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Conformal invariance of gluon amplitudes

  1. Nov 30, 2011 #1
    Hi,

    I'm very ashamed to not understand how even the simplest gluon amplitudes are conformally invariant. See eg http://arxiv.org/abs/hep-th/0312171 pages 11-12.
    [tex]
    M(1^-,2^-,3^+)=\delta(\sum_i \lambda_i\tilde{\lambda}_i)\frac{\langle12\rangle^4}{\langle12\rangle \langle 23\rangle\langle31\rangle}
    [/tex]
    the dilation operator is:
    [tex]
    D\sim \lambda\frac{\partial}{\partial \lambda}+\tilde{\lambda}\frac{\partial}{\partial \tilde{\lambda}}+2
    [/tex]
    First, I assume the dilation operator contains a sum over all particles. Next, Witten says the delta function carries weight -4 under D. Ok. Then he says that [tex]\langle 12\rangle^4[/tex] has weight 4. This I don't get. Doesn't it have weight 4 under just eg [tex]\lambda_1\frac{\partial}{\partial \lambda_1}[/tex]
    So
    [tex]D \langle12\rangle^4=\sum_i (\lambda\frac{\partial}{\partial \lambda}+\tilde{\lambda}\frac{\partial}{\partial \tilde{\lambda}}+2)\langle12\rangle^4=[(4+0+2)+(4+0+2)+(0+0+2) ] \langle12\rangle^4?=14\langle12\rangle^4[/tex]

    Thanks for any help:)
     
  2. jcsd
  3. Nov 30, 2011 #2
    Conformal invariance is obvious, because you're talking about the tree amplitude which simply reflects the classical behaviour of the theory. The Yang-Mills Lagrangian is classically conformal invariant. Only loop graphs break this symmetry. But if you want to explicitly verify conformal invariance from the MHV amplitude expression, I haven't studied enough to answer that.
     
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