Calculation trafo chiral multiplet

[hendriko373]

Hello,

I'm trying to deduce the suspersymmetry transformation of the chiral multiplet out of the superfield formalism. In doing this I got stuck with this:

$$\theta^{\alpha}\sigma^{\mu}_{\alpha\dot{\alpha}}\overline{\sigma}^{\dot{\alpha}\beta}_{\nu}\theta_\beta(\partial_\mu\psi_\gamma\sigma^{\nu}_{\gamma\dot{\gamma}}\overline{\xi}^{\dot{\gamma}})$$

Here I wrote all the indices explicitly, with the undotted and dotted respectively left and right handed 2D Weyl indices (theta's anti commuting).

What bothers me is the part in front of the paragraphs, I would like to get something like this:

$$\theta^{\alpha}\epsilon_{\alpha\beta}\theta^\beta(\partial_\mu\psi_\gamma\sigma^{\mu}_{\gamma\dot{\gamma}}\overline{\xi}^{\dot{\gamma}})$$

Notice that there is the contraction now in the paragraph on space time indices. I think it's not that difficult, one has to use the clifford algebra commutation relation I guess, but I don't see it coming out. Maybe tracing over it would also work. Any help would be appreciated much.

greetz

hendrik

 

[hendriko373]

Ok I solved it. For those interested, use the defining equation for the clifford algebra and the fact that the antisymmetric part in spacetime indices is symmetric in its weyl indices, such that this term is zero when combined with the antisymmetry of the theta's. This gives a kronecker delta for both weyl and space time indices, giving the desired result.

cheers

 

[sir-pinski]

Hi Hendrik,

Thank you for sharing your question about the calculation of the supersymmetry transformation of the chiral multiplet. It seems like you are on the right track and just need a little nudge in the right direction. Here are some steps that may help you reach your desired result:

1. Use the Clifford algebra relation to rewrite the part in front of the parentheses as \theta^{\alpha}\sigma_{\alpha\dot{\alpha}}^{\mu}\overline{\sigma}^{\dot{\alpha}\beta}_{\nu}\theta_{\beta} = \theta^{\alpha}\epsilon_{\alpha\beta}\theta^{\beta}\sigma^{\mu}_{\nu}. This follows from the identity \sigma^{\mu\nu}_{\alpha\beta} = \epsilon_{\alpha\beta}\sigma^{\mu\nu} + \epsilon_{\alpha\beta}\epsilon^{\mu\nu}.

2. Substitute this into your original expression to get \theta^{\alpha}\epsilon_{\alpha\beta}\theta^{\beta}(\partial_{\mu}\psi_{\gamma}\sigma^{\mu}_{\gamma\dot{\gamma}}\overline{\xi}^{\dot{\gamma}}).

3. Use the fact that \epsilon_{\alpha\beta}\theta^{\beta} = \theta_{\alpha} to simplify the expression further.

4. Finally, use the property \theta^{\alpha}\theta_{\alpha} = 0 (since theta's are anti-commuting) to get rid of the remaining \theta's and arrive at your desired result.

I hope this helps and good luck with your calculations!