- #1
Adgorn
- 133
- 19
- 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?
$$-i \partial^\mu \phi \epsilon^\dagger \sigma^2 \partial_\mu \chi^* =-i \partial^\mu \phi \partial_\mu \chi^\dagger \sigma^2 \epsilon^*.$$
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?