# Tag Info

## Hot answers tagged functional-determinants

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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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 ...

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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 ...

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