# Decomposition of SL(2,C) Weyl Spinors

## Homework Statement

Using

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

show that

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

## The Attempt at a Solution

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

I can get

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

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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fzero
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Gold Member
I find

$$(\sigma^{\mu \nu} \epsilon^{T})_{\alpha \beta} (\Psi \sigma_{\mu \nu} X) = \Psi_{\alpha} X_{\beta}+ \Psi_{\beta} X_{\alpha}.$$

dextercioby
Homework Helper
The 1/2 must come from the symmetrization

$$\Psi_{\beta}X_{\alpha} = \frac{1}{2} \left(\Psi_{\beta} X_{\alpha}+\Psi_{\alpha} X_{\beta}\right) + \frac{1}{2} \left(\Psi_{\beta} X_{\alpha}-\Psi_{\alpha} X_{\beta}\right)$$

The a-symmetric part must be proportional to the spinor metric, the symmetric one is what's left.

I find

$$(\sigma^{\mu \nu} \epsilon^{T})_{\alpha \beta} (\Psi \sigma_{\mu \nu} X) = \Psi_{\alpha} X_{\beta}+ \Psi_{\beta} X_{\alpha}.$$
how did you get that without a factor of two and if it is correct then symmetrising solves it

fzero
Homework Helper
Gold Member
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 $$\epsilon$$ contractions.

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

fzero
Homework Helper
Gold Member
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.

I've been stuck on this for a day

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

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

even looking at double epsilon identity in two dimensions you get

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

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

at least give me the starting point of that calculation

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fzero
Homework Helper
Gold Member
I've been stuck on this for a day

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

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

even looking at double epsilon identity in two dimensions you get

$$(\sigma^{\mu \nu})^{\beta}_{\alpha} (\sigma_{\mu \nu})^{\delta}_{\gamma} = \delta^{\beta}_{\alpha} \delta^{\delta}_{\gamma}$$
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
$$(\sigma^{\mu \nu} \epsilon^{T})_{\alpha \beta} (\Psi \sigma_{\mu \nu} X) = {(\sigma^{\mu \nu})_\alpha}^\gamma \epsilon_{\beta \gamma} \Psi^\delta {( \sigma_{\mu \nu} )_\delta}^\epsilon X_\epsilon .$$

Use

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

and $$\Psi^\alpha = \epsilon^{\alpha\beta}\Psi_\beta$$.

thank you it seems I made the mistake of having three repeated indicies

but this is correct

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

otherwise you can't derive the rest

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