Oscillation experiments

A neutrino oscillation experiment is a wager against quantum mechanics that quantum mechanics keeps winning. Fire a beam of one flavour — say muon neutrinos from pion decay — let it travel, and ask a detector far away what flavours it now contains. If the weak eigenstates $\nu_e,\nu_\mu,\nu_\tau$ were identical to the states of definite mass, the answer would be boring: what you make is what you catch. But the flavour states are superpositions of mass eigenstates $\nu_1,\nu_2,\nu_3$ that travel with slightly different phases, so the flavour content sloshes back and forth along the baseline. Catching that sloshing is the whole game, and there are two ways to play it. In an appearance experiment you look for charged leptons of the wrong flavour — an $e^-$ or a $\tau^-$ born in a beam that started as pure $\nu_\mu$ — because such a lepton can only be made by a neutrino that changed identity in flight. In a disappearance experiment you count the right-flavour leptons and look for a deficit: fewer $\mu^-$ than a non-oscillating beam would deliver. Both rest on the same subtlety — a neutrino is invisible until it undergoes a charged-current weak interaction and converts into its partner lepton, and that conversion has a kinematic threshold. The threshold to make a $\tau^-$ on a nucleon sits at $3.45\ \mathrm{GeV}$ , which is why a low-energy beam can oscillate into $\nu_\tau$ and still show nothing. The story of the field — Homestake, Super-Kamiokande, SNO, KamLAND, Daya Bay, T2K — is the story of choosing a source, a baseline, and an energy so the sloshing lands inside the detector.

Key takeaways

Two signatures: appearance of wrong-flavour leptons ( $e^-$ or $\tau^-$ from a $\nu_\mu$ beam) and disappearance of right-flavour leptons (fewer $\mu^-$ than expected).
The oscillation probability is $P=\sin^2 2\theta\,\sin^2(1.27\,\Delta m^2 L/E)$ in practical units eV², km, GeV — so an experiment is designed by matching its $L/E$ to the target $\Delta m^2$ .
A neutrino is only seen when it makes its partner lepton in a charged-current reaction, and that reaction has a threshold $E_\nu^{\rm thr}=[(m_\ell+m_p)^2-m_n^2]/2m_n$ : essentially zero for $e$ , $110\ \mathrm{MeV}$ for $\mu$ , $3.45\ \mathrm{GeV}$ for $\tau$ .
Sources span the ladder in $L/E$ : reactors (km, MeV), the Sun and atmosphere, and accelerator long-baselines (hundreds of km, GeV) — each tuned to a different $\Delta m^2$ .
CP violation needs the complex phase $\delta$ of the PMNS matrix, which only survives for three or more generations; a two-flavour mixing matrix is real and CP-conserving.
Key idea: you detect oscillation either by seeing wrong-flavour leptons appear or by counting a shortfall of right-flavour leptons — but whether a flavour can appear at all is fixed by the kinematic threshold to make its heavy charged-lepton partner.

The core insight

You detect oscillation either by seeing wrong-flavour leptons appear or by counting a shortfall of the right-flavour leptons — but whether a given flavour can appear at all is dictated by the kinematic threshold to produce its heavy charged-lepton partner.

Where this lives today

Oscillation experiments have moved from discovery to precision. Two long-baseline projects are under construction: DUNE fires a wide-band beam 1300 km from Fermilab to a liquid-argon detector in South Dakota, while Hyper-Kamiokande scales up the water-Cherenkov technique behind a beam from J-PARC. Both target the leptonic CP phase $\delta$ — comparing $\nu_\mu\to\nu_e$ against $\bar\nu_\mu\to\bar\nu_e$ to see whether neutrinos and antineutrinos oscillate differently, which could underlie the matter–antimatter asymmetry of the universe. They will also resolve the mass ordering by reading how passage through the Earth's matter distorts the appearance probability via the MSW effect. In parallel, JUNO measures the solar parameters and the ordering with a 20-kton scintillator, short-baseline programmes hunt for a sterile neutrino, and beta-decay endpoints and neutrinoless double-beta decay probe the absolute mass scale. The phase $\delta$ that equation (4) hides behind a small $s_{13}$ is now the single most sought-after parameter in the field.

