Peskin and Schroeder problem 3.5(a): Figuring out the cancellation

[Adgorn]

Homework Statement

Given the Lagrangian:
$L=\partial_\mu \phi \partial \phi^* +\chi^\dagger i \bar{\sigma}^\mu \partial_\mu \chi +F^*F$

Where $\phi$ is a complex vector field, $\chi$ is a free Weyl Fermion and $F$ is an auxiliary complex scalar field, show that the Lagrangian is symmetric under the transformation:

$\delta \phi = -i \epsilon^T \sigma^2 \chi$
$\delta \chi = \epsilon F +\sigma^\mu \partial_\mu \phi \sigma^2 \epsilon^*$
$\delta F = -i \epsilon^\dagger \bar{\sigma}^\mu \partial_\mu \chi$

where $\epsilon$ is a two component spinor of Grassmann numbers.

Relevant Equations

$\chi^T \sigma^2 \epsilon = \epsilon^T \sigma^2 \chi$
$\sigma^2 \sigma^\mu \sigma^2 = (\bar{\sigma}^\mu)^T$
$(\sigma^\mu \cdot \partial) (\bar{\sigma}^\mu \cdot \partial) = \partial^\mu \partial_\mu$

I've managed to account for all the terms except for two, which seem to have a minus sign I cannot get rid of. When expanding the variation, one of them comes from the $\phi$ variation:

$$-i \partial^\mu \phi \epsilon^\dagger \sigma^2 \partial_\mu \chi^* =-i \partial^\mu \phi \partial_\mu \chi^\dagger \sigma^2 \epsilon^*.$$

The other comes from the $\chi$ term and is:

$$i {\sigma}^\mu \bar{\sigma}^\nu \partial_\mu \partial_\nu\phi \chi^\dagger \sigma^2 \epsilon^*=i \partial_\mu \partial^\mu \phi \chi^\dagger \sigma^2 \epsilon^*.$$

If all is right in the world, these two terms should combine to make a total derivative, but they differ by a sign. The online solution manual writes the first term without the minus sign, but I don't see how that can be the case. Is there some rule with Grassmann numbers that I'm not aware of here?

 

[Adgorn]

Never mind, I immediately realized upon hitting "post" that complex conjugation switches the order of Grassmann numbers. Better late then never.

 

[Orodruin]

Hitting the post button is one of the absolutely best ways of realizing things …