# Fermions mass terms

## Main Question or Discussion Point

Hi all,

I have some points not so clear for me about the fermions mass terms in SM; first, why
$\bar{\psi}\psi = \bar{\psi}_L \psi_R + \bar{\psi}_R \psi_L$, that since

$\bar{\psi} = \bar{\psi}_R + \bar{\psi}_L$ and
$\psi = \psi_R + \psi_L$

Where are such terms: $\bar{\psi}_R \psi_R , \bar{\psi}_L \psi_L$

Second: Why in SM the Majorana mass terms has not taken into account, that since term as:
$m~ \bar{\psi}_L \psi_R$ is not $SU(2)_L$ invariant, a term as

$m~ \bar{\psi}_L \bar{\psi}^c_L, ~ m ~\bar{\psi}_R \bar{\psi}^c_R$ are gauge invariant. I mean could not Majorana mass terms replace the Higgs mechanism ?

Last: I read in literatures that $m~ \bar{\psi}_L \psi_R$, is not a re-normalizable term, but I don't see why , if $\psi$ has dimensions of mass equals 3/2, then the the mass dimension of this term is only 4 !

Thanks,
S.

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ChrisVer
Gold Member
Your first question is answered by the fact that $P_L P_R$ or $P_R P_L$ are equivalent to the zero-operation:

$(1 \pm \gamma_ 5) (1 \mp \gamma_5) = 1 - \gamma_5^2 = 0$

now your $\bar{\psi}_R \psi_R = \psi^\dagger P_R \gamma_0 P_R \psi = \bar{\psi} P_L P_R \psi =0$
where I used that $P_R^\dagger = P_R$ and that $P_R \gamma_0 = \gamma_0 P_L$

For the second, first of all, the $\psi_L$ combination you have given is not SU(2)_L invariant, because both your fields transform as doublets under SU(2) transformation. It's the same reason that in the Sea-Saw mechanism you don't add such a term for the left handed neutrinos.
For the rights I am not sure, but which fermion in the SM is Majorana? appart from neutrinos there is no other chargeless fermion.... And in neutrinos you can have such a term [in Seesaw mechanism]...

Finally for the mass term. Can you give some "literature" you found it in? In general the fermionic mass dimension doesn't have to be 3/2 (in general dimensions).
Then the mass term won't necessarily have the dimension 1.
However I don't understand why someone would call it non-renormalizable. You can renormalize the bare mass by adding mass regularizing terms (self energy diagrams).

• vanhees71
For the second, first of all, the $ψ_L$ combination you have given is not SU(2)_L invariant, because both your fields transform as doublets under SU(2) transformation.
Do you mean that $\psi_L$ transforms by $\psi_L' \to e^{-iT_i a^i(x) } \psi_L$, and $\bar{\psi_L}^c$ transforms also by $\bar{\psi_L'}^c \to e^{-iT_i a^i(x) } \bar{\psi_L}^c$ ?

Nevertheless, in a talk as
http://www.ippp.dur.ac.uk/~dph3tcl/NuNotes/TLiSeeSaw.pdf [Broken]

slide 5, $\bar{\nu_l}^c \nu_l$ term has taken in Seesaw mechanism ..

What do you think about good and clear reference for Seesaw mechanism ?

Last edited by a moderator:
Orodruin
Staff Emeritus
Homework Helper
Gold Member
Your original post is a bit unclear since you have not defined how the fields transform under the gauge group $SU(2)_L$.

The Majorana mass term for Standard Model neutrinos is breaking $SU(2)_L$ due to the reason given by ChrisVer, the neutrino field is part of a weak doublet together with the corresponding left handed charged lepton. However, we know that $SU(2)_L$ is broken and at low energies it is possible to have such mass terms. In the seesaw mechanisms, they are generally introduced by integrating out a heavy field which gives rise to a Weinberg operator. Upon the Higgs taking a vev and breaking $SU(2)_L$, this results in a neutrino Majorana mass. This is the effective regime that Tracey is looking at.

• vanhees71
I don't ask about neutrino specifically, but as in my second question it isn't clear why a combination as $\bar{\psi_L}^c \psi_L$ is not allowed under SU(2)_L .

Orodruin
Staff Emeritus
You cannot get an answer to this question unless you specify how $\psi$ transforms under $SU(2)_L$. If $\psi$ is a neutrino field $\nu_L$, it is part of an $SU(2)_L$ doublet and therefore cannot appear in the Lagrangian without its $SU(2)_L$ partner. As I said in the previous post, it can appear after electroweak symmetry breaking once $SU(2)_L$ is broken, generally by the Higgs field in the Weinberg operator $\overline{L^c}\phi \tilde \phi^\dagger L /\Lambda + h.c.$ taking a vev and thus singling out the neutrino component of the $SU(2)_L$ doublet $L = (\nu_L e_L)$, where $e_L$ is the left handed charged lepton field.
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