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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?

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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. $\endgroup$ – DanielSank Aug 7 '15 at 0:56
  • $\begingroup$ 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. $\endgroup$ – DanielSank Aug 7 '15 at 0:58
  • $\begingroup$ 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. $\endgroup$ – user81619 Aug 7 '15 at 1:23
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    $\begingroup$ Have you looked at this? en.wikipedia.org/wiki/Grassmann_number Grassmann numbers are more like matrices than actual numbers. $\endgroup$ – user73352 Aug 7 '15 at 1:54
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    $\begingroup$ 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? $\endgroup$ – ACuriousMind Aug 7 '15 at 13:06
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Comments to the question (v3):

  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.

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