2
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ 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

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.