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1 SuSy identity

  1. Apr 8, 2014 #1


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    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?
    Last edited: Apr 8, 2014
  2. jcsd
  3. Apr 8, 2014 #2
    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}##.
  4. Apr 8, 2014 #3


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    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 ...
  5. Apr 8, 2014 #4
    don't you think you have missed a factor of 2 here.
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