# Tag Info

## Hot answers tagged functional-determinants

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.

2

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 ...

1

OP's underlying question is essentially the same as this Phys.SE post, although the detailed calculation is slightly different and interesting to compare. I) The action for a free non-relativistic point particle with mass $m=1$ reads: \tag{1} S ~=~\frac{1}{2}\int_0^T\! dt~ \dot{x}(t)^2~=~ \frac{1}{2}\langle x,Ax \rangle~=~\frac{1}{2}\sum_{n\in \mathbb{N}}...

1

1)They neglect higher powers of $\Delta \phi$ because this effective action desribes dynamics of the fluctuations $\Delta \phi$ above the background fields $\phi_0$. Namely $\Delta \phi$ is small 2)As you know $$\int d^n r\, e^{-r_i A_{ij}r_j}=\frac {(2\pi)^{n/2}}{(\mbox{det}\, A)^{1/2}},$$ where $A$ is a matrix with positive ...

1

I know this is a year old question, but I am going to attempt an answer. As far as I can tell, this is not really a caveat. The reason for this is that I can always set the overall phase of the quark mass determinant to be zero with a chiral U(1) transformation. For a discussion of this see for example the chapter on theta vacua in Weinberg's QFT book. The ...

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