- #1
RedX
- 970
- 3
Is the following expression Hermitian:
[tex]\theta \xi(x)+\theta^*\xi^{\dagger}(x)+\theta\sigma^{\mu}\theta^*\nu_{\mu}(x)[/tex]
[tex] \theta [/tex] is a left-handed spinor coordinate [(1/2, 0) representation of SO(4)], [tex] \xi [/tex] is a left-handed spinor field, and [tex] \nu_\mu[/tex] is a real vector field.
Normally:
[tex] (\theta \xi)^{\dagger}=\xi^{\dagger} \theta^{\dagger}[/tex]
However, since theta is a coordinate and not a field, it just gets complex conjugated instead of daggered and there is also no order change, so that the first two terms added together in the very top expression are Hermitian and transform into each other under the adjoint operation.
However, if this is the case, then the third term is not Hermitian but anti-Hermitian. The third term is only Hermitian is if you treat the anticommutating variable [tex] \theta [/tex] as a field (and hence changes order).
[tex]\theta \xi(x)+\theta^*\xi^{\dagger}(x)+\theta\sigma^{\mu}\theta^*\nu_{\mu}(x)[/tex]
[tex] \theta [/tex] is a left-handed spinor coordinate [(1/2, 0) representation of SO(4)], [tex] \xi [/tex] is a left-handed spinor field, and [tex] \nu_\mu[/tex] is a real vector field.
Normally:
[tex] (\theta \xi)^{\dagger}=\xi^{\dagger} \theta^{\dagger}[/tex]
However, since theta is a coordinate and not a field, it just gets complex conjugated instead of daggered and there is also no order change, so that the first two terms added together in the very top expression are Hermitian and transform into each other under the adjoint operation.
However, if this is the case, then the third term is not Hermitian but anti-Hermitian. The third term is only Hermitian is if you treat the anticommutating variable [tex] \theta [/tex] as a field (and hence changes order).