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I'll start by setting the tangential part of the question aside: In quantum field theory there are currents and charges which are conserved classically, but not quantum mechanically, that is, there are violations of these conservation laws to order $\hbar^2$, and hence the symmetries hold classically but not quantum mechanically. (See below for what that ...

2

There is a little confusion in your statements about the quantisation of the gravitational field, especially then at any time slice one can define the state of the universe as the probability of gravitons being at positions: is not correct. Let us start from the beginning instead. The standard way to quantise fields is to write down their path ...

2

Suppose $\left|\psi\right\rangle =\psi\left(0\right)\left|0\right\rangle$ is another primary state, created by the primary operator $\psi\left(z\right)$. Now you want to calculate \begin{align*} \phi\left(z\right)\left|\psi\right\rangle & =\phi\left(z\right)\psi\left(0\right)\left|0\right\rangle \\ & =\sum_{p}C_{\phi\psi ...

2

The holomorphic/coherent state path integral is explained in e.g. Ref. 1. Let us here highlight some of the points. Notation in this answer: In this answer, let $z,z^{\ast}\in \mathbb{C}$ denote two independent complex numbers. Let $\overline{z}$ denote the complex conjugate of $z$. Also Planck's constant $\hbar=1$ is put equal to one. It is customary to ...

2

Kallen-Lhemann representation is just a way to expand in the momentum basis the two point correlation function of a local operator $\hat{O}(x)$, it holds true for massive and massless theories alike except for non abelian gauge theory in which the situation is a bit more complicated. Let's demonstrate the K-L formula with a more general proof: let's start ...

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Given that the regular (non-covariant) derivative of the adjoint satisfies $$\nabla \psi^\dagger = \left(\nabla \psi \right)^\dagger$$ one likewise expects and defines $$\tilde{\nabla}\psi^\dagger = \left( \tilde{\nabla}\psi \right)^\dagger = \left( \nabla \psi +ie\bf{A}\psi \right)^\dagger = \nabla \psi^\dagger -ie\bf{A} \psi^\dagger$$

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The terms generated by the RG will all respect the symmetries of the microscopic action (though one have to be careful with anomalies). That's why people tend to directly write a low energy effective action, and do not bother calculating the RG flow of the parameters. However, this implies that you do not know how the effective parameters relates to the ...

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Not sure whether I understood the intention of your question correctly. In electrodynamics you usually use a 1-form $A=A_{\mu}dx^{\mu}$ to write down an action \begin{equation*} S = \int F \wedge \star F \end{equation*} with the 2-form field strength $F=dA$, which the gives the known (vacuum) Maxwell equations. Scalar fields $\phi$ (such as the Higgs) ...

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Let us look at this prolem from a (relativistic) field theory perspective. The Hamiltonian for $\psi$ must contain a term of the form $$\nabla\psi^\dagger\cdot\nabla\psi$$ due to Lorentz invariance. Assuming $\psi$ to transform in a representation of U(1). Resulting in the following simultanious transformations: \begin{aligned}\psi\rightarrow U(x)\psi ... 2 p^2 is p_1^2+p_2^2+p_3^2 - p_0^2. This integrand does have poles, the intuition behind the poles is that you expect a small amplitude when p^2 is very different than m^2, because then the particle has more (or less) momentum than you would conclude by looking at the energy. The intuition is that the particle has "borrowed'' this extra kinetic energy ... 2 Fields in QFT have a particle content too, so the field is not replacing particles, they are just described in a different way as the possible modes of vibration of the field. Now fields originate in classical physics, most notably with the work of Maxwell. It was then realised that a field solves many problems, like instantaneous interactions at a distance, ... 1 Consider for the shake of simplicity a free neutral scalar field \phi. Passing to the second quantization picture, it is a operator valued distributionC_0^\infty(M;\mathbb R) \ni f \mapsto \phi(f)$$where M is Minkowski spacetime and \phi(f) is a densely defined symmetric operator on the Hilbert space$$F_+(\cal H) = \mathbb C \oplus \cal H ...

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