Meaning of the fermion path integral?

Question

I'm trying to understand fermion fields with the Feynman integral. Is there an explicit matrix representation of the Grassmann numbers used in the field integral? Is there a Grassmann-valued measure that results in the Berezin integral? In my book, they just introduce the algebraic relations $ab = -ba$ with no construction of $a$ or $b$, and I'm having trouble visualizing it.

Answer 1

You are not alone in this. I tend to regard the Grassmann integral as tool for combinatorics, but if you want a deeper view and a discussion of analytic subtleties you might like to read Martin R. Zirnbauer, Riemannian symmetric superspaces and their origin in random-matrix theory, J. Math. Phys.\ 37 (1996) 4986; arXiv:math-ph/9808012.

Answer 2

Grassmann variables are just a convenient way of representing determinants in path integrals. Hence the anti-commutativity. There is some similarity with Slater determinants, which represent fermions in multi-particle quantum mechanics.

What is kind of strange, is that determinants also appear in gauge-fixed path integrals of Yang-Mills fields (i.e. bosons), where they are called Faddeev-Popov determinants and represent ghost fields.

If I remember correctly, the latter appear in the "numerator" of the path integrand, while the "regular" fermion determinants appear in the denominator. Hence they cannot be interpreted as an integral over a hypersurface of the configurations (unlike the gauge-fixed Yang-Mills fields, which are integrated over the hypersurface of the configurations that satisfy the chosen gauge condition).