
Nuclear, particle & quantum-information sequence · Interactive lesson
PMNS mixing and flavour oscillation
Neutrinos are made and detected by the weak charged current, which knows only flavour, and they travel as eigenstates of the free Hamiltonian, which knows only mass. The two bases are rotated relative to each other, so a beam changes identity in flight. Here that becomes arithmetic you can do on an envelope.
UPV-17.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Nuclear, particle & quantum-information sequence. These units continue into nuclear and particle physics and quantum information. Follow your university’s prerequisites, mathematical depth and assessment; this independent pathway is not a universal programme specification.
Build the model
Two Hermitian operators that refuse to share an eigenbasis.
The key idea
Neutrino oscillation is the periodic change of a neutrino's measured flavour with distance, because each flavour state is a coherent superposition of mass eigenstates that accumulate different phases as they propagate.
Example. A 0.60 GeV νμ crossing the 295 km from J-PARC to Super-Kamiokande picks up a phase 1.27 × 2.5×10⁻³ × 295/0.60 = 1.56 rad, one hair under π/2, so 99% of the beam has left the muon flavour on arrival.
The complete physical model and assumptions
A neutrino is born at a weak charged-current vertex beside a definite charged lepton, so production and detection define the flavour basis (|νₑ⟩, |νμ⟩, |ντ⟩). Propagation is governed by the free Hamiltonian, whose eigenbasis is the mass basis (|ν₁⟩, |ν₂⟩, |ν₃⟩) with eigenvalues Eᵢ = √(p² + mᵢ²). Two Hermitian operators share an eigenbasis only if they commute, and these two do not: the neutrino mass matrix, whatever generates it, is not diagonal in flavour. Diagonalising it costs a unitary rotation, the PMNS matrix U, with |να⟩ = Σᵢ U*αi |νᵢ⟩. Everything else is bookkeeping. Rotate into the mass basis, apply the diagonal propagator exp(−i mᵢ² L/2E) that the ultrarelativistic expansion Eᵢ ≈ E + mᵢ²/2E supplies, rotate back, and square. The interference between the three phases is the oscillation. The common factor exp(−i m₁² L/2E) is a global phase and cancels in |A|², which is the whole cost of the model: oscillation sees two squared splittings, three angles and one CP phase, and is structurally blind to the absolute mass scale, to the Majorana phases, and — in vacuum, at two flavours — even to the sign of Δm²₃₁.
1Two bases, and the rotation between them
The weak charged current couples νₑ to e, νμ to μ, ντ to τ. That vertex is what a source and a detector actually measure, so it defines the flavour basis. Insert the mass-basis resolution of the identity, Σᵢ |νᵢ⟩⟨νᵢ| = 1, and each flavour state is expanded with coefficients ⟨νᵢ|να⟩ = U*αi. The PMNS matrix is nothing more than that table of inner products between two orthonormal bases, which is why it is unitary, and it differs from the identity because the neutrino mass matrix, whatever mechanism generates it, is not diagonal in the basis where the charged-lepton masses are. With two flavours the table is a real rotation, |νₑ⟩ = cos θ |ν₁⟩ + sin θ |ν₂⟩ and |νμ⟩ = −sin θ |ν₁⟩ + cos θ |ν₂⟩, and one angle is the whole story. Note what propagation does not change: the mass content of the travelling state |ψ(L)⟩, |⟨νᵢ|ψ(L)⟩|² = |Uαi|², is the same at every L. Only the relative phases of the components evolve, and that moves the state relative to the flavour basis your detector is built to ask about.
2Expand Eᵢ and read the phase off
In the mass basis the propagator is diagonal, exp(−iEᵢt/ħ), which is exactly why you rotate there. Neutrinos in every terrestrial experiment are ultrarelativistic — mᵢ ≲ 0.1 eV against E ≳ 1 MeV — so Eᵢ = √(p² + mᵢ²) ≈ p + mᵢ²/2E, and with t ≈ L the phase is Eᵢt ≈ pL + mᵢ²L/2E. The pL piece is identical for all three components and factors out. Restoring units, the relative phase between components i and j is Δm²ᵢⱼc⁴L/(2ħcE), and half of it, Δm²c⁴L/(4ħcE), is what sits inside the sin². In the units experiments use — Δm² in eV², L in km, E in GeV — the numerical factor is 10³/(4 × 10⁹ × 1.973×10⁻⁷) = 1.267. That is all the famous 1.27 is: a unit conversion. Whether you assume the three components share a momentum or share an energy shifts the phase only at relative order m²/E² ≲ 10⁻¹⁴, which no experiment can see.
