# 1 SuSy identity

by ChrisVer
Tags: identity, susy
 P: 779 THREAD CHANGE *SPINOR IDENTITY*...although it's connected with SuSy in general, it's more basic... I am trying to prove for two spinors the identity: $θ^{α}θ^{β}=\frac{1}{2}ε^{αβ}(θθ)$ I thought that a nice way would be to use the antisymmetry in the exchange of α and β, and propose that: $θ^{α}θ^{β}= A ε^{αβ}$ where A is to be determined.... To do so I contracted with another metric ε so that: $ε_{γα}θ^{α}θ^{β}= A ε_{γα}ε^{αβ} = Α (-δ^{β}_{γ})$ So I got that: $θ_{γ}θ^{β}= Α (-δ^{β}_{γ})$ So for β≠γ I'll have that $θ_{γ}θ^{β}=0$ And for β=γ I'll have that $θ_{β}θ^{β}=-A=-θ^{β}θ_{β}$ or $A=(θθ)$ And end up: $θ^{α}θ^{β}= ε^{αβ} (θθ)$ Another way I could determine A, would be by dimensionaly asking for [spinor]^2 term, without indices which would lead me again in A=(θθ)...but the same problem remains Unfortunately I cannot understand how the 1/2 factor disappears...Meaning I counted something twice (I don't know what that something is).. Could it be that I had to write first: $θ^{α}θ^{β}=\frac{(θ^{α}θ^{β}-θ^{β}θ^{α})}{2}$ and then say that the difference on the numerator is proportional to the spinor metric ε? If so, why?
 Quote by ChrisVer And for β=γ I'll have that $θ_{β}θ^{β}=-A=-θ^{β}θ_{β}$ or $A=(θθ)$