How does QED at a finite temperature differ from QED at zero temperature? Currently, I do not have any knowledge of finite temperature field theory. But I have learnt ordinary QFT calculations and I am reasonably familiar with Statistical mechanics. With this background, I wonder if there is a simple way to get the basic idea of the following scenario.
Consider a $T=0$ QED process. To be specific, let me consider the $e^-e^-\to e^-e^-$ scattering. I know how to calculate the scattering amplitude of this process at tree-level. There are also corrections at one-loop, two-loop etc. Now, let us imagine the same process supposing that this system of photons and electrons are now in equilibrium in a thermal environment. 


*

*Why does the scattering amplitude of the same process at $T\neq 0$ differ (or get modified) from $T=0$, physically? 

*What is the concept or what is the physics distinguishes this scattering process at $T=0$ from $T\neq 0$? 
Please let me know if I have communicated the question well enough. As I do not have a fair knowledge of this, the question might be unclear.
 A: Note that at finite temperature we don't have scattering amplitudes in quite the same way that we do at $T=0$, because the notion of an asymptotic state does not quite exist. The natural observables are the partition function, and correlation functions of gauge invariant operators. 
But let's ignore this issue, and look at $ee$ scattering. At $T=0$ the leading diagram is one-photon exchange. There are loop corrections to this result. The divergent pieces just renormalize the charge, and there are small finite terms that correct the tree level result. At $T\neq 0$ there is still one-photon exchange, but it now receives corrections not only from vacuum polarization, but also from scattering off electrons and positrons already present in the plasma. In particular, the photon propagator receives a correction that corresponds to Debye screening. In terms of Feynman rules, it corresponds to replacing the $1/q^2$ in the Coulomb part of the interaction by $1/(q^2-m_D^2)$ with $m_D^2\sim e^2T^2$. This qualitatively changes the result, because even though the coupling is weak, at long distance (small $\vec{q}$) the Debye mass is always the dominant term. In particular, the static limit is no longer Coulomb scattering, but scattering from a screened Coulomb potential
$$
 V = -\frac{e^2}{r}\exp(-m_D r)\, . 
$$
The transverse part of the interaction receives a modification known as Landau Damping, which amounts to a self energy $\Sigma(\omega,q) \sim i\omega m_D^2/|\vec{q}|)$. Note that at finite $T$ Lorentz invariance is broken, so there is nothing wrong with the longitudinal and transverse self-energies being different.   
A: Thomas gave you a good answer, but here’s an additional perspective …
At finite temperature, the so-called vacuum isn’t all that empty.  It is suffused by blackbody radiation and a miasma of electron-positron pairs.  If a diagram of interest contains a on-shell boson propagator, and the not-so-empty environment contains a photon line, you must now add the exchange diagram.  The effect is to stimulate emission and enhance photon propagation by the factor ${{[1-\exp (-\beta E)]}^{-1}}$, familiar from the Planck distribution.  Likewise, if the diagram contains an on-shell fermion propagator, you must subtract a crossed diagram, and the effect will be to suppress propagation by the factor ${{[1+\exp (-\beta E)]}^{-1}}$.  
Such modifications do not affect tree diagrams because the propagators are off-shell.  If you cross an on-shell line (representing a particle in the thermal environment) with an off-shell propagator, the crossed diagram will vanish because it fails to conserve energy at its vertices.  Loop diagrams, on the other hand, are affected, because integrations over energy variables pick up poles in propagators.  The Debye-like shielding comes from  a one-loop diagram.  
There is an elegant algebraic formulation of FTFT involving periodic boundary conditions in imaginary time, discrete summation vice integration in Im(E), and Sommerfeld-Watson transformations, but the physical interpretation of the result is actually quite simple.  
