Majorana Neutrinos, and a Decay Nobody Has Seen

The question Majorana asked in 1937 is still open, and the only handle on it is a decay whose rate carries the square of a mass we already know to be nearly zero.

🎯 Why this matters

The width of every published limit is theoretical, not experimental. A matrix element known to a factor of three enters squared, so a factor of nine in sensitivity is available to whoever calculates it properly — with no detector involved at all.

The Standard Model says the neutrino is a Dirac particle with mass exactly zero. The mass part is dead — this chapter killed it. The Dirac part has been an open question since Majorana proposed the alternative in 1937, and after nearly a century there is still no experimental evidence either way.

A Majorana fermion is its own antiparticle. For a boson that is unremarkable — the photon and the $Z^0$ are their own antiparticles. For a fermion it requires every charge to vanish, and the neutrino is the only candidate in the Standard Model.

The condition, and the mass term it permits

\htmlClass{t-c}{\nu_i^C = \nu_i} \qquad\text{which permits}\qquad \htmlClass{t-m}{m_i\,\bar\nu_i^C \nu_i}

(10.74)

Bettini p. 470. The Majorana condition, and the mass term that becomes available once you impose it.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — the Majorana question and the mass-origin question are one question

This is why the question matters rather than being bookkeeping. The Standard Model gives neutrinos no mass because there is no right-chirality neutrino to pair a left one with. A Majorana mass term does not need one — it pairs the left field with its own charge conjugate, which is right-chiral for free.

So “is the neutrino Majorana?” and “where does the neutrino mass come from?” are the same question. And the price is lepton number: the term changes $L$ by two units, which is not a small modification of the Standard Model.

A second consequence, from §10.1: a Majorana field cannot be rephased, so two extra phases in the PMNS matrix become physical. They are invisible to everything in §§10.2–10.6 and they appear exactly once, in the quantity this section is about.

Racah’s test, and why its failure proves nothing

In 1937 Racah proposed a decisive-looking experiment. Take a neutrino from a $\beta^-$ decay and let a second nucleus absorb it. If neutrinos are Dirac particles, the one produced alongside an electron is an antineutrino, so it can only ever make a positron. If they are Majorana particles, the same particle should be able to make either sign of lepton.

Bettini Fig. 10.19. On the left, Racah’s 1937 test: emit a neutrino with an electron and see which sign of lepton it makes when absorbed. On the right, Furry’s 1939 improvement — put both vertices inside one nucleus, so that instead of catching a neutrino you only have to watch a macroscopic amount of matter and wait.

Racah’s process has never been seen. That does not prove neutrinos are Dirac particles, and the reason is the V−A structure of the weak interaction together with the smallness of the mass — neither of which Majorana or Racah knew about.

\nu \approx \htmlClass{t-w}{\tfrac12\tfrac{m}{E}}\,\nu^+ + \nu^- \qquad\qquad \bar\nu \approx \bar\nu^+ + \htmlClass{t-w}{\tfrac12\tfrac{m}{E}}\,\bar\nu^-

(10.75)

Bettini p. 471, from (2.63) and (2.65). A massive neutrino is not a helicity eigenstate — and the size of the contamination is the whole story.

Every symbol, one at a time

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💡 What this really says — the two cases differ only in what the wrong-helicity component does

The Dirac and Majorana cases differ only in what the wrong-helicity component does. Everything else is identical.

Take a $\pi^+ \to \mu^+\nu_\mu$ : the neutral particle comes out with predominantly $h = -1$ , and when it interacts it makes a $\mu^-$ . From a $\pi^-$ it has $h = +1$ and makes a $\mu^+$ . You never needed lepton number to explain that — helicity plus V−A does the whole job. Lepton number conservation and the Dirac hypothesis are, for every experiment ever done, unfalsifiable dressing on a result that helicity already explains.

