SUSY variation Wess-Zumino

Question

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?

Answer 1

Starting with $\mathcal{L}_f = i \psi^{\dagger} \bar{\sigma}^{\mu} \partial_{\mu} \psi $ we can compute the variation of this kinetic term by using $\delta \psi = -i \sigma^{\nu} \epsilon^{\dagger} \partial_{\nu} \phi$ (which implies $\delta \psi^{\dagger} = i \epsilon \sigma^{\nu} \partial_{\nu} \phi^*$, since the $\sigma^{\mu}/ \bar \sigma^{\mu}$ are Hermitian)

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

The goal is to simply obtain the terms needed to cancel variations of the complex scalar kinetic energy term. In the first term, insert the identity $[\sigma^{\mu} \bar \sigma^{\nu} + \sigma^{\nu} \bar \sigma^{\mu}]^{\beta}_{\alpha} = 2 \eta^{\mu \nu} \delta^{\beta}_{\alpha} $ and notice that symmetry of partial derivatives in the second term allows us to write $(\bar \sigma^{\mu} \sigma^{\nu})_{ \dot \alpha}^{ \dot \beta} \partial_{\mu} \partial_{\nu} \phi = \frac{1}{2} ( \bar \sigma^{\mu} \sigma^{\nu} + \bar \sigma^{\nu} \sigma^{\mu} )_{ \dot \alpha}^{ \dot \beta} \partial_{\mu} \partial_{\nu} \phi = \delta^{\dot \beta}_{\dot \alpha} \partial^{\mu} \partial_{\mu} \phi $ so we obtain

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

Half of the first term will cancel one piece of the $\phi$ KE term. We can obtain the second term we need by inserting the chain rule $\partial_{\mu}( \psi^{\dagger}_{\dot \alpha} \partial^{\mu} \phi) = \partial_{\mu} \psi^{\dagger}_{\dot \alpha} \partial^{\mu} \phi + \psi^{\dagger}_{\dot \alpha} \partial^{\mu} \partial_{\mu} \phi $ into the last term to obtain

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

where we note that $\psi^{\dagger} \epsilon^{\dagger} = \epsilon^{\dagger} \psi^{\dagger}$. Now we simply need to write the two unnecessary terms (half of the first and the second) as a total derivative. Inserting the chain rule into half of the first term gives $- \partial_{\mu} ( \epsilon \psi \partial^{\mu} \phi^* ) + \epsilon \psi \partial_{\mu} \partial^{\mu} \phi^* $ and also doing the same in the second term yields $\partial_{\mu} ( \epsilon \sigma^{\mu} \bar \sigma^{\nu} \psi \partial_{\nu} \phi^*) - \epsilon \sigma^{\mu} \bar \sigma^{\nu} \psi \partial_{\mu} \partial_{\nu} \phi^* = \partial_{\mu} ( \epsilon \sigma^{\mu} \bar \sigma^{\nu} \psi \partial_{\nu} \phi^*) - \epsilon \psi \partial^{\mu} \partial_{\mu} \phi^*$ where we've again used symmetry of the partial derivatives. The result is then

\begin{align} \delta \mathcal{L}_f & = - \epsilon \partial^{\mu} \phi^* \partial_{\mu} \psi - \partial_{\mu} ( \epsilon \psi \partial^{\mu} \phi^* ) + \epsilon \psi \partial_{\mu} \partial^{\mu} \phi^* + \partial_{\mu} ( \epsilon \sigma^{\mu} \bar \sigma^{\nu} \psi \partial_{\nu} \phi^*) - \epsilon \psi \partial^{\mu} \partial_{\mu} \phi^* + \partial_{\mu} ( \psi^{\dagger} \epsilon^{\dagger} \partial^{\mu} \phi ) - \epsilon^{\dagger} \partial_{\mu} \psi^{\dagger} \partial^{\mu} \phi \nonumber \\ & = - \epsilon \partial^{\mu} \phi^* \partial_{\mu} \psi - \epsilon^{\dagger} \partial_{\mu} \psi^{\dagger} \partial^{\mu} \phi + \partial_{\mu} ( \epsilon \sigma^{\mu} \bar \sigma^{\nu} \psi \partial_{\nu} \phi^* - \epsilon \psi \partial^{\mu} \phi^* + \psi^{\dagger} \epsilon^{\dagger} \partial^{\mu} \phi ) .\nonumber \end{align}