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5

Your question seems to come in two parts. First, it seems the use of functional determinants to represent (formally) the result of taking a path integral may be new to you. Is that so? If it is, then I would suggest reading about this idea first. It's broader than zeta regularization. Once you're comfortable with that, then I would suggest thinking ...


5

This is because the path integral ${\cal Z}$ is an infinite-dimensional version of a Grassmann-odd Gaussian integral $$\int \!\mathrm{d}^n \bar{\theta} ~\mathrm{d}^n\theta ~e^{\sum_{i,j=1}^n\bar{\theta}_i ~M^i{}_j ~\theta^j}~\propto~\det(M), $$ where the indices $i,j$ can be interpreted as DeWitt's condensed notation.


1

I take it you mean the determinant with the straight bars...? If so, the only way to compute it for general background gauge field is, as nervxxx mentioned in the comments, to expand the determinant in $A_\mu$. If you are considering a specific background gauge field (like a constant magnetic field) you should look whether you can find the eigenvalues of ...



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