Fermions mass terms

  • #1
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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.
 

Answers and Replies

  • #2
ChrisVer
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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).
 
  • #3
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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 ?
 
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  • #4
Orodruin
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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.
 
  • #5
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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 .
 
  • #6
Orodruin
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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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