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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 ?

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2 Answers 2

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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 $$(\eta\xi)^*=[(a+ib)(c+id)]^*=(ac-bd+ibc+iad)^*=ca-db-icb-ida=(c-id)(a-ib)=\xi^*\eta^*$$

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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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  • $\begingroup$ Please use TeX for typing your formulae. See the editing help on how to do this. $\endgroup$
    – Mårten W
    Commented Apr 18, 2013 at 15:04

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