Dirac vs Majorana Neutrinos: Why Oscillations Alone Can't Decide

1 point by nadermx 1 hour ago

The previous post ended on a fact and a gap: neutrinos have mass, the minimal Standard Model says they cannot, and repairing it takes either a new field or a new operator. The obvious next question is which repair nature actually used — is the neutrino a Dirac particle like every other fermion, or a Majorana particle that is its own antiparticle?

The tempting answer is that this is a measurement problem, and that enough oscillation data will eventually settle it. It will not. Not because the experiments are too coarse, but because oscillations are the wrong observable in a way that can be made exact.

Same masses. Same mixing. Same oscillations. Opposite neutrino nature.

What an oscillation actually sees

An oscillation probability is built from two things and nothing else: the mass-squared splittings $\Delta m^2_{ij}$ and the mixing matrix $U$ . Write the pair as the descriptor $(m_i, U)$ .

If neutrinos are Majorana, $U$ carries two extra physical phases that a Dirac $U$ does not have. Those Majorana phases cancel out of every oscillation probability identically — they multiply the mass eigenstates by phases that the modulus-squared then discards. So an oscillation experiment, however precise, measures $(m_i, U)$ and is blind to the rest.

That is the standard textbook statement. What follows makes it sharp: two complete, anomaly-free theories with identical $(m_i, U)$ and opposite neutrino nature.

One skeleton, two branches

Start from the anomaly-free charge

X = 3(B - L).

Per generation, augmented with one right-handed neutrino, the charges are

X(Q_L,\, u_R,\, d_R,\, L_L,\, e_R,\, N_R) = (1,\, 1,\, 1,\, -3,\, -3,\, -3).

With three right-handed neutrinos, every local cubic, mixed-gauge and mixed-gravitational anomaly cancels. Now note how the two mass mechanisms sit under this charge:

X\big(\bar L \widetilde{H} N_R\big) = 0, \qquad X(N_R N_R) = X\big((LH)(LH)\big) = -6.

The Dirac Yukawa is $X$ -neutral. Both $\Delta L = 2$ objects — the right-handed Majorana bilinear and the Weinberg operator — carry charge $-6$ .

Let a single scalar $\Phi_N$ of integer charge $N$ take the only $X$ -charged vacuum expectation value. The gauge group does not vanish; it leaves a discrete remnant:

U(1)_X \longrightarrow \mathbb{Z}_N.

Multiplying a $\Delta L = 2$ bilinear by any number of powers of $\Phi_N$ or $\Phi_N^{\dagger}$ shifts its charge only by multiples of $N$ . So a Majorana mass survives exactly when

\boxed{\;N \mid 6\;}

and that single divisibility condition is the whole fork in the road.

Take $N = 6$ . Then $-6 \equiv 0 \pmod 6$ . The operator $\Phi_6 N_R N_R$ is allowed, an ordinary type-I seesaw runs, and the light neutrinos are Majorana.

Take $N = 12$ . Then $-6 \not\equiv 0 \pmod{12}$ , and since $-6 + 12k$ is never zero for integer $k$ , every Majorana self-energy and every $\Delta L = 2$ local operator is forbidden to all orders. The Dirac Yukawa is still neutral and still allowed. The light neutrinos are exactly Dirac.

The counterexample

Both branches can be made to reproduce any prescribed set of positive masses $m_i$ and the same mixing matrix $U$ . On the Dirac side, in a suitable right-handed basis, take

m_D^{(D)} = U \operatorname{diag}(m_i).

On the Majorana side take $M_R = M \mathbb{1}$ and

m_D^{(M)} = i\, U^{*} \sqrt{M \operatorname{diag}(m_i)},

so that the seesaw returns

-\,m_D^{(M)} M_R^{-1} \big(m_D^{(M)}\big)^{T} = U^{*} \operatorname{diag}(m_i)\, U^{\dagger}.

Identical light masses, identical oscillation probabilities. The new gauge and scalar sectors decouple by taking their scales high enough. Which gives

\begin{aligned} -6 &\equiv 0 \pmod 6 &&\Rightarrow\; \mathbb{Z}_6 : \text{ Majorana allowed},\\ -6 &\not\equiv 0 \pmod{12} &&\Rightarrow\; \mathbb{Z}_{12} : \text{ Majorana forbidden} \Rightarrow \text{Dirac},\\ (m_i, U)_{\mathbb{Z}_6} &= (m_i, U)_{\mathbb{Z}_{12}} &&\Rightarrow\; \boxed{\text{masses} + \text{oscillations cannot determine Dirac vs Majorana.}} \end{aligned}

Why this is stronger than "we haven't measured it yet"

There is a familiar objection that would kill a Dirac neutrino: exact global symmetries are believed to be inconsistent with quantum gravity, and a Dirac neutrino seems to need lepton number as an exact global symmetry.

That objection does not touch this construction. Both branches protect themselves with an exact gauged symmetry — a discrete $\mathbb{Z}_N$ remnant of a local $U(1)_X$ , not a global one. So the whole theory-only route closes: the Standard Model action, anomaly cancellation, all oscillation data, the absence of a long-range $B-L$ force, and the prohibition on exact global symmetries, taken together, still do not select the branch.

No manipulation of the present descriptor helps either. Sums, inversions, limits, transforms and infinite compositions are all functions of the same collided input $(m_i, U)$ , and that input is identical on both sides.

If several fields condense, nothing changes in spirit — replace $N$ by $d = \gcd\{X(\langle \Phi_a \rangle)\}$ and the criterion becomes $d \mid 6$ .

What would actually decide it

Exactly two kinds of evidence close this, and neither exists today.

A Majorana certificate. A source- and external-state-authenticated nonzero vacuum amplitude with $\Delta L = 2$ , with backgrounds and invisible external carriers accounted for, establishing

\Sigma_{\nu\nu}(0) \neq 0.

That is what neutrinoless double beta decay is for. A detector topology on its own is not sufficient — the inventory has to be closed.

A Dirac certificate. Dirac is harder, because it is the absence of something, and a finite null search cannot prove an exact zero: pseudo-Dirac theories with tiny Majorana components accumulate continuously at the Dirac locus, so "we looked and saw nothing" never converges on it. It needs a positive certificate that the residual sector is non-self-conjugate. A $\mathbb{Z}_N$ flux string carries holonomy $W_q = \exp(2\pi i q / N)$ for charge $q$ , and the antiparticle carries $W_q^{-1}$ , so

W_q \neq W_q^{-1} \iff 2q \neq 0 \pmod N,

which for the neutrino at $q = -3$ means a measured $W_\nu^2 \neq 1$ would be a direct Dirac certificate. Alternatively, a genuinely diagonal C-odd electric or magnetic dipole form factor of a resolved, non-degenerate mass eigenstate — note not the anapole form factor, which a Majorana particle is perfectly entitled to have.

No neutrino holonomy has ever been measured.

What this is not

This is a no-identifiability result, not a discovery. It does not claim nature has a $U(1)_{B-L}$ , or a $\mathbb{Z}_{12}$ , or a discrete flux string, or a Majorana mass. It does not report a $\Delta L = 2$ event. It identifies neither nature's hidden gauge group nor its condensates.

What it does is remove a hope: that the answer is already latent in the oscillation data and merely needs to be extracted more cleverly. It is not. The descriptor is degenerate, provably, and the branch has to be read off some other observable entirely.

Nature's neutrinos have not been shown to be exactly Dirac, and have not been shown to carry a Majorana mass component. That remains open — and it is open in a specific way, which is worth more than being open in a vague one.