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Within the Standard Model (SM) each Lagrangian term has to have a mass dimension of [L] =4. While the mass dimensions of scalar fields $[\Phi] = 1$, Dirac fields $[\Psi] =3/2$ and Vector fields $A_{\mu} = 1$ are usually explained in books, I haven't found any description on the mass dimensions of e.g. lepton or quark doublets $L= \begin{pmatrix} \nu_L \\ e_L \end{pmatrix}$ , $Q= \begin{pmatrix} u_L \\ d_L \end{pmatrix}$ or singlets $e_R$, $u_R$... which are also represented in the SM Lagrangian. How can I determine their mass dimensions?

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    $\begingroup$ They are plain Dirac Fermions, aren't they?? $\endgroup$ – Cosmas Zachos Feb 24 '18 at 21:15
  • $\begingroup$ That's what I thought first too, but looking at the SM Lagrangian term $\bar{d_R} \bar{M}^d \bar{\Phi} Q$ the mass dimensions don't add up to 4. If I assume that the mass dimension for Dirac fields is the same as for fermion doublets and singlets then the term gives me: 3/4 + 1 +1 + 3/4=5, unless the mass matrix (which should collapse to the mass) would be 0, but masses have a mass dimension of 1. $\endgroup$ – Alex Feb 24 '18 at 21:28
  • $\begingroup$ The expression you wrote is horrible nonsense. The M you wrote is unwarranted--it should be a dimensionless Yukawa coupling, as you should have learned in your course: the 2nd job of the Higgs. Mass arises out of the v.e.v. of the Higgs you did write down. Had you written down the correct Yukawa coupling in your question, it would have made itself superfluous. $\endgroup$ – Cosmas Zachos Feb 24 '18 at 21:44
  • $\begingroup$ Well, I didn't take a course, that might explains why I don't have all the knowledge. However, this term I have found in several sources, so I guess it can't be that non-sense unless the sources are not good: e.g.: einstein-schrodinger.com/Standard_Model.pdf or quora.com/What-is-the-Lagrangian-of-the-Standard-Model. How would the correct Yukawa coupling look like? And does that mean, that doublets and singlets have the same mass dimension of 3/2? Also correction for my previous comment 3/4 => 3/2! $\endgroup$ – Alex Feb 24 '18 at 22:06
  • $\begingroup$ You misread and are misquoting your link. It has M/v s instead, which amount to dimensionless Yukawa couplings. All Dirac fermions have dimension 3/2 --their transformation properties under the gauge groups do not matter. $\endgroup$ – Cosmas Zachos Feb 24 '18 at 22:12
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The fact that the SM contains, for example, fermion doublets is simply a consequence of their representation under the gauge group $SU(2)$. However the lepton doublet is a singlet under $SU(3)$ for example, while the quarks would be written as the triplet $$q_L=\begin{pmatrix} q_{L,\text{ red}} \\ q_{L,\text{ green}}\\ q_{L,\text{ blue}} \end{pmatrix} \, .$$.

These are just the different degrees of freedom associated to that term in the lagrangian. But the fact that the fermion has different representations under each of the groups doesn't change its mass dimension. In four spacetime dimensions the fermion doublet will still have mass dimension $3/2$.

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