I have found in the literature at least two different definitions of $\bf{super}$ Hilbert spaces:

  1. Definition 1: A super Hilbert space is a complex super-vector space $\mathcal{H}=\mathcal{H}_0\oplus \mathcal{H}_1$ with a inner product such that $$\overline{\langle v,v'\rangle}=(-1)^{|v||v'|}\langle v',v\rangle.$$ Here $v$ and $v'$ are homogeneous elements. It implies that $\mathcal{H}_0$ and $\mathcal{H}_1$ are perpendicular and that the norm square of an even vector is real and for an odd one is pure imaginary.
  2. Definition 2: A super Hilbert space is a complex super-vector space $\mathcal{H}=\mathcal{H}_0\oplus \mathcal{H}_1$ with an inner product such that $\mathcal{H}$ is a usual Hilbert space and $\mathcal{H}_0$ is perpendicular to $\mathcal{H}_1$.

It is clear that there is a bijective correspondence between this two kind of super Hilbert spaces. If $\mathcal{H}$ satisfies definition 1, then only changing the inner product in the odd elements as $\langle v',v''\rangle' = \overline{\langle v',v''\rangle}$ you get a super Hilbert space of definition 2.

My question is: Which is the most used definition in physics and why?

  • 1
    $\begingroup$ If the definitions are equivalent, then how could you say that we use one, but not the other? $\endgroup$
    – ACuriousMind
    Jun 14, 2015 at 16:47
  • $\begingroup$ It sounds like def 2 is more natural, at least I'm more used to norm square of any states being positive definite. $\endgroup$
    – Meng Cheng
    Jun 14, 2015 at 17:51
  • $\begingroup$ Why should $H_0$ and $H_1$ be perpendicular w.r.t. the inner product? $\endgroup$
    – Phoenix87
    Jun 14, 2015 at 18:02
  • $\begingroup$ Which literature? $\endgroup$
    – Qmechanic
    Jun 17, 2015 at 16:20
  • $\begingroup$ @ACuriousMind Worth talking about nevertheless, I think. Confused not only me, but people who write books on the stuff as well (compare with first link). $\endgroup$ Mar 5, 2018 at 15:25


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