3
$\begingroup$

I'm following John Terning book on Supersymmetry and in particular I'm trying to check the susy variations of the Wess-Zumino model given by

$\mathcal{L}_s = \partial^\mu \phi^* \partial_\mu \psi \,, \quad \mathcal{L}_f=i \psi^\dagger \overline{\sigma}^\mu \partial_\mu \psi$

In order to compute the variation of the fermionic part it's necessary to consider:

$[\sigma^\mu \overline{\sigma}^\nu + \sigma^\nu \overline{\sigma}^\mu]_\alpha^\beta = 2\eta^{\mu \nu} \delta_\alpha^\beta \,, \quad [\overline{\sigma}^\mu {\sigma}^\nu + \overline{\sigma}^\nu {\sigma}^\mu]_{\dot{\alpha}}^{\dot{\beta}} = 2\eta^{\mu \nu} \delta_{\dot{\alpha}}^{\dot{\beta}} \, , \quad \delta \psi = -i(\sigma^\nu \epsilon^\dagger)_\alpha \partial_\nu \phi \,$

Following the book one only needs to use the mentioned properties of the $\sigma$'s, by doing so

$\delta \mathcal{L}_f = -\epsilon \sigma^\nu \partial_\nu \phi^* \overline{\sigma}^\mu \partial_\mu \psi + \psi^\dagger \overline{\sigma}^\mu \sigma^\nu \epsilon^\dagger \partial_\mu \partial_\nu \phi \\ \delta \mathcal{L}_f= -2\epsilon \partial^\mu \phi^* \partial_\mu \psi + 2 \psi^\dagger \epsilon^\dagger \partial^\mu \partial_\mu \phi + \epsilon \sigma^\mu \overline{\sigma}^\nu \partial_\nu \phi^* \partial_\mu \psi-\psi^\dagger \overline{\sigma}^\nu \sigma^\mu \epsilon^\dagger \partial_\mu \partial_\nu \phi$

While the right answer is given by

$\delta \mathcal{L}_f = - \epsilon \partial^\mu \psi \partial_\mu \phi^* - \epsilon^\dagger \partial^\mu \psi^\dagger \partial_\mu \phi + \partial_\mu (\epsilon \sigma^\mu \overline{\sigma}^\nu \psi \partial_\nu \phi^* - \epsilon \psi \partial^\mu \phi^* + \epsilon^\dagger \psi^\dagger \partial^\mu \phi)$

that exactly cancels the bosonic part of the action. Performing some integration by parts in my computation gives

$\delta \mathcal{L}_f = -2\epsilon \partial^\mu \phi^* \partial_\mu \psi - 2 \partial^\mu \psi^\dagger \epsilon^\dagger \partial_\mu \phi + \partial^\mu (2\psi^\dagger \epsilon^\dagger \partial_\mu \phi) + \epsilon \sigma^\mu \overline{\sigma}^\nu \partial_\nu \phi^* \partial_\mu \psi-\psi^\dagger \overline{\sigma}^\nu \sigma^\mu \epsilon^\dagger \partial_\mu \partial_\nu \phi $

While the first 3 terms resemble the right answer there are problems with the factor of 2 and if I integrate by parts the remaining 2 terms I get 2 total derivatives plus 2 terms that don't seem to cancel between each other. What's the thing that I'm missing in order to obtain the right result?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.