# Regularization of functional determinant over an Instanton background

I am reading the paper "ABC of instantons" and meet some problems at section 8. I simplify this problem a little bit as follows.

First, we have a Euclidean path integral like $$$$Z=\int \mathcal{D}A\; {\rm e}^{-S},~~~S=\int d^4x \mathcal{L}_0(A^{a}_{\mu})$$$$ An instanton is a solution of equation of motion that makes $$S$$ finite.

Now expand this action at the instanton solution $$A^{ins}$$ up to 2nd order: $$$$A=A^{ins}+a,~~~S=S(A^{ins})+\int d^4 x ~a^{j}_{\mu}\hat{L}^{jk}_{\mu \nu}(A^{ins})a^{k}_{\nu}.$$$$ Here $$\hat{L}^{jk}_{\mu \nu}(A^{ins})$$ is an operator depending on $$A^{ins}$$. One also needs to add a gauge-fixing term and ghosts to the action $$S$$, these are $$$$\Delta S=\int d^4 x a^{j}_{\mu}\Delta\hat{L}^{jk}_{\mu \nu}(A^{ins})a^{k}_{\nu}$$$$ for gauge-fixing and $$$$\Delta S_{gh}=\int d^4x \bar{\Phi}^a \hat{L}^{ab}_{gh}\Phi^b$$$$ for ghost.

Combining everything, one has $$$$Z=e^{-S(A^{ins})} det(\hat{L}+\hat{\Delta L})^{-1/2} det(\hat{L}_{gh})$$$$ Now since the operator $$\hat{L}+\hat{\Delta L}$$ has zero modes (eigenfunction of vanishing eigenvalue), the expression $$det(\hat{L}+\hat{\Delta L})^{-1/2}$$ is ill-defined. This paper claims we have to regularize it with a cutoff $$M^2$$ (eq 74): $$$$\bigg[\frac{det(\hat{L}+\hat{\Delta L})}{det(\hat{L}+\hat{\Delta L}+M^2)}\bigg]^{-1/2} \frac{det(\hat{L}_{gh})}{det(\hat{L}_{gh}+M^2)}$$$$ My question is: How does this cutoff come into the current calculation? I know the infinity of $$det(\hat{L}+\hat{\Delta L})^{-1/2}$$ is from the integral $$$$\int dc \exp[-\frac{1}{2}\lambda c^2]$$$$ for $$\lambda=0$$. But how is the cutoff introduced and how does it work?

• Aug 25, 2020 at 5:28
• Aug 25, 2020 at 5:50

This is Pauli-Villars regularization. Pauli-Villars introduces a new field into the action with the same quantum numbers as $$A$$ but opposite statistics, and a large mass $$M$$. In the factor $$det(\hat L + \hat{\Delta L} + M^2)^{1/2}$$ the exponent $$+1/2$$ comes about since the Pauli-Villars field is Grassman-valued and the $$+M^2$$ is just its Gaussian mass term. It seems they did the same for the ghost. At the end they will take $$M\to \infty$$ in which case the field does not affect physics at lower energy scales.
• I am still confused. In Grassman field integral, the power of determinant should be $+1$, where is this $+1/2$ from? Aug 25, 2020 at 7:07
• Not always; sometimes it is $+1/2$, see en.wikipedia.org/wiki/Berezin_integral Aug 25, 2020 at 13:38