# Tag Info

The complex world sheet (WS) coordinate $z\in\mathbb{C}$ contains a WS time and a WS space coordinate. It is important to realize that the time derivatives inside the Boltzmann factor in the path integral should respect the underlying time slicing procedure. See e.g. this & this Phys.SE answers, and my Phys.SE answer here. Thus it is implicitly ...
You're right that the Feynman propagator for a spinor field is indeed $(i \gamma^\mu D_\mu - m)^{-1}$. The tricky part is interpreting exactly what the "inverse" means. It doesn't just mean that you invert the gamma matrices (although you do do that). The derivative operator is also being inverted. That is, if we define a function $G(y, x) := (i ... 0 Hello handsome poster (why thank you kind stranger). The answer to your question it turns out is in Rovelli's "Quantum Gravity", at least insofar as the free scalar field is concerned. This is done in the following way. As you may recall (from Feynman and Hibbs), through various arguments about doing the path integral as a perturbation on the classical path, ... 1 We do not start from the gauge fixed path integral in the BRST construction. What you describe (once one adds the missing Faddeev-Popov determinant) is the original Faddeev-Popov trick to get the ghosts, not the systematic BRST construction. The (Hamiltonian) BRST construction crucially first introduces the ghosts as parts of the extended phase space, and ... 1 The BRST symmetry cannot be seen without introducing auxiliary variables. The fastest way to realize the BRST symmetry is to "exponentiate" the delta function $$\delta(G)~=~\int \!{\cal D} B ~\exp\left[iB_{\alpha}G^{\alpha}\right]$$ and the Faddeev-Popov (FP) determinant $$\det\Delta ~=~\int \!{\cal D} c ~{\cal D} \bar{c} ~\exp\left[\bar{c}_{\alpha} ... 3 Jean Zinn-Justin has a great way of teaching path integral techniques starting with finite dimensional random variables (sometimes called "0-dimensional fields"). Here, you should think of these as discrete lattice approximations to continuous fields. In the spirit of Zinn-Justin's approach, I'll describe how this is done for the 1D free particle system you ... 0 The coherent state path integral is basically a recipe for converting a Hamiltonian into a Lagrangian. In condensed matter, we often start with a "microscopic" Hamiltonian description of a material at the level of individual atoms/electrons, and want to convert that into a Lagrangian so that we can more easily do QFT. In high energy, it's usually easier to ... 2 Short answer to your question: yes. There's no reason for the path integral and the operator formalism to be equivalent. A simple example are all the non-Lagrangian theories. We know (some of) them through their operator algebra, but there's no corresponding Lagrangian. This might sound strange to most people, like myself not long ago, but it is not hard ... 2 General comments to the question (v1): Any textbook derivation of the correspondence between$$\tag{1} \text{Operator formalism}\qquad \longleftrightarrow \qquad \text{Path integral formalism}$\$ is just a formal derivation, which discards contributions in the process, cf. e.g. this Phys.SE post. Rather than claiming complete understanding and existence ...