The physical picture

Picture a pure $\nu_\mu$ beam leaving a decay tunnel. Nothing about the neutrinos changes as they fly — no force acts, no energy is lost — yet the label you would assign them drifts, because the three mass components of a $\nu_\mu$ accumulate phase at different rates. After some distance the superposition that was "muon-like" has rotated part-way into "electron-like" and "tau-like". A detector there has two handles: hunt for the newcomers (appearance) or audit the survivors (disappearance). The first figure lays the two readouts side by side and reminds you that appearance is gated by whether the beam can make the wrong-flavour lepton at all.

Figure

Two ways to see oscillation. A pure ν_μ beam travels a baseline L. Appearance looks for wrong-flavour leptons e⁻ or τ⁻ — but making a τ⁻ needs E_ν > 3.45 GeV. Disappearance counts the surviving μ⁻ and reads the ratio N_obs/N₀ = P(ν_μ→ν_μ) = 1 − sin²2θ sin²(1.27 Δm²L/E)

Experiments come in a bewildering variety — reactors a kilometre away, the Sun 150 million kilometres away, accelerator beams through 300 km of rock — because the phase driving the sloshing depends only on $L/E$ and $\Delta m^2$ . An experiment is sensitive only when its $L/E$ is comparable to $1/\Delta m^2$ : too short and the oscillation has not developed, too long and it smears into an average. The second figure plots the $\nu_\mu$ survivor fraction versus $L/E$ and shows where T2K is deliberately parked — on the first oscillation minimum, where the deficit is deepest.

Figure

The muon-neutrino survival probability versus L/E for the atmospheric splitting. P(ν_μ→ν_μ) = 1 − sin²2θ₂₃ sin²(1.27 Δm²₃₂ L/E) with maximal mixing sin²2θ₂₃ = 1 and Δm²₃₂ = 2.5×10⁻³ eV². The first and deepest minimum sits at L/E ≈ 496 km/GeV — exactly where T2K parks its 295 km baseline at 0.6 GeV

The core equations

Everything an experiment measures reduces to one probability. For two flavours mixing through a single angle $\theta$ , the amplitude to appear in the other flavour after a distance $L$ at energy $E$ interferes as the two mass eigenstates dephase, giving a clean $\sin^2$ envelope modulated by the mixing strength.

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

where: $\theta$ — the two-flavour mixing angle; $\Delta m^2=m_2^2-m_1^2$ — the mass-squared splitting (natural units); $L$ — baseline; $E$ — neutrino energy. The prefactor $\sin^2 2\theta$ sets how much mixing is possible; the $\sin^2$ argument sets where along the baseline it happens. Appearance and disappearance are one physics read with opposite sign. Experimentalists never work in natural units, so equation (1) is rewritten with $\hbar c$ absorbed into a single number. Restoring units with $\Delta m^2$ in eV², $L$ in km and $E$ in GeV produces the coefficient every talk in the field quotes.

P(\nu_\alpha\to\nu_\beta)=\sin^2 2\theta\,\sin^2\!\left(1.27\,\frac{\Delta m^2[\mathrm{eV}^2]\;L[\mathrm{km}]}{E[\mathrm{GeV}]}\right) \qquad (2)

where: the constant $1.267$ equals $10^3/(4\times10^9\,\hbar c)$ with $\hbar c=1.973\times10^{-7}\ \mathrm{eV\,m}$ . The first oscillation maximum lands at $1.27\,\Delta m^2 L/E=\pi/2$ , i.e. $L/E=\pi/(2.53\,\Delta m^2)$ — the baseline design equation once $\Delta m^2$ is fixed. Whether the beam can be seen at all is a separate, purely kinematic question. A neutrino reveals itself through a charged-current reaction $\nu_\ell\, n\to\ell^-\, p$ , which must supply the rest energy of the final-state lepton and proton; working in the invariant $s$ with the target at rest gives the minimum beam energy.

