I'm currently reading about spontanous symmetry breaking. In particular about a Lagrangian that is invariant under $SU(2)\times U(1)$, in other words pretty standard QFT stuff. I know that the generators of $SU(2)$ are the Pauli matrices.

I was wondering how one would go about determining the generators of the group $SU(2)\times U(1) $? In particular the generator of the associated $U(1)$ symmetry. The reason for this is that I want to determine which of the generators of the symmetry $SU(2)\times U(1) $ are broken by the ground state.

A generator $T$ is said to be broken by the ground state $ \phi = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 0 \\ v \end{array}\right)$

if $$Tv \neq 0.$$

I think that the generator for the corresponding $U(1)$ symmetry is $\frac{1}{2}(1+\sigma_3)$, but I dont understand how I can show this.

  • $\begingroup$ Related: physics.stackexchange.com/q/16354/2451 $\endgroup$
    – Qmechanic
    Jul 1 '16 at 17:14
  • $\begingroup$ @MJ, are you asking, why the $U(1)$ of $SU(2)\otimes U(1)$ is associated with weak hypercharge instead of $U(1)$ that associated with QED? $\endgroup$
    – Mass
    Jul 1 '16 at 17:26
  • $\begingroup$ The generator you wrote down does not commute with SU(2), the Pauli matrices, as it should. What makes you unsatisfied with the identity? $\endgroup$ Jul 1 '16 at 22:56

It seems like you're fixing the representation of $\mathrm{SU}(2)$ to be $T^i=\frac{1}{2}\sigma^{i}$ (i.e., the fundamental representation). This makes sense if you're talking about the Higgs mechanism.

Now you want to find the generator $Y$ for the $\mathrm{U}(1)$ part of $\mathrm{SU}(2)\times\mathrm{U}(1)$. Let's call that generator $Y$. Now, any generator needs to satisfy the commutation relations (i.e., the structure constants) of the Lie algebra to which it belongs. In the case of a $\mathrm{U}(1)$, it needs to commute with all the other elements of the Lie algebra, which are spanned by the $\{T^i\}$. A suitable choice (and oftentimes, like here, your only choice) is the identity matrix times an overall constant: $Y=\tilde{Y}\,\mathbb{I}$.

For the $\mathrm{SU}(2)$ generators, the normalization is fixed by the commutation relations of the group. However, since $Y$ commutes with everything, its normalization is unconstrained, and needs to be chosen. We usually refer to the choice of the overall constant $\tilde{Y}$ (which we often just call $Y$, leaving the identity matrix implied) as the choice of representation.

For example, the definition of a field (like the Higgs doublet, for example) requires one to specify its representation under the gauge group of the theory we're considering. If the gauge group is $\mathrm{SU}(2)\times\mathrm{U}(1)$, then one would specify something like the fundamental representation of $\mathrm{SU}(2)$, and $Y=\frac{1}{2}$ under $\mathrm{SU}(2)$. (This is the representation of the Higgs doublet.)


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.