5

My original answer was going to be much longer, but everything started to be a bit too long and complex, so here's some rough elements to give you how to derive the idea : Formally, the Hadamard function (which is related to the Green's function by some differing contour integral) is defined on curved spacetimes by $$G^{(1)}(x, y) = \sum_{\gamma} \frac{\...


3

Yes, it can be easily derived. First some notation. We denote by $\left< \Omega \right>$ (the quantum expectation value) the path integral $$ \left< \Omega \right> = \int \mathcal{D} \phi \, e^{i S[\phi]} \Omega[\phi]. $$ Here $\Omega[\phi]$ is an arbitrary functional of $\phi$, though we shall not promise that all possible functionals give ...


3

Lagrangian theories are indeed obiquitous because those are the ones we can understand better and ultimately those with which we can do computations. However, sadly, they only make sense when the theory admits a weakly coupled description. The reason is the very existence of the couplings. The fact that your theory has couplings means that there is a fine ...


3

The 'in' and 'out' states, as Weinberg defined in Chapter 3, are eigenstates of the full Hamiltonian $H$, not the free Hamiltonian $H_0$. $a_{\text{in}}$ and $a_{\text{out}}$ are the annihilation operators for the 'in' and 'out' states, respectively. They are not free-field operators. Any proof of the LSZ formalism would involve constructing such operators,...


2

First of all you find an orthogonal matrix that diagonalizes $\mathbf{B}$ $$\mathbf{B} = \mathbf{O}^T\mathbf{DO} \qquad B_{ij} = (O_{ik})^TD_{kl}O_{lj}$$ where $\mathbf{D}$ is a diagional matrix. Putting it in the integral $$\left(\prod_{k} \int d \xi_{k}\right) \exp \left[-\xi_{i} B_{i j} \xi_{j}\right]=\left(\prod_{k} \int d\xi_k\right) \exp \left[-...


2

I preface my answer by saying that this is really a non-answer and more of a long comment or just a rewrite of the relevant parts of the article that you cited. So I will start by reformulating the question slightly differently, and, for simplicity, in 1D. So we have a system $x_t$ and a control $u_t$, with the system evolving as: $$\mathrm{d}x_t = (b(t, ...


2

I believe they are just describing in words the following equation $$\frac{1}{p^2- m^2 \pm i\epsilon} = P\frac{1}{p^2- m^2} \,\mp\,i\pi \delta(p^2- m^2) $$ where $P$ denotes the principal part. By the counting that you describe you can show that the factors of $i$ coming from propagators and vertices always balance out (you also need to remember that ...


2

Dijkgraaf-Witten theories (Chern-Simons theory for a finite gauge group) are an example of a quantum theory without a Lagrangian, which nevertheless has an action (an exponentiated action to be precise). It is neither odd nor difficult to deal with, in fact, toy models like these are often used precisely because they are easier to handle than the quantum ...


2

The idea is that the quadratic part of the action should be invertible and the inverse gives the propagator of the field. Eq. 9.52 is trying to solve for the inverse $\tilde{D}^{\nu\rho}_F$, but you cannot because $-k^2g_{\mu\nu} + k_\mu k_\nu$ is singular. That's the exact same as saying that the quadratic action vanishes for too many field configurations ...


1

I believe that the answer to your question lies in thinking about the whole path integral, rather than just the Lagrangian of QCD. The purpose of using the Faddeev-Popov prescription is not necessarily to modify the form of the Lagrangian itself, but to ensure that the path integral is not summing over field configurations which are "equivalent", i.e. field ...


1

You mean the (equilibrium) statistical mechanics textbooks, which deal mainly with the statistical sum, and therefore need only the imaginary path integral. The real-time path integral is usually derived in quantum mechanics textbooks, where it is sometimes also shown to be the direct solution of the Schrödinger equation with a source term, i.e. a true ...


1

All right so after days of looking textbooks I finally get a feel of how things are arranged, I’ll try to put all things together to give a clear distinction for the people who are also confused by this. So basically it is the difference between the operator language and the path integral language, and it uses the fact that the real-time green function is ...


1

Instead of numerical simulations, consider to explore Feynman's checkerboard model for (massive) fermions in a discretized 1+1D spacetime. Much of the "sum over histories" of the FPI and the causal structure of the lightcone can be recovered from purely analytical considerations.


1

There are an uncountably infinite number of paths, because spacetime is continuous, so direct computation of the path integral is impossible without some discretization, outside of some very special cases. You might want to look into lattice field theory, which is precisely the restriction of quantum field theory to a discrete spacetime lattice.


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