Why does fermion have the expansion with Grassmann-numbers?

Question

I learn the chiral anomaly by Fujikawa method. The text book "Path Integrals and Quantum Anomalies, Kazuo Fujikawa", in the page 151, says that

…one can define a complete orthonormal set $\{\phi_n\}$and the expansion of fermionic variables $$D\phi_n(x)=\lambda_n\phi_n(x)$$ $$\int d^2x \phi_n^\dagger(x)\phi_m(x)=\delta_{nm}$$ $$\phi(x)=\sum_na_n\phi_n(x)$$ where $a_n$ is the Grassmann numbers.

Here $\phi$ is any fermionic matter field, and the $D$ is the Dirac operator twisted by gauge field. Since it is self-dual, the eigenvalues $\lambda_n$ are real numbers.

My question is, what is the reason why the expansion $\phi(x)=\sum_na_n\phi_n(x)$ with Grassmann coefficients holds? In mathematics, fermion (interacted by gauge field) $\phi$ is interpreted by a section of the Dirac bundle $S\otimes E$ that is the spinor bundle $S$ twisted by the vector bundle $E$. Then the orthonormal basis $\phi_n$ is complete in the $L^2$-space $L^2(S\otimes E)$. So I understand the expansion $\phi(x)=\sum_nc_n\phi_n(x)$ with complex coefficient $c_n$, in the sense of $L^2$-norm. But why Grassmann numbers? I know that the bundle $S\otimes E$ has a Grassmann algebra module structure, canonically. Possibly, do the $\phi_n$ form a basis with respect to the Grassmann module? Or, are there operators $a_n$ such that $a_n\phi_n=c_n\phi_n$ and that $\{a_n,a_m\}=0$?

Answer 1

You need Grassmann's so that the classical Fermi field anticommutes with itself. Once you insert the Grassmann field expansion into the action you can perform integral over space time and then the Berezin "Gaussian" integral to get the Matthews-Salam determinant. If you used ordinary complex numbers in field expansion, you would get the inverse of the Matthews-Salam determinant, or, equivalently, you would miss the $-1$ factor for each closed fermion loop.