E_\nu^{\rm thr}=\frac{(m_\ell+m_p)^2-m_n^2}{2m_n} \qquad (3)

where: $m_\ell$ — charged-lepton mass; $m_p,m_n$ — proton and neutron masses. Because $m_n>m_p$ , the electron channel is essentially threshold-free ( $E^{\rm thr}<0$ ); the muon channel opens at $110\ \mathrm{MeV}$ and the tau channel only at $3.45\ \mathrm{GeV}$ — this single formula makes appearance flavour-dependent. Real neutrinos have three flavours, and the mixing is packaged in the unitary Pontecorvo–Maki–Nakagawa–Sakata matrix $U$ relating flavour to mass states via $|\nu_\alpha\rangle=\sum_i U_{\alpha i}^*|\nu_i\rangle$ . It factorises into three rotations carrying the angles measured by atmospheric, reactor, and solar experiments, plus one CP-violating phase.

U=\begin{pmatrix}1&0&0\\0&c_{23}&s_{23}\\0&-s_{23}&c_{23}\end{pmatrix}\!\begin{pmatrix}c_{13}&0&s_{13}e^{-i\delta}\\0&1&0\\-s_{13}e^{i\delta}&0&c_{13}\end{pmatrix}\!\begin{pmatrix}c_{12}&s_{12}&0\\-s_{12}&c_{12}&0\\0&0&1\end{pmatrix} \qquad (4)

where: $c_{ij}=\cos\theta_{ij}$ , $s_{ij}=\sin\theta_{ij}$ ; the phase $\delta$ appears only alongside the small $s_{13}$ . Measured values are $\theta_{12}\approx33^\circ$ (solar), $\theta_{23}\approx45^\circ$ (atmospheric), $\theta_{13}\approx8.6^\circ$ (reactor), with $\Delta m^2_{21}\approx7.5\times10^{-5}\,\mathrm{eV}^2$ and $|\Delta m^2_{31}|\approx2.5\times10^{-3}\,\mathrm{eV}^2$ . The golden appearance channel of accelerator experiments is $\nu_\mu\to\nu_e$ . In three flavours it is driven by $\theta_{13}$ at the atmospheric frequency, with the leading term below dominating away from matter and CP effects.

P(\nu_\mu\to\nu_e)\approx\sin^2\theta_{23}\,\sin^2 2\theta_{13}\,\sin^2\!\left(\frac{\Delta m^2_{31}L}{4E}\right) \qquad (5)

where: the amplitude is set by the once-unknown $\theta_{13}$ — nonzero (Daya Bay, 2012), which opened the door to hunting for CP violation through the sub-leading terms that carry $\delta$ .

Derivation AFrom flavour mixing to the oscillation probability and the 1.27 factor

The goal of this derivation is to obtain equation (1) from first principles and then pin down the practical coefficient in equation (2), so the whole experimental programme rests on a transparent calculation rather than a memorised formula. We write a flavour state as a superposition of two mass eigenstates, evolve each with its own quantum phase, project onto the other flavour and square; the ultrarelativistic limit collapses the phase, and restoring $\hbar$ and $c$ yields $1.267$ .

Step 1. Write the produced flavour state as $|\nu_\alpha\rangle=\cos\theta\,|\nu_1\rangle+\sin\theta\,|\nu_2\rangle$ and the orthogonal flavour as $|\nu_\beta\rangle=-\sin\theta\,|\nu_1\rangle+\cos\theta\,|\nu_2\rangle$ . Why: the two-flavour mixing matrix is a single rotation, so one angle $\theta$ fully specifies how the detectable flavour states are built from the propagating mass states.

Step 2. Propagate each mass eigenstate as a plane wave, $|\nu_i(t)\rangle=e^{-iE_i t}|\nu_i\rangle$ , so after time $t$ the state is $\cos\theta\,e^{-iE_1 t}|\nu_1\rangle+\sin\theta\,e^{-iE_2 t}|\nu_2\rangle$ . Why: the mass eigenstates are the states of definite energy, hence the only ones with a simple time evolution — flavour states have none, which is why flavour changes.

