This question comes from Srednicki's textbook "Quantum Field Theory". On page 532, the left-handed Weyl fields $\ell$ (a single lepton family, electron and its neutrino) and $\overline{e}$ are in the representations $(2, -\frac{1}{2})$ and $(1, +1)$ of $SU(2) \times U(1)$. It is stated:

We cannot write down a mass term involving $\ell$ and/or $\overline{e}$ because there is no gauge-group singlet contained in any of the products \begin{equation} (2, -\frac{1}{2}) \otimes (2, -\frac{1}{2}), \\ (2, -\frac{1}{2}) \otimes (1, +1), \\ (1, +1) \otimes (1, +1) .\tag{88.4} \end{equation}

I calculate the first product as follows:

  1. Using Young tableaux to do the calculation for the first entry in $SU(2)$ gives \begin{equation} 2 \otimes 2 = 1 \oplus 3 \end{equation}
  2. For the second entry, I simply use addition \begin{equation} -\frac{1}{2} - \frac{1}{2} = -1 \end{equation}
  3. Combining 1. and 2., I get the result \begin{equation} (2, -\frac{1}{2}) \otimes (2, -\frac{1}{2}) = (1, -1) \oplus (3, -1) \end{equation} Is this correct? If yes, isn't $(1, -1)$ a singlet, which will refute the statement in the text?
  • 2
    $\begingroup$ Please explain your notation for the representations. I would expect the label to be $(s,q)$, where $s$ is dimension of the weak isospin representation and $q$ the weak hypercharge, but that doesn't mesh with your label since the electron and the neutrino both have weak hypercharge $-1$, and the trivial $\mathrm{U}(1)$ representation would be the one denoted by 0 in that case. $\endgroup$ – ACuriousMind Mar 27 '17 at 12:36
  • $\begingroup$ When you talk about a singlet, what it is a singlet of? Surely by your notation $(1,-1)$ is an $su(2)$ singlet but is this enough? Can you clarify how you see the second product as a product of a represenation of $su(2)$ and a representation of $u(1)$? Why would it be different from the first product? $\endgroup$ – ZeroTheHero Mar 27 '17 at 12:40
  • $\begingroup$ This is not my notation, but Srednicki's notation. I just quoted his text from his book. It is true that $\it{q}$ is the weak hypercharge, but using the equation $Q = I_{3} + Y$, we have $Y = -\frac{1}{2}$ for both the electron and the neutrino; whereas using the equation $Q = I_{3} + \frac{1}{2}Y$, we have $Y = -1$ for both the electron and the neutrino. You probably used the second equation to get the weak hypercharge. $\endgroup$ – Shen Mar 27 '17 at 13:01
  • $\begingroup$ @Shen They are all representations of $SU(2)\times U(1)$. $\endgroup$ – ZeroTheHero Mar 27 '17 at 14:32
  • $\begingroup$ @ZeroTheHero I made a mistake. You reminded me. Thanks. So the second and third products should be calculated in the same way as the first product. $\endgroup$ – Shen Mar 27 '17 at 14:51
  1. Srednicki is merely trying to convey that terms (such as, e.g., a mass term) in the Lagrangian should be gauge-invariant under the electroweak gauge group $SU(2)\times U(1),$ i.e. belong to the trivial representation $(1,0).$ In particular, the three tensor products mention in eq. (88.4) do not contain the trivial representation $(1,0).$

  2. OP's question seems to be spurred by the fact that all irreducible representations of an abelian Lie group (such as, e.g., $U(1)$) are 1-dimensional. So shouldn't they all be called singlets? However, Srednicki seems to adapt the opposite convention that the trivial $U(1)$ irreducible representation is the only $U(1)$ singlet.

  • $\begingroup$ @Qmechanic- The first entry of $(1, 0)$ is in $SU(2)$. So, is $(1, 0)$ an $SU(2)$ singlet instead of a $U(1)$ singlet? $\endgroup$ – Shen Mar 28 '17 at 14:59
  • $\begingroup$ $(1,0)$ is a $SU(2)\times U(1)$ singlet, while the first entry $1$ is a $SU(2)$ singlet and the second entry $0$ is a $U(1)$ singlet. $\endgroup$ – Qmechanic Mar 28 '17 at 15:02
  • $\begingroup$ @Qmechanic- How about $(1, -1)$? Is it only an $SU(2)$ singlet? Is the second entry $-1$ a $U(1)$ singlet or not? If not, is it a $U(1)$ multiplet? $\endgroup$ – Shen Mar 28 '17 at 15:28
  • $\begingroup$ Srednicki would not call the second entry $-1$ a $U(1)$ singlet, as mentioned in the answer. $\endgroup$ – Qmechanic Mar 28 '17 at 15:33

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