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Take a real Grassmann variable, by which I mean $\theta=\theta^*$. We have

$$\int d\theta~ \theta =1,\qquad \frac{\partial}{\partial\theta}\theta=1$$

If I define the conjugation of Grassmann variables to invert their order, $$(\eta\theta)^*= \theta^*\eta^*,$$ should I then have $$(d\theta\theta)^*=\theta^*d\theta^*~?$$ But this means

$$(\int d\theta~\theta)^*=1 \quad \Rightarrow \quad\int \theta^* d\theta^*=-\int d\theta^* ~\theta^*=1,$$

so if $\theta=\theta^*$, I should have $$d\theta^*=-d\theta.$$ The same can be found for the derivative of $\theta$.

Is this the usual convention to take, or should I instead choose $$(d\theta \theta)^*=d\theta^* \theta^*~?$$

Qmechanic
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user121664
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1 Answers1

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Yes. OP is right. There is a minus. Since by convention the complex conjugation obeys

$$ (z w)^{\ast} ~=~ w^{\ast}z^{\ast}~=~(-1)^{|z|~|w|} z^{\ast}w^{\ast} \tag{1}$$

for any two supernumbers $z$, $w$ (of definite Grassmann parities $|z|$,$|w|$), we should also have

$$ (A f)^{\ast} ~=~(-1)^{|A| ~|f|} A^{\ast}f^{\ast} \tag{2}$$

for complex conjugation of an operator $A$ and a function $f$, cf. e.g. Refs. 1 & 2. Eq. (2) reduces to eq. (1) if $A$ is a left multiplication operator. It is easy to check that eq. (2) implies that$^{1}$

$$ \left(\frac{\partial_L}{\partial z}\right)^{\ast}~\stackrel{(2)}{=}~ (-1)^{|z|} \frac{\partial_L}{\partial (z^{\ast})}.\tag{3}$$

Since Berezin integration is the same as left differentiation

$$ \int \!d\theta ~=~\frac{\partial_L}{\partial \theta}, \qquad \int \!d\theta^{\ast} ~=~\frac{\partial_L}{\partial (\theta^{\ast})},\tag{4} $$

we derive that complex conjugation of Grassmann-odd differentiation produces a minus

$$ \left( \int \!d\theta \right)^{\ast} ~\stackrel{(4)}{=}~\left(\frac{\partial_L}{\partial \theta}\right)^{\ast} ~\stackrel{(3)}{=}~-\frac{\partial_L}{\partial (\theta^{\ast})} ~\stackrel{(4)}{=}- \int \!d\theta^{\ast}.\tag{5} $$

References:

  1. B. DeWitt, Supermanifolds, Cambridge Univ. Press, 1992; eq. (2.2.19).

  2. S.J. Gates, M.T. Grisaru, M. Rocek & W. Siegel, Superspace, or One thousand and one lessons in supersymmetry, arXiv:hep-th/0108200; eq. (3.1.9).

--

$^{1}$ The subscript $L$ ($R$) denotes left (right) differentiation, i.e. acting from left (right), respectively. For completeness, let us mention that left and right differentiation are connected via the formula

$$ \frac{\partial_L f}{\partial z}~=~(-1)^{(|f|+1)|z|}\frac{\partial_R f}{\partial z},\tag{6} $$

so that complex conjugation satisfies

$$\left(\frac{\partial_L}{\partial z}\right)^{\ast}f~\stackrel{(3)+(6)}{=}~ (-1)^{|z||f|} \frac{\partial_R f}{\partial (z^{\ast})}, \qquad \left(\frac{\partial_R}{\partial z}\right)^{\ast}f~\stackrel{(3)+(6)}{=}~ (-1)^{|z||f|} \frac{\partial_L f}{\partial (z^{\ast})}.\tag{7} $$

Qmechanic
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  • Notes for later: $\langle v, w \rangle~:=~ v^{\dagger}w~=~\langle w, v \rangle^{\ast}$; $\langle v, Aw \rangle~=~\langle A^{\dagger}v, w \rangle$; $(AB)^{\dagger}~=~B^{\dagger}A^{\dagger}$; $A^{\dagger}~=~(A^T)^{\ast}~=~(A^{\ast})^T$; $(AB)^T~=~(-1)^{|A||B|}B^TA^T$; – Qmechanic Nov 18 '17 at 14:29