Grassmann-even action

Question

I am currently studying supersymmetric quantum mechanics with the help of the book Mirror Symmetry by Kentaro Hori (and others). On page 155 where they introduce Grassmann variables they say that the action is Grassmann-even without an explanation. But i do not quite understand why this is the case and a Grassmann-odd action is not allowed?

Answer 1

The Grassmann-odd path integrals are special cases of Grassmann-even path integrals. This follows since

$$ \int [d\psi]e^{-S}=-\int [d\psi]S=-\int[d\psi]d\chi (\chi S) = \int [d\psi]d\chi e^{-\chi S} $$

for $S$ being Grassmann-odd and $\chi$ a fermion.

Let us work with an instructive example. If you have a Grassmann-odd action of the type:

$$ S=A^{ijk}\psi_i\psi_j\psi_k $$

The number of fermionic integrations mus be equal or lower than $3$ such that the path integral does not vanishes. Note that

$$ \exp\left(A^{ijk}\psi_i\psi_j\psi_k\right)=1+A^{ijk}\psi_i\psi_j\psi_k $$

since

$$ (A_{ijk})(A_{lmn})=+(A_{lmn})(A_{ijk}), \qquad (\psi_i\psi_j\psi_k)(\psi_l\psi_m\psi_n) = -(\psi_l\psi_m\psi_n)(\psi_i\psi_j\psi_k) $$

There will be no products of $A_{ijk}$ like in the general Grassmann-even case since the expansion stops here. This will always happen for the Grassman-odd case. The result of the path integral will be proportional to $A_{ijk}$ with some indices contracted. In particular, no Pfaffians or determinants.

This type of action will not survive in the continuum limit $N\rightarrow\infty$ since the number of fermionic integrals increase in with $N$.

This Grassmann-odd path integral can be easily promoted to be a Grassmann-even by introducing an extra variable $\chi$:

$$ S=\chi A^{ijk}\psi_i\psi_k\psi_k $$

and integrate over this variable.

Answer 2

1. At the classical level, the action $S$ can have any Grassmann-parity.

2. At the quantum mechanical level in a path integral/partition function, it would be rather strange/exotic$^1$ to consider a Boltzmann factor $$\exp(\frac{i}{\hbar}S)~=~1 +\frac{i}{\hbar}S$$ for a Grassmann-odd action $S$. In particular, the Boltzmann factor would no longer have definite Grassmann parity.

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$^1$ For an example of a Grassmann-odd action in the Literature, see e.g. Ref. 1.

References:

1. H. Hata & B. Zwiebach, arXiv:hep-th/9301097; Section 3.2.