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Typically, you need an integral over a scalar field each time you want to know the total amount of an extensive value fiven in the field: total mass, total load, etc from the density field... To compute how a light ray is absorbed as traversing a semi-opaque media, you how have to write the integral giving the optical depth along the ray. etc.


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Thanks to Qmechanic I got convinced that maybe it's better to compute that product from scratch: $$ P_{a,b} = \frac{1}{b}\prod_{n=0}^\infty(an+b) = b^{-1}\exp\left\{\sum_{n=0}^\infty\log(an+b) \right\}; $$ so the problem is to evaluate the infinite sum in term fo Hurwitz zeta function: $$ \zeta(s;z) := \sum_{n=0}^\infty(n+z)^{-s}, $$ and its analytic ...


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In this answer, we give a heuristic explanation for the formula (18) in Ref. 1. Consider two zeta-function regularized infinite products $$\tag{1a} F_a(b)~:=~\prod_{\lambda \in \mathbb{N}+b} a~=~a^{-\frac{1}{2}-b} $$ and $$\tag{1b} G(b)~:=~\prod_{\lambda \in \mathbb{N}+b}\lambda~=~\frac{\sqrt{2\pi}}{\Gamma(b+1)} $$ over a half-lattice $\Lambda= ...


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Since this has become a community wiki... I think that the following two references (authors mentioned below) are among the standard recommendations for thermal field theory: Kapusta & Gale Le Bellac If I recall correctly, the former uses the imaginary-time formalism while the latter also has a treatment of the real-time formalism.


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There are, of course, many ways you could go about evaluating path integrals. The process described below is just an example of how I would proceed. To make it easier to read this answer quickly, here's an outline: In the introduction, I explain how the problem of "evaluating path integrals" for Gaussian Lagrangians essentially reduces to the problem of ...


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OP is basically asking (v4) the following. How the equation $$ \int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) \langle F^r\left[x,\Phi\right] \rangle_J\right) ~=~0, \tag{4.5} $$ or equivalently, $$ \int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) F^r\left[x,\frac{\delta}{i \delta J}\right]\right)Z[J] ~=~0, \tag{4.5'} $$ ...


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The derivation in the given reference indeed seems confused and inconsistent. The crucial error seems to me that $$ S[\phi + \epsilon\delta\phi] = S[\phi]$$ is just not true for an infinitesimal symmetry. The definition of a symmetry is that $S[\phi']=S[\phi]$ (modulo boundary terms) for the finite transformation $\phi\mapsto \phi'$. Writing this ...


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There is a simple proof that the cancellation is impossible (at least unless you are willing to add to the classical a term proportional to $\hbar$), I am reformulating an answer by @Qmechanic in a simpler language: The anomaly, or the measure variation which is the typical source of the anomaly, contributes a term which is independent of $\hbar$, while any ...


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Comments to the question (v2): Traditionally, the classical action $S$ sits in the Boltzmann factor $\exp\left[\frac{i}{\hbar} S\right]$ behind an inverse power of $\hbar$ in the path integral, while the path integral measure is independent of $\hbar$. In the conventional way of counting, we say that the Jacobian $J$ from the path integral measure is a ...


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If you are interested in an path integral with the action: $$ \mathcal{S}= \int_{t_0}^{t_1} dt \langle \Phi(t) | i \hbar\partial / \partial t - \hat{H}(t) | \Phi(t) \rangle \tag{1} $$ then $\Phi(k, t)=\langle k| \Phi(t) \rangle$ is now an operator or a mude variable inside the path integral. The bridge between the operator and path integral linguage is: ...


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At last I think I managed to rearrange the terms in the sums... (which wasn't a monumental task after all ;-) ) \begin{equation} \sum_{k=0}^{N-1}\bar{z}_{k+1}z_k-\sum_{k=1}^{N-1}z_k\bar{z}_k=(*) \end{equation} In the notation here: $\lambda=0$: \begin{equation} ...


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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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