Decomposition of SL(2,C) Weyl Spinors

[sgd37]

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)

Homework Equations

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

 

[fzero]

I find

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

 

[dextercioby]

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.

 

[sgd37]

I find

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

how did you get that without a factor of two and if it is correct then symmetrising solves it

 

[fzero]

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.

 

[sgd37]

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]

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.

 

[sgd37]

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}.<br /> <br />

even looking at double epsilon identity in two dimensions you get

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

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

at least give me the starting point of that calculation

 

[fzero]

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}.<br /> <br />

even looking at double epsilon identity in two dimensions you get

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

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

<br /> {(\sigma^{\mu \nu})_{\alpha}}^{\beta} {(\sigma_{\mu \nu})_{\gamma}}^{\delta} = \epsilon_{\alpha \gamma} \epsilon^{\beta \delta} + \delta^{\delta}_{\alpha} \delta^{\beta}_{\gamma}<br />

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

 

[sgd37]

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

but this is correct

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

otherwise you can't derive the rest