What is the isospin of the Higgs particle

1. Jul 8, 2012

johanw

I had always assumed that the simplest possible standard model Higgs particle had isospin 0, but when I tried to verify this assumption I read some conflicting opinions (not to mention discussions about more complicated Higgs mechanisms).

I also read conflicting views about the conservation of isospin, I know it is conserved under CPT but which processes violate it?

2. Jul 8, 2012

fzero

There are two types of isospin commonly used in particle physics. What is usually called just "isospin" is the approximate SU(2) flavor symmetry of the up and down quarks that would be exact in the absence the mass difference between the quarks. The corresponding charge is the eigenvalue of the diagonal ($I_3$) generator on the individual particle states. Only the hadrons have nonzero isospin, so it is associated with the strong interaction.

The other type is "weak isospin," which is the eigenvalue of diagonal generator ($T_3$)for the SU(2) subgroup of the electroweak gauge group $SU(2)\times U(1)_Y$. In the case of the up and down quarks, this coincides with the (strong) isospin, but this charge is carried by all left-handed fermions, including leptons.

The Higgs field is a doublet under weak isospin, with the electrically neutral component usually assigned $T_3 = -1/2$. This includes the physical Higgs boson, but this excitation is defined around the vacuum state, for which $T_3$ is no longer a symmetry. The conserved charge remaining from the electroweak symmetry is the electric charge $Q = T_3 + Y/2$, where $Y$ is the weak hypercharge of $U(1)_Y$. Weak isospin and hypercharge are not conserved below the EW symmetry breaking scale.

Several facts correlate with nonconservation of the weak charges in the vacuum. The mass eigenstates of a Dirac fermion are combinations of left and right-handed components, so the mass eigenstate of, say, the electron, is not in a definite weak eigenstate. The Dirac mass terms explicitly break weak isospin.

Futhermore, the physical Higgs boson is its own antiparticle, so by CPT invariance cannot carry a nonzero U(1) charge. This would suggest the assignment $T_3=0$, but the above discussion of symmetry breaking should already convince you that it doesn't make sense to assign any particular number to a broken symmetry.

3. Jul 9, 2012

johanw

Sorry, but I don't understand that. If $Y\neq0$ and $Q = 0$, how can then be $T_3=0$ if $Q=T_3+Y/2$?

That makes sense of course.

Last edited: Jul 9, 2012