1
$\begingroup$

I am doing a question that asks me to identify the gauge groups of a Lagrangian with the field strength tensors

$$\bf{F}_{\mu \nu} = \partial_{\mu}\bf{W}_{\nu} - \partial_{\nu} \bf{W}_{\mu} - g\bf{W_{\mu}} \times \bf{W_{\nu}},$$

$$B_{\mu \nu} = \partial_{\mu}B_{\nu} - \partial_{\nu}B_{\mu}.$$

I understand that the second one of these corresponds to a $U(1)$ gauge theory because it is clearly abelian. I'm a little confused about why the first one has to give us an $SU(2)$ gauge theory. I understand that it can do because the structure constants of an $SU(2)$ Lie algebra are $f^{abc}= \epsilon^{abc}$ in the fundamental of $SU(2)$, but aren't these also the structure constants of $SO(3)$?

Why do we know it's not $SO(3)$?

$\endgroup$
0
$\begingroup$

The Higgs field $\phi$ is a complex scalar, not a real one. This is crucial to its spontaneous symmetry breaking. It's also why the field's full symmetry is of the form $U(N)$, not $O(N)$. We introduce gauge bosonic fields to localize an otherwise global symmetry. For example, a "what's the weak force? I've never heard of that!" theory of electromagnetism has a Lagrangian containing $|(\partial_\mu+iqA_\mu)\phi|^2$ terms, an $F_{\mu\nu}F^{\mu\nu}$ term with $F_{\mu\nu}:=\partial_\mu A_\nu-\partial_\nu A_\mu$, and a scalar potential we may as well write as $V(|\phi|^2)$ (with a $-$ sign in front of it in the Lagrangian density, of course). This takes $N=1$, whereas to understand electroweak theory we need $N=2$ so $\phi_1,\,\phi_2$ are both complex scalars. Spontaneous symmetry breaking splits $U(2)$ into $SU_L(2)$ and $U_Y(1)$, a slightly more complicated option than a hypothetical one to $SU(2)$ and $U(1)$; either way, this gives $SU(N)$ish factors rather than $SO(N)$ish ones.

With a bit of effort, we can split each complex scalar into two real scalars and rewrite any $U(1)$ symmetry as $O(2)$ (although we'd still only need one auxiliary vector field, not two), but $SU(2)$ isn't quite so easy. While important, the structure constants don't tell the whole story. As explained here, the vector representations of $SO(3)$ are a proper subset of the representations of $SU(2)$, which also includes spinor representations. This allows the Higgs field to couple to fermions. The universe could theoretically have used completely different gauge groups from the one it did, but we have to work with whatever fits the data. In particular, there are so few electroweak gauge bosons it's best to think in terms of how a complex multiplet transforms instead of a real one.

$\endgroup$

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.