1
$\begingroup$

As I understand it, when we say that the $SU(2)_{L} \times U(1)_{Y}$ is broken via the Higgs mechanism, this is because the symmetry acts on the Higgs mass in a way that would change it's value. If we want to pick a particular model we need to pick a fixed value of the Higgs mass, and this is only possible if we say $SU(2)_{L} \times U(1)_{Y}$ is broken to $U(1)_{em}$. The Lagrangian is always invariant to $SU(2)_{L} \times U(1)_{Y}$ (even if the Higgs mass changes under $SU(2)_{L} \times U(1)_{Y}$ the Lagrangian is such that it remains invariant). The symmetry breaking is just neccesary in choosing a theory with one particular value of the Higgs vev.

Real life does appear to have a fixed Higgs mass so we require that $SU(2)_{L} \times U(1)_{Y}$ breaks to $U(1)_{em}$. But then aren't we saying that observable physics is described by $SU(3)_{C} \times U(1)_{em}$, why does the weak interaction still work in our universe that contains a fixed Higgs mass?

$\endgroup$
4
$\begingroup$

The symmetry breaking just breaks the symmetry in the electroweak sector of the standard model, it doesn't turn off the interaction--neutrinos become different from electrons, the gauge bosons get mass, etc, but all of those things are still there.

In essence, what the symmetry breaking does is make the weak interaction short-ranged.

$\endgroup$

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.