4
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – Trimok Nov 4 '13 at 17:51
  • $\begingroup$ We are using Pauli matrices for any SU(2) group. Isotopic space is just an internal degree of freedom, allowing any (isospin 1/2) particle to be in "up" or "down" state. For a nucleon it means to be either proton or neutron, for a quark $(u,d)$, for a lepton $(\nu_e,e)$. Halzen & Martin's book contains very good explanations of that. $\endgroup$ – firtree Aug 23 '14 at 9:25
  • $\begingroup$ @Trimok Why don't you post your comment as an answer? It is hard to answer something better than that. $\endgroup$ – Diracology May 14 '16 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.