Simple Quantum Electrodynamics question First of all, what causes electrons to emit photons in QED? Why wouldn't electrons just not interact in that way
Second question, how do electrons even get close enough to interact/exchange photons. Shouldn't they just be repelled due to being the same charge?
 A: QED is the theory of Quantum Electrodynamics and not of electrons alone. Sure you can have noninteracting electrons theory but it is a bit boring in that case... The reason for the emission/absorption of photons by electrons is the coupling between them. Indeed, imposing the local $\text{U}(1)$ invariance imposes the presence of the interaction term in the Lagrangian density. The latter is of the following form:
\begin{equation}
\mathcal{L}=\underbrace{-\frac{1}{4}F_{\mu \nu}F^{\mu \nu}}_{\text{EM field term}}+\underbrace{\overline{\Psi}(i\gamma^\mu \partial_\mu-m)\Psi}_{\text{Fermions' term}} \underbrace{-e\overline{\Psi}\gamma^\mu A_\mu \Psi}_{\text{Coupling term}}
\end{equation}
Note that the minus sign in the interaction term is included by convention. So the local $\text{U}(1)$ invariance, that is a symmetry of the theory tells us that electrons must interact through electromagnetism. Now one can ask "But why $\text{U}(1)$ invariance and what does it mean?". $\text{U(1)}$ is the group associated with the charge of complex-valuated objects. Our fermions are Dirac bi-spinors. So they are some sort of complexification of spinors. But this symmetry is a Global one. Namely, the "fermions' term" in the Lagrangian density is invariant by the change $\Psi \leadsto \Psi\times e^{i\theta}$ with $\theta$ some constant phase. The theory is then non-interacting. In order to correctly recover electromagnetism, because our Lagrangian density is devoted to this, one has two choices:

*

*Adding "by hands" an interaction term.


*Searching for a way for an interaction term to pop out from the maths
These two methods give the same answer and it is the local $\text{U}(1)$ symmetry. Then it means that our fermions are bathed in the electromagnetic field and that they are in some sense free in it: like stones in a water flow, they follow paths guided by the electromagnetic field.
For your second question, well roughly speaking, this is because electrons exchange virtual photons that they get repelled. In some sense, the repelling of electrons is the exchange of virtual photons. Virtual particles are mathematical artifacts that arise from the perturbative expansion of what we call the $S$ matrix: an operator that makes evolve a state far from the past to a state far into the future (asymptotic states). When we do this expansion at the lowest order, the virtual photon alone is represented by a "propagator": as per Wikipedia, a propagator is "a function that specifies the probability amplitude for a particle to travel from one place to another in a given period of time, or to travel with a certain energy and momentum". In fact, this very propagator for the electromagnetic field gives (it is in fact no coincidence) exactly the Coulomb potential.
A: In Quantum Electrodynamics the electromagnetic field is quantized. So the fields as we are used to are actually represented by the quanta (therefore the name Quantum Electrodynamics) -- namely the photons.
So according to this concept the electron will not feel any attraction or repulsion if it does not meet a quantum of the  electromagnetic field, the photon. The interaction is local, not over "remote interaction" as this would contradict the concept of locality of a(n electromagnetic) field theory.
Attraction and repulsion can be correctly described by the formalism of Quantum Electrodynamics but is rather complicated and large, too large to be posted here. But I can later give an argument for the attraction of 2 opposed charges based on QED-formalism as an extension of this post.
