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1 SuSy identity

by ChrisVer
Tags: identity, susy
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Apr8-14, 12:58 PM
P: 1,014
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?
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Apr8-14, 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}##.
Apr8-14, 03:05 PM
P: 1,014
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 ...

Apr8-14, 03:28 PM
P: 1,020
1 SuSy identity

Quote Quote by ChrisVer View Post
And for β=γ I'll have that
[itex] θ_{β}θ^{β}=-A=-θ^{β}θ_{β}[/itex]
or [itex]A=(θθ)[/itex]
don't you think you have missed a factor of 2 here.

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