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Decomposition of SL(2,C) Weyl Spinors

  • Thread starter sgd37
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  • #1
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Homework Statement



Using

[tex] (\sigma^{\mu \nu})^{\beta}_{\alpha} (\sigma_{\mu \nu})^{\delta}_{\gamma} = \epsilon_{\alpha \gamma} \epsilon^{\beta \delta} + \delta^{\delta}_{\alpha} \delta^{\beta}_{\gamma}[/tex]

show that

[tex] \Psi_{\alpha} X_{\beta} = \frac{1}{2} \epsilon_{\alpha \beta} (\Psi X) + \frac{1}{2} (\sigma^{\mu \nu} \epsilon^{T})_{\alpha \beta} (\Psi \sigma_{\mu \nu} X) [/tex]

Homework Equations






The Attempt at a Solution



if I do [tex] (\sigma^{\mu \nu})^{\beta}_{\alpha} (\sigma_{\mu \nu})^{\delta}_{\gamma} \Psi _{\beta} X_{\delta} [/tex]

I can get


[tex]\Psi_{\beta} X_{\alpha} = \epsilon_{\alpha \beta} (\Psi X) + (\sigma^{\mu \nu} \epsilon^{T})_{\alpha \beta} (\Psi \sigma_{\mu \nu} X) [/tex]

so i don't know where the factors of a half come from and how to get the right index order
 
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Answers and Replies

  • #2
fzero
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I find

[tex]
(\sigma^{\mu \nu} \epsilon^{T})_{\alpha \beta} (\Psi \sigma_{\mu \nu} X) = \Psi_{\alpha} X_{\beta}+ \Psi_{\beta} X_{\alpha}.
[/tex]
 
  • #3
dextercioby
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The 1/2 must come from the symmetrization

[tex] \Psi_{\beta}X_{\alpha} = \frac{1}{2} \left(\Psi_{\beta} X_{\alpha}+\Psi_{\alpha} X_{\beta}\right) + \frac{1}{2} \left(\Psi_{\beta} X_{\alpha}-\Psi_{\alpha} X_{\beta}\right) [/tex]

The a-symmetric part must be proportional to the spinor metric, the symmetric one is what's left.
 
  • #4
213
8
I find

[tex]
(\sigma^{\mu \nu} \epsilon^{T})_{\alpha \beta} (\Psi \sigma_{\mu \nu} X) = \Psi_{\alpha} X_{\beta}+ \Psi_{\beta} X_{\alpha}.
[/tex]
how did you get that without a factor of two and if it is correct then symmetrising solves it
 
  • #5
fzero
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how did you get that without a factor of two and if it is correct then symmetrising solves it
That follows directly from the stated identity. Just put in the correct indices and keep track of the [tex]\epsilon[/tex] contractions.
 
  • #6
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sorry for being dense but if I would have known how to manipulate this spinor algebra i wouldn't be asking, so please be more explicit
 
  • #7
fzero
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sorry for being dense but if I would have known how to manipulate this spinor algebra i wouldn't be asking, so please be more explicit
You should really give it try first or at least point out exactly which term you don't understand.
 
  • #8
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8
I've been stuck on this for a day

what I don't understand is why it isn't

[tex] \bold 2 (\sigma^{\mu \nu} \epsilon^{T})_{\alpha \beta} (\Psi \sigma_{\mu \nu} X) = \Psi_{\alpha} X_{\beta}+ \Psi_{\beta} X_{\alpha}.

[/tex]

even looking at double epsilon identity in two dimensions you get

[tex]
(\sigma^{\mu \nu})^{\beta}_{\alpha} (\sigma_{\mu \nu})^{\delta}_{\gamma} = \delta^{\beta}_{\alpha} \delta^{\delta}_{\gamma}
[/tex]

which gives you a factor of two in that calculation you did

at least give me the starting point of that calculation
 
Last edited:
  • #9
fzero
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I've been stuck on this for a day

what I don't understand is why it isn't

[tex] \bold 2 (\sigma^{\mu \nu} \epsilon^{T})_{\alpha \beta} (\Psi \sigma_{\mu \nu} X) = \Psi_{\alpha} X_{\beta}+ \Psi_{\beta} X_{\alpha}.

[/tex]

even looking at double epsilon identity in two dimensions you get

[tex]
(\sigma^{\mu \nu})^{\beta}_{\alpha} (\sigma_{\mu \nu})^{\delta}_{\gamma} = \delta^{\beta}_{\alpha} \delta^{\delta}_{\gamma}
[/tex]
You're missing part of this expression.

which gives you a factor of two in that calculation you did

at least give me the starting point of that calculation
[tex] (\sigma^{\mu \nu} \epsilon^{T})_{\alpha \beta} (\Psi \sigma_{\mu \nu} X) = {(\sigma^{\mu \nu})_\alpha}^\gamma \epsilon_{\beta \gamma} \Psi^\delta {( \sigma_{\mu \nu} )_\delta}^\epsilon X_\epsilon .[/tex]

Use

[tex]
{(\sigma^{\mu \nu})_{\alpha}}^{\beta} {(\sigma_{\mu \nu})_{\gamma}}^{\delta} = \epsilon_{\alpha \gamma} \epsilon^{\beta \delta} + \delta^{\delta}_{\alpha} \delta^{\beta}_{\gamma}
[/tex]

and [tex]\Psi^\alpha = \epsilon^{\alpha\beta}\Psi_\beta[/tex].
 
  • #10
213
8
thank you it seems I made the mistake of having three repeated indicies

but this is correct

[tex]

(\sigma^{\mu \nu})^{\beta}_{\alpha} (\sigma_{\mu \nu})^{\delta}_{\gamma} = \delta^{\beta}_{\alpha} \delta^{\delta}_{\gamma}

[/tex]

otherwise you can't derive the rest
 
Last edited:

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