-E_2)t-(p_1-p_2)L$ . Neutrinos are ultra-relativistic, $m_k\ll E$ , and both components share essentially one path, so $t\approx L$ and $E\approx p$ . Why: no honest simplification is available until we commit to $m\ll E$ , which is spectacularly well satisfied — sub-eV masses against MeV–GeV energies.

Step 2. Take the equal-momentum assumption $p_1=p_2=p$ . Then $\Delta\phi=(E_1-E_2)t$ , and expanding $E_k=\sqrt{p^2+m_k^2}\approx p+m_k^2/2p$ gives $E_1-E_2=(m_1^2-m_2^2)/2p$ . With $t\approx L$ and $p\approx E$ , $|\Delta\phi|=\Delta m^2 L/2E$ . Why: this is the textbook shortcut; the momentum term dies and the energy splitting is purely the mass term.

Step 3. Take instead the equal-energy assumption $E_1=E_2=E$ . Then $\Delta\phi=-(p_1-p_2)L$ , and $p_k=\sqrt{E^2-m_k^2}\approx E-m_k^2/2E$ gives $p_1-p_2=(m_2^2-m_1^2)/2E$ , so again $|\Delta\phi|=\Delta m^2 L/2E$ . Why: the two assumptions look physically different but land on the same phase, because to leading order in $m^2/E^2$ only the on-shell relation $E^2=p^2+m^2$ matters, not how we split the small difference between the $E$ and $p$ channels.

Step 4. Note the robustness. The apparent factor-of-two ambiguities that plague careless "equal time" arguments cancel once $t\approx L$ is imposed consistently, and a treatment with genuine wave packets — where each mass eigenstate is a localized bump moving at its own group velocity — reproduces $\Delta\phi=\Delta m^2 L/2E$ and additionally predicts loss of coherence once the packets separate. Why: the plane-wave result is not an accident of a convenient assumption; it is the leading term of the correct wave-packet calculation, which is why it survives.

Go deeper · Derivation C · 1 minRestoring ħ and c to get the practical factor 1.267

Derivation CRestoring ħ and c to get the practical factor 1.267

The purpose here is to explain where the mysterious $1.267$ in equation (5) comes from, because it is not a physical constant but a bookkeeping factor that lets you feed in eV², kilometres, and GeV and read out a pure number in radians. In natural units the oscillation phase is $\Delta m^2 L/4E$ ; we reinsert the powers of $\hbar$ and $c$ needed to make the three mismatched units cancel, then evaluate the resulting constant. The exercise also fixes, once and for all, that the same constant works for metres with MeV.

Step 1. Reinstate the constants. Dimensionally the phase must be $\dfrac{\Delta m^2 c^4\,L}{4\,\hbar c\,E}$ , since $\Delta m^2 c^4$ is an energy-squared, $\hbar c$ carries the energy times length, and $E$ is an energy, leaving a dimensionless argument. Why: only this combination makes the units cancel; $\hbar c$ is the single conversion constant between the natural-unit answer and laboratory units.

Step 2. Insert $\hbar c=1.973\times10^{-7}\ \mathrm{eV\,m}$ and choose lab units $\Delta m^2$ in $\mathrm{eV}^2$ , $L$ in km $=10^3$ m, $E$ in GeV $=10^9$ eV. The numerical prefactor is $\dfrac{10^3}{4\,(1.973\times10^{-7})\,10^9}=1.267$ . Why: this is just arithmetic on the powers of ten and the value of $\hbar c$ — nothing physical, but it is the number you cannot afford to get wrong.

Step 3. Read off equation (5): $P=\sin^2 2\theta\,\sin^2\!\big(1.267\,\Delta m^2[\mathrm{eV}^2]L[\mathrm{km}]/E[\mathrm{GeV}]\big)$ . Because km/GeV and m/MeV carry the same ratio of powers of ten, the identical $1.267$ serves for $L$ in metres and $E$ in MeV — the combination reactor and short-baseline experiments live in. Why: recognizing the scale invariance saves re-deriving the constant for every experiment; one number covers reactors, accelerators, and atmospheric neutrinos alike.

