Proving the "Thread Change: Spinor Identity

[ChrisVer]

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?

 

[andrien]

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,
$-\frac{1}{2}ε^{AB}(θθ)=-\frac{1}{2}ε^{AB}ε_{CD}θ^Cθ^D=-\frac{1}{2}[δ^{B}_{C}δ^{A}_{D}-δ^{A}_{C}δ^{B}_{D}]θ^Cθ^D=-\frac{1}{2}[θ^Bθ^A-θ^Aθ^B]=θ^Aθ^B$

 

[ChrisVer]

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 ...

 

[andrien]

And for β=γ I'll have that
$θ_{β}θ^{β}=-A=-θ^{β}θ_{β}$
or $A=(θθ)$

don't you think you have missed a factor of 2 here.