Great question. The experimental situation remains inconclusive. However, theoretically, there exists a damn good reason to think that the neutrinos have Majorana masses - and, consequently, the double beta decays should be possible. It's called the seesaw mechanism.
The mechanism is justified by an intriguing observation:
$$ m_{\nu}:m_{Higgs} \approx m_{Higgs}:m_{GUT}$$
Both ratios are about $10^{-14}$: note that $m_\nu, m_{Higgs}, m_{GUT}$ are about $10^{-12},10^2,10^{16}$ GeV. The light, millielectronvolt-scaled neutrinos, seem to be on the other side of this geometric sequence.
This fact can be explained by a rather robust mechanism. The light neutrino masses - written as a $3\times 3$ mass matrix - may be generated as
$$ m = -m_D M_R^{-1} m_D^T. $$
The Dirac masses $m_D$, linking the known left-handed neutrinos with the heavy, unknown, right-handed neutrinos, are comparable to the Higgs mass - the electroweak scale. Meanwhile, the Majorana masses for the right-handed neutrinos, $M_R$, are comparable to the GUT scale. Right-handed neutrinos are automatically predicted e.g. as the "invisible" 16th component of the 16-dimensional spinor multiplet in $SO(10)$ grand unified theories, among others.
The formula above automatically produces the "seesaw": the resulting masses $m$ are as many times smaller than $m_D$ as $m_D$ is smaller than $m_R$.
It's very natural for the right-handed neutrinos to acquire Majorana masses: they're pretty much neutral and nothing prevents them from being coupled in pairs. The Dirac masses comparable to the electroweak scale may also be justified.
As a consequence, one automatically obtains the small masses for the known, left-handed neutrinos: and they're inevitably Majorana masses, too. They can't be Dirac masses simply because Dirac masses would require the right-handed parts of the particles to be as light as the left-handed one. But the right-handed neutrinos are superheavy and may be neglected.
The seesaw formula may be obtained by diagonalizing a matrix $((0,m_D),(m_D,M_R))$. One eigenvalue is inevitably close to $M_R$ (which are really three eigenvalues); the other is much smaller than $m_D$. Alternatively, one may view $1/M_R$ in the formula as a suppression factor from more general higher-dimensional operators.
It should be emphasized that the Dirac masses are used in the seesaw formula. However, because they're much smaller than the right-handed Majorana masses, the Dirac masses don't dictate the basic character of the fermions.
Of course, there may be surprises, for example both Majorana and Dirac masses may occur at low energies, too. That would require some mixing of the three generations of the neutrinos - a scenario that is problematic for many reasons but it is not excluded.
In the text above, I assumed that the reader understands that the fields needed to describe a Weyl fermion and a Majorana fermion are the same fields: a 2-component complex spinor. Automatically, its Hermitean conjugate (one with the opposite chirality) also exists as an operator on the Hilbert space.
Whether it is more natural to call this field "Weyl" or "Majorana" depends on the mass terms (or the interaction terms). Those that dominate decide about the name. If the Majorana mass terms are much larger than the Dirac masses, the particles are essentially Majorana particles, and the lepton number violation may be substantial while parity is kind of conserved. If the Dirac masses are much larger, the parity is qualitatively preserved - but one needs a doubled number of excitations (left-handed and right-handed particles of the same or similar mass).
There can't be any mass terms that would both preserve the lepton number (distinguished particles and particles) and that would also keep the neutrinos purely left-handed.
This post imported from StackExchange Physics at 2014-05-14 19:45 (UCT), posted by SE-user Luboš Motl
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