3Square the amplitude: why Δm² and never m²
Write the amplitude and factor out the first phase: A(α→β) = exp(−i m₁²L/2E) Σᵢ U*αi exp(−i Δm²ᵢ₁ L/2E) Uβi. The prefactor is a global phase, and P = |A|² is blind to it. So the observables are the two independent splittings Δm²₂₁ and Δm²₃₁, with Δm²₃₂ = Δm²₃₁ − Δm²₂₁ fixed by them — never the masses. Expanding the square gives P(α→β) = δαβ − 4 Σ(i>j) Re(U*αi Uβi Uαj U*βj) sin²(Δm²ᵢⱼL/4E) + 2 Σ(i>j) Im(U*αi Uβi Uαj U*βj) sin(Δm²ᵢⱼL/2E). The Re term is symmetric under α ↔ β; the Im term is antisymmetric and flips sign for antineutrinos. That second term is CP violation, and it survives only if all three splittings and a non-zero δ are present, which is why two-flavour physics can never show it.
4Depth and spacing are set by different numbers
Collapse to two flavours and P(α→β) = sin²2θ sin²(1.27 Δm² L/E). The mixing angle appears only in the prefactor and the splitting only inside the sine, so the two are fitted from different features of the same data: the depth of the dip gives sin²2θ, the L/E at which the dip sits gives Δm². Two real experiments show how far apart that can put you. Daya Bay watches reactor ν̄ₑ at E ≈ 4 MeV over L ≈ 1.65 km, which is L/E ≈ 410 km GeV⁻¹, and reads sin²2θ₁₃ = 0.086 — a floor only 8.6% below one. T2K watches νμ at E = 0.60 GeV over L = 295 km, L/E ≈ 492 km GeV⁻¹, and reads sin²2θ₂₃ ≈ 0.99 — a dip that removes 99% of the beam. Nearly the same L/E, nearly the same Δm², utterly different depths, because the two channels read different entries of the third column of U: |Uₑ₃|² ≈ 0.02 for νₑ, |Uμ₃|² ≈ 0.54 for νμ.
5Three flavours: one fast phase, one frozen
With Δm²₃₁/Δm²₂₁ = 2.5×10⁻³/7.5×10⁻⁵ = 33, the two phases in Pₑₑ run at very different rates, and which term you can see is a design choice. At Daya Bay's L/E = 412 km GeV⁻¹ the atmospheric phase is 1.31 rad and the solar phase is 0.039 rad, so sin²Φ₃₁ = 0.933 against sin²Φ₂₁ = 1.5×10⁻³, a factor 605. Folding in cos⁴θ₁₃ sin²2θ₁₂ = 0.814 and sin²2θ₁₃ = 0.086 gives a solar contribution of 0.0013 against an atmospheric 0.080 — still a factor 64, and still too big to throw away at 0.2% systematics. KamLAND, at L/E ≈ 5×10⁴ km GeV⁻¹, has the opposite problem: Φ₃₁ ≈ 159 rad oscillates far faster than its energy resolution, so sin²Φ₃₁ averages to ½ and the fast term becomes a constant offset rather than a fringe.
6What the plane wave hides, and what oscillation cannot reach
Two approximations were smuggled in. Treating each mass eigenstate as a plane wave gives it infinite extent, so the three components never stop overlapping; real wave packets separate after a coherence length, beyond which the interference terms damp and P settles at its average. And vacuum propagation is not what solar or long-baseline neutrinos do: coherent forward scattering off electrons adds V = √2 GF nₑ to the νₑ diagonal of the effective Hamiltonian only, and that term flips sign between ν and ν̄, so the ordering of the states — the sign of Δm²₃₁, invisible to a two-flavour vacuum fit because sin² is even — becomes measurable in matter. What no oscillation experiment can reach is the absolute scale: m₁ cancelled as a global phase, and the Majorana phases cancel because they multiply U on the right and leave every product U*αi Uβi untouched.
Key equations
Read the units, assumptions and limits before substituting.
Flavour basis and mass basis
|να⟩ = Σᵢ U*αi |νᵢ⟩, U†U = 1
Names both bases and the unitary map between them. Unitarity is what keeps Σβ P(α→β) = 1 at every L.
