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This is because there are 1 less position integration due to the Dirichlet boundary conditions $$q(t_0)~=~q_0\quad\text{and}\quad q(t_N)~=~q_N,$$ and the fact that the insertion of complete sets of position resp. momentum eigenstates in phase space path integral alternates temporally $$p(t_{1/2}),\quad q(t_1),\quad p(t_{3/2}),\quad q(t_2),\quad \ldots,\... 2 Hint: Instead of Wick's theorem for free fields, one can alternatively use the Schwinger-Dyson equation on a (n\!-\!1)-point function \left\langle\phi(x_2)\ldots\phi(x_n)\right\rangle. Next apply (-\nabla_1^2+m^2)^{-1} on both sides. 2 It's only true in the limit as \beta increases. Your Z' contains no dependence on \beta, so it can not give you the full function of F(\beta). Z'=\lim_{\beta\rightarrow \infty}\langle \Omega| e^{-iH\beta}|\Omega\rangle=\lim_{\beta\rightarrow \infty}e^{-iE_0\beta}, where \Omega is the vacuum and E_0 is the vacuum energy (which is infrared ... 2 The target space for a path integral \int\!{\cal D}\phi ~e^{\frac{i}{\hbar}S} could be a complex manifold. Example: The coherent state path integral. But say that the target space is a real manifold and the corresponding classical field theory is manifestly real. In other words, all the virtual paths/histories in the path integral are assumed to be real. ... 2 The QFT path integral has, in general, no known mathematically rigorous formulation. So all explanations of it are neither "necessary" nor "sound" - they are heuristics that essentially define the path integral of fields to behave as we want it to, but they are not derivations in the traditional mathematical sense. Using the path integral ... 1 If I understand correctly your answer there are two confusions going on here. The first is that in the calculation of the tree matrix element for the e^+e^-\to \mu^+\mu^- reaction there are actually no fermionic propagators involved. You can see it already in the expression for iM that you provided. The only propagator involved (inverse powers of momenta,... 1 One needs to impose appropriate^1 boundary or fall-off conditions at both temporal and spatial infinity in field theory. For starters, to ensure that the action S has a mathematically well-defined variational/functional derivative \delta S/ \delta\phi. -- ^1 To assume that the field \phi vanishes might be too strong to describe relevant physical ... 1 TL;DR: Yes, OP is correct. Let's discretize time t. In other words, assume that we have n Grassmann-odd variables \theta^1, \ldots, \theta^n. In this language OP's differential operator \partial_t is replaced with some Grassmann-even matrix A^i{}_j. Define$$\theta^{\prime i}= \sum_{j=1}^nA^i{}_j \theta^j,$$so that$$\frac{\partial}{\partial\theta^...
The constraint $\delta\left(z^{2}-1\right)$ is actually added for any time and space, rather than just $\tau=0$, thus, it need to be written as: $$\prod_{x, \tau} \delta\left(|z(x, \tau)|^{2}-1\right)$$ since $$\delta\left(|z(x, \tau)|^{2}-1\right)=\int_{-\infty+i c}^{+\infty+i c} d \lambda(x, \tau) e^{-i \lambda(\tau, x)\left(|z(x, \tau)|^{2}-1\right)}$$ ...