The difference lives entirely in the small component. For a Dirac neutrino it still makes the lepton that lepton number demands. For a Majorana one it makes the opposite sign — and violates $L$ by two units.

Bettini Fig. 10.20. Long arrows are velocities, short thick ones spins. Column (a) is the massless case — one helicity each, no ambiguity. In (b) and (c) a small wrong-helicity component appears with amplitude m/2E, and the only difference between Dirac and Majorana is the red label in the last row. Everything measured so far is column (a) to a part in 10²⁰.

🔢 Worked example — how small is “too small to see”?

Rates and suppressions

import numpy as np
NA, ln2 = 6.02214076e23, np.log(2)

print("how many 0nu2beta decays could 1 t of isotope give in 1 yr, at the current limits?")
for iso, A, T in (("136Xe", 136, 2.3e26), ("76Ge", 76, 1.8e26), ("130Te", 130, 2.2e25)):
    N = 1e6/A*NA                       # nuclei in 1 t
    print(f"  {iso:6s} A={A:3d}   N = {N:.2e} nuclei   T_1/2 > {T:.1e} yr"
          f"   ->  fewer than {N*ln2/T:5.1f} decays / t / yr")

print("\nagainst a background target of below 1 count / keV / t / yr and a few-keV")
print("peak, so the whole search window holds only a handful of counts.\n")

E, m = 1.0e9, 0.100                    # eV: a 1 GeV neutrino of mass 100 meV
print(f"wrong-helicity amplitude m/2E   = {m/(2*E):.2e}   [Eq. (10.75)]")
print(f"lepton-number violation goes as its square = {(m/(2*E))**2:.2e}")
print(f"the book's figure and prose quote (m/E)^2  = {(m/E)**2:.2e}, a factor 4 larger")

prints

how many 0nu2beta decays could 1 t of isotope give in 1 yr, at the current limits?
136Xe  A=136   N = 4.43e+27 nuclei   T_1/2 > 2.3e+26 yr   ->  fewer than  13.3 decays / t / yr
76Ge   A= 76   N = 7.92e+27 nuclei   T_1/2 > 1.8e+26 yr   ->  fewer than  30.5 decays / t / yr
130Te  A=130   N = 4.63e+27 nuclei   T_1/2 > 2.2e+25 yr   ->  fewer than 146.0 decays / t / yr

against a background target of below 1 count / keV / t / yr and a few-keV
peak, so the whole search window holds only a handful of counts.

wrong-helicity amplitude m/2E   = 5.00e-11   [Eq. (10.75)]
lepton-number violation goes as its square = 2.50e-21
the book's figure and prose quote (m/E)^2  = 1.00e-20, a factor 4 larger

Two parts in $10^{21}$ . That is the fraction of pion decays in which a Majorana neutrino would betray itself by making the wrong-sign muon. No beam ever built comes within fifteen orders of magnitude of it.

Which is why the search moved from catching a neutrino to waiting — put both weak vertices inside one nucleus, take a tonne of the stuff, and see whether it ever decays. The suppression is the same; the number of chances is $10^{27}$ .

Erratum — the wrong-helicity amplitude is m/2E, and Fig. 10.20 says m/E

Eq. (10.75) prints the amplitude as $\tfrac12 (m/E)$ . Fig. 10.20’s three captions and the paragraph beneath it — “a (small) fraction, with amplitude $m/E$ , of positive helicity (see Eq. (10.75))” — drop the factor two, and the later estimate of the lepton-number-violating fraction quotes $(m/E)^2$ .

(10.75) is the right one. §2.8 derives the helicity content of a chiral spinor exactly, and its own PyBlock checks the exact expression against $(m/2E)^2$ over four decades of $E/m$ . The factor matters not at all for an order-of-magnitude argument — $2.5\times10^{-21}$ against $10^{-20}$ — but it is worth being consistent about, because the site has already taught $m/2E$ two chapters earlier.

