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#1
Apr814, 12:58 PM

P: 762

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: [itex] θ^{α}θ^{β}=\frac{1}{2}ε^{αβ}(θθ)[/itex] I thought that a nice way would be to use the antisymmetry in the exchange of α and β, and propose that: [itex] θ^{α}θ^{β}= A ε^{αβ} [/itex] where A is to be determined.... To do so I contracted with another metric ε so that: [itex] ε_{γα}θ^{α}θ^{β}= A ε_{γα}ε^{αβ} = Α (δ^{β}_{γ})[/itex] So I got that: [itex] θ_{γ}θ^{β}= Α (δ^{β}_{γ})[/itex] So for β≠γ I'll have that [itex] θ_{γ}θ^{β}=0[/itex] And for β=γ I'll have that [itex] θ_{β}θ^{β}=A=θ^{β}θ_{β}[/itex] or [itex]A=(θθ)[/itex] And end up: [itex] θ^{α}θ^{β}= ε^{αβ} (θθ)[/itex] 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: [itex] θ^{α}θ^{β}=\frac{(θ^{α}θ^{β}θ^{β}θ^{α})}{2}[/itex] and then say that the difference on the numerator is proportional to the spinor metric ε? If so, why? 


#2
Apr814, 02:58 PM

P: 1,020

I think there is a minus sign on right hand side.Anyway ,you should use the identity ##ε_{AB}ε^{CD}=δ^{D}_{A}δ^{C}_{B}δ^{C}_{A}δ^{D}_{B}##.
So, ##\frac{1}{2}ε^{AB}(θθ)=\frac{1}{2}ε^{AB}ε_{CD}θ^Cθ^D=\frac{1}{2}[δ^{B}_{C}δ^{A}_{D}δ^{A}_{C}δ^{B}_{D}]θ^Cθ^D=\frac{1}{2}[θ^Bθ^Aθ^Aθ^B]=θ^Aθ^B## 


#3
Apr814, 03:05 PM

P: 762

Thanks... although I'm also trying to understand how/where I did the "mistake" in my approach :)
The minus, at least for the notations I'm following, is for when you have the conjugate spinors ... 


#4
Apr814, 03:28 PM

P: 1,020

1 SuSy identity



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