I am glad to hear back from the inventor @Dharam Vir Ahluwalia of ELKO! And I really appreciate the chance to have a in-depth discussion with the true experts of the ELKO theory.

There is a growing body of literature on the ELKO spinors (see references here), which are alleged to be mass dimension one fermions and can be a dark matter candidate.

But is the ELKO spinor a red herring? Is the mass dimension one fermion term irrelevant at the standard model energy scale?

A Dirac type fermion Lagrangian can be written as $$ L = i\bar{\psi}\not D\psi - m\bar{\psi}\psi $$ while the Lagrangian for ELKO type fermion is $$ L = \bar{\psi}\partial^\mu\partial_\mu\psi - m'^2\bar{\psi}\psi $$ Actually the most general fermion Lagrangian should read $$ L = i\bar{\psi}\not D\psi + M^{-1}\bar{\psi}\partial^\mu\partial_\mu\psi- m\bar{\psi}\psi \tag{1} $$ (or equivalently: $$ L = iM\bar{\psi}\not D\psi + \bar{\psi}\partial^\mu\partial_\mu\psi- m'^2\bar{\psi}\psi \tag{2} $$ where $m'^2 = Mm$. it's just a matter of re-scaling the fermion field.)

The ELKO kinetic term $M^{-1}\bar{\psi}\partial^\mu\partial_\mu\psi$ observes the Lorentz symmetry, so ideally it should be included in the modern effective quantum field theory framework.

The key question here is the magnitude of the ELKO term $M^{-1}\bar{\psi}\partial^\mu\partial_\mu\psi$. How large should $M^{-1}$ be? The naturalness principle tells us that $M$ should be of the Planck scale $$ M \sim M_{Planck} $$ so that the ELKO term is drastically suppressed by the order of $$ \frac{p}{M_{Planck}} $$ where $p$ is the momentum/energy scale of the physics process in concern.

Additionally, the ELKO term $M^{-1}\bar{\psi}\partial^\mu\partial_\mu\psi$ breaks the axial symmetry $$ \psi \rightarrow e^{\theta i\gamma_5}\psi $$ Hence this term is further suppressed due to t' Hooft's technical naturalness argument, analogous to the suppression of the axial-symmetry-breaking fermion mass term $m\bar{\psi}\psi$.

With that, shall we regard the ELKO term as irrelevant, unless you are dealing with Planck scale quantum processes where all bets are off.

Reply to @Dharam Vir Ahluwalia comment: "the presented Lagrangian has dimensionality mismatch between various terms".

As to "dimensionality mismatch", that is why I included the mass parameter $M$ in equation (1) and (2). This parameter $M$ plays the central role in my argument that the ELKO term is diminishingly small compared with the normal Dirac spinor term. And therefore, the ELKO term can be regarded as virtually nonexistent at sub-Planck energy scales.

  • $\begingroup$ I think this elko does not qualify as a (local) QFT, as it trivially violate the spin-statistic theorem (for example set M to zero to see it neatly). I would not personally go that far to explain dark matter when much less radical QFT-based models can be found easily. $\endgroup$
    – TwoBs
    Dec 23, 2019 at 23:52

1 Answer 1


Following, "Actually the most general fermion Lagrangian should read" the presented Lagrangian has dimensionality mismatch between various terms. Please refer to my monograph "Mass Dimension One Fermions" (Cambridge Monographs on Mathematical Physics, Cambridge University Press, 2019) for a detailed construction of Elko and the new fermions. The darkness of the new fermions comes about because these new fermions, because of their mass dimensionality, cannot enter the standard model doublets (where the fermions have mass dimension of 3/2). The new fermions, unlike their Dirac and Majorana counterparts, carry a dimension four quartic coupling -- similar coupling is suppressed by two powers of unification scale for the SM fermions. Please refer to literature that supports quartic self interaction for dark matter.

Since I have written at length on the subject I invite a seriously interested scientist to my CUP monograph mentioned above. There are now numerous papers on Elko and cosmology. The reader can find them by following citations to my papers at https://old.inspirehep.net/search?ln=en&ln=en&p=a+d.v.ahluwalia.1&of=hb&action_search=Search&sf=earliestdate&so=d&rm=&rg=250&sc=0 for a good discussion.

Please consult, D.V.Ahluwalia et al. Nuclear Physics B 987 (2023) 116092 for rotational symmetry of the formalism, and required Wigner degenercay.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.