🌀 Helicity is not chirality

energy in units of the mass, E/mE/m = 5.012 · β = 0.979892

A left-chiral fermion — what helicity will you measure?

negative helicity (the “right” one)0.989946

positive helicity (the “wrong” one)0.010054

amplitudes  a∓ = 1 ∓ p/(E+m) = 0.1831, 1.8169
wrong-helicity probability = 0.010054
the (m/2E)² estimate       = 0.009953
⟨h⟩ = (a₊² − a₋²)/(a₊² + a₋²) = -0.979892
−β                            = -0.979892

⟨h⟩ = −β exactly, at every energy — Eq. (2.66). A left-chiral fermion is a definite-helicity particle only in the limit β → 1; at rest the two helicities are equally likely and chirality says nothing about spin direction at all.

Now run alongside it — what survives the boost?

observer's velocity β_obs0.0000

particle in the lab       β  = 0.979892
seen by the observer      β' = 0.979892
   the observer is still behind it

helicity	negative, unchanged
chirality	left, unchanged — always

Because the particle is massive you can always find a frame moving faster than it — set β_obs above 0.9799 and its momentum reverses while its spin does not, so the helicity changes sign. Helicity is not Lorentz invariant. Chirality, an eigenvalue of γ⁵, is.

exact wrong-helicity probability
the (m/2E)² limit

The wrong-helicity admixture falls as (m/2E)², so it is a 1 % effect at E = 5m and a part in ten thousand at E = 50m. This is why 'left-handed' and 'left-chiral' can be used interchangeably at collider energies and must not be at low ones — and why neutrino mass, however small, means the weak interaction is not quite chirally pure (Ch. 10).

	polarization	helicity	chirality
eigenstate of	σ_z, along an axis you chose	σ·p̂, along the motion	γ⁵
needs	an external field to define the axis	the particle to be moving	nothing
Lorentz invariant?	no — the axis is frame-dependent	NO, unless massless	YES, always
conserved for a free particle?	no	YES	NO, the mass term mixes L and R
directly measurable?	yes	yes	no — you measure helicity and infer it
what the weak interaction couples to	—	—	THIS (Ch. 7)

Read the two middle rows of the invariance/conservation pair: helicity and chirality have exactly opposite properties, which is why neither alone is the whole story and why the literature confuses them. The book's convention, followed here: left and right always mean chirality, never helicity.

💡 What this really says — the frame-dependence puzzle

The widget’s second slider answers an objection Bettini raises and disposes of in a paragraph. Helicity is not Lorentz invariant: overtake a particle of $h = -1$ and you see $h = +1$ . So does a Majorana neutrino make an electron or a positron depending on who is watching?

Obviously not, and the resolution is worth having. What enters the interaction is the relative velocity between the neutrino and the target, which is frame independent. Equivalently: the amplitude is a product of two currents, each of which is frame dependent, and their scalar product is not. Chirality, which is what the V−A current actually couples to, is invariant — drag the observer slider and watch the helicity row flip while the chirality row does not.

Furry’s idea: put both vertices in one nucleus

(Z,A) \to 2e^- + \htmlClass{t-n}{2\bar\nu_e} + (Z+2,A) \qquad 2n \to 2e^- + \htmlClass{t-n}{2\bar\nu_e} + 2p \qquad \htmlClass{t-q}{2d \to 2e^- + 2\bar\nu_e + 2u}

(10.76, 10.77, 10.78)

Bettini p. 473. Two-neutrino double-beta decay — allowed in the Standard Model, and observed.

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💡 What this really says — a nuclear accident is what makes the whole programme possible

Double beta decay happens at all only because of a nuclear accident: in even–even nuclei, single beta decay is energetically forbidden while the double step is allowed. So a nucleus that cannot decay once can decay twice, and it takes $10^{20}$ years to get round to it.

That accident is what makes the whole programme possible. Without it there would be no candidate isotopes at all.

