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 \begin{equation} Z=\int \mathcal{D}A\; {\rm e}^{-S},~~~S=\int d^4x \mathcal{L}_0(A^{a}_{\mu}) \end{equation} 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: \begin{equation} 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}. \end{equation} 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 \begin{equation} \Delta S=\int d^4 x a^{j}_{\mu}\Delta\hat{L}^{jk}_{\mu \nu}(A^{ins})a^{k}_{\nu} \end{equation} for gauge-fixing and \begin{equation} \Delta S_{gh}=\int d^4x \bar{\Phi}^a \hat{L}^{ab}_{gh}\Phi^b \end{equation} for ghost.

Combining everything, one has \begin{equation} Z=e^{-S(A^{ins})} det(\hat{L}+\hat{\Delta L})^{-1/2} det(\hat{L}_{gh}) \end{equation} 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): \begin{equation} \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)} \end{equation} 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 \begin{equation} \int dc \exp[-\frac{1}{2}\lambda c^2] \end{equation} for $\lambda=0$. But how is the cutoff introduced and how does it work?


1 Answer 1


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.

  • $\begingroup$ I am still confused. In Grassman field integral, the power of determinant should be $+1$, where is this $+1/2$ from? $\endgroup$
    – Sven2009
    Aug 25, 2020 at 7:07
  • $\begingroup$ Not always; sometimes it is $+1/2$, see en.wikipedia.org/wiki/Berezin_integral $\endgroup$
    – Dwagg
    Aug 25, 2020 at 13:38

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