1 SuSy identity

  • Thread starter ChrisVer
  • Start date
  • #1
ChrisVer
Gold Member
3,381
462
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:

Answers and Replies

  • #2
1,024
32
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##
 
  • Like
Likes 1 person
  • #3
ChrisVer
Gold Member
3,381
462
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
1,024
32
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.
 

Related Threads on 1 SuSy identity

  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
15
Views
4K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
4K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
15
Views
4K
Top