(Z,A) \to \htmlClass{t-e}{2e^-} + (Z+2,A) \qquad\qquad \htmlClass{t-q}{2d \to 2e^- + 2u}

(10.79, 10.80)

Bettini p. 473. Neutrinoless double-beta decay: the same nuclear transition with nothing else coming out.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

Fig. 10.22 — the two double-beta decays at quark level

Click a vertex or an internal line.

Left, (a): two independent W emissions, two electrons and two antineutrinos out. Right, (b): the two neutrino lines have joined. The propagator carries U_ei at both ends and the mass insertion in the middle — click a line.

Erratum — p. 473 cites Fig. 10.21 twice where Fig. 10.22 belongs

“The corresponding lowest-order diagram is shown in Fig. 10.21(a)”, and seven lines later, “The lowest-order diagram is shown in Fig. 10.21(b)”. Both should be Fig. 10.22, which is captioned “Lowest-order diagrams at quark level for (a) the 2ν2β decay and (b) the 0ν2β decay”. Fig. 10.21 is the two-panel sketch of helicity changing under a boost.

The same page then cites Fig. 10.22(a) and 10.22(b) correctly in its last paragraph — so it has it wrong twice and right twice.

Why the rate goes as the mass squared, and why it is observable anyway

💡 What this really says — one power of the mass in the amplitude, two in the rate

Follow the neutrino line in the right-hand diagram. The lower vertex creates it as a $\bar\nu_e$ , exactly as the Standard Model says. The upper vertex has to absorb it as a $\nu_e$ — and can only do so through its small wrong-helicity component, amplitude $m/2E$ . So the amplitude carries one power of the mass and the rate carries two.

That suppression is brutal — and it is fought by something pulling the other way. The neutrino in the two-neutrino decay is real, on the mass shell; the one in the neutrinoless decay is virtual, and its momentum is set by the nuclear scale, of order 100 MeV, not by the decay energy. Furry showed in 1939 that this enhances the rate by many orders of magnitude.

Net result: the two decay rates would be comparable for neutrino masses around 10 eV. They are not — the masses are below 0.1 eV — so the neutrinoless mode is suppressed by at least four more orders of magnitude, and that is the size of the problem.

\frac{1}{T_{1/2}} = \htmlClass{t-g}{G}\,\htmlClass{t-m}{M_{ee}^2}\,\htmlClass{t-n}{|M_{\text{nucl}}|^2} \qquad \htmlClass{t-m}{M_{ee}} = \left|\sum_{i=1}^{3}U_{ei}^2 m_i\right| \approx \left|0.68\,m_1 + 0.30\,\htmlClass{t-p}{e^{2i(\eta_2-\eta_1)}}m_2 + 0.02\,\htmlClass{t-p}{e^{-2i(\eta_1+\delta)}}m_3\right|

(10.81, 10.82)

Bettini p. 474. The half-life, and the quantity it measures — the only place the Majorana phases of (10.2) ever appear.

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the two phases are unknown, so the last two arrows can point anywhere — and the sum can vanish

Bettini Fig. 10.23. Three phasors, added head to tail, with two free angles. If the longest arrow is shorter than the other two together the resultant can be made zero; if it is longer, it cannot. That single geometric fact is what separates the two mass orderings in the plot below. (The book’s figure labels the arrows 0.67, 0.3 and 0.03 where its own (10.82) has 0.68, 0.30 and 0.02.)

What the experiments look for, and what they have found

The signature of neutrinoless double-beta decay is unmistakable in principle: the two electrons carry the whole $Q_{\beta\beta}$ , so the summed energy is a peak sitting on the continuum of the two-neutrino mode.

2ν2β — always present, continuous
0ν2β — a peak at Q_ββ, if it exists at all

Bettini Fig. 10.24, idealised. The continuum is drawn with an illustrative phase-space shape ∝ E²(Q−E)⁵ — not the exact 2ν2β distribution — normalised to the same height as the peak so both are visible. In reality the peak, if it is there at all, is smaller than the continuum by more than four orders of magnitude, which is why energy resolution matters more than anything else: a peak a few keV wide sits on a continuum that a poor resolution would smear straight over it.

