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In many text books, in the mean-field treatment of weakly interacting Fermi gas, the standard operation is to first write down the interacting action in terms of fermion densities

$$ S_1 \propto \int D[\psi] \sum_q \rho_q \rho_{-q} $$

and then use Hubbard-Stratonovich transformation to decouple the density.
Here Hubbard-Stratonovich rely on Gaussian integral for complex variable, and most textbooks argued that

$$ \rho_q = \sum_p \bar{\psi}_{p+q} \psi_p $$

is a real variable because of the commutation relation. But strictly following the rule of Grassmann number, Grassmann number in the exponential is

$$ \exp{\bar{\psi} \psi} = 1 + \bar{\psi} \psi $$

How could we explain this discrepancy?

Jay Ren
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1 Answers1

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The main point seems to be that if a path integral involves both Grassmann-even and Grassmann-odd fields, then the Grassmann-even fields are supernumber-valued fields, i.e. they have both a body and a soul. However, the result of the path integration produce ordinary numbers without soul parts. See e.g. this related Phys.SE post.

Qmechanic
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