How can QFT perturbation be used for electron-positron scattering? I'm studying scattering and perturbation theory in QFT from Peskin and Schroeder book. After all the calculations and theory developing, they made a calculation for $$e^-e^+\to\mu^-\mu^+$$ scattering. Even though I understood the calculations, one thing I don't agree on is using perturbation approach for this kind of problem. How can the interacting Hamiltonian be considered small when the force is attractive between them, and gets bigger and bigger the closer they get? for $e^-e^-\to\mu^-\mu^-$ scattering I can somewhat understand that they don't get close enough for the interacting energy to be big enough, but what about the case I mentioned above?
 A: First of all, the description with perturbation theory is possible because the coupling constant $e^2$ is small: $\frac{1}{137}$. That makes it possible to use $e j^\mu A_\mu$ as an interaction Hamiltonian (or Lagrangian). If it were a strong interaction process with an energy below 200MeV such a process could not be considered as perturbation since the strong coupling constant is larger than $1$. Perturbation theory could not be applied. But the coupling constant of QED is substantially smaller than 1, which allows for perturbation theory.
Second of all, from a larger perspective of view the processes $e^-e^-\rightarrow\mu^-\mu^-$ and  $e^-e^+\rightarrow\mu^-\mu^+$ are based on the same Feynman diagram which is shown page 156 respectively page 157 of P& S. The "only" difference is that they are turned by $90^\circ$ with respect to each other. So $e^-e^+\rightarrow\mu^-\mu^+$ is actually in the $s$-channel, whereas $e^-e^-\rightarrow\mu^-\mu^-$ is in the t-channel. This similarity is based on  crossing (symmetry) which says that an outgoing line can be interpreted as an outgoing electron but as well as an in-going positron. Furthermore an in-going line can be interpreted as an in-going electron or an outgoing positron.
So both processes only distinguish by the values of the 4-momenta of the particles attributed to it, otherwise there is no difference.
This means the physics these diagrams are based on is exactly the same.
Third point is the role of the Coulomb interaction which might play a role in such diagrams. The short answer is that the Coulomb interaction is not involved in the process since the photon propagator is actually only based on the transverse degrees of freedom of the photon field.
In other words, the Coulomb interaction actually is cancelled out in the photon propagator.
It is a bit complicated to see that so I will not demonstrate it in all details.The full computation is shown in Srednicki's book chapter 56 and as well in Bjorken & Drell's Quantum Field Theory in section 17.9.
It requires to carry out the calculation in a particular gauge of the EM-field. This is probably the reason that P & S are not showing it, after making the form of the photon propagator plausible in section 4.8 they derive it in
Lorentz-covariant and gauge-independent way according to Faddeev & Popov, a method that is very abstract. In particular the cancellation of the Coulomb interaction cannot be perceived in this approach.
I outline here Srednicki's demonstration which I recommend to read in order to understand the proof completely. Srednicki's book is accessible on the web: http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf
The demonstration is done in Coulomb gauge which also requires the use of a particular reference system, i.e. is not manifestly Lorentz covariant.
The Coulomb interaction in the action is given by
$$S_{coul} = -\frac{1}{2}\int d^4x d^4 y \delta(x^0-y^0) \frac{J^0(x) J^0(y)}{4\pi | \mathbf{x}- \mathbf{y} |}$$
This term can actually be added to the transverse part of the propagator which consists of the sum over the transverse degrees of freedom of the polarisation vectors (normalized by $k^2$) of the intermediate photon which together yields:
$$\tilde{\Delta}^{\mu\nu}(k)= - \frac{1}{\mathbf{k^2}}\delta^{\mu 0}\delta^{\nu 0} + \frac{1}{k^2-i\epsilon}\sum\limits_{\lambda=\pm} \epsilon^{\ast\mu}_\lambda(\mathbf{k})\epsilon^\nu_\lambda(\mathbf{k})$$
The sum of the degrees of freedom of the polarisation vectors can be expressed as
$$\sum\limits_{\lambda=\pm} \epsilon^{\ast\mu}_\lambda(\mathbf{k})\epsilon^\nu_\lambda(\mathbf{k})  =g^{\mu\nu} + t^\mu  t^\nu - z^\mu z^\nu $$
where $t^\mu =(1,\mathbf{0})$ an unit vector in time direction and $z^\mu$ is
$$z^\mu = \frac{k^\mu + (t\cdot k)t^\mu}{[k^2 + (t\cdot k)^2]^{1/2}}$$
Actually as the $k^\mu$ cancel out against the conserved electron currents it the actual $z^\mu$ can be substituted by:
$$ z^\mu \rightarrow \frac{(t\cdot k)t^\mu}{[k^2 + (t\cdot k)^2]^{1/2}}$$
We can then write the propagator:
$$\tilde{\Delta}^{\mu\nu}(k) = \frac{1}{k^2-i\epsilon}\left[g^{\mu\nu} + \left(- \frac{k^2}{k^2 + (t\cdot k)^2} + 1 - \frac{(t\cdot k)^2}{k^2 + (t\cdot k)^2}\right)t^\mu t^\nu\right]$$
Further algebra shows that the round braket expression completely cancels out and we are left with
$$ \tilde{\Delta}^{\mu\nu}(k) = \frac{g^{\mu\nu}}{k^2-i\epsilon}$$
Actually the $1$ in the round braket expression is due to the Coulomb interaction which completely cancels out.
Bottom line is:
In Feynman diagram QED interactions only the transverse degrees of freedom of the 4-vector potential play a role, the longitudinal degrees do not have any influence and the Coulomb interaction is related to the longitudinal degrees of the photon field. Therefore the  Coulomb interaction does not intervene in such processes although it might appear not intuitively.
A: According to the path integral approach (qualitatively), they take all possible paths, which presumably includes arbitrary closeness...which may mean up to some UV limit in the renormalization scheme. How those infinities cancel is a question for experts.
Moreover: Feynman diagrams are an expansion in momentum space, so spatial information isn't available. From the point-of-view of a former Deep Inelastic Scattering experimentalist, elastic electron scattering from a charged object is factored into a point-point particle scattering cross-section (Rutherford scattering) times a structure function $G_E(Q^2)$, something like:
$$ \frac{d\sigma}{d\Omega} = \Big(\frac{m^2\alpha^2}{4p^2\sin^4(\theta/2)} \Big)^2 \times G_E(Q^2) $$
(which only includes the charge scattering, not magnetic moments). $Q^2$ is the squared 4-momentum transfer of the virtual photon.
The structure function $G_E(Q^2)$ can be related to the Fourier transform of the radial charge distribution at wavelength $\lambda =\hbar c/\sqrt{Q^2}$.
In say electron-muon point-on-point scattering (which I choose to avoid $s$-channel annihilation or $u$-channel exchanges), the structure function is:
$$ G_E(Q^2) = 1 $$
which means all length scales contribute equally,
including impact parameter $b \rightarrow 0$. No divergence, and this only at tree-level.
