All Questions
14 questions
0
votes
1
answer
97
views
(Basic) confusion about value of scalar-fermion vertex $\phi\psi_i\psi_j$ for Majorana fermions
If I have a toy model with $N$ Majorana fermions $\lbrace\psi_i\rbrace_{i=1,\ldots,N}$ and a scalar field $\phi$ where the interaction among the fields is
$$
\mathcal L_\text{int}= \sum_{i,j=1}^N
a_{...
3
votes
1
answer
580
views
$Z$-boson decay into two neutrinos depends on the Dirac or Majorana nature of the neutrinos?
The decay rate of the $Z$-boson into two active neutrinos $Z \rightarrow \nu \overline{\nu}$ can be calculated straightforwardly and I obtained the same as in the literature. However, I was wondering, ...
2
votes
0
answers
79
views
Number and masses of right-handed neutrinos in the seesaw mechanism
In the Wikipedia article about the seesaw mechanism, it is argued that the seesaw mechanism "extends the Standard Model by assuming two or more additional right-handed neutrino fields", with ...
3
votes
0
answers
80
views
Zee QFT book on the neutrino mass
in p.102 of Zee QFT book,
"As of this writing, it is not known whether the neutrino mass is Dirac or Majorana. We will see in chapter VII.7 that a Majorana mass for the neutrino arises naturally ...
21
votes
3
answers
4k
views
Why do people say that neutrinos are either Dirac or Majorana fermions?
The question of whether a given particle "is" a Dirac or Majorana fermion is more subtle than is sometimes presented. For example, if we just consider the "old" Standard Model with massless neutrinos, ...
2
votes
2
answers
827
views
When considering a general mass term with Majorana masses, how do you exclude the existence of the left-handed neutrino mass?
Considering a general mass term describing the neutrino masses:
$$
-\frac{1}{2}
\begin{pmatrix}
\bar\nu_{R} & \bar\nu_L
\end{pmatrix}
\begin{pmatrix}
M_R & m \\
m & M_L
\end{pmatrix}
\...
6
votes
2
answers
1k
views
Can a Majorana field $\psi$ be charged under some $U(1)$ with a charge other than zero?
I know Majorana particles have to be electrically neutral because electric charged is conserved.
My question, however, is whether at all a Majorana field $\psi$ be charged under any $U(1)$ (other ...
5
votes
1
answer
421
views
Photons are self-conjugate but neutrinos may or may not: why is that?
Caution: This may be a very naive question but I find it confusing. Moreover, I believe this question is based on potential misconception. I would like it to be clarified.
Although the neutrinos are ...
1
vote
1
answer
901
views
A question about the Dirac mass and Majorana mass
I am sorry if my question seems to be naive.
For the free Dirac field, the Lagrangian is
$$\mathcal{L}=\bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m_D)\psi$$
or expressed in the Weyl spinor, the mass term ...
3
votes
1
answer
335
views
What are the quantum numbers of Majorana neutrinos?
I have a question about majorana neutrinos.
Majorana particles are particles that are their own antiparticle.
From this I would argue that they need to have all quantum numbers equal to zero. My ...
0
votes
1
answer
230
views
Neutrino mass and the Majorana equation
I can't seem find this on the Internet. What does the Majorana equation predict neutrino masses to be (if they were their own antiparticle), and how?
(I have little understanding of spinors, btw...)
...
8
votes
3
answers
839
views
Why aren't purely Dirac neutrinos ruled out?
It is common knowledge that in neutrinos can be Dirac particles without any Majorana masses as given a mass matrix,
\begin{equation}
\left( \begin{array}{cc}\nu _L & \nu _R \end{array} \right)
\...
5
votes
1
answer
197
views
What sorts of complications do massive neutrinos bring to the Standard Model?
Naively, I'd just think of considering them as any other massive fermions (but without electric charge), including the appropriate chiralities (and neutrino-higgs coupling when necessary). ...
32
votes
5
answers
3k
views
Why do or don't neutrinos have antiparticles?
This was inspired by this question. According to Wikipedia, a Majorana neutrino must be its own antiparticle, while a Dirac neutrino cannot be its own antiparticle. Why is this true?