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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
    [itex]
    F^{\mu\nu}
    [/itex]
    that satisfies the further irreducibilitry 'duality' conditions
    [itex]
    F^{\mu\nu}= \pm {{i}\over{2}} \epsilon ^{\mu\nu\lambda\rho}F_{\lambda \rho}
    [/itex]
    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

    Bill_K

    User Avatar
    Science Advisor

    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
    Sir(Bill_K),
    can you give me some reference on this topic.
     
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