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Question on Fierz identity 
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#1
Mar2714, 12:17 PM

P: 306

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=VVAA[/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
Mar2714, 12:26 PM

Sci Advisor
Thanks
P: 4,160

http://hepwww.px.tsukuba.ac.jp/~yuj...fierzTrans.pdf http://onlinelibrary.wiley.com/doi/1...48887.app5/pdf 


#3
Mar2714, 12:30 PM

P: 306

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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