α = e, μ, τ labels the weak vertex; i = 1, 2, 3 labels the free-Hamiltonian eigenvalues; U is dimensionlessMass-basis propagator
|νᵢ(L)⟩ = exp(−i mᵢ² L / 2E) |νᵢ(0)⟩
Evolution is diagonal only in this basis, so rotate before you propagate and rotate back after.
From Eᵢ ≈ E + mᵢ²/2E with t ≈ L and ħ = c = 1, up to a phase common to all i; with mᵢ² in eV², L in km and E in GeV the exponent is −2.53i mᵢ²L/E, twice the 1.27Oscillation amplitude
A(α→β) = Σᵢ U*αi exp(−i mᵢ² L / 2E) Uβi
Three matrix operations — rotate, phase, rotate back. Every formula below is |A|² expanded.
Dimensionless complex amplitude; P(α→β) = |A|², and A = δαβ at L = 0 by unitarityTwo-flavour reduction
P(α→β) = sin²2θ sin²(1.27 Δm² L / E)
sin²2θ sets how deep the dip goes, Δm² sets where in L/E it falls. Fit them from different features.
Δm² in eV², L in km, E in GeV; the 1.27 is 10³/(4×10⁹ × 1.973×10⁻⁷), a unit conversionChoosing a baseline
1.27 Δm² L/E = π/2 → L/E ≈ 1237 / (Δm²/10⁻³ eV²)
This is how an experiment is designed: pick the splitting you want to see, then buy that L/E.
L/E in km GeV⁻¹. Δm²₃₁ = 2.5×10⁻³ eV² puts the first minimum at 495 km GeV⁻¹Standard PMNS parametrisation
U = R₂₃(θ₂₃) Ũ₁₃(θ₁₃, δ) R₁₂(θ₁₂) diag(1, e(iα₂₁/2), e(iα₃₁/2))
Three angles and one CP phase are all oscillation can see; the two Majorana phases survive rephasing only if neutrinos are their own antiparticles.
θ₁₂ ≈ 33.6°, θ₂₃ ≈ 47.9°, θ₁₃ ≈ 8.5°, δ ≈ 200°; the Majorana phases α cancel in every PInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set the depth to 0.09 for the reactor θ₁₃ channel, then raise it to 0.99 for the accelerator θ₂₃ channel: the dip goes from barely visible to near-total while the first minimum stays at L/E = {1236.85/max(dm,0.1)|0} km/GeV, because only Δm² decides where it falls.
Live results
- PHASE Φ = 1.27Δm²L/E
- 1.59 rad
- SURVIVAL P
- 0.100
- FIRST MINIMUM AT L/E
- 495 km/GeV
- BASELINE AT E = 0.6 GeV
- 300 km
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Mixing depth sin²2θ: 0.90; Splitting Δm²: 2.50 ×10⁻³ eV²; Detector L/E: 500 km/GeV
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Students read "oscillation proves neutrinos have mass" and then quote a mass. The fit returns Δm²₃₁ = 2.5×10⁻³ eV², and √(2.5×10⁻³) = 0.050 eV is not m₃; with Δm²₃₁ > 0 it is a lower bound on m₃, because m₃² = m₁² + Δm²₃₁ ≥ Δm²₃₁ whatever m₁ is. The lightest state could sit at 0.3 eV, making all three nearly degenerate at 0.3 eV, and every oscillation experiment ever run would return exactly the same probabilities — the common phase exp(−i m₁²L/2E) drops out of |A|² whatever m₁ is. Absolute mass comes from elsewhere: the β-decay endpoint (KATRIN, mβ < 0.45 eV), cosmology, or neutrinoless double β decay. And Δm² is not (Δm)²: with m₁ = 0.3 eV the same splitting gives m₃ − m₁ ≈ 0.004 eV, twelve times smaller than 0.050 eV.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A reactor supplies ν̄ₑ at E = 4.0 MeV to a detector 1.0 km away. Take Δm²₃₁ = 2.5×10⁻³ eV² and sin²2θ₁₃ = 0.086, and treat the channel as two-flavour. Find the survival probability P(ν̄ₑ→ν̄ₑ), then say how far the detector would have to sit for the deepest possible deficit.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Put the baseline and energy into the units the 1.27 rule expects: L = 1.0 km, E = 4.0 MeV = 4.0×10⁻³ GeV, so L/E = 250 km GeV⁻¹.
- Phase: Φ = 1.27 Δm² L/E = 1.27 × 2.5×10⁻³ × 250 = 0.794 rad.
