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?

  • 3
    $\begingroup$ Possible duplicate of Who chooses the representation of SM particles? $\endgroup$ 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
  • 1
    $\begingroup$ 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 '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. $\endgroup$ Feb 19 '19 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.