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7

(I henceforth assume $c= \hbar=1$.) It is forbidden by the four-momentum conservation law. Put yourself in the centre of mass reference frame of the couple of massive particles (electron and positron). There $P_{e\overline{e}} = (2E,\vec{0})$ with $E\geq m_e>0$. Just because four momentum is conserved, this four-momentum must be the same as the one of the ...

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 ...

2

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 $$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p ... 1 The kinetic term$$\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.

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} ...

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