Step 3. For fixed momentum $p\gg m_i$ , expand $E_i=\sqrt{p^2+m_i^2}\approx E+m_i^2/2E$ , so the relative phase is $\Delta\phi=(E_2-E_1)t\approx\Delta m^2 L/2E$ , using $t\approx L$ for relativistic neutrinos. Why: the common $p$ term is an overall phase that cancels in the probability; only the tiny mass-dependent difference survives, growing linearly with baseline.

Step 4. Project onto $|\nu_\beta\rangle$ and square: $P=|\langle\nu_\beta|\nu_\alpha(t)\rangle|^2=\sin^2 2\theta\,\sin^2(\Delta\phi/2)=\sin^2 2\theta\,\sin^2(\Delta m^2 L/4E)$ . Why: the interference between the two mass components, weighted by the mixing, produces the $\sin^2$ envelope — oscillation is literally a two-path interference experiment run on a single particle.

Step 5. Restore units. The phase $\Delta m^2 L/4E$ carries $c$ and $\hbar$ as $\Delta m^2 c^4 L/(4\hbar c\,E)$ ; inserting $\hbar c=1.973\times10^{-7}\ \mathrm{eV\,m}$ and converting $L\to$ km, $E\to$ GeV gives $1.27\,\Delta m^2[\mathrm{eV}^2]L[\mathrm{km}]/E[\mathrm{GeV}]$ . Why: the innocuous $1.27$ is the bridge between theorist units and the detector units used to fit every experiment. *The result is $P=\sin^2 2\theta\,\sin^2(1.27\,\Delta m^2 L/E)$ , with the coefficient following from $10^3/(4\times10^9\times1.973\times10^{-7})=1.267$ .*

Go deeper · Derivation B · 1 minThe charged-current threshold that decides which flavours can appear

Derivation BThe charged-current threshold that decides which flavours can appear

The aim here is to derive equation (3) — the minimum neutrino energy to produce a charged lepton of mass $m_\ell$ on a nucleon — and evaluate it for the electron, muon and tau, so the flavour-dependence of appearance experiments follows as arithmetic. The tool is the Mandelstam invariant $s$ , the squared centre-of-mass energy, which must at least equal the squared sum of the final-state rest masses for the reaction $\nu_\ell\, n\to\ell^-\, p$ to proceed.

Step 1. Compute $s$ for a massless neutrino of energy $E_\nu$ striking a nucleon of mass $m_n$ at rest: $s=(p_\nu+p_n)^2=m_n^2+2E_\nu m_n$ . Why: $s$ is Lorentz-invariant, so evaluating it in the lab frame gives a frame-independent condition.

Step 2. Impose the threshold condition that the final state be produced at rest in the centre-of-mass frame, $s_{\min}=(m_\ell+m_p)^2$ . Why: at threshold every scrap of available energy goes into rest mass with nothing left for kinetic motion, which is exactly the minimum-energy configuration.

Step 3. Equate and solve for the beam energy: $m_n^2+2E_\nu^{\rm thr}m_n=(m_\ell+m_p)^2$ , hence $E_\nu^{\rm thr}=[(m_\ell+m_p)^2-m_n^2]/2m_n$ . Why: this isolates the single quantity the accelerator controls — the beam energy — in terms of fixed particle masses.

Step 4. Evaluate. For the electron, $(0.511+938.27)^2-939.57^2<0$ , so the channel is open at essentially any energy (the neutron being heavier than the proton makes it nearly exothermic). For the muon, $E^{\rm thr}=[(105.66+938.27)^2-939.57^2]/(2\times939.57)=110\ \mathrm{MeV}$ . For the tau, $E^{\rm thr}=[(1776.9+938.27)^2-939.57^2]/(2\times939.57)=3.45\ \mathrm{GeV}$ . Why: the enormous tau threshold is why low-energy beams that oscillate into $\nu_\tau$ produce no visible $\tau^-$ , and why only high-energy beams like CNGS could run a tau-appearance search. *The three results are: $E^{\rm thr}_e<0$ (threshold-free), $E^{\rm thr}_\mu=110\ \mathrm{MeV}$ , and $E^{\rm thr}_\tau=3.45\ \mathrm{GeV}$ .*

Go deeper · Derivation C · 1 minWhy CP violation demands three neutrino generations

Derivation CWhy CP violation demands three neutrino generations

This derivation counts the physically meaningful parameters of an $N\times N$ unitary mixing matrix to show that a CP-violating phase can exist only for $N\ge 3$ — the reason a measurement of $\delta$ probes three-flavour structure, and the reason the two-flavour picture used everywhere else here is exactly CP-symmetric. We start from the free parameters of a unitary matrix, subtract the rotation angles, then subtract the phases removable by redefining the particle fields, leaving the irremovable phases.

