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?

  • $\begingroup$ It’s not a real variable, where did you get that idea? $\endgroup$
    – knzhou
    Feb 28, 2020 at 5:06
  • $\begingroup$ @knzhou Sorry maybe I did not say it clearly. In the mean-field treatment of interacting fermi gas, the standard Hubbard-Stratonovich transformation seems to treat a "fermion density field" as a complex field, but strictly speaking it is a product of two Grassmann field, and that’s my question. $\endgroup$
    – Jay Ren
    Feb 28, 2020 at 5:55
  • $\begingroup$ Which textbook? Which page? $\endgroup$
    – Qmechanic
    Feb 28, 2020 at 6:37
  • $\begingroup$ @Qmechanic For example in the book Condensed Matter Field Theory by Alexander Altland, on page 244, where the author showed the case for interacting fermi gas as an example of H-S transformation. In his treatment, he shift the bosonic field $\phi$ by $\rho_q=\sum_q \bar{\psi}_p \psi_{p_q}$, which is legit if $\rho$ is a real/complex number. But in fact it is a product of two Grassmann number. $\endgroup$
    – Jay Ren
    Feb 28, 2020 at 7:17
  • $\begingroup$ "$\exp{\bar{\psi} \psi} = 1 + \bar{\psi} \psi$", not correct. For non-relativistic fermions in your case, one should have $\exp{\bar{\psi} \psi} = 1 + \bar{\psi} \psi + 1/2 (\bar{\psi} \psi )^2$ since there are spin up/down freedoms. And in relativistic setting, one should have $\exp{\bar{\psi} \psi} = 1 + \bar{\psi} \psi + 1/2 (\bar{\psi} \psi )^2 +1/3! (\bar{\psi} \psi )^3 + 1/4! (\bar{\psi} \psi )^4$ since a Dirac fermion column has 4 components. $\endgroup$
    – MadMax
    Feb 28, 2020 at 15:06

1 Answer 1


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.


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