# Integral representation of functional determinant

I'm studying the proof of a theorem and, being not very expert in QFT, I'm having problems understanding a couple of equalities that my professor said to be useful in order to understand said proof.

The first one is $$\det{\left( -\Delta+V''[\varphi] \right)} = \int\mathscr{D}\psi\,\mathscr{D}\bar{\psi}\,\,\exp{\left[ -\int d^D x\, \bar{\psi}(x)\left( -\Delta+V''[\varphi] \right)\psi(x)\right]}$$ where $V$ is a generic potential, $\psi$ and $\bar{\psi}$ two fermionic fields.

The second one is $$\delta[F] = \int\mathscr{D}\omega\,\,\exp{\left[ \int d^D x\,\omega(x)\,F \right]}$$ with $F$ a functional of the type $F=F[\varphi(x)]$, with $\varphi(x)$ and $\omega(x)$ bosonic fields.

Any help would be much appreciated, as I really don't know where to start!

• Since the path integrals of fields don't really exist rigorously to begin with, these assertions are "proved" by showing them for ordinary integrals. Do you know how that would work if $\psi,\omega$ were not fields but ordinary variables? Mar 15 '17 at 15:44
• I am able to understand that, in the first one, $\int d^D x\, \bar{\psi}(x)\left( -\Delta+V''[\varphi] \right)\psi(x)$ corresponds somehow to calculating the operator on the base $\psi$ but I can't undertand how taking the exponential of this (with a minus sign) and integrating over the fields gives the determinant! About the second one I have no clue... Mar 15 '17 at 15:49
• That doesn't really answer to the comment above. If instead of a functional integral you had a standard integral, would you be able to derive these equalities? Mar 15 '17 at 15:59
• @AccidentalFourierTransform I'm sorry, thought that was clear: I wouldn't! While I think (hope?) I would still be able to understand it... Mar 15 '17 at 16:03

Since the path integral is in general ill-defined, these expressions are not rigorously derived but simply thought to be true as extensions of their validity of ordinary integrals. Here's how you'd show that they hold in the cases where $\psi,\omega$ are not fields but merely variables.
1. Let $A$ be an $n\times n$ matrix. The integral goes over fermionic variables $\psi$, in a manner known as Berezin integration. In $\mathrm{e}^{\bar{\psi}A\psi}$, only the term with $2n$ variables when expanding as a power series contributes, the rest vanishes due ot the properties of the Berezin integral. Using that $\int \psi_{i_1}\dots\psi_{i_n}\mathrm{d}\psi = \epsilon_{i_1\dots i_n}$, it follows almost immediately that $$\int \bar{\psi}_{i_1}\dots\bar{\psi}_{i_n} A_{i_1 j_1}\dots A_{i_n j_n}\psi_{j_1}\dots \psi_{j_n}\mathrm{d}\psi\mathrm{d}\bar{\psi} = n!\det(A).$$
2. Your expression is missing a $\mathrm{i}$ in the exponent. Then it's a standard property of the Fourier transform that $\delta$ is the Fourier transform of the constant function, i.e. $$\delta(x) = \mathscr{F}(1)(x) = \int \mathrm{e}^{\mathrm{i}px}\mathrm{d}p$$