Limiting cases

Case	Result	Meaning
No mixing, $\theta=0$	$P=0$	flavour and mass bases coincide; each flavour propagates unchanged
Degenerate masses, $\Delta m^2=0$	$P=0$	the two phases stay locked; no interference, even with full mixing
Maximal mixing, $\theta=45^\circ$	$P=\sin^2\!\big(\Delta m^2 L/4E\big)$	amplitude reaches 1; complete flavour conversion at the crest (atmospheric $\nu_\mu\to\nu_\tau$ )
Very short baseline, $\Delta m^2 L/4E\ll1$	$P\approx\sin^2 2\theta\,\big(\Delta m^2 L/4E\big)^2$	probability rises quadratically from zero; near-detectors see no oscillation
Oscillation maximum, $\Delta m^2 L/4E=\pi/2$	$P=\sin^2 2\theta$	first and largest peak; experiments are tuned to sit here
Long baseline / broad spectrum (averaged)	$\langle P\rangle=\tfrac12\sin^2 2\theta$	fast oscillations wash out with energy resolution; only the mean amplitude survives

Common trap“An oscillation experiment measures the neutrino mass.”

It does not — and the distinction matters. Every term in the vacuum two-flavour formula depends on the mass-squared difference $\Delta m^2=m_2^2-m_1^2$ and on $\sin^2 2\theta$ , and on nothing else. Oscillations are therefore blind to the absolute mass scale: two neutrinos of $0.001$ eV and $0.009$ eV oscillate identically to two of $1.001$ eV and $1.009$ eV, because both pairs share the same $\Delta m^2$ . They are blind to the individual masses. And in vacuum, at two flavours, they are blind even to the sign of $\Delta m^2$ , because $\sin^2$ is an even function — resolving that sign is exactly what matter effects (the MSW resonance) and three-flavour interference are brought in to do. A closely related trap is to speak of "the mass of the electron neutrino": the flavour state $\nu_e$ has no definite mass at all, being a superposition of $\nu_1$ and $\nu_2$ . What oscillations prove is only that the masses are unequal and not both zero — a momentous result, but a strictly weaker one than a mass measurement.

Worked examples

1 · Choosing a baseline from the oscillation length (confidence builder)

Reactor antineutrinos of energy $E=4$ MeV are studied with the solar mass splitting $\Delta m^2_{21}=7.5\times10^{-5}\ \mathrm{eV}^2$ . Over what distance does the oscillation develop, and why did KamLAND sit its detectors near 180 km?

Step 1. Use equation (6) with $E=4\ \mathrm{MeV}=0.004\ \mathrm{GeV}$ : $L_\mathrm{osc}=2.48\times(0.004)/(7.5\times10^{-5})=2.48\times53.3=\mathbf{132\ km}$ . The first appearance maximum, at $L_\mathrm{osc}/2$ , sits near 66 km.

Step 2. A flux-weighted baseline of roughly 180 km is about $1.4\,L_\mathrm{osc}$ , so KamLAND's reactor antineutrinos arrive well past the first minimum, where the survival probability is strongly suppressed and — crucially — energy-dependent, imprinting an oscillatory distortion across the spectrum rather than a flat deficit. A single deficit could be absorbed into flux uncertainty; a spectral wiggle is the fingerprint that fixed $\Delta m^2_{21}$ .

Step 3. The same equation explains the atmospheric case: for $\Delta m^2_{32}=2.5\times10^{-3}\ \mathrm{eV}^2$ and $E=1$ GeV, $L_\mathrm{osc}=2.48/(2.5\times10^{-3})=\mathbf{992\ km}$ — comparable to distances through the Earth, which is precisely why upward-going GeV atmospheric $\nu_\mu$ that cross the planet are seen to disappear while downward-going ones are not.

2 · The T2K disappearance at the oscillation maximum (fully worked)

The T2K experiment fires a $\nu_\mu$ beam of mean energy $E=0.6$ GeV to a detector $L=295$ km away. Taking the atmospheric parameters $\Delta m^2_{32}=2.5\times10^{-3}\ \mathrm{eV}^2$ and near-maximal mixing $\sin^2 2\theta_{23}=1.0$ , find the $\nu_\mu$ survival probability.

Step 1. Evaluate the phase argument of equation (5): $1.267\times\dfrac{(2.5\times10^{-3})(295)}{0.6}=1.267\times\dfrac{0.7375}{0.6}=1.267\times1.229=\mathbf{1.557\ rad}$ .

Step 2. This sits almost exactly at $\pi/2=1.5708$ — T2K's baseline and beam energy were chosen so the neutrinos reach the detector at the first oscillation maximum. Then $\sin^2(1.557)=(0.99991)^2=\mathbf{0.99982}$ .

Step 3. The transition probability is $P(\nu_\mu\to\nu_\tau)=\sin^2 2\theta\cdot\sin^2(1.557)=1.0\times0.99982=0.9998$ , so the survival probability is $P(\nu_\mu\to\nu_\mu)=1-0.9998=\mathbf{0.0002}$ .

