Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fermions mass terms

  1. Dec 31, 2014 #1
    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.
     
  2. jcsd
  3. Dec 31, 2014 #2

    ChrisVer

    User Avatar
    Gold Member

    Your first question is answered by the fact that [itex]P_L P_R[/itex] or [itex] P_R P_L [/itex] are equivalent to the zero-operation:

    [itex] (1 \pm \gamma_ 5) (1 \mp \gamma_5) = 1 - \gamma_5^2 = 0 [/itex]

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

    For the second, first of all, the [itex] \psi_L [/itex] 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).
     
  4. Jan 1, 2015 #3
    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: May 7, 2017
  5. Jan 1, 2015 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    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.
     
  6. Jan 1, 2015 #5
    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 .
     
  7. Jan 1, 2015 #6

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook