# Why are gauge fields bosons?

1. Aug 14, 2010

### metroplex021

Got a quick question on gauge particles: why are they always spin-1? Is it because they are introduced into theories in the form \partial_mu +cA^mu, and hence must be vectors (given that the derivative they have to compensate is a vector?)

2. Aug 14, 2010

### bcrowell

Staff Emeritus
Gauge fields are needed because without them, we would get spurious effects arising from the interchangeability of identical fermions. If you had a gauge field that was a fermion, then a vertex of the Feynman diagram would be an intersection of three fermions' world-lines, which would make it impossible to conserve angular momentum.

I don't think it's true that gauge fields are always spin-1. For example, the gauge transformations of GR are smooth coordinate transformations, and the gauge field is a rank-2 tensor, which, if we knew how to quantize it, would be spin-2.

3. Aug 14, 2010

### Dickfore

A 4-vector field $A^{\mu}$ is in the $(\frac{1}{2}, \frac{1}{2})$ - representation. To see this, we note that the 4-vector field has 4 components that all transform between each other under a general Lorentz transformation, thus the vector field is in an irreducible representation. A field in the $(m, n)$-representation has $(2m + 1)(2n + 1)$ components. The number 4 factors as $4 \times 1 = 2 \times 2$.

Therefore, the 4-vector field has to be in either $(\frac{3}{2}, 0)$. $(\frac{1}{2}, \frac{1}{2})$ or $(0, \frac{3}{2})$ representations. But, according to the vector addition model, the first two representations allow for $J = \frac{3}{2}$ angular momentum, only, while the second one allows for $J = 0. 1$. as it should be, because, under ordinary rotations, the time component of the 4-vector behaves as a scalar ($J = 0$), while the spatial components behave like an ordinary vector ($J = 1$).

There is a dictionary that transforms the components $A_{a \. \dot{a}}$ to the components $A^{\mu}$:

$$A^{\mu} = \sigma^{\mu}_{a \, \dot{a}} \, A_{a \, \dot{a}}$$

where, numerically it turns out that $\sigma^{\mu}_{a \, \dot{a}} = (I, \vec{\sigma})$, where $\vec{\sigma}$ is a Cartesian vector whose components are the Pauli matrices.

4. Aug 18, 2010

### metroplex021

That is a fantastic answer: it is appreciated. The only thing I'm not sure of is the subscript $$\alpha$$ on the a in the part: does that refer to the 1/2 that labels the irrep of the Poincare group?