On page 527 of Srednicki's textbook "Quantum Field Theory", the Standard Model is described as follows:

It can be succinctly specified as a gauge theory with gauge group $SU(3) \times SU(2) \times U(1)$, with left-handed Weyl fields in three copies of the representation $(1, 2, -\frac{1}{2}) \oplus (1, 1, +1) \oplus (3, 2, + \frac{1}{6}) \oplus (\overline{3}, 1, -\frac{2}{3}) \oplus (\overline{3}, 1, +\frac{1}{3})$, and a complex scalar field in the representation $(1, 2, -\frac{1}{2})$. Here the last entry of each triplet gives the value of the $U(1)$ charge, known as $\it{hypercharge}$.

I am puzzled by the group representation $(1, 2, -\frac{1}{2}) \oplus (1, 1, +1) \oplus (3, 2, + \frac{1}{6}) \oplus (\overline{3}, 1, -\frac{2}{3}) \oplus (\overline{3}, 1, +\frac{1}{3})$. How does it come about? What are the steps (if any) to get this representation?

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    $\begingroup$ Possible duplicate of Who chooses the representation of SM particles? $\endgroup$ – AccidentalFourierTransform Mar 25 '17 at 8:39
  • $\begingroup$ I want to know the matchematical steps to derive this representation, which is not provided in the question and answer that you mentioned. $\endgroup$ – Shen Mar 25 '17 at 12:27
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    $\begingroup$ What do you want to derive that from? $\endgroup$ – ACuriousMind Mar 25 '17 at 13:22
  • $\begingroup$ I don't know from what and how to derive this representation. I just saw it in Srednicki's textbook. That's why I am puzzled. I guess it is derived from a representation of $SU(3) \times SU(2) \times U(1)$. I want to know the mathematical steps to derive it. $\endgroup$ – Shen Mar 25 '17 at 13:44
  • $\begingroup$ The specific representations that are chosen in the SM are a physical input, and can't be derived from the group. Are you asking how to see that the numbers $(1,2, -\frac{1}{2})\oplus\cdots$ correspond to the particle content of the model? $\endgroup$ – user47224 Mar 25 '17 at 20:16

Your text assumes you are familiar with the quantum number content of the elementary particle fermions, determined by the Millikan oil-drop experiment, structure functions of the light quarks, V-A structure of the weak currents, etc. These are experimental inputs and they come from out there, your world.

It helps you summarize the self-evident logic of their apparently diverse quantum numbers so you could write a compact QFT for them, that's all. I assume you seek an appreciation of the manifest logic involved.

It gives you the SU(3) color rep, singlet for leptons, color 3 for quarks, or color anti triplet for antiquarks. Likewise, their SU(2) weak isospin, vanishing for right handed singlets, and doublet for left-handers. (No separate 2-bars, of course, as SU(2) is pseudoreal.) And, of course, mutatis mutandis for their CPT conjugates. You only have singlets and fundamental reps, since these are fundamental fermion building blocks of our world.


  • (1,2,-1/2) , e.g. for $e _L$
  • (1,1,1), e.g. for $\overline {e _R}$
  • (3,2,1/6) e.g. for $u_L, d_L$
  • ($\bar 3$,1,-2/3) for $\overline{u_R}$
  • ($\bar 3$,1,1/3) for $\overline{d_R}$.

The hypercharge in the third entry is dross -- an error-correction number, if you wish, given by $Y_W\equiv Q-T_3$, once you input the charge, in the "minority usage", but actually modern mainstream definition, so you might have to multiply it by 2 to agree with hidebound historical listings, like those linked here. It is the eigenvalue of U(1), as your particles are all singlets, of course, under it, and multiplies the B coupling charge of the fermion currents. The sooner you get used to its Golden Mnemonic, the better: It is the average charge of isomultiplets.

There is nothing more to it. Given these numbers you may completely, and concisely specify the fermion sector of the SM QFT.

  • $\begingroup$ The right-handed electron should have an electric charge -1, but you put a + on it. Is this an error? $\endgroup$ – Shen Feb 19 at 12:43
  • $\begingroup$ The right-handed electron indeed has charge and hypercharge -1. But what I have in there is the positron in the weak isospin isosinglet, which is L, conjugate to the electron R. This is the gist of Srednicki's convention: he writes down only left-handed fields, so for the R electron he writes down the L positron. Perhaps I should have written an overbar. Let me do that. $\endgroup$ – Cosmas Zachos Feb 19 at 15:19

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