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Wyatt's Notes — University

Neutrino Physics

8.1 Neutrino Oscillations

Neutrinos are produced and detected in flavour eigenstates (νe,νμ,ντ)(\nu_e, \nu_\mu, \nu_\tau)But propagate As mass eigenstates (ν1,ν2,ν3)(\nu_1, \nu_2, \nu_3) related by the PMNS mixing matrix UU:

∣να⟩=∑iUαi∗∣νi⟩|\nu_\alpha\rangle = \sum_i U_{\alpha i}^* |\nu_i\rangle

As a neutrino of flavour α\alpha propagates, the mass eigenstates acquire different phases: exp⁡(−imi2L/(2E))\exp(-im_i^2 L/(2E))Leading to oscillations.

Two-flavour oscillation probability:

P(να→νβ)=sin⁡2(2θ)sin⁡2(Δm2L4E)P(\nu_\alpha \to \nu_\beta) = \sin^2(2\theta)\sin^2\left(\frac{\Delta m^2 L}{4E}\right)

Where Δm2=m22−m12\Delta m^2 = m_2^2 - m_1^2, θ\theta is the mixing angle, LL is the distance, and EE is the Energy.

Evidence: The Solar Neutrino Problem (deficit of νe\nu_e from the Sun, resolved by νe→νμ,ντ\nu_e \to \nu_\mu, \nu_\tau oscillations) and atmospheric neutrino oscillations (Super-Kamiokande, 1998).

8.2 Neutrino Masses

Neutrino oscillations imply that neutrinos have mass, but the masses are extremely small: ∑mν<0.12\sum m_\nu \lt 0.12 eV (Planck 2018).

In the Standard Model, neutrinos are massless. Their masses require physics beyond the Standard Model, most commonly via the seesaw mechanism:

mν∼mD2Mm_\nu \sim \frac{m_D^2}{M}

Where mDm_D is a Dirac mass and M≫mDM \gg m_D is the mass of a heavy right-handed neutrino.

Example 8.1: Atmospheric neutrino oscillation calculation

Atmospheric neutrinos are produced when cosmic rays strike the upper atmosphere, creating Pions that decay: π+→μ++νμ\pi^+ \to \mu^+ + \nu_\muFollowed by μ+→e++νˉμ+νe\mu^+ \to e^+ + \bar{\nu}_\mu + \nu_e.

Super-Kamiokande (1998) observed that upward-going muon neutrinos (travelling through the Earth, L∼104L \sim 10^4 km) were significantly depleted relative to downward-going ones (L∼10L \sim 10 km), while electron neutrinos showed no such deficit.

Using the two-flavour formula with the atmospheric parameters Δm322≈2.5×10−3\Delta m^2_{32} \approx 2.5 \times 10^{-3} eV2^2 and sin⁡2(2θ23)≈1\sin^2(2\theta_{23}) \approx 1 (maximal mixing):

For upward-going νμ\nu_\mu with E=1E = 1 GeV and L=10 000L = 10\,000 km:

\frac{\Delta m^2 L}{4E} = \frac{2.5 \times 10^{-3}\;\mathrm{eV}^2 \times 10^4\;\mathrm{km}{4 \times 1\;\mathrm{GeV}}}

Converting to natural units (ℏc=1.973×10−7\hbar c = 1.973 \times 10^{-7} eV⋅\cdotM): L=107L = 10^7 m, so L/E=107/109=10−2L/E = 10^7 / 10^9 = 10^{-2} eV−1^{-1}.

Δm2L4E=2.5×10−3×10−24=6.25×10−6  eV2⋅eV−1\frac{\Delta m^2 L}{4E} = \frac{2.5 \times 10^{-3} \times 10^{-2}}{4} = 6.25 \times 10^{-6}\;\mathrm{eV}^2\cdot\mathrm{eV}^{-1}

Wait --- we need to be more careful with units. Using the practical formula:

Δm2[eV2]⋅L[km]4E[GeV]=2.5×10−3×1044×1=254=6.25  rad\frac{\Delta m^2 [\mathrm{eV}^2] \cdot L [\mathrm{km}]}{4E [\mathrm{GeV}]} = \frac{2.5 \times 10^{-3} \times 10^4}{4 \times 1} = \frac{25}{4} = 6.25\;\mathrm{rad}

P(νμ→νμ)=1−sin⁡2(2θ23)sin⁡2(6.25)=1−1×sin⁡2(6.25)≈1−0.018≈0.98P(\nu_\mu \to \nu_\mu) = 1 - \sin^2(2\theta_{23})\sin^2(6.25) = 1 - 1 \times \sin^2(6.25) \approx 1 - 0.018 \approx 0.98

Hmm, this gives almost no oscillation. Let me reconsider. Actually:

P(νμ→ντ)=sin⁡2(2θ)sin⁡2(Δm2L4E)=sin⁡2(6.25)≈0.018P(\nu_\mu \to \nu_\tau) = \sin^2(2\theta)\sin^2\left(\frac{\Delta m^2 L}{4E}\right) = \sin^2(6.25) \approx 0.018

This seems small. But at E=0.5E = 0.5 GeV:

Δm2L4E=252=12.5  rad\frac{\Delta m^2 L}{4E} = \frac{25}{2} = 12.5\;\mathrm{rad}

sin⁡2(12.5)≈sin⁡2(0.35)≈0.12\sin^2(12.5) \approx \sin^2(0.35) \approx 0.12

And at the first oscillation maximum, L/E=2π/(Δm2)=2π/(2.5×10−3)≈2513L/E = 2\pi/(\Delta m^2) = 2\pi/(2.5 \times 10^{-3}) \approx 2513 km/GeV. For E=1E = 1 GeV, Losc≈2513L_{\mathrm{osc} \approx 2513} km, which is comparable to the Earth”s diameter (∼12 700\sim 12\,700 km). The observed deficit is an average over many oscillations and energies, Giving roughly ⟨P⟩≈1/2\langle P\rangle \approx 1/2 for maximal mixing, consistent with the Super-Kamiokande observation of approximately half the expected upward-going νμ\nu_\mu flux.