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Why are gauge fields bosons?

  1. Aug 14, 2010 #1
    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. jcsd
  3. Aug 14, 2010 #2

    bcrowell

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    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.
     
  4. Aug 14, 2010 #3
    A 4-vector field [itex]A^{\mu}[/itex] is in the [itex](\frac{1}{2}, \frac{1}{2})[/itex] - 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 [itex](m, n)[/itex]-representation has [itex](2m + 1)(2n + 1)[/itex] components. The number 4 factors as [itex]4 \times 1 = 2 \times 2[/itex].

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

    There is a dictionary that transforms the components [itex]A_{a \. \dot{a}}[/itex] to the components [itex]A^{\mu}[/itex]:

    [tex]
    A^{\mu} = \sigma^{\mu}_{a \, \dot{a}} \, A_{a \, \dot{a}}
    [/tex]

    where, numerically it turns out that [itex]\sigma^{\mu}_{a \, \dot{a}} = (I, \vec{\sigma})[/itex], where [itex]\vec{\sigma}[/itex] is a Cartesian vector whose components are the Pauli matrices.
     
  5. Aug 18, 2010 #4
    That is a fantastic answer: it is appreciated. The only thing I'm not sure of is the subscript [tex]\alpha[/tex] on the a in the part: does that refer to the 1/2 that labels the irrep of the Poincare group?
     
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