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Self-dual and Anti Self-dual antisymmetric tensor

  1. Nov 24, 2012 #1
    I am working on Chapter 5.6 of Weinberg QFT book vol1.

    In page 231 and 232, he said (1,0) or (0,1) field corresponds to an antisymmetric tensor
    that satisfies the further irreducibilitry 'duality' conditions
    F^{\mu\nu}= \pm {{i}\over{2}} \epsilon ^{\mu\nu\lambda\rho}F_{\lambda \rho}
    for (1,0) and (0,1) fields, respectively.

    I check that Self-dual antisymmetrc tensor has three independent components and so does Anti Self-dual antisymmetric tensor. So, I can guess that there might be correspondence between Self-dual(or Anti Self-dual) antisymmetric tensor and (1,0) field (or (0,1)field ).

    But, I could not show how they are related to each other mathematically.

    I would like you guys to show me that how we correspond Self-dual antisymmetric tensor with (1,0) field and
    Anti Self-dual antisymmetric tensor with (0,1) field.

    Thank you so much.
  2. jcsd
  3. Nov 24, 2012 #2


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    Given Fμν, we define its dual as Fμν* = ½εμνστFστ. The self-dual and anti-selfdual tensors are complex combinations,
    SFμν = Fμν + iFμν* and AFμν = Fμν - iFμν*.

    It's easier to represent the F's as a pair of 3-vectors, Fμν = (E, B). Then Fμν* = (B, -E), SFμν = (E + iB, B - iE), and AFμν = (E - iB, B + iE).

    This shows that SFμν and AFμν each have only three complex components.

    Under a space rotation, all of these 3-vectors behave the same way. Under an infinitesimal rotation about an axis with unit vector ω, a 3-vector V transforms as dV = ω x V, and this applies equally well to E, B and E ± iB. The generators for spatial rotations are denoted J.

    Under a finite boost, E' = γ(E + β x B), B' = γ(B - β x E), and the infinitesimal form of this is dE = β x B, dB = - β x E. Then d(E + iB) = β x (B - iE) = -i β x (E + iB) and d(E - iB) = i β x (E - iB). The generators for boosts are denoted K.

    The Lorentz group is a direct product of two groups generated by J + iK and J - iK. I'm getting tired of writing down all the details, but if you combine the above results and calculate how (E ± iB) transform under (J ± iK), you'll find the changes add in one case and cancel in the other, indicating that these quantities transform exactly as the (1,0) and (0,1) representations must.
    Last edited: Nov 24, 2012
  4. Nov 20, 2013 #3
    can you give me some reference on this topic.
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