# Convert Grassmann numbers to real numbers [closed]

We know Grassmann numbers are complex numbers. Hence Grassmann integrals are also complex. How can we convert a Grassmann integral into real one, ie is there any transformation of converting complex Grassmann numbers to real grassmann numbers?

## closed as unclear what you're asking by ACuriousMind♦, Kyle Kanos, Martin, John Rennie, user10851 Aug 9 '15 at 9:46

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• Welcome to Physics Stack Exchange. Please note that clarity helps people understand your question. Part of good clarity is proper English punctuation, such as using a space between sentences. I edited this question to fix the punctuation. Please do pay attention to these important details in future posts. – DanielSank Aug 7 '15 at 0:56
• In what sense are Grassman numbers complex numbers? I'm pretty sure these are not the same thing. I think Grassman numbers are more like the Fermionic $a^\dagger$ operator. Perhaps you can convert path integrals involving Grassman numbers into expressions involving complex numbers. – DanielSank Aug 7 '15 at 0:58
• This is way above me, but on the off chance it helps, (and you probably are aware) Wikipedia calls them c-numbers, which was very confusing to me till I read that this was Dirac's notation, I would have automatically taken c-numbers to mean complex numbers until I read that Dirac meant classical numbers. Wikipedia is not well written in this section, imo. – user81619 Aug 7 '15 at 1:23
• Have you looked at this? en.wikipedia.org/wiki/Grassmann_number Grassmann numbers are more like matrices than actual numbers. – user73352 Aug 7 '15 at 1:54
• The question is unclear: Grassmann number aren't real number, Grassmann integrals aren't actually integrals (complex or real), so what are you actually trying to ask? – ACuriousMind Aug 7 '15 at 13:06

1. A Grassmann-odd number is not a complex number. It is a complex supernumber $z=x+iy$, which can be decomposed in real and imaginary supernumbers, cf. e.g. this and this Phys.SE posts.
2. The Berezin integral $\int\! d\theta~f(\theta)$ over supernumbers is an ordinary complex number $c=a+ib\in\mathbb{C}$, which can be decomposed in real and imaginary numbers.