Step 1. A general $N\times N$ unitary matrix has $N^2$ real parameters. Why: unitarity imposes $N^2$ real constraints on the $2N^2$ reals of a complex matrix, leaving $N^2$ free.

Step 2. Of these, $N(N-1)/2$ are rotation angles (the parameters of the real orthogonal subgroup), so the remaining $N^2-N(N-1)/2=N(N+1)/2$ are phases. Why: the angles are the CP-conserving mixing; any excess beyond them are complex phases that could in principle violate CP.

Step 3. Rephasing the $N$ charged-lepton and $N$ neutrino fields removes $2N-1$ phases (one overall phase is common and does not act). Why: physics is invariant under $\nu_i\to e^{i\alpha_i}\nu_i$ and $\ell_\alpha\to e^{i\beta_\alpha}\ell_\alpha$ , so any phase that can be absorbed this way is unphysical and cannot be observable.

Step 4. The physical, irremovable phases number $N(N+1)/2-(2N-1)=(N-1)(N-2)/2$ : $0$ for $N=2$ , $1$ for $N=3$ . Why: two flavours admit no surviving phase, so their mixing matrix is real and CP-conserving; three flavours admit exactly one phase $\delta$ , the sole source of Dirac CP violation in the lepton sector. *The formula $(N-1)(N-2)/2$ gives 0 physical phases for $N=2$ , 1 for $N=3$ , and 3 for $N=4$ .*

Limiting cases

Case	Result	Meaning
Near detector, $\Delta m^2 L/4E\to0$	$P\to\sin^2 2\theta\,(\Delta m^2 L/4E)^2$	oscillation grows quadratically; a near detector measures the un-oscillated flux
First maximum, $1.27\,\Delta m^2 L/E=\pi/2$	$P=\sin^2 2\theta$	deepest disappearance / strongest appearance — where experiments are parked
Long baseline / broad spectrum	$\langle P\rangle=\tfrac12\sin^2 2\theta$	rapid oscillations average out; only the mixing amplitude survives (solar, atmospheric averaging)
No mixing, $\theta\to0$ , or degenerate masses, $\Delta m^2\to0$	$P\to0$	oscillation needs both mixing and a mass splitting — a null result bounds the product
Electron appearance	$E^{\rm thr}\approx0$	$\nu_e$ CC is threshold-free; $\nu_e$ always detectable if present
Tau appearance	$E^{\rm thr}=3.45\ \mathrm{GeV}$	only high-energy beams can show a $\tau^-$ ; low-E $\nu_\tau$ are invisible in appearance
Two flavours	$\delta$ absent, $P(\nu)=P(\bar\nu)$	no CP violation possible; needs the three-flavour phase

Common trapAssuming any beam that oscillates into ν_τ will show a τ⁻ appearance signal

It will not, and the reason is kinematic, not a detector inefficiency. Producing a $\tau^-$ through the only detectable process, $\nu_\tau\, n\to\tau^-\, p$ , requires a beam energy above $E_\nu^{\rm thr}=3.45\ \mathrm{GeV}$ on a nucleon (and far higher on an atomic electron, where the tiny target mass makes $s$ minute). A megawatt beam tuned to $0.6\ \mathrm{GeV}$ for a $\nu_\mu\to\nu_e$ search can be full of oscillated $\nu_\tau$ and still register exactly zero tau events. This is why the direct observation of $\nu_\tau$ appearance required a dedicated high-energy beam — CERN's CNGS at $\langle E\rangle\approx17\ \mathrm{GeV}$ feeding OPERA over 730 km. The lesson generalises: oscillation probability and observability are different questions, and the second is answered by equation (3), not equation (1).

