- #1
karlzr
- 131
- 2
I mean Goldstone bosons in the title. Sorry I don't know how to edit the title.
Goldstone's Theorem says that there is a massless Goldstone mode for each breaking symmetry. For instance symmetry of a theory is broken from SU(N) to SU(N-1), the # of Goldstone bosons is [itex] (N^2-1)-((N-1)^2-1)=2N-1[/itex] . The Goldstone modes are from some scalar fields. Take the electroweak theory in SM as an example, the [itex]SU(2)_L*U(1)[/itex] is broken to [itex]U(1)[/itex] where three degrees of freedom from the Higgs scalar H become Goldstone modes.
Then comes my question. Let's consider that the scalar fields live in the fundamental representation of SU(N) and thus there are [itex] 2N[/itex] degrees of freedom. If all [itex] N^2-1[/itex] symmetries in the gauge group are broken, how can we have [itex] N^2-1[/itex] Goldstone modes since there are only [itex] 2N[/itex] scalars in total?
Goldstone's Theorem says that there is a massless Goldstone mode for each breaking symmetry. For instance symmetry of a theory is broken from SU(N) to SU(N-1), the # of Goldstone bosons is [itex] (N^2-1)-((N-1)^2-1)=2N-1[/itex] . The Goldstone modes are from some scalar fields. Take the electroweak theory in SM as an example, the [itex]SU(2)_L*U(1)[/itex] is broken to [itex]U(1)[/itex] where three degrees of freedom from the Higgs scalar H become Goldstone modes.
Then comes my question. Let's consider that the scalar fields live in the fundamental representation of SU(N) and thus there are [itex] 2N[/itex] degrees of freedom. If all [itex] N^2-1[/itex] symmetries in the gauge group are broken, how can we have [itex] N^2-1[/itex] Goldstone modes since there are only [itex] 2N[/itex] scalars in total?