Let $\theta$ and $\eta$ denote odd complex supernumbers (also known as Grassmann numbers), $a$ and $b$ arbitrary complex supernumbers. Say that $a$ is $\circ$-real (resp. $\circ$-imaginary) if $a^\circ = a$ (resp. $a^\circ = -a$).

There are two candidates for the role of $(-)^\circ$, i.e. two ways to define conjugation of supernumbers: with $(\theta\eta)^* = \theta^*\eta^*$ and with $(\theta\eta)^\dagger = \eta^\dagger\theta^\dagger$. (Notation from Cartier and DeWitt-Morette, who define both, see Eq. (9.13).) Note that:

  • the product $\theta\eta$ is $*$-real when $\theta$ and $\eta$ are $*$-real, while $\theta^*\theta$ is $*$-imaginary;
  • the product $\theta\eta$ is $\dagger$-imaginary when $\theta$ and $\eta$ are $\dagger$-real, while $\theta^\dagger\theta$ is $\dagger$-real.

Physicists usually use $(-)^\dagger$, motivating this by the correspondence with Hermitian conjugation of operators through quantization. Two places that use $(-)^*$ are the draft chapter 9PDF of Cartier and DeWitt-Morette (again) and Quantum fields and strings by Deligne et al., see their “Sign manifestoPDF. The latter say about their choice (notation adjusted to match):

This is a consequence of the sign rule if we assume that $\theta\mapsto\theta^*$ is a $*$-operation and $\theta$ (super)commutes with $\eta$. (A $*$-operation satisfies $(ab)^* = (-1)^{\lvert a\rvert\lvert b\rvert} b^*a^*$.) Notice that the classical statement (34) is consistent with the quantum statement (19), since the adjoint operation on linear operators is also a $*$-operation.

Eq. (34) is the definition of $(-)^*$, and (19) is the statement that quantizing complex conjugates yields Hermitian conjugates.

That is, both camps say that their choice is the one that behaves properly under quantization! What am I missing?

Edit: To add insult to injury, Rogers section 3.3 defines a conjugation similar to $(-)^\dagger$, but with an $i$ before odd terms in the reality condition; see discussion in Cook section 2.10. She also references Kleppe and Wainwright, who define what they call a “pseudo-conjugation” alongside $(-)^\dagger$ and get some interesting group theory from there—it is not involutive, though. Pellegrini, also doing group theory, compares the same “pseudo-conjugation” with $(-)^*$.

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    $\begingroup$ Related questions: this (reversed order for $(-)^\dagger$) and this (left derivatives become right under $(-)^\dagger$). $\endgroup$ Mar 2, 2017 at 2:42

1 Answer 1


The answer lies in how the Hermitian product and conjugation are defined in the quantized theory. The usual physicist’s definitions do not use the super structure on the Hilbert space and imply $$ \langle u, v\rangle = \overline{\langle v, u\rangle},\quad \langle u, u\rangle\geq 0,\quad \langle u, Tv\rangle =\langle T^\dagger u, v\rangle. $$ The definitions used by Deligne et al., on the other hand, involve the parity $\lvert-\rvert$ explicitly: $$ (u, v) = (-1)^{\lvert u\rvert\lvert v\rvert}\overline{(v, u)},\quad (-i)^{\lvert u\rvert}(u, u)\geq 0,\quad (u, Tv) = (-1)^{\lvert u\rvert\lvert T\rvert}(T^*u, v). $$

Now let $e$ and $f$ be even vectors (bosonic states), $\omicron$ and $\pi$ odd vectors (fermionic states). Then the correspondence between the products is $$ (e,f) = \langle e,f\rangle,\quad (\omicron, \pi) = i\langle\omicron,\pi\rangle. $$ As for the conjugation, let $P$ be an even operator and $Q$ an odd operator. Then \begin{align} \langle P^\dagger e, f\rangle =\langle e,Pf\rangle = (e, Pf) &= +(P^*e, f) = +\langle P^* e, f\rangle,\\ i\langle P^\dagger\omicron,\pi\rangle = i\langle\omicron, P\pi\rangle = (\omicron, P\pi) &= +(P^*\omicron,\pi) = +i\langle P^*\omicron,\pi\rangle;\\ \langle Q^\dagger e,\omicron\rangle =\langle e,Q\omicron\rangle = (e,Q\omicron) &= +(Q^*e,\omicron) = +i\langle Q^*e,\omicron\rangle,\\ i\langle Q^\dagger\omicron, e\rangle = i\langle\omicron, Qe\rangle = (\omicron, Qe) &= -(Q^*\omicron,e) = -\langle Q^*\omicron,e\rangle; \end{align} whence $$ P^\dagger = P^*,\quad Q^\dagger = iQ^*, $$ and therefore $$ (ST)^* = (1/i)^{\lvert ST\rvert}(ST)^\dagger = (1/i)^{\lvert ST\rvert}T^\dagger S^\dagger = (1/i)^{\lvert ST\rvert}i^{\lvert S\rvert}i^{\lvert T\rvert}T^*S^* = (-1)^{\lvert S\rvert +\lvert T\rvert}T^*S^*. $$

This indeed agrees with the proper definition (Leites et al., 2011Russian PDF Sect. 1.11; Berezin and Shubin, 1999English DjVu Supplement 3 [MR]) of a real structure in a complex superalgebra, $$ (ab)^* = (-1)^{\lvert a\rvert +\lvert b\rvert}b^* a^*, $$ which coincides with $(ab)^* = a^*b^*$ in the supercommutative (classical) case.

The peculiar definitions of Rogers are due to the fact that she uses $(-)^\dagger$ as the conjugation but her real elements are the $*$-real ones (using the correspondence $a^\dagger = i^{\lvert a\rvert}a^*$ established for operators above). The “pseudo-conjugation” of Kleppe and Wainwright and of Pellegrini is what Leites et al. call a quaternionic structure; they also establish that these are the only interesting semilinear (super)involutions of this type up to a real superalgebra automorphism.

Finally, note that while $\zeta^*\zeta$ is not real for odd complex $\zeta$, $\{\zeta^*,\zeta\}/2\equiv\bigl(\zeta^*\zeta + (-1)^{\lvert\zeta^*\rvert\lvert\zeta\rvert}\zeta\zeta^*\bigr)/2$ is.


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