Worked examples

1 · The muon and tau charged-current thresholds (fully worked)

Using $m_p=938.27\ \mathrm{MeV}$ , $m_n=939.57\ \mathrm{MeV}$ , $m_\mu=105.66\ \mathrm{MeV}$ , $m_\tau=1776.9\ \mathrm{MeV}$ , find the minimum neutrino energy to produce a $\mu^-$ and a $\tau^-$ on a neutron, and comment on the electron channel.

Step 1. Muon numerator: $(m_\mu+m_p)^2-m_n^2=(1043.93)^2-(939.57)^2=1{,}089{,}790-882{,}792=206{,}998\ \mathrm{MeV}^2$ .

Step 2. Divide by $2m_n$ : $E^{\rm thr}_\mu=206{,}998/1879.1=\mathbf{110\ MeV}$ . A beam below this cannot make a muon no matter how much $\nu_\mu$ it contains.

Step 3. Tau numerator: $(m_\tau+m_p)^2-m_n^2=(2715.17)^2-(939.57)^2=7{,}372{,}148-882{,}792=6{,}489{,}356\ \mathrm{MeV}^2$ ; divide by $1879.1$ to get $E^{\rm thr}_\tau=\mathbf{3.45\ GeV}$ , more than thirty times the muon value.

Step 4. Electron: $(m_e+m_p)^2-m_n^2=(938.78)^2-(939.57)^2<0$ , so $E^{\rm thr}_e<0$ — essentially threshold-free because the neutron outweighs the proton. The three thresholds — 0, 110 MeV, 3.45 GeV — are the key constraint on which appearance channels a given beam can access.

2 · Parking T2K on the first oscillation maximum (fully worked)

T2K fires a $\nu_\mu$ beam 295 km from J-PARC to Super-Kamiokande, with $\Delta m^2_{32}=2.5\times10^{-3}\ \mathrm{eV}^2$ . At what beam energy is $\nu_\mu$ disappearance strongest, and how deep is the dip for maximal mixing?

Step 1. The first oscillation maximum is where the phase equals $\pi/2$ : $1.27\,\Delta m^2 L/E=\pi/2$ , so $L/E=\pi/(2\times1.27\times\Delta m^2)=\pi/(2.54\times2.5\times10^{-3})=\mathbf{496\ km/GeV}$ .

Step 2. Solve for energy at $L=295\ \mathrm{km}$ : $E=L/496=295/496=\mathbf{0.595\ GeV}$ . This is exactly why T2K designed an off-axis beam with a narrow spectrum peaked near $0.6\ \mathrm{GeV}$ .

Step 3. Depth of the dip: at the maximum $P(\nu_\mu\to\nu_\mu)=1-\sin^2 2\theta_{23}$ , and with near-maximal mixing $\sin^2 2\theta_{23}\approx1$ the survival probability drops essentially to zero. A sharply peaked 0.6 GeV beam at 295 km lands on the deepest point of the survival curve, maximising the disappearance signal.

3 · The reactor deficit that measured θ₁₃ (fully worked)

Daya Bay compares reactor $\bar\nu_e$ rates at a far detector ( $L\approx1.6\ \mathrm{km}$ , $E\approx4\ \mathrm{MeV}$ ) with $\Delta m^2_{31}=2.5\times10^{-3}\ \mathrm{eV}^2$ and $\theta_{13}=8.6^\circ$ . Estimate the survival deficit.

Step 1. Oscillation phase: $1.27\,\Delta m^2 L/E=1.27\times2.5\times10^{-3}\times1.6/0.004=1.27\ \mathrm{rad}$ , so $\sin^2(1.27)=\sin^2(72.7^\circ)=\mathbf{0.910}$ — near the first maximum, which is why the far detector sits at $1.6\ \mathrm{km}$ .

Step 2. Mixing amplitude: $2\theta_{13}=17.2^\circ$ , $\sin 17.2^\circ=0.296$ , so $\sin^2 2\theta_{13}=\mathbf{0.087}$ — small, because $\theta_{13}$ is the smallest mixing angle.