- Oscillating factor: sin(0.794) = 0.713, so sin²Φ = 0.508 — the detector sits at about half of the first fringe.
- Survival: P = 1 − sin²2θ₁₃ sin²Φ = 1 − 0.086 × 0.508 = 1 − 0.0437 = 0.956.
- Deepest deficit needs Φ = π/2, i.e. L/E = 1.5708/(1.27 × 2.5×10⁻³) = 495 km GeV⁻¹, so L = 495 × 4.0×10⁻³ = 1.98 km at this energy.
Answer. P = 0.956, a 4.4% deficit. The full 8.6% dip would need a baseline near 2.0 km, which is why short-baseline reactor experiments are built at roughly that distance.
MediumExample 2
T2K sends νμ 295 km from J-PARC to Super-Kamiokande, and the beam is tuned so that its 0.60 GeV peak sits exactly at the first disappearance maximum. Extract Δm²₃₂, find the energy of the second maximum at the same baseline, and with sin²θ₂₃ = 0.55 give both the νμ survival probability and the leading-order νₑ appearance probability at the peak.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- First maximum means Φ = π/2. Here L/E = 295/0.60 = 491.7 km GeV⁻¹.
- Solve 1.27 Δm² × 491.7 = π/2: Δm² = 1.5708/(1.27 × 491.7) = 1.5708/624.4 = 2.52×10⁻³ eV².
- The second maximum is Φ = 3π/2 at the same L, so the phase must triple, which means the energy must fall by three: E = 0.60/3 = 0.20 GeV.
- Disappearance depth: sin²2θ₂₃ = 4 sin²θ₂₃ cos²θ₂₃ = 4 × 0.55 × 0.45 = 0.99, and sin²Φ = 1 at the maximum, so P(νμ→νμ) = 1 − 0.99 = 0.010.
- Appearance at leading order: P(νμ→νₑ) ≈ sin²θ₂₃ sin²2θ₁₃ sin²Φ = 0.55 × 0.086 × 1 = 0.047.
Answer. Δm²₃₂ = 2.5×10⁻³ eV²; the second maximum falls at E = 0.20 GeV; P(νμ→νμ) = 0.010 and P(νμ→νₑ) = 4.7%. The appearance channel is small because θ₁₃ is small, not because θ₂₃ is.
HardExample 3
A reactor experiment at L = 1.65 km measures ν̄ₑ at E = 4.0 MeV. Using Pₑₑ = 1 − cos⁴θ₁₃ sin²2θ₁₂ sin²Φ₂₁ − sin²2θ₁₃ sin²Φ₃₁ with sin²θ₁₂ = 0.307, sin²θ₁₃ = 0.0220, Δm²₂₁ = 7.5×10⁻⁵ eV² and Δm²₃₁ = 2.5×10⁻³ eV², compute Pₑₑ and decide whether the solar term may be dropped at 0.2% systematics.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- L/E = 1.65/4.0×10⁻³ = 412.5 km GeV⁻¹, the same number for both terms.
- Solar phase: Φ₂₁ = 1.27 × 7.5×10⁻⁵ × 412.5 = 0.0393 rad, so sin²Φ₂₁ = 1.54×10⁻³. Atmospheric phase: Φ₃₁ = 1.27 × 2.5×10⁻³ × 412.5 = 1.310 rad, so sin²Φ₃₁ = 0.933 — a ratio of 605.
- Mixing prefactors: sin²2θ₁₂ = 4 × 0.307 × 0.693 = 0.851, cos⁴θ₁₃ = 0.978² = 0.9565, so the solar prefactor is 0.814; sin²2θ₁₃ = 4 × 0.0220 × 0.978 = 0.0861.
- Solar term = 0.814 × 1.54×10⁻³ = 1.26×10⁻³. Atmospheric term = 0.0861 × 0.933 = 0.0803.
- Pₑₑ = 1 − 0.00126 − 0.0803 = 0.918.
- Dropping the solar term returns 0.9197 instead of 0.9184 — an absolute error of 0.0013, or 0.14% of the rate. That is comparable to the 0.2% systematic, so it must be kept.
Answer. Pₑₑ = 0.918. The atmospheric term costs 8.03 percentage points and the solar term 0.13, a factor 64 apart — small, but not negligible at 0.2% systematics.
Quick check
Select an answer and explain it before checking.
Review & continue
Neutrino oscillation is the periodic change of a neutrino's measured flavour with distance, because each flavour state is a coherent superposition of mass eigenstates that accumulate different phases as they propagate.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.