To generate three light neutrino mass eigenstates via type-I seesaw, we include a set of $n$ (need not be equal to 3) heavy$^1$ right-handed fields $N_R$ in addition to three $\nu_L$ fields to the Standard Model. Then type-I seesaw mechanism gives 3 light and $n$ heavy mass eigenstates which are linear combinations if $\nu_L$ and $N_R$ fields.

The $N_R$ fields are gauge singlets, and therefore do not have any Standard Model (SM) interactions. Sterile neutrinos $$ also do not have SM interactions. This reference says $\nu_s$ fields are sterile neutrinos without defining it or distinguishing it from $N_R$ (or $\nu_R$) fields. While this reference says $\nu_R$ are sterile without talking about $\nu_s$ notation at all.

$\bullet$ Then how are the $N_R$ fields different from sterile neutrinos?

$\bullet$ Existing answers claim that $N_R$ is same as $\nu_s$. Indeed both are defined to be having no SM interactions. But we know that sterile neutrinos $\nu_s$ taking part in oscillation with active neutrinos must be light. On the other hand, the $N_R$ fields must be heavy for type-I seesaw to work.

$^1$ Heavy in the sense that $M$ in the term $M\bar{N_R^c}N_R$ is around the GUT scale. But of course $M_R$ aren't mass eigenstates.


They are not different. There is only a lot of different (and sometimes confusing) terminology on the market. For instance, in the case of Dirac neutrinos, you also need the right-handed neutrinos (for the mass term, for instance). Some people will call these right-handed components sterile, some won't.

UPDATE: There can be arbitrarily many right-handed (or sterile) neutrinos, see e.g. this paper and references therein. A priori, they can be light or heavy, and they may or may not mix with the active neutrinos. The notion "right-handed" refers to the chirality of a state that carries the same lepton number as an active neutrino particle (as opposed to antiparticle). Of course, if lepton number is broken, this notion can be confusing. It might therefore be better to distinguish between "active neutrinos" and "singlet neutrinos". But the bottom-line, or the answer to your question, is that right-handed and sterile neutrinos are not different.

  • $\begingroup$ Thanks for the reply. Can you tell me what is meant by active neutrinos? The $\nu_L$ fields? Are you aware of the notation $\nu_s$ (please see here arxiv.org/abs/hep-ph/0307359)? Is $\nu_s$ in appendix Eq.(A.2) same as one of the $N_R$ (or $\nu_R$) fields? @marmot $\endgroup$ – SRS Jun 6 '18 at 12:13
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    $\begingroup$ @SRS Active neutrinos are those which couple to the $Z$ boson. In the standard model, these are part of the lepton doublets. And yes, I am aware of the notation $\nu_s$. Whether or not they are exactly the same as $\nu_R$ depends on your model. For instance, in the case of Dirac neutrinos, lepton number might be conserved, and you may distinguish between $\nu_R$ and their charge conjugates. In eq. (A.2) all states are left-handed, so $\nu_s$ is something like $\nu_R^C$. $\endgroup$ – user178876 Jun 6 '18 at 13:39

Before spontaneous symmetry breaking, we have the lepton doublets $l_L$ (which contain the left-handed neutrinos), and the neutral singlets $N_R$. For the latter, we have $D_\mu N_R = \partial_\mu N_R$, so they don't have interactions with the gauge fields (which usually come from $\bar{\psi}\gamma^\mu D_\mu\psi$). This is the reason they're called sterile, they can't interact electromagnetically or through the weak or the strong forces. However, nothing forbids them to have Yukawa interactions $\sim \bar{l}_L \phi N_R$ and Majorana masses $\sim \bar{N}_R^c N_R$, so this terms should be included in the Lagrangian.

After spontaneous symmetry breaking, $l_L$ splits into the electron and neutrino $\nu_L$. There's a non-diagonal mass matrix for $(\nu_L, N_R^c)$. The fields $(\nu_{\text{light}}, \nu_{\text{heavy}})$ that diagonalize the mass matrix are linear combinations of the ones that come from the components of the gauge eigenstates. Therefore, we can't talk about them being sterile because they are not singlets, as they don't even have well-defined transformation laws under the gauge group.

  • $\begingroup$ I did not say the heavy and light mass eigenstates are sterile. @coconut $\endgroup$ – SRS Dec 23 '17 at 21:03
  • $\begingroup$ @SRS Sorry, I misunderstood you, then. Does this answer clarify the issue in some way? I’m still not sure what the problem is, maybe it’s just a matter of definitions? $\endgroup$ – coconut Dec 23 '17 at 21:24
  • $\begingroup$ @SRS The $N_R$ appear inside the linear combination $\nu_{\text{light}}$, so in some sense they're still in the picture. It's perhaps clearer to say that they provide a mechanism to generate the masses for $\nu_{\text{light}}$, which are necessary to explain neutrino oscillations. The Yukawa couplings and Majorana masses associated to the $N_R$ control the oscillation probabilities. $\endgroup$ – coconut Dec 28 '17 at 10:31

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