⚙️ Engineer’s bridge — resolution buys signal-to-noise quadratically

The design pressure here is unlike anything earlier in the chapter, and it is a detection problem you would recognise.

The signal is a line — zero intrinsic width, since the final state is fully determined. The dominant irreducible background is a smooth continuum from the two-neutrino mode. So the background inside the search window is proportional to the window width, which is set by your resolution $\sigma$ , while the signal is not.

That gives $S/\sqrt{B} \propto 1/\sqrt{\sigma}$ — and because the continuum falls steeply as $E \to Q$ , halving $\sigma$ actually buys more than the naive factor. It is why germanium detectors, with a few keV at 2 MeV, are competitive against xenon experiments with far more isotope: resolution substitutes for mass, and the exchange rate is favourable.

Where it breaks: only down to a point. Once the 2ν2β leakage is below the external radioactive background — a flat rate per keV per tonne per year that better resolution shrinks proportionally, not faster — you are back to the ordinary counting problem of §3.6, and the limit goes as √exposure.

0ν2β search · mass → nuclei → exposure → half-life limit

fiducial mass M = 0.75 trunning time Δt = 20 yrdetection efficiency ε = 44 %background rate = 0.002 events / t·yr

1	nuclei in the fiducial volume	3.32 × 10²⁷	¹³⁶Xe: N = M × N_A / 136 nuclei per gram of isotope
2	exposure M Δt	15 t·yr	the figure of merit — mass and time are interchangeable
3	nuclei-years N Δt	6.64 × 10²⁸	the number that actually enters the limit
4	expected background b	0.03	small — the search is still background-free
5	90 % CL limit on the signal	2.30	the Poisson rule of 2.3 for zero observed events
6	half-life limit T₁/₂	8.80 × 10²⁷ yr	ln2 × N Δt ε / s, at 90 % confidence

with background, 0.002 events / t·yr
background-free ideal (∝ exposure)

The two regimes. While the expected background stays below a few events the limit grows in proportion to the exposure — the dashed line. Once background dominates it grows only as its square root, and the curve bends away permanently. Doubling the run time then buys 40 %, not 100 %.

The signal limit is approximated as s ≈ max(2.30, 1.28√b): the Poisson rule of 2.3 for zero observed events, crossing over to the Gaussian one-sided 90 % point once the background dominates. On the Super-K preset this gives 2.9 × 10³⁴ yr against the published 2.4 × 10³⁴ — the real analysis is a likelihood fit carrying systematic uncertainties, so read this as the arithmetic of a search, not as a reproduction of one.

💡 What this really says — every advance here has been a cleaner tonne, not a bigger one

This is §3.6’s proton-decay arithmetic with the target swapped for an enriched isotope and the answer quoted as a half-life. The bend in the curve is the whole design problem: while the search is background-free the reachable half-life grows in proportion to exposure, and once background bites it grows only as its square root.

Drag the background rate down and watch the bend move to the right. That is what every advance in this field has been — not more mass, but a cleaner tonne.

🔢 Worked example — the two orderings, and the funnel

The effective Majorana mass is a sum of three phasors with two free angles. Whether it can vanish is a question about triangles: if the longest arrow exceeds the other two combined, it cannot.