Step 4. In this idealized maximal-mixing, single-splitting picture almost every muon neutrino has converted by the time it reaches Super-Kamiokande. At the oscillation maximum the $\nu_\mu$ beam essentially vanishes — the dramatic disappearance that long-baseline experiments are built to catch. (Real T2K mixing is close to but not exactly maximal, and the finite beam spectrum softens the dip, so the measured survival minimum is a few percent rather than zero.)

Physics of the portrait

Neutrino oscillations were a theorist's conjecture for forty years before an experiment forced them on the world. The idea belongs to one restless, brilliant, and politically improbable figure who kept asking whether the neutrino might be less simple than it looked.

Origins · 1 minBruno Pontecorvo and how this was discovered

Bruno Pontecorvo

1913–1993 · the man who proposed neutrino oscillations

Bruno Pontecorvo trained in Enrico Fermi's Rome group in the 1930s — one of the "Via Panisperna boys" — and made his early name on slow-neutron physics and on the idea, later industrialized, of detecting neutrinos by inverse beta decay. In 1957, inspired by the newly discovered $K^0$ – $\bar K^0$ oscillations, he proposed that neutrinos too might oscillate; with only one known neutrino species at the time he first framed it as a $\nu\leftrightarrow\bar\nu$ transition, then, after the muon neutrino was found in 1962, recast it as the flavour oscillation we now use. In parallel, Maki, Nakagawa and Sakata wrote down the flavour-mixing matrix in 1962, and the modern mixing matrix carries all their names as the PMNS matrix. Pontecorvo also foresaw, in the 1960s, that the flux of neutrinos from the Sun would arrive depleted if oscillations were real — the interpretation eventually vindicated when the solar neutrino deficit measured by Davis was resolved. His life was as singular as his physics: in 1950, at the height of the Cold War, he defected from the West to the Soviet Union, where he continued to work at Dubna and where his oscillation idea was long treated as a curiosity. He did not live to see the 1998 Super-Kamiokande result that turned his conjecture into established fact, but the effect is, by any fair reading, his.

History

1957

— Pontecorvo proposes that neutrinos oscillate, by analogy with kaon mixing

1962

— Maki, Nakagawa and Sakata formulate neutrino flavour mixing

1968

— Davis and Bahcall report a large deficit of solar neutrinos at Homestake

1998

— Super-Kamiokande discovers atmospheric neutrino oscillations

2015

— Kajita and McDonald share the Nobel Prize for the discovery of oscillations

Going further

Texts. Thomson, Modern Particle Physics, §13.4–13.6 (neutrino oscillations and the PMNS matrix) · Griffiths, Introduction to Elementary Particles, 2nd ed., §11.6 (neutrino oscillations) · Halzen & Martin, Quarks and Leptons, Ch. 12 (weak mixing) · Perkins, Particle Astrophysics, Ch. 9 (solar and atmospheric neutrinos) · Giunti & Kim, Fundamentals of Neutrino Physics and Astrophysics (the comprehensive reference, including the wave-packet treatment). The key results of this article are: $P=\sin^2 2\theta\,\sin^2(\Delta m^2 L/4E)$ from the rotate–evolve–project–square chain; $\Delta\phi=\Delta m^2 L/2E$ from both the equal-momentum and equal-energy routes; the constant $1.267$ from $\hbar c=1.973\times10^{-7}\ \mathrm{eV\,m}$ ; $L_\mathrm{osc}=2.48\,E[\mathrm{GeV}]/\Delta m^2[\mathrm{eV}^2]$ km; and the T2K phase argument $1.267\times2.5\times10^{-3}\times295/0.6=1.557\approx\pi/2$ . The ultra-relativistic expansion $E_k\approx p+m_k^2/2p$ underpins the phase difference. Benchmark values used throughout are the standard global-fit numbers: $\Delta m^2_{21}\approx7.5\times10^{-5}\ \mathrm{eV}^2$ , $\Delta m^2_{32}\approx2.5\times10^{-3}\ \mathrm{eV}^2$ , $\sin^2 2\theta_{12}\approx0.85$ , $\theta_{23}\approx45^\circ$ .

Prerequisites. Mass and Weak Eigenstates · The Weak Charged Current · Non-relativistic Quantum Mechanics Review · Special-Relativistic Kinematics. Leads to. Three-flavour PMNS Mixing · Matter Effects and the MSW Resonance · Oscillation Experiments · Neutrino Mass Mechanisms.

Ask the tutor: Ask me to plug your own baseline, energy, and $\Delta m^2$ into the 1.267 formula, or to show how the two-flavour picture grows into the full three-flavour PMNS matrix with CP violation.

