# Determinant as a fermionic path integral

I understand that the determinant of a matrix can be written in terms of a fermionic path integral. The expression is:

$$Z = \int D\bar{\psi}D\psi e^{-\iint d^4x' d^4x \bar{\psi}(x')B(x',x)\psi(x)}\tag{1}$$

The proof proceeds by rewriting the complex Grassmann fields in terms of basis functions:

$$\psi(x) = \sum_n c_n\chi_n(x)\qquad \text{and}\qquad \bar{\psi}(x) = \sum_n \bar{c_n}\chi_n^\dagger(x)\tag{2}$$

This induces a change in the measure in the form: $$D\bar{\psi}D\psi = \prod_nd\bar{c_n}dc_n.\tag{3}$$

I am failing to prove that the measure indeed changes in the above manner under the stated transformation. An explicit calculation would be much appreciated.

• Is this from a book? The rhs. of the last formula (3) is likely just a definition of how to normalize the lhs. path integral measure in the first place. Nov 8 '16 at 11:49
• Yes, but how to show it starting from the basis expansion? Nov 8 '16 at 13:39
• It's not clear to me where exactly the problem here lies. Can you prove the statement for a finite-dimensional matrix, i.e. ordinary Graßmann variables rather than fields? Nov 8 '16 at 14:30
• The problem here is in the change of basis, to me. I can't show that the said COV changes the measure in that fashion. Nov 13 '16 at 4:30

there is a quick and dirty way of proving $$\mathcal{D}\psi\mathcal{D}\bar{\psi} = \prod_n dc_nd\bar{c}{}_n,$$ although i admit that the method used is still subject of research. First, we have to see how the integral $\mathcal{D}\psi$ is defined. You can find yourself that, $$\mathcal{D}\psi = \prod_x d\psi(x)$$ so, consider the change of variables $\psi(x) = \sum_n c_n \chi_n(x)$, then $$\frac{\partial \psi(x)}{\partial c_n} = \chi_n(x)$$ so, using, $$d\psi(x) = \frac{\partial \psi(x)}{\partial c_n} dc_n$$ and the fact that volume measures change under a change of variables with the determinant of the Jacobian, we can write the identity, $$\mathcal{D}\psi\mathcal{D}\bar{\psi} = \det\left (\frac{\partial \psi(x)}{\partial c_n}\right )\det\left (\frac{\partial \bar{\psi}(x)}{\partial \bar{c}_n}\right )\prod_n dc_nd\bar{c}{}_n=\det\left (\chi_m(x)\right )\det\left (\bar{\chi}_n(x)\right )\prod_n dc_nd\bar{c}{}_n$$ since $\chi_n(x)$ satisfies completeness relations, $$\sum_n \chi_n(x) \bar{\chi}_n(x') = \delta(x-x')$$ $$\int dx \chi_n(x) \bar{\chi}_m(x) = \delta_{nm}$$ we can write that, as a matrix with indices $x$ and $n$, $\chi_n(x)$ satisfies, $$\chi \cdot \chi^\dagger = \mathbb{1}$$ and so $\det \chi \det \bar{\chi}= 1$.