While dealing with Grassmann Variables, the complex conjugate is defined as $ (\phi \psi)^{\dagger} = \psi^{\dagger} \phi^\dagger $ and why not $ \phi^{\dagger} \psi^\dagger $. I want to know the motivation behind this. In the case of multidimensional vectors, it makes sense, but why in the case of a Grassmann Variable ?
Grassmann Variables and Complex Conjugate
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0Maybe this is relevant:https://en.wikipedia.org/wiki/Grassmann_number#Matrix_representations – 2012-09-29
2 Answers
If $\eta$ is a complex Grassman variable then we require $\eta^*\eta=x$ to be a real (non-Grassmanian) variable. In particular it means that $(\eta^*\eta)^*=x^*\overset{!}{=}x=\eta^*\eta$ Write $\eta$ in terms of two real Grassman variables $\eta=a+ib$, then $\eta^*\eta=(a-ib)(a+ib)=iab-iba$ and $(\eta^*\eta)^*=(iab-iba)^*=-i(ab)^*+i(ba)^*\overset{!}{=}iab-iba$ It follows immedeately that for two real Grassman variables $a$ and $b$ we must have $(ab)^* = ba$ It also follows that for two complex Grassman variables $\eta=a+ib$ and $\xi=c+id$ we must have $(\eta\xi)^*=[(a+ib)(c+id)]^*=(ac-bd+ibc+iad)^*=ca-db-icb-ida=(c-id)(a-ib)=\xi^*\eta^*$
Lets use * for complex conjugation (for numbers dagger = *). you have a*a=|a|^2 if a=bc then |a|^2=(bc)*(bc)=c*b*bc=c*cb*b=|c|^2|b|^2. if the order was not reversed |a| = b*c*bc=-|b|^2|c|^2 which is inconvenient when you deal with moduli I suppose.
So the swap guarantees |ab|=|a||b| as opposed to -|a||b|.
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0Please use TeX for typing your formulae. See the [editing help](http://math.stackexchange.com/editing-help) on how to do this. – 2013-04-18