M_ee bands from (10.82), with the phases free

import numpy as np
dm2, Dm2 = 73.4, 2455.0                     # meV^2
U2 = np.array([0.68, 0.30, 0.02])           # |U_ei|^2, from (10.73) and (10.82)

def masses(lightest, order):
    if order == "NO":
        m1 = lightest; m2 = np.sqrt(m1**2+dm2); m3 = np.sqrt((m1**2+m2**2)/2+Dm2)
    else:
        m3 = lightest; half = m3**2+Dm2
        m1 = np.sqrt(half-dm2/2); m2 = np.sqrt(half+dm2/2)
    return np.array([m1, m2, m3])

def Mee_band(m):                            # free Majorana phases -> a band, not a point
    a = U2*m
    hi = a.sum()
    lo = max(0.0, 2*a.max() - hi)           # |a1 e^ia + a2 e^ib + a3| can vanish unless one dominates
    return lo, hi

print(" ordering  lightest   Sum m    M_ee range (meV)")
for order in ("NO", "IO"):
    for l in (0.0, 5.0, 20.0, 100.0):
        m = masses(l, order); lo, hi = Mee_band(m)
        print(f"   {order}     {l:6.1f}   {m.sum():6.1f}    {lo:6.2f} - {hi:6.2f}")

# where can normal ordering cancel completely?
l = np.linspace(0.01, 30, 30000)
lo = np.array([Mee_band(masses(x, "NO"))[0] for x in l])
z = l[lo < 1e-9]
print(f"\nnormal ordering can give M_ee = 0 for a lightest mass between "
      f"{z.min():.2f} and {z.max():.2f} meV — the 'funnel'")
m = masses(0.0, "IO"); lo, hi = Mee_band(m)
print(f"inverted ordering never cancels: its floor is M_ee > {lo:.1f} meV")
print(f"\nbest present limits: M_ee < 36-156 meV (136Xe), 79-180 (76Ge), 90-305 (130Te)")

prints

 ordering  lightest   Sum m    M_ee range (meV)
 NO        0.0     58.5      1.57 -   3.57
 NO        5.0     65.1      0.00 -   7.38
 NO       20.0     95.5      6.00 -  21.20
 NO      100.0    312.1     35.65 - 100.35
 IO        0.0     99.1     18.46 -  48.41
 IO        5.0    104.6     18.46 -  48.76
 IO       20.0    126.9     19.57 -  52.63
 IO      100.0    323.2     40.25 - 111.31

normal ordering can give M_ee = 0 for a lightest mass between 2.46 and 6.13 meV — the 'funnel'
inverted ordering never cancels: its floor is M_ee > 18.5 meV

best present limits: M_ee < 36-156 meV (136Xe), 79-180 (76Ge), 90-305 (130Te)

Two results that between them define the whole experimental programme, and that the book states only in words.

Inverted ordering has a floor: $M_{ee} > 18.5$ meV. With the lightest state massless, $0.68m_1 = 33.5$ meV against $0.30m_2 + 0.02m_3 = 15.0$ meV, so the longest phasor cannot be cancelled by the other two whatever the phases do. That is why every next-generation experiment quotes a target near 15 meV: reaching it either finds the decay or excludes inverted ordering entirely.

Normal ordering can cancel exactly. For a lightest mass between 2.5 and 6.1 meV the three phasors can close on zero — the “funnel”. So a null result at any sensitivity is compatible with Majorana neutrinos in normal ordering, and no experiment can ever settle the question by not seeing anything.

That asymmetry of inference is the defining feature of the field: a single event is a discovery; a null result is never a refutation.

inverted ordering — has a floor
normal ordering — can cancel to zero

Bettini Fig. 10.25, computed from (10.8) and (10.82) with the Majorana phases scanned over their full range. The inverted band never descends below 18.5 meV; the normal band pinches to zero near Σm ≈ 65 meV, which is the funnel. Everything to the right of the dashed vertical line is disfavoured by cosmology (§10.6), so the target region is the narrow window where the two bands are widest apart and lowest.

🔬 Experiment card — the 0ν2β searches

Apparatus

Source and detector are the same object. Germanium diodes grown from isotopically enriched $^{76}$ Ge (GERDA, Majorana Demonstrator); xenon dissolved in a liquid scintillator or run as a TPC (KamLAND-Zen, EXO); tellurium oxide crystals read out as cryogenic bolometers at 10 mK (CUORE). Hundreds of kilograms of active isotope, deep underground, in materials screened for radioactivity at levels no other field requires.

