THREAD CHANGE *SPINOR IDENTITY*...although it's connected with SuSy in general, it's more basic...(adsbygoogle = window.adsbygoogle || []).push({});

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 forβ≠γI'll have that

[itex] θ_{γ}θ^{β}=0[/itex]

And forβ=γI'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?

**Physics Forums - The Fusion of Science and Community**

# 1 SuSy identity

Have something to add?

- Similar discussions for: 1 SuSy identity

Loading...

**Physics Forums - The Fusion of Science and Community**