Fields with SO(3) diagonal subgroup symmetry

Question

I read about a Higgs field $\vec{\phi}=\frac{1}{2}a\hat{r}\cdot \vec{\sigma}$ (in the context of 't Hooft-Polyakov monopole) with SO(3) diagonal subgroup symmetry consisting of simultaneous and equal rotations in real and isotopic space, wherein $\sigma^i$ is the Pauli matrix, $\hat{r}$ is the spatial unit vector.

I only know a little bit of introductory group theory for physics student. Unfortunately, I cannot see how this form of field $\phi$ comes out. Does it mean that we are using Pauli matrices as kinda bases of the isotopic space? If so, why? I've learned nothing about such thing. And by the way, what does isotopic space mean

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user huotuichang

Comments

The interest is to mix internal degrees of freedom of fields ($\Phi$ is in the adjoint representation of $SO(3)$ or $SU(2)$, so has $3$ components : $\Phi = \Phi^a \sigma_a$ ), and space time degrees of freedom (we have 3 spatial dimensions $r^i$).Writing $\Phi^a = r^a$, may seem very curious, because we mix freedom degrees, which seem to have no relation. The interest is to establish topological structures, for instance, we may consider the unit vector $\hat \phi$, which lives in $S^2$, and a asymptotic sphere in space, also $S^2$, and we could consider mapping between these 2 spheres.

This post imported from StackExchange Physics at 2014-08-22 05:06 (UCT), posted by SE-user Trimok

Answer 1

"Isotopic space" means the internal space you move around in when you do Isospin transformation. For a Nucleon, "Isotopic space" consists of all superpositions of proton and neutron, for a pion, it's all superpositions of pi0 pi+ and pi- and so on. Because of the history of gauge theory, Yang and Mills were gauging the Isospin transformation, any gauge group is sometimes called isotopic space even if it has nothing to do with isospin.

The Pauli matrices are SU(2) generators, SU(2) and SO(3) are the same Lie algebra. The 'tHooft Polyakov monopole, you have an isospin vector (it's a collection of 3 scalars which form a vector under isospin rotations) which is "radial" (or a "hedgehog" in Polyakov's terminology), meaning that

$V_1 = x f(r)$

$V_2 = y f(r)$

$V_3 = z f(r)$

where $V_i$ are the three vector isospin components (they rotate into each other when you do an Isospin rotation, nothing happens to these components when you do a spatial rotation), and x,y,z are space indices (they rotate when you do an actual rotation, nothing happens to space when you do an isospin rotation).

When you do a rotation of space on a configuration of scalar fields like this, the field is not invariant, but you can fix that if you do the same isospin rotation also, independently, at the same time. That means the solution breaks two independent SO(3)s to one SO(3).

Comments

Nice explanation, +1. Minor comment/question: you say SO(3) and SU(2) are the same, but this is only true locally, right? If so, I would suggest you make it clear in the answer, especially since the OP "only knows a little bit of introductory group theory" and in the context of magnetic monopoles as solitons this distinction is very important.

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Fixed, I agree.