# 't Hooft - Polyakov monopole at large distance

1. Jan 29, 2014

### karlzr

According to 't Hooft - Polyakov monopole solution, SO(3) is spontaneously broken down to U(1) and the unbroken mode works very well as the electromagnetic field. In this case we do not need dirac string but just some scalar field. At very large distance , the two massive gauge modes can be integrated out. So I guess the only degrees of freedom left is the massless unbroken mode and the scalar field. This looks like there is a magnetic monopole at the core. My question is :
Can we relate this picture with those of Dirac or Schwinger where magnetic monopole was effectively regarded as some fundamental fermion field? I mean, how do we interpret the origin of the fermion field from the original SO(3) group?

2. Jan 31, 2014

### samalkhaiat

The original tHooft-Polyakov model which was based on the gauge group $SO(3)$ had no fermions in it. In 1976, Jackiw and Rebbi [1], Hasenfratz and tHooft [2] have (independently) shown that if an isodoublet scalar field is added to tHooft-Polyakov model (making it an $SU(2)$ gauge theory), then the bound state of monopole plus scalar has spin 1/2. This state has statistics of a fermion. Thus one find fermions in an $SU(2)$ gauge theory of Lorentz scalars and vectors. This result is the non-Abelian generalization of the well-known fact that the bound state of a Dirac monopole (with minimum magnetic charge) and a particle (with minimum electric charge) has angular momentum (1/2) stored in its electromagnetic field.
This demonstrates beautifully the fact that one cannot get spin-1/2 from SO(3). It can be understood if we compare the Dirac relationship between magnetic charge $g$ and the minimum electric charge $Q_{min}$(i.e., Dirac quantization condition):
$$g = \frac{ n }{ 2 Q_{ min } } ,$$
with the string removal condition:
$$g = \frac{ n }{ e } .$$
If $e$ is the minimum electric charge in the theory, then these conditions differ by a factor of 2 and allowed Dirac monopoles with $g = ( 2n + 1 ) / 2 e$ cannot have their strings removed. However, if doublet representations couple to the gauge fields, then their charge $e / 2$ becomes $Q_{ min }$ and both conditions are identical. These two possibilities distinguish between gauge theories with $SO(3)$ gauge group from those with $SU(2)$ gauge group (the global properties of the gauge group is essential for monopole solutions). So, only for $SU(2)$ gauge theory can all Dirac monopoles have their strings removed.

Sam

[1] Jackiw, R., and Rebbi, C. (1976b). Phys. Rev. Lett., 36, 1116.

[2] Hasenfratz, P., and tHooft, G. (1976). Phys. Rev. Lett., 36, 1119.