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Question on Fierz identity

  1. Mar 27, 2014 #1
    Hi everyone, I have a doubt on Fierz identities. If we define the following quantities: [itex] S=1,\; V=\gamma_\mu,\; T=\sigma_{\mu\nu},\; A=\gamma_\mu\gamma_5,\;P=\gamma_5[/itex], then we have the identity:

    $$
    (\Gamma_i)_{\alpha\beta}(\Gamma_i)_{\gamma\xi}=\sum_j F_{ij}(\Gamma_j)_{\alpha\xi}(\Gamma_j)_{\gamma\beta},
    $$
    where [itex]\Gamma_i[/itex] are the matrices define before. Moreover:
    $$
    F_{ij}=\frac{1}{8}\left(\begin{array}{ccccc}
    2 & 2 & 1 & -2 & -2 \\
    8&-4&0&-4&-8 \\
    24&0&-4&0&24 \\
    -8&-4&0&-4&8 \\
    2&-2&1&2&2
    \end{array}\right)
    $$
    Therefore, if we take the VV+AA combination it turns out that [itex]VV+AA=-VV-AA[/itex] with exchanged indices.

    However I usually read the Fierz transformation to be:
    $$
    (\psi_1\Gamma P_L\psi_2)(\psi_3\Gamma P_L\psi_4)=(\psi_1\Gamma P_L\psi_4)(\psi_3\Gamma P_L\psi_2).
    $$

    Without any minus sign. Does anyone knows why?
     
  2. jcsd
  3. Mar 27, 2014 #2

    Bill_K

    User Avatar
    Science Advisor

    Because the ψ's anticommute? I think it matters whether you just give the relation between matrices, as Wikipedia does, or include the ψ's. Both of these references give the table for Fij including the ψ's, with the opposite sign.
    http://hep-www.px.tsukuba.ac.jp/~yuji/mdoc/fierzTrans.pdf
    http://onlinelibrary.wiley.com/doi/10.1002/9783527648887.app5/pdf
     
  4. Mar 27, 2014 #3
    I think you are right. Once we write the identity for the matrices then we need to switch the two field and this should give an extra minus sign.
     
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