Step 3. Deficit: $1-P(\bar\nu_e\to\bar\nu_e)=\sin^2 2\theta_{13}\,\sin^2(1.27\,\Delta m^2 L/E)=0.087\times0.910=\mathbf{0.080}$ . About an 8 % disappearance of reactor antineutrinos at the far detector — a small but unmistakable deficit whose 2012 measurement fixed the last mixing angle and made the search for leptonic CP violation possible.

Physics of the portrait

Neutrino oscillation is one of the rare cases where the theory arrived decades before any experiment could test it.

Origins · 1 minBruno Pontecorvo and how this was discovered

Bruno Pontecorvo

1913–1993 · who first proposed that neutrinos could oscillate between states

Bruno Pontecorvo was Enrico Fermi's youngest student in the Rome group of the 1930s, and he carried Fermi's instinct for the weak interaction through a life that crossed most of twentieth-century physics and its politics. Working from ideas about the mixing of neutral kaons, in 1957 he proposed that neutrinos might oscillate — at first between neutrino and antineutrino, since only one flavour was then known. When the muon neutrino was discovered in 1962, and Maki, Nakagawa and Sakata wrote down a flavour-mixing scheme the same year, Pontecorvo generalised his idea to oscillation between flavours, the form tested today; the mixing matrix carries his initial in the acronym PMNS. He also invented the chlorine radiochemical method Ray Davis would use at Homestake to find the solar-neutrino deficit — the first hint, unrecognised for decades, that his oscillations were real. Pontecorvo did not live to see the 1998 Super-Kamiokande result that vindicated him, nor the 2015 Nobel Prize, but every oscillation experiment tests the proposal he made when there was no way to check it.

History

1957

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

1968

— Davis's Homestake chlorine experiment finds only about a third of the expected solar neutrinos

1998

— Super-Kamiokande observes the zenith-angle dependence of atmospheric ν_μ, the first firm oscillation signal

2002

— SNO shows solar neutrinos change flavour and KamLAND confirms it with reactor antineutrinos

2015

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

Going further

Texts. Thomson, Modern Particle Physics, §13.6 (oscillation experiments) and §13.6.1 (interaction thresholds), with the PMNS matrix at Eq. (13.33) · Griffiths, Introduction to Elementary Particles, 2nd ed., §11.6 (neutrino oscillations) · Halzen & Martin, Quarks and Leptons, Ch. 12 (weak interactions of leptons) · Peskin & Schroeder, An Introduction to Quantum Field Theory, §20.3 (fermion masses and mixing) · Schwartz, Quantum Field Theory and the Standard Model, Ch. 29 · Zuber, Neutrino Physics, 3rd ed. (the dedicated monograph). Key numerical checks across this article: $P=\sin^2 2\theta\,\sin^2(1.27\,\Delta m^2 L/E)$ with coefficient $1.267$ ; $E^{\rm thr}_\mu=110\ \mathrm{MeV}$ , $E^{\rm thr}_\tau=3.45\ \mathrm{GeV}$ , $E^{\rm thr}_e<0$ ; first-maximum $L/E=496\ \mathrm{km/GeV}$ and T2K energy $0.60\ \mathrm{GeV}$ at 295 km; Daya Bay deficit $\approx8\%$ ; physical CP phases $(N-1)(N-2)/2=0,1$ for $N=2,3$ .

Prerequisites. Neutrino flavours · Two-flavour oscillations · Three-flavour PMNS mixing · Mandelstam variables. Leads to. Reactor neutrinos · Long-baseline experiments · Matter effects (MSW) · CP violation in the Standard Model.

Ask the tutor: Ask me to walk through how a single beam runs appearance and disappearance at once, or to work the $L/E$ and threshold numbers for any source from reactors to the atmosphere.

What is the real difference between appearance and disappearance experiments?