If the neutrino is emitted as one flavour, what is actually "waving" as it travels?

Not the flavour itself, but the relative phase of the two mass eigenstates that make it up. At production the two components $\nu_1$ and $\nu_2$ are locked in a particular phase relationship that spells one flavour. Because they have different masses they advance their quantum phases at slightly different rates, so that relationship rotates smoothly as the neutrino flies. When the relative phase has turned by $\pi$ the superposition spells the other flavour; when it has turned by $2\pi$ it spells the original flavour again. Nothing about the individual mass states changes flavour — flavour is a property of how they interfere, and the interference is what oscillates.

Why does only the mass-squared difference appear, and not the masses themselves?

Because the observable is a phase difference. Each mass eigenstate contributes a phase proportional to its energy, and in the ultra-relativistic limit that energy is $E_k\approx p+m_k^2/2p$ . The part common to both states, $p$ , contributes an overall phase that is identical for both components and cancels when you take their difference — an overall phase is unobservable in quantum mechanics. What remains is the difference of the $m_k^2/2p$ terms, which is exactly $\Delta m^2/2p$ . So the mass scale and the individual masses live in the common phase that drops out, and only their squared difference survives to drive the oscillation.

Do antineutrinos oscillate the same way as neutrinos?

In two-flavour vacuum oscillations, yes, identically. The two-flavour formula depends only on $\sin^2 2\theta$ and $\Delta m^2 L/4E$ , none of which changes under charge conjugation, so $P(\nu_e\to\nu_\mu)=P(\bar\nu_e\to\bar\nu_\mu)$ . Any difference between neutrino and antineutrino oscillation — CP violation in the lepton sector — requires at least three flavours and a nonzero complex phase in the mixing matrix, which cannot appear in a single real rotation angle. It can also be mimicked by matter effects, since ordinary matter contains electrons but not positrons and so treats $\nu_e$ and $\bar\nu_e$ differently. Both of those lie beyond this article, but their absence here is precisely why the two-flavour vacuum case is CP-symmetric.

How do you actually see an oscillation rather than just a missing flux?

A flat deficit is weak evidence: the source flux is often uncertain at the ten-percent level, so a uniform shortfall can always be blamed on the prediction. An oscillation leaves a sharper signature — the survival probability varies with $L/E$ in a specific sinusoidal way, so the deficit is energy-dependent and distance-dependent in the pattern $\sin^2(1.267\,\Delta m^2 L/E)$ . KamLAND saw the reactor $\bar\nu_e$ spectrum dip and recover across energy; Super-Kamiokande saw atmospheric $\nu_\mu$ disappear as a function of zenith angle, hence path length. Reconstructing that wiggle, not just the total count, is what distinguishes genuine oscillation from a mundane flux error and what lets a single dataset measure $\Delta m^2$ and $\theta$ together.

Is two flavours ever enough, or is it just a warm-up for three?

It is a genuine and often excellent approximation, not merely a toy. Nature has three neutrinos, but the two mass-squared splittings are very different — the solar splitting is about thirty times smaller than the atmospheric one — so in many experiments one splitting dominates the relevant $L/E$ range and the third state decouples. Atmospheric and long-baseline $\nu_\mu$ disappearance is well described by a two-flavour $\nu_\mu\to\nu_\tau$ formula in $(\Delta m^2_{32},\theta_{23})$ ; solar and KamLAND data reduce to two-flavour $(\Delta m^2_{21},\theta_{12})$ . The full three-flavour PMNS treatment is needed for sub-leading effects — $\nu_e$ appearance driven by $\theta_{13}$ , CP violation, and the interference of the two splittings — but the two-flavour result captures the leading behaviour of most measurements to good accuracy.

What happens to the oscillation over very long distances or with a broad energy spread?

The clean sinusoid washes out into a constant. Real beams have a spread of energies, and a detector has finite energy resolution, so instead of a single value of $\Delta m^2 L/4E$ you average $\sin^2$ over a range of phases. Once the phase varies by more than about $\pi$ across that range — which happens at large $L$ or with a broad spectrum — the oscillatory factor averages to $\tfrac12$ , and the survival probability settles at $1-\tfrac12\sin^2 2\theta$ . This is exactly the regime of solar neutrinos observed at Earth: the enormous baseline and the broad production region wash out the vacuum wiggle, leaving an averaged survival that (with matter effects included) explains the solar deficit. The information in $\Delta m^2$ is not lost, but it is read from where the averaging sets in, not from a resolved fringe.