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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.

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  • $\begingroup$ 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? $\endgroup$ – AMS 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? $\endgroup$ – Cosmas Zachos Jul 1 '16 at 22:56
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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.)

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