They are two ways of reading the same oscillation. In a disappearance experiment you produce a beam of one flavour, predict how many should reach the detector if nothing happened, and look for a shortfall — Super-Kamiokande counting fewer atmospheric $\nu_\mu$ , or a reactor experiment counting fewer $\bar\nu_e$ . In an appearance experiment you look for the flavour that was not in the beam — an $e^-$ or $\tau^-$ from a $\nu_\mu$ beam — which can only come from a neutrino that changed identity. Disappearance is usually statistically cleaner because the signal starts large; appearance is more direct evidence but limited by thresholds and backgrounds. Modern long-baseline experiments run both channels at once from the same beam.

Why does oscillation prove that neutrinos have mass?

The oscillation phase in equation (1) is proportional to $\Delta m^2=m_2^2-m_1^2$ . If all neutrinos were massless, every splitting would vanish, the phase would be zero, and the flavour-change probability would be identically zero at every baseline. So the mere observation of any flavour change — a solar deficit, an atmospheric asymmetry, a reactor shortfall — requires at least two neutrinos to have different, and therefore nonzero, masses. This was the headline significance of the 1998 and 2002 results: the Standard Model as originally written has massless neutrinos, so oscillation is direct evidence for physics beyond it. Oscillation measures only differences of squared masses, not the absolute scale.

Why do different experiments use such wildly different baselines?

Because the oscillation depends on $L/E$ matched to a particular $\Delta m^2$ , and nature handed us two very different splittings. The atmospheric splitting $\Delta m^2_{32}\approx2.5\times10^{-3}\ \mathrm{eV}^2$ needs $L/E\approx500\ \mathrm{km/GeV}$ , met by an accelerator with a few-hundred-km baseline at GeV energies, or by the atmosphere over thousands of km. The solar splitting $\Delta m^2_{21}\approx7.5\times10^{-5}\ \mathrm{eV}^2$ is about thirty times smaller, so it needs thirty times the $L/E$ — met by reactor antineutrinos over ~180 km (KamLAND) or by the Sun over an astronomical unit. Picking a baseline is really picking which $\Delta m^2$ you want to see; too short and the phase is tiny, too long and it averages to a constant.

Where does the factor 1.27 in the oscillation formula come from?

It is pure unit conversion, not physics. In natural units the phase is $\Delta m^2 L/4E$ , but $\Delta m^2$ is measured in eV², baselines in kilometres, and beam energies in GeV, so you must reinstate the constants $\hbar$ and $c$ that natural units had set to one. Carrying $\hbar c=1.973\times10^{-7}\ \mathrm{eV\,m}$ and converting metres to kilometres and eV to GeV collapses all the powers of ten into the single coefficient $1.267$ , usually rounded to $1.27$ . So equation (2) is exactly equation (1) rewritten in the units an experimenter reads off dials — the quantum mechanics is unchanged.

Why can't two-flavour mixing produce CP violation?

CP violation in mixing shows up as a complex phase that cannot be rotated away by redefining the particle fields. An $N\times N$ unitary matrix has $(N-1)(N-2)/2$ such physical phases: for two flavours that is zero, so the matrix can always be made completely real — a single rotation angle — and a real matrix treats $\nu$ and $\bar\nu$ identically. For three flavours the count is one, the phase $\delta$ of the PMNS matrix, which is why leptonic CP violation is intrinsically a three-generation phenomenon. So any measured difference between $\nu_\mu\to\nu_e$ and $\bar\nu_\mu\to\bar\nu_e$ appearance is a direct probe of three-flavour structure that no two-flavour approximation can mimic.

Do antineutrinos oscillate the same way as neutrinos?

In vacuum, and if CP were exactly conserved, yes — the probabilities would be identical, because the antineutrino mixing matrix is the complex conjugate of the neutrino one and the leading $\sin^2$ terms are unchanged by conjugation. Two things break the symmetry. First, the CP phase $\delta$ enters the sub-leading appearance terms with opposite sign for $\nu$ and $\bar\nu$ , so a genuine difference between $P(\nu_\mu\to\nu_e)$ and $P(\bar\nu_\mu\to\bar\nu_e)$ is the CP-violation signature DUNE and Hyper-Kamiokande hunt for. Second, matter affects them differently — the Earth has electrons but not positrons — so the MSW effect enhances one and suppresses the other, a "fake" asymmetry that must be subtracted from the true CP signal.