What is measured

The summed energy of the two electrons. The two-neutrino mode gives a continuum; the neutrinoless mode would give a line at $Q_{\beta\beta}$ , between 1.5 and 3 MeV depending on the isotope, whose width is entirely the detector resolution. Everything is a counting experiment in a window a few keV wide.

The result

No peak, anywhere, in any isotope. The best half-life limits are $2.3\times10^{26}$ yr for $^{136}$ Xe, $1.8\times10^{26}$ yr for $^{76}$ Ge and $2.2\times10^{25}$ yr for $^{130}$ Te, at 90 % confidence — fewer than about ten decays per tonne per year, against backgrounds pushed below one count per keV per tonne per year.

What it proved

That $M_{ee}$ is below 36–305 meV, where the spread is the nuclear matrix element and not the measurement. It cannot prove neutrinos are Dirac particles and never will: normal ordering with a lightest mass near 4 meV cancels $M_{ee}$ exactly. What the next generation can do is cover the inverted-ordering band entirely, since that band has a floor at 18.5 meV — and covering it either finds the decay or excludes one of the two mass orderings.

The best present limits, from the isotopes with the most exposure, are $T_{1/2}(^{136}\mathrm{Xe}) > 2.3\times10^{26}$ yr, $T_{1/2}(^{76}\mathrm{Ge}) > 1.8\times10^{26}$ yr and $T_{1/2}(^{130}\mathrm{Te}) > 2.2\times10^{25}$ yr. Translated through (10.81) they give $M_{ee} <$ 36–156, 79–180 and 90–305 meV respectively — and the spread in each is not experimental. It is the nuclear matrix element, known to a factor 2–3 and entering squared.

🔑 If you remember only three things

Ninety years on, the question is binary and still open. Nothing measured since 1937 separates the two hypotheses, because every measurement so far sits in the massless limit.
A line on a continuum rewards resolution more than exposure. Germanium’s few keV at 2 MeV keeps it competitive against experiments carrying many times more isotope.
The observable can vanish while the hypothesis holds. Two unmeasured phases can arrange the three terms to cancel, so this field could be right and see nothing forever.

Where this goes next

The chapter closes with the largest open question in it. Neutrinos have mass, and nobody knows what kind. If the mass is Dirac, there is a right-chirality neutrino that has never been seen and a Yukawa coupling $10^{-12}$ times the electron’s, with no explanation for either. If it is Majorana, lepton number is not conserved, two more phases are physical, and the door opens to the seesaw mechanism and to leptogenesis — a way of making the Universe’s matter–antimatter asymmetry that the quark sector cannot supply.

One experiment decides it, and the decisive version has not been built. The current generation reaches tens of meV; the inverted-ordering floor is 18.5 meV; and if nature chose normal ordering with a lightest mass near 4 meV, the answer may be permanently out of reach by this route.

Chapter 11 turns to the one interaction the Standard Model leaves out altogether — gravity, and the gravitational waves that are now measured. The Epilogue then surveys everything else the model does not explain, and the origin of neutrino mass is on that list.

✅ Check yourself — Majorana neutrinos and 0ν2β

0/6 answered · 0 correct

1.Racah's test — emit a neutrino with an electron, see which lepton it makes on absorption — has never given a positive result. Why does that not prove neutrinos are Dirac particles?
2.Why is the 0ν2β amplitude proportional to the neutrino mass, and what fights the suppression?
3.M_ee sums U²_ei m_i, while §10.6's beta-decay observable sums |U_ei|² m_i². What follows from that difference?
4.Which of these follow from the computed M_ee bands? (Select all that apply.)
5.The published limits are quoted as ranges — M_ee < 36–156 meV for ¹³⁶Xe. What is the range?
6.Why does energy resolution matter more in this search than raw isotope mass, up to a point?