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The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles. Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is $$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p ... 2 There is the (non-genetal) relation between the free energy of interacting of two currents J^{a}, J^{b} and the propagator:$$ U = -\frac{1}{2} \int d^{4}xd^{4}y J^{a}(x) D_{ab}(x - y)J^{b}(y). $$It's not general, but it realizes the simple example which can help you to understand how to get the expression for force. The structure of field which causes ... 1 As @RobinEkman mentioned, the kinetic term changes as well. This can be easily computed \begin{split} {\cal L}_D = i {\bar \psi} \gamma^\mu \partial_\mu \psi &\to i {\bar \psi} e^{- i \alpha} \gamma^\mu \partial_\mu \left( e^{i \alpha} \psi \right) = i {\bar \psi} \gamma^\mu \partial_\mu \psi - \partial_\mu \alpha ( {\bar \psi} ... 1 The kinetic term$$\mathcal{L}_\text{fermion} = i\overline{\psi}\gamma^\mu \partial_\mu \psi$$changes too, by precisely the right quantity to cancel the change in the interaction term. Thus, the total Lagrangian is invariant, and this is what matters. 0 Wonderful question! And it relies on one central concept: virtual particles. Particles that do not necessarily exist, but explain (or would explain) phenomena wonderfully. By assuming that the particles (electrons, protons, etc) are exchanging photons, we can predict their behavior with great accuracy. Imagine that two protons are people, and they are ... 1 At the classical level, the global gauge invariance leads via Noether's theorem to electric charge conservation, cf. e.g. this Phys.SE post. The Ward-Takahashi identity (WTI) can roughly speaking be thought of as a quantum version of this. In particular, we stress that the WTI is intimately tied to electric charge conservation. OP's observation that ... 3 We can write the Fourier transform of \langle 0|\mathcal{T}A_{\nu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle as$$S(p) D_{\nu\alpha}(q) \ e\,\Gamma^{\alpha}(p,q,p+q)S(p+q) where $S(p)$ is the full fermion propagator, $D_{\nu\alpha}(q)$ is the full photon propagator, $\Gamma^{\alpha}(p,q,p+q)$ is the proper vertex function, and an overall momentum conservation ...

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We already know that QED is incomplete. That's why we needed to develop the standard model that includes the electroweak force and QCD. We know that the standard model is incomplete and that we may need supersymmetry or other similar extensions to make it better, and even then gravity would be missing on the level of current quantum field theory. How to ...

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I'll give you a draft of the answer to put you on the right track. You should then be able to fill the hole and complete the details. I think you just need to take the non-relativistic limit of a field theory where instead of the tree-level propagator you use the full 1PI propagator $\frac{1}{p^2-m^2-\Pi(p^2)}$ where $\Pi(\pi^2)$ is the self-energy. If the ...

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Once you have open systems you loose unitary time evolution in general and hence Hermitian Hamiltonians. You can see this by taking the full system+environment and then tracing over the degrees of freedom of the environment. This can be made rigorous with several assumptions - Lindblad equation. In your case it is more of an effective description for a ...

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Indeed, do not take Feynman diagrams as literal representations of what is happening in a particle picture. Only the external lines of a diagram correspond to real particles - the internal lines, though called virtual particles, are little more than artifacts of the perturbative expansion we do to calculate QFT amplitudes, and there is little reason to ...

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No. In the electron case the self-energy makes the (bare) mass an ill-defined quantity unless you specify a renormalization scheme. Usually one takes the dressed electron mass (i.e. the bare mass plus all radiative corrections) to be equal to the measured electron mass in the limit of vanishing external momentum. All the radiative corrections then do is ...

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Today we know that Collins is wrong. He appears to be unaware of Newton's finding, and of course, advances made after he wrote his book.

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